Austmaths-quest-11c-teacher-s-addition.pdf

11

YEAR

MATHS Quest

MATHS C FOR QUEENSLAND

D SECON

EDITION

TEACHER EDITION

11

YEAR

MATHS Quest

MATHS C FOR QUEENSLAND

T II O T INO N DEI D E D D N N S ESCEOC O

TEAC HE R EDI TI ON Nick Simpson Catherine Smith Peter Posetti Sue Campbell CONTRIBUTING AUTHOR

Robert Rowlan

Second edition published 2009 by John Wiley & Sons Australia, Ltd 42 McDougall Street, Milton, Qld 4064 First edition published 2001 Typeset in 10.5/12.5pt Times © John Wiley & Sons Australia, Ltd 2001, 2009 The moral rights of the authors have been asserted. National Library of Australia Cataloguing-in-Publication data Author:

Simpson, N. P. (Nicholas Patrick), 1957–

Title:

Maths Quest: Maths C Year 11 for Queensland/Simpson, Smith and Posetti.

Edition:

2nd ed.

ISBN:

978 0 7314 0814 6 (pbk.) 978 0 7314 0868 9 (web.)

Notes:

Includes index.

Maths Quest: Maths C Year 11 for Queensland, Teacher edition/Simpson, Smith and Posetti.

978 0 7314 0831 3 (pbk.) 978 0 7314 0866 5 (pdf)

Target Audience: For secondary school age. Subjects: Other Authors/ Contributors:

Mathematics — Textbooks. Smith, Catherine, 1969– Posetti, Peter.

Dewey Number: 510 Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL). Reproduction and communication for other purposes Except as permitted under the Act (for example, a fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher. Illustrated by Paul Lennon, Liz Sawyer and the Wiley Art Studio Cover photograph and internal design images: © Digital Vision Printed in Singapore by Craft Print International Ltd 10 9 8 7 6 5 4 3 2 1

Contents Introduction ix About eBookPLUS xi Acknowledgements xii

CHAPTER 1


Number systems: the Real Number System 1

Introduction 2 The Real Number System 3 Classification of numbers: rational and irrational 3 Exercise 1A 9 Recurring decimals 11 Exercise 1B 14 Investigation — Real number investigations 15 Investigation — Other number systems 16 Surds 17 Exercise 1C 20 Simplifying surds 21 Exercise 1D 22 Addition and subtraction of surds 24 Exercise 1E 25 Multiplication of surds 27 Exercise 1F 30 The Distributive Law 32 Exercise 1G 35 Division of surds 36 Exercise 1H 38 Rationalising denominators 40 Exercise 1I 42 Rationalising denominators using conjugate surds 44 Exercise 1J 48 Further properties of real numbers — modulus 49 Exercise 1K 50 Solving equations using absolute values 51 Exercise 1L 54 Solving inequations 55 Exercise 1M 62

Investigation — Approximations for p 63 Investigation — Real numbers — application and modelling 64

Summary 66 Chapter review 69

CHAPTER 2


Number systems: complex numbers 75

Introduction to complex numbers 76 Exercise 2A 79 Investigation — Complex numbers in quadratic equations 80 Basic operations using complex numbers 80 Investigation — Plotting complex numbers 84 Exercise 2B 86 Conjugates and division of complex numbers 87 Exercise 2C 91 Radians and coterminal angles 95 Exercise 2D 96 Complex numbers in polar form 96 History of mathematics — Abraham de Moivre 108 Exercise 2E 109 Basic operations on complex numbers in polar form 111 Investigation — Multiplication in polar form 111 History of mathematics — William Rowan Hamilton 118 Exercise 2F 119 Investigation — Complex numbers: applications 120 Summary 122 Chapter review 123

CHAPTER 3


Matrices

127

Introduction to matrices 128 Operations with matrices 130 Exercise 3A 135

vi Multiplying matrices 137 Exercise 3B 140 History of mathematics — Olga TausskyTodd 142 Powers of a matrix 143 Investigation — Matrix powers 143 Exercise 3C 144 Investigation — Applications of matrices 145 Multiplicative inverse and solving matrix equations 146 Exercise 3D 152 The transpose of a matrix 154 Exercise 3E 154 Applications of matrices 155 Exercise 3F 160 Investigation — Matrix multiplication using a graphics calculator 162 Dominance matrices 164 Investigation — Dominance matrices — another application of matrices 165 Exercise 3G 169 Summary 170 Chapter review 172

CHAPTER 4


An introduction to groups 177

Introduction 178 Investigation — Algebraic structures

178

Modulo arithmetic 179 Exercise 4A 180 The terminology of groups 180 History of mathematics — Niels Henrik Abel 183 Exercise 4B 184 Properties of groups 184 Exercise 4C 188 Cyclic groups and subgroups 189 Exercise 4D 191 Investigation — Application of groups — permutations 191 Further examples of groups — transformations 192 History of mathematics — Arthur Cayley 194 Exercise 4E 195

Investigation — Some applications of group theory 197 History of mathematics — Cryptography 199

Summary 201 Chapter review 202

CHAPTER 5


Matrices and their applications 205

Inverse matrices and systems of linear equations 206 Exercise 5A 208 Gaussian elimination 209 Exercise 5B 215 History of mathematics — Carl Friedrich Gauss 216 Investigation — Performing Gaussian elimination using a graphics calculator 217 Introducing determinants 222 Exercise 5C 224 Properties of determinants 224 Exercise 5D 227 Inverse of a 3 ¥ 3 matrix 228 Exercise 5E 232 Cramer’s Rule for solving linear equations 234 Exercise 5F 237 Investigation — Solving simultaneous equations 242 Investigation — Applications of determinants 243 Summary 244 Chapter review 246

CHAPTER 6


Transformations using matrices 249

Geometric transformations and matrix algebra 250 Exercise 6A 258 Linear transformations 259 Exercise 6B 262

vii Linear transformations and group theory 263 Exercise 6C 269 Rotations 270 Exercise 6D 275 Reflections 276 Exercise 6E 283 Dilations 284 History of mathematics — Maurits Cornelius Escher 290 Exercise 6F 291 Shears 291 Exercise 6G 295 Investigation — Transformations 295 Summary 296 Chapter review 297

CHAPTER 7


Introduction to vectors

CHAPTER 8 Vector applications

353

Introduction 354 Force diagrams and the triangle of forces 354 Exercise 8A 361

Newton’s First Law of Motion 364 Exercise 8B 371 Momentum 374 Investigation — Conservation of momentum using i and j notation 378 ˜ 8C ˜ 379 Exercise Investigation — Collision momentum 381 Relative velocity 382 Exercise 8D 384 Using vectors in geometry 385 Investigation — Three-dimensional non-zero vectors 387 Investigation — Vector geometry 388 Exercise 8E 388 Summary 390 Chapter review 391

299

Vectors and scalars 300 Exercise 7A 305 Position vectors in two and three dimensions 308 Exercise 7B 320 Multiplying two vectors — the dot product 324 Exercise 7C 329 History of mathematics — Charles Lutwidge Dodgson 331 Resolving vectors — scalar and vector resolutes 332 Exercise 7D 337 Investigation — Vectors and matrices 337 Time-varying vectors 339 Exercise 7E 344 Summary 346 Chapter review 348




History of mathematics — Sir Isaac Newton 363

CHAPTER 9


Sequences and series

395

Introduction 396 Arithmetic sequences 396 Exercise 9A 404 Geometric sequences 406 Exercise 9B 414 Applications of geometric sequences 418 Exercise 9C 424 Finding the sum of an infinite geometric sequence 427 Exercise 9D 431 Contrasting arithmetic and geometric sequences through graphs 432 Exercise 9E 438 Investigation — Reward time 440 Investigation — Changing shape 441 Fibonacci Sequence 442 Investigation — Fibonacci numbers 445 The Mandelbrot Set 446 Investigation — Draw the Mandelbrot Set 449 Summary 450 Chapter review 453

viii CHAPTER 10


Permutations and combinations 459

Introduction 460 The addition and multiplication principles 460 Exercise 10A 465 Factorials and permutations 467 Exercise 10B 474 Arrangements involving restrictions and like objects 476 Exercise 10C 480 Combinations 482 Exercise 10D 488 Applications of permutations and combinations 490 Exercise 10E 495 Pascal’s triangle, the binomial theorem and the pigeonhole principle 497 Investigation — Counting paths 498 Exercise 10F 504

History of mathematics — Blaise Pascal 506

Summary 507 Chapter review 509

CHAPTER 11


Dynamics

513

Displacement, velocity and acceleration 514 Exercise 11A 521 Projectile motion 524 Exercise 11B 536 Motion under constant acceleration 540 Exercise 11C 544 Summary 547 Chapter review 548 Appendix 553 Answers 591 Index 621

Introduction Maths Quest Maths C Year 11 for Queensland 2nd edition is one of the exciting Maths Quest resources specifically designed for the Queensland senior Mathematics syllabuses beginning in 2009. It has been written and compiled by practising Queensland Maths C teachers. It breaks new ground in Mathematics textbook publishing. This resource contains: • a student textbook with accompanying student website (eBookPLUS) • a teacher edition with accompanying teacher website (eGuidePLUS) • a solutions manual containing fully worked solutions to all questions contained in the student textbook.

Student textbook Full colour is used throughout to produce clearer graphs and headings, to provide bright, stimulating photos and to make navigation through the text easier. Clear, concise theory sections contain worked examples, graphics calculator tips and highlighted important text and remember boxes. Worked examples in a Think/Write format provide clear explanation of key steps and suggest how solutions can be presented. Exercises contain many carefully graded skills and application problems, including multiple-choice questions. Cross-references to relevant worked examples appear beside the first ‘matching’ question throughout the exercises. Investigations, often suggesting the use of technology, provide further discovery learning opportunities. Each chapter concludes with a summary and chapter review exercise containing questions that help consolidate students’ learning of new concepts. As part of the chapter review, there is also a Modelling and problem solving section. This provides students with further opportunities to practise their skills. Technology is fully integrated within the resource. To support the use of graphics calculators, instructions for two models of calculator are presented in worked examples and graphics calculator tips throughout the text. The two models of graphics calculator featured are the Casio fx-9860G AU and the TI-Nspire CAS. (Note that the screen shots shown in this text for the TI-Nspire CAS calculator were produced using OS1.4. Screen displays may vary depending on the operating system in use.) For those students using the TI-89 model of graphics calculator, an appendix containing matching instructions has been included at the back of the book. The Maths Quest for Queensland series also features the use of spreadsheets with supporting Excel files supplied on the student website. Demonstration versions of several graphing packages and geometry software can also be downloaded via the student website.

x Student website — eBookPLUS The accompanying student website contains an electronic version of the entire student textbook plus the following additional learning resources: WorkSHEETs — editable Word 97 documents that may be completed on screen, or printed and completed later. SkillSHEETs — printable pages that contain additional examples and problems designed to help students revise required concepts. Test yourself activities — multiple-choice quizzes for students to test their skills after completing each chapter.

Programs included Graphmatica: an excellent graphing utility Equation grapher and regression analyser: like a graphics calculator for PCs GrafEq: graphs any relation, including complicated inequalities Poly: for visualising 3D polyhedra and their nets

Teacher edition The teacher edition textbook contains everything in the student textbook and more. To support teachers assisting students in the class, answers appear in red next to most questions in the exercises and investigations. Each chapter is annotated with relevant syllabus information.

Teacher website — eGuidePLUS The accompanying teacher website contains everything in the student website plus the following resources: • two tests per chapter (with fully worked solutions) • fully worked solutions to WorkSHEETs • a syllabus planning document • assessment tasks (and answers) • fully worked solutions to all questions in the student textbook.

Solutions manual Maths Quest Maths C Year 11 for Queensland Solutions Manual contains the fully worked solutions to every question and investigation in the Maths Quest Maths C Year 11 for Queensland 2nd edition student textbook. Fully worked solutions are available for all titles in the Maths Quest for Queensland senior series. Maths Quest is a rich collection of teaching and learning resources within one package.

xi

Next generation teaching and learning About eBookPLUS Maths Quest Maths C Year 11 for Queensland 2nd edition features eBookPLUS: an electronic version of the entire textbook and supporting multimedia resources. It is available for you online at the JacarandaPLUS website (www.jacplus.com.au).

Using the JacarandaPLUS website To access your eBookPLUS resources, simply log on to www.jacplus.com.au. There are three easy steps for using the JacarandaPLUS system.

Step 1. Create a user account The first time you use the JacarandaPLUS system, you will need to create a user account. Go to the JacarandaPLUS home page (www.jacplus.com.au) and follow the instructions on screen. Step 2. Enter your registration code Once you have created a new account and logged in, you will be prompted to enter your unique registration code for this book, which is printed on the inside front cover of your textbook.

LOGIN Once you have created your account, you can use the same email address and password in the future to register any JacarandaPLUS books.

Step 3. View or download eBookPLUS resources Your eBook and supporting resources are provided in a chapter-by-chapter format. Simply select the desired chapter from the drop-down list and navigate through the tabs to locate the appropriate resource.

Minimum requirements

Troubleshooting

• Internet Explorer 7, Mozilla Firefox 1.5 or Safari 1.3 • Adobe Flash Player 9 • Javascript must be enabled (most browsers are enabled by default).

• Go to the JacarandaPLUS help page at www.jacplus.com.au • Contact John Wiley & Sons Australia, Ltd. Email: [email protected] Phone: 1800 JAC PLUS (1800 522 7587)

Acknowledgements The authors and publisher would like to thank the following copyright holders, organisations and individuals for their assistance and for permission to reproduce copyright material in this book.

Illustrative material Screenshots: TI–Nspire CAS and TI89 screenshots reproduced with permission of Texas Instruments; • Casio fx-9860G AU screenshots reproduced with permission of Casio. Images:

Dot points and Suggested Learning Experiences (SLEs) appearing throughout the textbook (overprinted in red) have been taken from the Mathematics C Senior Syllabus (2008), reproduced with permission from the Queensland Studies Authority, www.qsa.qld.edu.au.

• © Author’s Image, p. 540; • © Corbis, pp. 2 (bottom)/Matthias Kulka/zefa, 118/ Hulton-Deutsch Collection, 183/Bettmann, 194/Bettmann; • © Corbis Corporation, pp. 169, 299, 320, 456, 495 (flag), 550 (middle); • © Digital Stock/ Corbis Corporation, pp. 206, 307, 406, 425, 498, 544; • © Digital Vision, pp. 75/Jaroslav, 177/Jaroslav, 200 (top left), 202/Jaroslav, 205, 276, 352, 482; • © Emerald City Images p. 161/John Carnemolla; • © Flat Earth, p. 192; • © Getty Images, p. 130/Photonica/Loungepark, 300/Allsport/Nick Wilson; • © imageaddict.com.au, p. 383; • © Image Disk Photography, pp. 444 (left), 455 (bottom); • © Creative Cohesions, p. 446; • © John Wiley & Sons Australia, pp. 259/Kari-Ann Tapp, 368/Jennifer Wright, 455 (top); • © Photodisc, p. 1, 2 (top), 16, 17, 74, 127, 141, 145 (bottom, top), 146, 164, 174 (bottom), 175, 178, 200 (middle right, top right), 249, 250 (top), 270, 284, 322, 337, 350, 353, 354, 355 (right, left), 365, 366, 372, 380, 405 (jelly beans), 444 (right), 457, 459, 460, 466, 481, 495 (roulette wheel), 496, 503, 509, 510, 512, 513, 546, 550 (bottom), 552; • © Photolibrary, p. 89/SPL/Gregory Sams, 216/Photo Researchers/Photo Inc, 323/Bay Hippisley, 331/Science Photo Library, 363/Science Photo Library, 393/Vince Cavataio, 395/Alfred Pasieka, 405 (batsman)/Tommy Hindley, 506/Sydney/ SPL; • © Purestock, p. 174 (top); • © Stockbyte, p. 416 (eight images), 463.

Software The authors and publisher would like to thank the following software providers for their assistance and for permission to use their materials. However, the use of such material does not imply that the providers endorse this product in any way. Third party software — registered full version ordering information Full versions of third party software may be obtained by contacting the companies listed below. GraphEq and Poly Evaluation copies of GraphEq and Poly have been included with permission from Pedagoguery Software, Inc. email: [email protected] Web: www.peda.com Graphmatica Reproduced with permission of kSoft, Inc. 345 Montecillo Dr., Walnut Creek, CA 94595-2654. email: [email protected] Web: www.graphmatica.com

xiii Software included is for evaluation purposes only. The user is expected to register share-ware if use exceeds 30 days. Order forms are available at www.graphmatica.com/ register.txt Equation Grapher with Regression Analyser Reproduced with permission of MFSoft International. email: [email protected] Web: www.mfsoft.com Microsoft Excel Screen shots reproduced by permission of Microsoft Corporation. Note: Microsoft Software has been used only in screen dumps. Microsoft Excel is a registered trademark of the Microsoft Corporation in the United States and/or other countries.

Every effort has been made to trace the ownership of copyright material. Information that will enable the publisher to trace the copyright holders or to rectify any error or omission in subsequent reprints will be welcome. In such cases, please contact the Permission Section of John Wiley & Sons Australia, who will arrange for the payment of the usual fee.

5_61_08144_MQ11C2E_Prelim Page xiv Monday, November 10, 2008 1:51 PM

Number systems: the Real Number System

1 syllabus reference Core topic: Real and complex number systems

In this chapter 1A 1B 1C 1D 1E 1F 1G 1H 1I 1J 1K 1L 1M

Classification of numbers Recurring decimals Surds Simplifying surds Addition and subtraction of surds Multiplication of surds The Distributive Law Division of surds Rationalising denominators Rationalising denominators using conjugate surds Further properties of real numbers — modulus Solving equations using absolute values Solving inequations

2

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

• structure of the real number system including rational numbers and irrational numbers • simple manipulation of surds

Introduction The number systems used today evolved from a basic and practical need of primitive people to count and measure magnitudes and quantities such as livestock, people, possessions, time and so on. Early cultures and societies used their body parts, such as fingers and toes, as a basis for their numeration systems. As the need for larger numbers grew, symbols were developed to represent them. Ancient Egyptians, for example, used the symbol of the lotus flower to represent the number 1000, and Romans used the letter M to represent 1000. Roman numerals can be seen today on some clock and watch faces. At the end of movie credits Roman numerals are often used to indicate the year in which the movie was made. For example, MCMXCIX represents the year 1999 and MMIX represents the year 2009. As societies grew and architecture and engineering developed, number systems became more sophisticated. Number use developed from solely whole numbers to fractions, decimals and irrational numbers. We shall explore these different types of numbers and classify them into their specific groups. Consider solutions to equations such as: 2x = 10, 3x = 15, 20x = 100 What do they have in common? Each of the statements is true for a whole-number value of x. This type of equation represents many real-life situations; for example, how many people will I need to collect $2 from to cover the cost of hiring a $10 game? The first types of numbers to evolve were the whole numbers. As you work through this chapter on the Real Number System and Chapter 2 you will be introduced to types of numbers that evolved to fill other, more sophisticated needs.

3

Chapter 1 Number systems: the Real Number System

The Real Number System The Real Number System contains the set of rational and irrational numbers. It is denoted by the symbol R. Real numbers R

Rational numbers Q

Irrational numbers I (surds, non-terminating and non-recurring decimals, π ,e)

Negative Z–

Integers Z

Non-integer rationals (terminating and recurring decimals)

Zero (neither positive nor negative)

Positive Z+ (Natural numbers N)

The set of real numbers contains a number of subsets which can be classified as shown in the chart above.

Classification of numbers: rational and irrational Rational numbers (Q) A rational number (ratio-nal) is a number which can be expressed as a ratio of two a integers in the form --- where b ≠ 0 and a and b have no common factors. b Rational numbers are given the symbol Q. Examples are: --1- , 5

--2- , 7

3----, 10

--9- , 4

7

Rational numbers may be expressed as terminating decimals. Examples are: 7----10

= 0.7,

1 --4

= 0.25,

5 --8

= 0.625,

9 --5

= 1.8

These decimal numbers terminate after a specific number of digits. Rational numbers may be expressed as recurring decimals (non-terminating or periodic decimals). For example: . 1 --- = 0.333 333 . . . or 0.3 3 .. 9 ------ = 0.818 181 . . . or 0.8 1 11 . --5- = 0.833 333 . . . or 0.83 6 . . 3 ------ = 0.230 769 230 769 . . . or 0.2 30769 13

These decimals do not terminate, and the specific digit (or number of digits) is repeated in a pattern. Recurring decimals are represented by placing a dot or overscore above the repeating digit or pattern.

–3.743 3 –2 –4 –4 –3 –2 –1

1– 2

0

1.63 1

2

3.6 3 4Q

4

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Rational numbers are defined in set notation as: Q = set of rational numbers a Q = --- , a, b ∈ Z, b ≠ 0, g.c.d (a, b) = 1 where ∈ means ‘an element of’ and b g.c.d. (a, b) = 1 means greatest common divisor of (a, b) = 1. Rational numbers may be represented on the number line (as illustrated on page 3) and include whole numbers, fractions, and terminating and recurring decimals. Whole numbers form a set of integers (which is a subset of the set of rational numbers).

{

}

Integers (Z) The set of integers consists of positive and negative whole numbers, and 0 (which is neither positive nor negative). They are denoted by the letter Z and can be further divided into subsets. That is: Z = {. . . −3, −2, −1, 0, 1, 2, 3, . . .} Z + = {1, 2, 3, 4, 5, 6, . . .} Z − = {−1, −2, −3, −4, −5, −6 . . .} Positive integers are also known as natural numbers (or counting numbers) and are denoted by the letter N. That is: N = {1, 2, 3, 4, 5, 6, . . .} Integers may be represented on the number line as illustrated below. –3 –2 –1 0 1 2 3 Z The set of integers

1 2 3 4 5 6 N The set of positive integers or natural numbers

Z – –6 –5 –4 –3 –2 –1 The set of negative integers

Note: Integers on the number line are marked with a solid dot to indicate that they are the only points in which we are interested.

Irrational numbers (I) Numbers that cannot be expressed as a ratio between two integers are called irrational numbers. Irrational numbers are denoted by the letter I. Numbers such as surds (for example 7 , 10 ), decimals that neither terminate nor recur, and π and e are examples of irrational numbers. The numbers π and e are examples of transcendental numbers; these will be discussed briefly later in this chapter. Irrational numbers may also be represented on the number line with the aid of a ruler and compass. An irrational number (ir-ratio-nal) is a number which cannot be expressed as a ratio of two integers in the form --a- where b ≠ 0. b Irrational numbers are given the symbol I. Examples are: 7 13 , 5 21 , ------- , π, e 9 Irrational numbers may be expressed as decimals. For example: 7,

5 = 2.236 067 977 5 . . . 0.03 = 0.173 205 080 757 . . . 18 = 4.242 640 687 12 . . . 2 7 = 5.291 502 622 13 . . . π = 3.141 592 653 59 . . . e = 2.718 281 828 46 . . . These decimals do not terminate, and the digits do not repeat themselves in any particular pattern or order (that is, they are non-terminating and non-recurring).

Chapter 1 Number systems: the Real Number System

5

Once the decimal approximation for an irrational number is obtained, it can be shown on a number line. For example, 18 ≈ 4.24. This approximation is between 4 and 5, but closer to 4. –2 7

– 5

0.03

e

18

–6 –5 –4 –3 –2 –1 0 1 2 3 4 5

Irrational numbers in surd form can also be represented on the number line exactly, as follows. Consider an isosceles right-angled triangle of side length 1 unit. By Pythagoras’ Theorem, (OB)2 = (OA)2 + (AB)2; B therefore the length of the hypotenuse is 2 units. By using a compass, we can transfer the length of the 2 units hypotenuse OB to the number line (labelled C). This distance 1 unit can now be measured using a ruler. Although this distance A C O will be inaccurate due to the equipment used, there is an exact 2 2 R 0 1 unit 1 point on the number line for each irrational number. This geometric model can be extended to any irrational number in surd form.

π (pi)

The symbol π (pi) is used for a particular number; that is, the circumference of a circle whose diameter length is 1 unit. It can be approximated as a decimal which is nonterminating and non-recurring. Therefore, π is classified as an irrational number. (It is also called a transcendental number and cannot be expressed as a surd.) In decimal form, π = 3.141 592 653 589 793 23 . . . It has been calculated to trillions of decimal places with the aid of a supercomputer. Rational (Q) and irrational (I) numbers belong to the set of real numbers denoted by the symbol R. They can be positive, negative or 0. The real numbers can be represented on a number line as shown (irrational numbers above the line and rational numbers below the line). 2 – 12

–4 –3 –2 –1 –3.62

–π 4

–1– 2

– 3– 2

0

1

3

2

0.2 1.75

Relationship between subsets

3e — 2

9 3

4 3 5

3–

ε =R

The relationship which exists between the subQ (Rational numbers) sets of the Real Number System can be illustrated in a Venn diagram as shown on the right. Z (Integers) I We can say N ⊂ Z, Z ⊂ Q, and so on, where N (Irrational ⊂ means ‘is a subset of’. (Natural numbers) To classify a number as either rational or numbers) irrational: 1. Determine whether it can be expressed as an integer, a fraction, or a terminating or recurring decimal. 2. If the answer is yes, the number is rational; if the answer is no, the number is irrational.

6

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

WORKED Example 1 Specify whether the following numbers are rational or irrational. 1 --4

a

b

c

16

11

d 2π

e 0.28

THINK a b

1--4

f

3

64

g

3

22

h

3 1 --8

WRITE a

is already in rational form.

1 --4

16 = 4

b

1

Evaluate

16 .

2

The answer is an integer, so classify

is rational.

16 is rational.

16 . c

Evaluate

11 .

2

The answer is a non-terminating and non-recurring decimal; classify

d

11 = 3.316 624 790 36 . . .

c

1

11 is irrational.

11 . d 2π = 6.283 185 307 18 . . .

1

Use your calculator to find the value of 2π.

2

The answer is a non-terminating and non-recurring decimal; classify 2π.

2π is irrational.

e 0.28 is a terminating decimal; classify it accordingly.

e 0.28 is rational.

f

2

The answer is a whole number, so 3

3

64 = 4

3

64 is rational.

3

22 = 2.802 039 330 66 . . .

3

22 is irrational.

64 . 3

g

22 .

1

Evaluate

2

The result is a non-terminating and non-recurring decimal; classify

h

f

64 .

Evaluate

classify g

3

1

3

22 .

3 1 --- . 8

1

Evaluate

2

The result is a number in a rational form.

h

3 1 --8

=

3 1 --8

is rational.

1 --2

7

Chapter 1 Number systems: the Real Number System

Graphics Calculator tip! Square, cube and nth roots A graphics calculator can be used to find the square root, cube root or higher root of a number.

For the Casio fx-9860G AU 1. Press MENU and select RUN-MAT (highlight RUN-MAT and press EXE ). 2. To calculate the square root of a number (for example, 8 ), press SHIFT [ followed by the number (8 in this case) and press EXE .

]

3. To calculate the cube root of a number (for example, 3 8 ), press SHIFT [ 3 ] followed by the number (8 in this case) and press EXE . 4. To calculate a higher root of a number (for example, 4 81), first enter the type of root (4 in this example), then press SHIFT [ x 8] followed by the number (81 in this case) and press EXE .

For the TI-Nspire CAS 1. From the Home screen (press c), highlight 1: Calculator and press ·. Alternatively, open a new Calculator document. Press / N and follow the prompts as to whether you wish to save the previous document. Then press 1 to select 1: Add Calculator. 2. To calculate the square root of a number (for example, 8 ), press / and q, followed by the number (8 in this case) and press ·. For the decimal approximation to this answer, press / and ·. 3. To calculate the cube root of a number (for example, 3 8 ), press / and l. Complete the gaps in the expression on the screen. First enter 3 for the type of root (n) and then enter the number (8 in this case for x). Use the tab key to move to the appropriate place in the expression. Press ·. 4. To calculate a higher root of a number (for example, 4 81), repeat the steps used for the cube root. In this example, enter 4 for the type of root (n). (Note that the square and nth root functions can also be accessed from the Catalog menu. Press k to access the catalog. Select Option 5 (by pressing 5) then highlight the required symbol and press ·.)

Summary of set notation The following symbols are used to describe relationships in sets. Consider a group of numbers from 1 to 9 (i.e. 1, 2, 3, 4, 5, 6, 7, 8, 9). These numbers can be referred to as a set and denoted by A such that A = {1, 2, 3, 4, 5, 6, 7, 8, 9}. We can say that 2 is an element of set A and write this as 2 ∈ A. Similarly 0 is not an element of set A and this is written as 0 ∉ A. The elements 2 and 4 both belong to set A and this can be written as {2, 4} ⊂ A, where 2 and 4 are a subset of A.

8

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

WORKED Example 2

12 p 13 Classify each of the following elements of the set 5, ------ , −3.9, − ------ , 23 , --- into the 5 2 2 smallest subset in which it belongs, using Q, I, Z, Z + and Z −.

{

THINK 1 The number 5 is a positive whole number; classify it accordingly. 13 2 (a) Change ------ into a decimal.

WRITE 5 ∈ Z+

2

13 -----2

3 4

5

6

(b) The fraction can be expressed as a terminating decimal; therefore it can be classified as a rational number. The fraction a is in the form --- , b ≠ 0, so it is rational. b The number −3.9 is a terminating decimal, so classify it accordingly. ------ . (a) Simplify − 12 2

(b) The result is a negative whole number, ------ accordingly. so classify − 12 2 (a) Use your calculator to find the value of (b) The result is a non-terminating and non-recurring decimal, so 23 can be classified as an irrational number. (a) Change --π- into a decimal. 5 (b) The resulting decimal is neither terminating nor recurring, --- is an irrational number. so π 5

}

= 6.5

13 -----2 13 -----2

∈Q

−3.9 ∈ Q ------ = −6 − 12 2

− ------ ∈ Z − 12

2

23 .

23 = 4.795 831 523 31 . . . 23 ∈ I

--π- = 0.628 318 530 718 . . . 5 π --- ∈ I 5

remember remember 1. The real number system (R) contains the set of rational numbers (Q) and the set of irrational numbers (I). 2. Rational numbers are those that can be written as a ratio of two whole numbers a in the form --- where b ≠ 0. Rational numbers include whole numbers, b fractions, and terminating and recurring decimals. 3. The set of rational numbers includes the set of integers (Z). 4. The set of integers consists of positive whole numbers (Z +), negative whole numbers (Z −) and 0. Positive whole numbers (Z +) are also called natural numbers (N). 5. Irrational numbers cannot be expressed as a ratio of two whole numbers in the a form --- where b ≠ 0. Irrational numbers include surds, non-terminating and b non-recurring decimals, and numbers such as π and e.

9

Chapter 1 Number systems: the Real Number System

1A WORKED

Example

1

Classification of numbers

1 Specify whether the following numbers are rational (Q) or irrational (I). 4

f

0.04 Q g 2 1---

Q

b

4 --5

a

Q

2

k −2.4

Q

25 -----9

p

Q

u – 81 Q

l

7 --9

c

Q

Q

100

d

Q

2 9 --4

e

I

h

5

I

i

m

14.4 I

n

1.44

j

Q

Q o

7

I

0.15 Q

π

I

q 7.32 Q

r

– 21 I

s

1000 I

t

7.216 349 157 . .I .

v 3π

w

3

x

1 -----16

y

3

I

62 I

Q

0.0001 I

2 Specify whether the following numbers are rational (Q), irrational (I) or neither. a f

1 --8 3

b

Q

Q

625

81 I

g – 11 I

21 I

l

k

3

p

64 -----16

u

22 π --------7

Q

q

I

v

π --7

3

h m

I 2----25

c

I

r

11 ------ Q 4 1.44 ---------- Q 4 3 ( –5 )2 I

i

------62

s

I

– 1.728 Q w 6 4

Q

d

0 --8

π

3n − -----

11

x

3

e −6 1---

Q I

j

Q

o

27 Q

( 2)

Q

7

4

Q

8 --0

Undefined 1 --------100

t

1-----4

y

4 6

Q Q I

3 multiple choice Which of the following best represents a rational number? A π

4 --9

B

C

9----12

D

3

3

4 multiple choice Which of the following best represents an irrational number? A − 81 B 6--C 3 343 D 0.0676 5

E none of these

E

22

5 multiple choice Which of the following statements regarding the numbers −0.69, 7 , π--- , 49 is correct? 3 A π--- is the only rational number. 3

B

7 and

C −0.69 and

49 are both irrational numbers. 49 are the only rational numbers.

D 7 , --π- and 49 are all irrational numbers. 3 E −0.69 is the only rational number. 6 multiple choice 11Which of the following statements regarding the numbers 2 --1- , − ----, 2 3 correct? ------ and 624 are both irrational numbers. A − 11 3

624 is an irrational number and B C 2 1--- is the only rational number.

3

99 is a rational number.

2

D 624 and 3 99 are both irrational numbers. 11E 2 --1- is a rational number and − ----is an irrational number. 2

3

624 ,

3

99 is

10

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

7 Classify each of the following into the smallest subset in which it belongs, using Q, Z + and Z −. . 1a 5 Z+ b 0.621 Q c d 0.26 Q e 3 + 16 Z + ----Q

WORKED

Example

2

f Z−

Z−

Q

8 Z

l

0 --4

Q

q

------8Z+ 2 4 --Q 3

0.515 151 . . . g

k 9 – 144 . . p 0.4 21 Q u – 8× 2

+

3

v

h

81 42 ------ Z + 6

i 2

m –( – 4 ) Z − n r

5

– 32

Z−

6

Z+

w ( 2)

Z+

7 4 9 --3

j o

Z+

s

(– 6)

x

100-----------2

2

Z

Z+ +

−

27 -----3 2 −  3--- 2

Z− Q

t

− 6--- Q

y

3

5

– 343 Z −

8 Classify each of the following into the smallest subset in which it belongs using Q, I, Z + and Z −. 9- Q a 6 Z+ b 0.3415 . .I . c d ----e – 2 25 Z − 7 I 16

f

6× 2

k

144 --------9

p 5π u −

I g

Z

– 64 Z −

h 21 × 5

v

I

m −

11 -----19

r

( 3)

Z

q 16 × 3 – 27

I 16 -----8

3

l

Z+

– 49

−

Q

8 × 12.5 w − 1--5

Z+

I i

3

I

Q

Q

0.612 612 . . . j

n

9-------144

s

7× 5 I t

x (π)2

Q

I

0.25 Q 50 -----2

o

y

Z+

– 6 × 3 16 Z − – 3 – 125

Z+

9 multiple choice The smallest subset in which 7 + 2  3 1--- belongs is: 8

A Q

C Z+

B I

10 multiple choice The smallest subset in which A Q

×

144 --------9

B I

3 512

--------8 +

D Z

E Z−

belongs is:

C Z

D Z

E Z−

11 multiple choice Which of the following statements regarding numbers 16, −3 2 , 0, π,

{

A 16 and 0 are the only rational numbers. B 16, 0 and

8 --2

may be expressed as rational numbers.

C 16 and 0 are positive integers. D −3 2 is the only irrational number. E π is the only irrational number. 12 multiple choice Which of the following statements regarding the given set of numbers { 2 + 9 , 11 , 16 2 , 32 , 81 } is correct? A All of the above numbers in the set are irrational. B

2 + 9 and

C

2 + 9 is a rational number of the set.

D

81 is the only rational number of the set.

E

11 and

11 are the only irrational numbers of the set.

32 are the only irrational numbers of the set.

8 --2

} is correct?

Chapter 1 Number systems: the Real Number System

11

Recurring decimals A rational number may be converted to a decimal by dividing the numerator by the denominator. The resulting decimal may be a terminating decimal containing a specific number of digits, that is: 7 --5

= 1.4

1 --8

or

= 0.125

or it may be a recurring decimal containing a repeating digit or pattern, that is: 2 --9

= 0.222 . . .

or

7----13

= 0.538 461 538 461 . . .

For convenience, recurring decimals are represented by placing a dot over the repeating digit, for example: . 1. 0.777 777 7 . . . can be written as 0.7. . 2. 0.26666 . . . can be written as 0.26. If two or more digits repeat the same pattern, then dots or the overscore ( used as shown: .. 1. 0.454 545 . . . can be written as 0.4 5 or alternatively 0.45 . . . 2. 0.752 137 521 3 . . . can be written as 0.7 5213 or alternatively 0.75213 .

) are

Note: When using the overscore, place it over the whole pattern. The dots, however, are placed over the first and the last digits only of the repeating pattern.

WORKED Example 3 State which of the following rational numbers can be expressed as recurring decimals. a

2 -----27

b

5 --8

THINK a

b

WRITE 2----27

1

To convert by 27.

2

Use the overscore to indicate the repeating pattern.

3

Write your conclusion.

1

Convert by 8).

2

The resulting decimal terminates, so state your conclusion.

5--8

to a decimal, divide 2

to a decimal (divide 5

a

2----27

= 0.074 074 074 . . . = 0.074

2----27

can be written as a recurring decimal. b

5 --8 5 --8

= 0.625

cannot be written as a recurring decimal.

12

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Whole numbers and terminating decimals such as 3, 0.25 and 6.731 can easily be expressed as rational numbers. For example: 1. we may write 3 as

3 --1

2. we may write 0.25 as

25 --------100

=

1 --4

7313. we may write 6.731 as 6 ----------or 1000

6731 ------------ . 1000

a In each of these cases, the whole number and decimals are expressed in the form --- . b Recurring decimals are rational numbers. Therefore they can be converted to the a form --- . b

WORKED Example 4 Express the following recurring decimals as rational numbers in their simplest form. . .. a 0.4 b 0.2 1 c 1.285 THINK

WRITE

a

a

b

x = 0.444 444 . . .

[1]

We need to multiply both sides of the equation by a power of 10. The number of zeros in the power of 10 should be equal to the number of repeated digits. Since 1 digit is repeated, multiply both sides of equation [1] by 10. Label the new equation [2].

10x = 4.444 444 . . .

[2]

Subtract equation [1] from equation [2]. This removes all the repeating digits after the decimal point.

[2] − [1]:

1

Let x represent the recurring decimal. This is equation [1].

2

3

4

Divide both sides of the equation by 9.

5

Verify the answer using a calculator and you will . obtain the original value, 0.4.

1

Let x represent the recurring decimal. This is equation [1].

2

Since 2 digits are repeated, multiply both sides of equation [1] by 100 and label the new equation [2].

10x − x = 4.444 444 . . . − 0.444 444 . . . 9x = 4 9x 4 ------ = --9 9 4 x = --9

b

x = 0.212 121 21 . . . 100x = 21.212 121 21 . . .

[1] [2]

Chapter 1 Number systems: the Real Number System

THINK

c

13

WRITE

3

Subtract equation [1] from equation [2]. This removes all the repeating digits after the decimal point.

4

Divide both sides of the equation by 99.

5

Cancel to the simplest form; that is, divide through by 3.

6

Verify the answer using a calculator.

1

Let x represent the recurring decimal. This is equation [1].

2

[2] − [1]: 100x − x = 21.212 121 21 . . . − 0.212 121 21 . . . 99x = 21 99x 21 --------- = -----99 99 21 x = -----99 7 x = -----33

x = 1.285 285 285 . . .

[1]

Since 3 digits are repeated, multiply both sides of equation [1] by 1000 and label the new equation [2].

1000x = 1285.285 285 285 . . .

[2]

3

Subtract equation [1] from equation [2]. This removes all the repeating digits after the decimal point.

[2] − [1]: 1000x − x = 1285.285 285 285 . . . − 1.285 285 285 . . . 999x = 1284

4

Divide both sides of the equation by 999.

5

Cancel to the simplest form; that is, divide through by 3.

6

Verify the answer using a calculator.

c

999x ------------ = 1284 -----------999 999 1284 x = -----------999 428 x = --------333

remember remember 1. Rational numbers can be converted to decimals by dividing the numerator by the denominator. The resulting decimal can be either terminating or recurring. 2. Terminating decimals contain a specific number of digits. 3. Recurring decimals contain a repeating digit or a repeating pattern of digits. 4. Recurring decimals are represented by placing dots over the first and the last digits of the repeating pattern. Alternatively, an overscore can be placed over the whole pattern that repeats. 5. Recurring decimals are rational numbers and may be expressed as a ratio of two integers.

14

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

1B WORKED

Example

3

1 State which of the following rational numbers can be expressed as recurring decimals. 61a --1b --1c --1d ----e ----f

1 c f g j k l n oq r t w x y

k p u

WORKED

Example

4

Recurring decimals

8 4----11 2 --3 2----31 8----17

g l q v

2 5 --9 1 --6 2 --9 7----23

h m r w

3 7----16 3 --4 41-------333 7----15

i n s x

19 9----25 3----13 5 --8 3----22

j o t y

17 5 --7 5----21 17 -----18 7----33

2 Express the following recurring decimals as rational numbers in their simplest form. . 2 . . 8 . 5 . a 0.2 --9b 0.7 --79c 0.8 --9d 0.5 --9e 0.4 --49. . 17 . 19 . 31 . 32 f 0.16 1--6g 0.37 ----h 0.42 ----i 0.68 ----j 0.71 ----45 45 45 45 . . . 53 .. 4 . . 34 . . 36728 495 k 2.62 2 ----l 0.53 ----m 0.1 2 ----n 1.3 4 1 ----o 3.74 1 3 -------99 45 99 33 . . 361 . . 427 . . 868 . . 323 . . 152 -----------------------3 -----------------p 0.3 61 999 q 0.4 27 999 r 0.52 13 1665 s 0.3 23 999 t 3.4 56 333 . . 157 . . 1237 . 5611 . . 2 -------------------- x ----------------- y u 0.72 13 v 0.523 w 0.624 7 0.6234 0.1 53846 ----300 1980 13 18 9000 3 multiple choice

.. The recurring decimal 0.7 8 can be expressed as: -----C 77 ----------A 71 B 78 99 90

90

D

71 -----99

E

78 -----99

D

527-------990

E

532-------990

4 multiple choice

.. The recurring decimal 0.53 2 can be expressed as: 266532479A -------C -------B -------495

900

999

5 multiple choice Which statement regarding the fractions 1--- , 3--- , 2

A B C D E

7

11 1 4 ------ , --- , --13 3 5

is correct?

1 --- , 3 --- and 11 ------ are the only fractions which represent terminating decimals. 2 7 13 3 --- and 1 --- are the only fractions which represent terminating decimals. 7 3 3 --- , 11 ------ and 1 --- are fractions which represent recurring decimals. 7 13 3 11 ------ and 1 --- are the only fractions which represent recurring decimals. 13 3 1 4 --- and --- are fractions which represent recurring decimals. 2 5

6 multiple choice

. . The recurring decimal 0.369 can be expressed in its simplest form as the following fraction: 407 41D -----------------------------C 123 --------B 3663 E -------A 369 1100 999

9900

333

111

7 Irene and Bella are arguing about the correct way of writing the recurring decimal 0.020 20 . . . . Irene says it should be written as 0.020, while Bella thinks it is 0.020. Which of the girls is right? Irene. It can also be written as 0.02.

Chapter 1 Number systems: the Real Number System

15

Real number investigations A real number can be defined as a number that can be plotted on a number line. Even if the position of the number on the line is only an approximate value, as long as a number can be represented by one point on a line it can be regarded as real. This is not so with the numbers you will deal with in Chapter 2. The following steps will enable you to plot irrational numbers such as surds (for example 2 and 5 ) on a number line. Materials needed: ruler, a set of compasses, set square. Step 1

Step 2

Draw a number line approximately 10 cm long, with unit divisions of 2 cm. How can we draw a line segment exactly 2 units long? Using Pythagoras’ Theorem we can obtain the triangle shown at right which shows us that 2 = 1 + 1 .

–1

1

1

0

1

2

3

Use a set of compasses to transfer the length of the hypotenuse to the number line.

√2

–1

Step 4

1

Use a set square to construct a right-angled triangle as shown below:

√2

Step 3

√2

1

0

√2

1

2

3

If a second right-angled triangle (of height 1 cm) were constructed on this hypotenuse, what would be the length of its hypotenuse?

√3 √2

1

–1

Step 5

0

1

√2

√3

2

3

Continue constructing in this way to plot 7 on the original number line. Use your number line to give an approximate value for 7 .

16

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Other number systems Introduction Throughout early civilisations, numbers have been represented and recorded in a variety of ways. Our numeration system uses the 10 digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and combinations of these. It is called the decimal or base 10 system (possibly influenced by the fact that we have 10 fingers). Past civilisations have used base 5 and base 20 systems (again influenced by the fingers on one hand and the total number of digits). Mesopotamians used a base 60 system, which is still used today for units of time (60 seconds in 1 minute and 60 minutes in 1 hour). Numeration systems that are used today include a binary or base 2 system and a modular or base 12 system.

Place value The place value system was introduced as a means of recording numbers. Look at the number 285. In our numeration system — the base 10 (decimal) system — we interpret the number 285 (base 10) or 28510 as: plus 2 lots of 100 or 2 × 102 plus 8 lots of 10 or 8 × 101 plus 5 lots of 1 or 5 × 100 Using the base 5 system Numbers in the base 5 system use the digits 0, 1, 2, 3 and 4 only. The number 285 (base 10) can be written as a base 5 number in the following way: 2 lots of 125 or 2 × 53 plus 1 lot of 25 or 1 × 52 plus 2 lots of 5 or 2 × 51 plus 0 lots of 1 or 0 × 50 So 28510 = 21205

Using the base 2 system Numbers in the base 2 system use the digits 0 and 1 only. The number 285 (base 10) can be written as a base 2 number in the following way: 1 lot of 256 or 1 × 28 plus 0 lots of 128 or 0 × 27 plus 0 lots of 64 or 0 × 26 plus 0 lots of 32 or 0 × 25 plus 1 lot of 16 or 1 × 24 plus 1 lot of 8 or 1 × 23 plus 1 lot of 4 or 1 × 22 plus 0 lots of 2 or 0 × 21 plus 1 lot of 1 or 1 × 20 So 28510 = 100 011 1012

Notice how we need to use zeros to hold each place value. See Solutions Manual.

1 Investigate the following points relating to non-base 10 numbers, giving examples in each case: a How could numbers of different bases be compared to each other? b How are numbers of the same (non-base 10) system added and subtracted? c How are numbers of the same (non-base 10) system multiplied and divided? d How are fractions and decimals of a non-base 10 system represented?

Chapter 1 Number systems: the Real Number System

17

Binary systems As the name suggests, this numeration system is based on 2. In this system, 0 and 1 are the only two digits used. The binary system is used in computers. 2 Investigate how the binary system is used in computers, circuits or compact discs. Devise a situation which calls for the use of a binary system.

Modular arithmetic Modular arithmetic involves ‘clock’ arithmetic where, instead of saying that the time is 14 o’clock, we say it is 2 o’clock. This is called modular (mod 12) arithmetic. Any integer can be converted to modular (mod 12) arithmetic by subtracting 12 or any multiple of 12 from the integer. The remainder is called the residue. For example: 32 = 2 × 12 + 8 ≡ 8 (mod 12)

68 = 5 × 13 + 3 ≡ 3 (mod 13)

29 = 4 × 6 + 5 ≡ 5 (mod 6)

The remainders or residues in this case are 8, 3 and 5 respectively. 3 Investigate the purpose, usefulness and limitations of modular arithmetic. Include illustrations of how numbers of modular arithmetic are represented via a clock pattern.

Surds A surd is an irrational number which can only be represented exactly using a root sign or radical, for example: Examples of surds include:

,

3

4

,

5,

3

11 ,

16 ,

3

125 ,

7,

4

15

Examples that are not surds include: 9,

4

81

Numbers that are not surds can be simplified to rational numbers, that is: 9 = 3,

16 = 4 ,

3

125 = 5 ,

4

81 = 3

18

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

WORKED Example 5 Which of the following numbers are surds? a

25

b

10

c

1 --4

d

3

e

11

4

59

f

3

343

THINK

WRITE

a

Evaluate 25 . The answer is rational (since it is a whole number), so state your conclusion.

a

25 = 5 25 is not a surd.

b

2

Evaluate 10 . The answer is irrational (since it is a non-recurring and non-terminating decimal), so state your conclusion.

10 = 3.162 277 660 17 . . . 10 is a surd.

1

Evaluate

c

1 --4

=

2

The answer is rational (a fraction); state your conclusion.

1 --4

is not a surd.

1 2

b

c

d

1

1 2

e

1 2

f

1 2

1 --- . 4

Evaluate 3 11 . The answer is irrational (a nonterminating and non-recurring decimal), so state your conclusion.

d

Evaluate 4 59 . The answer is irrational, so classify 4 59 accordingly.

e

Evaluate 3 343 . The answer is rational; state your conclusion.

f

3 3

4 4

3 3

1 --2

11 = 2.223 980 090 57 . .. 11 is a surd.

59 = 2.771 488 002 48 . . . 59 is a surd.

343 = 7 343 is not a surd.

So b, d and e are surds.

Proof that a number is irrational As part of your Mathematics C course you are required to study a variety of types of proofs. One such method is called ‘Proof by contradiction’. This method is so named because the logical argument of the proof is based on an assumption that leads to contradiction within the proof. Therefore the original assumption must be false. a An irrational number is one that cannot be expressed in the form --- (where a and b b are integers). The next worked example sets out to prove that 2 is irrational.

Chapter 1 Number systems: the Real Number System

19

SLE 8: Use a proof by contradiction to show that 2 is irrational.

WORKED Example 6 Prove that

2 is irrational.

THINK 2 is rational; that is, it a can be written as --- in simplest form. b We need to show that a and b have no common factors.

WRITE a 2 = --- where b ≠ 0 b

1

Assume that

2

Square both sides of the equation.

3

Rearrange the equation to make a2 the subject of the formula.

4

If x is an even number then x = 2n.

∴ a2 is an even number and a must also be even; that is, a has a factor of 2.

5

Since a is even it can be written as a = 2r.

∴ a = 2r

6

Square both sides.

2

a 2 = ----2 b a2 = 2b2

But 7

Equating [1] and [2]

[1]

a2 = 4r2 a2 = 2b2 from [1]

[2]

∴ 2b2 = 4r2 2

4r b2 = -------2 = 2r2 2 ∴ b is an even number and b must also be even; that is, b has a factor of 2. 8

State your conclusion.

Both a and b have a common factor of 2. This contradicts the original assumption that a 2 = --- where a and b have no common factor. b ∴ 2 is not rational. ∴ It must be irrational.

The ‘dialogue’ included in the worked example should be present in all proofs and is an essential part of the communication that is needed in all your solutions. Note: An irrational number written in surd form gives an exact value of the number; whereas the same number written in decimal form (for example, to 4 decimal places) gives an approximate value.

remember remember A number is a surd if: 1. it is an irrational number (equals a non-terminating, non-recurring decimal) 2. it can be written exactly only by using a radical (or root sign).

20

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

1C WORKED

Example

5 1 b d f g h i l m o q s t w

Surds

1 Which of the numbers below are surds? a

81

b

48

c

f

11

g

3 --4

h

k 4 100

l

p

3

125

q

u

4

16

v

d

1.6

e

0.16

3 3

-----27

i

1000

j

1.44

m

3

32

n

361

o

6+ 6

r

2π

s

169

t

2

w

3

x

0.0001

y

5

3

27 ,

2 + 10

( 7)

16

33

3

3

100 7 --8

32

2 multiple choice

{

The correct statement regarding the set of numbers A B C D

3

27 and 6 --9 6 --9

6 --- , 9

20 ,

54 ,

}

9 is:

9 are the only rational numbers of the set.

is the only surd of the set. and

20 and

20 are the only surds of the set. 54 are the only surds of the set.

E All of the numbers of the set are surds. WORKED

Example

6

3 Prove that the following numbers are irrational, using a proof by contradiction: a

b

3

c

5

7

4 multiple choice Which of the numbers of the set A

3 1

-----27

B

21 only

{

C

1 1--- , 3 ----, 4 27 1 --8

1 --- , 8

only

21 , 1 --8

D

}

3

8 are surds?

and

3

8

E

1 --8

and

21 only

5 multiple choice

{

Which statement regarding the set of numbers π, A

12 is a surd.

C π is irrational but not a surd. E

1----49

1----, 49

12 ,

16 ,

}

3 + 1 is not true?

B

12 and

16 are surds.

D

12 and

3 + 1 are not rational.

when simplified is a rational number.

6 multiple choice

{

Which statement regarding the set of numbers 6 7 , not true? A

144 --------16

when simplified is an integer.

C 7 6 is smaller than 9 2 . E 6 7 , 7 6 , 9 2 and

18 are surds.

B

144 --------16

144 --------- , 16

and

7 6, 9 2,

18 ,

25 are not surds.

D 9 2 is smaller than 6 7 .

}

25 is

Chapter 1 Number systems: the Real Number System

7 If a is a multiple of 4, find the smallest, non-zero rational value of 8 Find the smallest value of m, where m is a positive integer, so that

6 3

21

a . 2 (when a = 64) 16m is not a surd. m = 4

Simplifying surds To simplify a surd means to make a number (or an expression) under the radical ( ) as small as possible. To simplify a surd (if it is possible), it should be rewritten as a product of two factors, one of which is a perfect square, that is, 4, 9, 16, 25, 36, 49, 64, 81, 100 and so on. We must always aim to obtain the largest perfect square when simplifying surds so that there are fewer steps involved in obtaining the answer. For example, written as

4 × 8 = 2 8 ; however,

32 = 2 × 2 2 ; that is

32 could be

8 can be further simplified to 2 2 , so

32 = 4 2 . If, however, the largest perfect square had been

selected and 32 had been written as would be obtained in fewer steps.

16 × 2 =

16 ×

2 = 4 2 , the same answer

WORKED Example 7 Simplify the following surds. Assume that x and y are positive real numbers. a

384

b 3 405

c − 1--- 175 8

d 5 180 x 3 y 5

THINK

WRITE

a

a

1

384 =

64 × 6

64 × 6

Express 64 × 6 as the product of two surds.

=

3

Simplify the square root from the perfect square (that is, 64 = 8).

=8 6

1

Express 405 as a product of two factors, one of which is the largest possible perfect square.

2

Express surds.

3

Simplify

4

Multiply together the whole numbers outside the root (3 and 9).

2

b

Express 384 as a product of two factors where one factor is the largest possible perfect square.

81 × 5 as a product of two 81 .

b 3 405 = 3 81 × 5

= 3 81 × 5 = 3×9 5 = 27 5 Continued over page

22

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

THINK

WRITE

c

c − 1--- 175 = − 1--- 25 × 7

d

1

Express 175 as a product of two factors where one factor is the largest possible perfect square.

2

Express surds.

3

Simplify

4

Multiply together the numbers outside the square root sign.

1

Express each of 180, x3 and y5 as a product of two factors where one factor is the largest possible perfect square.

2

Separate all perfect squares into one surd and all other factors into the other surd. Simplify 36x 2 y 4 .

= 5 × 36x 2 y 4 × 5xy

Multiply together numbers and the pronumerals outside the square root sign.

= 30xy 2 5xy

3 4

8

8

25 × 7 as a product of 2

= − 1--- × 8

25 × 7

= − 1--- × 5 7

25 .

8

= − 5--- 7 8

d 5 180x 3 y 5 = 5 36 × 5 × x 2 × x × y 4 × y

= 5 × 6 × x × y 2 × 5xy

remember remember 1. To simplify a surd means to make a number (or an expression) under the radical as small as possible. For example, 2 5 is equal to, but simpler than, 20 . 2. To simplify a surd, write it as a product of two factors, one of which is the largest possible perfect square.

1D WORKED

Example

Simplifying surds

1 Simplify the following surds. a

12

2 3

b

18

f

75

5 3

g

125 5 5 h

99 3 11 i

54

Digital docs:

k

112 4 7 l

m

68 2 17 n

SkillSHEET 1.1 Simplifying surds

p

338

13 2 q

88 2 22 r

u

245

7 5 v

320

7a eBook plus

EXCEL Spreadsheet Simplifying surds

98

3 2

7 2

c

8 5 w

56 2 14 e

27

3 3

3 6

60

2 15

150 5 6 o

180

6 5

135 3 15 s

162 9 2 t

200

10 2

8 7 x

735 7 15 y

405

9 5

24 2 6

448

d

j

23

Chapter 1 Number systems: the Real Number System

WORKED

Example

2 Simplify the following surds.

24 7

– 30 3

7b, c

a 2 8 4 2

b 3 50 15 2 c

10 17

f

g 7 54 21 6 h 10 32 40 2 i

– 6 75

– 28 5

k – 7 80

l

9 120

625 5

q

135

v

1 --9 5 --2

1 --3 1 --4 7 --8

18 30 --13

15

WORKED

Example

7d

p u

5 68

1 --5 1 --9

8 90 24 10 d 6 112

m 16 48 64 3 n

162

r

2

w

320 20 5

2 --3 3 -----10

s

54 2 6 175

--32

7 x

90

9 80

j

3 252

o

1 --7 1 --6

192 2 3 t --72

176

11 y

36 5 18 7

392

2 2

288

2 2

− 4--- 108 3

−8 3

3 Simplify the following surds. Assume that a, b, c, d, e, f, x and y are positive real numbers. a

16a 2 4a

b

81a 2 b 2 9ab

c

72a 2 6a 2

d

54a 2 b 2 3ab 6

e

90a 2 b

f

48a 3 b 4a 3ab

g

338a 4

h

150a 4 b 2

5a 2 b 6

i

338a 3 b 3 13ab 2ab

j

12a 5 b 7 2a 2 b 3 3ab

k

68a b

2ab 2 17ab

l

80x 6 y 4x 3 5y

m

125x 6 y 4 5x 3 y 2 5

n 3 64x 2 y

24x y

o 5 80x 3 y 2 20xy 5x

13a 2 2

3a 10b

3 5

p 2 343x 3 y 3 14xy 7xy q 6 162c 7 d 5 54c 3 d 2 2cd r 18c 3 d 4 5cd

10

e

s

2 405c 7 d 9

v

1 --3

120e 4 f 6

y

1 -----27

54x 3 y 9

3 126c 4 d 5 9c 2 d 2 14d

t

4 294c 10 d 10 28c 5 d 5 6 u

1 --2

2 --- 2 3

e f 3 30

w

1 --2

3 -----20

1 --9

xy 4 6xy

z

1 -----18

392e 11 f 11 7e 5 f 5 2ef x 108x 10 y 12

1--- 5 6 3

x y

88ef

22ef

175e 12 f 5

3--- 6 4

e f2 7f

3

4 multiple choice When expressed in its simplest form, A 3 15

B 5 9

45 is equal to:

C 5 3

D 9 5

E 3 5

5 multiple choice When expressed in its simplest form, 3 128 is equal to: A 6 32

B 12 8

C 24 2

D 16 2

E 32 3

6 multiple choice When expressed in its simplest form, A 49 11

B 7 11

1 --7

539 is equal to: 77

C

D

11

E 11

7 multiple choice 1 - 325x 4 y 3 when expressed in its Assuming that x and y are positive real numbers, – ----15 simplest form is equal to:

A – 1--- xy 13x 2 y

1 2 - x y 13y B – -----

D – 3x 2 y 13y

E – 3xy 13x 2 y

3

15

C – 1--- x 2 y 13y 3

24

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Addition and subtraction of surds Surds may be added or subtracted only if they are alike. Examples of like surds include

7 , 3 7 and – 5 7 . Examples of unlike surds

include 11 , 5 , 2 13 and – 2 3 . In some cases surds will need to be simplified before you decide whether they are like or unlike, and then addition and subtraction can take place. The concept of adding and subtracting surds is similar to adding and subtracting like terms in algebra.

WORKED Example 8 Simplify each of the following expressions containing surds. Assume that a and b are positive real numbers. a 3 6 + 17 6 – 2 6 b 5 3 + 2 12 – 5 2 + 3 8 c

1 --2

100a 3 b 2 + ab 36a – 5 4a 2 b

THINK

WRITE

a All 3 terms are alike, since they contain the same surd ( 6 ) , so group like terms together and simplify.

a 3 6 + 17 6 – 2 6 = ( 3 + 17 – 2 ) 6

b

b 5 3 + 2 12 – 5 2 + 3 8

1

Simplify surds where possible.

= 18 6

= 5 3+2 4×3–5 2+3 4×2 = 5 3+2×2 3–5 2+3×2 2 = 5 3+4 3–5 2+6 2

c

2

Add like terms to obtain the simplified answer.

1

Simplify surds where possible.

= 9 3+ 2

c

1 --2

100a 3 b 2 + ab 36a – 5 4a 2 b =

1 --2

× 10 a 2 × a × b 2 + ab × 6 a – 5 × 2 × a b

=

1 --2

× 10 × a × b a + ab × 6 a – 5 × 2 × a b

= 5ab a + 6ab a – 10a b 2

Add like terms to obtain the simplified answer.

= 11ab a – 10a b

Chapter 1 Number systems: the Real Number System

25

WORKED Example 9

Determine the perimeter of a rectangle whose length is ( 17 – 2 50 ) m and width is ( 5 + 32 ) m. THINK

WRITE P = 2l + 2w

1

Write down the rule for the perimeter of a rectangle where l is the length and w is the width.

2

Substitute the values of l and w into the rule.

3

Expand and simplify where possible.

= 34 – 4 50 + 10 + 2 32

4

Simplify surds where possible.

= 34 – 4 25 × 2 + 10 + 2 16 × 2

P = 2 ( 17 – 2 50 ) + 2 ( 5 + 32 )

= 34 – 4 × 5 2 + 10 + 2 × 4 2 = 34 – 20 2 + 10 + 8 2 = 44 – 12 2

5

Collect like terms.

6

State the answer, including the appropriate unit.

P = ( 44 – 12 2 ) m

remember remember 1. Only like surds may be added and subtracted. Examples of like surds: 7 , 3 7 and −5 7 . Examples of unlike surds: 3 , 5 and 2 13 . 2. Surds may need to be simplified before adding and subtracting.

1E WORKED

Example

8a

Addition and subtraction of surds

1 Simplify the following expressions containing surds. Assume that x and y are positive real numbers. a 3 5+4 5 7 5

b 6 2 + 11 2 17 2

c

2 3+5 3+ 3 8 3

d 6 7 + 8 7 + 5 7 19 7

e

8 5 + 3 3 + 7 5 + 2 3 15 5 + 5 3 f

2 6 + 9 2 + 6 2 + 5 6 15 2 + 7 6

26

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

g 6 11 – 2 11 4 11 i

h 12 13 – 5 13 – 2 13 5 13

7 2 + 9 2 – 3 2 13 2

k 9 6 + 12 6 – 17 6 – 7 6 17 3 – 18 7

WORKED

Example

8b

–3 6

j

3 7 – 2 5 + 7 7 – 9 5 10 7 – 11 5

l

5 2 – 12 2 – 3 6 + 8 6

–7 2 + 5 6

m 12 3 – 8 7 + 5 3 – 10 7

n

xy + 7 xy – 3 xy 5 xy

o 2 x+5 y+6 x–2 y 8 x+3 y

p 3 x + 4 y + 7 xy – 2 x – 9 y

x – 5 y + 7 xy

2 Simplify the following expressions containing surds. Assume that a and b are positive real numbers. a

200 – 300 10 ( 2 – 3 )

b

18 + 50 – 72

c

125 – 150 + 600 5 ( 5 + 6 )

d

96 – 5 24 + 12 – 6 6 + 2 3

e

27 – 3 + 75 7 3

f

8 + 18 + 50 10 2

h

45 + 20 5 5

g 2 20 – 3 5 + 45 4 5

2 2

14 3 + 3 2

i

6 12 + 3 27 – 7 3 + 18

j

44 – 99 + 121 – 3 11 11 – 4 11

3 6+6 3

k

150 + 24 – 96 + 108

l

98 – 2 50 + 5 32 17 2

m 3 90 – 5 60 + 3 40 + 100

n 2 99 – 44 – 176 0

– 8 11 + 22

o 5 11 + 7 44 – 9 99 + 2 121

p 5 3 + 8 27 – 4 3 + 2 147 39 3

12 30 – 16 15

q 2 30 + 5 120 + 60 – 6 135

r

15 10 − 10 15 + 10

12 ab + 7 3ab

WORKED

Example

8c 34 a – 6 2a

4ab ab + 3a 2 b b – 6ab 2a + 4a 2 b 3 3a

s

6 ab – 12ab + 2 9ab + 3 27ab

u

1 --2

98 +

w

1 --8

32 – 7--- 18 + 3 72

1 --3

48 +

1 --3

12

6

7 --2

2+2 3

15 2

20 – 50 – 80 – 120 + 60

t

1 --5

v

1 -----16

x

1 --8

–2 5 – 5 2 – 2 30 + 2 15

50 + 2--- 98 – 3--- 32 7

4

5 --8

512 –

128 +

1 --6

0

72 –3 2

7 5 - 12 – ------ 48 27 + ----16

32

--58

3

3 Simplify the following expressions containing surds. Assume that a and b are positive real numbers. 52 a – 29 3a a 7 a – 8a + 9 9a – 32a

b 10 a – 15 27a + 8 12a + 14 9a

c

150ab + 96ab – 54ab 6 6ab

d 16 4a 2 – 24a + 4 8a 2 + 96a

e

8a 3 + 72a 3 – 98a 3

f

g

9a 3 + 3a 5

i k

a 2a

3a a + a 2 3a

ab ab + 3ab a 2 b + 9a 3 b 3 32a 3 b 2 – 5ab 8a + 48a 5 b 6

1 --2

36a + 1--- 128a – 1--- 144a 4

6

h 6 a5b + a3b – 5 a5b

a + 2 2a

( a 2 + a ) ab

j

a 3 b + 5 ab – 2 ab + 5 a 3 b 3 ab ( 2a + 1 )

l

4a 2 b + 5 a 2 b – 3 9a 2 b

– 2a b 32a + 2 6a + 8a 2

27

Chapter 1 Number systems: the Real Number System

4 multiple choice When expressed in its simplest surd form, A 5 7 B 5 C 1

112 – 63 is equal to: D 7 E none of these

5 multiple choice When expressed in its simplest surd form, 2 40a – 6 72ab 2 is equal to: A – 32b 12a

B – 32b 8a

D – 4b 32a

E 4 10a – 36b 2a

6 multiple choice When expressed in its simplest surd form, A 5a + b 2 B 5 + 2b C 5a

C 4 10a – 12b 18a

7 -----10

7 multiple choice When expressed in its simplest surd form, B 3ab 3 3a

D 3ab 3

E 3 3a ( 3ab 3 – 1 )

A 6cd 6 Example

9

243a 3 b 6 – 27a is equal to:

A 6ab 3 3a

8 multiple choice When expressed in its simplest surd form,

WORKED

100a 2 – 2--- 25a 2 + 1--- 72b 2 is equal to: 5 6 D 9a + 2b E 9a a + 2 b

B – 2cd 6

C 6ab 3

150c 2 d 2 – cd 96 – c 54d 2 is equal to:

C 4cd 6

D – 2cd

E – 6cd 6

9 Find the perimeter of the following shapes, giving answers in the simplest surd form. Specify the units. a

b

eBook plus

18 cm

c

6 cm 48 cm

9 5 + 2 cm a 12 2 cm b ( 6 6 + 8 3 ) cm c ( 18 – 2 3 + 2 5 ) cm

Digital docs:

27 + 54 cm

SkillSHEET 1.2 Substitution using surds 1

7 – 3 cm 24 + 3 cm

WorkSHEET 1.1

e

d

d 3π 5 m e ( 18 2 + 2 5 ) m f 21 11 m

f 3 44 – 99 m

5 2– 5m

2 44 m

45 m 5+2 2m

4 44 + 2 99 m

Multiplication of surds To multiply surds, multiply together the expressions under the radicals. For example, a × b = ab , where a and b are positive real numbers. When multiplying surds it is best to first simplify them (if possible). Once this has been done and a mixed surd has been obtained, the coefficients are multiplied with each other and then the surds are multiplied together. For example, m a ¥ n b = mn ab

28

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

WORKED Example 10 Multiply the following surds, expressing answers in the simplest form. Assume that x and y are positive real numbers. b 5 3¥8 5

11 ¥ 7

a

d 6 12 ¥ 2 6

e

3 --5

1

70 ¥ --4- 10

c

5 ¥ 10

f

15 x 5 y 2 ¥ 12 x 2 y

THINK

WRITE

a Multiply surds together, using a × b = ab (that is, multiply expressions under the roots). Note: This expression cannot be simplified any further.

a

b Multiply the coefficients and then multiply the surds.

b 5 3×8 5 = 5×8× 3× 5 = 40 × 3 × 5 = 40 15

c

c

d

e

11 × 7 = =

5 × 10 = =

11 × 7 77

5 × 10 50

1

Multiply the surds.

2

Simplify the product surd if possible.

1

Simplify

2

Multiply the coefficients and multiply the surds.

= 24 18

3

Simplify the product surd.

= 24 9 × 2 = 24 × 3 2 = 72 2

1

Multiply the coefficients and multiply the surds.

2

3

12 .

Simplify the product surd.

Simplify by dividing both 10 and 20 by 10 (cross-cancel).

= 25 × 2 =5 2 d 6 12 × 2 6 = 6 4 × 3 × 2 6 = 6×2 3×2 6 = 12 3 × 2 6

e

3 --5

70 × 1--- 10 = 4

3 --5

× 1--- × 70 × 10 4

=

3 -----20

700

=

3 -----20

100 × 7

=

3 -----20

× 10 7

=

3 --2

3 7 7 or ---------2

29

Chapter 1 Number systems: the Real Number System

THINK

WRITE

f

f

1

Simplify each of the surds.

15x 5 y 2 × 12x 2 y 4

2

2

= 15 × x × x × y × 4 × 3 × x × y = x2 × y × 15 × x × 2 × x × 3 × y = x2y 15x × 2x 3y 2

3

Multiply the coefficients and multiply the surds.

= x2y × 2x 15x × 3y

Simplify the product surd.

= 2x3y 9 × 5xy

= 2x3y 45xy

= 2x3y × 3 5xy = 6x3y 5xy

When working with surds, we sometimes need to multiply surds by themselves; that is, square them. Consider the following examples:

( 2 )2

=

( 5 )2

2× 2

= 4 =2

=

5× 5

= 25 =5

We observe that squaring a surd produces the number under the radical. This is not surprising, since squaring and taking the square root are inverse operations and, when applied together, leave the original unchanged. When a surd is squared, the result is the number (or expression) under the radical; that is,

( a )2

= a , where a is a positive real number.

WORKED Example 11 Evaluate the area of a square of length simplest form. THINK

( 1--2-

28 xy ) m, expressing the answer in the WRITE

1

Write the rule for the area of a square.

A = l2

2

Substitute the value for l into the rule.

A=

3

Simplify, using

( a )2

= a.

( 1--2-

28xy )

=

( 1--2- ) × (

=

1 --4

2

× 28xy

= 7xy 4

Write the answer, including an appropriate unit.

A = 7xy m2

2

28xy)

2

30

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

remember remember 1. When multiplying surds, simplify the surd if possible, then apply the following rules: (a)

a× b =

ab

(b) m a × n b = mn ab , where a and b are positive real numbers. 2. When a surd is squared, the result is a number (or an expression) under the radical:

( a )2

= a , where a is a positive real number.

1F WORKED

Example

10a–e

1 Multiply the following surds, expressing answers in the simplest form. a

2× 7

d g j

WORKED

Example

10f

Multiplication of surds b

5 × 11

2 × 12 2 6

e

8× 6

10 × 10 10

h k

14

2 8× 5

4 10

55

c

6× 7

42

4 3

f

12 × 6

6 2

5 × 75 5 15

i

21 × 3

3 7

27 × 3 3

l

45 × 60

27

30 3

m 5 3 × 2 11 10 33

n 6 2 × 4 48 96 6

o 10 15 × 6 3 180 5

p 9 2 × 7 2 126

q 4 20 × 3 5 120

r

6 18 × 2 8 144

u

1 --4

48 × 2 2

x

2 --3

4 × 1--- 125

s

10 6 × 3 8 120 3

t

9 20 × 4 15

v

1 --2

w

1 --9

48 × 2 3

y

1 -----10

z

3 --4

30 × 2--- 10

72 × 1--- 3 3

60 × 1--- 40 5

6 2 --5

6

360 3 2 2--3-

5

2 6 4--3

5

3 3

5

2 Simplify the following expressions with surds. Assume that a, b, x and y are positive real numbers. a

xy × x 3 y 2

c

3a 4 b 2 × 6a 5 b 3

e g i

1 --2

b

x3 y4 × x2 y2 x2 y3 x

3a 4 b 2 2ab

d

5a 2 b 3 × 10ac 5 5abc 2 2abc

12a 7 b × 6a 3 b 4

6a 5 b 2 2b

f

18a 4 b 3 × 2a 2 b 5

15x 3 y 2 × 6x 2 y 3

3x 2 y 2 10xy

h 3 10x 7 y × 5x 5 y 3

x2 y y

15a 3 b 3 × 3 3a 2 b 6

9 --2

a 2 b 4 5ab

j

1 --3

12a 4 b 2 × 1--- 6a 3 b 3 4

6a 3 b 4 15x 6 y 2 2 --12

a 3 b 2 2ab

31

Chapter 1 Number systems: the Real Number System

WORKED

Example

11

3 Find the area of the following shapes. Answers must be expressed in the simplest surd form and the appropriate units specified. a

b

eBook plus

c

7 2 cm

2 4m

5 3 cm

Digital doc: SkillSHEET 1.3 Substitution using surds 2

5 11 m

d

3 a 98 cm2 b 75π cm2

e

f 5 10 m

6 5m 3 3m

c 20 11 m2

8 8m

3 6m 2 10 m

d 6 6 m2 e ( 45 π + 96 10 ) m2

3 6m

2 8m

f 72 15 m2

4 multiple choice The product of 3 30 × 5 6 expressed in its simplest form is: A 15 36

C 15 180

B 90

D 45 20

E 90 5

5 multiple choice 8x 5 y 2 × 5x 6 y 3 expressed in its simplest form is:

The product of A 2x 2 y 5 10xy

B

40x 11 y 5

C 2x 5 y 2 10xy D

13x 11 y 5

E x 5 y 2 13xy

6 multiple choice The product of A

3 -----16

x5 y7

x 7 y 2 × 1--- x 4 y 3 expressed in its simplest form is:

3 --8

2

B

3 -----16

x7 y5

C

3 ------ x 3 y 2 64

xy

D

3 ------ x 5 y 2 16

xy

E

3 ------ x 2 y 3 16

xy

7 multiple choice The area of the triangle expressed in its simplest form is:

5 3m

4 6m

A 30 2 m2

B 60 2 m2

C 24 12 m2

D 48 3 m2

E 20 18 m2

8 The height of a square-based pyramid is 20 8 units and the length of the side of its base is 12 8 units. Find the volume of the pyramid, expressing the answer in the simplest surd form. (Volume = --1- area of base × height) 15 360 2 3

32

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

The Distributive Law The Distributive Law states that a(b + c) = ab + ac. When multiplication of surds involving brackets is required, the Distributive Law is applied as in the case with algebraic terms. That is: a ( b + c) =

ab + ac

If there is a negative number outside the bracket, then every term inside the bracket will undergo a sign change since it has been multiplied by the negative number.

WORKED Example 12 Expand and simplify the following where possible. 5 ( 6 + 11)

a

b

7 ( 18 – 3)

c – 2 3 ( 10 – 5 3)

THINK

WRITE

a

a

b

c

5 ( 6 + 11)

1

Write the expression.

2

Apply the Distributive Law: Multiply the term outside the bracket with the first term inside the bracket, then multiply the term outside the bracket with the second term inside the bracket.

=

5 × 6 + 5 × 11

3

Simplify.

=

30 + 11 5

1

Write the expression.

2

Simplify

3

Apply the Distributive Law to remove the brackets.

=

4

Simplify.

= 3 14 – 3 7

1

Write the expression.

2

(a) Expand the brackets, using the Distributive Law. (b) Be sure to multiply through with the negative.

= – 2 3 × 10 – 2 3 × – 5 3

Simplify.

= – 2 30 + 10 × 3

3

18 .

b

7 ( 18 – 3 )

= 7 (3 2 – 3 ) 7 × 3 2 + 7 × –3

c – 2 3 ( 10 – 5 3) = – 2 30 + 10 9

= – 2 30 + 30 When expanding two binomial brackets the FOIL method is applied; that is, 4 pairs of terms must be multiplied in the order First, Outer, Inner and Last.

Chapter 1 Number systems: the Real Number System

33

WORKED Example 13 Expand

(

5 + 3 6 )( 2 3 – 2 ) . Write your answer in its simplest form.

THINK 1 Write the expression.

WRITE F

( 2

3

Apply FOIL. Multiply the first terms of each bracket. Multiply the outer terms of each bracket. Multiply the inner terms of each bracket. Multiply the last terms of each bracket.

L

5 + 3 6 ) (2 3 – 2 ) I O

=

5×2 3+ 5×– 2+3 6×2 3 + 3 6×– 2

= 2 15 – 10 + 6 18 – 3 12

Simplify.

= 2 15 – 10 + 6 × 3 2 – 3 × 2 3 = 2 15 – 10 + 18 2 – 6 3

Recall the perfect square identities: (a + b)2 = a2 + 2ab + b2 (a − b)2 = a2 − 2ab + b2 The perfect square identities can be applied to surds as follows:

(

a + b) =

(

b) =

2

( a)2 + 2

a b + ( b)

2

= a + 2 ab + b a–

2

( a)2 – 2

a b + ( b)

2

= a – 2 ab + b

WORKED Example 14

Expand ( 19 – 6 ) . Write your answer in its simplest form. 2

THINK

WRITE

(

19 – 6)

2

1

Write the expression.

2

Apply the perfect square identity.

=

3

Simplify.

= 19 – 2 114 + 6

(

19) – 2 19 × 6 + 2

( 6) 2

= 25 – 2 114 Note that the expansion of ( 19 – 6 ) in the previous example could also be done by writing it as a product of two repeated factors, ( 19 – 6 )( 19 – 6 ) , and applying FOIL. Naturally, the result would be the same, but the solution would take longer. 2

34

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Recall the difference of two squares (DOTS) identity: (a − b)(a + b) = a2 − b2 The DOTS identity can be applied to surds as follows:

(

a – b )( a + b ) = ( a ) – ( b ) =a-b 2

2

WORKED Example 15 Expand

(

5 y – 3 2 x )( 5 y + 3 2 x ) .

THINK

WRITE

(

5y – 3 2x )( 5y + 3 2x )

1

Write the expression.

2

Use DOTS identity for expansion.

=

3

Simplify.

= 5y – 9 × 2x = 5y – 18x

(

5y ) – ( 3 2x ) 2

2

In the above example the binomial factors which were multiplied together are a conjugate pair (that is, one bracket contains a sum and the other a difference of the same terms). Although the terms of the factors are irrational, the answer is not a surd, but an expression with rational terms. The product of a conjugate pair of surds (irrational numbers) yields a rational number. Note that to find the product of a conjugate pair (as in Worked example 15), FOIL could be used as an alternative to the DOTS identity. The latter, however, leads to the answer much more quickly.

remember remember 1. When expanding brackets, the Distributive Law is applied: a ( b + c ) = ab + ac 2. When expanding binomial brackets, FOIL is applied: ( a + b )( c + d ) = ac + ad + bc + bd 3. Perfect square identities: 2 2 ( a + b)2 = ( a) + 2 a b + ( b)

(

= a + 2 ab + b

a – b) = ( a) – 2 a b + ( b) 2

2

= a – 2 ab + b 4. DOTS identity:

(

a – b )( a + b ) = ( a ) – ( b ) 2

2

=a−b 5. The product of a conjugate pair of surds is rational.

2

Chapter 1 Number systems: the Real Number System

1G WORKED

Example

12 21 + 6 3 6 + 10 72 + 14 30 WORKED

Example

13

a

3( 7 + 6)

b

5 ( 18 – 7 )

d

2( 3 + 5)

e

7 ( 3 72 – 12 )

a

e g

h – 5 12 ( 3 5 – 4 8 ) – 30 15 + 80 6

i

( 18 – 5 ) ( 5 + 3 ) ( 4 8 + 2 6 )( 8 – 3 6 ) – 4 – 40 ( 7 8 + 6 3 )( 4 2 – 5 6 ) (2 7 – 3 2(5 5 + 7 2) ( 5 x + 2 y )( 3 x + 4 y ) 10 35 + 14 14 – 15 10 – 42

WORKED

Example

3 Expand and simplify where possible.

14

a

53 + 10 6

d

104 + 60 3

g j

WORKED

Example

15

c

5(2 – 2)

eBook plus

f

6 ( 5 14 – 4 )

Digital doc:

( 2 + 5 ) 2 27 + 10 2 b ( ( 3 + 5 2)2 e ( 2 (3 6 + 5 2) h ( 2 ( 2 8 – 5 ) 37 – 8 10

SkillSHEET 1.4 Algebraic expansion

–2 3 ( 4 6 – 2 3 )

i

126 2 – 14 3

– 24 2 + 12 10 21 – 4 6

( 7 + 5 )( 2 5 – 3 7 ) ( 3 6 – 2 5 )( 4 2 – 3 20 ) ( 11 – 2 3 )( 2 5 – 8 ) 2 55 – 2 22 – 4 15 + 4 ( 5 18 – 3 3 )( 2 18 – 6 )180 – 30 3 – 18 6 + 9 ( 8x – 10y )( 2x + 10y ) 4x + 2 5xy – 10y

b

– 35 – 11

d

3

2 5 – 10

f h j

112 – 140 3 + 24 6 – 90 2

6 + 10 ) 16 + 4 15 c 2

8 + 3 3) 5 – 3)

2

2

35 + 12 6 f

14 – 6 5

i

( 3 + 15 ) 2 18 + 6 5 ( 2 2 + 3 5 ) 2 53 + 12 ( 7 – 3 ) 2 10 – 2 21

10

4 Expand and simplify the following where possible. a c e g i k m o q s u w

xy(49x − 9y)

3 10 – 7 5

2 Expand and simplify where possible. 24 3 – 18 30 – 8 10 + 60

3 10 + 9 2 – 5 5 – 15 c

15x + 26 xy + 8y

The Distributive Law

1 Expand and simplify the following, where possible.

g 2 2 ( 6 18 + 7 15 )

35

y

( 3 + 7 )( 3 – 7 ) −46 ( 2 5 + 3 )( 2 5 – 3 ) 11 ( 8 + 2 )( 8 – 2 ) 6 ( 13 – 3 )( 13 + 3 ) 10 ( 2 3 – 5 )( 2 3 + 5 ) 7 ( 2 10 + 14 )( 2 10 – 14 ) 26 ( 13 – 6 )( 13 + 6 ) 7 ( 6 3 – 3 5 )( 6 3 + 3 5 ) 63 ( 7 2 – 3 5 )( 7 2 + 3 5 ) 53 ( 6 3 + 2 8 )( 6 3 – 2 8 ) 76 ( x – y )( x + y ) x − y ( 3 x – 4 y )( 3 x + 4 y ) 9x − 16y ( 7x y – 3y x )( 7x y + 3y x )

b d f h j l n p r t v x z

( 19 + 1 )( 19 – 1 ) 18 ( 3 11 + 7 )( 3 11 – 7 ) 50 ( 10 + 12 )( 10 – 12 ) −2 ( 7 – 12 )( 7 + 12 ) −5 ( 3 7 + 12 )( 3 7 – 12 ) 51 ( 18 – 19 )( 18 + 19 ) −1 ( 3 5 + 2 7 )( 3 5 – 2 7 ) 17 ( 5 2 – 6 )( 5 2 + 6 ) 44 ( 11 3 + 2 5 )( 11 3 – 2 5 ) 343 ( 7 2 – 3 9 )( 7 2 + 3 9 ) 17 ( 2x – 3y )( 2x + 3y ) 2x − 3y ( 2x x + 5 y )( 2x x – 5 y ) 4x − 25y ( 9 x 2 y – 5 xy 2 )( 9 x 2 y + 5 xy 2 ) xy(81x − 25y) 3

6 2

36

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

5 multiple choice When expressed in its simplest form, A 5 3–3 5 D

30

15 ( 5 – 3 ) is equal to:

B 5 5–3 3

C

75 – 45

E 2 2

6 multiple choice When expressed in its simplest form, ( 5 8 + 2 7 )( 6 5 – 3 3 ) is equal to: A 30 40 – 15 24 + 12 35 – 6 21 B 60 10 – 15 24 + 12 35 – 6 21 C 60 10 – 30 6 + 12 35 – 6 21 D 30 40 – 30 6 + 12 35 – 6 21 E 60 10 – 30 6 + 12 21 7 multiple choice When expressed in its simplest form, ( 7 5 – 2 3 ) is equal to: 2

A 49 25 – 4 9

B 245

D 269

E 257 – 28 15

C 257 + 28 15

8 multiple choice When expressed in its simplest form, ( 15 x 2 y + 4 xy )( 15 x 2 y – 4 xy ) is equal to: A 225x 2 y – 120 xy + 16xy B 15x 2 y – 4xy C 225 x 4 y 2 – 16 x 2 y 2 D 225x 2 y – 16xy E 225x 2 y – 120xy x + 16xy 9 Given that x = 3 5 – 2 3, find: a x 2 57 – 12 15 b x 2 + 3x + 2 59 – 12 15 + 9 5 – 6 3

Division of surds To divide surds, divide the expressions under the radicals; a a that is, ------- = --- , where a and b are positive real numbers. b b When dividing surds it is best to simplify them (if possible) first. Once this has been done, the coefficients are divided next and then the surds are divided.

Chapter 1 Number systems: the Real Number System

37

WORKED Example 16 Divide the following surds, expressing answers in the simplest form. Assume that x and y are positive real numbers. 55 a ---------5

48 b ---------3

9 88 c ------------6 99

36 xy d -----------------------25 x 9 y 11

THINK a

1 2

3

b

c

d

WRITE

a a Rewrite the fraction, using ------- = --- . b b Divide numerator by the denominator (that is, 55 by 5).

2 3

Evaluate

1

a Rewrite the surds, using ------- = b

a --- . b

11

48 b ---------- = 3 =

48 -----3 16

=4

16 . a --- . b

2

Simplify the fraction under the radical by dividing both numerator and denominator by 11.

3

Simplify the surds.

4

Multiply the whole numbers in the numerator together and those in the denominator together.

5

Cancel the common factor of 18.

1

Simplify each surd.

2

55 -----5

Check if the surd can be simplified any further. a Rewrite the fraction, using ------- = b Divide 48 by 3.

1

55 a ---------- = 5 =

Cancel any common factors — in this case xy .

c

9 88 9 ------------- = --6 6 99 9 = --6

88 -----99 8 --9

9×2 2 = ------------------6×3 18 2 = ------------18 =

2

36xy 6 xy d ----------------------- = -------------------------------------------25x 9 y 11 5 x 8 × x × y 10 × y 6 xy = -----------------------5x 4 y 5 xy 6 = -------------5x 4 y 5

38

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

WORKED Example 17

Find the perpendicular height of a triangle, given that its area is 27 15 cm2 and its base length is 6 3 cm. The answer must be expressed in the simplest surd form and the appropriate unit specified. THINK

WRITE

1

Write the rule for the area of a triangle.

A = 1--- bh

2

Substitute the values for A and b into the rule.

27 15 =

3

Cancel the 2 and the 6.

27 15 = 3 3 × h

4

Transpose the equation to make h the subject.

5

Divide numerator and denominator by 3 (cancel down).

6

Simplify and write the answer, using the appropriate unit.

2

1 --2

×6 3×h

27 15 h = ---------------3 3 9 5× 3 = -----------------------3 h = 9 5 cm

remember remember When dividing surds, simplify the surd if possible, then apply the following rule: a b = ------- = b where a and b are positive real numbers. a ÷

1H

a --b

Division of surds

1 Simplify the following surds, expressing answers in the simplest form. Assume that x and y are positive real numbers. 16 15 14 8 72 5 a ---------b ---------- 7 c ------- 2 d ---------- 2 3 3 2 2 6

WORKED

Example xample

e

60 ---------10

6

f

90 ---------6

i

18 ---------- ------34 4 6

j

2 24 4---------2------------3 3 3

15

128 g ------------- 4 8

h

45 ------------125

65 ------------2 13

l

5 72 ------------- 5 6 12

k

5 ------2

3--5

39

Chapter 1 Number systems: the Real Number System

96 m ---------- 2 3 8

n

2 63 1 ------------- 1 --55 7

7 44 o ---------------- 1 14 11

9 63 q ------------15 7

1 --5-

4

r

540 ------------20

s

x4 y3 --------------x2 y5

x -y

v

x 6 y 11 1 ------------------- ---------3 y2 x x 12 y 15

u

3 3

336 p ------------14

2040 ---------------- 2 17 30

16xy w -----------------8x 7 y 9

t

2 6

12 99 ---------------15 11

2

2 --5-

72x 4 y 3 x --------------------- 6x xy 2xy 2

2 ---------x3 y4

2 Simplify the following. Assume that all pronumerals are positive real numbers. 2xy 3y

2 2b 3b -----------------2a a

xy 12x 8 y 12 a --------------- × ----------------------x2 y3 x5 y7

6x 2 y 3 3x 7 y 2 b --------------------- × -----------------27x 4 y 4 3xy 3

3ab 5 6a 7 b 3 d ------------------ × -------------e 5b 2a 2 6 2 2x 2a b ------------

2mn 3 3m 4 n 6 -------------------- ÷ -------------------6m 5 n 2 8mn 3

3y

c

2 2a 2 b 4 10a 9 b 3 ---------------------- × --------------------5a 3 b 6 3 a7b

4 a ---------3

f

5 3m 3 n 2m 5 n 8 ----------------------- ÷ -------------------2 6m 3 n 2 6 mn 5

15 --------------2m 2 n 2

2 2 ---------------------3m 3 n m

3 multiple choice 75 Expressed in its simplest form, ---------- is: 5 A

70

B

15

C

13

5 3 D ---------5

E

9 6 D ---------7

3 6 E ---------7

x 2 y 4 10 D --------------------xy 20x

x2 y6 E -------------2

17

4 multiple choice 9 18 Expressed in its simplest form, ------------- is: 21 3 9 6 A ---------21

54 B ---------7

9 2 C ---------7 3

5 multiple choice 10x 5 y 8 Expressed in its simplest form, --------------------- is: 20x 3 y 2 xy 3 2 A ---------------2

xy 3 10 B ------------------20

x3 y 2 C ---------------2

6 multiple choice 2x 4 y 6x 7 y 3 Expressed in its simplest form, ---------------- × ------------------ is: 9xy 4x 3 y 5 y3 y A -----------x 3

x 12 x B -------------y 3

x 3 3x C ---------------3y

x7 D ---------y 3

2x 3 3x E ------------------6y

40

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

WORKED

Example

17

7 Find the length of the unknown side in each of the following. Answers must be expressed in the simplest surd form and the appropriate units specified. a

7 a 4 13 m

b

A = 28 39 m2

A = 12 30 cm2

c

A = 21 55 m2

b 4 6 cm w

c 7 11 m

h

3 5 cm

d 3 7m

f

15 -----2

b

7 3m

e 5 13 cm 5 cm

d

6 5m

V = 90 21 m3

e

V = 315π 13 cm3

f

V = 60π 75 cm3 h

h h 3 6m

3 7 cm

5 2m

Area of base = 24π 15 cm2

2E ------- , where E is the kinetic m energy of the object and m is the mass of the object. Express v as the simplest surd, if:

8 Velocity v of the object can be found using the formula v = a E = 80 J, m = 2 kg 4 5 5-----3

b E = 250 J, m = 60 kg c

E = 480 J, m = 120 kg 2 2

9 A rectangular fish tank has a base 20 3 cm by 30 6 cm and the height h. When the tank is filled, the volume of water is 84 L. Find:

2 --3

of

a the height of the tank (give the answer as the simplest surd) 35 2 cm b the full capacity of the tank in litres. 126 L (Remember that 1 cubic centimetre holds 1 mL of water.)

Rationalising denominators If the denominator of a fraction is a surd, it can be changed into a rational number. In other words, it can be rationalised. As we discussed earlier in this chapter, squaring a surd (that is, multiplying it by itself) results in a rational number. This fact can be used to rationalise denominators as follows. a b ab b ------- ¥ ------- = ---------- , (where ------- = 1) b b b b If both numerator and denominator of a fraction are multiplied by the surd contained in the denominator, the denominator becomes a rational number. The fraction takes on a different appearance, but its numerical value is unchanged, because multiplying the numerator and denominator by the same number is equivalent to multiplying by 1.

Chapter 1 Number systems: the Real Number System

WORKED Example 18 Express the following in their simplest form with a rational denominator. 6 a ---------13 THINK a

b

2 12 b ------------3 54

17 – 3 14 c ----------------------------7 WRITE

1

Write the fraction.

2

Multiply both the numerator and denominator by the surd contained in the denominator (in this case 13 ). This has the same effect as multiplying 13 the fraction by 1, since ---------- = 1 . 13

1

Write the fraction.

2

Simplify the surds. (This avoids dealing with large numbers.)

6 a ---------13 6 13 = ---------- × ---------13 13 78 = ---------13 2 12 b ------------3 54 2 12 2 4 × 3 ------------- = ------------------3 54 3 9 × 6 2×2 3 = ------------------3×3 6 4 3 = ---------9 6

3

Multiply both the numerator and denominator by

6.

(This has the same effect as multiplying the fraction by 6 1 since ------- = 1.) 6 Note: We need to multiply only by the surd part of the denominator (that is, by 6 , rather than by 9 6 ). 18 .

4

Simplify

5

Divide numerator and denominator by 6 (cancel down).

4 3 6 = ---------- × ------9 6 6 4 18 = ------------9×6

4 9×2 = ------------------9×6 4×3 2 = ------------------54 12 2 = ------------54 2 2 = ---------9 Continued over page

41

42

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

THINK

WRITE

c

17 – 3 14 c ----------------------------7

1

Write the fraction.

2

Multiply both the numerator and denominator by 7 . Use brackets so you realise the whole numerator must be multiplied by 7 .

3

Apply the Distributive Law in the numerator. a(b + c) = ab + ac

( 17 – 3 14 ) 7 = ------------------------------------ × ------7 7 17 × 7 – 3 14 × 7 = ---------------------------------------------------------7× 7 119 – 3 98 = -------------------------------49

4

Simplify

119 – 3 49 × 2 = -----------------------------------------7 119 – 3 × 7 2 = --------------------------------------7 119 – 21 2 = -------------------------------7

98 .

remember remember To rationalise the surd denominator, multiply the numerator and denominator by the surd contained in the denominator. This has the effect of multiplying the fraction by 1 and thus the numerical value of the fraction remains unchanged, while the denominator becomes rational: a ab a b ------- = ------- × ------- = ---------b b b b

1I WORKED

Example

18a, b

Rationalising denominators

1 Express the following in their simplest form with a rational denominator. 4 11 5 7 4 8 7 3 4 6 5 2 12 ---------- e --------a ------- ---------b ------- --------c ---------- -----------11 d ------3 3 2 2 3 11 6 7 f

15 ---------6

k

5 14 ------------7 8

10 ---------2

2 3 g ---------5

5 7 ---------- l 14

2 15 ------------- h 5

16 3 ------------6 5

3 7 ---------5

8 3 m ---------7 7 8 15 ------------15

3 35 ------------- i 5

5 2 ---------2 3

8 21 ------------- n 49

8 60 ------------28

5 6 ---------- j 6

4 3 ---------3 5

2 35 o ------------3 14 8 105 ---------------7

2 21 ------------7 4 15 ------------15 10 ---------3

43

Chapter 1 Number systems: the Real Number System

WORKED

Example

18c 2+2 3 10 + 6 14 --------------------------------4 14 – 5 2 ---------------------6

2 Express the following in their simplest form with a rational denominator. 6 + 12 a ----------------------3

15 – 22 b -------------------------6

e

3 5+6 7 --------------------------8

f

i

7 12 – 5 6 -----------------------------6 3

j

6 2 – 15 2 18 + 3 2 -------------------------d -----------------------------5–5 6 10 12 5 ------------------------------

c

4 2 + 3 8 5---------63 11 – 4 5 --------------------------- 3 g -----------------------------2 3 18

6 2– 5 ----------------------k 4 8 12 – 10 3 10 – 2 33 -----------------------------------------------------6

9 10 ------------5

10

6 3–5 5 --------------------------7 20

h

2 7–2 5 --------------------------12

6 15 – 25 ------------------------70

21 – 15 -------------------------3

3 22 – 4 10 --------------------------------6

16

3 multiple choice 12 When expressed in its simplest form, ------- is equal to: 3 4 3 A ---------3

B 4 3

12 3 D ------------3

C 48

E 6 3

4 multiple choice 8 5 When expressed in its simplest form, ------------- is equal to: 9 12 40 16 15 4 5 4 15 A --------B ---------------C ---------D ------------108 108 9 27

320 E --------972

5 multiple choice 7 5–6 7 When expressed in its simplest form, --------------------------- is equal to: 12 6 7 15 – 6 21 A ------B 7 15 – 21 C --------------------------------6 6 7 60 – 6 84 D --------------------------------12

E none of these

6 multiple choice 5 5–3 3 When expressed in its simplest form, --------------------------- is equal to: 8 8 5 10 – 3 6 A -----------------------------32

B 10 10 – 6 6

40 40 – 24 24 D --------------------------------------64

10 10 – 6 6 E --------------------------------8

80 10 – 48 6 C -----------------------------------64

7 Solve for x, giving the answers as the simplest surds with rational denominators: a x2 =

3 --7

21 ± ---------7

b 3x2 = 5

15 ± ---------3

c

6x2 − 4 = 12

2 6 ± ---------3

44

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Rationalising denominators using conjugate surds As shown earlier in the chapter, the product of pairs of conjugate surds results in a rational number. (Examples of pairs of conjugate surds include

6 + 11 and

6 – 11 ,

a + b and a – b , 2 5 – 7 and 2 5 + 7 .) This fact is used to rationalise denominators containing a sum or a difference of surds. To rationalise the denominator which contains a sum or a difference of surds, we multiply both numerator and denominator by the conjugate of the denominator. Two examples are given below: 1 a– b 1. To rationalise the denominator of the fraction -------------------- , multiply it by -------------------- . a+ b a– b 1 a+ b 2. To rationalise the denominator of the fraction -------------------- , multiply it by -------------------- . a– b a+ b A quick way to simplify the denominator is to use the DOTS identity:

(

a – b )( a + b ) = ( a ) – ( b ) =a−b 2

2

WORKED Example 19

Rationalise the denominator and simplify the following. 1 a ---------------4– 3 THINK a

6+3 2 b -----------------------3+ 3

1

Write the fraction.

2

Multiply the numerator and denominator by the conjugate of the denominator.

WRITE 1 a ---------------4– 3 (4 + 3) 1 = --------------------- × --------------------(4 – 3) (4 + 3)

 4 + 3 (Note that  ---------------- = 1 .)  4 + 3 3

Apply the Distributive Law in the numerator and the DOTS identity in the denominator.

4+ 3 = -----------------------------2 ( 4 )2 – ( 3 )

4

Simplify.

4+ 3 = ---------------16 – 3 4+ 3 = ---------------13

Chapter 1 Number systems: the Real Number System

THINK

WRITE

b

6+3 2 b -----------------------3+ 3

1

Write the fraction.

2

Multiply the numerator and denominator by the conjugate of the denominator.

45

( 6 + 3 2) (3 – 3) = ----------------------------- × --------------------(3 + 3) (3 – 3)

 3 – 3 (Note that  ---------------- = 1 .)  3 – 3 3

Apply FOIL in the numerator and DOTS in the denominator.

6×3+ 6×– 3+3 2×3+3 2×– 3 = ----------------------------------------------------------------------------------------------------------2 ( 3 )2 – ( 3 )

4

Simplify.

3 6 – 18 + 9 2 – 3 6 = -----------------------------------------------------------9–3 – 18 + 9 2 = ------------------------------6 – 9×2+9 2 = ------------------------------------6 –3 2+9 2 = ------------------------------6 6 2 = ---------6 = 2

You might wish to use a calculator to check if the final answer is correct. To do that, evaluate the original fraction and the final one (the one with the rational denominator) and check whether they both equal the same number.

WORKED Example 20 1 1 Rationalise the denominators and simplify the following. ----------------------- + --------------------------2 6– 3 3 6+2 3 THINK WRITE 1 We will rationalise the denominator of 1 ----------------------each term and then add them. 2 6– 3 Write the first fraction. 2

Multiply the numerator and denominator by the conjugate of the denominator.

(2 6 + 3) 1 = ---------------------------- × ----------------------------(2 6 – 3) (2 6 + 3) Continued over page

46

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

THINK

WRITE 2 6+ 3 = ---------------------------( 2 )2 × 6 – 3

3

Apply the Distributive Law in the numerator and DOTS in the denominator. (Note that when squaring 2 6 , we need to square both 2 and 6 .)

4

Simplify the denominator.

5

Write the second fraction.

6

Multiply the numerator and denominator by the conjugate of the denominator.

(3 6 – 2 3) 1 = -------------------------------- × ------------------------------(3 6 + 2 3) (3 6 – 2 3)

7

Apply the Distributive Law in the numerator and DOTS in the denominator.

3 6–2 3 = ----------------------------------2 3 × 6 – 22 × 3

8

Simplify the denominator.

3 6–2 3 = --------------------------42

9

Add the two fractions together. Bring them to the lowest common denominator first.

2 6+ 3 = -----------------------21 1 --------------------------3 6+2 3

2 6+ 3 3 6–2 3 ------------------------ + --------------------------21 42 3 6–2 3 2 6+ 3 2 = ------------------------ ×  --- + -------------------------- 2 42 21 4 6+2 3 3 6–2 3 = --------------------------- + --------------------------42 42

10

Add the numerators.

7 6 = ---------42

11

Simplify where appropriate.

6 = ------6

The following worked example demonstrates the rationalisation of the denominator when it is a trinomial (has three terms).

WORKED Example 21 1 Simplify ----------------------------- . 2+ 2– 3 THINK 1

Use a set of brackets to group the trinomial into a binomial.

WRITE 1 ---------------------------------(2 + 2) – 3

Chapter 1 Number systems: the Real Number System

THINK

WRITE [(2 + 2) + 3] 1 = ---------------------------------- × --------------------------------------(2 + 2) – 3 (2 + 2) + 3

2

Multiply the numerator and denominator by the conjugate of the denominator; that is, (2 + 2 ) + 3 . Use brackets around both factors so that you will recognise that all terms need to be multiplied.

3

Use FOIL to expand the denominator.

(2 + 2) + 3 = ----------------------------------2 (2 + 2) – 3

4

Expand the squared terms of the denominator.

(2 + 2) + 3 = -------------------------------------4+4 2+2–3

5

Group and simplify the denominator.

6

Rationalise the denominator as shown previously. Use brackets as in step 2.

7

8

47

2+ 2+ 3 = -----------------------------3+4 2 (2 + 2 + 3) (3 – 4 2) = ----------------------------------- × -----------------------(3 + 4 2) (3 – 4 2)

6–8 2+3 2–4 4+3 3–4 6 Expand the numerator, making sure that = -------------------------------------------------------------------------------------9 – 16 × 2 every term in the first set of brackets is multiplied by every term in the second set. Group like terms and simplify.

6–8–8 2+3 2+3 3–4 6 = ------------------------------------------------------------------------------9 – 32 –2 – 5 2 + 3 3 – 4 6 = -------------------------------------------------------–23

9

Multiply numerator and denominator by –1 to eliminate the negative denominator.

2+5 2–3 3+4 6 = ----------------------------------------------------23

remember remember 1. To rationalise the denominator containing a sum or a difference of surds, multiply both the numerator and denominator of the fraction by the conjugate of the denominator. This eliminates the middle terms and leaves a rational number. 2. To simplify the denominator quickly, use the DOTS identity:

(

a – b )( a + b ) =

( a)2 – ( b)2

= a–b 1 a– b 3. To rationalise the denominator of the fraction -------------------- , multiply it by -------------------- . a+ b a– b 1 a+ b 4. To rationalise the denominator of the fraction -------------------- , multiply it by -------------------- . a– b a+ b

– ( 10 3 + 15 6 + 9 2 + 27 ) 1 p --------------------------------------------------------------------42

48

5 14 + 2 10 – 25 7 – 10 5 1 i -----------------------------------------------------------------------155 – 20 2 + 9 10 + 4 30 – 9 6 m -------------------------------------------------------------------------2

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Rationalising denominators using conjugate surds

1J WORKED

Example

19

1 a ---------------5+2

eBook plus Digital docs:

4 ----------------------------2 11 – 13

i

1 8 11 + 4 13 e --------------------------------31 2 21 – 35 f ----------------------------14 15 15 – 20 6 g -----------------------------------13 19 – 4 21 k ------------------------5 3 3+2 6 o --------------------------18 115 + 31 21 s ------------------------------148 u 18 2 + 10 6 – 9 3 – 15

3+ 6 ---------------- c 3

1 -------------------8– 5

f

7 -----------------------------2 12 + 2 5

2–5 -------------------------5 7 – 20

j

8–3 ---------------8+3

5– 3 m --------------------------------4 10 + 3 18

n

2 8–3 2 -----------------------------3 24 – 2 6

4 5 + 10 q -----------------------------6 15 + 20

r

4 15 – 2 3 -----------------------------2 30 – 5 2

s

u

4 12 – 3 8 -----------------------------3 6–5 2

v

3 8+6 3 --------------------------7 2– 3

3 11 – 2 7 w --------------------------------3 14 + 4 11

y

3 7–5 2 --------------------------35 + 2 2

z

3 6 – 15 -------------------------6+2 3

102 + 48 6 v ---------------------------95

5 3 g --------------------------3 5+4 2 k 3 ------12

12 – 7 ----------------------12 + 7

3 6 + 2 12 o -----------------------------4 18 + 3 8 2 7+5 3 --------------------------5 7–3 3

– 6 + 6 2 + 10 – 2 5 --------------------------------------------------------2

21 5 – 6 14 – 5 70 – 20 -----------------------------------------------------------------27

2 Rationalise the denominator and simplify.

20

16 210 – 12 14 -----------------------------------------77

– ( 45 + 15 14 + 9 10 + 6 35 ) ---------------------------------------------------------------------------5

3 + 7 65 – 16 11 --------------------------------------------28

1 b ---------------3– 6

1 d ----------------------2 6– 7 h

9 3 ----------------------------2 33 – 12

l

11 + 7 -------------------------22 – 14

12 2 – 17

SkillSHEET 1.6 Applying DOTS to expressions with surds

Example

5–2

e

SkillSHEET 1.5 Conjugate pairs

WORKED

2 2+ 5 -----------------------3

1 Rationalise the denominator and simplify.

1 1 a ---------------- + ------------------8–2 2 8–2

9 2+8 ------------------14

t

2 11 – 3 3 -----------------------------2 11 + 3 3

4 15 + 2 5 x -----------------------------3 5 – 15

7 3+9 ------------------3

1 1 b --------------------------- – -----------------------2 7+2 3 3 7+ 3

9 7 – 13 3 -----------------------------120

6–7 2

e

3 5 7–2 2 -------------------- ÷ ----------------------7+ 2 5+ 2

f

2 2+ 3 2 2+ 3 ------------------------ ÷ -----------------------------2 2 – 3 12 2 + 6 3

h

3 7+2 5 7–2 -------------------------- + --------------------------2 7 – 11 7 + 2 11

j

5+ 6 2 6–2 5 --------------------------- – --------------------------4 5–4 6 3 6–3 5

13 + 5 11 + 2 ----------------------- – ------------------13 – 5 11 – 2

71 – 12 33 ---------------------------17

12 3 – 4 + 3 6 – 2 1 q --------------------------------------------------52 2 + 10 30 – 6 10 – 5 6 r 60 -----------------------------------------------------------------------35

2 3 4 6+ 3 d ----------------------- × --------------------------6–2 3 2 6+3 3

i

9 2 + 154 -----------------------------4

– 9 154 + 132 + 42 2 – 8 77 --------------------------------------------------------------------------50

3 7 4 8 ----------------------- × -----------------------3 5– 3 5+3 3

5 – 4 14

9 11 + 9 ---------------------20

5 2+3 3 p -----------------------------2 6 – 3 12

c

7+ 8 2 8–2 7 g --------------------------- + --------------------------3 7–3 8 3 8+3 7

2 6+ 7 -----------------------17

959 + 281 77 + 182 7 + 6 11 ----------------------------------------------------------------------------629

66 + 24 6 ------------------------5

– ( 41 + 6 30 ) --------------------------------12

Chapter 1 Number systems: the Real Number System

WORKED

Example

21

49

3 Rationalise the denominator and simplify. 2+ 3– 5 a -----------------------------------3+2 3+2 5

5– 3– 2 b ---------------------------------5+ 3+ 2

–6 + 2 15 + 3 10 – 5 6 -------------------------------------------------------------6

230 + 257 3 – 137 5 – 80 15 4 multiple choice ---------------------------------------------------------------------------431

If x =

7 +

1 11 , then x + --- when simplified with a rational denominator is equal to: x

3 7 + 5 11 A -----------------------------4

5 7 + 3 11 B -----------------------------4

3 7 + 5 11 D -----------------------------–4

3 7 + 3 11 E -----------------------------4

5 7 + 3 11 C -----------------------------–4

7–3 5 5 Given that x = ------------------------ find each of the following, giving the answer in surd form 7+3 5 with a rational denominator: 1 1 1 2 35 35 14------------------b x – --- 6-----------c x – ----- – 312 a x + --- −2 ----19 361 19 x x x2 6 Given that x = 5 2 – 3 find with a rational denominator: 1 2 – 120 b a x + --- 210 ------------------------------41 x 1 99 120 – 50 460 2 d x 2 – ------------------------------------------------e 1681 x2 x 2 + 3x g ----------------- 7 2 + 4 h x–2

each of the following, giving the answer in surd form 1 x – --x

200 2 – 126 ------------------------------41

x 2 + 6x + 3 44 x 2 – 3x ----------------x+2

c

1 238 – 50 400 2 --------------------------------------------x 2 + ----- 99 1681 x2

f

x 2 – 12x + 8 103 – 90 2

295 2 – 382 ------------------------------49

7 Is x = 5 + 3 2 a solution for the equation x 2 – 10x + 7 = 0 ? Show all working. Yes 8 Solve for x giving answers in surd form with rational denominators. 11 + 5 ----------------------6

a

11x + 7 = 8 + 5x

b 3 5x – 10 = 2 7x + 4

42 5 + 28 7 --------------------------------17

Further properties of real numbers — modulus The modulus or absolute value of a number is the magnitude of that number. It represents the distance of the number from the origin (that is, 0 on a number line). The modulus of x is denoted by |x| and is always positive. Note: Do not confuse the modulus of a number with modular arithmetic (see page 17). For example, |−2| = 2 |2| = 2 |0| = 0

50

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

WORKED Example 22

Evaluate the following where a, b, c, d ≥ 0. a

– 50

b –6 ¥ – 3

c – 6a 2 b ¥ – 2a 3 b

THINK a 1 Write the expression. 2 The modulus sign indicates that we want only the magnitude of a number and not the sign of it. So the negative in front of the number should be omitted. b 1 Write the expression.

c

d

2

Evaluate each modulus separately and then simplify.

1

Write the expression.

2

Evaluate each modulus separately, then simplify.

1

Write the expression.

2

Evaluate each modulus separately, then simplify.

– 4cd ¥ – – 6cd d ---------------------------------------– 12 WRITE a – 50 = 50 b –6 × – 3 = 6 × −3 = −18 c 6a 2 b × – 2a 3 b = 6a 2 b × 2a 3 b = 12a 5 b 2 – 4cd × – – 6cd d ---------------------------------------– 12 4cd × – 6cd = ---------------------------– 12 – 24c 2 d 2 = -------------------– 12 = 2c 2 d 2

remember remember 1. The modulus (or absolute value) of a number is the magnitude of that number. It tells us how far the number is from zero, and is always positive. 2. The modulus of x is denoted by |x|.

1K WORKED

Example

22

Further properties of real numbers — modulus

1 Evaluate the following where a, b, c, d ≥ 0: a 19 19 b 1--- 1--44 15 d – 15 e –8 8 1 h – --- 1--2g 3 – 4 12 2 2 k – – --- − 2--3j 0 0 3

c

0.75

0.75

f

– 2a

2a

i

– 3.21

l

–7 – –3

3.21 4

51

Chapter 1 Number systems: the Real Number System

m –2 + –8

−3a

10 −16

n

– 2 – 8 10

q

–3 × –9 – – 18 × – 3

p

– 21 – – 37

s

12 × – – 6

−72

t

v

– 24 ÷ – 3

−8

w – 6 – 3 – 20

y

– 2a × – 6b ÷ – 4 – b

27 −54 −11

4cd × 3cd ÷ – 2 cd

z

a2b2

o

–a 2 b 2

r

5 × – 3 15

u

7 ÷ – 14

x

9 × 10 ÷ – 3

1 --2

30

−6cd

2 multiple choice When simplified, – 2 × 3 becomes: B −6 C 6 A –6

E −5

D –1

3 multiple choice When simplified, – ab 2 × a 3 b ÷ – a 2 b 4 becomes: a a a4b3 a2 C – ----D ----B – ----------A ----2 2 4 b b b2 a b

a2 E – ----b

4 multiple choice When simplified, – – 8 × – 2 + 5 – 3 becomes: A −8 B 14 C −18

D −14

E 8

D − 9---

E − 9---

5 multiple choice When simplified, – 6 2 × 2 – 5 ÷ – 8 becomes: A

15 -----8

-----C − 15

9 --2

B

8

2

8

6 a Fill in the table below for the function y = 2x − 4.

x

–2

–1

0

1

2

3

4

5

6

y

–8

–6

–4

–2

0

2

4

6

8

8

6

4

2

0

2

4

6

8

|y| b

y

y = |2x – 4|

y = 2x – 4

4

2

x

b Use the table to plot (on the same set of axes) the graph of y = 2x − 4 and y = 2x – 4 . c State the range of each of the two functions. R and y ≥ 0 d Compare the ranges of the two functions and their graphs. Explain the difference.

–4

Solving equations using absolute values If | x | = 3 then, by definition of absolute values, there are 2 values of x that satisfy this equation that states that x is three units from 0. That is, x = 3 or x = −3 3 units

–4

–3

–2

–1

3 units

0

1

2

3

4

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Therefore, two separate cases need to be considered when solving equations involving absolute values. In general: Case 1: | x | = x when x > 0 (that is, when x is positive) Case 2: | x | = −x when x < 0 (that is, when x is negative) (Note: There is also a trivial case of | x | = 0 when x = 0.) So for | x | = 3, the two cases to consider are x = 3 and −x = 3 giving a solution of x = 3 or x = −3.

WORKED Example 23 Solve:

a

| 4x | = 16

b | 4 - 3x | = 3.

THINK a 1 Write the equation. 2 Remove the absolute value symbols and write the positive (+ve) and negative (−ve) cases to be considered. 3 Work the two cases side by side. 4 Verify your solution by substituting into the original expression. Start with the left-hand side (LHS) and ensure that it equals the right-hand side (RHS).

b

5

State the solution. x can equal either 4 or −4, written ±4.

1

Write the equation. Remove the absolute value symbols and write the +ve and –ve cases. Solve for x in both cases.

2 3

4

5

Verify your solution by substituting into the original expression. Start with the LHS and ensure that it equals the RHS.

State the solution.

WRITE a | 4x | = 16 Case 1: 4x = 16

Case 2: −4x = 16 −4x = −16

or

x=4 x = −4 Check: Using x = 4 Using x = −4 LHS = | 4 × 4| LHS = | 4 × −4| = | 16| = | −16| = 16 = 16 = RHS = RHS Solutions are correct for both cases. The solution is x = ±4. b | 4 − 3x | = 3 Case 1: 4 − 3x = 3 or 4 − 3 = 3x 1 = 3x x = 1--3 Check: Using x = 1--3 LHS = | 4 − 3 × 1--- | 3 = |4 − 1| =3 = RHS

Case 2: −(4 − 3x) = 3 −4 + 3x = 3 3x = 7 x = 2 1--3

Using x = 2 1--3 LHS = | 4 − 3 × = |4 − 7| = |−3| =3 = RHS Solutions are correct for both cases. The solution is x = 1--- or x = 2 1--- . 3

3

7 --3

|

Chapter 1 Number systems: the Real Number System

WORKED Example 24

Solve | x - 3 | = 3x + 8.

THINK 1 Write the equation. 2 Remove the absolute value symbols and write the +ve and –ve cases. Use brackets for the LHS of the –ve case. 3 Solve for x.

WRITE | x – 3 | = 3x + 8 Case 1: x – 3 = 3x + 8

Case 2: –(x – 3) = 3x + 8

or

−3 – 8 = 3x – x −11 = 2x x = −5 1---

−x + 3 = 3x + 8 3 − 8 = 3x + x −5 = 4x x = −1 1---

2

4

4

Verify your solution for both cases by substituting into the original equation. As the RHS should always be +ve, the solution x = −5 1--- is not suitable 2 and should be ignored. Notice how important this verification step is. We have followed all the correct steps but logically arrived at an answer that is not possible. Verify all results for these questions.

Check: Using x = −5 1--2 LHS = | −5 1--- − 3 |

– 11 RHS = 3 × --------- + 8 2 – 33 = --------- + 8 2 = −16 1--- + 8

2

= | −8 1--- | 2

=

8 1--2

2

= −8 1--2 (Not the correct solution since LHS ≠ RHS) Using x = −1 1--4 –5 LHS = | −1 1--- − 3 | RHS = 3 × ------ + 8 4 4 – 15 1 = | −4 --- | = --------- + 8 4 4 = 4 1--= −3 3--- + 8 4

5

State the solution.

4

= 4 1--4 (Correct solution since LHS = RHS) The solution is x = −1 1--- . 4

WORKED Example 25

Solve | x – 1 | = | 2x + 3 |. THINK 1 Write the equation. 2

Remove all absolute value symbols and write the +ve and –ve cases. Reassure yourself that there are only two possible cases. −(x – 1) = −(2x + 3) is the same as (x – 1) = (2x + 3) and (x – 1) = −(2x + 3) is the same as −(x – 1) = (2x + 3)

WRITE | x – 1 | = | 2x + 3 | Case 1: x − 1 = 2x + 3

or

Case 2: x − 1 = −(2x + 3)

Continued over page

53

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

THINK 3 Solve for x for both cases.

4

WRITE −1 − 3 = 2x − x −4 = x x = −4

Verify the solutions with respect to the original equation.

x − 1 = −2x − 3 x + 2x = −3 + 1 3x = −2 2 x = − --3

Check: Using x = −4 LHS = | −4 −1 | = | −5 | =5

RHS = | 2(−4) + 3 | = | −8 + 3 | = | −5 | =5 (Correct solution since LHS = RHS) 2 Using x = − --3 2 2 LHS = – --- – 1 RHS = 2 × – --- + 3 3 3 2 1 = | −1 --- | = | −1 --- + 3 | = 1 2---

3

3

= | 1 2--- |

3

5

3

= 1 2--3 (Correct solution since LHS = RHS) 2 Therefore x = −4 and x = − --- are both suitable 3 solutions.

State the solution.

remember remember To solve equations with absolute values: 1. remove the absolute value symbols and state the equation as positive and negative cases 2. verify your solutions by substituting your answer into the original equation.

1L

Solving equations using absolute values

1 Solve for x. Example xample a | 2x | = 10 x = ±5 23 d | 3x + 2 | = 4

x = 4 or x = −6

WORKED

x=

2 --3

b |x + 1| = 5

c

e | 1 + 2x | = 0 x =

or x = −2

− 1--2-

f

x = 1 1--2- or x = − 1--2-

| 2x − 1 | = 2 x --- = 3 x = ±9 3

eBook plus Digital doc: SkillSHEET 1.7 Solving equations

2 Solve for x. a | x + 1 | = 2x − 1 x = 2 24 c | 2x + 3 | = x − 5 No solutions

b | 3x + 5 | = x − 3 No solutions d | x − 2 | = 2x − 7 x = 5

3 Solve the following for x. 1 a | 2x − 5 | = | x + 1 | x = 6 or x = 1 --325 c | 3x − 1 | = | 2x + 2 | x = 3 or x = − --15-

b | 3x − 6 | = | 2x + 4 | x = 10 or x = 2--5d | x − 5 | = | 3x − 8 | x = 1 --12- or x = 3 --14-

WORKED

Example xample

WORKED

Example xample

Chapter 1 Number systems: the Real Number System

55

Solving inequations You have graphed inequations on a number line in your junior mathematics studies. These examples require more care and you will notice that the verification step is essential to test the values you obtain.

WORKED Example 26

Solve and graph (x - 1)(x + 2) > 0. THINK 1 2

3

4

5

WRITE/DRAW

Write the inequation.

(x − 1)(x + 2) > 0

If a × b > 0 then either a and b are both positive (+ve) or a and b are both negative (−ve). This gives rise to 2 cases. Rewrite the terms of the inequation. Note: > 0 means +ve, and Note: < 0 means −ve Solve each inequation.

Case 1: If a and b > 0 (x − 1) > 0 and

(x + 2) > 0

x>1

x > −2

Graph both these inequations and decide which part of the graph satisfies both inequations. Note that the region graphed in the last graph (x > 1) satisfies both parts of case 1. Repeat steps 2 and 3 for case 2. Graph both these inequations and decide which part of the graph satisfies both inequations. Note that the region graphed in the last graph (x < −2) satisfies both parts of case 2. Either case 1 is true or case 2 is true at the one time but not both, as they are contradictory. Combine both cases on one number line so that either x > 1 (from case 1) or x < −2 (from case 2).

–3 –2 –1

0

1

2

3

4

–3 –2 –1

0

1

2

3

4

Case 2: If a and b < 0 (x − 1) < 0 and x<1

(x + 2) < 0 x < −2

–3 –2 –1

0

1

2

3

4

–3 –2 –1

0

1

2

3

4

–3

–2

–1

0

1

2

3

Either x > 1 or x < −2

–3

–2

–1

0

Graphed Not graphed

1

2

3

Graphed Continued over page

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THINK

WRITE

Use a tabular form to verify this solution. Note how the number line is divided into 3 regions. When completing this table choose a number that falls in each region and work out the sign only of each expression. Because the original product was greater than 0 (or positive) the table has verified the results on the graph. We do not want to include those values between −2 and 1.

x < −2 −2 < x < 1 (let x = −3) (let x = 0) x−1 x+2 (x − 1)(x + 2)

− − + graphed OK

x>1 (let x = 2)

− + − not graphed OK

+ + + graphed OK

The solution is either x > 1 or x < −2.

WORKED Example 27 Solve and graph: 3 a --- < 6 where x ≠ 0 x

b

x–2 ------------ < 0 where x ≠ −1. x+1

THINK

WRITE/DRAW

a

3 a --- < 6 x

1 2

Write the inequation. x can be either +ve or −ve. When x is −ve and multiplied across the inequality sign, the sign must be reversed.

Case 1:

Case 2:

3 If x > 0, --- < 6 x 3 < 6x

3 If x < 0, --- < 6 x 3 > 6x

1 --2

or x > 3

4

Draw separate graphs for both these inequalities, but remember that in case 2, x < 0 so the only part that should be graphed is where x < 0. Combine these graphs on the one number line and state your answer. Remember that an initial condition of the problem was that x ≠ 0 so that has been satisfied also.

0

–3

1– 2

1

–2

1 --2

<x 1 --2

or x <

2

–1

x < 0 and x >

–1

0 1 --2

>x

1– 2

1

2

0

3

1 --2

1

Chapter 1 Number systems: the Real Number System

THINK b

1 2

3

WRITE/DRAW

x–2 b ------------ < 0 x+1 a Remember that < 0 means −ve. If --- < 0 Case 1: b If x − 2 < 0 then either a < 0 or b < 0, but not both at x<2 the one time. Graph both these inequalities.

Write the inequation.

0 4

5

6

7

Since both results from case 1 occur at the one time combine the two different graphs that satisfy both parts. Repeat steps 2 to 4 for case 2.

1

–3

–2

2 –1

Case 2: If x − 2 > 0 x>2

Since both results of case 2 occur at the one time think about how to combine the two different graphs that satisfy both results. Note that x can’t be greater than 2 and less than −1 at the same time. Therefore this solution is impossible. Reject this solution. Graph and state the final solution (that was obtained in step 4 above).

1

x+1>0 x > −1

and

0

1

–2

–1

2

3

–1

–2

0

0

x+1<0 x < −1

and

2 3 Case 1

–3

–2

1

2

–1 Case 2

0

3

The solution is −1 < x < 2.

WORKED Example 28

Solve and graph | 2x − 3 | < 2. THINK

WRITE/DRAW

1

Write the inequality.

| 2x − 3 | < 2

2

Consider the positive and negative cases of the absolute value.

Case 1: 2x − 3 < 2

or

Case 2: −(2x − 3) < 2

Solve for x. In case 2, note that multiplying both sides of an inequality by a negative number reverses the inequality sign.

2x − 3 < 2 2x < 5 x < 5---

or or or

2x − 3 > −2 2x > 1 x > 1---

3

2

2

x < 2 1--2

4

Graph both solutions on separate number lines.

1

1

2 2 –2 3 Case 1

0

1– 2

1 Case 2

2

Continued over page

57

58

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

THINK 5

Graph this combined solution and state the solution.

WRITE/DRAW –2

–1

0

The solution is 6

Verify the 3 regions of this solution.

1– 2

1 --2

1

1

2 2 –2 3

< x < 2 1--- . 2

Check: For x < 1--- (let x = 0) 2 | 2 × 0 − 3 | < 2 (Not valid and not graphed) For 1--- < x < 2 1--- (let x = 2) 2 2 |2 × 2 − 3| < 2 | 1 | < 2 (Valid and graphed) For x > 2 1--- (let x = 3) 2 |2 × 3 − 3| < 2 | 3 | < 2 (Not valid and not graphed)

WORKED Example 29 1 Solve and graph ----------------- < 2. x–3 THINK 1

Write the inequality.

2

As with Worked example 28, state the two cases that are possible. Solve for case 1.

Draw the graph for this solution. 3

Solve for case 2. Reverse the inequality sign when you multiply by a negative.

WRITE/DRAW 1 ----------------- < 2 x–3 Case 1: Case 2: 1 1 -------------- < 2 ------------------- < 2 or x–3 –( x – 3 ) For case 1 if x − 3 > 0 (x > 3) 1 < 2(x − 3) 1 < 2x − 6 7 < 2x 7 --- < x 2 x > 3 1--2 Since x > 3 (initial condition) x > 3 1--2 satisfies this condition. 0

1

2

Case 2 1 -------------- > −2 x–3 If x − 3 < 0 (x < 3) 1 < −2(x − 3) 1 < −2x + 6 −5 < −2x

1

3 3 –2 4

5

Chapter 1 Number systems: the Real Number System

THINK Reverse the inequality sign when you divide by a negative. Remember to always check with the initial condition.

WRITE/DRAW 2 1--- > x 2

x < 3 from the initial condition; therefore x < 2 1--- satisfies this condition. 2

Draw the graph for this solution. 4

Draw the combined graph for these two solutions and state the answer.

0

1

2 2 –2 3

1

–1

0

1

4 1

2 2 –2 3 3 –2 4

1

The solution is x < 2 1--- or x > 3 1--- . 2

5

Verify the results for the three regions on the graph.

2

Check: For x < 2 1--- (let x = 0) 2 1 ----------------- < 2 0–3 1 ------------ < 2 ( Valid and graphed) –3 For 2 1--- < x < 3 1--2

2

(let x = 3 1--- ; remember x ≠ 3) 4

1 -------------------- < 2 1 3 --4- – 3 1 ------- < 2 1 --4

All solutions have been verified. Even though this verification is a fairly lengthy step it gives you confidence that your solutions are more than likely going to be correct.

| 4 | < 2 (Not valid and not graphed) For x > 3 1--- (let x = 4) 2 1 ----------------- < 2 4–3 1 -------- < 2 (Valid and graphed) 1

WORKED Example 30 x–2 Solve and graph ------------ < 2 where x ≠ −3. x+3 THINK WRITE/DRAW x–2 ------------ < 2 1 Write the inequality. x+3

Continued over page

59

60

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

THINK 2

3

WRITE/DRAW

Remember x has the same value in the numerator and denominator at any one time. Separate the solution into the +ve and −ve cases for the absolute value.

Case 1: x–2 ------------ < 2 x+3

If the denominator is –ve, then the sign must be reversed when it is multiplied across. Therefore, we will have to make 2 sub-cases for each of case 1 and case 2.

Case 1 (+ve): If x + 3 > 0 (that is, x > −3) x − 2 < 2(x + 3) x − 2 < 2x + 6 −8 < x x > −8

or

Case 2: x–2 –  ------------ < 2  x + 3 x–2 ------------ > −2 x+3

But the initial condition is that x > −3, therefore x > −3 is a valid solution. Graph this first solution. Reverse the sign due to the –ve denominator.

–4

–3

–2

–1

0

1

Case 1 (−ve): If x + 3 < 0 (that is, x < −3) x–2 ------------ < 2 x+3 x − 2 > 2(x + 3) x − 2 > 2x + 6 −8 > x x < −8 But the initial condition is that x < −3, therefore x < −8 is a valid solution.

Graph this solution. 4

Determine case 2 as for case 1.

–9

–8

–7

–6

–5

x–2 ------------ > −2 x+3 Case 2 (+ve): If x + 3 > 0 (that is, x > −3) x − 2 > −2(x + 3) x − 2 > −2x − 6 3x > −6 + 2 –4 x > -----3 But x > −3 from initial condition so x > −1 1--- . 3

Graph this solution.

–2

–1

0

1

2

61

Chapter 1 Number systems: the Real Number System

THINK

WRITE/DRAW

Determine case 2 for the –ve denominator. Remember to reverse the sign when you multiply by the –ve denominator.

Graph this solution. 5

6

Since the denominator must be either positive or negative at any one time it is case 1 (+ve) and case 2 (+ve) that we need to combine as well as case 1 (−ve) and case 2 (−ve) to produce the final graph.

Case 2 (−ve): If x + 3 < 0 (that is, x < −3) x–2 ------------ > −2 x+3 x − 2 < −2(x + 3) x − 2 < −2x − 6 3x < −6 + 2 4 x < − --3 But x < −3 from the initial condition, so x < −3. –5

–4

–3

–2

–1

Combining case 1 (+ve) and case 2 (+ve) gives: –4

–3

–2

–1

0

1

Combining case 1 (−ve) and case 2 (−ve) gives: –10

–9

–8

–7

–6

–5

Combine the two solutions. –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

The solution is x < −8 or x > 7

0

Verify the solutions by testing within the 3 regions, using the original inequality.

0

1

2

−1 1--- . 3

Check: For x < −8 (let x = −9) –9–2 ---------------- < 2 –9+3 – 11 --------- < 2 (Valid and graphed) –6 For −8 < x < −1 1--- (let x = −2) 3 –2–2 ---------------- < 2 –2+3 –4 ------ < 2 (Not valid and not graphed) 1 For x > −1 1--- (let x = 0) 3 0–2 ------------ < 2 0+3 –2 ------ < 2 (Valid and graphed) 3

This type of problem demonstrates higher level reasoning for this study of numbers. All possibilities need to be carefully considered and examined in a thoughtful, methodical manner.

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

remember remember 1. When solving inequations reverse the inequality sign when you multiply or divide by a negative expression. 2. In any equation if the product of a and b (that is a × b) is positive then a and b are either both positive or both negative. 3. If a product a × b is negative then either a or b is negative. 4. If x − a is positive then x − a > 0 and x > a. 1 a 1<x<2

5. If x − a is negative, then x − a < 0 and x < a.

0

1

2

3

4

b x > 2 or x < −2

1M

–3 –2 –1

Solving inequations

–1

1 Solve and graph the following inequations.

WORKED

Example xample

0

2

x–1 d ----------- < 3 (where x ≠ 2) x–2

–1

3 ----------- < 2 (where x ≠ 3) x–3

0

−5 < x < −3 2

1

1 3 4

5

10

1 b -------------------- < 2 (where x ≠ 1 1--- ) 2 2x – 3 d

x–1 ------------ < 2 (where x ≠ −1) x+1

f

x ----------- > 4 (where x ≠ 4) x–4

x < −3 or x > − 3--1-

0

1

1 2--3- < x < 7, x ≠ 3

0 112– 2 3 4 5 6 7 8 3

3 1--5- < x < 5 1--3- , x ≠ 4

3 31–5

4

2

x+1 ------------ > 2 (where x ≠ 3) x–3

2

1

(where x ≠ 6)

–5

3

1 --3

3

e

2 3 4 5 6 7 8 9 10

1 ----------------- < x–6

14–3 2

c

WorkSHEET 1.2

x < 3 or x > 9

–10

–4 –3 –2 –1 _ 1– 0

1 a ----------------- < 3 (where x ≠ −1) x+1

3

4 Solve and graph the following.

4

Digital doc:

−9 < x < 11

5

1 1–1

eBook plus

4

0

29, 30

3

x < 1 4--1- or x > 1 4--3-

Example xample

2

2

WORKED

1

1 13–

2 0

1

1

4 42– 5

0

x–1 d ----------------- < 2 5

| 2x − 5 | < 1 2 < x < 3

3

–1 –11–3 – 2–3

c

4

–2

b |x + 4| < 1

–5 –4 –3 –2 –1 0 1 2 3 4 5

x < −1 1--3- or x > − 2--3–3

0 1

a |x| < 4

28

–6 –5 –4 –3 –2 –1

−4 < x < 4

2 22– 3

Example xample

4

2 a x < 0 or x > 2

1 b ----------- < 3 (where x ≠ 1) x–1

3 Solve and graph the following.

WORKED

3

2

b x < 1 or x >

c

4

d 3x − 10x + 4 < 2x − 5x − 2

(2x − 3)(x − 2) < 0

2

4 a --- < 2 (where x ≠ 0)) x

SLE 2: Solve simple inequality statements such as |z - a | > b in the real system, and be able to give a verbal description of the meaning of the mathematical symbolism.

1

1 3--1-

b (x − 2)(x + 2) > 0

0

27

3

1

Example xample

3

d 2<x<3

2 Solve and graph the following inequations.

WORKED

1

1 1–2 2

0

c x < 3 or x > 4 2--1-

c

2

0

a (x − 1)(x − 2) < 0

1

d x > 2 2--1- or x < 2

26

0

c 1 --12- < x < 2

5 51–3

6

7

Chapter 1 Number systems: the Real Number System

63

Approximations for p SLE 9: Investigate some of the approximations to p which have been used.

Research the following historical approximations for π and present your findings in concise form. 1 3000 BC Egypt: The pyramids are built. The sides and heights of the pyramids of Cheops and Sneferu at Giza are constructed in the ratio of 11:7. Hence the 22ratio of one perimeter to 2 heights is 22:7. The value of π is approximately ----. 7

2 2000 BC Egypt: The Rhind Papyrus, the oldest mathematical text in existence, gives the following rule for constructing a square having the same area as a given circle: Cut one-ninth off the circle’s diameter and construct a square on the 2 ------ ) . remainder. Using this method, π is found to equal ( 16 9

3 240 BC Greece: Archimedes, engineer, architect, physicist and mathematician, ------ < π < 3 1 --- . constructs polygons of 96 sides to show that 3 10 71

7

4 20 BC Italy: Vituvius, architect and engineer, measures distances using a wheel and determines that π is equal to 3 1--- . 8

5 AD 125 Greece: Ptolemy writes his famous work on astronomy, Syntaxis 830 Mathematica. He finds that π is equal to 3 + ----+ -------. 2 60

60

6 AD 480 China: Tsu Ch’ung-chih, expert in mechanics and interested in --------- . machinery, gives the value of π as 355 113

7 AD 1150 India: Bhaskara writes on astronomy and mathematics and gives ------------ . several values of π, the most accurate is 3927 1250

8 AD 1579 France: Vieta finds π correct to nine decimal places by considering polygons of 6.516 = 393 216 sides. He also discovers that 2 2+ 2 2+ 2+ 2 2 --- = ------- × -------------------- × ---------------------------------- . 2 2 2 π In this and the next two series, examine how the approximation improves as the number of terms is increased. 9 AD 1650 England: John Wallis uses a complicated and difficult method to 4 3 × 3 × 5 × 5 × 7 × 7 × 9 × 9… obtain π from --- = -------------------------------------------------------------------------2 × 4 × 4 × 6 × 6 × 8 × 8… π 10 AD 1699 England: Sharp calculates π to 72 decimal places by evaluating the

π 1 series --- = ------- − 6 3

1- 3  ----- 3 ---------------

3

+

1- 5  ----- 3 ---------------

5

−

1- 7  ----- 3 ---------------

7

…

11 AD 1913 India: The mathematician Ramanujan presents the following as an approximation for π : π =

4

2

2

--------- . 9 + 19 22

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Real numbers — application and modelling 1 Solve the following inequality and graph the solution on a number line 2x + 1 ---------------- < 1. −1 − 2 < x < 2 –3 –2 –1 0 1 2 x– 2 √2 –2.4 2 Rationalise the denominator for

a– b ---------------- . a+ b

2

(a – b) a – b ---------------------------------------2 a –b

2x – 5 3 Solve the inequation --------------- > 0. x ∈ R, x ≠ −3 1--2- , x ≠ 2 1--22x + 7 4 By noting the expansion of ( a +

b )2 and the fact that

12 + 2 35 = 7 + 5 + 2 7 × 5 , determine

--14

12 + 2 35 .

Hence, determine 17 + 6 8 . 5 If the integer points, n, and the points midway between them, n + --1- , are 2 mapped on a number line, how far away from the nearest of these points can any point on the number line be? Find an integer m such that | 2 – 1--- m | < 1--- and an integer k such that 2 4 | 5 – 1--- k | < 1--- . 2 4 Explain the significance of these results with respect to the topic of approximation of irrationals to rationals. (Your response should not rely on calculator computations.) 6 The most common way students learn to find the greatest common divisor of g.c.d. = 225 two integers is to factorise both numbers into their prime factors and take the common prime factors. For example, to find the greatest common divisor of 45 024 and 5712 we can write: 45 024 = 25 × 3 × 7 × 67 5712 = 24 × 3 × 7 × 17 So the greatest common divisor is 24 × 3 × 7 = 336. However, finding the prime factors is not always that straightforward and Euclid developed an algorithm that produces the greatest common divisor. To apply it we divide the smaller integer into the larger integer. Consider the integers 45 024 and 5712 again. When 45 024 is divided by 5712 the result is 7 with a reminder of 5040. Thus 45 024 = 7 × 5712 + 5040 Now divide the remainder (5040) into the divisor (5712) 5712 = 1 × 5040 + 672 and the new remainder into the previous remainder, and so on: 5040 = 7 × 672 + 336 672 = 2 × 336 + 0 The last non-zero remainder is the greatest common divisor, because it divides both 45 024 and 5712 and every divisor of both 45 024 and 5712 must also divide it.

, m = 3, k = 4

1 7 + ----------------------------------------1 1 + ------------------------------1 3 + --------------------1 1 + -----------1 2 + ----13

7 + 5, 3 + 2 2

Chapter 1 Number systems: the Real Number System

We can rewrite the previous equations as 45 024 1 5712 1 ---------------- = 7 + --------------- ------------ = 1 + --------------5712 5712 5040 5040 ----------------------5040 672

5040 1 ------------ = 7 + --------------672 672 --------336

65

672 --------- = 2 336

45 024 1 which can be combined as a continued fraction as ---------------- = 7 + ---------------------- and can 5712 1 1 + -----------1 7 + --2 also be written as 7, 1, 7, 2 (whole numbers obtained at each division step). Express the quotient 327 600 ÷ 42 075 as a continued fraction and hence state the greatest common divisor of these two integers.

7 Show that

2+ 2 +

2– 2 <2 2

(Do not use your calculator.)

The next 2 terms would be 1 2 = 1 + -------------------------------------------1 1 8 Show that 2 = 1 + ---------------- and hence deduce that 2 + ----------------------------------1 1+ 2 2 + -------------------------1 2 + ---------------1 1 2 = 1 + -------------------------- = 1 + ----------------------------------1+ 2 1 1 and 2 + ---------------2 + -------------------------1 1 1+ 2 2 = 1+ ----------------------------------------------------2 + ---------------1 -------------------------------------------2 + 1+ 2 1 2 + ----------------------------------1 Now write the next two terms in such a sequence. 2 + -------------------------1 2 + ---------------1+ 2 9 Diophantus of Alexandria, born between AD 75 and AD 250, was the author of

Arithmetica and wrote many papers investigating only whole numbers. Diophantine equations are involved in many real-life situations, due to their whole number solutions. The solutions are often found by trial and error. The following problem, which you might like to try, is based on Diophantine equations. (The answer is given at the end of the problem.) Five men and a monkey were shipwrecked on an island and they set out to collect as many coconuts as they could on their first day so that they would have provisions until they were rescued. After many hours collecting coconuts they piled them all together and went to sleep. Later that night, while the others slept one man woke up, and started to worry about whether the coconuts would be divided fairly. He decided to take his one-fifth share of the nuts and gave the one remaining coconut to the monkey. Throughout the night all of the men did likewise; they took their fifth share of what was left of the nuts and gave the one remaining nut to the monkey. In the morning they met to divide the nuts into five equal shares. Each man knew that there were some nuts missing but none admitted he had some extras, of his own, hidden away. How many nuts were there to start with? (Solution) The Diophantine equations that lead to the solution of this problem simplify to 1024x – 8404 = 15 625y where x is the original amount collected and y is the number each is given in the morning. The smallest possible value for x is 3121 with other values at intervals of 15 625.

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summary The Real Number System • The set of real numbers (R) is divided into two main sets: rational and irrational numbers. These sets may be further divided into smaller subsets as illustrated on the chart and Venn diagram below. Real numbers R

Irrational numbers I (surds, non-terminating and non-recurring decimals, π ,e)

Negative Z–

Rational numbers Q

Integers Z

Zero (neither positive nor negative)

Non-integer rationals (terminating and recurring decimals) Positive Z+ (Natural numbers N)

• Rational numbers (Q) can be expressed in ε =R a the form --- , where a and b are integers Q (Rational numbers) b with no common divisors and b ≠ 0. They Z (Integers) I may also be expressed as whole numbers, N (Irrational terminating decimals and recurring (Natural numbers) decimals. numbers) • Irrational numbers (I) cannot be expressed in the form a--- . They can be expressed only b as non-terminating and non-recurring decimals and include surds and numbers such as π and e. • Recurring decimals are rational numbers which may be expressed as a ratio of two integers. • Surds are irrational numbers which can only be represented exactly using a root symbol (or radical), that is: , 3 , 4 and so on. • π and e are examples of irrational numbers which may be converted to nonterminating and non-recurring decimals; however, they are not surds.

Set notation • Set notation is used when defining the Real Number System. • The following symbols are useful when working with sets: { }‘set’ ∈ ‘is an element of’ ∉ ‘is not an element of’ ⊂ ‘is a subset of’

Chapter 1 Number systems: the Real Number System

67

Working with surds • To simplify a surd, it should be written as a product of two factors, one of which is the largest perfect square. • Like surds may be added and subtracted; surds may need to be simplified before adding and subtracting. • Surds may be multiplied according to the rules: a× b =

ab

m a × n b = mn ab • When a surd is multiplied by itself (squared), the result is the number under the radical:

( a)2

• Multiplication involving brackets:

a( b + c) =

1. The Distributive Law: 2. FOIL:

= a

(

a + b )( c + d ) =

3. Perfect squares:

4. Difference of two squares DOTS:

( a + b)2 ( a – b)2 ( a + b )(

ab + ac ac + ad + bc + bd

= a + 2 ab + b = a – 2 ab + b a – b) = a – b

• The product of a conjugate pair of surds is rational. • Surds may be divided according to the rule: a a ÷ b = ------b =

a --b

• Rationalising denominators: 1. If the denominator contains a surd, multiply both numerator and denominator by the surd part of the denominator: a a b ------- = ------- × ------b b b ab = ---------b 2. If the denominator is a sum or difference of surds, multiply both the numerator and the denominator by the conjugate of the denominator: 1 1 a+ b -------------------- = -------------------- × -------------------a– b a– b a+ b a+ b = -------------------a–b

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Modulus • The modulus (or absolute value) of a number is the magnitude of that number and is always positive. • The modulus of x is denoted by | x |. • | x | = −x if x < 0 = 0 if x = 0 = x if x > 0

Solving equations using absolute values • First remove the absolute value symbols and state the equation as positive and negative alternative cases. • Verify your solutions for all these questions by substituting your answer into the original equation.

Solving inequations • Remember that if x > 0 then x is positive, and if x < 0 then x is negative. • If a product of two factors is greater than 0 then both factors must be either positive or negative. • Likewise, if a product of two factors is less than 0 then only one of the factors must be positive and the other must be negative, • Organise your solution into two cases that will develop arguments for all possible values. • The two values that result for each case are values that should occur at the one time. The graph you draw must be a combination of these two solutions for each case. • Verify your solutions by choosing values that fall in each of the regions of your graph. • When you multiply or divide by a negative factor across an inequality sign, remember to reverse the sign.

Chapter 1 Number systems: the Real Number System

2 a Irrational, since equal to non-recurring and non-terminating decimal b Rational, since can be expressed as a whole number c Rational, since given in a rational form d Rational, since it is a recurring decimal e Irrational, since equal to non-recurring and non-terminating decimal

CHAPTER review 1 multiple choice Which of the given numbers, A

. 0.81 , 5, −3.26, 0.5 and

C

6----, 12

0.81 and

6- , ----12 3----12

69

. π 0.81 , 5, −3.26, 0.5 , --- , 5 π 6 B ------ and --12 5

3----12

D 5, −3.26 and

3----12

are rational?

1A

6----12

E 5 2 For each of the following, state whether the number is rational or irrational and give the reason for your answer: . 12 121 a b c 2--d 0.6 e 3 0.08

1A

3 multiple choice Which of the following statements is not correct?

1A

9

A

9----81

∉Q

8 --4

B

∉Q

C

0 --4

∉ Z+

D

3

– 125 ∈ Z − E (−5)2 ∈ Z +

4 multiple choice Which. of the following statements regarding the given set of numbers, {5, 0.7, 64 , 21, 8 , 20 }, is correct? 64 ,

A 5, B

8,

D

64 ,

1A

20 , ∈ Z +

20 cannot be expressed as rational numbers. . C 5, 0.7 and 21 are the only rational numbers of the set. 8 and

20 cannot be expressed as rational numbers.

E None of the above. 5 Classify each of the following into the smallest subset in which they belong using Q, I, Z, Z + and Z −. (Simplify first where possible.) 3

a 4×

1 – --------- – 0.2 Z − 125

4 b -----------------2 0.01

6 multiple choice Which of the following fractions, decimals? A

1- 1--- 2------, , 17 5 3

B

3- --------, 5- , 2--13 12 3

Z+ c

15 15 – ------ × ------ Q 8 2

1- ----5- 1 ----, 3- , ----, --- , 2--- , 17 13 12 5 3

C

3- --------, 513 12

d

8 2 --- + -----9 25

1 --5

E

18 -----25

B

73 --------100

C

73 -----99

D

4 --5

E

1B

1- 1 3----, --- , ----17 5 13

7 multiple choice .. The recurring decimal 0.7 2 can be expressed as a fraction in its simplest form as: A

I

cannot be expressed as recurring D

1A

8 -----11

1B

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1B

8 Express the following recurring decimals as fractions in the simplest form. .. 62 a 0.6 2 ----. 99337 b 0.374 -------900 . . 157 c 0.95 1 -------165

1C

9 multiple choice Which of the following numbers of the given set,

{3

2 , 5 7 , 9 4 , 6 10 , 7 12 , 12 64 }, are surds? A All of these B 9 4 , 12 64 C 3 2 and 7 12 only D 3 2 , 5 7 and 6 10 only E 3 2 , 5 7 , 6 10 and 7 12

1C 1D 1D

10 Which of

2m ,

25m ,

m ------ , 16

20 ------ , m

1E

3

8m are surds 10 a b

2m , 25m ,

20 ------ , m

3

m ------ , 16

11 multiple choice The expression

250 may be simplified to:

A 25 10

B 5 10

C 10 5

D 10 25

E 5 50

12 multiple choice The expression

392x 8 y 7 may be simplified to: C

14x 4 y 3 2y D 14x 4 y 3 2

E 14x 4 y 3 2xy

13 Simplify the following surds. Give the answers in the simplest form. a 4 648x 7 y 9

72x 3 y 4 2xy

2 25 b – --- ------ x 5 y 11 5 64

– 1--4- x 2 y 5 xy

14 multiple choice When expressed in its simplest form, 2 98 – 3 72 is equal to: A –4 2

1E

m,

a if m = 4? b if m = 8?

A 196x 4 y 3 2y B 2x 4 y 3 14y

1D

3

B –4

C –2 4

D 4 2

15 Simplify the following, giving answers in the simplest form. a 7 12 + 8 147 – 15 27 b

1 --2

25 3

1 64a 3 b 3 – 3--- ab 16ab + --------- 100a 5 b 5 4 5ab

3ab ab

E None of these

m, 20 -----m

3

8m

Chapter 1 Number systems: the Real Number System

71

16 Determine the length of the unknown side, giving the answer in the simplest form and specifying the appropriate unit. a

b

16 a 5 m b ( 17 – 4 6 ) cm c ( 26 – 4 2 ) m d 22 cm

x l

10 cm P = 44 – 8 6 cm

1E 1F

P = 80 m

c

d

6+2 3m

c

y

11 3 cm 11 cm

P = 64 + 4 3 – 8 2 m

17 multiple choice

1F

When expressed in its simplest form, 9 12 × 3 5 is equal to: A 27 60

B 15 54

C 18 5

D 54 15

E 6 15

18 Simplify the following, giving answers in the simplest form. a

1 --5

1F

675 × 27 27

b 10 24 × 6 12

720 2

19 multiple choice When expanded and expressed in its simplest form, 12 ( 8 – 6 ) is equal to: A 4 6–6 2

B

96 – 72

C 4 3–6

D

24

E 2 6–3 2

20 multiple choice When expanded and expressed in its simplest form, ( 9 x 2 y – 7x ) ( 9 x 2 y + 7x ) is equal to: A 9x 2 y – 7x 2

B 32y

When expanded and expressed in its simplest form, ( 2 8 – 7 ) is equal to: 2

B 39 – 8 14

C 25 – 8 14

D 39

When expressed in its simplest form, x3 B --------4

1H

8x 3 -------- is equal to: 32

x3 C --------2

x x D ---------4

1G

E 25 – 16 14

22 multiple choice

x x A ---------2

1G

C 32x 2 y – 7x 2 D 81x 2 y – 49x 2 E 2x 2 y

21 multiple choice

A 39 – 16 14

1G

x3 E ----4

72 1H

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

23 multiple choice

10x A -------------25x 2 y

1H 1I

1I

and its width is

When expressed in its simplest form with a rational denominator, 2 ------7

B

C

14 ---------7

D

7 56 – 3 126 -----------------------------------32

7 ------14

is equal to: E

2

5 7 ---------4

27 Express the following with a rational denominator, giving answers in the simplest form. Assume that a, b, x and y are positive real numbers. x 5y ------------2

20x 5 y 3 b --------------------10xy

x2 y 2 c

9a 2 b ---------------b

3a

28 multiple choice 1 When expressed in its simplest form with a rational denominator, ------------------------ is equal to: 3 8+ 5 6 2– 5 B ----------------------77

3 8– 5 C ----------------------67

3 3 D ---------67

6 2– 5 E ----------------------67

D 22 – 5 11

E 23

29 multiple choice 11 – 3 then x 2 + 8x + 5 is equal to:

A 1 – 6 11

1J

2 --7

18 -----63

26 Express the following with a rational denominator, giving the answer in the simplest form.

If x =

1J

10x E ------------5x 2 y

x D --------5xy

2 + 3 cm. ( 23 6 – 48) cm

3 8– 5 A ----------------------77

1J

x 10 C ------------5x 2 y

25 multiple choice

5x 3 y a ---------------2 x

1J

10 x B ------------5x 2 y

24 Determine the length of the unknown side of a rectangle, given its area is 7 18 – 2 3 cm2

A

1I

6x 2 y 3 8xy ----------------- × -------------- is equal to: 4 5 12x y 10x 2 y

When expressed in its simplest form,

B 2 11 + 22

30 Given that x = 2 7 – 3 2 find: 1 1 b x 2 – ----a x 2 + ----2 x x2

C 2 11 + 1

c

x 2 – 9x + 5

2323 – 594 14 30 a ------------------------------------50 2277 – 606 14 b ------------------------------------50 c 51 – 12 14 – 18 7 + 27 2

31 Express the following with a rational denominator, giving the answer in the simplest form. 1 1 7– 3 --------------------------- – ----------------------- 3----------------------40 2 7–2 3 3 3– 7

Chapter 1 Number systems: the Real Number System

32 Determine the area of the triangle at right, expressing the answer with a rational denominator in the simplest surd form. Measurements are in metres.

73 1J

1 — 2+ 5

3 ------- m2 2 6 + 15

33 multiple choice –2 3 × –7 × –3 When expressed in its simplest form with a rational denominator, -------------------------------------------- is equal to: 3 6 A –7 2

14 C ------2

B 2 7

7 18 E ------------3

D 7 2

34 Simplify the following. a – 7 – – 4 −11

b

–3 × – 5 × 4 ------------------------------------ – – 8 –6 + 6

1K

−3

35 multiple choice

1L

Which of the given values of x solve the equation | 3 − 2x| = 2? A x=

B x=

1--2

C x = 1--- ,

5 --2

2

D x = 2,

5 --2

E x=2

2 --5

36 multiple choice

1L

Which of the given values of x solve the equation | x + 3 | = 2x + 7? A x = −4, −3 1--3

B x = −4, 1 1--3

C x = −3 1---

D x = −1, −4

3

E x = −2

37 multiple choice

1L

Which of the following values of x solve the equation | x − 3 | = | 2x − 1 |? A x = −2,

2--3

B x = −2, −1

C x = −2, 1 1--3

D x = 2--- , 1 1--3

E x=

3

1 --3

38 multiple choice

1M

Which values of x are a solution for the equation (x + 1)(x − 2) < 0? A −1 < x < 2

B x < −1 or x > 2 C −1 < x < 0

D −1 < x < 1

E x>2

39 multiple choice

1M

2 Which values of x are a solution for --- < 4? x A 0 < x < 1--B 0 < x < 1--C x < 0 or x > 3 D x < 0 or x > 4

1 --2

2

E x>1

40 multiple choice Which of the following values of x are a solution for A

--12

<x<2

B x<

--12

or x > 2 C 1 < x < 2 --12

1K

x+1 ------------ < 3? x–1 D x < 1, x > 2 --12

1M E x>1

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Modelling and problem solving 1 The electrical current I in amps, delivering electrical power P, through resistor R, is given by the rule I =

P --- . Express the current as a surd in the simplest form when: R

a P = 500, R = 18 b P = 425, R = 6 c

5 10 ------------- amps 3

P = 729, R = 0.38

d P = 1700, R = 8

5 34 ------------- amps 2

5 102 ---------------- amps 6 135 38 ------------------- amps 19

2 An ice-cream cone with measurements as shown is completely filled with ice-cream, and has a hemisphere of ice-cream on top. 27 cm

175 cm

a Determine the height of the ice-cream cone in simplest surd form. b Determine the volume of the ice-cream in the cone. c

2 37 cm

18 π 37 cm3

3 Determine the volume of the ice-cream in the hemisphere. 54 π 3 cm

d Hence, find the total volume of ice-cream.

18 π ( 37 + 3 3 ) cm3

3 A gold bar with dimensions of 5 20 , 3 12 and 2 6 cm is to be melted down into a cylinder of height 4 10 cm. eBook plus Digital doc: Test Yourself Chapter 1

a Find the volume of the gold, expressing the answer in the simplest surd form and specifying the appropriate unit. 360 10 cm3 b Find the radius of the cylinder, expressing the answer in the simplest surd form and specifying the appropriate unit. c

3 10 π ----------------- cm π

If the height of the cylinder was 3 40 cm, what would be the new radius? Express your answer in the simplest surd form. 2 15 π ----------------- cm π

Number systems: complex numbers

2 syllabus reference Core topic: Real and complex number systems

In this chapter 2A Introduction to complex numbers 2B Basic operations using complex numbers 2C Conjugates and division of complex numbers 2D Radians and coterminal angles 2E Complex numbers in polar form 2F Basic operations on complex numbers in polar form

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Introduction to complex numbers • definition of complex numbers including standard and trigonometrical (modulusargument) form • algebraic representation of complex numbers in Cartesian, trigonometric and polar form • geometric representation of complex numbers — Argand diagrams • operations with complex numbers including addition, subtraction, scalar multiplication, multiplication of complex numbers, conjugation • simple, purely mathematical applications of complex numbers

In 1545, the Italian mathematician Girolamo Cardano proposed (what was then) a startling mathematical expression: 40 =

(5 +

– 15 )( 5 – – 15 )

This was a valid expression, yet it included the square root of a negative number, which seemed ‘impossible’. What is a number such as – 1 or – 15 and how does it relate to a real number, and what does it signify in mathematics? Back in Chapter 1, you may recall that our definition of real numbers included whole numbers, fractions, irrational and rational numbers as subsets of the real number set. Whenever the square root of a negative number was encountered was this classified as real? Where did we sometimes encounter such numbers in calculations? Solution of quadratic equations sometimes brought these numbers to the foreground. What was the difference between these two situations: x2 + 3x − 6 = 0 and x2 + 3x + 1 = 0? How did the solutions to these equations relate to properties of the associated parabolas? In terms of the mathematics that you have studied so far, these square roots of negative numbers have some significance. But why did the square roots of negative numbers become central to the study of a new set of numbers called the complex numbers? It was partly curiosity and partly because mathematicians such as Diophantus (the Greek mathematician) and Leibniz (the German mathematician) found that real numbers could not solve all equations. Eventually scientists and engineers discovered their uses. Complex numbers are now used extensively in the fields of physics and engineering in areas such as electric circuits and electromagnetic waves. Combined with calculus theory, complex numbers form an important part of the study of mathematics known as complex analysis.

Square root of a negative number The quadratic equation x2 + 1 = 0 has no solutions for x in the Real Number System R because the equation yields x = ± – 1 and there is no real number which, when squared, gives −1 as the result. If, however, we define an ‘imaginary number’ denoted 2 by i such that i 2 = −1, then x = ± – 1 becomes x = ± i = ±i. For the general case x2 + a2 = 0, with a ∈ R, we can write: 2 x = ± –a = ± –1 × a 2

=± i ×a

2

2

= ±ai Powers of i will produce ±i or ±1. We have i 2 = −1, i 3 = i 2 × i = −1 × i = −i, i 4 = i 2 × i2 = −1 × −1 = 1, i 6 = (i 2)3 = (−1)3 = −1 and so on. The pattern is quite obviously that even powers of i result in 1 or −1 and odd powers of i result in i or −i.

Definition of a complex number A complex number (generally denoted by the letter z) is defined as a quantity consisting of a real number added to a multiple of the imaginary unit i. For real numbers x and y, x + yi is a complex number. This is referred to as the standard or Cartesian form.

Chapter 2 Number systems: complex numbers

77

C = {z: z = x + yi where x, y ∈ R} defines the set of complex numbers. The real part of z is x and is written as Re (z). That is, Re (z) = x. The imaginary part of z is y and is written as Im (z). That is, Im (z) = y. Note: Every real number x can be written as x + 0i and so the set of real numbers is a subset of the set of complex numbers. That is, R ⊂ C.

WORKED Example 1 Using the imaginary number i, write a simplified expression for: – 16 a b –5 . THINK a

1 2 3

b

1 2 3

WRITE

Express the square root of −16 as the product of the square root of 16 and the square root of −1. Substitute i 2 for −1. Take the square root of 16 and i2. Express the square root of −5 as the product of the square root of 5 and the square root of −1. Substitute i2 for −1. Simplify.

16 × – 1

a

– 16 =

b

= 16 × i = 4i –5 = 5 × –1 = 5 × =i 5

i

2

2

WORKED Example 2 Write down the real and imaginary parts of the following complex numbers, z. a z = −3 + 2i b z = − 1--2- i THINK

WRITE

a

a Re (z) = −3

b

1

The real part is the ‘non-i’ term.

2

The imaginary part is the coefficient of the i term. The real part is the ‘non-i’ term.

1 2

The imaginary part is the coefficient of the i term.

Im (z) = 2 b Re (z) = 0 Im (z) = − 1--2

WORKED Example 3

Write i 8 + i 5 in the form x + yi where x and y are real numbers. THINK 1 2

Simplify both i 8 and i 5 using the lowest possible power of i. Add the two answers.

WRITE i 8 = (i 2)4 = (−1)4 = 1 i 5 = i 4 × i = (i 2)2 × i = (−1)2 × i = 1 × i = i i8 + i5 = 1 + i

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

WORKED Example 4 Simplify z = i 4 − 2i 2 + 1 and w = i 6 − 3i 4 + 3i 2 − 1 and show that z + w = −4. THINK

WRITE

1

Replace terms with the lowest possible powers of i (remember i 2 = −1).

2

Add the two answers.

i 4 − 2i 2 + 1 = (i 2)2 − 2 × −1 + 1 = (−1)2 + 2 + 1 =4 6 4 i − 3i + 3i 2 − 1 = (i 2)3 − 3(i 2)2 + 3 × −1 − 1 = (−1)3 − 3(−1)2 − 3 − 1 = −1 − 3 − 3 − 1 = −8 z + w = i 4 − 2i 2 + 1 + i 6 − 3i 4 + 3i 2 − 1 z+w=4−8 z + w = −4

WORKED Example 5 Evaluate each of the following. a Re (7 + 6i)

b Im (10)

c Re (2 + i − 3i 3)

1 – 3i – i 2 – i 3 d Im  ----------------------------------   2

THINK

WRITE

a The real part of the complex number 7 + 6i is 7.

a Re (7 + 6i) = 7

b The number 10 can be expressed in complex form as 10 + 0i and so the imaginary part is 0.

b Im (10) = Im (10 + 0i) =0

c

d

1

Simplify 2 + i − 3i 3.

2

The real part is 2.

1

Simplify the numerator of – 3i – i 2 – i 3-  1--------------------------------.   2

2

Simplify by dividing the numerator by 2.

3

The imaginary part is −1.

c Re (2 + i − 3i 3) = Re (2 + i − 3i × i 2) = Re (2 + i + 3i) = Re (2 + 4i) =2 1 – 3i – i 2 – i 3 1 – 3i + 1 + i d Im  ---------------------------------- = Im  -------------------------------     2 2 2 – 2i = Im  --------------  2  2(1 – i) = Im -----------------2 = Im (1 − i) = −1

Chapter 2 Number systems: complex numbers

79

remember remember 1. 2. 3. 4.

The ‘imaginary number’ i has the property that i 2 = −1. A complex number z is of the form z = x + yi where x, y ∈R. The real part of z is x and is written as Re (z). The imaginary part of z is y and is written as Im (z).

2A

Introduction to complex numbers

1 Using the imaginary number i, write down expressions for: – 9 3i – 25 5i – 49 7i a b c 1 – 11 11i – 7 7i e f – 4--- 2--3- i g

WORKED

Example

9

d

–3

3i

h

-----– 36 25

6 --5

i

2 Write down the real and imaginary parts, respectively, of the following complex numbers, z. 2 a 9 + 5i 9, 5 b 5 − 4i 5, −4 c −3 − 8i −3, −8 d 11i − 6 −6, 11 e 27 27, 0 f 2i 0, 2 g –5 + i –5, 1 h –17i 0, –17

WORKED

Example

3 Write each of the following in the form x + yi, where x and y are real numbers. b i 9 − i 10 1 + i c i 12 + i 15 1 − i d i 7 − i 11 0 + 0i a i 9 + i 10 −1 + i 3 e i 5 + i 6 − i7 f i(i 13 + i 16) −1 + i g 2i − i 2 + 2i 3 h 3i + i 4 – 5i 5 1 − 2i −1 + 2i 1 + 0i 6 7 10 − 3 and w = 4i 8 − 3i 11 + 3 and show that z + w = 5. WORKED 4 Simplify z = i + 3i − 2i WORKED

Example

Example

z = −2 − 3i, w = 7 + 3i

4

5 Evaluate each of the following. a Re (−5 + 4i) −5 b Re (15 − 8i) 15 5 d Im (1 − 6i) −6 e Im (3 + 2i) 2

WORKED

c f

Example

9 – 5i 14 – 2i 7 g Re (i 5 − 3i 4 + 6i 6) −9 h Im  4i -----------------------------------   3

Re (12i) 0 Im (8) 0

2

i3 – i + 2 6 Write 3 – -------------------- in the form x + yi, where x and y are real numbers. 4 − i i2 – i4 7 multiple choice a The value of Re (i + i 3 + i 5) is: A 2 B −1 C 3 D 1 b The value of Im [i(2i 4 − 3i 2 + 5i)] is: A 0 B −5 C 5 D 10 2 3 4 5 6 c The expression i + i − i + i − i + i simplifies to: A i B 0 C i−1 D i+2

E 0 E 4 E −i

1+i+ +…+ d If f ( i ) = ---------------------------------------------- which one of the statements below is true? 4 A f(i) = 2 + i B Re [f(i)] = 5 C Im [f(i)] = − 1--4 D f(i) = 1 − i E f(i) = 0 i2

i 11

n ---

8 If n is an even natural number show that ( – 1 ) 2 = i n . Check with your teacher.

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SLE 1: Solve quadratic equations whose discriminant is negative.

Equation A: y = x2 − 2x −3 1 Roots are −1 and 3, hence there are two x-intercepts at x = −1 and x = 3. The discriminant is positive (b2 − 4ac = 16). 2 (1, −4) 3 y 2 y = x − 2x − 3

Complex numbers in quadratic equations In your junior mathematics studies you graphed quadratics and found the real roots of the expressions using the formula for the solution of a general quadratic equation of the form ax2 + bx + c. Sometimes the values for a, b and c meant that the value under the radical sign was negative; that is, it had a negative discriminant (for example, – 16 ). You might have been told that this meant there were ‘no real roots’ for this quadratic. That was correct, but only half the answer. Follow the steps below and you will hopefully develop a better understanding of the results you obtain. The following formulas are included for your assistance: 2

−1 0

1

–b ± b – 4ac –b x = ------------------------------------- , turning point occurs where x = -----2a 2a

3

−3 (1, −4)

Equation B: y = x2 − 2x + 1 1 Root is 1, hence there is one x-intercept at x = 1. Discriminant is 0 (b2 − 4ac = 0). 2 (1, 0) 3 y y = x2 − 2x + 1

1 0

x

1

Step 1 Use the formula for the solution of a quadratic equation to find the roots of: y = x2 − 2x − 3 ....................(A) Interpret this result. Step 2 Use the formula for the x-coordinate of the turning point and substitute this into the original quadratic to find the y-coordinate of this turning point. Step 3 Graph this quadratic equation using the information from steps 1 and 2. Repeat steps 1–3 with the following quadratic equations. Note the effect of the negative discriminant in equation (C). y = x2 − 2x + 1 ..................................(B) y = x2 − 2x + 2 ..................................(C) Graphically, we can see that there are no real values of x that satisfy the equation x2 − 2x + 2 = 0.

Basic operations using complex numbers

Equation C: y = x2 − 2x + 2 1 No real roots, hence there Complex numbers can be added, subtracted, multiplied and divided. In general, the are no x-intercepts. solutions obtained when performing these operations are presented in the standard form Discriminant is negative z = x + yi. (b2 − 4ac = −4). 2 (1, 1) Argand diagrams 3 2 y y = x − 2x + 2

2 1 0

(1, 1) 1

x

In the complex system, x2 − 2x + 2 = 0 has roots 1 + i and 1 − i.

We know that an ordered pair of real numbers (x, y) can be represented on the Cartesian plane. Similarly, if we regard the complex number x + yi as consisting of the ordered pair of real numbers (x, y), then the complex number z = x + yi can be plotted as a point (x, y) on the complex number plane. This is also referred to as the Argand plane or an Argand diagram in recognition of the work done in this area by the Swiss mathematician Jean-Robert Argand.

81

Chapter 2 Number systems: complex numbers

The horizontal axis is referred to as the Real axis and the vertical axis is referred to as the Imaginary axis. The points A, B, C, D and E shown on the Argand diagram at right represent the complex numbers 3 + 0i, 0 + 2i, −4 + 5i, −3 − 4i and 2 − 2i respectively. This method of representation is a useful way of illustrating the properties of complex numbers under the operations of addition, subtraction and multiplication. SLE 4: Use geometry to demonstrate the effect of addition, subtraction and multiplication of complex numbers.

C

Im (z) (Imaginary axis) 5 4 3 2 B 1 A

–4 –3 –2 –1 –1 –2 –3 D –4

1 2 3 4 Re (z) (Real axis)

E

Addition of complex numbers Addition is performed by adding the real and imaginary parts separately. If z = a + bi and w = c + di then z + w = (a + c) + (b + d)i where Re (z + w) = Re (z) + Re (w) and Im (z + w) = Im (z) + Im (w).

Geometric representation If z1 = x1 + y1i and z2 = x2 + y2i then z2 + z1 = (x2 + x1) + (y2 + y1)i. If a directed line segment connects the origin (0 + 0i) to each of the points z1, z2 and z1 + z2, then the addition of two complex numbers can be associated with standard methods of addition of the directed line segments. Im (z) The figure at right illustrates the situation for z2 + z1, with, say, positive values for x1, x2, y1, y2 and x1 < x2 and (y1 + y2) z1 + z 2 y1 < y2. Note: The origin, z1, z2 and z2 + z1 form a parallelogram. You will use this concept later in this course when you study vector addition.

y2 y1 0

z1

x1

z2 Re (z) x2 (x1 + x2)

Subtraction of complex numbers If we write z − w as z + −w we can use the rule for addition of complex numbers to obtain z + −w = (a + bi) + − (c + di) = a + bi − c − di = (a − c) + (b − d)i If z = a + bi and w = c + di then z − w = (a − c) + (b − d)i.

Geometric representation If z1 = x1 + y1i and z2 = x2 + y2i then z2 − z1 = (x2 − x1) + (y2 − y1)i. If a directed line segment connects the origin (0 + 0i) to each of the points z1, z2 and z2 − z1 then the subtraction of two complex numbers can also be associated with standard methods of the addition of directed line segments. The figure at right illustrates the situation for z2 − z1, again with positive values for x1, x2, y1, y2 and x1 < x2 and y1 < y2.

Im (z)

y2 (y2 – y1) y1

z2 z1

z2 – z 1 Re (z)

0

x1 (x2 – x1) x2

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WORKED Example 6

For z = 8 + 7i, w = −12 + 5i and u = 1 + 2i, calculate: a z+w b w−z c u − w + z. THINK WRITE a Use the addition rule for complex numbers. a z + w = (8 + 7i) + (−12 + 5i) = (8 − 12) + (7 + 5)i = −4 + 12i b Use the subtraction rule for complex b w − z = (−12 + 5i) − (8 + 7i) numbers. = (−12 − 8) + (5 − 7)i = −20 − 2i c Use both the addition rule and the c u − w + z = (1 + 2i) − (−12 + 5i) + (8 + 7i) subtraction rule. = (1 + 12 + 8) + (2 − 5 + 7)i = 21 + 4i

Multiplication by a constant (or scalar) If z = x + yi and k∈R Im (z) ky kz then kz = k(x + yi) = kx + kyi z y For k > 1, the product kz can be illustrated as shown at right. The ratio of corresponding sides of the two triangles is k:1. 0 x kx Re (z) A similar situation exists for k < 1. So when a complex number is multiplied by a constant, this produces a directed line segment in the same direction (or at 180 degrees if k < 0) which is larger in length if k > 1 or smaller if 0 < k < 1. Geometrically this is called a transformation or dilation, which means magnifying or decreasing by a constant factor.

WORKED Example 7

If z = 3 + 5i, w = 4 − 2i and v = 6 + 10i, evaluate: a 3z + w b 2z − v c 4z − 3w + 2v. THINK a 1 Calculate 3z + w by substituting values for z and w. 2 Use the rule for adding complex numbers. b

1 2

c

1 2

WRITE a 3z + w = 3(3 + 5i) + (4 − 2i) = (9 + 15i) + (4 − 2i) = (9 + 4) + (15 − 2)i = 13 + 13i b 2z − v = 2(3 + 5i) − (6 + 10i)

Calculate 2z − v by substituting values for z and v. Use the rule for subtraction of complex numbers. Calculate 4z − 3w + 2v by substituting values for z, w and v. Use the addition rule and the subtraction rule to simplify.

= 6 + 10i − 6 − 10i = 0 + 0i =0 c 4z − 3w + 2v = 4(3 + 5i) − 3(4 − 2i) + 2(6 + 10i) = 12 + 20i − 12 + 6i + 12 + 20i = 12 + 46i

Chapter 2 Number systems: complex numbers

83

Multiplication of two complex numbers So far we have shown that complex numbers can be plotted on an Argand diagram; adding and subtracting them is geometrically equivalent to adding and subtracting directed line segments and multiplication by a positive constant is equivalent to extending or shrinking the directed line segment without altering the direction. What geometrical interpretation, if any, can be given to multiplication of two (or more) complex numbers? The multiplication of two complex numbers also results in a complex number. If z = a + bi and w = c + di then z × w = (a + bi)(c + di) = ac + adi + bci + bdi 2 = (ac − bd) + (ad + bc)i (since i 2 = −1) If z = a + bi and w = c + di then z × w = (ac − bd) + (ad + bc)i.

WORKED Example 8

If z = 6 − 2i and w = 3 + 4i express zw in standard form. THINK 1 Expand the brackets. 2

Express in the form x + yi by substituting −1 for i 2 and simplifying the expression using the addition and subtraction rules.

WRITE zw = (6 − 2i)(3 + 4i) = 18 + 24i − 6i − 8i 2 = 18 + 24i − 6i + 8 = 26 + 18i

WORKED Example 9

Simplify (2 − 3i)(2 + 3i).

THINK 1 Expand the brackets. 2 2 Substitute −1 for i and simplify the expression.

WRITE (2 − 3i)(2 + 3i) = 4 + 6i − 6i − 9i 2 = 4 − 9 × −1 = 13

WORKED Example 10

Determine Re (z2w) + Im (zw2) for z = 4 + i and w = 3 − i. THINK 2 1 Express z w in the form x + yi.

2

The real part, Re (z2w) is 53.

WRITE z2w = (4 + i)2(3 − i) = (16 + 8i + i 2)(3 − i) = (16 + 8i − 1)(3 − i) = (15 + 8i)(3 − i) = 45 − 15i + 24i − 8i 2 = 53 + 9i Re (z2w) = 53 Continued over page

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THINK

WRITE

3

Express zw in the form x + yi.

zw2 = (4 + i)(3 − i)2 = (4 + i)(9 − 6i + i 2) = (4 + i)(8 − 6i) = 32 − 24i + 8i − 6i 2 = 38 − 16i

4

The imaginary part, Im (zw2), is −16.

Im (zw 2) = − 16

Calculate the value of Re (z2w) + Im (zw2).

Re (z2w) + Im (zw 2) = 53 − 16 = 37

5

2

Plotting complex numbers You will need 1 cm square grid paper, a ruler and a protractor. For z = −3 + 2i and w = 5 + i

1

1 Accurately plot z and w on an Argand diagram.

Im (z) zw = –17 + 7i

0

2 Find zw and plot this on the same diagram.

β

z = –3 + 2i

w=5+i

α

γ

Re (z)

2 zw = −17 + 7i (see figure at left) 3 Length of z = 13 , length of w = 26 , length of zw = 338 γ = 11.31°, β = 146.31°, α = 157.62° Length of zw = length of z × length of w; α=γ+β

3 Measure each length and angle from the positive end on the Real axis. Do you notice any pattern between the numbers you started with and your result? 4 Try this with some other complex numbers. Plot your original complex numbers accurately and plot the product. Test your original hypothesis.

Equality of two complex numbers If z = a + bi and w = c + di then z = w if and only if a = c and b = d. The condition ‘if and only if’ (sometimes written in short form as iff ) means that both of the following situations must apply. 1. If z = w then a = c and b = d. 2. If a = c and b = d then z = w.

WORKED Example 11

Find the values of x and y that satisfy (3 + 4i)(x + yi) = 29 + 22i. THINK 1 Write the left-hand side of the equation.

WRITE LHS = (3 + 4i)(x + yi)

2

Expand the left-hand side of the equation.

LHS = 3x + 3yi + 4xi + 4yi 2

3

Express the left-hand side in the form a + bi.

LHS = (3x − 4y) + (4x + 3y)i

85

Chapter 2 Number systems: complex numbers

THINK

WRITE

4

Equate the real parts and imaginary parts to create a pair of simultaneous equations.

3x − 4y = 29 4x + 3y = 22

[1] [2]

5

Simultaneously solve [1] and [2] for x and y.

9x − 12y = 87

[3]

Multiply equation [1] by 3 and equation [2] by 4 so that y can be eliminated.

16x + 12y = 88

[4]

6

Add the two new equations and solve for x.

Adding equations [3] and [4]: 25x = 175 x=7

7

Substitute x = 7 into equation [1] and solve for y.

Substituting x = 7 into equation [1]: 3(7) − 4y = 29 21 − 4y = 29 −4y = 8 y = −2

8

State the solution.

Therefore x = 7 and y = −2.

9

Check the solution by substituting these values into equation [2].

Check: 4 × 7 + 3 × −2 = 22.

Multiplication by i Let us examine the effect on z = x + yi after multiplication by i, i 2, i 3 and i 4. z = x + yi iz = i(x + yi) = −y + xi i 2z = −1z = −x − yi = −z i 3z = i(i 2z) = y − xi = −iz i 4z = i(i 3z) = x + yi = z The five points are shown on the complex plane at right. It is observed that multiplying z by i n, n ∈ N produces an anticlockwise rotation of 90n degrees.

remember remember

If z = a + bi and w = c + di for a, b, c, d∈R then: 1. z + w = (a + c) + (b + d)i 2. z − w = (a − c) + (b − d)i 3. kz = ka + kbi, for k∈R 4. z × w = (ac − bd) + (ad + bc)i 5. z = w if and only if a = c and b = d. (Note: ‘If and only if’ can be written as iff.)

Im (z) iz

x z or i 4z

–x

–y

–y

i 2z i 3z

x

Re (z)

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f

2B a

Basic operations using complex numbers

Im (z) –8 + i 3

2 1 0 Re (z)

–8

Im (z)

3+i

1 0

1

3 Re (z)

2

WORKED

Example xample

6 WORKED

Example xample

7

WORKED

Example xample

8 WORKED

Example xample

9

1 Represent each of the following complex numbers on an Argand diagram. a 3+i b 4 − 5i c −2 − 6i d 3i + 7 e f –8+i 3 5 – 2i 2 For z = 5 + 3i, w = −1 − 4i, u = 6 − 11i and v = 2i − 3 calculate: a z + w 4−i b u − z 1 − 14i c w + v −4 − 2i d u − v 9 − 13i e w − z − u −12 + 4i f v+w−z 3 If z = −3 + 2i, w = −4 + i and u = −8 − 5i, evaluate: a 3w −12 + 3i b 2u + z −19 − 8i c d 3z + u + 2w −25 + 3i e 2z − 7w + 9u −50 − 48i f

Example xample

10

WORKED

Example xample

11

Digital doc: SkillSHEET 2.1 Operations with complex numbers

4z − 3u 12 + 23i 3(z + 2u) − 4w −41 − 28i

4 Using z, w, u and v from question 2 express each of the following in the form x + yi. a zw 7 − 23i b uv 4 + 45i c wu −50 − 13i f u(wv) 176 − 61i d zu 63 − 37i e u 2 −85 − 132i 5 Simplify the following. 111 + 33i a (10 + 7i)(9 − 3i) b (3 − 4i)(5 + 4i) 31 − 8i c d (5 + 6i)(5 − 6i) 61 e (2i − 7)(2i + 7) −53 f 6 For z = −1 − 3i and w = 2 − 5i calculate z 2w.

WORKED

−9 − 5i

eBook plus

(8 − 2i)(4 − 5i) 22 − 48i (9 − 7i)2 32 − 126i

14 + 52i

7 Determine Re (z 2) − Im (zw) for z = 1 + i and w = 4 − i. −3 8 For z = 3 + 5i, w = 2 − 3i and u = 1 − 4i determine: a Im (u 2) −8 b Re (w 2) −5 c Re (uw) + Im (zw) −9 d Re (zu) − Im (w 2) 35 f Re (u2w) + Im (zw 2) −115 e Re (z 2) − Re (zw) − Im (uz) −30 9 Find the values of x and y that satisfy each of the following. 21 , - y = – 16 -----a (2 + 3i)(x + yi) = 16 + 11i x = 5, y = −2 b (5 − 4i)(x + yi) = 1 − 4i x = ----41 41 c (3i − 8)(x + yi) = −23 − 37i x = 1, y = 5 d (7 + 6i)(x + yi) = 4 − 33i x = −2, y = −3

10 multiple choice

a

Im (z) 4

0

e

3 + 4i

3

Re (z)

C 5

D 11

E 52

C 105

D 56

E −32

C 18 − 29i

D 24 − 13i

E 18

11 If z = 2 + i and w = 4 − 3i then represent each of the following on an Argand diagram. b zw c z+w d w−z a z2 2 e 3z + w f 2w − 4z h (w − z)3 Im (z) g (z + w)

Im (z)

0

If z = 8 − 7i and w = 3 + 4i, then: a Re (zw) is equal to: A −4 B 4 2 2 b Im (w ) + Re (z ) is equal to: A 76 B 39 c 3z − 2w is equal to: A 30 − 13i B 30 − 29i

10 Re (z)

12 If z = 3 + 2i, represent each of the following on the same Argand diagram. z, iz, i 2z, i 3z, i 4z, i 5z, −iz, −i 2z

iz, i5z

3 2 1

z, i4z, –i2z

–3 –2 –1–10 1 2 3 Re (z) i2z

–2 –3

i3z, –iz

Chapter 2 Number systems: complex numbers

87

Conjugates and division of complex numbers The conjugate of a complex number The conjugate of a complex number is obtained by changing the sign of the imaginary component. If z = x + yi, the conjugate z of z is defined as z = x − yi. Conjugates are useful because the multiplication (or addition) of a complex number and its conjugate results in a real number. zz– = (x + yi)(x − yi) = x2 + y2 where x, y ∈ R, and x − yi and x + yi are conjugates. You will use this result when dividing complex numbers. Multiplication:

Note: Compare this expression with the formula for the difference of two squares (a − b)(a + b) = a 2 − b 2 Addition:

z + z– = x + yi + x − yi = 2x

Graphing complex conjugates As seen earlier, z and its conjugate can be written as z = x + yi and z– = x − yi Im (z) The geometrical representation of z and z– is shown at right. Notice that the conjugate z– appears as a reflection of z in the Real axis. x Other properties of conjugates include: – 1. z = z 2. z1 ± z2 = z–1 ± z–2 3. z1z2 = z–1z–2 z1  z1 - where z2 ≠ 0. 4.  ------= ------ z2  z2

WORKED Example 12 Write the conjugate of each of the following complex numbers. a 8 + 5i

b −2 − 3i

c 4+i 5

THINK

WRITE

a Change the sign of the imaginary component.

a 8 − 5i

b Change the sign of the imaginary component.

b −2 + 3i

c Change the sign of the imaginary component.

c 4–i 5

z = x + yi y y z = x – yi

Re (z)

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WORKED Example 13

If z = 5 − 2i and w = 7 − i, show that z + w = z + w . THINK

WRITE

1

Add the conjugates z– and w .

z– + w = ( 5 + 2i ) + ( 7 + i ) = 12 + 3i

2

Add z to w.

z + w = ( 5 – 2i ) + ( 7 – i ) = 12 – 3i

3

Write down the conjugate of z + w.

4

–

The conjugate of z + w equals zz + w .

z + w = 12 + 3i z + w = z– + w

Division of complex numbers The application of conjugates to division of complex numbers will now be investigated. z Consider the complex numbers z = a + bi and w = c + di. To find ---- in the form x + yi w we must multiply both the numerator and denominator by the conjugate of w to make the denominator a real number only. (You might need to review rationalisation of the denominator which was discussed in Chapter 1.) a + bi z- --------------= w c + di a + bi c – di = -------------- × ------------c + di c – di ( ac + bd ) + ( bc – ad )i = -----------------------------------------------------c2 + d 2 ( bc – ad )iac + bd -----------------------= -----------------+ 2 2 2 c + d2 c +d

Multiply by the conjugate of c + di. Simplify the expressions in the numerator and in the denominator. Express in the form x + yi.

Thus we can state: z If z and w are complex numbers in the form x + yi, then ---- can also be expressed w in the form x + yi by simplifying: z- w --¥ ---w w

WORKED Example 14

2+i Express ----------- in standard form. 2–i THINK 1

Multiply both the numerator and denominator by the conjugate of 2 − i to make the denominator real.

WRITE 2---------+ i (2 + i) (2 + i) = --------------- × --------------2–i (2 – i) (2 + i)

Chapter 2 Number systems: complex numbers

THINK 2

3

89

WRITE

Expand the expressions obtained in the numerator and denominator.

4 + 4i + i 2 = ----------------------4 – i2 4 + 4i – 1 = ----------------------4+1 3 + 4i = -------------5 3 4i = --- + ----5 5

Substitute −1 for i 2 and simplify the expression.

Multiplicative inverse of a complex number Given a non-zero complex number z, there exists a complex number w such that zw = 1, 1 with w being the multiplicative inverse of z denoted by w = z –1 = --- . z

WORKED Example 15

If z = 3 + 4i, determine z−1. THINK

1 = --z

1

Write z–1 as a rational expression: z –1

2

Multiply both the numerator and denominator by the conjugate of 3 + 4i.

3

Write the expression in the form x + yi.

WRITE 1 1 z−1 = --- = -------------z 3 + 4i 1 (3 – 4i) = ------------------- × -----------------(3 + 4i) (3 – 4i) 3 – 4i = -------------25 3 4i = ------ – -----25 25

a – bi -. This example shows that if z = a + bi then z –1 = ---------------a2 + b2 Complex numbers can be used to generate fractal patterns such as the ‘Julia Set’ shown.

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WORKED Example 16 2 If z = 3 + i and w = ---------- , determine Im (4z − w). 4–i THINK 1 2

Substitute for z and w in 4z − w. Express 4z − w with a common denominator.

WRITE 2 4z − w = 4 ( 3 + i ) – ---------4–i 4(3 + i)(4 – i) – 2 = -------------------------------------------4–i 4 ( 13 + i ) – 2 = ------------------------------4–i 50 + 4i = ----------------4–i

3

(50 + 4i) (4 + i) = ---------------------- × --------------(4 – i) (4 + i)

Remove i from the denominator by multiplying the numerator and denominator by the conjugate of 4 − i.

4

Simplify the expression so that it is in the form x + yi.

5

State the imaginary component of 4z − w.

196 + 66i = ----------------------17 196 66i = --------- + -------17 17 66 Im (4z − w) = -----17

WORKED Example 17

Prove that z1z2 = z–1z–2. THINK 1

2

3

WRITE z1 = a + bi z–1 = a − bi z2 = c + di and z–2 = c − di

When asked to ‘Prove’ you should not use actual values for the pronumerals. State the general values of z1, z2, z–1 and z–2.

Let

Generally, in a proof do not work both sides of the equation at once. Calculate the LHS first.

LHS = (a + bi) × (c + di)

Calculate the RHS and show that it equals the LHS.

Let

LHS = ac + adi + bci + bdi2 LHS = (ac − bd) + (ad + bc)i LHS = (ac − bd) − (ad + bc)i RHS = (a − bi)(c − di) RHS = ac − adi − bci + bdi2 RHS = (ac − bd) − (ad + bc)i RHS = LHS Hence z1z2 = z–1z–2

91

Chapter 2 Number systems: complex numbers

remember remember

If z = a + bi and w = c + di, for a, b, c, d ∈ R, then: 1. The conjugate of z is z– = a − bi. bc – ad )i z ac + bd (------------------------. + 2 2. ---- = -----------------w c + d2 c2 + d 2 b 1 a -i . 3. The multiplicative inverse of z is z–1 = --- = ----------------- – ---------------2 2 2 2 z a +b a +b

2 a

Im (z) z=3+i Re (z) z= 3 – i

b

2C WORKED

Example

12

z = –1 + 3i

Im (z)

Conjugates and division of complex numbers

Re (z)

z = –1 – 3i

1 Write down the conjugate of each of the following complex numbers. a 7 + 10i 7 − 10i b 5 − 9i 5 + 9i c 3 + 12i 3 − 12i 7 – 3i 7 + 3i d e 2i + 5 5 − 2i f −6 − i 11 – 6 + 11 i

2 Graph the following complex numbers and their conjugates on an Argand diagram. z = – 4 + 5i a 3−i b −1 + 3i c −4 − 5i WORKED

Example

3 If z = 6 + 3i and w = 3 − 4i, show that z – w = z– – w .

Example

14

2+i 4 Express ---------- in the form x + yi. 1--2- + 1--2- i 3–i 5 Express each of the following in the form x + yi. 1–i 3 – 2i a ---------- 0 − i b -------------- 0 − i 1+i 2 + 3i 4 – 3i d -------------5 + 2i

WORKED

Example

15

14 -----29

−

23 -----29

i

e

7 If 676z = 10 − 24i, express z Example

16

4------------– 5i2 – 7i

6 Determine z−1 if z is equal to: a 2−i b 3+i 5 – 4i d 5 + 4i ------------e 2i − 3 41 −1

WORKED

Re (z)

Check with your teacher.

13 WORKED

Im (z)

43 -----53

+

18 -----53

z = – 4 –5i

i

– 3 – 2i -----------------13

c

2------------+ 5i- − -----7- + 25 4 – 3i

f

2 + i 3--------------------5–i 2

c f

4 − 3i 3–i 2

in the form x + yi. 10 + 24i

17 -----2

+ 9--2- i

( 2 + 5i ) 2 ( 5i – 2 ) 10 Simplify --------------------------------------------------- . –29 3 ( 4 + 7i ) – 2 ( 5 + 8i ) 11 Determine the conjugate of (5 − 6i)(3 − 8i). −33 + 58i

i

2 + 15 2 5 – 6 2--------------------------i ----------------------- + 7 7 2+i 6 a ----------5 4 + 3i c -------------25

3 + 2i ---------------------5

1 8 If z = 2 − i and w = ---------- determine each of the following: 3+i 9-----a Re (z + w) 23 b Im (w − z) ----c Re (z−1 + w−1) 10 10 16 14 d Im (3z + 2w) − -----5e Re (4w − 2z) − -----52 + i 9 – 2i 7 + i 9 Write ---------- + -------------- + ---------- in the form x + yi. 1+i 2–i 1–i

26 -----25

17 -----5

3–i b ---------10

92

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

12 multiple choice If z = 5 − 12i, w = −9 − i and u = 15 − 6i, then: a Re (z−1) is: 12 C --------A 5 B 12 169 b Im ( zw ) is equal to: A −33 B 103 C 113 c

The expression 2z – w + 3u is equal to: A 26 − 7i B 64 + 41i C 46 − 29i

5 D --------169

12i E --------169

D 70

E 0

D 34 − 41i

E −64 − 19i

13 If z = 6 + 8i and w = 10 − 3i: Check with your teacher. a show that zw = z × w b generalise the result by letting z = a + bi and w = c + di. eBook plus

14 Use the result zw = z × w to prove that z n = ( z )n. Check with your teacher. 15 If z = 4 + i and w = 1 + 3i

Digital doc: SkillSHEET 2.2 Complex numbers and their conjugates

Check with your teacher.

z z a show that  ---- = --- w w b generalise the result by letting z = a + bi and w = c + di. 16 If z = −5 − 4i and w = 2i, calculate Re (z w + z w). −16 17 If z1 = 2 + 3i, z2 = −4 − i and z3 = 5 − i find: a 2z1 − z2 − 4z3 −12 + 11i b z1 z 2 + z2 z 3 −30 − 19i

c

z1 z2 z3 − z1 z2 z3 0

18 If z1 = a + bi and z2 = c + di show that (z1z2)−1 = z1−1z2−1. Check with your teacher. 19 a If z = 1 + i, find z4, z8 and z12. –4, 16, –64 b Deduce from your results in a that z4n = (2i)2n, n ∈ N. z–1 20 If z = x + yi, find the values of x and y such that ----------- = z + 2 . x = −1, y = ± 2 z+1 z+i 21 Find values for a and b so that z = a + bi satisfies ----------- = i . a = − 1--2- , b = z+2 22 If z = x + yi, determine the values of x and y such that z =

1 --2

3 + 4i . x = 2, y = 1; x = −2, y = −1

23 If z = 2 − 3i and w = 1 − 2i a find i zz– 13 ii ww– 5 17 b show that iii z + w = z + w

WORKED

Example xample

eBook plus Digital doc: WorkSHEET 2.1

c

find

iii

zw = z– × w

iii

z- --z  --=  w w

1 i  ---  z 2----13

1 ii  ----  w −

3----13

i

d z2 + w2 −8 + 16i e z + zw −2 + 10i f --15

− --25- i

– 4 + 7i z–1w–1 -----------------65

Chapter 2 Number systems: complex numbers

93

algebra of Graphics Calculator tip! Simple complex numbers Operations with complex numbers, finding the real and imaginary parts of a complex number and finding the complex conjugate can be achieved with a graphics calculator. You may not need to use a graphics calculator with simple complex numbers but it can be useful in more complicated questions.

For the Casio fx-9860G AU Operations with complex numbers 1. Press MENU to display the MAIN MENU. Use the cursor keys to highlight RUN-MAT. Select it by pressing EXE . To perform simple algebra on complex numbers, use the standard keys to enter the expression. To enter i, press SHIFT [i ]. (The i is located above the zero key.) Press EXE to obtain the answer. For example: (a) Input (2 − 2i)(1 + 3i) and then press EXE . (b) Input (2 − 2i) ÷ (1 + 3i) and then press EXE . (c) Input (2 − 2i)^3 and then press EXE . Notice that we include a multiplication symbol when entering 2i and 3i in these examples. The calculator will generally assume that 2i means 2 × i. However, with some terms, the calculator may not read it as you intend so it is good practice to press the multiplication key each time the multiplication operation is needed. Try 2 calculating ( 1 – 2i ) with and without including a multiplication sign for 2i . 2. A complex number can be stored and then retrieved if a number of operations need to be performed on it. Input 2 − 2i and then press Æ followed by ALPHA [Z] to store this expression as the variable z. Press

EXE .

Input 1 + 3i and then press Æ followed by ALPHA [W] to store this expression as the variable w. Press

EXE .

3. We can now calculate expressions involving z and/or w; for example, z × w, z ÷ w and z^3. Press EXE to obtain each answer. Features of a complex number

s

1. To find the complex conjugate, the real part or the imaginary part of a complex number or expression, first press OPTN then F3 (CPLX). You will see a row of complex number options. There is also a second row of options available. Press F6 ( ) to move between the two rows.

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

s

2. You may then select F4 (Conj) for the complex conjugate or F6 ( ) followed by F1 (ReP) for the real part or F6 ( ) followed by F2 (ImP) for the imaginary part as required. Type the complex number within brackets and press EXE . s

3. Alternatively, you can enter an expression, for example (2 − 2i)(1 + 3i), and then press Æ followed by ALPHA [A] to store the output as the variable A. Press EXE . 4. To access the complex number options, first press OPTN then F3 (CPLX). Remember that you can move between the two rows of complex number options by pressing F6 ( ). (a) To find ( 2 – 2i ) ( 1 + 3i ) , press F4 (Conj). Key in the variable assigned to the stored data (A in this case) and press EXE . (b) To find Re [ ( 2 – 2i ) ( 1 + 3i ) ], press F6 ( ) to access the second row of options and then press F1 (ReP). Key in the variable assigned to the stored data and press EXE . (c) To find Im [ ( 2 – 2i ) ( 1 + 3i ) ], press F6 ( ) to access the second row of options and then press F2 (ImP). Key in the variable assigned to the stored data and press EXE . s

s

s

94

For the TI-Nspire CAS Operations with complex numbers 1. Open a new Calculator document. To perform simple algebra on complex numbers, use the standard keys to enter the expression. Use the j button located on the left of the calculator. Press · to obtain the answer. For example: (a) input ( 2 – 2i ) ( 1 + 3i ) and then press ·. (b) input ( 2 – 2i ) ÷ ( 1 + 3i ) and then press ·. (c) input (2 − 2i)^3 and then press ·. Notice that a multiplication symbol (a dot) appears on the screen for any multiplication operations in these examples. The calculator has assumed that 2i means 2 × i. However, if there were two or more variables in a term, the calculator may not read it as you intend so it is good practice to press the multiplication key each time the multiplication operation is needed. 2. A complex number can be stored and then retrieved if a number of operations need to be performed on it. Input 2 − 2i and then press /h and Z to store this expression as the variable z. Press ·. Input 1 + 3i and then press /h and W to store this expression as the variable w. Press ·. We can now calculate expressions involving z and/or w; for example, z × w, z ÷ w and z^3. Press · to obtain each answer.

Chapter 2 Number systems: complex numbers

95

Features of a complex number 1. To find the complex conjugate, the real part or the imaginary part of a complex number or expression, press b and select 2: Number, then 8: Complex Number Tools. A list of complex number options will be displayed. 2. You may then select 1: Complex Conjugate or 2: Real Part or 3: Imaginary Part as required. Type in the complex number, close the brackets and press ·.

3. Alternatively, you can enter an expression, for example, (2 − 2i)(1 + 3i), and then press /h and A to store the output as the variable a. Press ·. (a) To find ( 2 – 2i ) ( 1 + 3i ) , press b and select 2: Number, then 8: Complex Number Tools and 1: Complex Conjugate. Key in the variable assigned to the stored data (A in this case) and close the brackets. Press ·. (b) To find Re [ ( 2 – 2i ) ( 1 + 3i ) ], press b and select 2: Number, then 8: Complex Number Tools and 2: Real Part. Key in the variable assigned to the stored data and close the brackets. Press ·. (c) To find Im [ ( 2 – 2i ) ( 1 + 3i ) ], press b and select 2: Number, then 8: Complex Number Tools and 3: Imaginary Part. Key in the variable assigned to the stored data and close the brackets. Press ·.

Radians and coterminal angles When a complex number is expressed in a geometrical representation, we use a directed line segment which has length (modulus) and which lies in a certain direction with respect to the positive Real axis. This angle formed with the positive Real axis is called the argument. The argument of a complex number z is written as arg (z) and arg (z) = θ. Before we look at complex numbers in polar form (in the next section), a new unit of measuring angles is needed, the radian.

Radian measure A radian is the angle subtended by an arc the length of the radius of a circle, as shown in the diagram on the right. Because the circumference of a circle is given by c = 2π r, there are 2π radians in one complete circle. Taken in an anticlockwise rotation from the positive end of the x-axis as shown, the common angles have radian equivalents.

r

θ 1 radian r

r

96

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

360° Therefore, if 2π radians = 360°, 1 radian = ----------- g 57.3°. 2π So an arc of 2π r subtends an angle of 2π radians.

π – 2

π – 4

π

Coterminal angles Consider the angle 420°. This angle is made up of a full revolution, 360° plus 60°. When using degrees as our unit of angle measure, 420° and 60° are said to be coterminal angles; that is, angles which differ by a multiple of 360°.

2D Im (z) 3— π 4

π – 4

π – 2

π – 6

5— π π 6 7— π 6 5— π 4

3— π 2

0 2π

7— π 4

Radians and coterminal angles

3

(b) (a) (d)

Re (z) (e)

3— π 2

1 Draw a circle with a set of axes through its centre. Mark the following on the circumference of the circle. π π 3π 5π 7π π 5π 7π 3π b --- , ------ , ------ , -----c --- , ------ , -----a 0, --- , π, ------ , 2π 2 2 4 4 4 4 6 6 6 Re (z) 2 Convert the following common angle measures to radians. π 3 ππ 5π 3π a 45° --4b 60° --c 135° ----d 270° ----e 150° -----2

Im (z) (c)

0 2π

6

4

3 Convert the following radian measures to degrees. 5π 4π 7π a ------ 210° b ------ 225° c ------ 240° 6 4 3 4 Draw the following sets of coterminal angles: a 30°, 390° b 60°, 420° π 13 π 11 π 23 π d --- , --------e --------- , --------6 6 6 6

5π d ------ 300° 3 c f

135°, 495° π 5 π- 13 ----, --------4 4

(f)

Complex numbers in polar form The modulus of z The magnitude (or modulus or absolute value) of the complex number z = x + yi is the length of the line segment joining the origin to the point z. It is denoted by z, x + yi or mod z. The modulus of z is calculated using Pythagoras’ theorem. z = x 2 + y 2 so that we have zz = z 2 .

WORKED Example 18

Find the modulus of the complex number z = 8 − 6i. THINK WRITE Calculate the modulus by rule. Because it forms the hypotenuse of a rightangled triangle, the modulus is always greater than or equal to Re (z) or Im (z).

z =

8 2 + ( –6 )2

= 100 = 10

Im (z) y z =

2

x +y

P(x, y) z = x + yi

2

y

0

θ x

x

Re (z)

Chapter 2 Number systems: complex numbers

97

WORKED Example 19

If z = 4 + 2i and w = 7 + 6i, represent the position of w − z on an Argand diagram and calculate w − z. THINK 1 Calculate w − z. 2

Represent it on an Argand diagram as a directed line segment OP.

3

Use Pythagoras’ theorem to determine the length of OP.

WRITE/DRAW w − z = 7 + 6i − (4 + 2i) = 3 + 4i

Im (z) w 6 5 w–z P 4 3 2 z 1 O Re (z) 0 1234567

OP 2 = 3 2 + 4 2 = 25 OP = 5 So w – z = 5

WORKED Example 20

Represent z1 = 2 + 3i, z2 = 5 − 2i and z3 = −4 − 2i on the complex number plane and calculate the area of the shape formed when the three points are connected by straight line segments. THINK 1 Show the connected points on the complex number plane.

WRITE/DRAW Im (z) 4 3 2 1

z1 Re (z)

–4–3–2 –1 1 2 3 4 5 z2 –2 –3

z3 2

Calculate the area of the triangle obtained. The length of the base and height can be found by inspection (base = 9, height = 5).

Area of triangle = 1--- × 9 × 5 2 Area of triangle = 22.5 square units.

The argument of z The argument of z, arg (z), is the angle measurement anticlockwise of the positive Real axis. In the figure at right, arg (z) = θ, where y y x sin θ = -------- and cos θ = -------- or tan θ = -z z x As seen in the previous section, for non-zero z an infinite number of arguments of z exist since, for a given z {θ :θ = ±2nπ, n ∈ N} also represents the position of point P in the figure at right because a clockwise or anticlockwise rotation consisting of multiples of 2π radians (or 360°) merely moves P to its original position.

Im (z) P(x, y) z = x + yi

y

z 

y

θ 0

x

x

Re (z)

98

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

To ensure that there is only one value of θ corresponding to z we refer to the principal value of θ and denote it by Arg (z). Note the capital A. Arg (z) is the angle θ in the range −π < θ ≤ π or

(−π, π].

π –π

Exact values π 1 cos --- = ------- , 4 2

π 1 sin --- = ------- , 4 2

π tan --- = 1 4

π 1 cos --- = --- , 3 2

π 3 sin --- = ------- , 3 2

π tan --- = 3

π 3 cos --- = ------- , 6 2

π 1 sin --- = --- , 6 2

π 1 tan --- = ------6 3

3

Some useful triangles It will be easier if you remember these 2 triangles only — not the ratios shown above. Draw a quick sketch and work out each trigonometric ratio when you need to. — 6 — 4

√2

— 4

1

√3

2

— 3

1

1

WORKED Example 21

Find the Argument of z for each of the following in the interval (−π, π ]. a z = 4 + 4i

b z=1−

3i

THINK a 1 Plot z. 2

WRITE/DRAW a Im (z)

Sketch the triangle that has sides in this 1:1 ratio. π – 4

√2

θ

Re (z) 4

1

π – 4

1

3

4

This result can be verified using an y inverse trigonometric ratio, θ = tan–1 -- . x

From the diagram π θ = --4 π ∴ Arg (z) = --4 Check: 4 θ = tan–1 --4 π θ = --4

Chapter 2 Number systems: complex numbers

THINK b 1 Plot z. 2

WRITE/DRAW b Im (z)

Sketch the triangle that has sides in this ratio. θ

π – 6

2

√3 √3

π – 3

1

3

This result can be verified using an y inverse trigonometric ratio, θ = tan–1 -- . x

WORKED Example 22 Convert each of the following into Arguments. THINK a 1 Sketch the angle.

From the diagram π θ = --3 π ∴ Arg (z) = − --3 Check: – 3 θ = tan–1 ---------1 π θ = − --3

7π a -----4 WRITE/DRAW a Im (z) 7— π 4

b

Re (z)

1

2

Since the given angle is positive, subtract multiples of 2π until it lies in the range (−π, π ].

1

Sketch the angle.

5π b – -----2

Re (z) – –4π

7π Arg (z) = ------ – 2 π 4 π = – --4 b Im (z) Re (z) – –2π – 5—2π

2

Since the given angle is negative, add multiples of 2π until it lies in the range (−π, π ].

5π Arg (z) = – ------ + 2 π 2 π = – --2

99

100

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

WORKED Example 23 Find the modulus and Argument for each of the following complex numbers. 3+i

a

b – 1 – 2i

THINK a 1 Plot z. 2

WRITE/DRAW a Im (z)

This triangle has sides in the same ratio as

π – 6

2

θ √3

1

b

These results can be verified by calculating the modulus and Argument by rule.

1

Plot z.

Re (z)

√3

z = 2 π Arg (z) = --6 Check: 2 2 z = ( 3 ) + 1 = 4 =2 1 π θ = tan–1 ------- = --3 6

π – 3

3

1

b

Im (z)

1

Re (z)

α θ

√2

2

Find the modulus.

3

The triangle in the third quadrant will be used to find α but the answer will be finally expressed as θ and Arg (z).

4

Remember Arg (z) can be thought of as the quickest way to get to z.

2

z =

( –1 ) + ( – 2 )

z =

1+2

z =

3

y α = tan–1 -x – 2 = tan–1 ---------–1 = 0.955 θ = −π + 0.955 Arg (z) = −2.187

Arg (z)

z

2

Chapter 2 Number systems: complex numbers

101

and Graphics Calculator tip! Modulus Argument Your graphics calculator is also able to calculate the modulus and Argument of a complex number. Consider 3 + i and – 1 – 2i from Worked example 23.

For the Casio fx-9860G AU Modulus (or absolute value) of a complex number 1. Press MENU to display the MAIN MENU and select RUN-MAT. Press OPTN then F3 (CPLX) to display the complex number options. To select the modulus or absolute value option, press F2 (Abs). 2. (a) Enter ( 3 + i ) and press EXE . (Press SHIFT [ ] and type in 3 for 3 .) (b) The modulus of –1 – 2i is also shown in the screen at right. (c) The modulus of a stored complex number can also be calculated. For example, using the expression from the previous graphics calculator tip where we assigned the variable A to ( 2 – 2i ) ( 1 + 3i ) , we can obtain the value as shown in the screen at right. Argument of a complex number 1. Decide whether you want the angle to be shown in radians or degrees. See the instructions below for changing the system settings for Angle. 2. As before, press OPTN then F3 (CPLX) to display the complex number options. To select the Argument option, press F3 (Arg). 3. Enter the required complex number 3 + i within brackets and press EXE to obtain the answer. If the calculator is set to radians, the answer will be shown π as the decimal equivalent to --- . If the calculator is set 6 to degrees, the answer will be shown as 30 for 30°. 4. (a) The Argument of –1 – 2i is shown in degrees as a decimal in the screen at right.

s

(b) To convert this angle to degrees, minutes and seconds, we can first press SHIFT [Ans] and EXE to display the decimal answer again. (This step is optional and used to show both answers on the screen.) Press OPTN and then F6 ( ) for more options. Press F5 (ANGL) followed by F5 ( ∞¢¢¢ ) to obtain the answer in degrees, minutes and seconds. (c) The Argument of a stored complex number can also be calculated. See the screen above for the Argument of the variable A where A is assigned to ( 2 – 2i ) ( 1 + 3i ) . The calculator is set to degrees for this example.

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Changing the system settings for Angle To change the settings for Angle, press SHIFT [SET UP] and use the down arrow to highlight Angle. Press F1 (Deg) to set degrees or press F2 (Rad) to set radians.

For the TI-Nspire CAS Modulus (or magnitude) of a complex number 1. Press b and select 2: Number, then 8: Complex Number Tools and 5: Magnitude. A pair of modulus signs appears with a box to enter the required complex number. 2. (a) To enter 3 + i , press / then q and type in 3 for 3 . Use the right arrow of the NavPad to move the cursor out from under the square root sign. Press + and j to complete the expression. Again use the right arrow of the NavPad to move the cursor to the right of the second modulus sign. Press · to obtain the answer. (b) The modulus of –1 – 2i is also shown in the screen at right. (c) The modulus of a stored complex number can also be calculated. For example, using the expression from the previous graphics calculator tip where we assigned the variable a to ( 2 – 2i ) ( 1 + 3i ) , we can obtain the magnitude as shown in the screen above. Argument of a complex number 1. Decide whether you want the angle to be shown in radians or degrees. See the instructions below for changing the system settings for Angle. 2. Press b and select 2: Number, then 8: Complex Number Tools and 4: Polar Angle. 3. Enter the required complex number 3 + i , close the brackets and press · to obtain the answer. If the calculator is set to radians, the answer will be shown π as --- . If the calculator is set to degrees, the answer 6 will be shown as 30 for 30°. 4. (a) The Argument of –1 – 2i is shown as an exact answer in radians in the screen at right. (b) For an approximate answer, press / then ·. The answer is now shown as a decimal in radians. (c) To convert this angle in radians to degrees, minutes and seconds, first press / then v and select DMS in the catalog (press k, select 1, scroll down to DMS and press ·). ▼

▼

Now that you are back in the calculator screen, press · to obtain the answer. Note: You can convert degrees to radians by selecting

Rad in the catalog.

▼

102

103

Chapter 2 Number systems: complex numbers

5. (a) The Argument of a stored complex number can also be calculated. See the screen at right for the Argument of the variable a where a is assigned to ( 2 – 2i ) ( 1 + 3i ) . The exact answer is shown. (b) For the approximate answer, press / then ·. The calculator is set to degrees for this example. Changing the system settings for Angle 1. To change the settings for Angle, press the home key (c) and select 8: System information and then 2: System Settings. Use the tab key (e) to move down to Angle and then use the down arrow to select Degree or Radian as required.

2. Press · to accept your selection. Tab to OK and press ·. A message will appear asking whether to apply these system settings to current document settings. Select Yes and press ·. Note that you can also change the settings for Angle by accessing the setup menu (press / [#]) and selecting 1: File and then 6: Document Settings.

Expressing complex numbers in polar form Suppose z = x + yi is represented by the point P(x, y) on the complex plane using Cartesian coordinates. Using the trigonometric properties of a rightangled triangle, z can also be expressed in polar coordinates as follows. We have: x cos θ = -- or x = r cos θ r y sin θ = -- or y = r sin θ r

Im (z)

P(x, y)

y r = z 

0

y Re (z)

θ x

where z = r = x 2 + y 2 and θ = Arg (z).

The point P(x, y) in polar form is shown at right. Now z = x + yi in Cartesian form becomes z = r cos θ + r sin θ i (after substitution of x = r cos θ, y = r sin θ ) = r (cos θ + i sin θ) = r cis θ, where cis θ is the abbreviated form of cos θ + i sin θ. (Note: The acronym ‘cis’ is pronounced ‘sis’.)

Im (z) y

P(rcosθ , r sinθ )

r O θ 0 r cosθ

r sin θ Re (z)

104

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

WORKED Example 24

Express each of the following in polar form, r cis q, where q = Arg (z). a z=1+i

b z=1−

3i

THINK a 1 Plot z.

WRITE/DRAW a Im (z) 1 Re (z) 1

2

The ratio of the sides of this triangle matches the following special triangle:

From the diagram z = 2 π θ = --4

π – 4

√2

1

π – 4

1 3

These results can be verified by rule:

Check: r = 12 + 12

(a) Find the value of r using r = z = x2 + y2 .

=

y (b) Determine tan θ from tan θ = -- and x hence find θ. 4

5

b

1

1 tan θ = --1 =1

π θ = --4

The angle θ is in the range (−π, π ], which is required. Substitute the values of r and θ in z = r cos θ + r sin θ i = r cis θ.

z= z= b

Sketch z.

2

π π 2 cos --- + 2 sin --- i 4 4 π 2 cis --4 Im (z)

θ

Re (z)

1 √3

2

The ratio of the sides of this triangle is the same as that in the following special triangle:

π – 6

2 π – 3

1

√3

From the diagram r=2 π θ = --3 π Arg (z) = − --3

Chapter 2 Number systems: complex numbers

THINK 3

105

WRITE/DRAW

These results can be verified by rule:

Check:

(a) Calculate the value of r.

r=

1 + ( 3)2

r=2 (b) Determine the appropriate value of θ.

3 tan θ = – ------1 =– 3

θ = – --π3 Substitute for r and θ in z = r cos θ + r sin θ i and write in the form r cis θ.

π π z = 2 cos  – --- + 2 sin  – --- i  3  3 π = 2 cis  – ---  3

Graphics Calculator tip! Expressing complex numbers in polar form Complex numbers in Cartesian form (also known as rectangular form) can be written in polar form if we know the modulus and the Argument. Consider 1 + i and 1 – 3i from Worked example 24.

For the Casio fx-9860G AU

s

1. Press MENU to display the MAIN MENU and select RUN-MAT. Press OPTN then F3 (CPLX) to display the complex number options. To access the second row of complex number options, press F6 ( ). 2. Enter the complex number in Cartesian form. To convert the number to polar form, press F3 ( r– q ). Press EXE to show the modulus and Argument. ▼

4

For 1 + i, we can see that the modulus is 1.41421… (which is the decimal equivalent to 2 ) and the Argument is 45° (as the calculator is set to degrees). For 1 – 3i , the modulus is 2 and the Argument is −60°. 3. If the calculator is set to radians, the Argument will be displayed in radians as a decimal.

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For the TI-Nspire CAS 1. From the home screen, select 1: Calculator. Enter the complex number in Cartesian form. Press b and select 2: Number, then 8: Complex Number Tools followed by 6: Convert to Polar. Press · to show the modulus and Argument. For 1 + i, we can see that the modulus is 2 and the Argument is 45° (as the calculator is set to degrees). For 1 – 3i , the modulus is 2 and the Argument is −60°. 2. If the calculator is set to radians, the answer is shown in a different format but we can still see the modulus and Argument. Another way of π expressing 2 cis --- (the polar form of 1 + i) is 4 (This is beyond the scope of this course.)

iπ -----

2e 4 .

Expressing complex numbers in Cartesian form

WORKED Example 25

π Express 3 cis --- in Cartesian (or standard a + ib) form. 4 THINK WRITE Im (z) 1 Sketch z. 3 π– 4

2 3

π Express 3 cis --- in Cartesian form. 4 Simplify using exact values from the following triangle: π – 4

√2

1

π – 4

1

Re (z)

π π π 3 cis --- = 3 cos --- + 3 sin --- i 4 4 4 1 1 = 3 × ------- + 3 × ------- i 2 2 3 3 = ------- +  ------- i  2 2 3 = ------- ( 1 + i ) 2

Graphics Calculator tip! Expressing complex numbers in Cartesian form Complex numbers in polar form can be written in Cartesian form by entering both the π modulus and the Argument into the calculator. Consider 3 cis --- from Worked 4 example 25. π The modulus is 3 and the Argument is --- or 45°. 4

Chapter 2 Number systems: complex numbers

For the Casio fx-9860G AU

s

1. Press MENU to display the MAIN MENU and select RUN-MAT. Press OPTN then F3 (CPLX) to display the complex number options. Access the second row of complex number options by pressing F6 ( ).

▼

2. Enter the modulus (3 in this case) and then press π SHIFT [–] and enter the Argument in radians ( --- in this case) or degrees (45°) as appropriate to the 4 calculator setting. The screen at right shows the angle in radians. To convert to Cartesian form (a + bi form), press F4 ( a + bi). Press EXE to show the complex number. 3. If the calculator is set to degrees with the Argument entered as 45, the same result is obtained.

For the TI-Nspire CAS 1. Open a new Calculator document. Enter the modulus (3 in this case) then access the symbol palette (press / then k) and highlight the angle symbol ∠.

2. Press ·. Complete the entry line by entering the π angle in radians ( --- in this case) or degrees (45°) as 4 appropriate to the calculator setting. The screen at right shows the angle in radians. Press ) to close the set of brackets.

3. Press b and select 2: Number, then 8: Complex Number Tools followed by 7: Convert to Rectangular. Press ·. 3 2 3 The fraction ------- is equivalent to ---------2 2 (shown with a rational denominator).

4. If the calculator is set to degrees with the Argument entered as 45, the same result is obtained.

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History of mathematics ABRAHAM DE MOIVRE (26 May 1667 – 27 November 1754) During his lifetime. . . Christopher Wren finishes St Paul’s Cathedral. Blackbeard the pirate is killed. The first successful appendectomy is performed. People are put to death, as witches, in Salem. Abraham De Moivre was born in the French town of Vitry but from the age of eighteen he lived in England. The son of a doctor, he was educated at the Protestant Academy at Sedan and then attended college in Paris. In 1685 his family emigrated to England to escape the growing tensions between Catholics and Protestants in France. De Moivre contributed to the development of analytic geometry and to the theory of probability. One of his most famous books, The Doctrine of Chances, was published in 1718 and contained major advances in probability theory. In 1725, after investigating mortality statistics, he published Annuities on Lives. Insurance companies of the day used his work to calculate the probabilities of various events. He is best known to students for his formula (r cis θ)n = r n cis nθ which can be used to work out the powers of complex numbers. It is said that De Moivre was inspired to further research by reading Isaac Newton’s book Principia. He had little spare time so he tore out pages and carried them around with him, studying them in any free moment. Later in life he became involved in the controversy about whether Newton or Leibniz

had been the first to discover calculus. He was appointed by the Royal Society to the commission set up to investigate the rival claims. De Moivre always had difficulty earning money, but was able to eke out a living by working as a private tutor and by writing books. Unlike many other mathematicians of the time, he could not find a rich patron to support him because he was a foreigner. Even though he was made a fellow of the Royal Society in 1697 and had famous friends such as Newton and Halley, he was always poor and eventually died in poverty. Apparently, De Moivre predicted the time of his own death. Near the end of his life he noticed that he needed to sleep for an extra 15 minutes each night. He calculated the date when the cumulative result of this would mean that he was asleep for 24 hours. He died in his sleep on that day. Questions 1. What was the subject of De Moivre’s book The Doctrine of Chances? Probability 2. Why couldn’t De Moivre find a patron? He was a foreigner. 3. How did De Moivre make a living? Tutoring students and writing books 4. Which famous mathematician played a major role in his life? Newton 5. What was unusual about the date he died? De Moivre predicted it.

Research Investigate how insurance companies use probability to work out how much each insurance policy costs you.

3 a i

109

Chapter 2 Number systems: complex numbers Im (z) z–w 4

–1 0

Re (z)

remember remember 1. The magnitude (or modulus or absolute value) of z = x + yi is the length of the line segment from (0, 0) to z and is denoted by z, x + yi or mod z.

b i Im (z) 6

u+z

2. z = x 2 + y 2 and z z = z 0 1

y y 3. arg (z) = θ where tan θ = -- . ∴ θ = tan–1 -- . x x n 4. z × i , n ∈ N produces an anticlockwise rotation of 90n degrees. 5. z = r cos θ + r sin θ i = r cis θ in polar form. 6. Arg (z) is the angle θ in the range −π < θ ≤ π.

Re (z)

c i Im (z) 0

Re (z)

6

2.

1 a

Im (z)

w–u

–8

z = 4 + 8i

8

Complex numbers in polar form

2E

0

3 d i

4 Re (z)

In the following questions give arg (z) or Arg (z) correct to 3 decimal places where the angle cannot be easily expressed as a common multiple of π.

Im (z)

0 –2

7 Re (z) w+z

1 a Represent z = 4 + 8i on an Argand diagram. b Calculate the exact distance of z from the origin. (Do not use your calculator.)

z = 4 5

3 e i WORKED

Example

18

2 Find the modulus of each of the following. a z = 5 + 12i 13 b z = 5 – 2i 3 d z = −3 − 6i

e z=

3 5

3 + 2i

65

5

c

z = −4 + 7i

f

z = (2 + i)

2

Im (z) 0

5 –7

WORKED

Example

19

WORKED

Example

20

4 a Im (z) 6 4 2

z1 –4 –2

z2

z3

2 4 6 8

Im (z) 6

0

4 a Show the points z1 = −3 + 0i, z2 = 2 + 5i, z3 = 7 + 5i and z4 = 9 + 0i on the complex number plane. b Calculate the area of the shape formed when the four points are connected by straight line segments in the order z1 to z2 to z3 to z4. 42.5 square units

5 a Show the points z = −1 + 3i, u = 3 and w = 3 + 12i on the complex number plane. b Calculate the area of the triangle produced by joining the three points with straight line segments. 24 square units (z) Re 10

Example

21 Im (z) 12 10 8 6 4 z 2

–4 –2 0

6 Find the Argument of z for each of the following in the interval (−π, π ]. (Give exact answers where possible.)

w

a z = 3 + 2i 0.588 b z = e z = −2 − 2 3i i

u 2 4 6 8 Re (z)

z = −6i

π – --2 π – 2-----3

π 3 + i --6-

f

z = 6 − 10i

j

z = 55 0 −1.030

π z = 5 − 5i – --4π g z = 3i --2-

c

z+w–u

f i

z4

WORKED

5 a

3 If z = 3 + i, w = 4 − 3i and u = −2 + 5i then: i represent each of the following on an Argand diagram ii calculate the magnitude in each case. a z − w ii 17 b u + z ii 37 c w − u ii 10 d w + z ii 53 e z + w − u ii 130 f z2 ii 10

9 Re (z)

d z = −4 + 8i 2.034 h z= – 7 π

z2

8 Re (z)

110 WORKED

Example xample

22

WORKED

Example xample

23

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

7 Convert each of the following into Arguments. π π 3π 11 π π 15 π a ------ – --2b − --------- --6c --------- – --82 6 8 20 π 6 π 18 π 2 π 19 π 5 π e --------- – -----f --------- -----g − --------- -----7 5 6 6 7 5

8 Find the modulus and Argument of each of the following complex numbers. a 3 − 3i 3 2, – π--- b −5 + 5i 5 2, 3-----π- c −1 − 3i 2, – 2-----π- d 4 3 + 4i 8, π--4 4 e −7 − 10i

f

6i − 2

2 10, 1.893

149, – 2.182

WORKED

Example xample

3π a 2 cis -----4 2 π e cis  – ---- 3

24

f

25

3

6

3

1 1 z = − --- + --- i 4 4 2 3 π------- cis ----4

Example xample

g ( 3 + i ) 2 4, π---

π 2 5 cis  – ---  3

9 Express each of the following in the polar form, z = r cis θ, where θ = Arg (z). a z = −1 + i b z = 6 + 2i c z = – 5 – 5i d z = 5 – 15i 3 e z = − 1--- – ------- i 2 2

WORKED

5 π 3 πd − ------ ----4 4 13 π 11 π h − --------- --------12 12

4

π b 2 2 cis --6

π 10 cis  – 3------  4

c

10 Express each of the following complex numbers in Cartesian (or standard a + ib) form. 2π a 2 cis -----3 e

7π 7 cis  − ------  4

π b 3 cis --4

c

π 8 cis --2

g

f

5π 5 cis -----6

π d 4 cis  − ---  3

a −1 + 3i

3 cis π

11 multiple choice If z = 3 − 50i and w = 5 + 65i, the value of z + w is: A 64 B 15 C 17 D 225

10

3 2 b ---------- (1 + i ) 2 5 c ------- ( − 3 + i ) 2

E 289

d 2–2 3 i 14

e ---------- (1 + i ) 12 multiple choice 2 The perimeter of the triangle formed by the line segments connecting the points f 8i 2 − 4i, 14 − 4i and 2 + i is: g – 3 A 13 B 30 C 10 D 17 E 25

13 multiple choice The Argument of 4 π A --B 6

3 – 4i is: π --3

14 multiple choice In polar form, 5i is: π A cis --B cis 5π 2 15 multiple choice The Cartesian form of 1 3 A --- + ------- i 2 2

5π C -----6

π D – --6

π E – --3

5π C cis -----2

D 5 cis 5π

π E 5 cis --2

3 3 D − --- + ------- i 2 2

3 1 E − ------- – --- i 2 2

7π 3 cis  – ------ is:  6

1 3 B − --- + ------- i 2 2

3 1 C − ------- + --- i 2 2

Chapter 2 Number systems: complex numbers

111

Basic operations on complex numbers in polar form Addition and subtraction In general there is no simple way to add or subtract complex numbers given in the polar form r cis θ. For addition or subtraction, the complex numbers need to be expressed in Cartesian form first.

Multiplication in polar form SLE 3: Use polar forms to demonstrate multiplication and division of complex numbers.

In earlier sections we performed multiplication and division on complex numbers in standard form. This is quite a lengthy process for both these operations. However, as is the case in many aspects of mathematics, patterns exist that make the job so much easier. Work through the following investigation that will form the basis of future work. 1 Given that z = 1 + 3 i and w = 2 + 2i: a find zw in standard form 2 − 2 3 + 2( 3 + 1)i 7π b express the product from part a in polar form 4 2 cis -----12 c verify that | zw | = | z || w | d verify that arg (zw) = arg (z) + arg (w). 2 Given that z = 1 – i and w = 2 – 2 3 i: a find zw in standard form 2 − 2 3 − 2( 3 + 1)i 7π b express the product from part a in polar form 4 2 cis  – ------ 12 c verify that | zw | = | z || w | d verify that arg (zw) = arg (z) + arg (w) 3 Given that z = a + bi and w = c + di a find zw in standard form b verify that | zw | = | z || w | c For z = r1 cis θ = r1 (cos θ + i sin θ) and w = r2 cis φ = r2 (cos φ + i sin φ), show that zw = r1 r2[(cos θ cos φ – sin θ sin φ) + i(cos θ sin φ + sin θ cos φ)] 4 Using the trigonometric identities: cos (A + B) = cos A cos B – sin A sin B sin (A + B) = cos A sin B + sin A cos B verify that zw = r1 r2[cos(θ + φ) + i sin (θ + φ)] = r1 r2 cis (θ + φ) This investigation illustrates the following useful facts concerning multiplication of complex numbers in polar form: If z and w are two complex numbers, then | zw | = | z || w | and arg (zw) = arg (z) + arg (w) Similarly, for division of complex numbers: zz z --= -------- and arg  ---- = arg (z) – arg (w)  w w w The proofs required to establish these rules are outside the Mathematics C syllabus and will not be included in this chapter on complex numbers.

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WORKED Example 26

π 5π Express 5 cis --- ¥ 2 cis ------ in the form r cis θ where θ ∈ (−π, π]. 4 6 THINK WRITE/DRAW π π 5π 5π 5 cis --- × 2 cis ------ = ( 5 × 2 ) cis  --- + ------ 1 Simplify using the multiplication rule  4 6 4 6 zw = r1r2 cis (θ + φ) (see part 4 on 13 π page 111). = 10 cis  ---------  12  2 Sketch this number. Im (z) 13 —π 12

Re (z) π –11— 12

3

Subtract 2π from θ to express the answer in the required form.

WORKED Example 27

Express z1z2 in Cartesian form if z 1 = THINK 1

Use z1z2 = r1r2 cis (θ1 + θ2).

13 π 11 π 10 cis  --------- = 10 cis  – ---------  12   12 

5π 2 cis ------ and z 2 = 6 WRITE

Write the result in standard form.

5π π 2 cis ------ × 6 cis  – ---  6 3

z1z2 =

π π 2 × 6 ) cis  5------ – ---  6 3 π = 2 3 cis --2 π π = 2 3 cos --- + 2 3 sin --- i 2 2 = 2 3 × 0 + 2 3 × 1i =

2

π 6 cis  – --- .  3

(

= 2 3i

WORKED Example 28 If z = 5 3 + 5i and w = 3 + 3 3i , express the product zw in polar form. THINK 1 Sketch z.

WRITE/DRAW Im (z)

θ

5 5 √3

Re (z)

Chapter 2 Number systems: complex numbers

THINK 2

WRITE/DRAW

Write z in polar form. Use the special triangle below: π – 6

2

Let z = r 1 cis θ1. r1 = 5 × 2 = 10 π θ1 = --6

√3

π – 3

1

The ratio of sides in z is 5 times that of the sides in this triangle. 3

Verify this by rule if you wish.

Alternatively:

(5 3)2 + 52

r1 =

= 10

5 π θ = tan–1 ---------- , so θ 1 = --6 5 3 π Therefore z = 10 cis --6 4

Sketch w. The ratio of sides in w is 3 times that of the sides in the special triangle shown in step 2.

Im (z)

3 √3

θ

Re (z) 3

5

Write w in polar form.

Let w = r2 cis θ2 r2 = 3 × 2 = 6 π θ2 = --3

6

Verify this by rule if you wish.

Alternatively: r2 =

32 + (3 3)2 = 6

π 3 , so θ 2 = --3 π Therefore w = 6 cis --3 3 3 tan θ 2 = ---------- = 3

7

Determine zw using z1z2 = r1r2 cis (θ1 + θ2).

π π zw = 10 cis --- × 6 cis --6 3 π π   = 60 cis --- + -- 6 3 = 60 cis --π2

113

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

WORKED Example 29

5π π Express 10 cis  – --- ÷ 5 cis ------ in the form r cis θ where θ ∈ (−π, π ].  3 6 THINK 1

2

WRITE/DRAW 5π π π π 10 cis  – --- ÷ 5 cis ------ = 2 cis  – --- – 5------  3  3 6 6

Simplify using the division rule. (See part 4 on page 111.)

7π = 2 cis – ------   6

Sketch this number.

Im (z) 2 –π 6

5— π 6

Re (z)

– 7—6π

3

State θ, the principal Argument.

5π Arg (z) = -----6

4

State the result in polar form.

5π Arg ()z = 2 cis -----6

Powers of complex numbers Whole powers of z As with real numbers, powers of complex numbers can be written as: zn = z × z × z × z × … × z to n factors. Since z = a + bi is a binomial (containing two terms) we can express zn using Pascal’s Triangle to generate the coefficients of each term. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 5th row → 1 5 10 10 5 1 and so on. (a + bi)5 can therefore be expanded using the elements of the fifth row of Pascal’s Triangle: (a + bi)5 = 1a5 + 5a4(bi)1 + 10a3(bi)2 + 10a2(bi)3 + 5a(bi)4 + (bi)5 (a + bi)5 = 1a5 + 5a4bi + 10a3b2i2 + 10a2b3i3 + 5ab4i4 + b5i5 (a + bi)5 = 1a5 + 5a4bi − 10a3b2 − 10a2b3i + 5ab4 + b5i (a + bi)5 = 1a5 − 10a3b2 + 5ab4 + 5a4bi − 10a2b3i + b5i (a + bi)5 = 1a5 − 10a3b2 + 5ab4 + (5a4b − 10a2b3 + b5)i grouped into standard form. Re [(a + bi)5] = 1a5 − 10a3b2 + 5ab4 Im [(a + bi)5] = 5a4b − 10a2b3 + b5

Chapter 2 Number systems: complex numbers

115

Graphics Calculator tip! Pascal’s Triangle coefficients The coefficients of each term of the expansion of (a + bi)n can be found using your graphics calculator. For example, the coefficients of the expansion of (a + bi)5 can also be written as: 5 C0 a5 + 5C1 a4(bi)1 + 5C2 a3(bi)2 . . . 5C5(bi)5 5 5 5 where C0, C1, . . ., C5 represent the coefficients. The following steps show how to calculate 5C3 using a graphics calculator.

For the Casio fx-9860G AU s

1. Press MENU and select RUN-MAT. Press OPTN and then F6 ( ) for more options.

2. Press F3 (PROB) and you will see the function nCr. Enter 5 (for n), then press F3 (nCr) and enter 3 (for r). Press EXE to display the value.

For the TI-Nspire CAS 1. Press k to access the catalog and press 1 for the list of functions. Scroll down to select nCr(. You can do this more quickly by first pressing N.

2. Press · and then complete the entry line to obtain nCr(5,3). Press · to display the value. Note that with this calculator, we can obtain the actual expansion of (a + bi)5. 1. Open a new Calculator document. 2. Press b and select 1: Actions and then 4: Clear a-z. This sets the variables a–z to their default values and makes them ready for use. Accept OK by pressing ·. 3. Enter (a + bi)^5 and press ·. You will see a small arrow at the end of the answer line indicating that there are more terms. Use the arrows on the NavPad to see more of the expansion. The full expansion is shown as a(a4 − 10a2b2 + 5b4) + (5a4 − 10a2b2 + b4)bi.

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WORKED Example 30

Use Pascal’s Triangle to expand (2 - 3i)3. THINK

WRITE

1

Use the third row of Pascal’s Triangle to expand (1 3 3 1). Use brackets to keep the negative sign of the second term.

(2 − 3i)3 = 1(23) + 3(2)2(−3i) + 3(2)(−3i)2 + (−3i)3

2

Simplify the expression.

= 8 − 36i + 54i2 − 27i3 = 8 − 36i − 54 + 27i = −46 − 9i

Negative powers of z 1 1 Your earlier studies have shown that z–1 = --- . Similarly, z–3 = ---3- . z z

WORKED Example 31

Evaluate (3 - i)-4. THINK

WRITE

First find the expansion with a positive power. Use the fourth row of Pascal’s Triangle to expand (1 4 6 4 1).

(3 − i)4 = 34 + 4(3)3(−i) + 6(3)2(−i)2 + 4(3)(−i)3 + (−i)4

(Use brackets to keep the negative sign with the second term.)

= 81 − 108i − 54 + 12i + 1

2

Simplify to obtain the standard form.

= 28 − 96i

3

Express this as the denominator then multiply by the complex conjugate.

4

Write the final expression in standard form.

1

1 ( 28 + 96i ) (3 − i)–4 = ------------------------ × ------------------------( 28 – 96i ) ( 28 + 96i ) 28 + 96i –4 (3 − i) = --------------------------784 + 9216 28 + 96i (3 − i) –4 = -------------------10 000 7 6 (3 − i)–4 = ------------ + --------- i 2500 625

Fractional powers of z Fractional powers of complex numbers generally follow the same rules as with real numbers. p ---

zq =

q

z

p

Our discussion here will deal only with the square root of z, where

1--2

z = z .

Chapter 2 Number systems: complex numbers

117

WORKED Example 32 Express

3 + 4i in standard form.

THINK

WRITE 3 + 4i = a + bi

1

Let 3 + 4i be a complex number such as (a + bi), where a, b ∈ R.

Let

2

All dialogue given in the ‘write’ column should appear as ‘communication’ in your working.

Square both sides: 3 + 4i = (a + bi)2 3 + 4i = a2 + 2abi − b2 Equating real and imaginary terms: 3 = a2 − b2 4 = 2ab

[1] [2]

4 2 a = ------ = --- from [2] [3] 2b b Substitute for a into [1] 2 2 3 =  --- − b2  b 4 3 = ----2- − b2 b 3b2 = 4 − b4 b4 + 3b2 − 4 = 0 (b2 − 1)(b2 + 4) = 0 Therefore, b2 = 1, b = ±1 or b2 = −4, b = ±2i Since a and b are real numbers discard b = ±2i. Substitute for b = ±1 into [3] 2 When b = 1, a = --- = 2 1 2 When b = −1, a = ------ = −2 –1 3 + 4i = 2 + i or −2 − i = ±(2 + i)

3

State the final result in standard form.

Therefore

4

Verify this result.

Check: [±(2 + i)]2 = 4 + 4i − 1 = 3 + 4i

5

Alternatively, you can use a graphics calculator (such as a TI-Nspire CAS calculator) to verify this result.

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History of mathematics W I L L I A M R O WA N H A M I LT O N ( 1 8 0 5 – 1 8 6 5 ) During his life . . . Charles Darwin developed his theory of evolution. Charles Babbage developed the first automatic digital computer. Gregor Mendel laid the mathematical foundation for the science of genetics. Sometimes considered the second greatest mathematician of the Englishspeaking world, after Sir Isaac Newton, William Hamilton was born in Dublin, Ireland, on 3 August 1805. Even the fact that Hamilton did not attend school before he entered college did not deter his thirst for knowledge. By the age of three he was skilled at reading and arithmetic, by the age of five he read and translated Latin, Hebrew and Greek, and by the age of 14 he could speak 14 languages. By the age of 21 he published a paper entitled ‘A Theory of Systems of Rays’, introducing and developing concepts that created the field of mathematical optics. Propelled by the success of this work, at 22 he was unanimously voted into the chair of the professor of astronomy at Trinity College (Dublin), even though he was still an undergraduate and had not applied for the position.

In 1833 Hamilton further developed his work on complex numbers and in 1843 he released what he considered to be his greatest discovery — the algebra of quaternions. With these ordered sets of four numbers, magnitude and direction in 3-dimensional space could be determined. The fact that multiplication of quaternions is not commutative led to the development of the first ‘ring’ in which the commutative property does not hold. This inspiration came to him while he was crossing the Brougham Bridge in Dublin and he left the inscription i2 = j2 = k2 = ijk = −1 in a stone in the bridge. A stamp featuring these quaternions was issued in Ireland in 1983. His work also led to the development of the concepts of vectors, scalars and tensors, which you will encounter later in your studies. Plagued throughout his life with alcoholism, he died in 1865.

Research 1. Find out more about quaternions. 2. Research the notion of ‘rings’.

remember remember 1. If z1 = r1 cis θ1 and z2 = r2 cis θ2, then: z1 × z2 = r1r2 cis (θ1 + θ2) z r ---1- = ----1 cis (θ1 − θ2). z2 r2 2. A complex number zn = (a + bi)n can be expanded using Pascal’s Triangle to generate the coefficients of each term. 1 3. Negative powers of z: z –n = ---nz p --q p q 4. Fractional powers of complex numbers: z = z

Chapter 2 Number systems: complex numbers

2F WORKED

Example

26

Basic operations on complex numbers in polar form

1 Express each of the following in the form r cis θ where θ ∈ (−π, π ]. π π 3π a 2 cis --- × 3 cis --- 6 cis -----4 4 2 2π π b 5 cis ------ × 4 cis  – ---  3 3 c

WORKED

Example

119

3π 6 cis ------ × 5 cis π 4

π 20 cis --3 π 6 5 cis  – ---  4

d

5π π 3 cis  – ------ × 2 cis  – ---  6  2

e

5π 7 π × 2 7 cis  – ----cis ----- 12  12

2π 6 cis -----3

π 2 7 cis  – ---  6

2 Express the resultant complex numbers in question 1 in standard form.

27 WORKED

Example

28

3 Express the following products in polar form.

c WORKED

Example

29

4 Express each of the following in the form r cis θ where θ ∈ (−π, π ]. 5π π a 12 cis ------ ÷ 4 cis --- 3 cis π--2 6 3 3π π b 36 cis ------ ÷ 9 cis  – ---  6 4

11 π 4 cis --------12

4π 11 π d 4 3 cis ------ ÷ 6 cis --------7 14 e

Example

30

WORKED

Example

31

3 π 2 cis  – ---- 10 

π 20 cis  – π --- ÷ 5 cis  – ---  2  5

c

WORKED

c 3 10 – 3 10 i 3 2 d – ------6- + ---------- i 2 2 e 21 – 7 i

5π 4 2 cis -----12 ( 3 − 3i)(2 3 − 2i) 8 3 cis  – π--- 2 π- (−4 + 4 3 i)(−1 − i) 8 2 cis  – ----12

a (2 + 2i)( 3 + i) b

2 a –3 2+ 3 2 i b 10 + 10 3 i

π 5 a i 3 3 cis --4 b i 16 cis π c i 9 cis π 3π d i 32 cis -----4

3 π- 2 2 cis  – ---- 14 

5π 7 π 3 5 cis  – ----- ÷ 2 10 cis ----- 12  6

3 2 7 π---------- cis ----12 4

3π π 3 cis ------ and w = 2 cis  – --- then express each of the following in:  4 4 i polar form ii standard form. 3 b w4 c z4 d w5 a z

5 If z =

6 If z = 1 − i and w = – 3 + i , write the following in standard form. 1 a z−4 − --4-

3 1 d w−5 ------- – ------ i 64

64

1 b w−3 − --8- i z3 e -----4- 0.171 – 0.046i w

c

1 1 z−3 − --4- + --4- i

f

z2w3 16

3 6 3 6 ii ---------- + ---------- i 2 2 ii −16 ii −9 ii – 16 2 + 16 2i

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

7 Determine ( 2 + 2i ) 2 ( 1 – 3i ) 4 in standard form.

– 64 3 – 64i

( 3 – i )6 8 Write ----------------------------3 in the form x + yi. 1 ( 2 – 2 3i ) 9 multiple choice π π a 5 cis  – --- × 8 cis  – --- is equal to:  3  6 A 6 2i b If z =

(

A 1+i

B – 2 10i

C –6 3

6 + 2 ) + ( 6 – 2 ) i then B

C

2i

1----64

D −6i

E 6 6

D

E –1 – 2i

2z –3 is: (1 − i)

2+i

w4

c If z = –1 – 3i and w = 2 + 2i then -----3- is equal to: z A −4 + 4i B 2 3 C 2 D −4i 3π 2 cis ------ and w = 4

10 If z =

E −8

π z6 3 cis --- , find the modulus and the argument of -----4- . 6 w

8 π ---, – --9 6

11 If z = 4 + i and w = −3 − 2i, determine (z + w)9. 16 − 16i 12 Find z6 + w4, if z = 13 If z 1 =

2 – 2i and w = 2 − 2i. −64 + 64i

3π π 5 cis – 2------  , z 2 = 2 cis ------ and z 3 =  5 8

π 10 cis ------ , find the modulus and 12

z 12 × z 23 2---, – -------π the argument of ------------------. 5 1204 z3 WORKED

Example xample

32

eBook plus

14 Express each of the following in standard form: a c

5 + 12i ±(3 + 2i) 2 + 2i

±( 1 + 2 + – 1 + 2 i )

b

5 – 12i ±(3 − 2i)

d

3 – 4i ±(2 − i)

Digital doc: WorkSHEET 2.2

Complex numbers: applications 1 Choose a complex number that falls in the first quadrant of the complex plane. Calculate the first 8 powers of this number and investigate any pattern that exists between the modulus of each of the powers. Plot each power on an Argand diagram. What do you notice?

2 a iii iii iii iv iv vi

z2 = r2 cis 2θ z3 = r3 cis 3θ z4 = r4 cis 4θ z5 = r5 cis 5θ z6 = r6 cis 6θ z7 = r7 cis 7θ

2 Let z = r cis θ, a complex number. Find, in terms of r and θ: a iii z2 = z.z ii z3 = z.z2 iii z4 = z.z3 iv z5 = z.z4 v z6 = z.z5 vi z7 = z.z6 GP, a = r r = r b Write the moduli of the powers of z as a sequence. Tn = ar n – 1 c What do you notice about the sequence given in part b? Geometric progression

3 Im (z)

Chapter 2 Number systems: complex numbers

z2

121

2— π 3

z1

0 –2 π —– 3

Re (z)

z3

z1 = 1 = cis 0 z2 = −1 +

2π 3 i = cis -----3

z3 = −1 −

2π 3 i = cis  – ------  3

3 As mentioned at the beginning of this chapter the equation z2 = 1 has two solutions, z = ±1, whereas the equation z = 1 has only one solution, z = 1. The 2π 2π equation z3 = 1 has 3 solutions, z = 1, cis ------ and cis  − ------ .  3 3 Graph these solutions on an Argand diagram. Express all solutions in both rectangular and mod–arg form. 2

2

4 Let z = x + yi. Therefore | z | = x + y , and | z |2 = x2 + y2, where this is the general equation of a circle, of radius | z |, about the origin. Graph this circle and fully label the path of the rotating z as it moves about the origin. Therefore, what is the meaning of the statement | z | < x + yi? Sketch | z | < 4 and | z | > 1. 5 Research the life of William Rowan Hamilton and his contribution to the study of complex numbers. 6 Research the area of mathematics called fractals. You will investigate this fascinating area in more detail later in your studies. 7 In Chapter 1, you were introduced to the term ‘transcendental numbers’ — irrational numbers that are not algebraic, that is, cannot be produced by the algebraic operations of addition, subtraction, multiplication and division, and by taking roots. Pi (π) is one such transcendental number and e is another, 1

2

2

3

4

1

2

3

3

4

1 1 1 1 where e = e1 = 10 + ----- + ----- + ----- + ----- . . . 1! 2! 3! 4! (and 3! = 3 × 2 × 1, and so on. The symbol 3! is referred to as factorial 3.) The function ex is referred to as the exponential function. 1

x x x x ex = x0 + ----- + ----- + ----- + ----- . . . 1! 2! 3! 4! The graph of the function ex is especially interesting because the slope of the curve at any point equals the value of the curve, at that point. That is, the slope of a tangent to the curve at x = e2 is e2. Euler discovered a special relationship between e and i, 4

i i i i where ei = i 0 + ----- + ----- + ----- + ----- . . . 1! 2! 3! 4! Write four expressions for ei, with increasing numbers of terms and simplify them where possible. The alternating positive and negative signs suggest that the expression is approaching a particular value as the number of terms in the series increases. You might find it more methodical to list the results as each new term is added as you ‘creep’ closer to the value. Can you suggest what that value might be? What is the modulus of this number? Use a graphics calculator to evaluate ei.

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summary Introduction to complex numbers • We define the ‘imaginary number’ i as having the property that i 2 = −1. • A complex number z = x + yi with x, y ∈ R and C = {z: z = x + yi, x, y ∈ R} defines the set of complex numbers. • The real part of z is x and is written as Re (z). • The imaginary part of z is y and is written as Im (z).

Basic operations using complex numbers • If z and w are two complex numbers such that z = a + bi and w = c + di for a, b, c, d ∈ R then: 1. z = w if and only if (i.e. iff) a = c and b = d 2. z + w = (a + c) + (b + d)i 3. z − w = (a − c) + (b − d)i 4. kz = ka + kbi, for k ∈ R 5. z × w = (ac − bd) + (ad + bc)i.

Conjugates and division of complex numbers • If z = a + bi and w = c + di for a, b, c, d ∈ R then: 1. The conjugate z of z is z = a − bi. 2. z. z = a2 + b2. bc – ad )iz ac + bd- (----------------------3. ---- = ----------------+ 2 2 2 2 w c +d c +d

Radians and coterminal angles • A radian is the angle subtended by an arc of the radius of a circle. That is, an arc of 2πr subtends an angle of 2π radians. • 2π radians = 360°. 1 radian ≈ 57.3°. • Coterminal angles differ by a multiple of 360°.

The polar form of complex numbers • The magnitude (modulus or absolute value) of z = x + yi is the length of the line segment from (0, 0) to z. It is denoted by z, x + yi or mod z. • z = x 2 + y 2 and z z = z 2 . • The argument of z, arg (z), is the angle measurement anticlockwise of the positive y Real axis and arg (z) = θ where θ = tan–1 --x . • z = x + yi can be expressed in polar form as z = r cos θ + r sin θi = r cis θ. • Arg (z) is the angle θ in the range −π < θ ≤ π and is called the principal argument.

Basic operations on complex numbers in polar form • If z1 = r1 cis θ1 and z2 = r2 cis θ2, then: 1. z1 × z2 = r1r2 cis(θ1 + θ2) r z 2. ---1- = ----1 cis(θ1 − θ2) z2 r2 • A complex number zn = (a + bi)n can be expanded using Pascal’s Triangle to generate the coefficient of each term. 1 • z−n = ---nz

Chapter 2 Number systems: complex numbers

123

CHAPTER review Questions 1 and 2 refer to the complex number z = 2 5 – 4i . 1 multiple choice

2A

The real and imaginary parts of z respectively are: A 2 5 and 4 B 2 5 and −4 C 4 and 2 5

D −4 and 2 5

E 2 5 and −4i

2 multiple choice

2A

The Argand diagram which correctly represents z is: A Im (z) B Im (z)

0

0

Re (z)

2 5

2 5

–4

D

C Re (z)

z

4

Re (z) 4

0

z

E

2 5 Im (z)

0

Re (z)

–4

z

2 5

Im (z)

z

Im (z) 2 5

z

–4

0

Re (z)

3 Simplify i 6 − i 3 (i 2 − 1). −1 − 2i

2A

Questions 4 and 5 refer to the complex numbers u = 5 − i and v = 4 + 3i. 4 multiple choice The expression 2u − v is equal to: A 1 − 4i B −3 − 7i

2B C 6 − 5i

D 5 + 8i

E 14 + i

5 multiple choice The expression uv is equal to: A 9 + 2i B 20 − 3i

2B C 20 + 3i

D 15 − 4i

6 multiple choice If z = 5 − 12i, decide which statement is true concerning −iz. A −iz = 13 B −iz = 12 − 5i C The point z is rotated 90° clockwise. D Re (−iz) = 0 E Im (−iz) = −i

E 23 + 11i

2B

124 2B 2B,C 2C

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

7 If z = 3 − 8i, then find: a Im (z2) −48

b a and b if z3 = a + bi. a = −549, b = 296

8 If z = 2 − 5i, u = −3 + i and w = 1 + 2i evaluate: a z − 2u + 3w 11 − i b z 29

2C

2C

1 + 2i Im  -------------- is equal to:  1–i A 2 B − 1---

C

3 --2

D

If z = 3i and w = 4 − i then z w is equal to: A 12 + 3i B 12 − i C 7 + 3i

E z = 4 − 2i

D 4 − 2i

E 3−i

11 multiple choice 2i 3 The expresion ----------- – ---------- simplifies to: 1+i 2–i 2--1 1 3 A – --- + i C --- – --- i B 3 + 7i 5 5 4 4

1 z - ( 12 – 14i ) 12 If z = 6 − 2i and w = 5 + 3i, express ---- in the form a + bi, a, b ∈ R. ----17 w

2D

14 Convert the following radian measures to degrees. 3π 7π a ------ 135° b ------ 210° 4 6

2D

15 multiple choice

2E

E −2

D 12 − 3i

13 Convert the following common angle measures to radians. π a 30° --6b 180° π

2E

2 --3

10 multiple choice

2D

2E

uz + w 6 5

9 multiple choice

2

2C

c

Of the following pairs of angles, the pair that is coterminal is: π 9π 3π 5π A 40°, 220° B --- , -----C 135°, 435° D ------ , -----4 4 2 2

E 180°, 360°

16 multiple choice Arg (2 − 2i) is equal to: π B --A π 4

3π C -----4

π D – --4

E 2π

3π C 3 cis -----4

π D 3 cis – ---   4

5π E 3 2 cis -----4

17 multiple choice The polar form of −3 + 3i is: π 3π A 3 2 cis --B 3 2 cis -----4 4

18 If z = −7 − 7i, express z in polar form.

3 π 7 2 cis  – ---- 4

Chapter 2 Number systems: complex numbers

125

19 multiple choice How many degrees apart are two consecutive roots of z8 = 1 on the unit circle? A 180

B 90

C 135

D 225

2E

E 45

20 multiple choice π π If z1 = 10 cis --- and z2 = 5 cis  – --- then z1z2 in polar form is:  6 4 5 π π π π- E 2 cis  – ----5π D 15 cis  – ----A 50 cis -----B 15 cis -----C 2 cis ----- 12  12  12 12 12

2F

21 multiple choice

2F

3π π In standard form, 12 2 cis ------ ÷ 3 cis  – --- is equal to:  2 4 A 4 + 4i B −4 − 4i C 4 − 4i D −4 + 4i

E 36 − 36i

22 If z = −3 − 4i, write the following in standard form. a z4 –527–336i

117 44 i - + --------------b z−3 --------------15 625 15 625

c

z ±(1 − 2i)

Modelling and problem solving π π 1 Let z = 2 cis --- and w = 2 cis --- . 3 4 π π z a Express ---- in the form r cos θ + r sin θ i. cos ------ + sin ------ i 12 12 w b Express z and w in Cartesian form. z = 1 + 3 i, w = 2 + 2 i z 2 + 6 + ( 6 – 2 )i c Express ---- in Cartesian form. ----------------------------------------------------4 w d Using the results of parts a and c, find the exact values for: 6+ 2 π i cos ------ -------------------4 12 π ii sin -----12

6– 2 -------------------4

π iii tan ------ . 2 – 3 12 π π z e By letting z = 2 cis --- and w = 2 cis --- and following parts a to c for zw instead of ---- , 4 6 w 5π deduce that tan ------ = 2 + 3 . Check with your teacher. 12 2 Let u = 1 – i. a

i Find u u . 2 ii Find Arg u + Arg(2 u ). 0

b Let z = x + yi, x, y ∈ R, and |z – u | = |z – 2u|. Find the value of x when y = 0. x = 3

2F

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3 Consider the complex number z such that z = 3 + 2i. a Find the value for iz, i2z, i3z and i4z. (Give answers in standard x + yi form.) iz = −2 + 3i, i2z = −3 − 2i, i3z = 2 − 3i, i4z = 3 + 2i b Comment on the value of i4z. i4z = z c Plot each number from part a on the same Argand diagram. d Use a pair of compasses to draw a circle whose centre is at the origin and which passes through each point on the diagram. e Find the radius of the circle, giving your answer in exact (surd) form. 13 f Carefully study the five points on your diagram. What transformation is required to transform: i, ii and iii One-quarter turn i point z into point iz? (rotation by 90°) in an ii point iz into point i 2z? anticlockwise direction. iii point i 2z into point i 3z? eBook plus g On the Argand diagram, what transformation takes place when a complex number is multiplied by i? One-quarter turn in an anticlockwise direction Digital doc: Test Yourself h For a complex number z such that z = x + yi, describe the curve that all points representing Chapter 2 numbers of the form zi n (that is, z, zi, zi2, zi3, and so on) would lie on an Argand diagram. 3 c

d

Im z iz

4 3 2 1

iz 4

i z

0 –4 –3 –2 –1 –1 1 2 3 4 Re z –2 i 2z –3 i 3z –4

Circle with centre at the origin and radius

Im z 4 3 2 1

r =

2

x +y

2

4

i z

0 Re z –4 –3 –2 –1 –1 1 2 3 4 –2 2 i z –3 i 3z –4

Im z z = x + yi zi r

y x

Re z zi

zi 2

3

Matrices

3 syllabus reference Core topic: Matrices and applications

In this chapter 3A 3B 3C 3D

Operations with matrices Multiplying matrices Powers of a matrix Multiplicative inverse and solving matrix equations 3E The transpose of a matrix 3F Applications of matrices 3G Dominance matrices

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Introduction to matrices • definition of a matrix as data storage and as a mathematical tool • dimension of a matrix • matrix operations — addition and subtraction, transpose, multiplication by a scalar, multiplication by a matrix • inverse of a matrix • solution of simple matrix equations • definition and properties of the identity matrix • singular and non-singular matrices • applications of matrices in both life-related and purely mathematical situations

Four towns are connected by roads as shown in the figure. There is one road connecting A and B, two roads connecting A and C and so on. This information may be represented as shown in the table. To B A D

A

B

C

D

A

0

1

2

0

B

1

0

0

1

C

2

0

0

3

D

0

1

3

0

From C

If the headings at the top and side of this display are removed, an array of numbers only is left: 0 1 2 0

1 0 0 1

2 0 0 3

0 1 3 0

This array of numbers is called a matrix (plural, matrices). The arrangement of numbers in matrices is an extension of our number system and, as we will see, the rules that govern matrix calculations have many similarities with the arithmetic of numbers. Matrices are particularly useful in solving complex problems in linear programming. A matrix is a rectangular array of numbers arranged in rows and columns. The numbers in the matrix are called the elements of the matrix. The matrix above is a 4 × 4 matrix as it has 4 rows and 4 columns. We say the order of the matrix is 4 by 4. 2 0 1 – 4 is a 3 × 2 matrix because it has 3 rows and 2 columns. Note the –1 2 square brackets used to enclose the array. The matrix

A matrix with m rows and n columns is called an m × n matrix. We say the order of the matrix is m × n. The dimensions of a matrix are always given as the number of rows multiplied by the number of columns. The elements of the matrix are referred to by the row and then by the column position. In the 3 × 2 matrix above, the row 1, column 1 element is 2, the row 3, column 1 element is −1 and the row 1, column 2 element is 0.

Chapter 3 Matrices

129

We often use capital letters as symbols for matrices. Thus we may write A=

2 0 1 –4 –1 2

In general, the elements of a matrix A are referred to as ai j where i refers to the row position and j refers to the column position. a 11 a 12 a 13 a 14 … a 1n That is, A =

a 21 a 22 a 23 a 24 … a 2n a 31 . . . a m1

a 32 . . . a m2

a 33 a 34 . . . . . . a m3 a m4

… ... ... ... …

a 3n , depending on the order of the matrix . . . a mn

where A is an m × n matrix. The row 1, column 1 element is a11. The row 3, column 1 element is a31 and so on.

WORKED Example 1 For each of the following give the order of the matrix, if it exists, and where possible write down the elements in row 2, column 1 and row 1, column 3. 2 5 A= 3 6 4 7

B=

1 2 3 –1 –2 –3

THINK

–1 C = –2 –3

D= 5 0 2 6

WRITE

1

A has 3 rows of numbers and 2 columns of numbers.

A is a 3 × 2 matrix.

2

B has 2 rows and 3 columns.

B is a 2 × 3 matrix.

3

C has 3 rows and 1 column.

C is a 3 × 1 matrix.

4

D is not a rectangular array of numbers as it does not have all positions filled. The row 2, column 1 element is the number in the second row and the first column. The row 1, column 3 element is the number in the first row and the third column. In A and C there is no row 1, column 3 element since there is no third column in either matrix.

D is not a matrix.

5

6

7

Row 2, column 1 element

Row 1, column 3 element

A

3

—

B

–1

3

C

–2

—

Matrix

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Operations with matrices Addition The sports coordinator at Mathglen State High School kept records of the number of first, second and third ribbons awarded to competitors in each house at the swimming and athletics carnivals and sports events. The results were: Swimming

Athletics and sports

House

1st

2nd

3rd

House

1st

2nd

3rd

Hamilton

25

28

24

Hamilton

35

35

27

Leslie

38

30

35

Leslie

33

34

39

Barnes

34

36

35

Barnes

30

33

36

Cunningham

35

38

38

Cunningham

34

34

30

To find the total number of first, second and third places for each house, the swimming, athletics and sports results may be added. The elements in corresponding positions are added to give the total number of first, second and third places for each house: House Hamilton Leslie Barnes Cunningham

1st

2nd

3rd

25 + 35 = 60 38 + 33 = 71 34 + 30 = 64 35 + 34 = 69

28 + 35 = 63 30 + 34 = 64 36 + 33 = 69 38 + 34 = 72

24 + 27 = 51 35 + 39 = 74 35 + 36 = 71 38 + 30 = 68

60 Adding the elements for each event results in the following matrix: 71 64 69 Addition of matrices is performed by adding corresponding elements.

63 64 69 72

51 74 71 68

Chapter 3 Matrices

131

Subtraction The subtraction of matrices is also performed by the usual rules of arithmetic on corresponding elements of the matrices. It follows that: 1. Subtraction of matrices is performed by subtracting corresponding elements. 2. Addition and subtraction of matrices can be performed only if the matrices are of the same order; that is, they have the same number of rows and columns. Furthermore, addition of matrices is commutative. That is, for two matrices A and B of the same order: A+B=B+A

WORKED Example 2 A= 1 2 34 find, if possible: a A+B b A−B THINK

If

B= 1 4 23

C= 2 2 0 220

c B − C.

a Add the numbers in the corresponding positions of each matrix.

WRITE a A+B= 1 2 + 1 4 34 23 = 26 57

b Subtract the numbers in the corresponding positions of each matrix.

b A−B= 1 2 – 1 4 34 23 = 0 –2 1 1

c Subtraction cannot be performed since the order c B − C cannot be calculated because B of B is 2 × 2 and the order of C is 2 × 3. and C are of different orders.

Multiplication by a scalar Consider the matrix B = 1 4 23 To find 3B we could use repeated addition: 3B = B + B + B = 14 + 14 + 14 23 23 23 = 3 12 6 9 3B could have been calculated more efficiently by multiplying each element of B by 3.

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Thus 3B = 3 1 4 = 3 × 1 3×2 23

3 × 4 = 3 12 3×3 6 9

The number 3 in the term 3B is called a scalar because it is a real number. Terms such as 3B refer to scalar multiplication of matrices. When a matrix is multiplied by a scalar, each element of the matrix is multiplied by the scalar.

WORKED Example 3 3 3 find: If A = 2 – 3 and B = 4 1 –3 –2 a 2A b 5B c 2A + 5B d 4(A + B)

e 2(B − A).

THINK

WRITE

a Multiply each element of A by 2.

a

2Α = 2 2 – 3 4 1 = 4 –6 8 2

b Multiply each element of B by 5.

b

5Β = 5 =

c Add the two matrices found in parts a and b.

d

1

Add A and B inside the brackets.

3 3 –3 –2

15 15 – 15 – 10

c 2Α + 5Β = 4 – 6 + 15 15 8 2 – 15 – 10 = 19 9 –7 –8   d 4(Α + Β) = 4  2 – 3 + 3 3   4 1 –3 –2  =4 5 0 1 –1

e

2

Multiply each element of the resulting matrix by 4.

= 20 0 4 –4

1

Subtract A from B (inside the brackets).

  e 2(Β − Α) = 2  3 3 – 2 – 3   –3 –2 4 1 =2

2

Multiply each element of the resulting matrix by 2.

=

1 6 –7 –3 2 12 – 14 – 6

Chapter 3 Matrices

133

There are some obvious but important features of scalar multiplication. If A and B are matrices of the same order and a, b are real numbers, then: 1. aA + bA = (a + b)A 2. aA + aB = a(A + B) 3. (ab)A = a(bA) Operations 1 and 2 are similar to the Distributive Law of Multiplication over Addition. Operation 3 is similar to the Associative Law of Multiplication. If aA = 0, then a = 0 or A is a zero matrix. A zero matrix is a matrix which has all elements equal to zero.

Equality of matrices This leads to an important principle about the equality of matrices. Two matrices are equal if they are of the same order and all corresponding elements are equal; that is, if A = a b and B = a b then A = B. c d c d

Simple matrix equations We know that to solve an algebraic equation such as 2x + 5 = 11, we: 1. subtract 5 from both sides to obtain 2x = 11 − 5 which gives 2x = 6 2. then, divide both sides by 2 (or multiply by 1--- ) to obtain x = 6 × 1--- or x = 3. 2 2 Simple matrix equations which require the addition or subtraction of a matrix or multiplication of a scalar can be solved in similar ways to those employed with algebraic equations.

WORKED Example 4 Solve the following matrix equations. a 5A =

50 35 – 15 20

b P+ 3 2 = 6 9 15 –2 4

c 2B – 1 2 – 3 = 3 4 7 20 1 –2 6 –5

THINK

WRITE

a

a 5Α =

1

To get A by itself multiply both sides by 1--- .

50 35 – 15 20

5

Α=

2

Simplify the matrix A.

1 --5

50 35 – 15 20

Α = 10 7 ⇒ –3 4 Continued over page

134

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

THINK b

1

WRITE

To get P by itself subtract 3 2 from 15 both sides.

b P+ 3 2 = 6 9 15 –2 4 P=

c

⇒P =

2

Simplify the matrix P.

1

First get 2B by itself by adding 1 2 – 3 to both sides. 20 1

69 – 32 15 –2 4 3 7 –3 –1

c 2B – 1 2 – 3 = 3 4 7 20 1 –2 6 –5 3 4 7 + 1 2 –3 –2 6 –5 20 1

2B =

2

Simplify the right-hand side.

3

Multiply both sides by

4

Simplify the matrix B.

--12

to get B by itself.

= 46 4 0 6 –4

Β=

1 --2

46 4 0 6 –4

Β= 2 3 2 0 3 –2

Matrix operations can also be performed using a graphics calculator. However, tips on how to do this appear later in the chapter and in Chapter 5, as you should have first practised performing these operations manually.

remember remember 1. 2. 3. 4. 5. 6. 7. 8. 9.

A matrix is a rectangular array of numbers arranged in rows and columns. An m × n matrix has m rows and n columns. The numbers in the matrix are called the elements of the matrix. Elements are referred to by the row and column position. For example, ai j refers to the ‘ith’ row and the ‘jth’ column of matrix A. Addition of matrices is performed by adding corresponding elements. Subtraction of matrices is performed by subtracting corresponding elements. Addition and subtraction of matrices can be performed only if the matrices are of the same order. When a matrix is multiplied by a scalar, each element of the matrix is multiplied by that scalar. Two matrices are equal if they are of the same order and all corresponding elements are equal.

135

Chapter 3 Matrices

3A WORKED

1

1 Using a table format, give the order of each of the following matrices and where possible write down the row 2, column 1 and row 1, column 3 elements of each. Matrix A B C D E

Example

2

33 09

WORKED

Example

–4 6 8 14

3

d C−A

–2 8

3 Using the matrices A, B and C from question 2 find: a 2A b 2A − B c 2A + 3B d 3(A + B) –9 6 12 12

4 multiple choice

eBook plus Digital doc:

3 –6 –2 –1

2 If A = – 2 3 , B = 5 0 and C = 1 – 3 , calculate: 47 –4 2 2 6 –7 3 6 –3 a A+B b A−B c B+C 85

Use

SkillSHEET 3.1 Operations with matrices

1 23 A = –2 –1 4 , 6 30

11 6 – 4 20

B = 5 –4 , 1 3

2, 1 1, 3 Order element element 2×2 8 — 3×1 5 — 1×4 — 10 2×3 4 4 3×3 1 2

6 5 02 A = – 5 2 , B = 5 , C = 1 8 10 20 , D = 4 4 4 , E = 1 1 8 84 444 7 0 –5 3 WORKED

e 2A + 3B − 4C

9 9 0 27

C = –4 3 , 27

7 18 – 12 – 4

D = 2 34 1 0 –2 7 5

and

0 5 –1 E = 2 3 9 to answer questions a to e. 6 4 –2

5 a

b

c

d

e

f

2 0 14 40 0 6 0 18 0 8 0 0 10 16 0 12 0 2 8 14 4 10 16 6 12 18 3 4 21 65 8 9 6 27 4 0 28 80 0 12 0 36 –1 4 –7 –2 5 8 –3 6 –9

a The order of D is: A 3×2 B 4×2 C 2×4 D 3×3 b Which one of the following cannot be calculated? A A+E B B+C C 4D D A+B c 2A + 3E is equal to: 3 16 7 A – 2 3 30 30 17 – 6

B

17 2 0 2 13 12 7 – 2

4 15 – 2 C 6 3 8 9 7 2

E 2×2 E 6(E − A)

2 16 3 D – 2 7 35 30 18 – 6

E

2 19 3 2 7 35 30 18 – 6

10 01

E

– 32 – 7 5 9

d 3C − 4B is equal to: A

– 32 25 2 9

B

8 –7 10 33

e The element e3 2 is equal to: A 9 B 4

C

C 3

– 31 0 5 19

D

D 7

147 1 –4 7 5 If C = 2 5 8 and D = 2 – 5 – 8 , calculate: 369 3 –6 9 a C+D b C−D c 2C d 2C + D

1

Example

Operations with matrices

E −1

e 2(C + D) f

−D

136 WORKED

Example

4

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

6 Solve the following matrix equations. a 3P = 6 0 9 –3

b Q+ 4 0 = 2 0 –5 6 14

2 0 3 –1

c 3M – – 2 0 3 = – 1 0 0 4 6 –1 2 3 –2

–1 0 1 d 2 3 –1

–2 0 –6 2

2 –2 0 4 – N = 6 –6 7 6 3 –3 5 –2

2 –6 – 1 – 12 –2 –1

7 Explain why the following matrix equation has no solution. Different orders 2A +

48 = 513 –4 0 –2 5 2

8 Write down the matrix representing the following maps in the form: Number of routes to Number of routes from Use alphabetical order for the sequencing of rows and columns. b J a B A 0131 D C

1022 3201 1210

0 0 1 1

K

0 0 1 2

1 1 0 3

1 2 3 0

L M 82 54 75 68 91 82

15 14 104 7 10 52 13 7 5 1 31 18 26 12 4 4 4 17 15 16 14 8 5 1 35 19 29 13 4 4 5 18 19 16

9 A mathematically inclined student has decided to keep a record of her test results in matrix form. Her results so far are Maths B tests: 82%, 75% and 91%; Maths C tests: 54%, 68% and 82%. Write these results in a 3 × 2 matrix. 10 Place the following sporting results in a suitable matrix format. a Brisbane Lions 15 goals 14 behinds 104 points defeated Geelong 7 goals 10 behinds 52 points. b Adelaide Crows have played 13 games for 7 wins, 5 draws and 1 loss; they have scored 31 goals for and 18 against; their points score is 26. Fremantle have played 12 games for 4–4–4; their goals are 17–15 and their points score is 16. 11 Adelaide Crows defeat Fremantle 4 goals to 1. Update the matrix in question 10b (note that 3 points are awarded for a win and 0 for a loss). 12 Write down any 2 × 2 matrices called A, B and C. Check if the following are true. a A + B = B + A True b (A + B) + C = A + (B + C) True c A − B = B − A False d 2A + 2C = 2(A + C) True

Chapter 3 Matrices

137

Multiplying matrices The sports results at Mathglen State High School were: Position House

1st

2nd

3rd

Hamilton

60

63

51

Leslie

71

64

74

Barnes

64

69

71

Cunningham

69

72

68

Position

Points

1st 2nd 3rd

5 3 1

5 To calculate the total points for each house, this matrix is multiplied by 3 since 5 points are awarded for first, 3 for second and 1 for third. 1 The result can be obtained using the following operations. Hamilton: Leslie: Barnes: Cunningham:

60 × 5 + 63 × 3 + 51 × 1 = 540 71 × 5 + 64 × 3 + 74 × 1 = 621 64 × 5 + 69 × 3 + 71 × 1 = 598 69 × 5 + 72 × 3 + 68 × 1 = 629

60 We can also write A × B = C, where A = 71 64 69

63 64 69 72

51 540 5 74 , B = 621 3 and C = 71 598 1 68 629

The order of A is 4 × 3, B is 3 × 1 and C is 4 × 1. Therefore, a 4 × 3 matrix multiplied by a 3 × 1 matrix gives a 4 × 1 matrix. Two matrices can be multiplied only if the number of columns of the first matrix equals the number of rows of the second matrix. In general, if A is of order m × n and B is of order n × p then A × B exists and its order is m × p. Such a matrix is said to be conformable where m × n multiplied by n × p results in a matrix of order m × p. The order of AB should be established before multiplying. The procedure for multiplying two 3 × 3 matrices is outlined below. a 11 a 12 a 13

b 11 b 12 b 13

If A = a 21 a 22 a 23 and B = b 21 b 22 b 23 a 31 a 32 a 33 then

b 31 b 32 b 33

138

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

a 11 × b 11 + a 12 × b 21 + a 13 × b 31

a 11 × b 12 + a 12 × b 22 + a 13 × b 32

a 11 × b 13 + a 12 × b 23 + a 13 × b 33

AB = a 21 × b 11 + a 22 × b 21 + a 23 × b 31

a 21 × b 12 + a 22 × b 22 + a 23 × b 32

a 21 × b 13 + a 22 × b 23 + a 23 × b 33

a 31 × b 11 + a 32 × b 21 + a 33 × b 31

a 31 × b 12 + a 32 × b 22 + a 33 × b 32

a 31 × b 13 + a 32 × b 23 + a 33 × b 33

The rows of the first matrix are multiplied by the columns of the second matrix. The sum of the products of the elements of row 1 multiplied by column 1 results in the row 1, column 1 element. The sum of the products of the elements of row 3 multiplied by column 2 results in the row 3, column 2 element.

WORKED Example 5 2 –1 A = 123 , B = 0 4 456 5 3 a Write down the order of the two matrices. b Which of these products exist? i AB ii BA c Write down the order for the products which exist. d Calculate the products which exist. THINK

WRITE

a

a A is a 2 × 3 matrix.

Matrix A has 2 rows and 3 columns. 2 Matrix B has 3 rows and 2 columns. b i AB is the product of a 2 × 3 and a 3 × 2 matrix so it exists. A and B are conformable. ii BA is the product of a 3 × 2 and a 2 × 3 matrix so it also exists. c i The product of AB is a 2 × 2 matrix. ii The product of BA is a 3 × 3 matrix. d

1

i

1

B is a 3 × 2 matrix. b

i AB exists since a 2 × 3 matrix multiplied by a 3 × 2 matrix results in a 2 × 2 matrix. ii BA exists since a 3 × 2 matrix multiplied by a 2 × 3 matrix results in a 3 × 3 matrix.

c

Multiply the rows of matrix d A by the columns of matrix B.

i AB is a 2 × 2 matrix. ii BA is a 3 × 3 matrix. 2 –1 i AB = 1 2 3 0 4 456 5 3 AB =

2

Simplify AB.

1×2 + 2×0 + 3×5

1 × –1 + 2 × 4 + 3 × 3

4×2 + 5×0 + 6×5

4 × –1 + 5 × 4 + 6 × 3

AB = 17 16 38 34

Chapter 3 Matrices

THINK ii

1

WRITE Multiply the rows of B by the columns of A.

2 –1 ii BA = 0 4 1 2 3 456 5 3 ii BA =

2

139

Simplify BA.

2 × 1 + –1 × 4 2 × 2 + – 1 × 5 2 × 3 + – 1 × 6 0×1+4×4 0×2+4×5 0×3+4×6 5×1+3×4 5×2+3×5 5×3+3×6

–2 –1 0 ii BA = 16 20 24 17 25 33

Note: In Worked example 5, AB is a 2 × 2 matrix but BA is a 3 × 3 matrix. In general, matrix multiplication is not commutative. That is, for two matrices A and B, AB ≠ BA. For the product AB we say that A is post-multiplied by B and B is pre-multiplied by A.

The identity matrix There is one circumstance in which matrix multiplication is commutative. Look at the following example.

WORKED Example 6 If A =

2 – 3 and I = 1 0 , calculate AI and IA. –5 4 01

THINK 1 A and I are both 2 × 2 matrices so both the products AI and IA exist and are of order 2 × 2. 2

Find AI using the procedure for multiplying matrices.

WRITE A 2 × 2 matrix multiplied by a 2 × 2 matrix results in a 2 × 2 matrix. AI = =

3

Find IA using the procedure for multiplying matrices.

2 –3 1 0 –5 4 0 1 2 –3 –5 4

IA = 1 0 2 – 3 0 1 –5 4 =

2 –3 –5 4

This example demonstrates the only case in which matrix multiplication is always commutative — that is, when AI = IA = A. Here, I is called the multiplicative identity matrix.

140

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

The multiplicative identity matrix, I, acts in a similar fashion to the number 1 when numbers are multiplied, where I is the multiplicative identity matrix. An identity matrix can be defined only for square matrices; that is, for matrices of order 1 × 1, 2 × 2, 3 × 3. The other feature of an identity matrix is that it has the number 1 for all elements on the leading diagonal and 0 for all other elements.

Leading diagonal

1 0 0 0 0

0 1 0 0 0

0 0 1 0 0

0 0 0 1 0

0 0 0 0 1

AI = IA = A where A is a square matrix and I is the multiplicative identity matrix. If A is not square (say it is 3 × 2), then A × I = A means I would have to be a 2 × 2 matrix because a 3 × 2 matrix multiplied by a 2 × 2 matrix results in a 3 × 2 matrix. But I × A = A means that I would be a 3 × 3 matrix because a 3 × 3 matrix multiplied by a 3 × 2 matrix results in a 3 × 2 matrix. However, I cannot be a 2 × 2 and a 3 × 3 matrix at the same time. Therefore I can be defined only for square matrices.

remember remember

1. In general, if A is of order m × n and B is of order n × p then A × B exists and its order is m × p; that is, A and B are conformable. 2. In general, for two matrices A and B, AB ≠ BA. 3. AI = IA = A where A is a square matrix and I is the multiplicative identity matrix.

3B WORKED

Example

24 1 A = 2 –3 , B = 1 1 , C = 6 8 , D = –2 4 , E = –2 3 –1 , I = 1 0 4 5 10 0 –4 2 01 01

2 – 3 1 1 – 8 – 21 14 15 4 5 1 0 28 13 – 24 – 30

a Write down the order of the six matrices. A (2 × 2), B (2 × 2), C (3 × 2), D (1 × 2), E (2 × 3), I (2 × 2) b Which of the following products exist? CA, DB, AE, AI, IA, IB, A2, EC i AC ii CA iii DB iv BD v AE vi AI 2 2 xi A xii EC vii IA viii IB ix EB x E (3 × 2), (1 × 2), (2 × 3), (2 × 2), c Write down the order of the products which exist. (2 × 2), (2 × 2), (2 × 2), (2 × 2) d Calculate those products which exist. 2 4 and N = 5 2 , calculate MN and NM. 10 20 8 26 – 5 10 – 4 12 04 –1 3 b Is matrix multiplication commutative? That is, does MN = NM? No

2 a If M =

d

20 14 – 4 18 – 8 2 – 3 44 22 2 – 2 –8 –8 6 4 5 4 5

5

Multiplying matrices

WORKED

Example

6

3 A = 2 – 1 , B = 2 0 , C = 5 – 2 , D = 3 2 , I = 1 0 and O = 0 0 0 3 0 –3 8 3 –8 5 01 00 Calculate the following products. a AB b AC c DO d DI e IB f BC g CD h CA i OI j ID 4 3 0 –9

2 –7 24 9

00 00

32 –8 5

2 0 0 –3

10 – 4 – 24 – 9

31 0 0 31

10 – 11 16 1

00 00

32 –8 5

141

Chapter 3 Matrices

4 a Calculate the following products. i

4 3 4 –3 5 4 –5 4

10 01

ii

–2 –3 –8 3 –5 –8 5 –2

10 01

iii

–1 –2 –5 2 –2 –5 2 –1

10 01

b What do you notice about all of the answers? All are I c What term could be given to these matrices? Multiplicative inverses 5 multiple choice Use the matrices below to answer questions a to d. 2 5 12 3 A = 3 2 , B = 2 –2 4 , C = 1 –3 , D = –2 0 2 , E = 5 2 , F = 01 1 36 –1 –3 0 4 4 1 –3 a Which one of the following products does not exist? A AD B AB C BC D FC E b The order of the matrix BD is: A 2×2 B 3×3 C 2×3 D 5×3 E c Which one of the following products gives a matrix of order 2 × 2? A BF B AB C DC D BC E d Which one of the following represents the matrix CE: 20 – 17 A –1 8 8 – 12

B

5 – 11 8 11 – 4 – 12

5 8 –4 C Does not D – 11 11 – 12 exist.

E

3 –2 4

CE 4×3 FD

20 – 1 8 – 17 8 – 12

6 The matrix below shows the number of wins, draws and losses for two soccer teams, the Sharks and the Dolphins. 10 2 5 8 72

3 1 0

Thus the Sharks have 10 wins, 2 draws and 5 losses. If 3 points are awarded for a win, 1 for a draw and 0 for a loss: a write down a 3 × 1 matrix for the points awarded Sharks have a total of 32 32 points. Dolphins have b use matrix multiplication to find the total points for the two teams. 31

a total of 31 points.

7 In Australian Rules Football, 6 points are awarded for a goal and 1 point for a behind. The scores in two games were: Southport 18–12 defeated Broadbeach 14–15 and Lions 10–14 defeated Eagles 9–16.

142 18 14 10 9

12 15 14 16

eBook plus Digital doc: WorkSHEET 3.1

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

The first number is for goals scored and the second is for behinds. a Write the results in a 4 × 2 matrix. 6 Southport 120, Broadbeach 99, Lions 74, Eagles 70 b Write down the 2 × 1 matrix for the points. 1 c Use matrix multiplication to find the total number of points scored by each team. 8 Two shops, A and B, are supplied with boxes of different brands of chocolates — Yummy, Scrummy and Creamy — as shown in this table: Yummy

Scrummy

Creamy

Shop A

20

20

10

Shop B

10

5

10

The cost of the boxes are Yummy $10, Scrummy $25 and Creamy $12. a Write down the costs in a 3 × 1 matrix. b Use matrix multiplication to find the total cost for each shop.

10 25 12

Shop A = $820, Shop B = $345

History of mathematics O L G A TA U S S K Y- T O D D ( 3 0 Au g u s t 1 9 0 6 – 7 O c t o b e r 1 9 9 5 ) During her life … Mt Everest is finally climbed. The Richter scale for measuring the strength of earthquakes is devised. Morse code is used by the Titanic when it sinks. Gandhi struggles to free India from British Rule. Olga Taussky-Todd worked in the fields of matrix theory and number theory. She was born in Olmütz, now part of the Czech Republic, but when she was three the family moved to Vienna and later to Linz. Her father died early so it became difficult for her to continue her studies. Her father, an industrial chemist, had encouraged her studies in mathematics. Olga went to the University of Vienna where she studied mathematics and chemistry. She completed a doctorate in 1930 with research into algebraic number fields. After completing her studies she was employed at the university of Göttingen as an assistant and worked with Helmut Ulm by editing his book on number theory. By 1932 Olga had been promoted to the position of tutor. In 1935 Olga moved to Cambridge where she undertook a research fellowship before moving to London in 1937 to take up a

teaching position. In London she met and later married Jack Todd. After the Second World War, the couple moved to America where Olga began work on the design of computers. In 1943, she moved to the Ministry of Aircraft where she conducted research into stability in matrices. This work encouraged her to look in more detail at matrix theory. Olga was awarded the Austrian Cross of Honour, which is Austria’s highest award; in 1964, she was named woman of the year by the Los Angeles Times. In 1970 she was awarded the Ford Prize for her publication on ‘The Sums of Squares’. In 1971 she was named Professor Emeritus at CalTech. Questions 1. What field of mathematics was Olga’s speciality? Matrix theory and number theory 2. What did Olga work on when she moved to America? Computer development 3. What award did Olga receive from the Austrian Government? Cross of Honour 4. Where was Olga a professor? Caltech Research Find out about the uses of matrices, especially in dynamic programming.

Chapter 3 Matrices

143

Powers of a matrix A logical extension of matrix multiplication is using the power of a matrix, where A1 = A A2 = AA A3 = A2A, and so on. In general form, An = An –1A, where n is a positive integer. But what dimension can matrix A have?

Matrix powers Investigate powers of matrices by completing the following steps. (Remember to use pronumerals for the elements of A, not constant values.) 1 a Let matrix A be any 3 × 2 matrix. b Find A2. x y x y 2 c What do you notice? Cannot multiply A × A if A is a 3 × 2 matrix b A = z w z w p q p q 2 a Let matrix A be any 2 × 2 matrix. 3 a A= x y z 2 w p q b Find A . Not conformable 3 a Let matrix A be any 2 × 3 matrix. x y z x y z b A2 = b Find A2. w p q w p q 2 multiply A × A if A is a c What do you notice? 2Cannot Not conformable a A= x y × 3 matrix z w 4 a What general conclusion can you make concerning the order of a matrix that is to be raised to a power? If a matrix is to be raised to a power it must be a square matrix. x y b A2 = x y b Justify your conclusion by referring to the dimensions of matrices involved z w z w in a product. 2 A2 = x + yz xy + yw From the above investigations we can conclude that A × A must be conformable; 2 zx + zw zy + w that is, the number of columns of the first factor in the product should be the same as the number of rows in the second factor. That is, A must be a square matrix where n × n is multiplied by n × n to get A2. Hence powers of matrices are only defined for square matrices.

1 a A=

x y z w p q

WORKED Example 7 If A =

1 3 , find: –1 2

a

A2

b

A3

THINK

WRITE

a Write the power as a product.

a A2 = A × A A2 =

1 3 –1 2

1 3 –1 2

A2 = – 2 9 –3 1 Continued over page

144

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

THINK

WRITE

b Write the power as the product of lesser powers.

b A3 = A2A A2 = – 2 9 –3 1

1 3 –1 2

A2 = – 11 12 –4 –7

remember remember

1. The power, n, of matrix A, in general form, is An = An – 1A, where n is a positive integer. 2. Powers of matrices are only defined for square matrices; that is, A has to be a square matrix to obtain An.

3C 1 If A = 2 – 1 , find: 0 0 7 2 a A

Powers of a matrix

WORKED

Example xample

4 –2 0 0

1 0 0 0 1 0 0 0 1

1 0 0 8 9 0 2 4 1

–1 –1 –1 1 1 2 –1 –1 –1

b A3

8 –4 0 0

1 0 0 2 If A = 0 1 0 , find: 0 0 1 2 a A

3

b A

1 0 0 0 1 0 0 0 1

1 0 0 3 If A = 2 3 0 , find: 0 1 1 2 a A

b A3

1 00 26 27 0 10 13 1

1 1 0 4 If A = – 2 – 2 – 1 : 1 1 0 a find A2

b confirm that A2A = AA2

c

A4

c

n

A

16 – 8 0 0

1 0 0 0 1 0 0 0 1

Chapter 3 Matrices

145

Palm

Indoor plant

Hanging basket

Geranium

0 1 1 2 0 2 5 5 2 3 0 2 1 1 0 0 1 0 3 2 Bank

T=

Hotel

2 0 0 3 Patio

Office

Hanging Camellia Geranium basket Fern

ii

Type and number:

Indoor plant

Palm

iii

Cost: C =

Camellia

Fern

22 15 8 12 10 18

Applications of matrices

iii

Patio

CD =

Office Bank Hotel

163.80 147.60 203.40 370.80

1 A garden supplier provides live plants for displays in 5 penthouse patios, 7 office foyers, 3 banks and 4 hotels. The plants in each different type of display are listed below. • The patio displays consist of 2 ferns, 1 camellia, 1 geranium and 2 hanging baskets. • The office foyer displays have 1 palm, 1 geranium, 3 hanging baskets and 2 indoor plants. • The bank displays have 1 palm, 3 camellias and 5 indoor plants. • The hotel displays have 2 palms, 3 ferns, 2 camellias, 2 hanging baskets and 5 indoor plants. The wholesale cost of each plant bought by the supplier is: ferns $22, palms $18, geraniums $8, camellias $15, hanging baskets $12 and a variety of indoor plants that cost $10 on average. The supplier needs to be able to use this information to calculate costs of displays, number of plants required and profits, but in this form, the information is difficult to handle. a Develop matrices to display the following information (labels outside the matrices will help clarify the meaning of the elements): Patio Office Bank Hotel iii the number of displays supplied to each type of venue Venue: V = [ 5 7 3 4 ] iii the number and variety of plants used in each display iii the cost of each type of plant. Quantity: Hanging Indoor b Use matrix operations to determine the following: Fern Camellia Geramium basket plant Palm Q = [ 22 22 12 39 49 18 ] iii the quantities of each plant needed to fill the orders iii the supplier’s total outlay to provide the displays TC = $2192 iii the charge for each type of display if the supplier adds 80% profit to the (iii) cost. 2 A home builder advertised three designs of ‘Ownit Homes’ to entice people to buy rather than rent their home — the ‘Taps’ for $155 per week, the ‘Avalon’ for $203 per week and the ‘Torana’ for $238 per week. The weekly payments were based on finance available from a public finance company. Ownit Homes received orders for 10 ‘Taps’ homes, 8 ‘Avalon’ homes and 12 ‘Torana’ homes. The materials (given in units as stated in their building guide) required for each home are listed below: • The ‘Taps’ home requires 9 units of steel, 11 of timber, 6 of glass, 7 of paint and 20 of labour. • The ‘Avalon’ home requires 12 units of steel, 14 of timber, 15 of glass, 12 of paint and 25 of labour. • The ‘Torana’ home requires 14 units of steel, 12 of timber, 12 of glass, 16 of paint and 24 of labour.

35] 65

Bread

150 45

Sugar rolls

1.20 1 1.20 1.50 1.20 Milk

Sugar

Flour

Shortening

Buns

Pastries

Cakes

A=

Bread loaves

S2 S1

1 0.80 1 1.20 1.20 Eggs

1 0 1 Q= 0 1

Milk Shortening

0.25 0.25 1 0.33 1 0.25 – 2 – 3

Sugar Flour

4 3 3 1 2 1 0 4 1 0

Eggs

Sugar rolls

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

To reduce costs all materials are purchased from one supplier. The prices per unit are steel $660, timber $1140, glass $1020, paint $660 and labour is priced at $1128 per unit. Use matrix methods to obtain the following information: a the amount of money the bank would receive per week from the repayments on these homes $5018 b the total cost of raw materials for all the constructions. $1 661 420

N=

[15

Cakes

Pastries

Buns

146

260 eggs, 780 units of flour, 198.75 units of sugar, 142.7 units of shortening and 95 units of milk Sugar rolls $11.30, Bread loaves $5.15, Cakes $20.50, Pastries $4.15, Buns $13.30

3 A small bakery sells 5 main items: sugar rolls, bread, cakes, pastry and buns. The major ingredients (given in applicable units) required to make one of each item are listed below. • Sugar rolls (1 dozen) require 1 egg, 4 units of flour, 0.25 of sugar, 0.25 of shortening and 1 of milk. • Bread (1 loaf) requires 3 units of flour, 0.25 of shortening. • Cake (1) requires 4 eggs, 3 units of flour, 2 of sugar, 1 of shortening and 1 of milk. • Pastry (1) requires 1 egg, 1 unit of flour, 0.33 of shortening. • Buns ( 1 dozen) require 2 units of flour, 3 of sugar, 1 of shortening and 1 of milk. Two suppliers (Supplier 1 and Supplier 2) provide quotes for the ingredients, given as ordered pairs with prices given in dollars: eggs (1, 1.20), flour (0.8, 1), sugar (1, 1.20), shortening (1.20, 1.50) and milk (1.20, 1.20). For one office function the following orders were received: 15 dozen sugar rolls, 150 loaves of bread, 45 cakes, 65 pastries and 35 dozen buns. a Represent all the above information in matrix form taking into account ingredients, orders, suppliers’ quotes. b Use these matrices to provide a list of the amounts of the ingredients required to fill the orders for the function. c Which supplier provides the cheapest total quote? What savings are made by using this supplier? Supplier 1 is cheaper by $290.56. d Provide a list of selling prices (to the nearest 5 cents) if a 90% markup on the cost prices is used to fix the price. e Calculate the total takings based on this information from part d. $2599.75

Multiplicative inverse and solving matrix equations In question 4 of exercise 3B, you should have found that the product of the matrices was I. This means that one matrix is the multiplicative inverse of the other. We use the symbol A−1 for the multiplicative inverse of A. If AA-1 = A-1A = I, then A−1 is called the multiplicative inverse of A. In working with numbers, a similar result would be 7 × 1--- = 1 or 4--- × 5--- = 1. Numbers 7 5 4 such as these are called reciprocals or multiplicative inverses of each other.

Chapter 3 Matrices

147

WORKED Example 8 If A = 4 1 and B = 3 – 1 , find AB and hence write down the multiplicative inverse 63 –6 4 of A. THINK 1

WRITE

AB will be a 2 × 2 matrix since A and B are both 2 × 2 matrices.

AB = 4 1 3 – 1 6 3 –6 4 = 60 06

2

3

6 0 = 6 1 0 = 6I since 6 is a common factor 06 01 of each element.

AB = 6I

To produce I we need to multiply both sides by 1--- .

A ( 1--6- B ) = I

6

So A−1 = 1--- B 6

= 4

Since A ( 1--6- B ) = I , the inverse of A is 1--- B. 6

=

Inverse of a matrix

3 --6

– 1--6-

– 6--6-

4 --6

1 --2

– 1--6-

–1

2 --3

Consider matrix A, a 2 × 2 matrix, such that A = a b . If a multiplicative inverse of A c d exists, then A × A–1 = I. If A–1 exists, let A–1 = x y u v That is, AA–1 = I LHS = a b c d

x y u v

LHS = ax + bu ay + bv cx + du cy + dv LHS = RHS LHS = 1 0 0 1

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Equating terms of the two matrices: [1] ax + bu = 1 [2] cx + du = 0 [3] ay + bv = 0 cy + dv = 1 [4] Solving this system of simultaneous equations: Use [1] and [2] to eliminate x by multiplying equation [1] by c and equation [2] by a. [5] acx + bcu = c acx + adu = 0 [6] Equation [6] minus equation [5] gives: adu − bcu = −c u(ad − bc) = −c –c u = -----------------This will replace u in A–1. ad – bc Continue in a similar fashion to arrive at: –b d a y = -----------------x = -----------------and v = -----------------ad – bc ad – bc ad – bc d ----------------ad – bc Therefore A = –c ----------------ad – bc –1

–b ----------------ad – bc a ----------------ad – bc

1 = ------------------ d – b where ad − bc ≠ 0 ad – bc – c a If ad – bc = 0 then this scalar is undefined, therefore A–1 does not exist. That is, there is no matrix that, when multiplied by A will yield I, the identity matrix. If A has no inverse then it is said to be singular. There is a relationship between A and A−1 which is outlined below. If A is the matrix a b , proceed as follows. c d 1. Swap the elements on the main diagonal of A

on the other diagonal by −1

and multiply the elements

. This gives the matrix

d –b . –c a

2. Evaluate ad – bc. 1 3. Divide each element by (ad − bc) (or multiply by ----------------------- ). ( ad – bc ) These steps demonstrate a clear method for finding the multiplicative inverse of a matrix. 1 The inverse of A = a b is A –1 = ------------------ d – b . ad – bc – c a c d The number (ad − bc) is called the determinant of the matrix A and is written as det A or |A|.

Chapter 3 Matrices

149

Note: Only square matrices have inverses. We will be concerned only with the inverse of 2 × 2 matrices at this stage in this course.

WORKED Example 9 If C = 2 – 3 find C −1. 1 5 THINK

WRITE 1 C –1 = ------------------ d – b ad – bc – c a

1

Write the general form of C and the general form of its inverse.

C= a b c d

2

Swap the elements on the main diagonal of C.

1 C –1 = ------------------------------------------- 5 3 ( 2 × 5 ) – ( –3 × 1 ) –1 2 1 = ----------------------- 5 3 10 – ( – 3 ) – 1 2

5 2 Multiply the elements on the other diagonal of C by −1. 3 –1 3

1 C –1 = ------ 5 3 13 – 1 2

Write down the inverse of C.

We can check that CC –1 = I and C –1 C = I . C–1 is the multiplicative inverse of C if C × C –1 = C –1 × C = I. CC –1 =

1----13

2 –3 5 3 1 5 –1 2

=

1----13

13 0 0 13

= I

C –1 C =

1----13

=

1----13

5 3 2 –3 –1 2 1 5 13 0 0 13

= I

Singular matrices Matrices for which the determinant equals 0 do not have an inverse, since undefined. Such matrices are called singular matrices.

1 --0

is

If det A = 0 then A is singular and an inverse does not exist. There are two special types of singular matrices: nilpotent and idempotent. A square matrix A is nilpotent if A2 = O where O is the zero matrix. The zero matrix is a square matrix with all elements equal to zero. For example, the 2 × 2 zero matrix is

00 . 00

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A square matrix A is idempotent if A2 = A. The only non-singular idempotent matrix is the identity matrix.

WORKED Example 10

Show that a

6 – 3 is nilpotent 12 – 6

b

5 2 is idempotent. – 10 – 4

THINK a

1

WRITE a A2 =

Nilpotent means that A2 = O. Find A2.

6 –3 12 – 6

6 –3 12 – 6

= 00 00

b

2

State your conclusion.

1

Idempotent means that A2 = A. Find A2.

2

A2 = 0; therefore A is nilpotent. b A2 =

5 2 – 10 – 4

=

5 2 – 10 – 4

5 2 – 10 – 4

A2 = A; therefore A is idempotent.

State your conclusion.

Further matrix equations Matrix equations of the type AX = B may be solved by using the properties of multiplicative inverses. A matrix equation AX = B is similar to the equation 3x = 7. To solve this we would divide both sides of the equation by 3 (or multiply by 1--- ). To solve the matrix equation 3 we multiply both sides by A−1. Since the order of multiplying matrices is important we must be careful in which position we multiply by the inverse. 1. For AX = B Pre-multiply by A−1:

A−1AX = A−1B

or

IX = A−1B since A−1A = I X = A−1B since IX = X

2. For XA = B Post-multiply by A−1: XAA−1 = BA−1 or

XI = BA−1 since AA−1 = I X = BA−1 since XI = X

1. If AX = B, then X = A−1B. 2. If XA = B, then X = BA−1.

Chapter 3 Matrices

151

Note: A–1 cannot be ‘inserted’ between 2 matrices. It can either pre- or post-multiply A on one side of a matrix equation.

WORKED Example 11 A = 1 2 and B = 2 5 03 –2 1 Find X if: a AX = B b XA = B. THINK a

1

WRITE

We require A−1 so first calculate det A.

a

A= 1 2 03

∴ det A = 3 − 0 = 3

b

2

Write down A−1.

A−1 =

3

Write the equation.

AX = B

4

Pre-multiply both sides of the equation by A−1.

5

Remember A−1A = I and IX = X.

6

Calculate the product of A−1 and B.

1

Write the equation.

2

Post-multiply both sides of the equation by A−1.

3

Calculate the product of B and A−1 using A−1 which was found in part a.

--13

3 –2 0 1

A−1AX = A−1B

b

X=

1 --3

3 –2 2 5 0 1 –2 1

=

1 --3

10 13 –2 1

XA = B X = BA−1 =

1 --3

2 5 3 –2 –2 1 0 1

=

1 --3

61 –6 5

In part a of Worked example 11 both sides of the equation were pre-multiplied by A−1; in part b both sides were post-multiplied by A−1. Remember that the matrix and its inverse must be next to each other so that AA−1 = I. Fractional scalars should be left outside the matrix unless they give whole numbers when multiplied by each element.

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remember remember

1. If AA−1 = A−1A = I, then A−1 is called the multiplicative inverse of A.

1 2. The inverse of A = a b is A−1 = ------------------ d – b ad – bc – c a c d The number (ad − bc) is called the determinant of the matrix A and is written as det A or | A |. 3. If det A = 0 then A is singular and an inverse does not exist. 4. (a) If AX = B, then X = A−1B. (b) If XA = B, then X = BA−1.

3D WORKED

Example

8 eBook plus Digital doc: SkillSHEET 3.2 Inverse of a 2 × 2 matrix

1 If A =

Multiplicative inverse and solving matrix equations

1 a A –1 = --- B 6

4 1 and B = 1 – 1 , find AB and hence write down: –2 1 2 4

a the inverse of A

1 b B –1 = --- A 6

b the inverse of B.

2 If M = 2 6 and N = – 1 – 6 , find MN. Hence write down M −1 and N −1. 0 –1 0 2 3 Calculate the determinants of the following matrices. a

A = 2 3 5 10

d D = 43 41 WORKED

Example

9

1 AB = 6 1 0 01

5

b B = –2 –3 4 0

−8

E = –2 –1 –3 –5

e

12

7

c

C = 2 6 0 –1

f

F = 2 –1 6 4

5 multiple choice Using the matrices below, select the correct answer in questions a to d.

b R

C −10

D −2

E 8

is equal to:

A − 1--8

1D − -----

40

2 –6 4 8 2 –6 4 8

B

E

1 --8

8 6 –4 2

1----40

2 –6 4 8

14

–1

1 = – --- M 2

1 b ------ 0 3 12 – 4 – 2 1 c --- 1 6 2 0 –2

P = 4 3 , Q = 2 – 3 and R = 8 6 2 –1 –1 0 –4 2

−1

N

1 4 a --- 10 – 3 5 –5 2

4 Write down the inverses of each matrix in question 3.

a Det P is equal to: A 10 B 2

2 MN = – 2 1 0 , 01 –1 1 −2 M = – --- N , 2

C

1 -----40

86 –4 2

1 d --- – 1 3 8 4 –4 1 e --- – 5 1 7 3 –2 1 f ------ 4 1 14 – 6 2

153

Chapter 3 Matrices

c

Det (PQ) is equal to: A −30 B 10 d If QX = R, then X is equal to: – 12 6 0 10

A − 1--3

1 7 a --- 1 6 2 0 –2 1 12 b --4–2 0 c

3

12 6 0 – 10

3

Answers will vary.

7 C = 2 6 , D = 0 –2 0 –1 2 1 Find: a C −1

1 d --- 1 2 8 – 2 – 12 1 e --- – 11 2 8 40

b D−1

c

d (CD)−1

CD

e C −1D−1

f

D−1C −1

8 Explain why these matrices do not have an inverse.

1 f --- 1 2 8 – 2 – 12 SLE 13: Research nilpotent matrices where the matrix, A, is nilpotent if it has the property A 2 = O, (O is the zero matrix).

C − 1---

6 28 2 8

E − 1---

6 Write down a 2 × 2 matrix which is singular.

12 2 –2 –1

SLE 14: Research idempotent matrices where the matrix, A, is idempotent if it has the property A2 = A .

E −20

D 30

12 6 0 – 10

1 --3

B

6 28 2 8

1 --3

D

C −10

a D = 21 42

9 If A = a AB WORKED

Example

10a

WORKED

Example

10b

WORKED

Example

11

D − det = 0 b

E = –2 –4 5 10

c E − det = 0

4 2 –8 –4

b

6 –3 10 – 5

b

– 10 20 – 5 10

c

c

AX =

2 –4

b BX = 15 7

10 50

1 b --- – 5 5 2 14 – 8 1 c --- – 6 2 6 –6 4 1 d --- 18 23 2 – 12 – 16 1 e ------ 78 103 30 – 24 – 34

d AX = C h A−1BX = C

x . y

f

10 01

1 g ------ 1 – 5 15 2 5

Solve these matrix equations. a

6 –9 4 –6

Check with your teacher.

–4 4 –5 5

3 4 , B = 6 1 and X = –1 2 21

1 b --- – 2 – 8 8 1 0

0 8 –1 –2

1 a --- – 31 – 22 2 24 18

Check with your teacher.

12 Let A = 2 3 , B = 5 5 , C = 0 – 1 . Find X if: 45 –2 1 6 6 a AX = B b XA = B c XC = A e ABX = C f CX = C g XB = I 13 A =

9

12

11 Show that the following matrices are idempotent. a

F − Not a square matrix

a

4 0 and B = 0 2 , find: –1 1 –1 0 −1 b (AB)

10 Show that the following matrices are nilpotent. a

14 F = 25 36

a

2 –1

b

2 3

1 h ------ – 132 – 114 15 186 162 or 1 – 44 – 38 --5 62 54

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

14 Find the value of x and y by solving these matrix equations. a

34 –1 5

c

–4 2 3 –2

x y

= –2 7 x y

=

x = −2, y = 1

14 x = −2, y = 3 – 12

b

2 3 4 –1

d

–1 3 2 –3

x y

= 8 2 x y

= 5 2

x = 1, y = 2

x = 7, y = 4

The transpose of a matrix The transpose of matrix A is A′, where A = a b and A′ = a c . The transpose of c d b d a matrix is an interchange of rows and columns (row 1 becomes column 1 and so on). Consider the following laws that apply to the transpose of matrices A and B: 1. (A¢)¢ = A 2. (A + B)¢ = A¢ + B¢ 3. (kA)¢ = kA¢ 4. (AB)¢ = B¢A¢ 5. AA¢ is a symmetric matrix [that is, (AA¢)¢ = AA¢]. The proofs of these laws are given as problems in exercise 3E.

remember remember 1. When required to prove a statement is true: (a) do not assume it is true and use the statement in your proof (b) work only one side of the statement at a time, not both together (c) do not use actual constant values for the elements, use pronumerals only. 2. If you are asked to show a statement is true, you are expected to use actual values as given.

3E

The transpose of a matrix Check with your teacher.

1 Prove that for any 2 × 2 matrix A, (A′)′ = A. 2 Show that for A = 1 3 and B = 0 – 1 , (A + B)′ = A′ + B′. 0 2 1 2 3 Show that for k = −2 and A = – 3 1 , (kA)′ = kA′. 0 1 4 Show that for A = 3 – 4 and B = 0 1 , (AB)′ = B′A′. 1 2 1 0 5 Show that for any 2 × 2 matrix, AA′ is symmetrical.

Chapter 3 Matrices

155

Applications of matrices Application 1: Simultaneous equations As we saw in questions 13 and 14 from exercise 3D, matrices may be used to solve linear simultaneous equations. The pair of equations may be written in the form AX = B x and B is y the matrix of the numbers on the right-hand side of the simultaneous equations. A is called the coefficient matrix. For example, the simultaneous equations: where A is the matrix of the coefficients of x and y in the equations, X =

ax + by = u cx + dy = v can be expressed as the matrix equation: a b c d

x = y

u v

which is of the form AX = B. Here a b is called the coefficient matrix, c d u the constant matrix. v As we have seen, this equation can be solved by using:

variable matrix and

A−1AX = A−1B X = A−1B

WORKED Example 12

Solve 3x − y = 16 and 2x + 5y = 5 by matrix methods. THINK

WRITE

1

Write the simultaneous equations under each other making sure the variables are in corresponding positions.

3x − y = 16 2x + 5y = 5

2

Write the matrix equation.

AX = B 3 –1 2 5

3

Rearrange the equation in general form so that X is the subject.

x y

= 16 5

A–1AX = A–1B IX = A–1B X = A–1B Continued over page

x y

the

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

THINK

WRITE

4

Calculate A−1.

5

Multiply A–1 by B.

1 A–1 = -----17

5 1 –2 3

1 X = -----17 1 = -----17 x y

Write the answers in the form x = . . . and y = . . . Note: The solution should be verified by substituting x = 5 and y = −1 into the original equations.

85 – 17

5 –1

∴ x = 5 and y = −1

Graphics Calculator tip! Solving matrix equations Most graphics calculators provide a facility for calculating inverses of matrices. To solve the equations in Worked example 12, follow these steps.

For the Casio fx-9860G AU For operations on matrices, press MENU to display the MAIN MENU. Use the arrow keys to highlight RUN-MAT. Select it by pressing EXE . 1. Set up the dimensions for matrix A. (a) Press F1 ( screen.

▼

6

=

5 1 16 –2 3 5

MAT) to enter the matrix editing

(b) Highlight Mat A and press EXE or F3 (DIM). (c) Specify the number of rows, 2 in this case, and then press EXE . (d) Specify the number of columns, 2 in this case, and then press EXE . 2. Press EXE again to display the 2 × 2 array for matrix A. 3. Enter the values for the elements of matrix A, pressing EXE after each number.

Chapter 3 Matrices

4. Exit the Matrix input screen by pressing EXIT .

5. Repeat steps 1 to 4 to create matrix B.

6. Press EXIT again to return to the MAT screen.

7. Pre-multiply matrix B by the inverse of matrix A. (a) Press OPTN then F2 (MAT) to bring up the matrix menu. (b) Press F1 (Mat) then ALPHA [A] and then SHIFT [x–1] to specify the matrix A–1. Press F1 (Mat) then ALPHA [B] to specify matrix B. 8. Press EXE to obtain the answer screen. (Press EXIT to leave.)

For the TI-Nspire CAS 1. Open a new Calculator document (press / N and select 1: Add Calculator). Press k to access the catalog. Select Option 5 (by pressing 5) then highlight the 2-by-2 matrix symbol.

2. Press ·. Use the arrow keys to move from one element to the next to fill in the 2 × 2 matrix.

3. Use the arrow keys to move the cursor outside the matrix, to the right. Press the power key (l) and type in the index (–1).

157

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

4. Press the right arrow to bring the cursor to the base then press the multiplication key (r). The multiplication symbol appears as a dot on the screen.

5. Press k to access the catalog. Select Option 5 and highlight the 2-by-1 matrix symbol.

6. Press ·. Fill in the values for the 2 × 1 matrix.

7. Move the cursor outside the matrix, then press · to obtain the answer.

Application 2: Summarising information We have already seen how matrices may be used to summarise information such as town–road connections. Information which can be summarised in tabular form may also be presented as a matrix.

WORKED Example 13

In a large country town, there are three major supermarkets. Customers switch from one to another due to advertising, better service, prices and for other reasons. A survey of 1000 customers has revealed the following information for the past month. Best Buys started with 40% of the market; 90% of its customers remained loyal to Best Buys but 5% changed to Great Groceries and 5% to Super Store. Great Groceries started with a 36% market share; 85% remained loyal, 10% transferred to Best Buys and 5% to Super Store. Super Store started with 24% of the customers; it lost 15% to Best Buys and 5% to great Groceries, but 80% remained. Summarise the information in matrix form and calculate the new market shares.

159

Chapter 3 Matrices

THINK The information may be summarised in a 3 × 3 matrix with the rows representing retention rates and gains and the columns representing retention rates and losses. This may be called a transition matrix. Row 1 shows that Best Buys retains 90% of its customers, gains 10% of Great Groceries’ customers and gains 15% of Super Store’s customers. Column 1 indicates that Best Buys retains 90% of its customers, loses 5% to Great Groceries and loses 5% to Super Store. Note that each column totals 100%.

2

Write the initial market shares as a 3 × 1 matrix. This information is found as the market share at the beginning of the month. Note: The values total 1.

3

The new market share will be the transition matrix, converted to decimal numbers, multiplied by the market share matrix.

4

Express the new market shares as percentages. Check the values add up to 100%.

Retention rates and losses (%) Best Buys Retention rates and gains (%)

1

WRITE

Best Buys

Great Super Groceries Store

90

10

15

Great Groceries

5

85

5

Super Store

5

5

80

0.40 The initial market share matrix is 0.36 0.24

0.432 0.90 0.10 0.15 0.40 0.05 0.85 0.05 0.36 = 0.338 0.230 0.05 0.05 0.80 0.24 The new market shares are Best Buys 43.2%, Great Groceries 33.8% and Super Store 23.0%.

remember remember 1. Matrices may be used to solve simultaneous equations: ax + by = u cx + dy = v The pair of equations may be written in the form AX = B, where A = a b , X = c d

x y

and B =

u . v

2. Matrices can also be used to summarise information which is in table form and solve related problems; however, care must be taken in setting up the matrices.

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3F

Applications of matrices

In the following exercise solve all problems manually then use a graphics calculator wherever appropriate to check your solutions. WORKED

Example

12

2 Consider these two pairs of simultaneous equations: i 3x − 2y = 4 ii 3x − 2y = 6 6x − 4y = 12 6x − 4y = 12

eBook plus Digital doc: SkillSHEET 3.3 Using matrices to solve linear equations

d i

y 0 –1

4– 3

2

3

1 Solve these simultaneous equations by matrix methods. a 2x − 3y = 13 and x + 2y = 3 (5, −1) b 3x + y = 9 and −2x + 5y = −6 (3, 0) c −x + 4y = −2 and x − 5y = 0 (10, 2) d 6x + 7y = 0 and 4x − 3y = 0 (0, 0) e 4x + y = 20 and x − y = 0 (4, 4) f 3x − 2y = 0 and x − y = 1 (−2, −3)

x

in ii there is only one line.

–2

3 multiple choice

–3

3x − 2y = 5 y + 2x = 8

Consider the simultaneous equations: ii

y

Both lines

0

2 x

–3

a and b Answers will vary.

a Show by algebraic means that the simultaneous equations in i have no solution. b Show that the simultaneous equations in ii have an infinite number of solutions. c Write the equations in matrix form and explain how these facts are related to the determinant of the matrix of the coefficients. det = 0 d Draw, on two sets of axes, graphs of the two lines in each of i and ii. e Explain how the graphs are related to parts a and b. In i there are parallel lines;

a The coefficient matrix is: A

3 –2 1 2

B

5 8

32 –2 1

C

D

b The solution to the simultaneous equations is: A x = 2, y = 3 B x = 3, y = 2 D x = −2, y = 2 E x = 4--- , y = 7--3

31 –2 2

3 –2 2 1

E

C x=

13 -----4

,y=

19 -----8

3

4 multiple choice In an alternative Australian Rules Football game, a team gains x points for a goal and y points for a behind. In one game Cairns obtained 66 points by scoring 10 goals and 8 behinds and Townsville obtained 70 points from 12 goals and 5 behinds. a This information is represented by which of the following matrix equations? A

8 10 5 12

D

8 5 10 12

x y

= 66 70 x y

= 66 70

b The value of x − y is: A 5 B 4

B

10 12 8 5

x y

= 66 70

E

12 10 5 8

x y

= 70 66

C 6

D 3

C

10 8 12 5

x y

E 2

= 66 70

Chapter 3 Matrices

161

5 The sum of two numbers is 20 and their difference is 12. Find the numbers by setting up simultaneous equations and solving by matrix methods. 16, 4 6 In a factory, two types of components are processed on two separate machines. The respective processing times on the first machine are 18 minutes and 21 minutes, while for the second machine the times are 4 minutes and 42 minutes. How many of each type of component, per machine, should be processed in an 8-hour shift so that both machines are fully occupied and the output of each machine is the same? 15, 10 WORKED

Example

13

7 In a swimming competition, 5 points are awarded for first place, 3 for second, 2 for third and 1 point for an unplaced result. The top competitors’ results were:

Name

Number of races competed in

First placings

Second placings

Third placings

Rania

6

2

—

2

Patricia

4

4

—

—

Anh

5

3

2

—

Mayssa

6

1

3

2

Rachel

6

2

3

—

Place the results and points in suitable matrices and use matrix multiplication to find the highest points scorer. Anh 8 Cyril’s circus arrived in town last week and during the week the number of adults, children and pensioners attending the circus was recorded for the first five shows (see table below).

Adults Children Pensioners

eBook plus Digital doc: WorkSHEET 3.2

Monday

400

200

20

Tuesday

450

350

50

Wednesday

370

410

45

Thursday

290

380

70

Friday

420

530

65

The entry cost is $20 for adults, $6 for children and $5 for pensioners. Set up the information in suitable matrices to find the total takings for the first five shows. $51 070

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Matrix multiplication using a graphics calculator Worked example 13 may be solved using a graphics calculator as follows.

For the Casio fx-9860G AU 1. Enter the 3 × 3 transition matrix as matrix A. (For more details, see the previous graphics calculator tip on page 156.) (a) Press F1 ( MAT) to enter the matrix editing screen. (b) Set the dimensions of A to 3 × 3 and press EXE . (c) Enter the values of A. ▼

162

2. Press EXIT . 3. Enter the 3 × 1 market share matrix as matrix B. (a) Scroll down to Mat B and press EXE . (b) Set the dimensions to 3 × 1 and press EXE . (c) Enter the values of B. 4. Exit the Matrix input screen by pressing EXIT . Press EXIT again to return to the MAT screen. 5. Multiply the matrices A and B (and store as matrix C). (a) Press OPTN then F2 (MAT) to bring up the matrix menu. (b) Press F1 (Mat) then ALPHA [A] to specify matrix A. Press F1 (Mat) then ALPHA [B] to specify matrix B. Press EXE to obtain the answer screen. Alternatively, press SHIFT [{], then Æ then F1 (Mat) and ALPHA [C] SHIFT [}] to store as matrix C. 6. Press EXE to obtain the answer screen. (Press EXIT to leave.)

For the TI-Nspire CAS 1. Open a new Calculator document (press / N and select 1: Add Calculator). Press k to access the catalog. Select Option 5 then highlight the m-by-n matrix symbol.

Chapter 3 Matrices

2. Press ·. Create the matrix with Number of rows: 3 and Number of columns: 3, pressing e to move from one box to the next. Press e until OK is hightlighted.

3. Press · to display the matrix template. Fill in the values in the matrix using the arrow keys or the tab key to move from one element to the next.

4. Move the cursor to the right of the matrix and press the multiplication (r) key. Press k to access the catalog and highlight the m-by-n matrix symbol.

5. Press · then select Number of rows: 3 and Number of columns: 1.

6. Highlight OK and press ·. Fill in the values in the 3 × 1 matrix.

7. Move the cursor outside the matrix and press · to obtain the answer.

163

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Questions Use a graphics calculator to find A × B for each of the following: 133

1 A= 5 0 2,

B=

628 2.5 2 A = 9.2 6.6 3.7 2 0 3 A= 5 –5 4

SLE 10: Investigate the use of matrices in dominance problems such as in predicting the next round results (rankings) for the national netball competition.

6.1 0.3 , 0.7 4.6 3 1 –3 1 4

0 –1 9 1 4

9 2 5

30 AB = 55 98

B = 3.7 0.4 9.4 5.1 5.9 2.2

–2 4 –2 9 4

7 –7 6, 6 4

Dominance matrices Have you ever wondered how tennis players are seeded or ranked? It obviously has something to do with their performance against past opponents. In a knock-out competition, one loss and you are out of the competition. Only the winners continue to play. Dominance matrices are often used to determine player rankings. The following investigation will explain how matrices are used to establish the seedings or rankings of players in round-robin situations where each player plays every other player, thereby creating a more just system of ranking.

–3 5 B= 4 –1 9

8 11 –7 –2 2

40.36 AB = 35.57 27.99 37.15

36.99 5.45 6.77 28.62

74 – 66 AB = 62 69 56

67 –4 – 40 – 42 48

36.92 87.14 63.58 44.9

Chapter 3 Matrices

165

Dominance matrices — another application of matrices Consider 4 players Alan, Brian, Carlo and Denis (A, B, C, D), who on past performances have shown that A defeats D and B, D defeats B, C defeats A and D, and B defeats C. This situation can be represented on a digraph — a network diagram that has arrows on the edges, where A → B indicates that A defeats B. A B

C D

The information from the digraph can be converted into matrix form (a dominance matrix) as below: defeats A M= B C D

A 0 0 1 0

B 1 0 0 1

C 0 1 0 0

D 1 0 1 0

where 1s are used to indicate ‘defeats’ and 0s to indicate otherwise. Obviously A can’t defeat A so a ‘0’ is used along the leading diagonal. Notice also that: (1) there are as many 1s as there are paths (2) corresponding elements occur on either side of the leading diagonal. That is, if A defeats B (1), then a 0 will be stored in the B defeats A element on the opposite side of the leading diagonal. When the elements of each of the rows are added they yield a dominance vector, showing how many players each has defeated. A V= B C D

2 1 2 1

This result can be readily checked from the original digraph by counting the number of arrows out of each node. Note, from now on the row/column labels will be omitted. It can be seen from this information that A and C are ranked equally, and B and D are ranked equally; this can be written as A   B       C  D 

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So we still need to distinguish between A and C, and B and D to establish the ranking. We assume in most ranking situations that if A defeats B and B defeats C then A will defeat C. This relationship is described as being transitive, where if variable a < b and b < c, then a < c. In our example, A defeats B who defeats C, and A defeats D who defeats B. This means that A has second-order influence over C and B but not D. (A doesn’t defeat anyone who defeats D). The matrix M2 can be used to investigate second-order influence. 0 M2 = 0 1 0

1 0 0 1

0 1 0 0

1 0 1 0

0 M2 = 1 0 0

1 0 2 0

1 0 0 1

0 1 1 0

0 0 1 0

1 0 0 1

0 1 0 0

1 0 1 0

Notice that the leading diagonal is still 0. It is impossible for a player to have second-order influence over themselves. Row 1 represents the second-order influence of player A over the other players. The element 2 in row 3 occurs because C defeats 2 players (A and D) who defeat B. 2 We can find the second-order dominance vector, V2 = 2 , but how much 3 importance should it be given? 1 If we assign equal importance, we calculate 0 M + M2 = 0 1 0

1 0 0 1

0 1 0 0

1 01 0 + 10 1 02 0 00

0 = 1 1 0

2 0 2 1

1 1 0 1

1 1 2 0

1 0 0 1

0 1 1 0

4 This gives a dominance vector 3 = V1 + V2 and allows us to rank the competitors 5 in the order C, A, B, D. 2

Chapter 3 Matrices

167

If we wanted to investigate third-order influence, we could calculate M 3 (= M 2M). 0 M M= 1 0 0 2

1 0 2 0

1 0 0 1

0 1 1 0

0 0 1 0

1 0 0 1

0 1 0 0

1 10 0 = 02 1 01 0 10

1 0 2 0

1 1 0 1

3 giving V3 = 3 3 2

At this stage, notice that the leading diagonal is no longer 0. If there were more players in the tournament, we could continue finding powers of M, but with 4 players, we stop at M 3. In general, if there are m players, we stop at M m – 1. In most scenarios, it is probably unfair to assign equal importance to first-, second- and third-order influence. We can allocate arbitrary constants to weight the influence; that is, M + xM 2 + yM 3. The resulting dominance vector can be found by calculating V1 + xV 2 + yV3. If we choose x = 0.5 and y = 0.25, the dominance vector would be 2 V1 + 0.5V2 + 0.25V3 = 1 + 0.5 2 1

2 2 + 0.25 3 1

3 3.75 3 = 2.75 3 4.25 2 2

This would rank the four players as C, A, B, D. When you compare the final seeding with the initial information, we can see that A and C both won 2 games, but the wins by A were against the lower ranked B and D. Players B and D both won 1 game but B managed to defeat the higher placed C. This justifies the seeding as produced.

Graphics Calculator tip!

Alternative method for adding the elements in each row in a dominance matrix

Multiplying a square matrix by a column vector with the same number of rows and all entries shown as 1 has the effect of adding the elements in each row of the matrix. In the example above, the dominance vector V1 could have been obtained using the following steps. (The main advantage is if the dominance matrix is 5 × 5 or larger. You don’t need to arrow across the screen to see the elements when you are adding them.)

For the Casio fx-9860G AU 1. Enter the 4 × 4 dominance matrix M. (Refer to the graphics calculator tip on page 156 if you are unsure how to do this.)

2. Enter the 4 × 1 vector N with 1 shown for every element.

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3. Evaluate M × N. Notice that this gives V1.

For the TI-Nspire CAS 1. Enter the 4 × 4 dominance matrix M. (Refer to the graphics calculator tip on page 157 if you are unsure how to do this.) 2. Move the cursor to the right of the matrix. Press the multiplication (r) key. 3. Enter the 4 × 1 vector N with 1 shown for every element.

4. Move the cursor outside the matrix and press · to obtain the answer to M × N. Notice that this gives V1.

remember remember 1. Dominance matrices are often used to determine player rankings in round-robin situations. 2. Information from a digraph that indicates the win–loss outcome of matches played in a tournament (for example, A defeats B, D defeats C, and so on) can be converted into matrix form. This matrix is called a dominance matrix. 3. A dominance vector shows how many players each has defeated. It is obtained by adding the elements of each of the rows of the dominance matrix. This allows you to rank the players. 4. For a dominance matrix, M, we can calculate the second-order influence of players by calculating M 2 (third-order influence by calculating M 3 and so on) and finding the resulting dominance vector, V2 (V3 and so on). 5. Arbitrary constants can also be allocated to weight the influence; for example, when considering four players, we calculate M + xM 2 + yM 3 where x and y are constants. The resulting dominance vector can be found by calculating V 1 + xV 2 + yV 3. This refines the ranking process.

Chapter 3 Matrices

3G

169

Dominance matrices

1 We want to seed 4 chess players, Breanna, Kayley, Teagan and Cameron. In past matches, Cameron defeated Breanna and Teagan, both Breanna and Teagan defeated Kayley, Kayley defeated Cameron, and Breanna defeated Teagan.

C T

K B

a Draw a digraph to represent this information. b By giving equal importance to first- and second-order influence, use dominance matrices to rank the players. Cameron, Breanna, Kayley, Teagan 2 Three friends have noticed that when they played chess, Mair defeated Ann and Janine, and Ann defeated Janine. Use dominance matrices to rank these players. Mair, Ann, Janine 3 A round-robin netball match was arranged for house competitions where Barnes lost to all but Cunningham, Cunningham lost to Leslie but defeated Hamilton. No teams went undefeated. a If it is decided to give equal importance to first- and second-order influence, use dominance matrices to rank the students’ houses. Hamilton, Leslie, Cunningham, Barnes b If house points are allocated as 20, 15, 10, 5 for the overall ranking, how many points did each house receive? 20 points to Hamilton, 15 to Leslie, 10 to Cunningham, 5 to Barnes 4 Five schools are debating in a round-robin tournament — the following table shows the results.

Clifton Warwick Goondiwindi Stanthorpe 0 1 M= 0 0 1

0 0 0 0 1

1 1 0 1 0

1 1 0 0 0

0 0 1 1 0

Clifton

Warwick

Goondiwindi

Stanthorpe

Ipswich

—

L

W

W

L

—

W

W

L

—

L

W

—

W

Clifton lose against Warwick and Ipswich and so on. a Construct a dominance matrix of this information. b A total of 15 points is divided in the ratio 5:4:3:2:1 and awarded according to the ranking of the schools at the end of the tournament. If this division of points is allocated according to the figures produced by the dominance matrix sum M + 0.8M 2 + 0.5M 3, list the number of points each school wins. 5 points to Warwick, 4 points to Ipswich, 3 points to Stanthorpe, 2 points to Clifton, 1 point to Goondiwindi

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summary Operations with matrices • A matrix (plural: matrices) is a collection of numbers arranged in rows and columns. • An m × n matrix has m rows and n columns. • The numbers in the matrix are called the elements of the matrix. Elements are referred to by the row and column position. • Addition and subtraction of matrices is performed by adding or subtracting elements in corresponding positions. These operations can be performed only if the matrices have the same order. • Scalar multiplication of a matrix is performed by multiplying each element of the matrix by a number. Thus kA means each element in matrix A is multiplied by the number k.

Multiplying matrices • Matrices are multiplied in the following way: a 11 a 12 a 13

b 11 b 12 b 13

If A = a 21 a 22 a 23 and B = b 21 b 22 b 23 a 31 a 32 a 33

b 31 b 32 b 33

then a 11 × b 11 + a 12 × b 21 + a 13 × b 31

a 11 × b 12 + a 12 × b 22 + a 13 × b 32

a 11 × b 13 + a 12 × b 23 + a 13 × b 33

AB = a 21 × b 11 + a 22 × b 21 + a 23 × b 31

a 21 × b 12 + a 22 × b 22 + a 23 × b 32

a 21 × b 13 + a 22 × b 23 + a 23 × b 33

a 31 × b 11 + a 32 × b 21 + a 33 × b 31

a 31 × b 12 + a 32 × b 22 + a 33 × b 32

a 31 × b 13 + a 32 × b 23 + a 33 × b 33

The orders are related as follows: (m × n) (n × p) = (m × p). • Matrix multiplication is usually not commutative. That is, AB ≠ BA.

Powers of a matrix • The power, n, of matrix A, in general form, is An = An – 1A, where n is a positive integer. • Powers of matrices are only defined for square matrices; that is, A has to be a square matrix to obtain An. • A matrix A is nilpotent if A2 = O where O is the zero matrix. The zero matrix is a square matrix with all elements equal to zero. For example, the 2 × 2 zero matrix is 00 . 00 • A matrix A is idempotent if A2 = A.

Chapter 3 Matrices

171

Multiplicative inverse and solving matrix equations • An identity matrix, I, is defined for square matrices such that AI = IA. • The multiplicative inverse of matrix A is A–1 such that AA−1 = A−1A = I. 1 If A = a b , then A –1 = ------------------ d – b . ad – bc – c a c d The number ad − bc is called the determinant of A and has the symbols det A or |A|. If det A = 0, then A−1 does not exist (A does not have an inverse) and A is said to be singular.

The transpose of a matrix • The transpose of a matrix is an interchange of rows and columns. • The transpose of a matrix A is A′ where A = a b and A′ = a c . c d b d

Applications of matrices: solving simultaneous equations and summarising information • Matrices may be used to solve simultaneous equations: ax + by = u cx + dy = v. The pair of equations may be written in the form AX = B, where A = a b , c d x u X= and B = . y v • Matrices can also be used to summarise information which is in table form and solve related problems but care must be taken in setting up the matrices. • Dominance matrices can be used to determine player rankings in sports competitions.

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CHAPTER review 3A

1 multiple choice The solution to

A

2 0 – 2 A = 4 2 is given by A equals: –2 0 02

31 –1 1

B

12 4 –4 4

C

–4 0 –4 –8

D

–1 –1 –1 –1

E

3A

2 A is a 3 × 2 matrix, B is 2 × 2 and C is 3 × 2. Which of the following may be calculated? a A+B b A+C c B+C

3B

3 multiple choice

3B

If A is a 3 × 2 matrix and B is 2 × 1, then the order of AB is: A 2×2 B 3×2 C 3×1 D 1×3

11 11

E 2×1

4 multiple choice 1 The product of 1 2 3 0 is: 4 5 6 1 A

4 10

B

4 10

C

1 0 3 4 0 6

D

14

E

1 0 3 0 0 0 4 0 6

3B

5 Using the same matrices as in question 2, which of the following may be calculated? a AB b AC c BA d BC e CA f CB g A + CB h A + BC i AB−1

3C

6 If A = –3 2 –2 –4

3C

1 2 , find –2 0

a A2

b A3

–7 –6 6 –4

c

A4

5 – 14 14 12

100 384 100

c

A4

1 00 7 16 8 1 00

1 0 0 7 If A = 0 2 1 , find 1 0 0 a A2

100 142 100

b A3

Chapter 3 Matrices

8 multiple choice

3D

Consider the following matrices. 10 01

A=

6 3 – 12 – 6

B=

173

C= 3 2 3 –2

a Which of the following are idempotent? B Β

A A

C Α and Β

D C

E A and C

D C

E A and C

b Which of the following are nilpotent? A Α

B Β

C Α and Β

9 multiple choice

3D

The determinant of – 2 0 is: 1 5 B −11

A 0

C −2

D −10

E 10

10 multiple choice

3D

If AB = 4I then B−1 is: A 4A

B A

C

1 --- A 4

D

1 --- B 4

E 4B

11 multiple choice

3D

Which of the following matrices is singular? A

1 0 0 1

B

1 0 0 –1

C

4 –2 –6 3

D

4 2 –2 1

E

10 2 5 0

12 multiple choice

3D

If AX = B then X is given by: A A−1B

B BA−1

B C --A

13 Find matrix A if A 0 1 = 6 6 . 2 –1 0 –6

D AB−1

E IA−1

93 –6 0

3D

14 multiple choice Using matrices, the solution to: A (2, 3)

B (3, 5)

4x − y = 7 y−x=2

3E is:

C (−3, 4)

15 a Write down the inverse of 4 – 2 3 1

1 ------ 1 2 10 – 3 4

b Hence solve 3x + y = 14 and 4x − 2y = 22. (5, −1)

D (1, 1)

E (5, 3)

3D,E

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3F

16 In a township, 25% of households own no pets, 40% of households own one pet, 20% have two pets and 15% own more than two pets. a Set up a 1 × 4 matrix to represent the percentage ownership of pets. [0.25 0.40 0.20 0.15] b Write an equation that will enable you to calculate the number of households for each category, given that there are 800 households in the town. A = 800B c Evaluate the number of households for each category as a 1 × 4 matrix. [200 320 160 120]

3F

17 The matrix below represents the prices (in dollars) of some mobile phone options. The first column displays the costs of two types of pre-paid mobile phones and the second column represents two types of 12-month-plan mobile phones. 249 29 680 49 The company wants to increase the price of the pre-paid mobile phones by 12% and decrease the cost of the 12-month-plan mobile phones by 5%. ˙ 0 a Show the matrix used to represent the 1.12 0 0.95 price changes (a 2 × 2 matrix). b Use matrix multiplication to calculate 278.88 27.55 the new prices. 761.60 46.55

3G

18 In a backgammon competition, four players — Glen, James, Cameron and William — competed with the following results: Glen’s only win was against James. James defeated both Cameron and William. Cameron also defeated William. James, Cameron, Glen, William Using dominance matrices and assigning a weighting of 1 to first-order influence and 0.5 to second-order influence, rank the players.

Chapter 3 Matrices

175

Modelling and problem solving 1 A company has two plants manufacturing components for different models of car. The time spent in hours per car is given in the following matrix. Standard model Deluxe model 4-wheel drive

Assembly Packaging Despatch 25 1 0.5 30 1.5 1 35 1.5 0.5

The wage rates ($ per hour) at the two sites are given by:

1: Despatch for Deluxe model takes 1 hour.

Assembly Packaging Despatch

Plant 1 16.50 14.00 13.50

Plant 2 16.00 14.00 13.00

14: Packaging at Plant 1 has a wage rate of $14 per hour.

a In the first matrix, write down the 2, 3 element and explain what it refers to. b In the second matrix, write down the 2, 1 element and explain what it refers to. c Write down the order of each matrix and the order of the matrix found by multiplying the first matrix by the second matrix. 3 × 3, 3 × 2, 3 × 2 433.25 420.50 d Find the product of the two matrices. 529.50 514.00 e Explain what the first row of the product matrix represents. 605.25 587.50 f Explain what the first column of the product matrix represents. The assembly costs for each model at Plant 1 g Write down the cost of producing the Deluxe model at: ii Plant 1 $529.50 ii Plant 2. $514.00 The total costs for the Standard model at Plants 1 and 2

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2 Tickets for a one-way trip on a Brisbane-to-Sydney passenger train can be purchased as either Adult, Child (under 15 years old) or Pensioner. The table below shows the number of passengers and the total takings for three trips.

Number of adult passengers

Number of child passengers

Number of pensioner passengers

Total takings ($)

145

103

121

20 260

130

110

90

18 400

142

115

80

19 200

145x + 103y + 121z = 20 260 a Let x = the cost of an adult’s ticket. 20 260 145 103 121 x 130x + 110y + 90z = 18 400 Let y = the cost of a child’s ticket. 130 110 90 y = 18 400 Let z = the cost of a pensioner’s ticket. 142x + 115y + 80z = 19 200 19 200 142 115 80 z Construct three equations in terms of x, y and z. 0.025 544 –0.093 523 0.066 579 b Using matrices, express the equations in the form AX = B. A−1 = – 0.039 222 0.091 991 –0.044 166 c Use your graphics calculator to find A–1. 0.011 042 0.033 767 – 0.042 189 d Use your graphics calculator to determine the costs of a train ticket for an adult, a child and a pensioner. The cost of an adult’s ticket is $75, a child’s ticket is $50 and a pensioner’s ticket is $35.

3 Use a and b to complete A where A =

4 Use a and b to complete A where A =

a

a

b

b

so that it is nilpotent (that is, A2 = O). A =

a

b

so that it is idempotent. A = a – a 2 -------------- 1 – a b

5 Prove that if a square idempotent matrix A is non-singular, then A must be the identity matrix. eBook plus Digital doc: Test Yourself Chapter 3

6 Prove that if A is idempotent a I – A is idempotent b A(I – A) = O.

a

b 2

a – ----- – a b

An introduction to groups

4

syllabus reference Core topic: Introduction to groups

In this chapter 4A 4B 4C 4D

Modulo arithmetic The terminology of groups Properties of groups Cyclic groups and subgroups 4E Further examples of groups — transformations

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Introduction

Concepts of: • closure • associativity • identity • inverse • definition of a group

Through your study of mathematics, you have developed an understanding of the rules that apply to numbers. You know that if you add two integers, the result is also an integer. However, if you divide two integers, you don’t always get an integer as the answer. You know that addition is associative; for example, 7 + (6 + 2) = (7 + 6) + 2; but subtraction is not; for example, 7 − (6 − 2) ≠ (7 − 6) − 2. Towards the end of the 19th century, mathematicians began to talk about the concept of groups. Essentially, a group is a set of elements, such as integers or matrices, that can be combined using an operation, like addition or multiplication, and which satisfy certain conditions. For example, integers form a group under addition but not under division (because dividing integers does not always result in an integer). Historically, group theory came from the study of number theory and the theory of algebraic equations at the end of the 18th century and the study of geometry at the beginning of the 19th century. Today, group theory is applied to many areas of science such as genetics, quantum theory, molecular orbits, crystallography and the theory of relativity.

Algebraic structures In algebra, symbols that can be manipulated are elements of some set and the manipulation is done by performing certain operations on elements of that set. The set involved is referred to as an algebraic structure. Research the topic of algebraic structures examining early algebraic systems that developed in ancient civilisations such as the Indian, Arabic, Babylonian, Egyptian and Greek. Highlight differences and similarities among the various forms. But first a new tool to help you deal with some notions used in groups.

Chapter 4 An introduction to groups

179

Modulo arithmetic Not to be confused with the modulus of a number (see Chapter 1 on real numbers, R, where the modulus of −4, written | −4 | = 4), modulo arithmetic uses a finite number system with a finite number of elements. This is sometimes referred to as ‘clock arithmetic’ because of the similarities with reading the time on an analog clock. Consider reading the time shown on the clock face to the right. Whether it is 2 am or 2 pm we would say it is 2 o’clock, but in 24-hour time the 2 pm would be 1400 hours. In effect 11 12 1 2 we have subtracted 12 hours from the 1400 (14 hours) to give 10 an answer of 2. In this case we say that 2 is the residue, or 9 3 what is left over when 12 hours is subtracted from the 14. 8 4 In modulo 12 arithmetic the same principle is used except 7 6 5 that the 12 is replaced by a 0. 5 + 6 ≡ 11 5+7≡0 5 + 8 ≡ 1 and so on. 11 0 1 In our normal decimal system 5 + 8 = 13, but in modulo 12 2 10 arithmetic the residue of 1 differs from 13 by 12 (or a mul9 3 tiple of 12) and 1 and 13 are said to be congruent. That is, in 8 4 modulo 5 arithmetic, the numbers 3, 8 and 13 are congruent 7 6 5 and in modulo 12 arithmetic, 2, 14, and 26 are congruent numbers. The symbol for congruency, ≡, is used. Using more precise terminology, addition modulo 10 is written 3 + 9 ≡ 2 mod 10, 5 + 5 ≡ 0 mod 10, and so on. (Note the abbreviation of modulo to mod.) In mod 12, the numbers 0 to 11 are referred to as residues, as with 0 to 5 in mod 6. This information can be stored in a table, known as a Cayley Table.

WORKED Example 1 Draw up a Cayley Table that shows the residues using addition modulo 4. THINK 1

2

WRITE/DRAW

Draw an empty table with 0, 1, 2, 3 in the first row and column and put a + sign in the top corner.

+

Start working across the first row. 0 + 0 = 0 etc. and do likewise with the first column.

0

1

2

3

+

0

1

2

3

0 1 2 3

0 1 2 3

1

2

3

0 1 2 3

Continued over page

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

THINK 3

WRITE/DRAW

The residues are the numbers left over when 4 is taken from the answer (if the answer is 4 or greater). As you complete the table note that the answers are less than 4. So, for 2 + 2 the residue is 0.

+

0

1

2

3

0 1 2 3

0 1 2 3

1 2 3 0

2 3 0 1

3 0 1 2

remember remember

1. Modulo arithmetic is like clock arithmetic where 5 + 9 ≡ 4 in mod 10. 2. The residues of modulo x are all the whole numbers less than x. 3. Congruent numbers in mod x all differ by multiples of x.

4A

Modulo arithmetic

1 List 4 numbers congruent to: a 4 in mod 8 4, 12, 20, 28, 36 … b 4 in mod 6. 4, 10, 16, 22

eBook plus Digital doc:

2 List the residues in: a mod 3 0, 1, 2

SkillSHEET 4.1 Modulo arithmetic

b mod 9 0, 1, 2, … 8

c

mod 11. 0, 1, … 10

3 Draw up a Cayley Table that shows the residues for each of the following: a addition mod 6 1 b multiplication mod 4 c multiplication mod 5.

WORKED

Example xample

3 a + 0 0 0 1 1 2 2 3 3 4 4 5 5

1 1 2 3 4 5 0

2 2 3 4 5 0 1

3 3 4 5 0 1 2

4 4 5 0 1 2 3

5 5 0 1 2 3 4

b × 0 1 2 3

0 0 0 0 0

1 0 1 2 3

2 0 2 0 2

3 0 3 2 1

c × 0 1 2 3 4

0 0 0 0 0 0

1 0 1 2 3 4

2 0 2 4 1 3

3 0 3 1 4 2

4 0 4 3 2 1

The terminology of groups In Chapter 1 you dealt with different sets of numbers within the Real Number System. Throughout your student life you have used the operations of addition, multiplication, subtraction and division, finding a square root, reciprocals, and so on. These are examples of operations performed on numbers that are part of a certain set. Operations (such as addition) that involve 2 input values, for example 2 + 3, are called binary operations. Those that involve only one input value, such as finding the square root of a number (for example 8 ) are called unary operations. Others that involve 3 input values are called ternary; for example, the principal, interest and term of a loan are the 3 input values involved in calculating the amount of interest due on a loan. (Strictly speaking the multiplication involved is still carried out on pairs of values.)

Definition of terms Groups that we will deal with consist of a system that involves a set of elements (often numbers) and a binary operation. Lower case letters, a, b, c …, are used to refer to elements of the set and the symbol ‘°’ denotes whatever operation is involved.

Chapter 4 An introduction to groups

181

For a non-empty set of elements S = {a, b, c, …} involved in the binary operation ‘°’ to be a group, G = [S, °], the following properties must hold.

1 Closure An operation is closed if the result of that operation is an element of the same set as the two inputs. That is, a ° b must be in S.

For example, consider 2 + 3 = 5 where S = {Real numbers} (or R) and ° is the operation of addition. The operation is closed because 5 ∈ R. But consider 2 − 3 = −1 where S = {Natural numbers} (or N) and ° is the operation of subtraction. Because the result (−1) is not a member of the set of natural numbers this operation is not closed. That is, the answer is not part of the initial set of natural numbers.

2 Associativity If an operation is associative, the order in which operations are performed does not affect the answer. That is, (a ° b) ° c = a ° (b ° c). Often brackets are employed to determine the order of operations. For example, consider (2 × 3) × 4 and 2 × (3 × 4): (2 × 3) × 4 = 6 × 4 2 × (3 × 4) = 2 × 12 = 24 = 24 In this case, both answers are the same. Note that only the position of the brackets changes and the order of the numbers remains the same. But consider the operation of division: (20 ÷ 2) ÷ 4 and 20 ÷ (2 ÷ 4) = 10 ÷ 4 = 20 ÷ 0.5 = 2.5 = 40 Here the answers are not the same. Division, like subtraction, is not associative. You would have realised this in your earlier junior mathematics studies.

3 Identity For all elements of a set, if a unique element exists in the set such that a ° u = u ° a = a then u is the identity element (IE) for that operation. That means that there is only one element that leaves every element unchanged when the operation ‘°’ has been applied. For example, 3 + 0 = 3, so 0 is the identity element for addition (IE+) for real numbers. However, 3 × 0 = 0 so 0 is not the identity element for real numbers under the operation of multiplication. Note: The one identity element must work for all elements of the set so 5 + 0 = 5 and −8 + 0 = −8. It must also work from both the left and right sides of the operation. For example, 6 + 0 = 6 and 0 + 6 = 6. That is, 6 + 0 = 0 + 6 = 6.

4 Inverse For each element of a set there is a unique element a–1 such that a ° a–1 = a–1 ° a = u where u is the identity element for that operation. Unique means that every element has only one inverse. 2×

1 --2

=

1 --2

Therefore

× 2 = 1 where 1 is the identity element for multiplication (IE×) 1 --2

is the multiplicative inverse of 2.

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Now consider 2 + −2 = −2 + 2 = 0 where 0 is IE+; in this case −2 is the additive inverse of 2. However, note that the set involved here would have to be integers (that is, both positive and negative) not just whole numbers because −2 ∉ {Whole numbers}. We can now restate the definition of a group. If the following 4 properties hold for a set of elements under a certain operation ‘°’: 1. closure 2. associativity 3. existence of an identity element 4. existence of an inverse then the system under investigation [S, °] is a group. If a fifth property, commutativity, also holds, then the group is an Abelian group.

Commutativity If the order of the elements involved has no effect on the outcome, then the operation is commutative. That is, a ° b = b ° a. For example, 2 × 5 = 10 and 5 × 2 = 10. Hence multiplication with real numbers is commutative. Note the stated condition, ‘with real numbers’, because you have already worked with matrices where multiplication is not commutative. However, consider 10 ÷ 2 = 5 and 2 ÷ 10 = 0.2. So division is not commutative. You would be familiar with other operations as well that are not commutative.

WORKED Example 2

Find a the identity element and b the inverse for the operation defined as a ° b = a + b + 2. THINK

WRITE

a

b

1

An identity element (IE) is an element that, when involved in an operation with another element, does not change the value of that element.

a Let a ° b = a (where b = IE) therefore a + b + 2 = a b = −2 so IE = −2

2

State the identity element.

The identity element is −2.

1

An inverse is an element that, when involved in an operation with another element, results in the IE for that operation.

b Let a ° b = −2 (where b is the inverse of a and −2 = IE from part a) therefore a + b + 2 = −2 a + b = −4 b = −4 − a

2

The inverse must work from the left as well as from the right of the operation.

Check b ° a where b = −4 – a b°a=b+a+2 = −4 – a + a + 2 = −2 = IE −1 Therefore, a ° a = a−1 ° a = −2 when a−1 = −4 − a.

3

State the inverse.

The inverse is −4 − a.

Chapter 4 An introduction to groups

WORKED Example 3

Find the identity element for the operation defined as a ˚ b = non-negative real numbers. THINK 1

2

3

2

183

2

a + b where a and b are

WRITE

An identity element (IE) is an element that, when involved in an operation with another element, does not change the value of that element.

Let a ˚ b = a (where b = IE)

Check that the identity element works from both sides of the operation.

Check b ˚ a where b = 0

State the identity element.

The identity element is 0.

Therefore

2

2

a +b =a

Squaring both sides: a2 + b2 = a2 b2 = 0 b=0 2

2

0˚a= 0 +a =a Therefore, 0 ˚ a = a ˚ 0 = a. Thus, IE = 0.

History of mathematics NIELS HENRIK ABEL (1802–1829) During his life . . . Lord Byron, the English poet, writes Don Juan. Napoleon Bonaparte becomes emperor of France. Jean-Baptiste Lamarck, the French biologist, proposes that acquired traits are inherited by individuals in a population. Niels Abel was one of the most productive mathematicians of the 19th century. Born in Norway on 5 August 1802, by the age of 16 he had started his private study of the mathematics of Newton, Euler, Gauss and

Lagrange. As the sole supporting male of his family, at 18 he tutored private pupils while continuing his own mathematical research. By the age of 19 he had proved that there was no finite formula for the solution of the general fifth degree polynomial. He died of tuberculosis on 6 April 1829, two days before the announcement of his posting as professor to the Berlin university. His life in poverty stands in contrast to the regard with which he is held in his field; the term Abelian group is used in honour of Abel. His studies on group theory were central to the development of abstract algebra. Questions: 1. How did Abel financially support his family? He tutored students. 2. Which property do groups bearing his name exhibit? Abelian groups are those that have the property of commutativity.

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remember remember A set S forms a group under the operation ‘°’ if and only if (iff) all of the following are true: 1. it is closed under ‘°’; that is, the result is an element of S 2. the order in which operations are performed has no effect on the results; that is, it is associative 3. there is only one identity element (IE), u, such that a ° u = u ° a = a 4. there is a unique inverse a–1 for every element such that a ° a–1 = a–1 ° a = u, where u = IE. 5. If the property of commutativity also holds, then it is an Abelian group.

SLE 2: Determine the identity element and inverses in a group table.

4B

The terminology of groups

a+b 1 Show that a ° b = ------------ is not closed with respect to whole numbers. (Remember you 2 only have to find one example where the operation is not closed to disprove a state+2 ment.) 3----------- = 2 1--- and 2 1--- is not an element of the set of whole numbers. 2

2

2

2

2

2 If an operation a ° b is defined as a + b determine whether this is closed if a and b are whole numbers. 1 ° 3 = 1 + 9 = 10 Not a whole number, ∴ not closed. 3 Find a the identity element and b the inverse for the operation ° on real numbers where a ° b = a + b − 1. a IE = 1 2 b b=2−a

WORKED

Example xample

4 What is the identity element of the operation a ° b = a + b − ab if a and b are real numbers? IE = 0 a ° 0 = a + 0 − a × 0 3

WORKED

Example xample

Assuming this operation has an identity then +b =a let a----------ab a + b = a2b a = a2b − b But a ≠ a2b − b therefore the operation has no identity.

5 The operation a ° b = 4ab2 is defined for positive real numbers a and b. Does the identity element for this operation exist? No identity. 4a × ( 1--2- )2 = a but 1--2- ° a ≠ a a+b 6 Develop a proof to show that a ° b = ------------ has no identity. ab 7 An operation is defined with respect to an ordered pair of integers as (a, b) ° (c, d) = (ad + bc, bd). Show that (0, 1) is the identity element for the operation. 8 Show that a ° b = (a + b)2 has no identity for real numbers.

Properties of groups Let (a + b)2 = a where b = IE Take the square root of

Let (0, 1) = (a, b) = IE. Therefore, (0, 1) ° (c, d) = (0 × d + 1 × c, 0 × c + 1 × d) = (c, d) and (a, b) ° (0, 1) = (a × 1 + b × 0, a × 0 + b × 1) = (a, b).

In the previous section, we looked at the conditions under which a set forms a group. To check whether a set S forms a group under the operation ‘˚’, that is, [S, ˚], there are four properties to be tested. both sides: a + b = ± a 1. Closure: the result of the operation is an element of S; that is, a ˚ b ∈ S. If a is negative then a ∉ R. 2. Associativity: the order in which the operation is performed has no effect on the Since an identity must be result; that is, (a ˚ b) c = a ˚ (b ˚ c). applicable to all elements of the set, there is no IE for a ° b.

Chapter 4 An introduction to groups

185

3. Existence of an identity element: there is only one identity element (IE), u, such that a ˚ u = u ˚ a = a. 4. Existence of an inverse: there is a unique inverse for every element such that a ˚ a−1 = a−1 ˚ a = u where u = IE.

Abelian groups If a set forms a group and the property of commutativity also holds, then it is an Abelian group. An operation is commutative if the order of the elements involved has no effect on the result. That is, a ˚ b = b ˚ a.

WORKED Example 4

a Verify that the set of integers forms a group under addition. b Is this group Abelian? THINK

WRITE

a

a Let Z = {a, b, c, …} be the set of integers; the operation is addition.

1

What numbers are involved? All positive and negative integers and 0 are involved so state the set and operation. While you can think of actual values for the integers (−1, 0 4 …) your answer should use only variables, with constants used as examples only.

2

Test each of the 4 properties in the same order each time to help you remember the 4 tests. iii The sum of any 2 integers is an integer. iii The order in which the operation is performed has no effect on the result. iii Since 0 ∈ Z, IE+ exists. iv Since Z contains all positive and negative whole numbers, the inverse is −a.

3

State that the system forms a group under the conditions stated.

b If the group is Abelian we need to show that this operation is commutative.

iii The operation is closed: a + b = c where a, b and c ∈ Z iii The operation is associative: (a + b) + c = a + (b + c) iii The identity element exists: a+0=0+a=a iv The inverse exists: a + −a = −a + a = 0 Thus the set of integers forms a group under addition. b Commutativity a+b=b+a Therefore the group is Abelian.

Note that the test for commutativity is performed last because the first 4 properties are necessary to state that it is a group in the first place, before it is shown to be Abelian. This group, G = [Z, +], is an infinite group, having an unlimited set of elements. You will also deal with finite groups which have a countable number of elements.

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WORKED Example 5 Verify that the set of odd integers does not form a group under addition. THINK 1 What numbers are involved? The set of odd integers includes −5, −3, −1, 1, 3, 5 … State the set and operation.

WRITE S = {a, b, c, …} is the set of odd integers. The operation is addition.

2

Test the 4 properties as shown in Worked example 4.

Closure: a + b ∈ S Let a = 3 and b = 5 3 + 5 = 8 and 8 ∉ S Therefore G ≠ [S, +]

3

There is no need to proceed any further with tests to verify the system is a group as it is not closed.

The set of odd integers does not form a group under addition.

WORKED Example 6

Construct a Cayley Table for [{1, i, −1, −i}, ×] and determine whether this constitutes a group. THINK 1 Set up the empty table.

WRITE ×

1

i

−1

−i

×

1

i

−1

−i

1 i −1 −i

1 i −1 −i

i −1 −i 1

−1 −i 1 i

−i 1 i −1

1 i −1 −i 2

Complete the table. Remember from Chapter 2 on complex numbers that i = – 1 and i × i = −1.

3

Test the 4 group properties — closed set, associative, identity element and multiplicative inverse. The answers can be obtained from the table. (Multiplication by 1 leaves all elements unchanged.)

1. All the results are members of the original set {1, i , −1, −i}. This is a closed set. 2. The set is associative e.g. (1 × i) × −i = i × −i = 1 and 1 × (i × −i) = 1 × 1 = 1 3. The identity element, IE× = 1 4. Multiplicative inverse: there is a 1 (IE×) in every row of the table so each element has a unique inverse.

4

State your conclusion.

Therefore, the system is a group.

187

Chapter 4 An introduction to groups

Note that the Cayley Table is symmetrical about the leading diagonal. The table could be flipped over on the leading diagonal and remain unchanged. This means that the order of operations will not affect the results; that is, that the operation is commutative. Therefore this group is also Abelian.

×

1

i

−1

−i

1

1

i

−1

−i

i

i

−1

−i

1

−1

−1

−i

1

i

−i

−i

1

i

−1

Leading diagonal

WORKED Example 7

Construct a Cayley Table for [{mod 5}, +] and determine whether it is an Abelian group. THINK 1

WRITE

Decide what numbers are present in mod 5 and complete a Cayley Table of residues.

+

0

1

2

3

4

0 1 2 3 4

0 1 2 3 4

1 2 3 4 0

2 3 4 0 1

3 4 0 1 2

4 0 1 2 3

2

Test for the 4 group properties.

1. All results are members of the original set. So, the set is closed. 2. Addition with whole numbers is associative. 3. The identity element, IE+ = 0 exists. 4. There is a 0 entry in each row because each element has a corresponding element that, when added, results in 0 (IE+). So, there is an additive inverse. Therefore the system forms a group.

3

Test for commutativity.

Addition mod 5 is commutative as shown by the symmetry about the leading diagonal. For example: 4 + 0 ≡ 4 and 0+4≡4 and 4 + 2 ≡ 1 and 2+4≡1 Therefore the group is Abelian.

+

0

1

2

3

4

0

0

1

2

3

4

1

1

2

3

4

0

2

2

3

4

0

1

3

3

4

0

1

2

4

4

0

1

2

3

Note: There are 9 axioms that relate to operations and whole numbers that require no proof: they are assumed to be true. The associativity statement in the example above relied on one of these axioms and you can state that these axioms have been used.

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6 a + 4 4 8 16 20 64 68 256 260 . . . . . .

They are given here with no explanation. 1. Closure Law of Addition 2. Commutative Law of Addition 3. Associative Law of Addition 4. Identity Law of Addition 5. Closure Law of Multiplication 6. Commutative Law of Multiplication 7. Associative Law of Multiplication 8. Identity Law of Multiplication 9. Distributive Law of Multiplication over addition, where a(b + c) = ab + ac 5

× 1 2 3 4

1 1 2 3 4

2 2 4 1 3

3 3 1 4 2

16 20 32 80 272 . . .

64 68 80 128 320 . . .

256 260 272 320 512 . . .

… … … … …

remember remember

4 4 3 2 1

1. To determine whether a set forms a group under an operation (°) test each of the four properties; that is, test whether it is closed and associative, whether there is an identity element and a unique inverse. 2. To determine whether the group is Abelian, show that the operation is commutative (e.g. a ° b = b ° a).

Closed, associative, IE× = 1 and there is an inverse; therefore it is a group.

4C WORKED

Example xample

4

SLE 1: Determine whether the elements of a set form a group under a binary operation. SLE 2: Determine the identity element and inverses in a group table. SLE 3: Use a small Cayley table to determine whether a set of elements under a binary operation forms a group.

Properties of groups SLE 4: Investigate when the integers modulo n form groups under addition or multiplication. SLE 9: Investigate commutativity and abelian groups.

1 a Verify that the set of real numbers, [R, +], forms a group under addition. [R, +] It is closed, associative, IE+ = 0, inverse = −a, therefore it is a group. b Is this group Abelian? It is Abelian.

Closed, associative, no IE since 0 ∉ {even numbers}, there is an inverse; therefore not a group.

2 a Consider the set of even numbers (2n) where n ∈ ±Z. b Does this form a group under addition? (Note: 0 ∉ {even numbers}) Closed, associative, no IE since 1 ∉ {even c Does it form a group under multiplication?

eBook plus Digital doc:

numbers}, no inverse; therefore not a group.

SkillSHEET 4.2 Properties of groups

3 Does the set of powers of 1 form a group under: a addition? 12 + 13 is not closed; not a group. b multiplication? 12 × 13 is closed, and associative, IE× = 1, there is an inverse; so it is a group.

WORKED

Example xample

4 Verify that the set of even integers does not form a group under division.

Check with your teacher.

5 WORKED

Example xample

6

5 Construct a Cayley Table for [{mod 5 excluding 0}, ×] and determine whether this constitutes a group. 6 a Draw up a Cayley Table for the set of even powers of 2 under addition. Under addition: not closed, associative, no IE+ since b Does this form a group under addition? 0 ∉ 22n, no inverse (always +ve); not a group. c Does this form a group under multiplication?

Under multiplication: closed, associative, IE× = 1 is not present as no 20 (0 ∉ {even numbers}), inverse is 2–2n; not a group. WORKED

Example xample

7

7 Construct a Cayley Table for [{mod 3}, ×] and determine whether it is an Abelian It is closed and associative, IE× = 1; inverse does not exist since there are no 1s in group. × 0 1 2 0 0 0 0 1 0 1 2 2 0 2 1

the first row or column. This is not a group; therefore, it is not Abelian, even though the commutative law does apply.

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Chapter 4 An introduction to groups

8 Determine whether each of the tables below forms a group. a 10 ° N L R A

N N L R A

L L A N R

R R N A L

A A R L N

Closed, associative, IE° = N, there is an inverse, N appears in every row and column.

c

°

a

b

c

a b c

c a b

a b c

b c a

°

a

b

c

a b c

a b c

b b b

c c a

b

Yes

No, no inverse d for b

°

a

b

c

a b c

a b c

b a d

c c a

°

a

b

c

a b c

b c a

c a b

a b c

No, not closed

Yes

° 5 10 20

9 a Construct a Cayley Table for the set a ° b = [{5, 10, 20}, lowest common multiple of a, b] b Does this set form a group? It is closed, associative, IE° = 5, no inverse; so not a group. 10 The movements of a robot are restricted to no change (N), turn left (L), turn right (R), turn about (A): {N, L, R, A}. Construct a Cayley Table and show that this set of movements under the operation ˚, meaning ‘followed by’, forms a group.

Cyclic groups and subgroups By completing the exercises so far, you would be aware that both real numbers and integers form a group under addition. However, the set of integers is contained within the set of real numbers (this is called a subset). Because of this, we can say that integers under addition is a subgroup of real numbers under addition. When identifying subgroups, it is not necessary to verify associativity, but the existence of closure, the identity element and inverses needs to be confirmed.

WORKED Example 8 Does the set of numbers {0, 2, 4} form a subgroup of addition modulo 6? THINK 1

To check for closure, construct a Cayley Table.

WRITE +

0

2

4

0

0

2

4

2

2

4

0

4

4

0

2

This is a closed set. 2

Is the identity included?

IE+ = 0. Therefore the identity is included.

3

Do inverses exist?

There is a 0 entry in each row, so the inverses are included.

4

State your conclusion.

The set of numbers {0, 2, 4} forms a subgroup of addition modulo 6.

5 5 10 20

10 20 10 20 10 20 20 20

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In modulo 6 addition, 2 + 2 = 2 × 2 ≡ 4 mod 6, 2 + 2 + 2 = 3 × 2 ≡ 0 mod 6, 2 + 2 + 2 + 2 = 4 × 2 ≡ 2 mod 6, 2 + 2 + 2 + 2 + 2 = 5 × 2 ≡ 4 mod 6. If we continue, we get 6 × 2 ≡ 0 mod 6, 7 × 2 ≡ 2 mod 6 and so on. This means that all members of the subgroup {0, 2, 4} can be a result of repeatedly adding the number 2. This subgroup is said to be cyclic (because the answers cycle through the set) and 2 is the generator of the cyclic subgroup. We can write this group as G = [{0, 2, 4}, + mod 6] = <2>, where <2> means that G is cyclic and 2 is the generator. From previous sections, we know that integer multiplication modulo 5 forms a group if zero is not included. We can write this set of numbers as Z5*. If we calculate powers of 3, we find that 30 ≡ 1 mod 5, 31 ≡ 3 mod 5, 32 ≡ 4 mod 5, 33 ≡ 2 mod 5, 34 ≡ 1 mod 5, 35 ≡ 3 mod 5 and so on. This means that all elements in Z5* can be found by repeatedly multiplying the number 3. The group is cyclic and 3 is a generator. We can write this group as G = [Z5*, ×] = <3>. Writing this more mathematically, let’s consider a group G = [S, ˚] and let a be one element of S. If ˚ represents addition, that is, G = [S, +], and all elements of S can be written as na, where n is an integer, then the group is cyclic and a is the generator. Similarly, if ˚ represents multiplication, that is, G = [S, ×], and all elements of S can be written as an, then the group is also cyclic and a is the generator. In both cases, we can write G = , meaning G is a cyclic group generated by a. Another group we can consider as an example is integers under addition. Every integer can be written as n1. This means that integers under addition is a cyclic group and 1 is the generator. Every integer can also be written as n(−1), therefore −1 is also a generator. We can write this group as G = [Z, +] = <1> = <−1>.

WORKED Example 9

a Compute the cyclic subgroups <1>, <2>, <3>, <4>, <5> of the group addition modulo 6. b Hence identify the generators of the group addition modulo 6. THINK a

1

2

We need to compute na for a = 1, 2, 3, 4 and 5. For integers modulo 6, there are 6 elements therefore n = 0, 1, 2, 3, 4, 5 will cover all elements.

Each column is the subgroup for that element. Write the subgroups.

b Identify the generators. A generator produces all elements of the group, that is, {0, 1, 2, 3, 4, 5}. Look for a column that has every element.

WRITE a

n

1n

2n

3n

4n

5n

0 1 2 3 4 5

0 1 2 3 4 5

0 2 4 0 2 4

0 3 0 3 0 3

0 4 2 0 4 2

0 5 4 3 2 1

<1> = [{0, 1, 2, 3, 4, 5}, + mod 6] <2> = [{0, 2, 4}, + mod 6] <3> = [{0, 3}, + mod 6] <4> = [{0, 2, 4}, + mod 6] <5> = [{0, 1, 2, 3, 4, 5}, + mod 6] b The generators are 1 and 5 for the cyclic group addition modulo 6.

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Chapter 4 An introduction to groups

remember remember

2 Closed, IE = 0 0 , 00

For the group G = [S, ˚]: 1. A subgroup exists if there is a subset of S that is closed and contains the identity element, and if inverses for all elements exist. 2. The group is cyclic if an element a can be found that generates the group. This can be written as G = .

the inverse of a 0 is – a 0 0b 0 –b which is a member of M, therefore a subgroup.

4D WORKED

Example

SLE 6: Construct a Cayley table and use it to identify subgroups (if any) such as the rotations of a square about its centre. SLE 7: Find the element(s) which generate(s) the group in a group table.

Cyclic groups and subgroups

1 Does the set of numbers {0, 3, 6, 9} form a subgroup of integer addition modulo 12? Closed, IE+ = 0, inverses exist (3 + 9 ≡ 0 mod 12, 6 + 6 ≡ 0 mod 12), therefore subgroup. Generators are 3 and 9.

8

2 The set M(2, Z) is the set of all 2 × 2 matrices with integers as entries. These matrices form a group under matrix addition. Prove that the subset of M containing only diagonal matrices (those with 0s in the upper right-hand and lower left-hand corners) form a subgroup. 3 a Compute the cyclic subgroups <3>, <6> and <9> of the group integer addition modulo 12. <3> = [{0, 3, 6, 12}, + mod 12]; <6> = [{0, 6}, + mod 12]; <9> = [{0,3,6,12}, + mod 12] 9 b Hence identify the generators of the subgroup defined in question 1. Generators are 3 and 9.

WORKED

Example

eBook plus Digital doc: WorkSHEET 4.1

4 Which of the following groups are cyclic? For each cyclic group, name all generators of the group. a [Z, +] Cyclic. Generators are 1 and −1. b [Q, +] (Q is the set of rational numbers) Not cyclic. c [6n where n is an integer, +] Cyclic. Generators are 6 and −6. d [3n where n is an integer, ×] Cyclic. Generators are 3 and --13- .

Application of groups — permutations P2 P3 P1 ° P1 P1 P2 P3 P2 P2 P3 P3 P1 Not closed ∴ Not a group 1 2 3 4 5 P2 ° P2 =   3 4 5 1 2

A symmetry of a square (or any other shape) may be written as a permutation by changing the positions of the vertices. For example, referring to the figure below right, we could write:

1 2 3 4 5 P2 ° P3 =   5 4 3 2 1

to the position of vertex 2, and so on.

1 2 3 4 5 P3 ° P2 =   2 1 5 4 3

    P1 =  1 2 3 4 5  and P3 =  1 2 3 4 5   1 5 4 3 2  1 2 3 4 5

1

  P2 =  1 2 3 4 5  , which means that vertex 1 goes  2 3 4 5 1

5

2

The only other two permutations allowed here are:

1 2 3 4 5 P3 ° P3 =  = P1  1 2 3 4 5

4

3

Determine whether these permutations form a group under the operation ° meaning ‘on’. P1 ˚ P3 means perform P1 on P3.

192

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Further examples of groups — transformations Consider all the transformations that a shape could undergo. Rotations about its centre and reflections about its axes of symmetry involve changes in the vertices only. Carefully examine the diagram below. Make sure you understand the symbols and the new positions of the vertices. Rotations anticlockwise: R90 → 90° R180 → 180° R270 → 270° Reflections: RV → in the vertical axis of symmetry RH → in the horizontal axis of symmetry RR → in the top right diagonal RL → in the top left diagonal R0 → no change. D

A

D D

A

C

A

R0 RL

RL

R 90 B

C

D

B

A C

B

C

B B

D

A

RV

RL

RR

RR C

D

B

C

D RV

A B

D

A C

A

RH

C R180

RH

B

C

D

B

A

D R 270

B

A

C

Therefore the set of all transformations or symmetries is given by the set {R90, R180, R270, RV, RH, RR, RL, R0} and the binary operation that combines any two of these transformations is referred to as composition, where one operation follows another.

Chapter 4 An introduction to groups

193

All the computerised movements involved in screen animations are based on similar compositions of transformations. As with permutations, the operations are performed right to left.

WORKED Example10 Find the result of Rv ° R180 . THINK 1

2

3

4

5

WRITE/DRAW

Draw the initial square with labelled vertices. Consider the order in which the operations are to be performed. Since the expression Rv ˚ R180 means that Rv follows R180, then R180 is performed first. Transform the square using R180 — 180° rotation anticlockwise. Locate vertex A and move it 180˚ anticlockwise. All other vertices follow in order around the square. For RV mark a vertical axis of symmetry in this figure (from step 2) and reflect or ‘flip’ the square about this axis.

Reposition the vertices one side at a time, B ↔ C and A ↔ D.

This matches with a single transformation representing RH.

D

A

C

B

B

C

A

D

B

C

A

D

C

B

D

A

The result is RH.

Functions

1 1 Consider functions f(x) = x, g(x) = −x, h(x) = --- and k(x) = − --- (where x ≠ 0). x x When these functions are involved in composition of functions such as g[h(x)], the function h(x) is substituted as the inner function into the outer function which is g(x). 1 1 That is, g[h(x)] = − --- where --- (the inner function) is substituted into g(x) which is −(x). x x 1 Similarly, k[g(x)] = − ---------- where g(x) = −x (the inner function) is substituted into (–x) 1--1 1 k(x) = − (the outer function). That is, k[g(x)] = − ---------- = --- = h(x). x (–x) x

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WORKED Example 11

1 1 Show that functions f(x) = x, g(x) = −x, h(x) = --- and k(x) = − --- form a group under x x composition. THINK 1 Complete a Cayley Table for these compositions.

WRITE

°

f

g

h

k

f g h k

f g h k

g f k h

h k f g

k h g f

2

Test the 4 group properties.

Closure: yes — all results are elements of the original set. Associative: yes — for example, (f ° g) ° h = g ° h = k f ° (g ° h) = f ° k = k Identity element is f(x). Inverse: yes — f(x) occurs in every row and column.

3

State your conclusion.

Composition of these functions forms a group.

History of mathematics A R T H U R C AY L E Y ( 1 8 2 1 – 1 8 9 5 ) During his life . . . Thomas Edison invents the phonograph. Slavery is officially abolished throughout the western world. Alfred Nobel invents dynamite. Arthur Cayley, a famous English mathematician, was born on 16 August 1821. His published mathematical papers are classics and include discussions on the concept of n-dimensional geometry. At the age of 25 he began practising law which he continued to do until 1863. In his spare time he wrote more than 300 mathematical papers. In 1863 he accepted a professorship in

mathematics at Cambridge University. One of his most famous non-mathematical accomplishments was his role in having women accepted at Cambridge. Like Niels Abel (see page 183), many of his research topics are now used in abstract algebra and group algebra, as well as in work with matrices and the theory of determinants. The Cayley Table is named after him. He died on 26 January 1895 having received many academic distinctions. His total works fill 13 volumes of about 600 pages each — a testimony to his prodigious life and study in mathematics. He worked towards having women

Questions accepted at Cambridge University. 1. What is one of Cayley’s most significant non-mathematical accomplishments? 2. List four fields of mathematics which feature in Cayley’s work. Abstract algebra, group algebra, n-dimensional geometry, matrices and determinants

195

Chapter 4 An introduction to groups

remember remember The binary operation that combines any two transformations (for example, rotation and reflection) is called composition, when one operation follows another. SLE 5: Investigate groups formed by geometric transformations such as the reflections of a rectangle in its axes of symmetry and rotations of an equilateral triangle.

4E

Further examples of groups — transformations

1 a Draw a Cayley Table for the rotation of an equilateral triangle. Label each vertex. b Does it form a group? Is it Abelian? IE = R0, Inverse exists for all elements. It is an Abelian group because the table is symmetrical about the leading diagonal.

2 a Draw a Cayley Table for the reflections of an equilateral triangle through each of the vertices R0, RV, RL, RR. b Does it form a group? 2

a

A

° R0 RV RL RR

RL

RR

C

R0 R0 RV RL RR

RV RV R0 — —

RL RL — R0 —

1

Does not form a group.

RR RR — — R0

C

RL

RR

° R0 R120 R240

B

RV

R0

a

C

B

A

B A R0 R0 R120 R240

R120 C B R120 R120 R240 R0

R240 A

R240 R240 R0 R120 4 c

RV Not Abelian.

2 2

31

2 2

4

2

1

2 1

33

2

3 3

RV 1

4

1

F

4 2

1

R240

3

2

R180

2

2

4 2

R0

R0 4

1 3 1

4

3

RH

4

R240

1

2 1 3

F

RH

RV 1 2

3

1

3

4

3

3 R180

3

1 4

2 4

4 a

2

10

3

Example

3 Explain what the following diagrams represent about the group shown below.

1

WORKED

1 3

1

4 Describe the symmetries of the following figures, using fully annotated diagrams. a a non-square rectangle b a non-square rhombus c

an ellipse

5 Consider an infinitely long strip of Hs, printed on transparent paper, as shown below …..H H H H H H …. Describe the axes of symmetry of this group.

… H H H H … RV … H H H H … RH R180 … H H H H … … H H H H … R0 … H H H H …

196

6 M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

6 Locate the axes of symmetry for the following figures. 7 a

°

f1

f2

f3

f4

f1 f2 f3 f4

f1 f2 f3 f4

f2 f1 — —

f3 f4 — —

f4 f3 — —

WORKED

Example xample

11

7 a Complete a Cayley Table for the composition of the following functions. 1 1 f1(x) = x f2(x) = --f3(x) = x − 1 f4(x) = ----------x x–1 where f1 ° f2 = f1[f2(x)]

8 a Closed: addition of b 2 × 2 matrices results in a 2 × 2 matrix. 8 a Associative: matrix addition is associative. b Identity exists: 0 0 0 0 is the identity element. Inverses exist: the inverse of A is −A. b i 12 ii 1 2 34 12 iii The set of 2 × 2, non-singular matrices. Identity = I. Inverse is present as I is present in each row and column. Closed and associative a IE+ = 0 0 0 0 (Remember 0 is a complex number.) Inverse =

Does this composition form a group?

Show that the set of all 2 × 2 matrices addition.

b Yes. Closed, associative, IE, Inverse of 0 0 is 0 0 1--- 1--z z z z

a b c d

forms a group under matrix

The set of all 2 × 2 matrices a b does not form a group under matrix multic d plication but it is possible to find a subset of 2 × 2 matrices that will form a multiplicative group. i Apart from the identity matrix, give an example of a 2 × 2 matrix that would be a member of this subset. ii Give one example of a 2 × 2 matrix that would not be part of this subset. iii Describe the largest subset of 2 × 2 matrices that would form a group under matrix multiplication.   9 Show that the set of matrices  1 0 , 1 0 , – 1 0 , – 1 0  forms a group 0 1 0 –1  under matrix multiplication.  0 1 0 – 1

z1 z2 10 Show that the set of matrices of the form , where z is a complex number, –z2 z1 forms a group a under matrix addition z1 –z2 1 b under matrix multiplication. IE× = I Inverse = --------------where the determinant is real. 2 2 z1 + z2 z2 z1 Assume z12 + z22 ≠ 0. The inverse exists if the determinant ≠ 0.

–z1 –z2 z2 –z1

Not a group

11 S is the set of all 2 × 2 matrices such that 0 0 , where z is a non-zero complex z z number. a Show that 0 0 is the identity element under matrix multiplication. 1 1 b Does the set form a group under matrix multiplication? 12 C =

i 0 , where i = 0 i

– 1 . The set T consists of positive powers of C such that

i 0 , –1 0 , –i 0 , 1 0 T = C n where n is a positive integer. 0 –1 0 –i 0 1 0 i a Find all the elements of set T. b Does the set T form a group under matrix multiplication? Yes.

Chapter 4 An introduction to groups

197

Some applications of group theory 1 Do the residues of {0, 1} mod 2 form a group under addition? Yes 2 A teacher of abstract algebra intended to give a typist a list of 9 integers that form a group under multiplication modulo 91. Instead, one of the 9 integers was omitted so that the list read: 1, 9, 16, 22, 53, 74, 79, 81. Which integer was left out? 29 3 Show that {1, 2, 3} multiplication mod 4 is not a group but {1, 2, 3, 4} multiplication mod 5 is a group. 4 Give an example of group elements a and b with the property that a−1ba ≠ b. 5 The integers 5 and 15 are two of 12 integers that form a group under multiplication mod 56. List all 12 integers. 1, 3, 5, 9, 13, 15, 19, 23, 25, 27, 39, 45 6 If the following table is that of a group, fill in the blank entries.

e a b c —

e

a

b

c

d

e — — — —

— b c d —

— — d — —

— — e a —

— e — b —

e a b c d

e e a b c d

a a b c d e

b b c d e a

c c d e a b

7 Prove that if G is a group such that the square of every element is the identity, then G is Abelian. 8 Examine whether a rotations and Yes b reflections No as stated earlier in this section, form Abelian groups. 9 Quaternions The concept of a set of elements called quaternions was first developed by the Irish mathematician William Hamilton (see page 118). Quaternions are ordered sets of four ordinary numbers, satisfying special laws of equality, addition and multiplication. Quaternions are useful for studying quantities having magnitude and direction in three-dimensional space and this has enabled great advances in quantum theory, relativity, number theory and group theory. The 4 numbers are 1, i, j and k and have the following properties: 12 = 1 i2 = j2 = k2 = ijk = −1 1i = i1 1j = j1 1k = k1 ij = −ji = k i(jk) = (ij)k = ijk All real and complex numbers do commute with i, j, and k but they are not commutative with each other.

d d e a b c

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Follow this example that shows that jk = i ijk = −1 from the definitions i × ijk = i × −1 multiply both sides by i on the left (or pre-multiply by i) associativity i2jk = −i −1 × jk = −i from the definitions −1 × −1 × jk = −1 × −i pre-multiply both sides by −1 jk = i Because multiplication between these elements is not commutative it is essential that all multiplication is done from a particular side of an expression and to perform this multiplication on both sides of the equals sign. You must respect the order of placement of terms in this system. a Show that i jk = −kj ii ki = j SLE 15 (Matrices and applications): b Show that i–1 = −i Investigate the group properties of matrices of the c If q = s + wi + vj + yk and p = m + ni + oj + jk, find the product of the two z1 z2 quaternions. form -------- ----- under both –z2 z1 10 Pauli Matrices addition and multiplication; find interesting subsets of The ideas introduced in the section on quaternions above can be extended to this class of matrices represent matrices. One 2 × 2 set is: (known as quaternions); in particular, show that the eight matrices

1= 1 0 0 1

1 0 , –1 0 , i 0 , 01 0 –1 0 i –i 0 , 0 1 , 0 –1 , 0 i –1 0 1 0 0 i , 0 – i form a group i0 –i 0

0 –i –i 0

W V U T P Q R S

i 0 0 –i

V T W U R P S Q

–i 0 0 i

U W T V Q S P R

10 a

–1

0

j=

0 1 –1 0

k=

0

–1

0 – –1 –1 0 While the matrices for i and k might look a little daunting, they can be simplified by replacing the – 1 elements with complex i. The last three of these matrices are used in the study of quantum theories to explain and predict the behaviour of electrons. They are called the Pauli Spin Matrices and students of chemistry will appreciate the importance of the spin of electrons in atomic bonding and the strength of different materials. A variation of these matrices used in the study of nuclear physics is shown below: P= 1 0 0 1

Q=

0 1 –1 0

R = 0 –1 1 0

S = –1 0 0 –1

P Q R S T U V W

Q S P R V T W U

R P S Q U W T V

S R Q P W V U T

T U V W S R Q P

T= 0 i V = –i 0 U= i 0 W = 0 –i i 0 0 i 0 –i –i 0 On examination of the first and second rows of the matrices above you will notice that the second row is a reflection of each matrix in the first row, multiplied by i. a Construct a Cayley Table to display the results of matrix multiplication using these 8 matrices. Arrange them in the order given, that is, from P to W. b Determine whether the total set forms a group. Yes c Mark off the top left-hand 4 × 4 corner. Examine this section of the table and show that this subset forms a group. This is an example of a subgroup, where a subset of a group forms a complete group of its own. 11 Internet search The real life applications of groups are quite complex. Use the internet to research this field of study. Include a list of distinct topics and a more detailed report that highlights the use of group theory.

P Q R S T U V W

01 –1 0 10 01

0 –1 1 0

–1 0 0 –1

0 i i0

W Q P

R

S

T

V

U

under multiplication.

i=

Chapter 4 An introduction to groups

199

History of mathematics CRYPTOGRAPHY Since World War II, mathematicians have played a large part in the development and attack on ciphers. A cipher is a way of scrambling text so it can be read only by the people who know how to unscramble it. Had the Government Code and Cipher School (at Bletchley Park, England) not been able to read the German Enigma ciphers, historians estimate that World War II may have lasted another two to three years. Until 1974, the contribution made by mathematicians to the war effort went largely unrecognised. Why include talk of ciphers and code breaking in a chapter on group theory? It is because, in the late 1970s, the two topics collided. In 1969, ARPAnet (the forerunner to the Internet) was born. Whitfield Diffie, a mathematician employed in a series of jobs related to computer security, could see the potential of ARPAnet but was concerned that people using email would be deprived of the right to privacy. At that time, if two people (say, Alice and Bob) wanted to encrypt a message, they needed to exchange a key. For Alice to send a secure message to Bob, encryption can n 2n be thought of as putting the message in a strong box and 0 1 locking it. The locked box can be sent securely. The problem is that Bob needs a key to open the box. What is known as the ‘key distribution problem’ is basically ‘How do Alice and Bob swap keys?’. Whitfield Diffie and a cryptographer called Martin Hellman worked on the key exchange problem from September 1974. In 1975, they came upon the ‘double padlock solution’. Essentially, Alice locks the box and mails it to Bob. Bob applies his own lock (leaving Alice’s in place) and returns it. Alice removes her lock and sends the box (still locked by Bob) back to Bob. Bob can now unlock the box. What was needed was a mathematical function that was the equivalent of a padlock (called a one-way function). Like a padlock, anyone could lock (encrypt), but without a key, the lock couldn’t be opened (decrypted). Group theory provides the solution to the one-way problem. Exponentiation (working out ab) is easy and, in the real number system, it is almost as easy to reverse using logarithms. In the group formed by multiplication modulo [largeprime-number], the logarithms are much harder to calculate. In the table at right, you can see that 2 is a generator for this group. Can you work out what modulo has been used? If you were given a number from the group formed from 2n, say 10, but you knew the modulo that had been used and that 2 was the generator, could you work out the logarithm or power (n) used? What if the modulo used was a much larger prime number?

1

2

2

4

3

8

4

16

5

13

6

7

7

14

8

9

9

18

10

17

11

15

12

11

13

3

14

6

15

12

16

5

17

10

18

1 (continued)

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

A solution to the key distribution problem is as follows. Alice and Bob agree to use mod 19 (in practice, it would be a massive prime number) and the generator 2 (anyone can know this information). Alice chooses her key, a (let’s make it 13). She evaluates 2a = 213 ≡ 3 mod 19 and sends this to Bob. Bob chooses his own key, b (let’s make it 16). He evaluates 2b = 216 ≡ 5 mod 19 and sends this to Alice.

When Bob receives 2a from Alice, he evaluates (2a)b ≡ 316 ≡ 17 mod 19. Alice, receiving 2b from Bob, can evaluate (2b)a ≡ 513 ≡ 17 mod 19. They can then use the key generated (2ab) to safely send messages. If an eavesdropper, Evan, intercepts their messages, he cannot calculate the key without first being able to solve what is called the ‘discrete log problem’. Publicly, three researchers from Massachusetts Institute of Technology (MIT), computer scientists Ron Rivest and Adi Shamir, and mathematician Leonard Aldeman, used modular arithmetic and properties of large primes to create the RSA algorithm for public key cryptography. (RSA stands for Rivest, Shamir, Aldeman.) If the primes used are large enough, this cipher is believed to be secure against attack. We now know that Diffie and Hellman were not the first people to develop the idea of public key cryptography. That honour must go to James Ellis, an employee of the British Government Communications Headquarters (formed from the remnants of Bletchley Park after World War II). Like Diffie and Hellman, Ellis was unable to find the one-way function. In 1969, he shared his idea with his bosses but it wasn’t until 1973 that a mathematician called Clifford Cocks found the answer using group theory. The discovery made by Ellis and Cocks was not publicly acknowledged until 1997.

Chapter 4 An introduction to groups

201

summary Modulo arithmetic • Modulo arithmetic is like clock arithmetic where 5 + 9 = 4 in mod 10. • The residues of modulo x are all the numbers less than x. • Congruent numbers in mod x all differ by multiples of x.

The terminology of groups For a set S = {u, a, b, c, …} and an operation ˚, we say that • the operation is closed if a ˚ b is an element of S • the operation is associative if (a ˚ b) ˚ c = a ˚ (b ˚ c) • u is the identity element (IE) if a ˚ u = u ˚ a = a. • a−1 is the inverse of a if a ˚ a−1 = a−1 ˚ a = u where u = IE.

Properties of groups • A set forms a group under an operation if elements of the set are closed and associative, and there is a unique identity element and every element has a unique inverse. • The group is an Abelian group if the operation is commutative (e.g. a ° b = b ° a).

Cyclic groups and subgroups For the group G = [S, ˚]: • a subgroup exists if there is a subset of S that is closed and contains the identity element, and if inverses for all elements exist. • the group is cyclic if an element a can be found that generates the group. This can be written as G = .

Transformations • The set of all transformations (for example, rotations and reflections) and the binary operation that combines any two of these transformations is referred to as a composition.

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1 a

CHAPTER review

+ 0 1 2 3 4

0 0 1 2 3 4

1 1 2 3 4 0

2 2 3 4 0 1

3 3 4 0 1 2

4 4 0 1 2 3

b

× 0 1 2 3 4 5

0 0 0 0 0 0 0

1 0 1 2 3 4 5

4A

1 Draw up a Cayley Table that shows the residues for each of the following: a addition mod 5 b multiplication mod 6.

4B

2 Determine whether the following are groups: a {1, −1} under multiplication Yes b {0, 1} under addition. No, not closed

4C

3 Determine whether the following are groups: a the set {1, 2, 4, 5, 7, 8} under multiplication modulo 9 Yes b the set {0, 1, 2, 3, 4} under multiplication modulo 5 No, 0 does not have an inverse c the set {2, 4, 6, 8} under multiplication modulo 10 Yes d the set {0, 1, 2} under addition modulo 3. Yes

2 0 2 4 0 0 4

3 0 3 0 3 0 3

4 0 4 2 0 4 2

5 0 5 4 3 2 1

4 Determine whether each of the following form groups: a the set of integers where p ° q = p + 2q No, no identity p b the set of positive rational numbers where p ° q = --- . No, not associative q c Show that the set of all integers forms an Abelian group under the operation a ° b = a + b − 3.

4C

5 There are two lights in a room, one on the ceiling and one on the wall with 4 possible states for the two lights — both on, both off, wall light on only, or ceiling light on only. There are 4 possible changes of state: no change, both change, ceiling light change and the wall light change. These changes are denoted by N, CW, C and W respectively. Show that the set {N, C, CW, W} forms a group with respect to the operation ‘followed by’. Check with your teacher.

4C

6 What property of a group is displayed in a Cayley Table if: There is only element x such that p ° x = q and a the elements are symmetrical about the leading diagonal Commutativity x ° q = p. b the same element does not appear more than once in any row or column c the identity element occurs only once in each row or column. Each element has a unique inverse.

Chapter 4 An introduction to groups

203

7 Determine whether the following are groups:

4C

a the set of integers, modulo n under addition Yes b the set of integers, modulo n under multiplication No, 0–1 doesn’t exist. c

the set of integers, modulo n, excluding 0, under multiplication No, inverses don’t always exist.

d the set of integers, modulo n, excluding 0, under multiplication, if n is prime. Yes 8 Determine whether the set of all moves that can be made by a knight on a chessboard forms a group or not. (The diagram shows three of the possible moves a knight can make on a chessboard. The knight can move two squares horizontally and one square vertically, or two squares vertically and one square horizontally — like a letter L.) Check with your teacher.

4C

9 Consider an operation defined by the following Cayley Table.

4D

˚

u

a

b

c

u

u

a

b

c

a

a

e

c

b

b

b

c

e

a

c

c

b

a

e

a Does the table define a cyclic group? No, there is no generator. b Does the subset {u, a, b} form a subgroup? No, the operation is not closed on these elements. c

Identify a subgroup and the generator of the subgroup. = {0, a}, = {0, b}, = {0, c}

Check with your teacher.

10 a Verify that the set

m 0 , where m ≠ 0 forms a group under matrix multiplication. 0 m

4E

b Verify that all p × q matrices form a group under matrix addition. 11 Show that the following set of matrices forms a group under multiplication. 1 0 0 1

–1 0 0 –1

i 0 0 i

–i 0 0 –i

0 1 1 0

0 –1 –1 0

0 i i 0

4E 0 –i –i 0

12 Determine whether or not the following functions form a group under composition of functions. Assume that they are associative. 1 1 x x+1 f3(x) = 1 + x f4(x) = -----------f5(x) = -----------f6(x) = -----------f1(x) = x f2(x) = --x 1+x x+1 x

4E

204

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Modelling and problem solving 1 If a group consists of 4 elements, the identity element u and a, b, c, and an operation ˚, then we can start to complete a Cayley Table as follows: Check with your teacher.

eBook plus Digital doc: Test Yourself Chapter 4

˚

u

a

b

c

u

u

a

b

c

a

a

✸

b

b

c

c

a Investigate the claim ‘The value indicated by ✸ can’t be filled with a’. b If ✸ is filled with the identity element, u, the table can be completed in two possible ways to describe a group. Find the two tables. (You need not check the associative law.) c If ✸ is filled with b (or c), there is only one way to complete the table that results in a group. Using b for ✸, find the table. (You need not check the associative law.) d If you consider the three tables you have produced, two of them describe the same structure. Determine which two tables these are and show how the elements in one table would have to be renamed to produce the other. 2 Earlier in this chapter, we found that G = [Z6, + modulo 6] is a cyclic group with two generators, 1 and 5. For two distinct primes p and q, investigate the number of generators for the group [Zpq, + modulo pq]. (Distinct primes means that p ≠ q.) Check with your teacher.

Matrices and their applications

5 syllabus reference Core topic: Matrices and applications

In this chapter 5A Inverse matrices and systems of linear equations 5B Gaussian elimination 5C Introducing determinants 5D Properties of determinants 5E Inverse of a 3 ¥ 3 matrix 5F Cramer’s Rule for solving linear equations

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Inverse matrices and systems of linear equations • inverse of a matrix • determinant of a matrix • solution of systems of homogeneous and nonhomogeneous linear equations using matrices • applications of matrices in both life-related and purely mathematical situations

The ideas you were introduced to in Chapter 3 on matrices provide a foundation for using matrices to solve simultaneous equations. The following section presents a simplified account of complex aspects of economies and industries and will serve to introduce the idea of matrix applications in a wide field of studies.

Input–output analysis

SLE 7: Use input-output (Leontief) matrices in economics.

To enable government bodies and companies to plan for future needs, input–output analysis models are employed. In these models the interaction between major components of an economy is analysed so that the effect of an increase in one component can be measured against the demand for one product in other industries. Input–output models were first developed by Nobel Prize winner, Wassily Leontief; they are used to describe economies where demand equals supply, or consumption equals production. These models can be applied to whole economies or to segments within economies. The model, as presented here, is based on the idea that there is a finite number of goods that are produced, consumed or used as input for the same finite number of industries. Each industry produces only one type of product and can use some of its own product. Each industry and its products are interdependent. The total demand of product is the sum of the demands throughout the entire production process as well as the demand for the product by consumers. Consider an economy comprising only 2 industries — coal and steel. One tonne of steel requires an input of 0.5 tonne of coal as well as 0.4 tonne of steel (perhaps in the form of machinery and plant). To produce 1 tonne of coal, 0.3 tonne of steel and 0.4 tonne of coal are required (perhaps to produce the steel needed for machinery). Suppose also that the final demand for steel is 18 tonnes and for coal 15 tonnes. Let x1 and x2 denote the total demands for coal and steel respectively. This information can be presented on an input–output table. User Coal

Steel

Final demand

Total demand

Coal

0.4

0.3

15

x1

Steel

0.5

0.4

18

x2

Producer

Chapter 5 Matrices and their applications

207

This information can be represented by a system of simultaneous linear equations: x1 = 0.4x1 + 0.3x2 + 15 x2 = 0.5x1 + 0.4x2 + 18 and put into matrix form: x1 x2

= 0.4 0.3 0.5 0.4

x1 x2

+ 15 18

which can be represented as X = AX + D where X =

x1

, A = 0.4 0.3 and D = 15 . 0.5 0.4 18 x2

Rearrangement of the matrices (as shown below) isolates X so it can be solved. X − AX = D (I − A) X = D where X post-multiplies other factors and I is the identity matrix 1 0 . 0 1 (I − A)–1(I − A)X = (I − A)–1D X = (I − A)–1D We call X the output matrix as it holds variables x1 and x2 that will state the total demand for steel and coal. We call A the technology matrix and D the final demand matrix. Because of its significance I − A is called the Leontief matrix. To calculate X we first need to find I − A. I − A = 1 0 − 0.4 0.3 0 1 0.5 0.4 I−A=

Therefore

x1 x2

0.6 – 0.3 – 0.5 0.6 = 0.6 – 0.3 – 0.5 0.6

–1

15 18

1 = --------------------------- 0.6 0.3 15 0.36 – 0.15 0.5 0.6 18 1 = ---------- 14.4 0.21 18.3 = 68.57 87.14 Thus, in order to provide the final demand of 18 tonnes of steel and 15 tonnes of coal in this economy, the steel industry must produce 87.14 tonnes of steel and the coal industry must produce 68.57 tonnes of coal. These values can also be regarded as equilibrium values — if these values are met then the input will match the output — thus eliminating both over- and underproduction.

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remember remember

1. A pair of equations, ax + by = u and cx + dy = v, may be written in the form

2. 3. 4.

5.

AX = B where A = a b , X = x and B = u . y v c d The matrix equation AX = B can be rearranged to X = A−1B so that values for x and y may be found. These values can also be called equilibrium values. Matrices can be used for input–output analysis. An input–output analysis matrix can be written as X = AX + D where: matrix X contains the variables to be determined and is called the output matrix matrix A contains information about the input details and is called the technology matrix matrix D contains information about the final demand and is called the final demand matrix. The matrix equation X = AX + D can be rearranged to make X the subject: X = (I − A)–1D

5A

Inverse matrices and systems of linear equations

1 The equilibrium state for 2 commodities is given by 2a − 3b = 25 3a − 2b = 60. 26 Find the equilibrium values for these goods. 9

2 Find the equilibrium values of P and Q in P = Q +Y Q = X + yP where Y = $6 million, X = $15 million and y = 0.5.

42 36

3 Find the equilibrium values of x = 45 − 2y 10.38 17.31 x = 0.6y. 400 370

4 Consider the simplified Keynesian system Y = C + 30 C = 50 + 0.8Y where Y denotes national income and C denotes consumption. Find the equilibrium values for Y and C. 5 Find the equilibrium values for the supply and demand model x = 21 − 4y (demand) 17 x = 14 + 3y (supply). 1 6 The technology matrix for an economy which produces only 2 commodities is given by x A = 0.2 0.6 and the final demand matrix D = 25 . Solve X = 1 . 0.1 0.5 36 x2

100.29 92.06

Chapter 5 Matrices and their applications

209

7 A simple economy is shown in the table. User p

q

Final demand

Total

p

0.25

0.4

30

x1

q

0.30

0.5

25

x2

Producer

98.04 108.82

3.18 0.681 1.81 1.81

Find the total demand matrix. 8 A certain economy consists of 2 industries, mining ore and manufacturing. The production of 1K dollars (K = 1000) of manufactured goods requires 0.6K dollars of manufactured goods and 0.15K dollars of ore, while the production of 1K dollars of ore requires 0.4K dollars of manufactured goods and 0.3K dollars of ore. The final demand for manufacturing and mining ore is 120K dollars and 145K dollars respectively. a Prepare a matrix table for this information. b Represent the information in matrix form, using X as the output matrix, A the technology matrix and D the demand matrix. c Find (I − A)–1. 480.68 d Find the total demand matrix (output matrix) X. 481.82 e Verify that D = (I − A) X.

Gaussian elimination SLE 1: Solve linear equations by using matrices and Gaussian elimination; solution of equations involving more than three variables will involve the use of graphing calculators.

The example based on the economy with only 2 industries used in the previous section is obviously a little unrealistic. However, if this economy were to involve more than 2 industries we have no tools to solve this type of problem, at this stage. This is because it would involve finding the inverse of a 3 × 3 matrix using the example of an economy with 3 industries, whereas we are limited to finding the inverse of a 2 × 2 matrix. A variety of methods can be used to solve a system of simultaneous equations, such as those generated in the previous section. We will concentrate at this stage on the method known as Gaussian elimination. In future sections of this chapter we will introduce 2 more methods, as well as graphics calculator techniques for solving simultaneous equations.

Using the Gaussian elimination method to solve simultaneous equations The Gaussian elimination method is described below. Consider a set of 2 simultaneous linear equations in 2 unknowns, x and y as given: ax + by = u cx + dy = v As we saw earlier, this system can be represented as A= a b B= u X= x c d y v where A is the coefficient matrix, B is the constant matrix and X is the variable or unknown matrix.

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We can combine matrices A and B as a b u where this is referred to as an augmented matrix, denoted by [A | B]. c d v The following example shows the familiar algebraic process involved in solving simultaneous equations, only this time it is in matrix form. These steps will work towards producing the identity matrix I on the left-hand side and then the solution will be on the right-hand side. [A | B] → [I | X]

WORKED Example 1 Solve this system of simultaneous linear equations: x + 2y = 3 2x + 3y = 5 THINK

WRITE

1

Set equations in matrix form.

A= 1 2 2 3

2

Convert to augmented matrix form.

[A | B] = 1 2 3 2 3 5

3

Aim to produce the identity matrix on the left-hand side of the augmented matrix. Element a11 is already 1. We need to eliminate the a21 element to produce a 0. Replace R2 with R2 − 2R1. That is, replace row 2 with row 2 minus twice row 1.

1 2 3 R2 − 2R1: 0 – 1 – 1

4

Next we want a22 to be 1. Multiply R2 by −1.

1 2 3 −R2: 0 1 1

Finally, we want to make a12 = 0. Replace R1 with R1 − 2R2.

R1 − 2R2: 1 0 1 0 1 1

[A | B] now resembles [I | X] and X = 1 . 1

[I | X] = 1 0 1 0 1 1

5

6

7

Verify your answers by substitution into the original equations.

B= 3 5

X=

x y

X = 1 , that is x = 1 and y = 1 1 Check: 1 + 2(1) = 3 (true) 2(1) + 3(1) = 5 (true)

All the communication given to the left of each matrix should be included in your own work as justification for how you have proceeded from one step to the other, and also for your own reference.

Chapter 5 Matrices and their applications

211

The following procedures are used in this method (they match operations you use when solving equations by elimination): 1. multiply a row (equation) by a constant 2. add and subtract one row (equation) to or from another 3. swap rows (equations). The object is to have ones in the leading diagonal and zeros under the leading diagonal. This is known as row-echelon form. Once this is achieved, the solution can be found either algebraically or by continuing with Gaussian elimination until all elements above the leading diagonal are zero. This is known as reduced row-echelon form. Let’s investigate these ideas by working with a 3 × 3 matrix.

WORKED Example 2 Solve this system of linear equations. x + 2y + z = 8 x−y+z=7 x + y + 2z = 4 THINK

WRITE

1

Set the equations in matrix form.

1 2 1 A = 1 –1 1 1 1 2

2

Convert to augmented matrix form.

1 2 1 8 [A | B] = 1 – 1 1 7 1 1 2 4

3

Element a11 is already 1. Eliminate a21 (make it become 0) by replacing R2 with R2 – R1.

1 2 1 8 R2 – R1: 0 – 3 0 – 1 1 1 2 4

4

Next eliminate a31 by replacing R3 with R3 − R1.

1 2 1 8 0 –3 0 –1 R3 – R1: 0 – 1 1 – 4

5

To obtain a 1 in a22, we could divide R2 by −3 (giving us 1--- in 3 b21) or swap R2 and R3. Let’s swap the rows and then multiply the new R2 by −1.

1 2 1 8 Swap R2 and R3: 0 – 1 1 – 4 0 –3 0 –1

Eliminate a32 by replacing R3 with R3 + 3R2.

1 0 R3 + 3R2: 0

6

8 B= 7 4

X=

x y z

1 2 1 8 – R2: 0 1 – 1 4 0 –3 0 –1 2 1 8 1 –1 4 0 – 3 – 11

Continued over page

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THINK 7

WRITE

To obtain a 1 in a33, multiply R3 by − 1--- .

1 2 1 0 1 –1

3

8 4

-----− 1--- R3: 0 0 1 – 11 3 3

8

Row-echelon form has now been achieved. We can solve for x, y and z using equations or continue to reduced row-echelon form. Using equations R3 gives z. Substituting the value for z in R2 gives y. Substituting the values for both y and z in R1 gives x. Continuing to reduced row-echelon form We first need to eliminate above a33. Eliminate a23 by replacing R2 with R2 + R3. Eliminate a13 by replacing R1 with R1 – R3.

-----z = − 11 3 ------ in y – z = 4 gives y = 1 --- . Substituting z = − 11 3

Substituting y = x = 11.

1 --3

3

------ in x + 2y + z = 8 gives and z = − 11 3

Alternatively: 1 2 1

8

R2 + R3: 0 1 0

1 --3

-----0 0 1 – 11 3

R1 − R3: 1 2 0

35 -----3

0 1 0

1 --3

-----0 0 1 – 11 3

Now eliminate above a22. Eliminate a12 by replacing R1 with R1 − 2R2.

R1 − 2R2: 1 0 0

11

0 1 0

1 --3

0 0 1

-----– 11 3

11 9

State the solution.

Therefore, X =

1 --3

------ . where x = 11, y = 1--- , z = − 11 3

-----– 11 3

10

Verify your results by substitution into at least one of the original equations. In this case, substitute values for x, y and z into the left side of the third equation.

Check: substituting into x + y + 2z gives ------ ) 11 + 1--- + 2(− 11 3

= 11 + =4

1 --3

−

3

22 -----3

Solution is verified.

3

Chapter 5 Matrices and their applications

213

This method can be quite frustrating if you ‘get lost’. Work on only one row at a time in any one step, unless the operation is quite straightforward and the line in question isn’t involved in any other procedures in that step. (For example, Step 5 in Worked example 2 would satisfy these criteria.) Use the following steps: 1. Start with a 1 in position a11, by swapping rows, dividing by a suitable number or both. 2. Eliminate the elements (make them become 0) below a11, starting with a21 and working down. 3. Work to get a 1 in position a22 by either swapping rows, dividing by a suitable number or both. 4. Eliminate the elements below a22, starting with a32 and working down. 5. Continue until there are ones in the leading diagonal and zeros below it (rowechelon form). 6. Once you have row-echelon form, decide if you are going to work algebraically or continue to reduced row-echelon form. 7. If you are working to reduced row-echelon form, start with the far right-hand column and work up and to the left.

Using the Gaussian elimination method to find an inverse In the previous section we used an augmented matrix [A | B] to find solutions for a system of linear equations. The Gaussian elimination method can also be used to find the inverse of matrix A. This is achieved by augmenting A with I — as in [A | I] and performing row reduction procedures to produce [I | A–1]. This is shown in the following example.

WORKED Example 3 Find the inverse of A = 1 1 . 1 2 THINK

WRITE

Set up the augmented matrix [A | I].

[A | I] = 1 1 1 0 1 2 0 1

2

Perform Gaussian elimination to obtain the identity matrix on the left side of the augmented matrix. Element a11 is already 1. Eliminate a21 by replacing R2 with R2 – R1.

1 1 1 0 R2 – R1: 0 1 – 1 1

3

Element a22 is already 1. Next eliminate a12 by replacing R1 with R1 – R2.

R1 – R2: 1 0 2 – 1 0 1 –1 1

4

This augmented matrix is now in the form [I | A−1].

1

1 0 2 – 1 = [I | A−1] 0 1 –1 1 A−1 =

5

Check this is so by verifying A−1A = I.

Check: A−1A =

2 –1 –1 1 2 –1 1 1 = 1 0 = I –1 1 1 2 0 1

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WORKED Example 4 1 2 1 Find A–1 (if it exists) for A = 1 0 1 . 0 1 3 THINK

WRITE

1

Set up the augmented matrix.

1 2 1 1 0 0 [A | I] = 1 0 1 0 1 0 0 1 3 0 0 1

2

Perform Gaussian elimination to obtain the identity matrix on the left side of the augmented matrix. Element a11 is already 1. Eliminate a21 by replacing R2 with R2 − R1. Note that a31 is already 0.

1 2 1 1 0 0 R2 − R1: 0 – 2 0 – 1 1 0 0 1 3 0 0 1

3

We need a 1 in a22. The easiest way is to swap R2 and R3.

1 2 1 1 0 0 Swap R2 and R3: 0 1 3 0 0 1 0 –2 0 –1 1 0

4

Eliminate a32 by replacing R3 with R3 + 2R2.

1 2 1 1 0 0 0 1 3 0 0 1 R3 + 2R2: 0 0 6 – 1 1 2

5

Make a33 = 1 by dividing R3 by 6.

1 2 1 0 1 3 1 --- R3: 6

6

7

Row-echelon form has now been achieved. We now need to eliminate above a33. Replace R1 with R1 − R3 and replace R2 with R2 − 3R3. Now eliminate above a22. Replace R1 with R1 − 2R2.

1 0 0 0 0 1

0 0 1 – 1--6-

1 --6

1 --3

R1 − R3: 1 2 0

7--6

– --61- – 1--3-

R2 − 3R3: 0 1 0

1 --2

– 1--2-

0

0 0 1 – 1--6-

1 --6

1 --3

R1 − 2R2: 1 0 0

1 --6

5 --6

– --13-

0 1 0

1 --2

– --12-

0

0 0 1 – --16-

1 --6

1 --3

Chapter 5 Matrices and their applications

THINK

8

WRITE

This is now in the form [I | A−1]. State A−1.

1 0 0

1 --6

5 --6

0 1 0

1 --2

– 1--2-

0 0 1 – 1--6-

1 --6

A−1 =

9

215

Verify that A−1A = I.

1 --6

5 --6

– 1--3-

1 --2

– 1--2-

0

– 1--6-

1 --6

1 --3

– 1--3−1 0 = [I | A ] 1 --3

Check: A−1A =

1 --6

5 --6

1 --2

– 1--2-

– 1--6-

1 --6

– 1--3-

1 2 1 0 1 0 1 1 0 1 3 --3

1 0 0 = 0 1 0 0 0 1 =I

remember remember For a set of simultaneous equations represented by matrices A, X, B such that AX = B, Gaussian elimination can be used to change an augmented matrix [A | X] to [I | X] and [A | I] to [I | A–1] where I is the identity matrix.

5B

Gaussian elimination

Use the Gaussian method of elimination to solve the following systems of linear equations. Example

1 2x − y = 1 x = 1, y = 1 3x + 2y = 5 1

2 3a + 2b = −11 a = −3, b = −1 a + 3b = −6

3 2y − z = 5 x = 2, y = 1, z = −3 Example x + z = −1 2 2x − y − z = 6

4 3x + 4y − z = 11 x − y + 2z = −1 5x + y − 3z = 7

WORKED

eBook plus Digital doc:

WORKED

5 3x + 4y + z = −10 2x + y + 2z = −5 x − 2y + 2z = 0

x = −2, y = −1, z = 0

x = 1, y = 2, z = 0

6 x + y − z = 6 x = −3, y = 5 1--4- , z = −3 3--4x + 2y + 2z = 0 −2x − y + z = −3

SkillSHEET 5.1 Using Gaussian elimination to solve linear equations

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7 Use Gaussian elimination to find the inverses of the following matrices, if they exist. WORKED

Example xample

3

a

1 2 2 3

c

1 2 2 0 1 3 1 3 2

WORKED

Example xample

4

–3 2 2 –1

b

2 1--3- – 2--3- – 1 1--3–1

0

1

--13

--13

– --13-

d

3 1 2 2 2 1 0 0 2 0 2 2 1

1 --2

– --14-

– 1--2-

3--4 1--2

0

– 1--4- 0 --12

0

– 1 – 1--2- 1

8 Write each of the systems of linear equations below in the matrix form AX = B and find inverses to solve the equations. a 2x + y = 6 b 3x + 2y = 9 c x + y − z = −6 x = 2.7, y = 0.6 1 x + 3y = 4 --x + 4y = 7 2x − y + 2z = 1 2

2x − z = 5 x=

2 1--5-

x + 2y = −7 ,y=

1 1--5-

, z = − --35-

x = −2 1--7- , y = −2 3--7- , z = 1 3--7-

History of mathematics CARL FRIEDRICH GAUSS (1777–1855) During his lifetime . . . Karl Marx and Friedrich Engels jointly publish ‘The Communist Party Manifesto’. The Frenchman Nicéphore Niepie produces the world’s first photographic image. Samuel Morse develops the ‘Morse Code’. You will come across the work of Carl Gauss in many fields of mathematics. His work is so diverse that he is considered by many to be the greatest mathematician of all times. Gauss was born in Brunswick, Germany, on 20 April 1777. From the age of three he had shown his superior skills by performing mental calculations and by the age of ten had progressed to algebra and analysis. While still a teenager he had developed the ‘least squares’ method used in statistical data, and had devised a proof that a regular 17-sided polygon could be constructed using a compass and ruler and his quadratic reciprocity theorem. At the age of 22 he received his PhD, proving the Fundamental Theorem of Algebra. The next year he published his work on number theory, organising previous work and introducing the notion of modular arithmetic.

In 1801, he used the information from three sightings of an asteroid, Ceres, to calculate its orbit. In the process of this work he showed that the variation involved in experimental data followed a bell-shaped curve, now called the Gaussian or normal distribution. In 1807 Gauss became professor of astronomy and director of the new observatory in Göttingen. His work involved branches of astronomy, mechanics, optics, geodesy and magnetism, and in collaboration with Weber, the first practical telegraph. His extensive use of complex numbers advanced the acceptance of them by fellow mathematicians, although he was generally not supportive of young, aspiring scholars. He died in Göttingen in 1855. His memorial bears the 17-point star of his early fame.

Questions 1. Try to reproduce the 17-point star with all angles and sides the same. 2. Research the Fundamental Theorem of Algebra. Science of determining the 3. What is geodesy? size and shape of the Earth. 4. Find out about the quadratic reciprocity theorem.

Chapter 5 Matrices and their applications

217

Performing Gaussian elimination using a graphics calculator Matrix row operations can be performed on your graphics calculator. To demonstrate this, we will repeat the steps of Worked example 4 to find the inverse of 1 2 1 A= 1 0 1 . 0 1 3

For the Casio fx-9860G AU ▼

1. Enter Matrix A. (a) Press F1 ( MAT) to enter the matrix editing screen. (b) Highlight Mat A and press EXE or F3 (DIM). (c) Specify the number of rows, 3 in this case, and press EXE . (d) Specify the number of columns, 3 in this case, and press EXE . (e) Press EXE again to display the 3 × 3 array for matrix A. (f) Enter the values for the elements, pressing EXE after each number. 2. Exit the Matrix input screen by pressing EXIT . Press EXIT again to return to the MAT screen. 3. Set up an augmented matrix B = [A | I]. Enter the augmented matrix [A | I] first and then save as matrix B.

s

(a) Press OPTN then F2 (MAT) to bring up the matrix menu. Press F5 (Aug), then F1 (Mat), then ALPHA [A] followed by , then press F6 ( ) and F1 (Iden). Enter 3 (for a 3 × 3 identity matrix) and press ) to close the set of brackets. s

(b) Press Æ then F6 ( ) then F1 (Mat) and ALPHA [B] to save the augmented matrix as matrix B. (c) Press EXE . Matrix B will be displayed. You can use the cursor keys to see more of the augmented matrix.

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

4. To perform row operations, we need to be in the matrix editing screen. (a) Press EXIT three times to return to the MAT screen as seen at right. (b) Press F1 ( MAT). Highlight Mat B and press EXE . Notice that this screen is different from the one obtained in Step 3. ▼

218

5. Replace R2 with R2 − R1 (or − R1 + R2). (a) Press F1 (R-OP) for the Row Operation menu. (b) Press F3 (×Rw+). This allows the addition of one row and the product of a specified row with a scalar to be found. (c) Input the scalar multiplier. Enter −1 for k and press EXE . (d) Specify the number of the row to be multiplied by the scalar. Enter 1 for m and press EXE . (e) Specify the number of the row that the result should be added to. Enter 2 for n and press EXE . (f) Press EXE . Compare this screen with the matrix obtained in Step 2 of Worked example 4.

6. Swap R2 and R3. (a) Press F1 (SWAP). (b) Specify the number of the rows to be swapped. Enter 2 for m and press EXE . Enter 3 for n and press EXE . (c) Press EXE . Compare this screen with the matrix obtained in Step 3 of Worked example 4.

Chapter 5 Matrices and their applications

7. Replace R3 with R3 + 2R2. (a) Press F3 (×Rw+). As before, this allows addition of one row and the product of a specified row with a scalar. (b) Enter 2 for k (this is the scalar multiplier) and press EXE . Enter 2 for m (this is the row to be multiplied) and press EXE . Enter 3 for n (this is the number of the row where the result should be added) and press EXE . (c) Press EXE . Compare this screen with the matrix obtained in Step 4 of Worked example 4.

8. Multiply R3 by 1--- . 6

(a) Press F2 (×Rw). This allows the product of a specified row and scalar to be found. (b) Using the fraction key, enter 1--- for k (this is 6 the scalar multiplier) and press EXE . Enter 3 for m (this is the row to be multiplied) and press EXE . (c) Press EXE . Compare this screen with the matrix obtained in Step 5 of Worked example 4.

9. Now that we have row-echelon form, continue performing operations until reduced rowechelon form is achieved.

10. Use the cursor keys to scroll across to see A−1.

Now complete the questions at the end of this investigation.

219

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For the TI-Nspire CAS 1. Enter Matrix A. (a) Open a new Calculator document. Press k to access the catalog. Select Option 5 (by pressing 5) then highlight the m-by-n matrix symbol. (b) Press ·. Create the matrix with Number of rows: 3 and Number of columns: 3, pressing e to move from one box to the next. (c) Highlight OK and press ·. Fill in the values in the matrix using the cursor keys to move from one element to the next. (d) Move the cursor to the right of the matrix. Press / and h, then press A to store the matrix as matrix A. Press · to display the matrix. 2. Set up an augmented matrix B = [A | I]. Enter the augmented matrix [A | I] first and then save as matrix B. (a) Press b and select 7: Matrix & Vector (by pressing 7 or highlighting 7: Matrix & Vector and pressing ·). Select 6: Create and then select 7: Augment. (b) Press A then press the comma key (,). To enter the identity matrix, press b and select 7: Matrix & Vector then 6: Create and 2: Identity. Press 3 (for a 3 × 3 identity matrix) and press ) twice to close the two sets of brackets. Press / and h, then press B to save the augmented matrix as matrix B. Press · to display the matrix. 3. Replace R2 with R2 − R1 (or − R1 + R2). (a) To proceed to the Row Operations menu, press b then select 7: Matrix & Vector and 9: Row Operations. For − R1 + R2, select 4: Multiply Row & Add. (b) Input the scalar multiplier, the matrix name, the row number to be multiplied and then the row number for the result to be added to, each separated by a comma. In this case, enter −1, B, 1, 2. Press ) to close the set of brackets, then press /, h and C to save the matrix as matrix C. (You can continue to save the new matrix formed as matrix B if you wish or use a new name after each row operation.) (c) Press ·. Compare this screen with the matrix obtained in Step 2 of Worked example 4.

Chapter 5 Matrices and their applications

221

4. Swap R2 and R3. (a) Proceed to the Row Operations menu and select 1: Swap Rows. (b) Input the matrix name and the two row numbers to be swapped, each separated by a comma. In this case, enter c, 2, 3. Press ) to close the set of brackets then press /, h and D to save the matrix as matrix D. (c) Press ·. Compare this screen with the matrix obtained in Step 3 of Worked example 4. 5. Replace R3 with R3 + 2R2. (a) As before, proceed to the Row Operations menu and select 4: Multiply Row & Add. (b) Enter 2, D, 2, 3 to represent 2 × R2 + R3 in matrix D. Press ) to close the set of brackets, then press /, h and E to save the matrix as matrix E. (c) Press ·. Compare this screen with the matrix obtained in Step 4 of Worked example 4. 6. Multiply R3 by 1--- . 6 (a) Proceed to the Row Operations menu and select 3: Multiply Row. (b) Enter 1--- , E, 3 to represent 1--- × R3 in matrix E. 6

6

Press ) to close the set of brackets, then press /, h and F to save the matrix as matrix F. (c) Press ·. Compare this screen with the matrix obtained in Step 5 of Worked example 4.

1 a x = 2.7, y = 0.6

7. Now that we have row-echelon form, continue performing operations until reduced row-echelon form is achieved.

b x = 2 1--5- , y = 1 1--5- , z = − 3--5c x = −2 --17- , y = −2 --37- , z = 1 --37-

QUESTIONS 1 Repeat the calculations for Exercise 5B Question 7 using a graphics calculator. 2 Solve the following system of equations by performing row operations using a graphics calculator. x = 1, y = 0, z = 3 x − y − 4z = −11 6x + 2y + z = 9 −3x − 2y + 2z = 3

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Introducing determinants As mentioned in the previous section, Gaussian elimination is just one method used to solve systems of linear equations. Other methods involving determinants of matrices were used as early as 1100 BC by the Chinese and more recently by Gabriel Cramer (1704–1752) and Augustine Cauchy (1789–1857). You were introduced to determinants in Chapter 3 on matrices where a quick test to determine whether a matrix was singular (that is, had no inverse) was to evaluate its determinant. 1 1 For example, for A = 2 2 A–1 = ------------------------------ 1 – 2 = --- 1 – 2 . 2 × 1 – 2 × 1 0 –1 2 –1 2 1 1 Because the determinant of A = 0, no inverse exists such that A–1A = I. By definition, if A = [a], then det A = | a | for a 1 × 1 matrix.

Determinant of a 2

¥

If A = a b then det A = c d

a b c d

2 matrix = ad - bc.

Therefore the determinant can only be found for a square matrix A and is denoted by straight brackets, | |, not the square brackets [ ] used for a matrix. Its value is a single numerical answer, not a table of values like a matrix.

WORKED Example 5 Evaluate the determinant of A = 2 1 . 3 2 THINK

WRITE

For A = a b , det A = ad − bc. c d

det A = 2 × 2 − 3 × 1 =4−3 =1

¥ 3 matrix As with the inverse of matrices, we need to be able to find determinants of matrices larger than 2 × 2.

Determinant of a 3

The determinant of a 3 × 3 matrix involves evaluating three 2 × 2 determinants. a b c e f d f d e -b +c d e f , det A = a h i g i g h g h i where the 3 sub-determinants are referred to as minors.

If A =

Therefore det A = a(ei - fh) - b(di - fg) + c(dh - eg) = aei - afh - bdi + bfg + cdh - ceg

Chapter 5 Matrices and their applications

223

Note that the coefficients of each sub-determinant are the elements of row 1 and the minor of a11 is formed by using elements not in row 1 or column 1. That is, mentally cross out the first row and first column. a b c d e f and similarly for the other minors. g h i The second coefficient is given a negative sign.

WORKED Example 6 Evaluate the determinant THINK 1 Use elements of row 1 as the coefficients of the minors.

2

Evaluate minors.

2 1 3 1 –1 2 . –1 2 0 WRITE 2 1 3 1 –1 2 –1 2 0 = 2 –1 2 − 1 1 2 + 3 1 –1 2 0 –1 0 –1 2 = 2(−1 × 0 − 2 × 2) − 1(1 × 0 − 2 × −1) + 3(1 × 2 − −1 × −1) = 2(0 − 4) − 1(0 + 2) + 3(2 − 1) = −8 − 2 + 3 = −7

If det A = 0 then the inverse of A does not exist. In this case A is said to be singular. A special set of simultaneous equations such as ax + by = 0 cx + dy = 0 is said to be homogeneous, where all the right-hand side constants are zero. The trivial solution to this system yields x = y = 0. If det A = 0 then an infinite number of non-trivial solutions exist. However, if det A ≠ 0 only the trivial solutions exist. You will encounter questions later in this chapter that deal with this situation.

remember remember

1. By definition, if A = [a], then det A = | a | for a 1 × 1 matrix. 2. If A = a b then det A = a c c d

b = ad − bc. d

a b c e f −b d f +c d d e f , det A = a h i g i g g h i where the 3 sub-determinants are referred to as minors.

3. If A =

e h

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5C

Introducing determinants

Evaluate the determinants of the following matrices: WORKED

Example xample

5

WORKED

Example xample

1

2 3 0 1

5 1 2 3

2

2

3

13

0 1 1 --2

− 1--2-

3

4

1 --2

1

1 --4

2

eBook plus --34

Digital docs: SkillSHEET 5.2 Determinants of 2 × 2 and 3 × 3 matrices

5 Evaluate the following determinants:

6

a

1 –3 2 –1 4 3 0 2 5

b

–1 –1 0 –3 4 2 2 3 5

d

1 1 1 2 2 3 0 3 4

−3

e

2 5 –1 3 –2 4 –1 –2 –3

g

1 1 1 2 2 2 3 3 3

0

h

1 2 3 0 4 5 0 0 6

j

4 10 – 2 3 –2 4 –1 –2 –3

−5

−33

c

0 2 –3 –3 1 2 6 3 2

61

f

0 0 0 2 3 –5 6 8 1

i

2 1 3 4 0 5 0 0 6

24

81

0

−24

eBook plus

122

Digital doc: WorkSHEET 5.1

Properties of determinants In question 5 of the previous exercise, review parts a and b, f, g and h. Do you notice any patterns that you think could occur in other situations? In fact there are 8 properties of determinants that can be very useful. We have already used one of them (see property 6 given on the next page) in using the Gaussian method to solve systems of linear equations. The 8 properties of determinants are given below.

Property 1

Determinant of a transpose

For every square matrix A, det A = det A′ where det A′ is the determinant that results from the transpose of A as seen in questions 5 a and b above.

Property 2

Identical rows

If 2 or more rows (or columns) of a matrix are identical or in proportion, then det A = 0 (this can be seen in question 5 g).

Chapter 5 Matrices and their applications

Property 3

225

Zero rows/columns

If all the elements of a row (or column) are zero, then det A = 0 (see question 5 f).

Property 4

Interchanging rows or columns

If 2 rows (or columns) of A are interchanged to give B, then det A = −det B (see questions 5 h and i).

Property 5

Multiples of rows/columns

If a row (or column) of matrix A is multiplied by a constant k (where k ≠ 0), to give matrix B, then det B = k det A (see questions 5 e and j).

Property 6

Adding rows/columns

If a non-zero multiple of a row or column of A is added to another row or column, then the determinant is unchanged.

Property 7

Zero lower-triangular matrix

If all the elements below the leading diagonal are zero, then det A equals the product of the elements on the leading diagonal (see question 5 h).

Property 8

Det I

If I is the identity matrix then det I = 1 (this property follows from property 7). Some of these properties are easier to identify than others. For instance, a matrix with a zero row or column is readily identified. As well as these properties, there are other short-cut methods that are very convenient for calculating determinants. The following expansion is one of these.

Expansion of a 3 or column

×

3 determinant using any row

The initial example of expansion of a 3 × 3 determinant (Worked example 6 on page 223) used the first row as coefficients of the minors. However, any row or column can be used — with the following adaptations. 1. A minor can be obtained by blocking out all the elements of the row and column of the coefficient. 2. Alternating signs are attached to the coefficients of each minor, as shown: + − + − + − + − + Note: The initial example (Worked example 6) used a (−) in front of the second coefficient because the first row elements were used as the coefficients of expansion.

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If the elements of the second row had been used as the coefficients of expansion, then the signs on the minors would have been −, +, −. The same would have happened if the second column had been used. The signed minors are called cofactors. Follow the next worked example to see how this alternative row or column can be used.

WORKED Example 7 1 2 1 Evaluate 1 3 6 . 1 4 9 THINK 1

WRITE

Because we can expand by any row or column, use column 1 as it will have coefficients of 1. (a) With a different colour pen write in the signs of the elements. Don’t get them confused and think a21 = −1. (b) Draw an arrow to indicate the row/column of expansion.

(+)

1 2 1

(–)

1 3 6

(+)

1 4 9

2

Write out the expansion with the signed minors (cofactors).

=1 3 4

6 −1 2 9 4

1 +1 2 9 3

1 6

3

Complete the expansion. Take care with the minus sign and the brackets.

= 1(27 − 24) − 1(18 − 4) + 1(12 − 3) = 3 − 14 + 9 = −2

In the example above, any row or column could have been used, but you can see that if column 1 is used, the coefficients will be 1 — a number that is easy to multiply by. Also, any row or column with mostly zeros allows you to complete the expansion faster, so it is useful to use that row or column to expand by. For square determinants greater than 3 × 3, this simple alternating pattern of signs is continued. For example, a 4 × 4 determinant can be evaluated using alternating signs of the coefficients. The signs to be used in front of the coefficients are shown below. + – + –

– + – +

+ – + –

– + – +

Chapter 5 Matrices and their applications

227

remember remember All determinants have the following properties: 1. For every square matrix A, det A = det A′ where det A′ is the determinant that results from the transpose of A. 2. If 2 or more rows (or columns) of a matrix are identical or in proportion, then det A = 0. 3. If all the elements of a row (or column) are zero, then det A = 0. 4. If 2 rows (or columns) of A are interchanged to give B, then det A = −det B. 5. If a row (or column) of matrix A is multiplied by a constant k (where k ≠ 0), to give matrix B, then det B = k det A. 6. If a non-zero multiple of a row or column of A is added to another row or column, then the determinant is unchanged. 7. If all the elements below the leading diagonal are zero, then det A equals the product of the elements on the leading diagonal. 8. If I = identity matrix then det I = 1.

5D

Properties of determinants

Evaluate the following determinants:

WORKED

Example

1 a

1 4

2 a

1 2 1 3 6 0 1 2 1

3 a

0 0 4 2 0 5 3 0 –2

4 a

2 3 6 1 –1 1 2 1 0

5 a

2 4

5 1

6 a

2 0

4 1

7 a

2 –1 3 0 4 7 0 0 1

7

2 5

−3

b

0

1 2

4 5

−3

c

State the property involved.

b State the property involved.

Property 2

0

b State the property involved.

Property 3

22

b

3 2 6 –1 1 1 1 2 0

−18

b

4 4

10 1

2

b

2 2

4 5

8

−22

Property 1

c

State the property involved.

Property 4

−36

c

State the property involved.

Property 5

2

c

State the property involved.

Property 6

b State the property involved.

Property 7

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

8 a

1 0 0 0 1 0 0 0 1

1

b State the property involved.

Property 8

9 Use any of the properties investigated in earlier sections to evaluate the following determinants: a

2 3 4 –1 1 1 0 1 0

d

1 0 0 –1 0 4 3 2 –1

−6

−8

b

4 3 1 –1 6 1 2 5 1

e

2 1 –1 2

1 0 0 0

1 0 0 1

−4

3 1 2 3

3

c

1 1 –1 0 1 1 2 1 0

f

4 0 0 0

3

2 1 –0 2 –1 1 –0 1 3 0 0 5

40

Inverse of a 3 × 3 matrix In Chapter 3 on matrices you were introduced to the idea of an inverse matrix A–1 where AA–1 = A–1A = I. For A =

a b c d

1 the inverse matrix A–1 = ------------------ d – b , where ad - bc ≠ 0 and ad – bc – c a

ad - bc = det A. This rule is limited to 2 × 2 matrices. However, in its most general form it can be used to find the inverse of any square matrix, if the inverse exists. The steps below demonstrate how the above formula can be used to find the inverse of a 3 × 3 matrix. 2 3 1 For matrix A = 4 6 5 9 7 8 1. Matrix C, the cofactor matrix of A, is obtained by replacing each element of A with its corresponding cofactor or signed minor.

C=

6 5 7 8

– 4 5 9 8

4 6 9 7

– 3 1 7 8

2 1 9 8

– 2 3 9 7

3 1 6 5

– 2 1 4 5

2 3 4 6

13 13 – 26 = – 17 7 13 9 –6 0

Chapter 5 Matrices and their applications

229

2. The adjoint of A is the transpose of the cofactor matrix and is written adj A. adj A =

13 – 17 9 13 7 –6 – 26 13 0

The adjoint matrix has the property that A(adj A) = (det A)I and since det A is a scalar we can divide by det A to produce adj A A -------------- = I det A therefore adj A------------= A–1. det A You can verify this result by showing A–1A = I. In its simplest form, this is the formula that was used to find the inverse of 2 × 2 matrices.

WORKED Example 8 1 –1 2 Use the cofactor/adjoint matrices to find the inverse of A = 0 1 2 . 0 1 3 THINK

1

Set up the cofactor matrix using signed minors and evaluate.

WRITE

C=

C=

2

Write the adjoint as the transpose of C.

1 2 1 3

– 0 2 0 3

0 1 0 1

– –1 2 1 3

1 2 0 3

– 1 –1 0 1

–1 2 1 2

– 1 2 0 2

1 –1 0 1

1 0 0 5 3 –1 –4 –2 1

1 5 –4 adj A = 0 3 – 2 0 –1 1 Continued over page

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

THINK

WRITE

3

Calculate det A.

det A = 1 1 1

2 − 0 + 0 = 1 (down column 1) 3

4

Set up A−1.

adj A A−1 = ------------det A

5

Calculate A−1.

1 5 –4 A−1 = 0 3 – 2 0 –1 1

6

Verify this matrix is A−1 by testing AA−1 = I.

Check: 1 –1 2 AA−1 = 0 1 2 0 1 3

1 5 –4 0 3 –2 0 –1 1

1 0 0 AA−1 = 0 1 0 0 0 1 AA−1 = I

This method can now be used to solve systems of linear equations involving a 3 × 3 matrix.

WORKED Example 9 Solve the system of linear equations given below. x−y−z=0 2x + y = 4 x+y+z=2 THINK

1

Set up the matrices in the form AX = B.

WRITE AX = B

X=

2

Rearrange to change X to be the subject.

x y z

1 –1 –1 where A = 2 1 0 1 1 1 and B =

A−1AX = A−1B X = A−1B

0 4 2

Chapter 5 Matrices and their applications

THINK

3

Use the cofactor matrix to find A−1.

231

WRITE

C=

1 0 1 1

– 2 0 1 1

2 1 1 1

– –1 –1 1 1

1 –1 1 1

– 1 –1 1 1

–1 –1 1 0

– 1 –1 2 0

1 –1 2 1

1 –2 1 C = 0 2 –2 1 –2 3

4

Find the adjoint of A.

5

Find the determinant of A.

1 0 1 adj A = – 2 2 – 2 1 –2 3 det A = 1 – 1 – 1 –1 1 – 1 +1 1 – 1 (across row 3) 1 0 2 0 2 1 det A = 1 × 1 − 1 × 2 + 1 × 3 det A = 2

6

Find the inverse.

adj A A−1 = ------------det A 1 0 1 1 A−1 = --- – 2 2 – 2 2 1 –2 3

A

−1

1 --2

= –1 1 --2

1 --2

7

Solve for X and check the values provided.

X = –1 1 --2

1 --2

0

1 –1 –1

1 1--2-

0

1 --2

1 –1 –1

1 1--2-

0 4 2

1 2 –1 So x = 1, y = 2, z = −1. X=

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remember remember 1 a i 1

ii

3 –2 –1 1

1. The determinant of a 3 × 3 matrix can be found using signed minors. The signs on the minors can be determined from the following: + − + − + − + − +

3 – 1 iv 3 – 1 –2 1 –2 1

iii

b i −12

ii

9 –5 –6 2

iii

iv

9 –6 –5 2 – 3--4-

5 -----12

--12

– --16-

adj A 2. A–1 = ------------- where adj A = C′, the transpose of the cofactor matrix of signed det A minors. 1 – 1 1--28 2 –4 1 c i 8 ii – 12 2 8 8 –2 –4

5E WORKED

Example xample

8

--14

– 1--2-

– --14-

1 – 1--2-

1 1 2 3

2 5 6 9

b

1 2 1 2 0 4 3 –2 5

c

d

–1 4 0 2 3 –1 –1 4 3

d

–1 2 2 –3 6 1 2 0 5

2 Find the inverse of the following matrices, if they exist. a

WORKED

Example xample

9

2 a

--14

Inverse of a 3 × 3 matrix

Digital doc: SkillSHEET 5.3 Using the cofactor/adjoint method

iv

1

1 Find the i determinant, ii cofactor matrix iii adjoint matrix and iv inverse of each of the following. a

eBook plus

8 – 12 8 iii 2 2 –2 –4 8 –4

–3 –1 5 1 0 –1 6 2 –9

1 –3 –2 b 0 0 0.5 – 1 4 2.5 – 0.16 0.08 0.44 c 0.52 0.24 – 0.68 0.28 0.36 – 0.52 – 1.5 0.5 0.5 d – 0.85 0.45 0.25 0.6 – 0.2 0

2 1 1 3 –3 2 2 0 1

4 1 3 1 –1 1 0 2 0

b

3 5 –4 2 –1 3 3 2 –2

c

1 2 –1 3 Find the inverse of 0 3 1 and use it to solve the system of linear equations given 2 0 –2 1 d i −33 by x + 2y – z = 0 13 – 5 3y + z = 9 2x − 2z = 8

– 1.5 1 1.25 0.5 0 – 0.25 – 1.5 1 0.75

19 X = –2 15

4 Solve the following systems of linear equations. a x+y−z=9 −2x − y + z = −11 x + 2y + 2z = 0 5 Find x if

2+x 1 1 2–x

b x − y − 4z = −11

2 3 –4

= 0.

6x + 2y + z = 9 −3x − 2y + 2z = 3 x=± 3

1 0 3

ii

11 – 12 – 3 0 – 4 – 1 – 11

iii

13 – 12 – 4 –5 –3 –1 11 0 – 11

iv

-----– 13 33

12 -----33

4----33

5----33

1----11

1----33

– 1--3-

0

1 --3

6 a b c d

(y − z)(v − u) xy2 − xz − x2y + yz2 + x2 − y2z (1 − x)(1 − y) + y + x − 1 2a3 − 3a2b + b3

Chapter 5 Matrices and their applications

233

6 Evaluate the following: a

x y z u 1 1 v 1 1

1 x yz 1 y x 1 z xy

b

1 1 1 1 1–x 1 1 1 1–y

c

b a a a b a a a b

d

7 Solve each of the following equations: 2

a

a = 0, 1, 2

d

1 a a 1 1 1 1 2 4

=0

1 a a a 1 0 = 0 a= ± a 0 1

a = 1, 2 b

a 0 0 =0 0 a–1 0 0 0 a–2

x–1 2

e

3 0 5 8 If A = 2 4 6 find A–1. 1 2 4

1--2

–3 = 0 x–6

1--3

5--6

– 1 2--3-

– 1--6-

7----12

– 2--3-

0 – 1--2-

1

c

a 2

a =0 a–1

f

x–3 0 0 =0 0 x 3 0 1 x–2

a = 0, 3

x = −1, 3

x = 3, 4

2 1 –1 9 A = 7 –9 3 2 4 5 0.3

0.05 0.03

a Find det A. −189 0.15 – 0.06 0.07 b Find A–1. – 0.24 0.03 0.13 c Use the result from b to solve the system: 2x + y − z = 2 --17x − 9y + 3z = −5 1--- 2 2 1 2x + 4y + 5z = 5 0

1 a 0 10 Show that 0 1 0 0 b 1

–1

1 –a 0 = 0 1 0 . Check with your teacher. 0 –b 1

1 2 1 11 If A–1 = 0 1 2 1 0 1 0.25 – 0.5 0.75 A= a find A and A′ 0.5 0 – 0.5 – 0.25 0.5 0.25 b verify that (A′)–1 = (A–1)′.

0.25 0.5 – 0.25 A′ = – 0.5 0 0.5 0.75 – 0.5 0.25

12 Evaluate for complex i: a

1 i–1

i 0

1+i

b

1 –i 1

i i i

0 1 0

0

c

1–i 1+i i

i 1–i 1+i

1+i − 4 − 7i i 1–i

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13 Verify that det (AB) = det A × det B for the following: 1 2 1 a A= 0 2 4 4 3 1 5 1 –1 b A = 1 –3 2 2 0 1

2 1 0 B = 3 2 –1 1 4 5

Check with your teacher.

3 0 2 B = 4 1 –2 0 –1 1

Cramer’s Rule for solving linear equations Determinants on their own can also be used to solve systems of linear equations. In fact, determinants were first studied in this context. This method of solving linear equations is known as Cramer’s Rule and will be used to solve systems of 2 linear equations. For the equations ax + by = u and cx + dy = v u b a u v d c v x = ----------------------- and y = ----------------------- provided ad - bc ≠ 0. a b a b c d c d Note the pattern of elements in the determinant of the numerators. Mathematics isn’t referred to as the study of patterns without good reason. The proof of this statement follows. Let the system of linear equations be ax + by = u and cx + dy = v. a b c d

Written in matrix form they appear as In matrix equation form, this is

x y

=

u . v

AX = B X = A–1B. x y

1 = ------------------ d – b ad – bc – c a 1 = -----------------ad – bc

du – bv – cu + av

u v

Chapter 5 Matrices and their applications

235

du – bv – cu + av Therefore, x = ------------------ and y = ---------------------- where ad − bc ≠ 0. ad – bc ad – bc The numerators and denominators of both these expressions can be written as determinants: u b a u v d c v x = ----------------------- and y = ----------------------- . a b a b c d c d

WORKED Example 10 Use Cramer’s Rule to solve the following linear equations. 2x + 2y = 3 x − y = 1--2THINK 1

Write the general equations in matrix form and apply Cramer’s Rule.

WRITE a b c d

For

x y

=

u v

u b a u v d c v x = ----------------------- and y = ----------------------a b a b c d c d

2

Substitute the given values for a, b, c, d and u, v.

2 2 1 –1

x y

3

=

–1 x = -------------------------2 2 1 –1

Verify these results by substitution into the original system of equations.

1 --2

2

1 --2

3

3

2 3 1 1--2------------------------and y = 2 2 1 –1

–4 = -----–4

–2 = -----–4

=1

=

--12

Check: 2(1) + 2( 1--- ) = 3 (Verified for the 1st equation) 1−

1 --2

=

2 1 --2

(Verified for the 2nd equation)

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Cramer’s Rule can readily be extended to find the solution of 3 equations in 3 unknowns.

WORKED Example 11 Solve: 2x + y + z = 3 x + 2y − z = −6 5x − 2z = −1. THINK

1

Write the system in matrix form.

2

Apply Cramer’s Rule.

Matrix B is used as column 1 for x, column 2 for y and column 3 for z.

WRITE 2 1 1 A = 1 2 –1 5 0 –2

B=

3 –6 –1

3 1 1 –6 2 –1 –1 0 –2 x = -------------------------------------2 1 1 1 2 –1 5 0 –2

2 3 1 1 –6 –1 5 –1 –2 y = -------------------------------------2 1 1 1 2 –1 5 0 –2

2 1 3 1 2 –6 5 0 –1 z = ----------------------------------2 1 1 1 2 –1 5 0 –2

3

Evaluate the determinants, don’t forget the signed minors.

– 21 x = --------- = 1 – 21 42 y = --------- = −2 – 21 – 63 z = --------- = 3 – 21

4

Verify these results by substituting into the 3 equations.

Check: 2(1) + −2 + 3 = 3 (Verifying the 1st equation) 1 + 2(−2) – 3 = −6 (Verifying the 2nd equation) 5(1) − 2(3) = −1

(Verifying the 3rd equation)

Chapter 5 Matrices and their applications

237

remember remember

Cramer’s Rule for solving linear equations: ax + by = u cx + dy = v a u u b c v v d x = ----------------------- and y = ----------------------a b a b c d c d

states that

5F

Cramer’s Rule for solving linear equations

Use Cramer’s Rule to solve the following: Example

10

WORKED

Example

11

1 2x + y = 1 x = 2, y = −3 3x − 2y = 12

2 2x + 4y = 14 3x − y = −7

3 −x − y = 7 x = −2, y = −5 4x − y = −3

4 3x + 2y = 6 −2x + y = 3

5 x + y + z = 4 x = 3, y = −1, z = 2 2x + y − z = 3 3x + 3y + 2z = 10

6 −x + 2y + 5z = −1 −2x + 3y − z = 7 x − 2y − 2z = 0

x = −1, y = 4

x = 0, y = 3

1

x = −11 1--3- , y = −5 --3- , z = − 1--3-

eBook plus Digital doc: WorkSHEET 5.2

Graphics Calculator tip! Matrix operations The graphics calculator can perform a number of matrix operations and can provide quick and reliable answers to some of the problems that you have encountered in this chapter. A number of operations are shown below, some of which have already been covered in earlier graphics calculator tips. 2 3 2 Consider the matrix A = 0 1 4 . 2 1 –1

For the Casio fx-9860G AU Entering a matrix 1. Select RUN-MAT from the Main Menu. 2. Press F1 ( MAT) to enter the matrix editing screen. 3. Highlight Mat A and press EXE or F3 (DIM). 4. Specify the number of rows, 3 in this case, and press EXE . 5. Specify the number of columns, 3 in this case, and press EXE . ▼

WORKED

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

6. Press EXE again to display the 3 × 3 array for matrix A. 7. Enter the values for the elements, pressing EXE after each number. (To exit the Matrix input screen, press EXIT . Press EXIT again to return to the MAT screen.) Inverse of A 1. Press OPTN then F2 (MAT) to bring up the matrix menu. 2. Press F1 (Mat) then ALPHA [A] and SHIFT [x -1] to specify the matrix A−1. 3. Press EXE to display the answer screen. (Press EXIT to leave or press EXE again to return to the matrix menu screen.)

Powers of A 1. Press OPTN then F2 (MAT) to bring up the matrix menu. (Ignore this step if you are already in the matrix menu screen from a previous operation.) 2. Press F1 (Mat) then ALPHA [A] to specify matrix A. 3. To find A2, press x 2 or press the power key Ÿ and type in the required index of 2. For other powers of A, press the power key and type in the required index. (For example, press Ÿ and then 3 to specify A3.) 4. Press EXE to display the answer screen. Determinant of A 1. From the matrix menu screen, press F3 (Det) then F1 (Mat) and ALPHA [A] to specify det A. 2. Press EXE to display the answer. The identity matrix s

238

1. From the matrix menu screen, press F6 ( ) then F1 (Iden) to specify the identity matrix. For a 3 × 3 identity matrix, press 3 .

Chapter 5 Matrices and their applications

2. Press EXE to display the answer screen.

Calculate (I – A)-1 and store it as B

s

s

1. From the matrix menu screen, press ( then F6 ( ) and F1 (Iden) 3 to specify the 3 × 3 matrix I. Press the subtraction key – and then press F6 ( ) followed by F1 (Mat) and ALPHA [A] to specify the matrix A. Press ) to close the set of brackets and then press SHIFT [x-1] to find the inverse matrix. 2. Press Æ then F1 (Mat) and ALPHA [B] to store the resulting matrix as matrix B. 3. Press EXE to display the resulting matrix B. 1 2 1 2 Calculate the product A-1 3 4 by entering 3 4 as a list 5 6 5 6 1. Enter A−1 as before. (Press F1 (Mat) then ALPHA [A] followed by SHIFT [x-1].) 2. For the second matrix, we enter each row of the matrix as a set of elements enclosed in square brackets. All of the row sets are then enclosed in another set of square brackets. Use SHIFT [ [ ] to open a set of square brackets and SHIFT [ ] ] to close the set. The keys to be pressed can be seen in the screen shown. 3. Press EXE to display the answer screen.

Fill cells of a matrix with a given value

s

1. Set the dimensions of matrix A. (Let’s use 3 × 3.)

s

2. From the matrix menu screen, press F6 ( ) then F3 (Fill). Enter the given value. (Let’s use 5.) Press the comma key , and then press F6 ( ) followed by F1 (Mat) and ALPHA [A] to enter matrix A. Press ) to close the set of brackets. Press EXE . ▼

3. To display matrix A, press EXIT twice to return to the MAT screen. Then press F1 ( MAT) to enter the matrix editing screen and highlight Mat A. Press EXE .

239

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

For the TI-Nspire CAS Entering a matrix 1. Open a new Calculator document. Press k to access the catalog. Select Option 5 then highlight the m-by-n matrix symbol. 2. Press ·. Create the matrix with Number of rows: 3 and Number of columns: 3, pressing e to move from one box to the next. 3. Highlight OK and press ·. Fill in the values in the matrix using the cursor keys to move from one element to the next. 4. Move the cursor to the right of the matrix. Press /, h, then A to store the matrix as matrix A. Press · to display the matrix. Inverse of A 1. With matrix A entered in the calculator, press A then the power key l and type in the index of −1 to specify the matrix A−1. 2. Press · to display the required matrix. Powers of A 1. With matrix A entered in the calculator, press A to recall matrix A to the screen. 2. To find A2, press q or press the power key l and type in the required index (2). For other powers of A, press the power key and type in the required index. For example, press l and then 3 to specify A3. 3. Press · to display the required matrix. Determinant of A 1. Press b and select 7: Matrix & Vector (press 8 or use the cursor keys to highlight 7: Matrix & Vector and press ·).

2. Select 2:Determinant. 3. Press A to specify matrix A and press ) to close the set of brackets. Press · to display the answer.

The identity matrix 1. Press b and select 7: Matrix & Vector. Select 6: Create and then select 2: Identity. 2. For a 3 × 3 identity matrix, press 3. Press · to display the required matrix.

Chapter 5 Matrices and their applications

Calculate (I – A)-1 and store it as B 1. Press ( to open a set of brackets. Press b and select 7: Matrix & Vector then 6: Create and 2: Identity. Press 3 and then ) to specify the 3 × 3 matrix I. 2. Press the subtraction key (-) and then press A to specify the matrix A. Press ) to close the set of brackets and then press the power key (l) and type in −1 as the index to specify the inverse matrix. 3. With the cursor placed after the full expression (press the right arrow key once), press / and h, then press B to store the matrix as matrix B. 4. Press · to display the resulting matrix B. Alternatively, you can store the resulting matrix after the calculation has been performed. Input (I – A)−1 and press · to display the resulting answer matrix. To store this as matrix B, press /, h then B.

-1

Calculate the product A as a list

1 2 1 2 3 4 by entering 3 4 5 6 5 6

1. Enter A−1 as before. (Press A then the power key (l) and type in the index of −1.) 2. Press the right arrow to expand the cursor to the baseline. Press the multiplication key (r). 3. Enter the second matrix. First press / and ( to set up a set of square brackets. Enter each row of the matrix. Each element in a row needs to be separated by a comma (press the , key). Each row of the matrix needs to be separated by a semi-colon. To insert a semi-colon, press k to access the Catalog and select Option 4 (by pressing 4). Highlight the ; symbol and press ·. The keys to be pressed can be seen in the screen shown. 4. Place the cursor at the end of the expression and press · to display the resulting matrix.

241

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Fill cells of a matrix with a given value 1. Set up a new matrix A with specified dimensions. (Let’s use 3 × 3.) Press b and select 7: Matrix & Vector. Select 6: Create and then 1: New Matrix. Enter the number of rows followed by a comma and then the number of columns. Press ) to close the set of brackets. Press /, h and A to store the new matrix as matrix A. Press · to display the matrix. 2. Press b and select 7: Matrix & Vector. Select 6: Create and then 5: Fill. Enter the given value. (Let’s use 5.) Press the comma key (,) and then press A to enter matrix A. Press ·. You will see a message on the screen that indicates that this command has been done. 3. To display matrix A, press A and ·.

Solving simultaneous equations Consider

x+y+z=4 2x − y − 2z = 6 3x − 2y = 2

This set of linear equations can be set up as a matrix equation of the form AX = B. Use a graphics calculator to solve this set of equations by each of the methods listed below. 1 Find A−1. Hence find X by calculating A−1B. 2 Set up the augmented matrix [A | B]. Perform Gaussian elimination (using row operations) to obtain [I | X] and hence find X. 3 Set up the augmented matrix [A | B]. Perform Gaussian elimination but this time use the Reduced Row-Echelon Form function of your calculator to find X. (Hint: For the TI-Nspire CAS calculator, look for the Reduced Row-Echelon Form in the 7: Matrix & Vector menu.) 4 Use the determinant function of your calculator to apply Cramer’s Rule to solve the set of equations. Present your work in a report. Clearly communicate how you performed each 34 20 method and discuss your findings. x = 30 ------ , y = ------ , z = − -----11

11

11

243

Chapter 5 Matrices and their applications

Applications of determinants Using determinants to find the equation of a line One of the many applications of determinants is in determining the equation of a straight line. You are familiar with the equation of a straight line through two points. Assume that the given line passes through points A (x1, y1) and B (x2, y2). Let P (x, y) be a point on this line. As P is a point on the line then the slope of AP equals the slope of AB. y–y y2 – y1 -------------1- = --------------x – x1 x2 – x1 Therefore (y − y1)(x2 − x1) = (y2 − y1)(x − x1) x2y − x1y − x2y1 + x1y1 = xy2 − x1y2 − xy1 + x1y1 (x1y2 − x2y1) − (xy2 − x2y) + (xy1 − x1y) = 0 1×

x1 x2 y1 y2

−1×

x x2 y y2

+1×

x x1 y y1

B(x2, y2) P(x, y) A(x1, y1)

=0

The multipliers of one have been included to display the determinant form more clearly and can be written as: 1 1 1 x x1 x2 y

=0

y1 y2

1 Find the equation of the straight line joining points (2, 4) and (4, −6) using determinants. y = −5x + 14 Using determinants to find the area of a triangle Similarly the area of a triangle can be represented in a determinant form: where the area of LABC equals the total area of the two trapeziums ADEB and BEFC less the area of trapezium ADFC.

B(x3, y3) C(x2, y2) A(x1, y1) D

1 2 Demonstrate that the area of LABC = --2

E

1 1 1 x 1 x 2 x 3 and use this y1 y2 y3

determinant to find the area of a triangle of your design.

F

244

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

summary Systems of linear equations • A pair of equations, ax + by = u and cx + dy = v, may be written in the form AX = B where A =

a b ,X= c d

x y

and B =

u . v

• The matrix equation AX = B can be rearranged to X = A−1B so that values for x and y may be found. These values can also be called equilibrium values.

Input–output (Leontief) matrices • An input–output analysis matrix can be written as X = AX + D where: matrix X contains the variables to be determined and is called the output matrix matrix A contains information about the input details and is called the technology matrix matrix D contains information about the final demand and is called the final demand matrix. • The matrix equation X = AX + D can be rearranged to make X the subject: X = (I − A)−1D.

Gaussian elimination • An augmented matrix [A | B] can be used to find the solution of a set of simultaneous equations when Gaussian row reduction changes [A | B] to [I | X]. • An augmented matrix [A | B] can be used to find the inverse of A when Gaussian row reduction changes [A | B] to [A–1 | I].

Determinants • The determinant of A is written det A; det A = ad − bc where A = • If A =

a b c e d e f , det A = a h g h i

f −b d i g

f +c d i g

a b . c d

e h

where the 3 sub-determinants are referred to as minors. • In general, the determinant of a 3 × 3 matrix is found by using the + − + alternating signs of − + − attached to the coefficients of each minor. + − +

Chapter 5 Matrices and their applications

245

Adjoint matrix adj A • A–1 = ------------- , where adj A is the transpose of the cofactor matrix of A, made up of det A the minors of A.

Cramer’s Rule for solving linear equations • Cramer’s Rule for solving equations: ax + by = u cx + dy = v u b a u v d c v states that x = ----------------------- and y = ----------------------- provided ad − bc ≠ 0. a b a b c d c d

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

CHAPTER review 5A

1 Find the equilibrium values of G and H in the following: G = 50, H = 10 G = H + 40 X = 15.15 (15.15 tonnes of aluminium, 14.1 tonnes of gold) H = 2G − 90 14.1

5A

2 A certain economy produces only two commodities, aluminium and gold. To produce 1 tonne of aluminium, 0.01 tonne of aluminium is required. To produce 1 tonne of gold, 0.005 tonne of gold and 0.2 tonne of aluminium are required. The final demand is for 15 tonnes of aluminium and 11 tonnes of gold. Find the output matrix.

5B

3 Use the Gaussian method of elimination to solve the following system of linear equations: 2x − y + z = −1 x = −1, y = 8, z = 9 3x + 2y − z = 4 x − 2y + 2z = 1

5B

4 Use Gaussian elimination to find the inverse of the following matrices: a

5C

0.2 – 0.4 0.2 0.6

b

3 0 1 2 0 1 –1 1 2

1 –1 0 5 –7 1 –2 3 0

5 Evaluate the following determinants: a

5D

3 2 –1 1

1 2 3 1 1 2 –2 0 0

−2

b

2 3 2 0 1 4 2 1 –1

10

c

1 0 1 0 1 1 1 3 2

−30

Property 7

−2

6 Evaluate the following determinants and list the property involved. b

3 1 4 0 –2 2 0 0 5

2 1 3 2 3 1 given that –3 2 1 –3 1 2 = 52 1 3 –2 1 –2 3

d

1 0 2 1 3 0 given that 3 1 0 0 1 2 = 13 0 2 1 2 0 1

e

1 1 1 1 1 1 given that 2 –2 2 1 – 1 1 = −4 –3 –3 3 – 1 –1 1

f

1 –1 2 3 4 0 –2 2 –4

g

1 2 2 –2 –3 4 given that 1 4 3

a

c −24 Property 5 (twice)

1 2 3 0 0 0 2 3 5

0

Property 3 −52

1 0 2 –2 1 4 1 2 3

Property 4

= −15

13

−15

Property 6

0

Property 2

Property 1

7 a i −2 ii

4 –3 –2 1

iii

4 –2 –3 1

iv

–2 1 1.5 – 0.5

247

Chapter 5 Matrices and their applications

7 Find the i determinant, ii cofactor matrix, iii adjoint matrix and iv inverse of the following: a

1 2 3 4

b

3 1 1 2 1 2 –1 2 –3

5E

b i −12 ii

–7 4 5 5 –8 –7 1 –4 1

iii

–7 5 1 4 –8 –4 5 –7 1

7 –5 –1 1 iv ------ – 4 8 4 12 –5 7 –1

8 Use the cofactor–adjoint method to solve the following system of linear equations: 2x − y + 3z = 4 2 ------ , y = −2 -----, z = x = 17 27 27 −x + 2y − z = −5 4x + y − 2z = 0

2 --9

9 a State Cramer’s Rule for solving two equations in two unknowns. a u u b c v v d b Use this rule to solve y = ----------------------- and x = ----------------------2x − 3y = 7 a b a b 3x + y = 5 c d c d x = 2, y = −1

5E 5F

where ax + by = u and cx + dy = v

Modelling and problem solving 1 Let A be a 4 × 3 matrix. Consider matrix B which is a pre-multiplier of matrix A, that is, BA. Find matrix B if it performs the following elementary row operations on A. 0 0 1 0 a Multiplies the second row of A by 4. B= 0 4 0 0 b Adds twice row 3 to row 4. 1 0 0 0 0 0 2 1 c Interchanges rows 1 and 3. 2 Let A be a 3 × 4 matrix. Consider matrix C which is a post-multiplier of A, that is, AC. Find matrix C if it performs the following elementary row operations on A. 0 3 0 1 a Adds 3 times the first column of A to the second column. C= 0 1 0 0 b Interchanges the first and fourth columns of A. 0 0 –2 0 1 0 0 0 c Multiplies the third column of A by −2.

3 Find the value of a if

a+1 1 2

−10 a−1 2a

a2 –0 −1

= 0. a = ±3

2 413 1 692 56 524 3 313 8 382 a D = 8 844 7 433 25 989 13 159 16 487 7 195

4 532 4 273 81 609 8 570 14 100 b X = 10 501 17 256 48 238 13 756 19 116 10 168

4 The table shown on the next page is part of Scotland’s ‘Aggregate Combined Use Matrix 2004 (Purchasers’ Prices), £millions’. The Industries’ intermediate consumption section shows the inputs of commodities used by Scottish industries in the production of their output. The Final demand section shows the purchases of each product by each category of final 3713 – 1160– 17 - --------------– 93 – 2 - --------------– 2 - --------------– 3 - --------------–6 ------------ 0 --------------0 --------------0 --------------demand (for example, consumers, government, exports). 4532 81 609 14 100 10 501 48 238 13 756 19 116 10 168 – 1716- –----------195- --------------– 285- --------------– 41 - --------------– 7 - --------------– 2 - --------------–3 –4 ------------ --------------0 3945 0 --------------a Find the final demand matrix D for this system. 4273 81 609 8570 14 100 10 501 17 256 48 238 13 756 10 168 –----------821- –----------383- 69 438- –----------107- –--------------1 918- –--------------1 982- --------------– 939- –--------------1 511- –--------------2 979- –--------------1 601- --------------– 672b Find the output matrix X. --------------4532 4273 81 609 8570 14 100 10 501 17 256 48 238 13 756 19 116 10 168 – 24- –----------106- --------------– 778- 4733 – 15 - --------------– 83 - --------------– 36 - --------------– 78 - --------------– 141- --------------– 127- --------------– 33 c Calculate the Leontief matrix (I – A) for this system. ---------------------- --------------c (I − A) =

4532 – 48----------4532 –----------1004532 – 130 -----------4532 – 209 -----------4532 –4 -----------4532 – 29 -----------4532 – 23----------4532

4273 –----------1364273 – 52----------4273 – 416 -----------4273 – 915 -----------4273 –9 -----------4273 – 18 -----------4273 – 42----------4273

81 609 – 136--------------81 609 – 97 --------------81 609 – 981 ---------------81 609 – 2 120 ---------------81 609 – 34 ---------------81 609 – 75 ---------------81 609 – 192--------------81 609

8570 –----------2098570 – 64----------8570 – 103 -----------8570 – 522 -----------8570 –9 -----------8570 – 38 -----------8570 – 44----------8570

14 100 16 713--------------14 100 – 82 --------------14 100 – 106 ---------------14 100 – 1 370 ---------------14 100 – 14 ---------------14 100 – 11 ---------------14 100 – 11 --------------14 100

10 501 – 78 --------------10 501 10 194--------------10 501 – 1 587 ---------------10 501 – 2 606 ---------------10 501 –7 ---------------10 501 – 39 ---------------10 501 – 88 --------------10 501

17 256 – 128--------------17 256 – 150--------------17 256 14 203 ---------------17 256 – 1 226 ---------------17 256 – 25 ---------------17 256 – 61 ---------------17 256 – 113--------------17 256

48 238 –--------------1 41548 238 – 373--------------48 238 – 2 140 ---------------48 238 39 454 ---------------48 238 – 452 ---------------48 238 – 349 ---------------48 238 – 423--------------48 238

13 756 – 628--------------13 756 – 216--------------13 756 – 649 ---------------13 756 – 2 302 ---------------13 756 13 719 ---------------13 756 – 365 ---------------13 756 – 438--------------13 756

19 116 – 286--------------19 116 – 152--------------19 116 – 384 ---------------19 116 – 848 ---------------19 116 –1 ---------------19 116 17 543 ---------------19 116 – 129--------------19 116

10 168 – 41 --------------10 168 – 64 --------------10 168 – 274 ---------------10 168 – 1 348 ---------------10 168 –5 ---------------10 168 – 71 ---------------10 168 8 699--------------10 168

383

106

136

52

416

915

9

18

42

2 406

821

24

48

100

130

209

4

29

23

2 207

Energy and water

Construction

Distribution and catering

Transport and communication

Finance and business

Public admin.

Education, health and social work

Other services

Total intermediate consumption

Mining

Agriculture, forestry and fishing

Manufacturing

Agriculture, forestry and fishing

328

Mining

0

Digital doc:

0

eBook plus

819

Product

Manufacturing

19 459

192

75

34

2 120

981

97

136

778

12 171

1 716

1 160

Energy and water 5 128

44

38

9

522

103

64

209

3 837

107

195

0

Construction 6 441

11

11

14

1 370

106

82

2 613

15

1 918

285

17

Distribution and catering 6 911

88

39

7

2 606

1 587

307

78

83

1 982

41

93

Transport and communication 5 738

113

61

25

1 226

3 053

150

128

36

939

7

0 2

2

15 530

423

349

452

8 784

2 140

373

1 415

78

1 511

Finance and business

Industries’ intermediate consumption

Public admin. 7 761

438

365

37

2 302

649

216

628

141

2 979

3

2

Education , health and social work 5 104

129

1 573

1

848

384

152

286

127

1 601

0

3

Other services 3 987

1 469

71

5

1 348

274

64

41

33

672

4

6

Total intermediate demand 80 671

2 973

2 629

596

22 250

9 823

1 657

5 718

5 257

25 085

2 581

2 101

Consumers 62 999

5 117

3 172

231

10 784

4 411

8 248

356

1 357

27 959

27

1 337

Government 26 393

1 057

12 499

12 837

0

0

0

0

0

0

0

0

Final demand

14 728

344

1

91

2 052

106

30

6 883

0

4 998

91

131

Gross capital formation

Test Yourself Chapter 5

32 225

654

663

0

10 712

2 278

462

1 067

1 931

12 577

1 126

756

Exports RUK

Exports RoW 15 104

23

151

0

2 440

639

104

75

25

10 991

448

207

Total final demand 151 449

7 195

16 487

13 159

25 989

7 433

8 844

8 382

3 313

56 524

1 692

2 431

232 120

10 168

19 116

13 756

48 238

17 256

10 501

14 100

8 570

81 609

4 273

4 532

Total demand for products

248 M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Transformations using matrices

6

syllabus reference Core topic: Matrices and applications

In this chapter 6A Geometric transformations and matrix algebra 6B Linear transformations 6C Linear transformations and group theory 6D Rotations 6E Reflections 6F Dilations 6G Shears

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

• group properties of 2 × 2 matrices • applications of matrices in both life-related and purely mathematical situations

Geometric transformations and matrix algebra In your junior mathematics studies you encountered the idea of translation, reflection, rotation and dilation and how these transformations changed the position, size and orientation of the original figure. However, at that stage your investigations were limited to identifying the type of transformation that had taken place, the position of the mirror line or centre of rotation, and perhaps the size of the image figure. However, now you have skills with matrices that will allow much greater detailed explanation of the position of images or, conversely, the transformation necessary to map point (x, y) onto point (z, w). The matrix algebra used is very straightforward and because we are limiting our discussion at this stage to 2-dimensional space, most of our matrices will be of order 2 × 2. Throughout this section you will be reminded of the properties of groups and how transformations involved in matrix algebra can be considered to be a group.

Transformations

y

A transformation t is an operation which maps each point of the Cartesian plane onto some other point on the plane. Consider point P(x, y). Under a transformation t this point is mapped onto P′(x′, y′). The point P(x, y) is referred to as the original or pre-image point and P′(x′, y′) is the image. This transformation can be written in its most general form as (x′, y′) = t(x, y).

P (x, y) t

WORKED Example 1

P'(x', y')

Find the coordinates of the image points of A(2, −1) and B(3, 0) under the transformation defined by the equations: x′ = 2x − 4xy + y2 − 4 y′ = 6 − 2xy + x − 2y2 THINK

WRITE

1

Think of x′ and y′ as functions of x and y. Substitute x = 2 and y = −1 into equations for x′ and y′.

For A(2, −1) x′ = 2(2) − 4(2)(−1) + (−1)2 − 4 x′ = 4 + 8 + 1 − 4 x′ = 9 y′ = 6 − 2(2)(−1) + (2) − 2(−1)2 x′ = 6 + 4 + 2 − 2 x′ = 10

2

Write the coordinates of the transformed image. The symbol → is used to denote ‘maps onto’.

A(2, −1) → A′(9, 10)

x

C h a p t e r 6 Tr a n s f o r m a t i o n s u s i n g m a t r i c e s

THINK

251

WRITE

3

Substitute x = 3 and y = 0 for B.

For B(3, 0) x′ = 2(3) − 4(3)(0) + (0)2 − 4 x′ = 6 − 0 + 0 − 4 x′ = 2 y′ = 6 − 2(3)(0) + (3) − 2(0)2 x′ = 6 − 0 + 3 − 0 x′ = 9

4

Write the coordinates of the transformed image.

B(3, 0) → B′(2, 9)

5

Sketch each original point and its image. Notice that the transformation of A seems quite unconnected with the transformation of B.

y

A'(9,10) B'(2, 9)

t t

B(3, 0) x

0 A(2, –1)

Translations The equations used in the previous example define a general transformation or mapping. A translation is a specific transformation that involves a shift of each point in the same direction. y P'(x', y') x′ = x + a y′ = y + b t Each x-coordinate is moved a units parallel to the b x-axis and each y-coordinate is moved b units parallel to the y-axis. a P(x, y) The image of P is written P′ and this translation can be expressed in matrix equation form as x¢ y¢

=

x y

+

a b

x

where

1.

x¢ is the vector holding the image coordinates (x′, y′) of point P′ y¢

2.

x represents the original coordinates (x, y) of point P y

3.

a b

is the translation vector and represents information about the horizontal and vertical displacement.

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Note t (lower case) denotes the translation itself and T (upper case) denotes the matrix of the translation. Therefore (x′, y′) = t(x, y) can be written in matrix form as x′ y′

= x y

+T

= x y

+

=

a b

x+a y+b

WORKED Example 2

Find the image of triangle PQR with vertices P(2, -3), Q(0, 1) and R(-1, -2) under the translation vector

5 . Sketch the original and image figures. –1

THINK 1

State the general translation matrix equation.

2

Substitute x- and y-values for each point in turn. Consider P(2, −3).

WRITE x′ y′

=

x y

+

a b

For P(2, −3) x′ y′

=

2 + 5 –3 –1

=

2+5 – 3 + –1

=

7 –4

3

Write the coordinates of the point P′, the image of P. P(2, −3) maps to P′(7, −4).

P(2, −3) → P′(7, −4)

4

Consider Q(0, 1).

For Q(0, 1) x′ y′

= 0 + 5 1 –1 =

0+5 1 + –1

= 5 0

C h a p t e r 6 Tr a n s f o r m a t i o n s u s i n g m a t r i c e s

THINK 5 Write the coordinates of the point Q′, the image of Q. Q(0, 1) maps to Q′(5, 0). 6 Consider R(−1, −2).

WRITE Q(0, 1) → Q′(5, 0) For R(−1, −2) x′ y′

7 8

253

= –1 + 5 –2 –1 =

–1+5 – 2 + −1

=

4 –3

R(−1, −2) → R′(4, −3)

Write the coordinates of the point R′, the image of R. R(−1, −2) maps to R′(4, −3). Sketch both the original and the image points for the triangle PQR.

y Q(0, 1) Q'(5, 0) x

R(–1, –2)

R'(4, –3) P(2, –3) P'(7, – 4)

Note that the image has been moved 5 units to the right and 1 unit down but remains unchanged in shape, area, size and orientation. Such a transformation is said to be congruent.

Successive translations The translation above could have been achieved by a succession of translations that have the final effect of 5 across and 1 down. Any number of successive translations could achieve this: 5 and 0 or the reverse order, 3 and 2 , and so on. 0 –1 2 –3

WORKED Example 3 Show that the translation T1 = 3 followed by T2 = 2 maps the point P(2, -3) from the 2 –3 previous example to the same point P′(7, -4) as found in Worked example 2, and that the order of the translation has no effect on the result. THINK 1

Set up the general matrix equation.

WRITE x′ y′

=

x y

+ T1 Continued over page

254

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

THINK 2

WRITE

Use T1 followed by T2.

For P(2, −3) x′ y′

2 + 3 –3 2

=

5 –1 P(2, −3) → P′(5, −1) =

3

x″ y″

x″ y″

is the image, P″ of image P′.

x′ y′

=

+ T2

5 + 2 –1 –3

=

7 –4 P′(5, −1) → P″(7, −4) Therefore P(2, −3) → P′(5, −1) → P″(7, −4) =

4

x′ y′

Use T2 followed by T1.

=

2 + 2 –3 –3

=

2+2 – 3 + –3

4 –6 P(2, −3) → P′(4, −6) =

x″ y″

4 + 3 –6 2

=

7 –4 P′(4, −6) → P″(7, −4) Therefore P(2, −3) → P′(4, −6) → P″(7, −4) =

5

Sketch the translated image in 2 stages.

y P1' (5, – 1) t1 P(2, –3)

x t2 P'' (7, –4)

t2

t1 P2' (4, – 6)

C h a p t e r 6 Tr a n s f o r m a t i o n s u s i n g m a t r i c e s

255

This example shows, but does not prove, that a set of translations is commutative; that is, the order of operation does not affect the final result.

Translation of a curve

WORKED Example 4

1 . –4

Find the equation of the curve y = x2 under the translation of T = Sketch the original and image curves on the same set of axes. THINK

WRITE x′ y′

x y

+

1 –4

x = x′ y y′ x = x′ − 1 y = y′ + 4

−

1 –4

=

1

Set up the general matrix equation.

2

To find the image of the curve we must express x and y as found in the original function in terms of x′ and y′.

3

Substitute for x and y in the original function to obtain the function in terms of the image coordinates.

y = x2 becomes y′ + 4 = (x′ − 1)2

Rearrange and expand this function.

y′ = x′2 − 2x′ + 1 − 4 y′ = x′2 − 2x′ − 3

4

State the equation of the image curve.

The equation of the image is y′ = x′ 2 − 2x′ − 3.

5

To assist in graphing the image curve, first find the intercepts with the x-axis and the y-axis.

x-axis intercepts occur when y = 0 0 = (x – 3)(x + 1) x-axis intercepts occur when x – 3 = 0 or x + 1 = 0 3−x=3 x = −1 y-axis intercepts occur when x = 0 y = 02 − 2(0) − 3 y = −3

6

Sketch the original and image functions. Note that the turning point (0, 0) maps to (1, −4) which was the translation vector.

y

y = x2 y' = x' 2 – 2x' – 3 (3, 0)

(–1, 0) –3 –4

0

x

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the original Graphics Calculator tip! Graphing and its image We can use a graphics calculator to draw the original function and its image on the same axes. Consider the function y = x2 and its image y ′ = x′ 2 − 2x′ − 3 found in Worked example 4.

For the Casio fx-9860G AU 1. Press MENU to display the MAIN MENU. Use the cursor keys to highlight GRAPH.

2. Press EXE . Enter the first equation y = x2 as Y1. Press X,q,T to enter X and then press x 2 to show the index of 2. Press EXE .

3. Enter the second equation y = x2 − 2x − 3 as Y2. Remember to press X,q,T to enter X. Press EXE .

4. To distinguish between the two graphs on the screen, we can change the appearance of Y2. Highlight the equation and press F4 (STYL). Select F2 , F3 or F4 for a different line style. In this example, F3 was chosen. 5. Press EXE to graph the curves.

6. To see the two curves more clearly, we need to set up a view window. Press SHIFT F3 (V-WIN) and enter values for Xmin and Xmax. From the screen at right you can see that −6 and 6 have been chosen. Press EXE after each entry. Scroll down to enter values for Ymin and Ymax. The values −6 and 6 have been chosen. Again, press EXE after each entry. 7. Press EXIT to return to the GRAPH function screen and then press F6 (DRAW) to see the graphs.

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For the TI-Nspire CAS 1. Open a new Graphs & Geometry document (press / N and select 2: Graphs & Geometry).

2. To draw the graph of y = x2, type in x2 (by pressing X and then q) into the function entry line next to f1(x) =.

3. Press · to obtain the curve labelled with its equation.

4. To enter the second equation, type x2 − 2x − 3 into the function entry line next to f2(x) = and press ·.

5. To see the two curves more clearly, we can alter the scale settings of the axes. Press b and then press 4 to select 4: Window.

6. Press 1 to select 1: Window settings. The values in the Window Settings box can now be changed. Enter the values as shown. Press e to move to the next setting. (To return to a previous setting, press the shift button (g) followed by e.)

7. Press e until OK is highlighted and press ·. The graphing window will appear with the new scale.

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8. If necessary, the label for each graph can be moved to a different position. Press d. A pointer (å) will appear in the work area. Use the NavPad to move the pointer so that it hovers over the equation. The pointer will then appear as an open hand (÷). Press / and then x. The hand will close ({) and the equation will flash. Use the NavPad to move the equation to the desired location and then press d. Note: When the function entry line is shaded, you cannot edit the function entry line because you are in the graph work area. To move between the graph work area and the function line, press e. Also, if the function entry line is covering too much of the graph and you wish to hide it, press / and then G. To bring the entry line back, press / and then G.

remember remember

1. A translation T can be written as x ′ = x + T in matrix equation form. y′ y x The matrix is the vector representing the coordinates of the point (x, y) y and x ′ represents the coordinates of the point (x′, y′) — the image of (x, y) y′ after translation. 2. A translation results in an image congruent to the original object. 3. A set of translations is commutative — the order of operation does not affect the final result.

6A

Geometric transformations and matrix algebra

1 Find the image of each of the following points under the transformation defined by x′ = 2xy − 3y + x2 1 y′ = xy + 4y − x a (0, 0) (0, 0) b (2, −4) (0, −26) c (1, 1) (0, 4) d (−5, −2) (51, 7) Sketch the original point and its image.

WORKED

Example xample

2 Find the image of each of the following points under the translation T = 2 . –5 a (0, 0) (2, −5) b (2, −4) (4, −9) 2 c (3, 5) (5, 0) d (−4, 1) (−2, −4)

WORKED

Example xample

3 The vertices of a triangle are given by A(0, 0), B(3, 5) and C(7, 2). Find the image of the vertices under each of the following translations: A′(4, 2), B′(7, 7), C′(11, 4)

a

4 2

b

4 0

A′(4, 0), B′(7, 5), C′(11, 2)

c

0 –2

A′(0, −2), B′(3, 3), C′(7, 0)

d

0 0

A′(0, 0), B′(3, 5), C′(7, 2)

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4 The line y = −x + 4 undergoes a succession of translations defined by T1 = 4 and 3 3 – 2 T2 = . Show that the order in which these translations take place has no effect on –1 the result. Check with your teacher.

WORKED

Example

5 The line y = 2x + 3 undergoes a translation defined by T = – 1 . Find the equation of 2 the image and sketch the original line and its image. 4

WORKED

Example

6 Find the equation of the image of each of the following curves under the following translations. Graph the original curve and its image using a graphics calculator. a y = −x2

y′ = −(x′ − 3)2 − 1

3 –1

b y = x2 − 4 y′ = (x′ − 5)2 − 5

d x2 + y2 = 4 2 2

5 –1

c

e y2 + x2 + 6y = 0 – 4 2

x′2 − 4x′ + y′2 − 4y′ + 4 = 0

y = x2 − x − 6 – 2 0

y′ = x′2 + 3x′ − 4

x′2 + 8x′ + y′2 + 2y′ + 8 = 0

7 Rearrange a matrix equation to find the translation vector that maps each of the pairs of points: –1 –4 a (2, 4) → (0, 1) – 2 b (4, −1) → (3, 5) c (6, 2) → (2, −5) 6

–3

–7

Linear transformations 5 y′ = 2x′ + 7 y 7

y = 2x + 3

y' = 2x' + 7

3 –3.5 –1.5 0

x

Have you ever wondered how programmers who develop computer games move and manoeuvre characters on a screen to get them to spin or shrink as they appear to move further away from the observer? The study of linear transformation forms the foundation for these changes of form and size — the warping of the plane on which the characters are mapped. There are many different ways in which the original, or pre-image, can be changed or moved so that it looks different, or is in a different place. A linear transformation l is a mapping of the pre-image P(x, y) onto the image P′(x′, y′) where: x′ = ax + by y′ = cx + dy for all real values of a, b, c, and d. In matrix form this system is written as: x′ y′

=

x′ y′

=L x y

a b c d

x y where L =

a b and is called the transformation matrix. c d

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WORKED Example 5

y

Find the images of the vertices of a unit square ABCD under the transformation given by L=

B (1, 1)

A(0, 1)

1 2 . –1 1 C (1, 0) x

D(0, 0)

THINK

WRITE

1

Set up the initial matrix equation where the image of P is given as P′.

2

Investigate the transformation of each point in turn. (Recall that the symbol → is used to denote ‘maps onto’.)

x′ y′

=

1 2 –1 1

x y

For point A(0, 1) x′ = 1 2 0 = 2 y′ –1 1 1 1 That is, A(0, 1) → A′(2, 1) For point B(1, 1) x′ = 1 2 1 = 3 y′ –1 1 1 0 That is, B(1, 1) → B′(3, 0) For point C(1, 0) x′ = 1 2 1 = 1 y′ –1 1 0 –1 That is, C(1, 0) → C′(1, −1) For point D(0, 0) x′ = 1 2 0 = 0 y′ –1 1 0 0 That is, D(0, 0) → D′(0, 0)

3

Plot the image on the same axes as the original.

y A(0, 1)

D(0, 0)

B(1, 1) A'(2, 1)

C(1, 0)

B' (3, 0)

x

C' (1, –1)

This type of transformation leaves the origin unchanged and therefore differs from a translation. The transformation matrix can also be extracted from information about the original and image points. An example of this is shown in the following worked example.

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WORKED Example 6

Find the matrix of the linear transformation that maps A(1, 1) onto A′(2, -1) and B(2, -1) onto B′(1, -1). THINK 1 2

WRITE

Set up the initial matrix equation. State matrix equations for points A, A′, B and B′.

x′ y′

=

x y

a b c d

For point A: 2 = a b 1 –1 c d 1 and for point B: 1 = –1

a b c d

2 –1

3

Multiply the matrices to arrive at 4 simultaneous equations for 4 unknowns, a, b, c and d.

From the equation for point A: 2=a+b −1 = c + d From the equation for point B: 1 = 2a − b −1 = 2c − d

4

This is really 2 sets of 2 equations in 2 unknowns that can be solved by elimination. If you wish to use your graphics calculator, enter this as

a+b=2 c + d = −1 2a − b = 1 2c − d = −1 [1] + [3]: 3a = 3 a=1 [2] + [4]: 3c = −2 c = − 2---

1 1 A= 0 0 2 –1 0 0

0 0 2 1 1 , B = –1 0 0 1 2 –1 –1

[1] [2] [3] [4]

3

and find A−1B. 5

Find b by substituting a = 1 into Equation [1] and find d by substituting c = − 2--- into Equation [2]. List the 3 values for a, b, c and d.

Sub. a = 1 into [1]:

1+b b Sub. c = − 2--- into [2]: − 2--- + d 3 3 d 2 1 a = 1, b = 1, c = − --- , d = − --3

6

Use these values to build L, the linear transformation matrix.

L=

1

1

– 2--3-

– 1--3-

=2 =1 = −1 = − 1--3

3

As hinted at in the introduction to this section, there are two ways to conceptualise a transformation. The more obvious way is to imagine that the points move to new positions on the Cartesian plane. The other less obvious notion is that it is actually the

y

B'(–2, 0)

A'(–1, 2) C(–3, 1)

remember remember A(1, –2)

0

A'

B(2, 0)

x

B'(4, 2)

ii y

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Cartesian plane on which the original points are plotted that undergoes distortions to yield the transformed image. Perhaps the former is more straightforward, but the end product will be the same.

0

B(2, 0)

C'(3, –1) A(1, –2)

x

262

x or 1. A linear transformation can be represented by x ′ = a b y′ y c d x ′ = L x where L = a b is the transformation matrix that maps y′ c d y point (x, y) onto the image (x′, y′).

x = a b y c d 3. The transformed image is not congruent to the object.

–1

Linear transformations

A′(0, −1), B′(4, 2), C′(−5, −2) A′(−1, 2), B′(−2, 0), C′(3, −1) A′(2, − 4), B′(4, 0), C′(−6, 2) A′(1, −2), B′(2, 0), C′(−3, 1)

6B

x′ . y′

2 a i ii iii iv

2 b i

C(–3, 1)

C'(–5, –2)

2. The transformed image can be found using

2 –3

1 ii Which of the following transformations are linear? 3 2 ii Write the transformation matrices for each of these. a x′ = x + y b x′ = x − 1 c x′ = 2x – 3y C 1 1 y′ = y + 2 y′ = 3x + 2y y′ = 2x + y 2 1 2 4 6 8 x A 2 d x′ = x + y e x′ = x 1 y′ = y2 A' y′ = 1 + --C' y WORKED 2 a Find the images of the points A(1, −2), B(2, 0) and C(−3, 1) under the following Example xample transformations: 5 i 2 1 ii – 1 0 iii 2 0 iv 1 0 1 1 0 –1 0 2 0 1 b Sketch the original triangle from a and its 4 different images. iv No change

C'(–6, 2)

C

y

0

A

B

A'(2, –4)

B'(4, 0) x

3

y B' 5 4 3 2 –3 1 –2 B –2 –3 –4 –5 –6 –7

3 Find the image of the points (given below) under the transformation defined by: x′ = x − 2y y′ = −2x + y a A(2, −3) A′(8, −7) b B(−3, −1) B′(−1, 5) c C(4, 1) C′(2, −7) Plot the original point and its image in each case.

2 b iii

4 Find the image of the pre-image points A(4, 1), B(−4, 1) and C(0, 5) under the transformation defined by: 4 A′(7, − 6), B′(−1, 2), C′(15, −10) x′ = x + 3y y y′ = −x − 2y B'(–1, 2) 5 C(0, 5) B(–4, 1) Plot the original and image points. A(4, 1) 5 Find the matrix of the linear transformation which maps: a (1, 2) → (−3, 1) and (3, 0) → (1, 4) 6 b (−2, 3) → (0, 0) and (−2, 4) → (1, 1) c (2, −1) → (1, 1) and (2, 1) → (3, 6) d (3, 4) → (5, 0) and (−3, −2) → (−2, 4) 1.5 1 2

–4 –1

WORKED

Example xample

1 1 1.75 2.5

– 1--3– 2 --23-

2

5 a

--3

– 1 --3-

1 1--3-

– 1--6-

45 7 10

15 x

–5 A'(7, –6) –10

b

1.5 1 1.5 1

C'(15, –10)

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Linear transformations and group theory Earlier in your Mathematics C course of study you were introduced to group theory (Chapter 4). You found that a system formed a group if the properties of closure and associativity applied and an identity element and inverse existed. These properties apply to many areas of mathematics — including linear transformations. In Chapter 4 we investigated whether matrices, in general, formed a group; now we will study groups that perform linear transformations.

Closure

If l1 is a linear transformation that maps (x, y) → (x′, y′) then (x′, y′) = l1 (x, y) If l2 is a linear transformation that maps (x′, y′) → (x″, y″) then (x″, y″) = l2 (x′, y′) Therefore it follows that (x″, y″) = l2 [l1(x, y)] where l1 is followed by l2 and maps (x, y) → (x″, y″). This double transformation can be represented as a single, where l = l2 l1. This is known as composition of transformations, where the order is significant. From the Mathematics B course you would be familiar with the idea of composition of functions, where g(x) = h(f(x)) indicates that f(x) is the ‘inner’ function within the structure and general shape of h(x). In matrix form x″ y″

= L2

x′ y′

  = L2  L 1 x   y  = L2 L1

x y

= L x where L is a 2 × 2 matrix and L = L2 L1 y We can verify this result by considering the image y of the point P(1, 2) after a linear transformation L1 = 1 2 followed by a linear transformation 2 1 3 – 3 . Show that following this double –3 4 transformation produces the point P″(3, 1). If we mapped P(1, 2) directly to P″(3, 1) in a single transformation, find the transformation matrix L. Is this transformation matrix L equivalent to L1L2 or L2L1? L2 =

5 P' 4 l1 3 l2 2 P l 1 P" 0 1 2 3 4 5 6

x

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WORKED Example 7 If l1 and l2 are 2 linear transformations such that L1 = 1 1 and L2 = 0 0 : 2 –2 2 1 a find P′, the image of P (1, 3) under l1 b find P≤, the image of P′ under transformation l2 c find the single transformation of P such that l = l2 l1 d verify that P≤ (as found in part b) is equal to LP THINK

WRITE

a Use matrix operation to find P′, the image of P(1, 3) under l1..

a

x′ y′

= 1 1 2 –2

b Find P≤, the image of P′ under transformation l2.

b

x″ y″

= 0 0 2 1

c Find the single transformation of P such that l = l2 l1.

c L= 0 0 2 1

d Verify that P≤ (as found in part b) is equal to LP.

d LP = 0 0 4 0

1 = 4 3 –4 4 = 0 –4 4

1 1 = 0 0 2 –2 4 0 1 = 0 = P″ 3 4

Therefore P(1, 3) → P″(0, 4)

Associativity As seen with matrix operations, matrix multiplication is associative; that is, (L1 L2)L3 = L1(L2 L3). Therefore linear transformations are associative; that is, (l1 l2)l3 = l1(l2 l3).

Identity Remember the identity element (IE) is one which leaves the original number unchanged. When dealing with linear transformation this means that matrix multiplication has been performed which leaves the original point unchanged. This is the identity transformation and is denoted by li — and the matrix is I. For a 2 × 2 matrix, I = 1 0 . 0 1

Inverse transformations An inverse transformation is one that maps the image back to the original point — where (x, y) → (x′, y′) → (x, y).

y

P'(x', y') l l–1

This transformation is denoted by l–1. As with other inverses ll–1 = l–1l As with other inverses ll–1 = li

P(x, y) x

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If L is the linear transformation matrix and A is the transformation matrix which returns the point to the original, then A = L–1. As with general matrix terminology, the transformation l is non-singular; that is, l has an inverse if it has a matrix l –1 that will map the image back to the original. Therefore, only linear transformations that have an inverse l –1 can be considered to form a group. If l is singular then the set of linear transformations does not form a group.

Abelian groups If the composition of linear transformations is commutative, then the set will form an Abelian group. But in general, multiplication of linear transformations is not commutative, that is l1 l2 ≠ l2 l1.

WORKED Example 8

a Find the image of the point P(2, 3) under l1 followed by l2 with L1 = 2 3 1 –1

L2 = 0.2 0.6 0.2 – 0.4

b Verify that l2 = l1–1 in 2 ways. THINK a

1

2

3

b

Set matrices in x ′ y′

WRITE = LP form.

P′ is the point (13, −1). Now find P″ using P″ = L2P′. State the image of P under l1 followed by l2.

1

Verify this by showing L2 L1 = I.

2

Verify by finding the inverse of L1.

a

x′ y′

= 2 3 1 –1

x″ y″

= 0.2 0.6 0.2 – 0.4

2 = 13 3 –1 13 = 2 –1 3

The image of the point P(2, 3) under l1 followed by l2 is (2, 3). Since P″(x″, y″) = P(x, y), L2 has mapped P′(x′, y′) back onto the original, therefore L2 is the inverse linear transformation of L1. b L2L1 = 0.2 0.6 0.2 – 0.4

2 3 = 1 0 1 –1 0 1

1 L1–1 = ------------------ d – b ad – bc – c a 1 L1–1 = ---------------- – 1 – 3 – 2 – 3 –1 2 L1–1 = 0.2 0.6 0.2 – 0.4 L1–1 = L2

3

State the conclusion.

Therefore L2 = L1–1

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WORKED Example 9 Determine whether the following linear transformation, l1, is singular or non-singular. x′ = 2x + y y′ = −2x + 3y THINK

WRITE

1

State l1 in matrix form.

2

Test to determine whether the determinant = 0.

| L1 | = ad − bc = 6 − −2 = 8

3

State your conclusion.

Since det L1 ≠ 0, L1 is non-singular, that is, it has an inverse.

L1 =

2 1 for L = –2 3

a b c d

Images of curves — non-singular transformations So far we have mainly considered only the images of individual points under linear transformation where

x′ y′

= L x . Now consider the image of a curve — essentially y

a set of points.

WORKED Example 10

Find the image of the line y = 2x − 3 under the linear transformation L = 2 1 . 0 3 Sketch the original line and its image. THINK 1

We need to express the original function in terms of the image points so we need to find x and substitute y image points for the original points x and y.

2

Evaluate the inverse.

WRITE x′ y′

=L x y

L–1 x ′ y′

= L–1L x y

x y

= L–1 x ′ y′

x y

1 = ------------ 3 – 1 6–0 0 2 =

1 --2

– 1--6-

0

1 --3

x′ y′

x′ y′

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THINK 3

267

WRITE x y

Express x′ and y′ in terms of the original points.

=

1 --- x′ 2

– 1--6- y′

0x ′ + 1--3- y ′

Therefore x = 1--- x′ − 1--- y′ 2

y= 4

Substitute for x and y in terms of the image points, into the original function. Simplify and rearrange the image equation.

y = 2x − 3 becomes 1 --- y′ 3

6

State the equation of the image. Some texts drop the ‘primes’ on x′ and y′ at this stage, but if they are left in it reminds us that the graph of this function is the image of the original.

= 2( 1--- x′ − 1--- y′) − 3 2

6

= x′ − 1--- y′ − 3 3 2 --- y′ = x′ − 3 3

5

6

1 --- y′ 3

y′ = 3--- x′ − 4 1--2

2

The image of y = 2x − 3 has the equation y′ = 3--- x′ − 4 1--- . 2

Sketch the original and image functions.

2

y

y = 2x – 3

0

11–2 3

x

y' = 3–2 x' – 4 1–2

–3 – 4 1–2

Images of curves — singular transformations If a linear transformation L is singular, then L does not have an inverse and the method shown in Worked example 10 cannot be used. We need to use a different approach as shown in the next worked example.

WORKED Example 11

Find the image of the circle x2 + y2 = 1 under the linear transformation L = 1 2 . 2 4 Sketch the original curve and its image. THINK 1

State the initial transformation in general matrix form.

WRITE x′ y′

=L x y = 1 2 2 4

x y

Continued over page

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THINK

WRITE

2

Find values for x′ and y′.

3

Notice that the equation for y′ equals twice the equation for x′. Therefore this should be stated as the function of the image.

4

State the equation of the image.

5

Sketch the original curve and its image.

x′ y′

=

x + 2y 2x + 4y

y′ = 2x′

The image of x2 + y2 = 1 has the equation y′ = 2x′. y y' = 2x' (0, 1) x2 + y2 = 1 (–1, 0)

(1, 0)

x

(0, –1)

remember remember 1. (a) Linear transformations are closed. (b) If (x′, y′) = l1(x, y) where l1 is a linear transformation that maps (x, y) → (x′, y′) and l2 is a linear transformation that maps (x′, y′) → ( x″, y″) then (x″, y″) = l2(x′, y′) = l2[l1(x, y)] where l1 is followed by l2 2. Linear transformations are associative; that is, (l1l2)l3 = l1(l2l3). 3. The identity transformation is denoted by li and is represented by the identity matrix I. 4. An inverse transformation is one that maps the image back to the original point where (x, y) → (x′, y′) → (x, y) and is denoted by l –1. As with other inverses ll –1 = l –1l = li Only linear transformations that have an inverse l –1 can be considered to form a group. 5. If linear transformations are commutative, then they will form an Abelian group. But in general, multiplication of transformations is not commutative, that is l1l2 ≠ l2l1.

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6C

Linear transformations and group theory

0 1 –2 1

1 If l1 and l2 are 2 linear transformations such that L1 = 2 1 and L2 = 0 1 : 0 1 –1 2 a find P′, the image of P(2, 5) under l1 P′(9, 5) b find P′′, the image of P′ under transformation l2 P′′(5, 1) c find the single transformation of P such that l = l2l1 d verify that P′′ is equal to LP. Check with your teacher.

WORKED

2 a Find the image of the point P(1, 4) under l1 followed by l2 with

WORKED

Example

7

Example

8

L1 = 3 – 1 L2 = 2 – 1 5 –2 5 –3 -1 b Verify that l1 = l1 in 2 ways.

Check with your teacher.

3 Determine whether, in each of the following linear transformations, l1 is singular or non-singular: a x′ = 3x − y Non-singular b x′ = 2x − y Singular y′ = −x + 2y y′ = 4x − 2y

WORKED

Example

9 5 y y = 3x + 2

d i y′ = --47- x′ 4 A linear transformation l1 is defined as x′ = 2x + 5y y′ = x + 3y 2------ x′ + ----ii y′ = 10 17 17 P′(31, 18) a What will the image of P(3, 5) be? x iii 10x′2 − 34x′y′ + 29y′2 = 2 b Is this linear transformation singular? det A = 1 (non-singular) c Show that l1–1(l1P ) = P. d Use this linear transformation to state the image of the following curves: i y=x ii y = 3x + 2 iii x2 + y2 = 2 5 Find the image of the line y = 3x + 2 under the linear transformation L = – 4 2 . 1 0 Sketch the original line and its image. y′ = 1--- x′ − 2

2

y' = _12 x' – 2 – _23 0

P′′(1, 4)

4

–2

WORKED

Example

10

2

6 Find the image of the circle x + y = 9 under each of the following transformations. 17x′2 − 26x′y′ + 10y′2 = 9

1 3 1 4

2

b

1 3 1 0

c 10y′2 + x′2 − 2x′y′ = 81

–1 –1 3 2

13x′2 + 10x′y′ + 2y′2 = 9

7 Find the image of the circle x2 + y2 = 9 under each of the following transformations.

WORKED

Example

11

a

9 a, c ii

2 0 2 0

y′ = x′

b

2 4 3 6

y′ = --32- x′

c

8 4 4 2

y′ = --12- x′

8 Show that under any linear transformation the image of a straight line is itself a straight line. Check with your teacher.

y y = –x + 4 y' = 4

4

0

a

2

4

x

9 a Sketch the following lines on separate axes. i y = 2x − 1 ii y = −x + 4 b Find the image of each line under the linear transformation 2 4 1 1 c Sketch each image with the original line.

3b i y′ = ----x′ + 10 ii y′ = 4

10 Find the image of each of the following functions under the linear transformation 5 3 . 2 1 a y = x2 x′2 − 6x′y′ + 9y′2 − 2x′ + 5y′ = 0 11y′ = 4x′ − 5 b y = 2x + 5

1--5

270

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Rotations

SLE 4: Demonstrate the use of the transformation matrices (rotation, reflection, dilation) as an application of 2 × 2 matrices to geometric transformations in the plane.

A rotation is a transformation in y which the plane rotates about a fixed point called the centre of B' C' rotation. This point is usually taken C as the origin. The rotation in an anticlockwise direction is conB A sidered to be a positive rotation and A' θ in a clockwise direction to be a x 0 negative rotation. Examine the diagram at right to note that the centre of rotation is the only point that doesn’t move. In a rotation: 1. each original point rotates through the same angle of rotation. 2. the image is congruent to the original — the length, angle and area remain unchanged in the image. This is referred to as a congruent transformation. 3. rq denotes rotation in a positive direction through an angle of θ and Rθ is the matrix of rotation. With all the transformations that will be discussed we will generate matrices based on where the points (1, 0) and (0, 1) are mapped to on the plane, as a result of the transformation. These points are represented by columns 1 and 2 of the identity matrix: ↓ ↓ 1 0 0 1

Special rotations In this section we will discuss transformations involving rotations of 90°, 180°, 270° and 360°, as well as general rotations.

Rotation of 90° Consider the figure at right.

y

y (0, 1) 90º

(0, 1) (1, 0)

(–1, 0) x

90º x

As the plane rotates through θ = 90° about the origin, point (1, 0) will map to point (0, 1) and point (0, 1) will map to point (−1, 0). ↓ ↓ Hence, the identity matrix, I, is altered to 0 – 1 to achieve a rotation of 90° about 1 0 the origin. It is most important that you recognise the pattern that is displayed by the columns in the matrix and the coordinates of the image points. This concept forms the basis of the next section of work and totally eliminates ‘remembering’ formulas so that you will be able to understand what is happening to the points.

C h a p t e r 6 Tr a n s f o r m a t i o n s u s i n g m a t r i c e s

271

y

Hence R90° = 0 – 1 and is the matrix of rotation. 1 0

P'(x', y')

In general terms

P(x, y)

(x, y) → (−y, x) x′ y′

= 0 –1 1 0

x

x y

x′ = −y y′ = x As mentioned earlier, these rotation matrices should not be learned. They are quite similar and can be too readily confused. Sketch the original (1, 0) and (0, 1) points and then use their images to build the rotation matrices.

Rotation of 180° In the diagrams below, notice that point (1, 0) is mapped onto point (−1, 0) and point (0,1) is mapped onto (0, −1). y

º

(1, 0) x

P(x, y) 180

180º

y

180º

(–1, 0)

y (0, 1)

x

x

0

P'(x', y')

(0, –1)

Therefore R180° = – 1 0 where (x, y) → (−x, −y). 0 –1

Rotation of 270° In the diagrams below, notice that point (1, 0) is mapped onto point (0, −1) and point (0, 1) is mapped onto point (1, 0). y

y

y (0, 1) (1, 0) 0

(0, –1)

Therefore R270° =

x

270º

x

270º

P(x, y) (1, 0)

270º 0

x P'(x', y')

0 1 where (x, y) → (y, −x). –1 0

Rotation of 360° R360° = 0 1 because R360° essentially leaves the original unchanged (or mapped onto itself). 1 0

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y

General rotation of θ Consider the points (1, 0) and (0, 1) that are rotated through angle θ about the origin.

B(0,1) B' cos θ

–sin θ

A'

Q 1

1

θ

θ 0

sin θ A(1, 0) cos θ

x

P

Careful examination of the diagram shows that point (1, 0) is mapped onto point (cos θ, sin θ) and point (0, 1) is mapped onto point (−sin θ, cos θ) where cos θ = x and sin θ = y

sin θ

(horizontal) (vertical)

θ

Rθ = cos θ –sin θ sin θ cos θ

y

R–θ , where θ is taken in a clockwise, negative rotation about the origin, and is shown in the diagram to the right. R−θ = cos ( – θ ) –sin ( – θ ) sin ( – θ ) cos ( – θ ) R−θ =

cos θ

P(x, y) –θ

x P'(x', y')

cos θ sin θ since cos (−θ) = cos θ and sin (−θ) = −sin θ –sin θ cos θ

Both Rθ and R−θ can be used to confirm the specific cases of R90°, R180° and R270°. R90° = cos 90° –sin 90° = 0 – 1 sin 90° cos 90° 1 0 Remember that when you need to evaluate a trigonometric ratio: 1. sketch the angle concerned in the correct quadrant 2. write the coordinates or length of the sides on the right-angled triangle 3. in the unit circle, the cosine ratio involves only the x-coordinate and the sine ratio involves only the y-coordinate. Verification of the other angle measures is left as a future exercise.

WORKED Example 12

πc Find the image of the point (2, -2) under a rotation of --- about the origin. Sketch the 4 original point and its image. THINK 1

Write the general rotation matrix and sketch the original point (shown on next page).

WRITE Rθ = cos θ –sin θ sin θ cos θ

C h a p t e r 6 Tr a n s f o r m a t i o n s u s i n g m a t r i c e s

THINK

WRITE c

2

3

273

π Substitute ----- for θ. 4 (Note: The small c is the symbol for circular or radian measure.)

Always use a sketch to develop the matrix. –π 4

√2

π π cos --- –sin --4 4 Rπ = --π π 4 sin --- cos --4 4 11-----– -----2 2 Rπ = --114 ----------2 2

1

–π 4

1

4

Set up the general matrix form for transformations.

x′ y′

11-----– -----2 2 = 11----------2 2

2 –2

2- + -----2-----2 2 = 2- – -----2-----2 2

5

Rationalise the denominator and simplify.

=

4-----2 0

=

4- ------2-----× 2 2 0

= 2 2 0 6

State the coordinates of the image point.

7

Sketch the original and the image points.

The image of the point (2, −2) is (2 2 , 0). y (2 √2, 0) 0

–π 4

x

(2, –2)

274

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WORKED Example 13

πc Find the image of the line y = −x + 4 under the rotation of --- about the origin. Sketch the 6 original line and its image. THINK

WRITE

1

Write the general Rθ matrix.

Rθ = cos θ –sin θ sin θ cos θ

2

πc Substitute --- for θ and evaluate using 6 the relevant triangle of ratios.

π π cos --- –sin --6 6 R --π- = 6 π π sin --- cos --6 6

–π 6

2

------3- – 1--2 = 2 1--- ------32 2

√3

–π 3

1 3

Set up the general transformation matrix model, rearranged so that is the subject.

4

x y

Evaluate the inverse of R.

x′ y′ x y

= R --π- x 6 y –1

= R --π6

x′ y′

1 = -----------1 3 --- – – --4 4

------32

3 – 1--- ------2 2

3 1 --------2 2 = 3 – 1--- ------2 2 5

Multiply out the matrices.

6

Substitute for x and y in the original function.

7

After applying the Distributive Law and rationalising the denominator, this expression can be simplified.

1--2

x′ y′

x′ y′

3 1 x = ------- x′ + --- y′ 2 2 1--3 y = − x′ + ------- y′ 2 2 y = −x + 4 becomes 1 3 3 1 − --- x′ + ------- y′ = − ------- x′ − --- y′ + 4 2 2 2 2 (1 – 3 )x′ 8 y′ = -------------------------- + ---------------3+1 3+1 y′ = ( 3 − 2)x′ + 4( 3 − 1)

275

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THINK 8 Use your calculator only at the end to simplify surds for sketching purposes.

WRITE y y' = (√3 – 2) x' + 4 (√3 – 2) 4 2.9 4

10.9

x

y = –x + 4

remember remember

1. For general rotation θ in an anticlockwise direction about the origin Rθ = cos θ –sin θ . sin θ cos θ 2. Use the special right-angled triangles to obtain the trigonometric ratios. 3. Rotation is a congruent transformation.

6D

Rotations

1 Construct matrices for the following anticlockwise rotations about the origin (the angles are given in radians). π 0 –1 3π –1 0 0 1 a --b π c -----d 2π 1 0 2 2 0 –1 1 0 –1 0 0 1 2 Find the image of the following points under the given anticlockwise rotations about the origin. 12 π c (1 − ------3- , 3 + 1--- ) a (2, 1) θ = --b (0, 4) θ = π c (0, −4) 2 2 3

WORKED

Example

3 b i y

c

1 − y = _x +√_2 2

0.35 –0.7

0

_1 3

(6, 3)

4

x

1 y = –3x + 1

e (2, 3)

πc θ = --4

2( 3--------, 2

9---------22

θ = 90° (−3, 2)

)

d (1, −3)

θ = −60°

f

πc θ = --6

(1, 1)

( 1--2- −

3---------32

3( -----− --12- , 2

--12

3, − -----− 3--2- ) 2

+

------32

)

3 a Find the equation of the image of the line y = −3x + 1 as a result of the following rotations: x′ 2 13 a i y′ = ---- + ------c c π π 2 4 i θ = 45° ii θ = --iii θ = − --2 2 ii y′ = --3x- + --13b Sketch each original line and its image. iii y′ = --3x- − 1--3-

WORKED

Example

eBook plus Digital doc: WorkSHEET 6.1

πc 4 Find the equation of the image of the circle x2 + y2 = 1 after a rotation of --- . What do 2 No change, rotation about the centre of the circle. you notice? Can you explain why this is so?

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Reflections

SLE 4: Demonstrate the use of the transformation matrices (rotation, reflection, dilation) as an application of 2 × 2 matrices to geometric transformations in the plane.

y

=

=

=

A reflection is a linear transformation in A' A which every point of the original is reflected through a straight line called a mediator. This line can be thought of as a mirror. The diagram at right shows LABC reflected through the mediator m, at x = 1. In a reflection: C' B' B 1. corresponding points of the image and original figures are equidistant from and x=1 perpendicular to the mediator m 2. length, angle and area of the image and original are unchanged, hence it is a congruent transformation 3. any points of the original on the mediator are left unchanged. =

276

C

We usually let m denote the reflection transformation and M the reflection matrix.

Reflection in the x-axis (where y

=

0)

↓ ↓ Again, sketch the points (1, 0) and (0, 1) from the identity matrix I = 1 0 . 0 1 y Under a reflection in the x-axis, point (0, 1) will map to (0, −1) and point (0, 1) (1, 0) will map onto itself because it is on the mediator. (1, 0)

Therefore My = 0 = 1 0 . 0 –1

x

0

(0, –1)

my = 0

x

C h a p t e r 6 Tr a n s f o r m a t i o n s u s i n g m a t r i c e s

Reflection in the y-axis (where x

=

277

0)

If you sketch the original points (1, 0) and (0, 1) you will notice that if these points are reflected in the y-axis then point (1, 0) will map to (−1, 0) and point (0, 1), which is on the mediator, will map onto itself.

mx = 0 y (0, 1) (1, 0)

(–1, 0) 0

Therefore, Mx = 0 = – 1 0 . 0 1

x

WORKED Example 14 Find the image of point (3, 1) under reflection My = 0. Sketch the original and its image. THINK 1

Sketch the diagram to construct your reflection matrix.

WRITE y (0, 1)

(1, 0)

my = 0

x

0

(0, –1)

My = 0 = 1 0 0 –1

2

Write the initial transformation matrix statement.

3

Substitute the necessary values and evaluate.

x′ y′

= My = 0 x y = 1 0 0 –1 =

4

Sketch the original and image points.

3 1

3 –1

The image is the point (3, −1). y P(3, 1) 0

x P'(3, –1)

my = 0

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

WORKED Example 15

Find the image of y = x under reflection in the y-axis. Sketch the original and its image. THINK 1

WRITE

Sketch the diagram to construct your reflection matrix.

mx = 0 y (0, 1) (–1, 0)

(1, 0) x

0

Mx = 0 = – 1 0 0 1

2

3

Write the initial transformation matrix statement and rearrange it to have the original points as the subject.

Substitute for Mx = 0 and evaluate the inverse.

x′ y′

= Mx = 0 x y x′ y′

x y

= M x–1= 0

x y

1 = ------ 1 0 –1 0 –1 = –1 0 0 1

4

Multiply to give expressions for x and y.

x = −x′ y = y′

5

Substitute for x and y into the original equation.

y = x becomes y′ = −x′

6

Sketch the original and image graphs. Note the origin is left unchanged.

mx = 0 y

x′ y′ x′ y′

y=x

0

x y' = –x'

279

C h a p t e r 6 Tr a n s f o r m a t i o n s u s i n g m a t r i c e s

Reflection in line y

=

x

To find this reflection, sketch the situation as described. Remember to note the main points from the introduction to this section:

y y=x =

(0, 1)

=

1. corresponding points of the image and original figures are equidistant from and perpendicular to the mediator

(1, 0)

0

2. length, angle and area of the image and original are unchanged, hence it is a congruent transformation 3. any points of the original on the mediator are left unchanged. We find that (1, 0) and (0, 1) map to each other, therefore My = x = 0 1 . 1 0

WORKED Example 16

Find the equation of the image y = x2 reflected in the line y = x. THINK 1

WRITE

Sketch the relevant diagram to establish the reflection matrix.

y y=x

=

=

(0, 1)

(1, 0) x

0

My = x = 0 1 1 0 2

Set up the initial matrix equation and rearrange to have x and y as the subject.

x x′ = M y=x y y′ x y

3

Find the inverse of My = x.

= My−1= x x ′ y′ 1 = ------ 0 – 1 –1 –1 0 = 0 1 1 0

x′ y′

x′ y′ Continued over page

x

280

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THINK

WRITE

4

Multiply matrices to determine x and y.

x = y′ y = x′

5

Substitute for x and y in the original expression.

y = x 2 becomes x′ = y′2 y′ = ± x′

6

Sketch the original and image curves. Note that the points (1, 1) and (0, 0) are unchanged as they are on the mediator.

y = x2

y y=x (1, 1)

y' = √x'

x

0 y=x

Reflection in the line y = x tan q

y' = –√x'

y

This line might be more easily recognised as y = mx, where m is the gradient of the line which passes through the origin. y2 – y1 Remember that the gradient m = --------------x2 – x1

0

y2 – y1 -. and tangent ratio = --------------x2 – x1

y

Note the following from these diagrams.

=

1

θ θ

y = x tanθ

= A(1, 0) x

B(0, 1)

Therefore the tangent and gradient ratios provide rise the same information: -------- . run Carefully examine these diagrams that illustrate reflection of the points (1, 0) and (0, 1) in the line y = x tan θ.

A' (cos 2θ , sin 2θ )

my = x tanθ = 90° – θ

θ 0

90° – 2θ

x

=

B'(cos (90° – 2θ ), – sin (90° – 2θ ))

For the point A(1, 0): 1. point A is reflected to a point equidistant from and perpendicular to the line 2. the angle from the x-axis to A′ is 2θ 3. the x-coordinate of the right-angled triangle is cos 2θ 4. the y-coordinate of this triangle is sin 2θ. 5. Hence point (1, 0) → (cos 2θ, sin 2θ). For the point B(0, 1): 1. point B is reflected to a point equidistant from and perpendicular to the line 2. ∠MOB = 90° − θ therefore ∠MOB′ = 90° − θ

C h a p t e r 6 Tr a n s f o r m a t i o n s u s i n g m a t r i c e s

3. 4. 5. 6. 7.

281

therefore ∠XOB′ = (90° − θ) − θ = 90° − 2θ the x-coordinate = cos (90° − 2θ) the y-coordinate = −sin (90° − 2θ) because the angle is in the fourth quadrant. Hence point (0, 1) → [cos (90° − 2θ), −sin(90° − 2θ)]. Using trigonometric ratios, this simplifies to yield (sin 2θ, −cos 2θ). (Remember that sin 30° = cos 60°, etc.)

Using all this information from the reflection of points (1, 0) and (0, 1) in the line sin 2 θ . y = x tan θ yields: My = x tan θ = cos 2 θ sin 2 θ –cos 2 θ

WORKED Example 17

Find the matrix for the reflection in the line y =

3 x.

THINK

WRITE

1

Note that the sign applies only to the 3. Use a sketch to express 3 as the tangent ratio of some angle.

–π 6

√3

2

–π 3

1

π tan --- = 3 2

State the general reflection matrix in the line π y = x tan θ, then substitute --- for θ. 3

sin 2 θ My = x tan θ = cos 2 θ sin 2 θ –cos 2 θ

My =

3

Evaluate these ratios using the following triangle.

√3

2 –π 3

–1

π 2— 3

3

3x

2π 2π cos -----sin -----3 3 = π π 2 2 sin ------ –cos -----3 3 1 3 − --- ------2 2 = 1--------32 2

282

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WORKED Example 18 Find the image of the line y = −x − 1 as reflected in the line y = THINK

1

Use the matrix from the previous example as My = 3 x.

2

Set up the initial matrix transformation and inverse statement.

3

Find the inverse and multiply the matrices.

3 x.

WRITE

My =

x′ y′

3 1 – --- ------2 2 3x = 3 1 --------2 2 = My =

x y

3x

x′ y′

x y

= My–1=

x y

1--3 − ------1 2 2 = ----------------– 1--- – 3--3 1 − --4 4 − ------2 2

3x

1 3 − --- ------2 2 = 1--------32 2

x′ y′

x′ y′

1 3 x = − --- x′ + ------- y′ 2 2 3 1 y = ------- x′ + --- y′ 2 2 4

Substitute for x and y into the original equation. Make sure you carry through the minus sign from the function.

y = −x − 1 becomes ------3- x′ + 1--- y′ = 1--- x′ − ------3- y′ − 1 2 2 2 2 1--3 1 3 y′ + ------- y′ = --- x′ − ------- x′ − 1 2 2 2 2

5

Simplify and rationalise the denominators to find the equation of the image line.

1--------------+ 31– 3 y′ = ---------------- x′ − 1 2 2 y′ = ( 3 − 2)x′ + 1 −

3

C h a p t e r 6 Tr a n s f o r m a t i o n s u s i n g m a t r i c e s

THINK

WRITE y

Sketch the original and its image. To assist in graphing the image, a calculator can be used to obtain y′ = −0.27x′ − 0.73.

6

283

y = √3x

0

x y' = (√3 – 2) x' + 1 – √ 3

1 a –1 0 0 1

y = –x – 1 2 a i (−3, −1) ii (−4, 2) iii (1, −3) iv (2, 4) v (−3, 0) vi (2, −1) y (iv)

(iv)' (ii)

(ii)' (v)'

(v) x

0 (i)' (vi) (iii)

(vi)' (i) (iii)' mx = 0

b 1

b i y′ = x′2 iii y′ = −x′2 iii y′ = ± – x′

3 a i y′ = −x′ iii y′ = −x′ iii y′ = x′ iv y′ = --12- ( 3 − 1)2x′ or y′ ≈ 0.268x′

3

i ( --2- −

0 –1 and

transformations matrices would be – 1 0 followed by the addition of 2 . Verify this 0 1 0 by checking that (0, 0) → (2, 0).

remember remember 2 d

0

0 –1 In the reflections covered so far, the mediator has always passed through the origin. If we return to the original reflection in the line x = 1, it needs to be broken into two trans0 1 formations: a reflection and a translation. Reflecting in the line x = 1 can be thought of c 1 0 as reflection in the y-axis (x = 0) followed by a shift 2 to the right (x = 1 is 1 unit to the right of the origin therefore the image would be 2 × 1 = 2 units to the right). Thus the d 1 0

------32

,

e

1. Reflection is a congruent transformation. 2. Reflection occurs through a mediator, m. sin 2 θ . 3. Reflection in the line y = x tan θ is represented by M = cos 2 θ sin 2 θ –cos 2 θ 3--3---------31--–-----13 3- –---------3--------2

+ 2)

ii (2 +

iv (−1 + 2 3 , − 3 − 2) v

( 3--2-

,

3 , 2 3 − 1) iii (

3 3 ---------2

6E

)

then 0 4

−

2

vi (−1 −

3 ------2

2

,

2

+ 2)

2 b

, − 3 + --12- )

f

– 1--2-

3 ------2

------32

--12

0 –1 –1 0

i (3, 1) ii (4, −2) iii (−1, 3) iv (−2, − 4) v (3, 0) vi (−2, 1) y

Reflections

(iv) (iii)' my = 0 (vi)'

1 Write the matrices for the following reflections: a mx = 0 b my = 0 d my = 2 e my = 3 x

0

(vi)

c f

my = x my = −x

(ii) (i)' (v)(v)' x (i) (ii)'

(iii) (iv)'

2 Find the images of each of the following points under the reflection given below. Sketch each original and its image. 14 3 a y-axis b x-axis c y = −x d y = ------- x 3 i (3, −1) ii (4, 2) iii (−1, −3) iv (−2, 4) v (3, 0) vi (−2, −1)

WORKED

Example

WORKED

Example 15,16,17,18

3 Find the image of the following curves under each of the reflections given below. a y=x b y = x2 c y = 2x2 + 1 d y = −x2 i y-axis

ii x-axis

–1 3 c i y′ = 2x′2 + 1 ii y = −2x′2 − 1 iii y′ = ± –x′ ----------------2 d i y′ = −x′2 ii y′ = x′2 iii y′ = ± x′

iii y = −x

3 iv y = ------- x (part a only) 3

2 c i ii iii iv v vi

(1, −3) (−2, − 4) (3, 1) (− 4, 2) (0, −3) (1, 2)

284

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Dilations

SLE 4: Demonstrate the use of the transformation matrices (rotation, reflection, dilation) as an application of 2 × 2 matrices to geometric transformations in the plane.

So far we have investigated 4 kinds of transformations. The translation shifted the figure on the plane; the general linear transformation produced an image that, on occasions, bore little resemblance to its original. The rotation and reflection transformations are congruent transformations with the original basically repositioned on the plane. A dilation is a transformation in which point P and image P′ are collinear from a fixed point, usually the origin O, as shown in the figure below. P'

P

O

The length OP′ = kOP where k is referred to as the dilation factor. If k > 0, a dilation may be an enlargement (for k > 1) or a reduction (for 0 < k < 1). A

A'

B

B'

A

A'

B

B' O

C k>1

C'

O

C' 0
C

If k < 0 then the image of the original has been mapped through the origin in a reverse direction. A

In this diagram, k = − 1--- , therefore the 2 image appears half the distance from the B' fixed point O and on the opposite side of C' O C O to the original points. In a dilation: 1. length and area are not preserved; the A' shape will appear similar, but not conk<0 B gruent to the original 2. the dilation d is denoted by the matrix Dk, x with the dilation factor of k given parallel to the x-axis and the anchor line being the y-axis.

C h a p t e r 6 Tr a n s f o r m a t i o n s u s i n g m a t r i c e s

285

Dilation parallel to the x-axis The dilation matrix Dk, x of the points (1, 0) and (0, 1) under the dilation dk, x is given by Dk, x = k 0 0 1

y (0, 1)

where

(1, 0) (k, 0) 1. (0, 1) is left unchanged since it is on the anchor line x 0 2. the x-coordinate is mapped k × 1 units away from the anchor line. This is shown graphically in the figure at right. Dilation parallel to the x-axis can be thought of as pulling the plane away from the fixed point or anchor line — in this case, the y-axis.

Dilation parallel to the y-axis This dilation pulls the plane away from the x-axis so all points on the x-axis are anchored. The figure at right shows that

y (0, k) (0, 1)

Dk, y = 1 0 0 k

(1, 0)

where 1. (1, 0) is left unchanged since it is on the anchor line 2. the y-coordinate is mapped k × 1 units away from the anchor line (the x-axis).

0

WORKED Example 19

Find the coordinates of the image of point (4, 3) under the dilation factor of -2 parallel to the x-axis. Sketch the original point and its image. THINK 1

WRITE

Sketch the dilation and construct the matrix from the sketch.

y (0, 1) (–2, 0)

(1, 0) 0

x

D–2, x = – 2 0 0 1 2

Write the general transformation matrix equation.

x′ y′

= D–2, x 4 3 Continued over page

x

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THINK 3

WRITE

Multiply the matrices to produce the image coordinates.

x′ y′

= –2 0 0 1

4 3

= –8 3 Point (4, 3) maps to image point (−8, 3) under a dilation of −2 parallel to the x-axis. 4

Sketch the original point and its image.

y P' (–8, 3)

P(4, 3)

x

0

WORKED Example 20

Find the equation of the image of y = 2x + 1 under the dilation d2, x. Sketch the original line and its image. THINK 1 Sketch the dilation and construct the matrix from the sketch.

WRITE y (0, 1) (1, 0)

(2, 0) x

0

D2, x = 2 0 0 1 2

Set up the general transformation matrix equation and rearrange to have x y

3

as the subject.

Find the inverse of D2, x and substitute it into the equation.

x′ y′

= D2, x x y

x y

= D2,–1x x ′ y′ 1 = --- 1 0 2 0 2 =

1 --2

0

0 1 4

Multiply the matrix equations.

x = 1--- x′ 2 y = y′

x′ y′ x′ y′

C h a p t e r 6 Tr a n s f o r m a t i o n s u s i n g m a t r i c e s

THINK

287

WRITE

5

Substitute x and y in the original equation and simplify.

6

Sketch the original and its image. Note that (0, 1) remains unchanged since it is on the anchor line of the y-axis.

y = 2x + 1 becomes y′ = 2( --1- x′) + 1 2 y′ = x′ + 1 y y = 2x + 1

y' = x' + 1 (0, 1) x

0

WORKED Example 21

Find the image of the circle x2 + y2 = 9 with a dilation factor of Sketch the original circle and its image. THINK 1

1 --3

parallel to the y-axis.

WRITE

Sketch the situation and use this to construct the dilation matrix.

y

(0, 1) (0, 1–3 ) (1, 0) x

0

D --1-, y = 3

2

Set up the general transformation matrix equation and rearrange to put x y

3

as the subject.

Calculate the inverse of D and substitute it into the equation.

1 0 1 --3

0

x′ y′

= D 1---, y

x y

x y

= D 1--–1 -, y

x′ y′

3

3

1 = ----1 --3

1 --3

x′ y′

0

0 1

= 1 0 0 3

x′ y′ Continued over page

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THINK

WRITE

4

Multiply the matrices and write expressions for x and y.

x = x′ y = 3y′

5

Substitute x and y into the original equation and rearrange to fit the general equation of an ellipse.

x2 + y2 = 9 becomes (x′)2 + (3y′)2 = 9 x′2 + 9y′2 = 9

2

2

x y This can be written as ----- + ----- = 1 9 1 2 2 x y since ----- + ----- = 1 is the general equation of an 2 2 a b ellipse about the origin. Therefore a = 3 So the length of the semi-major axis is 3. b = 1 So the length of the semi-minor axis is 1. 6

Sketch the original and its image.

y x2 + y2 = 9

x

0

x' 2

y' 2

— + — =1 9 1

If you think about the original shape and its image as shown in this example you will understand that the dilation factor of 1--- , in effect, shrinks the original shape, parallel to 3 the y-axis so the figure falls back towards the anchor line (the x-axis) and leaves all points on the x-axis unchanged.

Dilation about the origin, dk The previous dilations have been mapped parallel to an axis, where that axis has provided the anchor line for the stretching of the plane. However, a dilation about the origin does not anchor to a line, but rather to a point — the origin. The diagram at right shows this stretch that results in both x and y coordinates being mapped a dilation factor of k from the origin. Therefore, if the original point is on the origin it will map onto itself. Only the dilation factor is given in the dilation matrix: Dk = k 0 0 k

y

P'(kx , ky) (0, k) (0, 1)

P (x, y) (1, 0)

0

(k, 0) x

C h a p t e r 6 Tr a n s f o r m a t i o n s u s i n g m a t r i c e s

WORKED Example 22

289

Find the image of y = x2 under a dilation factor of −2 about the origin. Sketch the original and its image. THINK 1

WRITE

Sketch the situation to construct the matrix.

y (0, 1) (–2, 0) 0

(1, 0) x (0, –2)

D–2 = – 2 0 0 –2 2

Set up the initial transformation matrix equation.

x′ y′

= D–2 x y

x y 3

x′ y′

–1

= D –2

x′ y′

1 = --- – 2 0 4 0 –2

Evaluate the inverse and multiply.

=

– 1--2-

0

0

– 1--2-

x′ y′

x = − 1--- x′ 2

y = − 1--- y′ 2

4

Substitute for x and y into the original equation and simplify.

y = x becomes 2

− 1--- y′ = (− 1--- x′)2 2

2

=

1 2 --- x′ 4

y′ = − 1--- x′2 2

5

Sketch the original and its image. The minus sign results in the image reversing its position with respect to the origin, and the factor of 2 results in the broader parabola.

y y = x2

0

x

y' = – –21 x' 2

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History of mathematics M AU R I T S C O R N E L I U S E S C H E R ( 1 8 9 8 – 1 9 7 2 ) Portrait of M. C. Escher © 2000 Cordon Art, Baarn, Holland. All rights reserved.

During his life . . . World War I and World War II take place. Flight technology develops — from the Wright brothers first flight in 1903 to the moon landing in 1969. Israel is established.

Maurits Escher is quoted as having said ‘I never got a pass mark in math. … And just imagine — mathematicians now use my prints to illustrate their books. … I guess they are quite unaware that I am ignorant about the whole thing.’ Escher was born on 17 June 1898 in the Netherlands. His early work was mainly concerned with the representation of visible reality, such as landscapes and buildings. However, he gradually became more interested in studying the abstract space-filling patterns used by the Moors in mosaics found in Spain. He also studied a paper by Polya on 17-plane crystallographic groups; however, instead of using geometrical motifs, Escher used animals, plants or people to fill the space on his intricate prints. Even though he professed ignorance of all things mathematical, Escher incorporated many mathematical ideas in his works — infinity, Mobius strips, stellations, deformations, reflections, rotations, Platonic solids, spirals and the hyberbolic plane.

Original Escher prints are highly prized possessions now, but it was not until 1951 that he actually began to earn a reasonable income from his prints. Widely regarded as a graphic artist, his designs have appeared on postage stamps, bank notes, T-shirts, jigsaw puzzles, record album covers, and, as he remarked, in many scientific and mathematical publications.

‘Relativity’ by M. C. Escher © 2000 Cordon Art, Baarn, Holland. All rights reserved.

His work has been held in high regard by both artists and mathematicians. He died in 1972, in the Netherlands. Research 1. Research Mobius strips, stellations, deformations, reflections, rotations, Platonic solids, spirals and the hyberbolic plane. 2. Look through scientific and mathematical publications to see if any use Escher’s prints as covers or illustrative pages.

remember remember 1. Dilation occurs in relation to an anchor line or point. 2. Dilation is not a congruent transformation.

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6F

1 a i ii iii iv v vi

Dilations

y

4, 1) (−8, 3) (0, 3) (6, 0) (4, 5) (1, −3)

(ii)'

(i) 0

Example

19

i (2, 1)

ii (−4, 3)

iii (0, 3)

iv (3, 0)

v (2, 5)

vi

2 Find the image of the line y = 3x − 2 under the following dilations: a dilation factor 2 parallel to the x-axis y′ = 3--2- x′ − 2 20 b dilation factor −1 parallel to the y-axis y′ = −3x′ + 2

WORKED

(i)' (iv)' x

(iv)

(vi)

Sketch the original and its image for all questions. 1 Find the image of each of the following points under the dilations given: a 2, parallel to the x-axis see top right b −3, parallel to the y-axis

WORKED

(v) (v)' (iii) (iii)'

(ii)

(vi)'

see bottom left

( 1--- , 2

−3)

2

y

y = 3x – 2

Example

2

(a)

4 a

2

y = 2 x2

y

4

9

2

(b)

2

x y------ + ---=1 64 9

4 Find the image of y = 2x with a dilation factor of 4 a parallel to the x-axis y′ = 1--8- x′2 b parallel to the y-axis 2

y' = –18 x' 2

x

0

2

x y WORKED 3 Find the image of the ellipse ----- + ----- = 1 with a dilation factor of Example 4 9 2 2 21 x a 1--- , parallel to the y-axis ---b 4, parallel to the x-axis - + 4y -------- = 1 y′ = 8x′2

x

5 Find the image of the line y = 3x − 2 under a dilation factor of

WORKED

Example

22 6 a i (1, 1--2- )

ii (−2, 1 1--2- )

iii (0, 1 --12- ) iv (1 --12- , 0) v (1,

2 1--2-

) vi

( 1--4-

,

−1 1--2-

7 Find the image of the ellipse in question 3 with a dilation factor of a −2, about the origin b --1- , about the origin )

(vi)'

(v) (iii) (i) (iv) (iv)' x (vi)

(ii)'

(i)'

(iii)'

(v)'

y

(a)

x

4

2

2

x y 16y 7 a ------ + ------ = 1 b 4x2 + ----------- = 1 16 36 9

(b)

6 b i (−8, − 4) ii (16, −12) iii (0, −12) iv (−12, 0) v (−8, − 20) vi (−2, 12)

1b The final transformation discussed in this i (2, −3) ii (− 4, −9) chapter is that of shears, which can be thought of iii (0, −9) iv (3, 0) 1--as a push from one side that results in a change v (2, −15) vi ( 2 , 9) y

1--2

2

Shears

0

about the origin. y′ = 3x′ −

6 Find the image of each of the points in question 1 under the following dilations: a 1--- , about the origin b −4, about the origin 3

2

(ii)

1 --4

in shape. An example of this is seen when changing a rectangle into a parallelogram. Where a dilation ‘pulls’ the plane from a certain anchor point or line, a shear pushes from one side and any points on the anchor line again remain unchanged.

x2 — 4

y

Push

A shear parallel to the x-axis (see the figure shown at right) moves every point in the plane parallel to the x-axis by a distance proportional to its distance from the x-axis. That is, points on the x-axis remain anchored while points further away are pushed further from their original position.

x

0

y

0

x

y +— 9 =1 2

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Similarly a shear parallel to the y-axis (see the figure shown at right) moves every point in the plane parallel to the y-axis by a distance proportional to its distance from the y-axis. The shear transformation matrix uses similar notation to other linear transformations, where Sk, x denotes the shear with a shear factor of k parallel to the x-axis and Sk, y denotes the shear with a factor of k parallel to the y-axis.

y

x

0

That is, Sk, x = 1 k , Sk, y = 1 0 . 0 1 k 1 y

Shears parallel to the x-axis As can be seen from the figure at right, point (1, 0) remains unchanged because it is anchored to the x-axis while point (0, 1) is mapped to (k, 1). This means that point (2, 1) will map to (2k, 1).

(0, 1)

(k, 1) (1, 0) x

0

y

Shears parallel to the y-axis

(1, k)

The figure at right shows the point (0, 1) unchanged by the shear because it lies on the y-axis while point (1, 0) is mapped to (1, k).

(0, 1) (1, 0) 0

WORKED Example 23 The vertices of a triangle are O(0, 0), A(2, 0) and B(2, 3). Find the image of these points O′A′B′ under a shear factor of 2 parallel to the y-axis. Sketch the original and its image. THINK 1

Sketch the initial unit diagram and use this to determine the shear matrix.

WRITE y (1, 2) (0, 1) (1, 0) 0

S2, y = 1 0 2 1

x

x

C h a p t e r 6 Tr a n s f o r m a t i o n s u s i n g m a t r i c e s

THINK 2

Set up the initial general transformation matrix equation.

3

Substitute S2,y and solve for each point in turn.

WRITE x′ y′

= S2, y x y

For O(0, 0), the equation is: x′ y′

= 1 0 2 1

0 0

= 0 0 O(0, 0) is unchanged because it is on the anchor axis.

So O(0, 0) → O′(0, 0)

For A(2, 0), the equation is: x′ y′

= 1 0 2 1

2 0

= 2 4 So A(2, 0) → A′(2, 4) For B(2, 3), the equation is: x′ y′

Note that the y-coordinate is not actually multiplied by 2. 4

Sketch the image points with the original.

= 1 0 2 1

2 3

= 2 7 So B(2, 3) → B′(2, 7)

y B'(2, 7)

A'(2, 4) B(2, 3)

O(0,0)

A(2, 0)

x

293

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WORKED Example 24

Find the image of the parabola y = 2x2 under the shear factor of 3, parallel to the y-axis. Sketch the original and its image. THINK 1

Sketch the initial unit diagram and use this to determine the shear matrix.

WRITE y (1, 3)

(0, 1) 0

(1, 0)

x

S3, y = 1 0 3 1 2

Set up the initial general transformation matrix equation.

3

Rearrange in terms of x and y.

4

Find the inverse of S3, y .

x′ y′ x y

= S3, y x y –1

= S 3, y

x′ y′ x′ y′

1 = --- 1 0 1 –3 1 =

x′ – 3x′ + y′

5

Multiply to find expressions for x and y.

x = x′ y = −3x′ + y′

6

Substitute for x and y into the original equation.

y = 2x2 becomes −3x′ + y′ = 2x′2 y′ = 2x′2 + 3x′ = x′(2x′ + 3)

7

Sketch the image with the original. Find the intercepts to aid in sketching the image. Notice that ii(i) the origin is anchored i(ii) the positive x-values are pulled up from the x-axis (iii) the negative x-region seems to have slipped back from the x-axis.

y′ = 0 when x = 0 and x = −1 1--2

y

y=

2x2

y' = x'(2x' + 3)

0

x

1 i a y

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6

P'

P 0

1. A shear can be thought of as a push parallel to an anchor line that transforms all points on the plane by a distance proportional to their distance from the anchor line. 2. A shear is not a congruent transformation.

y

–7 0

P'

P x

x

WORKED

Example

1 Find the image of each of the following points under a shear factor of i 2 parallel to the y-axis

23 y P'

1

P

0

1 --4

eBook plus

parallel to the x-axis

a (3, 0) b (2, 1) c (−4, 1) d (3, −2) e (−2, −5) f (0, −5) 1 i a (3, 6) b (2, 5) c (−4, −7) d (3, 4) e (−2, −9) f (0, −5) Sketch each pair of original and image points.

1 i b

5

ii

x

2

2 A parallelogram has vertices A(0, 0), B(1, 3), C(6, 3) and D(5, 0). Sketch the original shape and its image under the shear factor of 2a a 3 parallel to the x-axis b 3 parallel to the y-axis c −1 parallel to the x-axis d −1 parallel to the y-axis

Digital docs: SkillSHEET 6.1 Building a transformation matrix SkillSHEET 6.2 Finding the image of a curve after transformation y

B'(10, 3) C(6, 3) C'(15, 3) B(1, 3) A = A' D(5, 0)

3 Find the image of each of the following curves under a shear factor of i 3 parallel to the y-axis ii 3 parallel to the x-axis 24 3 d ii y a y=x b y = −x 2 y = 2x + 5 c y=x d y = 2x + 5 y' = 2–7 x' + 5–7 Sketch the original and image curves.

x

WORKED

Example

eBook plus Digital doc: WorkSHEET 6.2

x

Transformations

1

2

2

x′ y′ ------- + -----=1 16 4

x2 — 16

2

y +— 4 =1

0

y

x2 — 4

2

y +— 1 =1

x

2

2

1 i c

–4

y

Shears

P

f (−1 1--4- , −5)

3

6G

e (−3 1--4- , −5)

P'

d (2 1--2- , −2)

d

c (−3 3--4- , 1)

4

b (2 1--4- , 1)

0 –2

1 ii a (3, 0)

remember remember

x

3

x y - = 1 under the dilation of d . Give the equation of 1 Transform the ellipse ----- + ---–2 4 1 the new ellipse, fully supporting your response with matrix operations and a fully labelled diagram. 2 Design a set of 4 or 5 transformations that map a shape of your own choosing to another shape on the plane. Your response should include all working and fully labelled diagrams. 3 With your current knowledge of transformations using matrix applications, investigate whether the following transformations are possible. You may need to consider a series of transformations. a A square into a straight line Possible; perform a singular linear transformation b A triangle into a square Not possible c A circle into a straight line Possible; perform a singular linear transformation d A square into a circle Not possible e A kite into a square Possible; one way is to perform a translation then a linear transformation

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summary Geometric transformations and matrix algebra • A general transformation maps each point of the Cartesian plane onto some other point of the plane. • A translation, t, moves each x-coordinate a units parallel to the x-axis and each y-coordinate b units parallel to the y-axis, such that x ′ = x + T, where y y′ (x′, y′) is the image of point (x, y) and T is the transformation matrix. The translation t results in a congruent transformation.

Linear transformations • A linear transformation, l, warps the plane such that

x′ y′

= L x . Linear y

transformations are not congruent transformations.

Rotations • A rotation, r, rotates the plane about a fixed point to result in a congruent transformation. The matrix representing the rotation is Rθ = cos θ –sin θ . sin θ cos θ

Reflections • A reflection, m, reflects every point of the original through a straight line called a mediator and results in a congruent transformation. The reflection matrix sin 2 θ . My = x tan θ = cos 2 θ sin 2 θ –cos 2 θ

Dilations • A dilation, d, transforms each point P to P′ where P and P′ are collinear with a fixed point O. The matrix Dk, x represents a dilation of k units parallel to the x-axis anchored from the y-axis. The matrix Dk represents a dilation factor of k units through the origin and D–k represents the same dilation in the reverse direction.

Shears • A shear, s, is a transformation like a ‘push’ from one side. The matrix Sk, x moves every point in the plane parallel to the x-axis by a distance proportional to its distance from the x-axis. Points on the x-axis remain unchanged.

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7 y′ = −2x′ + 3

9 y′ =

y

CHAPTER review

y y' = –2x' + 3

x′--2

+1

y' = 1–2 x' + 1

x

x y = –2x + 2

y = –x + 1

1 State all congruent transformations. Translations, rotations and reflections

6A

2 Find the image of the following points under the translation T = – 1 3

6A

a

(0, 0) (−1, 3)

b (3, 1) (2, 4)

c

(4, −2) (3, 1)

3 The line y = 2x − 1 undergoes a succession of translations defined by T1 =

1 and –4

6A

T2 = – 2 . Show that the order in which these occur has no effect on the result. 3 Check with your teacher. 4 Find the equation of the image of y = 2x2 under the translation of

2 . y′ = 2x′2 − 8x′ + 7 –1

5 Find the image of points A(0, 1) and C(3, 2) under the transformation

2 3 . –1 0

A′(3, 0) C′(12, −3)

– 2 --13- 2 --23– 2 2--3- 2 1--3-

6A 6B

6 Find the matrix of the linear transformation which maps (1, 2) to (3, 2) and (3, 3) to (1, −1).

6B

7 Find the image of y = −x + 1 under the linear transformation 1 2 . Sketch the original 1 –1 and image curves. y′ = −2x′ + 3 See graph top of page.

6C

8 Find the image of the following points under the given anticlockwise rotations about the origin. π 3b (−3, 0) where θ = --- ( – 3--2- , − 3--------a (2, −1) where θ = −π (−2, 1) ) 2 3 π 22, − 3--------) c (4, 1) where θ = − --- ( 5--------d (2, 5) where θ = −60° (1 + 5--2- 3 , − 3 + 2 1--2- ) 2 2 4

6D

π 9 Find the image of the line y = −2x + 2 through a rotation of --- . Sketch the original and the 2 x′ image. y′ = ---2- + 1 See graph top of page. 10 Find the image of each of the following points under the reflection as given: a (3, −1) in the y-axis (−3, −1) b (2, 2) in the line y = x (2, 2) 13------3 (− + , + 1) c (1, 2) in the line y = 3 x d (−2, 1) in the line y = 2 (−2, 3) 2 2

6D 6E

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6E

11 Find the image of the line y = −3x under reflection in the line y = −x. Sketch the original and the image.

6F

12 Find the image of each of the following points under the dilation factors given: a (2, 1), 2 units parallel to the y-axis (2, 2) b (2, 4), --1- unit about the origin (1, 2) 2 c (4, −1), −4 units about the origin (−16, 4) d (0, 2), −3 units parallel to the x-axis (0, 2)

6F 6G

13 Find the image of y = −3x under the dilation factor of 3 parallel to the x-axis. y′ = −x′

6G

15 Find the image of the curve y = 2x2 under a shear factor of 2 parallel to the y-axis. y′ = 2x′2 + 2x′

14 Find the image of each of the following points under the given shear factor: a

(3, 1),

1 --2

unit parallel to the y-axis (3, 2 1--2- )

b (−1, 1), 3 units parallel to the x-axis (2, 1)

Modelling and problem solving 1 Under a certain transformation, the circle (x − 2)2 + (y − 2)2 = 4 becomes (x + 2)2 + (y − 2)2 = 4. It is claimed that three different transformations could have achieved this outcome. Investigate this claim giving details of the possible transformations and their matrices of transformation. 2 A square, ABCD, formed by the points A(0, 0), B(1, 0), C(1, 1) and D(0, 1) is mapped to A′(0, 0), B′(−3, 0), C′(−3, 3) and D′(0, 3) after two successive transformations. Analyse this eBook plus mapping to determine the transformations involved, and find a single transformation matrix that could achieve this result. Digital doc: Test Yourself Chapter 6

1 Rotation of 90°: 0 – 1 1 0 or reflection in y-axis: – 1 0 0 1 or translation 4 left: – 4 0

2 Dilation of 3 about the origin followed by reflection in the y-axis: – 3 0 0 3

11 y′ = − 1--3- x′ y

x y' = – 1–3 x' y = –3x

7

Introduction to vectors

syllabus reference Core topic: Vectors and applications

In this chapter 7A 7B 7C 7D 7E

Vectors and scalars Position vectors in two and three dimensions Multiplying two vectors — the dot product Resolving vectors — scalar and vector resolutes Time-varying vectors

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Vectors and scalars • definition of a vector, including standard unit vectors i, j and k • relationship between vectors and matrices • two- and three-dimensional vectors and their algebraic and geometric representation • operations on vectors including addition, and multiplication by a scalar • unit vectors • scalar product of two vectors • resolution of vectors into components acting at right angles to each other • calculation of the angle between two vectors • applications of vectors in both life-related and purely mathematical situations

Introduction In mathematics, one of the important distinctions that we make is between scalar quantities and vector quantities. Scalar quantities have magnitude only; vector quantities have direction as well as magnitude. Most of the quantities that we use are scalar, and include such measurements as time (for example 1.2 s; 15 min), mass (3.4 kg; 200 t) and area (3 cm2; 400 ha). However, consider the measurement of velocity. A velocity of 20 km/h has both magnitude and direction. One of the boats shown below may travel 20 km/h north from Townsville, while the other one may travel 20 km/h east from the same point. Although they both are travelling at the same speed (magnitude) they are travelling in different directions; they do not end up in the same place!

Now consider the force involved in Daniel and Anna fighting over who gets to use the television remote control. Daniel exerts a force of 40 N and Anna exerts a force of 50 N and they apply these forces as shown. In what direction will the remote control move and what is the force in that direction? That is, what is the resultant force? The resultant force depends not only on the size of each force but the direction in which the forces are applied. In the following discussion we will develop techniques to find the resultant force. A vector is a quantity that has magnitude and direction.

40 N

150° 50 N

301

Chapter 7 Introduction to vectors

Vector notation A vector is shown graphically as a line, with a head (end) and B tail (start). The length of the line indicates the magnitude and the orientation of the line indicates its direction. In the figure at right, the head of the vector is at point B A (indicated with an arrow), while the tail is at point A. When writing this vector we can use the points A and B to indicate the start and end points with a special arrow to indicate that it is a vector: AB . Some textbooks use a single letter, in bold, such as w, but this is difficult to write using pen and paper, so w ˜ can also be used. The symbol (~) is called a tilde.

Equality of vectors Since vectors are defined by both magnitude and direction: two vectors are equal if both their magnitude and direction are equal. In the figure, the following statements can be made: u= v ˜ ˜ (directions are not equal) u≠w ˜ ˜ (magnitudes are not equal). u≠z ˜ ˜

w ~ ~u ~v ~z

Addition of vectors

Consider a vector u which measures the travel from A to B and another vector, v , ˜ which measures the˜ subsequent travel from B to C. The net result is as if the person travelled directly from A to C (vector w ). Therefore we can say that w = u + v . ˜ ˜ ˜ ˜ To add two vectors, take the tail of one vector and join it to the head of another. The result of this addition is the vector from the tail of the first vector to the head of the second vector. Returning to Daniel and Anna who are fighting over the television remote control (see page 300), we see that the forces they apply to the remote control unit can be represented as a sum of two vectors.

From this figure we are able to get a rough idea of the magnitude and direction of the resultant force. In the following sections, we will learn techniques for calculating the resultant magnitude and direction accurately.

The negative of a vector If u is the vector from A to B, then – u is the vector from ˜ A. ˜ B to We can subtract vectors by adding the negative of the second vector to the first vector.

–u ~

A ~u

B

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WORKED Example 1 Using the vectors shown at right, draw the result of: a u+v b –u c u–v d v – u. ˜ ˜ ˜ ˜ ˜ ˜ ˜ THINK WRITE a

1

2

Move v so that its tail is at the head of u . ˜ ˜

~v ~u

a ~v

~v ~u

Join the tail of u to the head of v to ˜ ˜ find u + v . ˜ ˜

~u + ~v

~v

~u

b Reverse the arrow on u to obtain – u . ˜ ˜

b

–u ~ ~u

c

1

Reverse v to get – v . ˜ ˜

c –v ~ ~u

2

d

1

2

Join the tail of – v to the head of u to get – v + u which˜ is the same as u˜– v ˜ ˜ or u +˜ ( – v˜ ) . ˜ ˜ Reverse u to get – u . The vectors are ˜ properly’ ˜ with the head of now ‘aligned – u joining the tail of v . ˜ ˜ Join the tail of – u to the head of v to ˜ ˜ get v – u . ˜ this is the same as ( – u + v ) . Note˜ that ˜ ˜

–v ~

–v u ~+~ ~u

d ~v –u ~ ~v – ~u

~v –u ~

Multiplying a vector by a scalar Multiplication of a vector by a number (scalar) affects only the magnitude of the vector, not the direction. For example, if a vector u has a direction of north and a magnitude of 10, ˜ 3u has a direction of north and magnitude then the vector ˜ of 30. If the scalar is negative, then the direction is reversed. Therefore, – 2u has a direction of south and a magnitude ˜ of 20.

N W

E S

~u 3u ~

–2u ~

Chapter 7 Introduction to vectors

303

WORKED Example 2

Use the vectors shown at right to draw the result of: a 2r + 3s b 2s – 4r . ˜ ˜ ˜ ˜ THINK WRITE a 1 Increase the magnitude of r by a factor a of 2 and s by a factor of 3.˜ ˜

2

~s

~r

3s ~ 2r~

Move the tail of 3s to the head of 2r . ˜ ˜ 2r to the head of Then join the tail of 3s to get 2r + 3s . ˜ ˜ ˜

3s ~

2r~ + 3s ~ 3s ~ 2r~

b

1

Increase the magnitude of s by a factor of 2 and r by a factor of 4.˜ ˜

b 2s~ 4r~

2

3

Reverse the arrow on 4r to get – 4r . ˜ ˜

2s~ –4r~ 2s~ – 4r ~

Join the tail of – 4r to the head of 2s . ˜ ˜

2s~ –4r~

WORKED Example 3 The parallelogram ABCD can be defined by the two vectors b ˜ and c . ˜ In terms of these vectors, find: a the vector from A to D b the vector from C to D c the vector from D to B. THINK

WRITE

a The vector from A to D is equal to the vector from B to C since ABCD is a parallelogram.

a AD = c ˜

b The vector from C to D is the reverse of D to C which is b . ˜ c The vector from D to B is obtained by adding the vector from D to A to the vector from A to B.

b CD = – b ˜ c DB = – c + b = b –˜ c ˜ ˜ ˜

D

C ~c

A

~b

B

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WORKED Example 4 A cube PQRSTUVW can be defined by the three vectors a , b and c ˜ ˜ ˜ as shown at right. Express in terms of a , b and c : a the vector joining˜ P ˜to V ˜ b the vector joining P to W c the vector joining U to Q d the vector joining S to W e the vector joining Q to T. THINK All of the opposite sides in a cube are equal in length and parallel. Therefore all opposite sides can be expressed as the same vector. a The vector from P to V is obtained by adding the vector from P to Q to the vector from Q to V. b The vector from P to W is obtained by adding the vectors P to V and V to W. c The vector from U to Q is obtained by adding the vectors U to P and P to Q. d The vector from S to W is obtained by adding the vectors S to R and R to W. e The vector from Q to T is obtained by adding the vectors Q to P, P to S and S to T.

T

W R

S ~c

U

P

V ~b ~a

Q

WRITE

a PV = a + b ˜ ˜ b PW = a + b + c ˜ ˜ ˜ c UQ = –b + a = a ˜– b ˜ ˜ ˜ d SW = a + b ˜ ˜ e QT = –a + c + b = b +˜ c –˜ a ˜ ˜ ˜ ˜

WORKED Example 5

A boat travels 30 km north and then 40 km west. a Make a vector drawing of the path of the boat. b Draw the vector that represents the net displacement of the boat. c What is the magnitude of the net displacement? d Calculate the bearing (from true north) of this net displacement vector. THINK a 1 Set up vectors (tail to head), one pointing north, the other west. 2 Indicate the distances as 30 km and 40 km respectively.

WRITE a

b Join the tail of the N vector with the head of the W vector. ˜ ˜

b

N

W ~ (40 km) N (30 km) W ~

E S

W ~ (40 km) ~N + W ~

N (30 km) ~

305

Chapter 7 Introduction to vectors

THINK c 1 Let R km = length of N + W . ˜ ˜

WRITE c W ~ (40 km) N (30 km) ~

R=~ N+W ~ ~ 2

d

1

The length (magnitude) of R can be calculated using Pythagoras’˜ theorem.

Indicate the angle between N and ˜ N + W as θ. ˜ ˜

2

Use trigonometry to find θ.

3

The true bearing is 360° minus 53.13°.

R =

30 2 + 40 2

= 900 + 1600 = 50 km d

W ~ (40 km) ~N + W ~

θ

N (30 km) ~

40 sin θ = -----50 = 0.8 θ = 53.13° Therefore the true bearing is: = 360° − 53.13° = 306.87°

remember remember 1. Definition: A vector is a quantity that has magnitude and direction. 2. Equality of vectors: Two vectors are equal if both magnitude and direction are equal.

a i

~s ~r

ii

–s ~

iii ~s

3. Addition of vectors: To add two vectors, take the tail of one vector and join it to the head of the other. The result of addition is the vector from the tail of the first vector to the head of the second.

r +~s ~ ~s

4. Subtraction of vectors: Subtract vectors by adding the negative of the second vector to the first vector.

~r

~s

~r – ~s –r ~

5. Multiplication of a vector by a scalar: Multiply the magnitude of the vector by the scalar; maintain the direction of the original vector.

~s – ~r

7A WORKED

Example

1 WORKED

Example

2

Vectors and scalars

1 a Draw the result of: i r+s ii r – s ˜ ˜ ˜ ˜

iii s – r ˜ ˜

b Draw the result of:

i 2r + 2s ˜ ˜

ii 2r – 2s ˜ ˜

iii 3s – 4r ˜ ˜

~s ~r 3s~ ~s

Same as 1 a i except scaled by a factor of 2.

Same as 1 a ii except scaled by a factor of 2.

~r

–4r ~ 3s~ – 4r ~

306 WORKED

Example

3

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

2 The pentagon ABCDE at right can be defined by the four vectors, s , t , u and v . ˜ these˜ 4 vectors: ˜ ˜ of Find in terms a the vector from A to D s + t ˜ ˜ b the vector from A to B s + t + u + v ˜ ˜ ˜ ˜ c the vector from D to A –s – t ˜ ˜ d the vector from B to E –v – u – t ˜ ˜ ˜ e the vector from C to A. –u – t – s ˜

˜

C

~v

~u

B

D ~t ~s

A

E

˜

3 multiple choice A girl travels 4 km north and then 2 km south. What is the net displacement vector? A 6 km north B 6 km south C 2 km north D 2 km south E −2 km north 4 In the rectangle ABCD, the vector joining A to B is denoted B by u and the vector joining B to C is v . Which pairs of points ˜ are ˜joined by: ~u a u + v ? A to C b u – v ? D to B ˜ ˜ ˜ ˜ A c v – u ? B to D d 3u + 2v – 2u – v ? A to C ˜ ˜ ˜ ˜ ˜ ˜

~v

C

D

5 multiple choice Consider the following relationships between vectors u , v and w . ˜ ˜ ˜ i u = 2v + w ˜ ˜ ˜ ii w = v – u ˜ ˜ ˜ Which of the following statements is true? A u = w B u = v E u = 3v C u = 2--3- v D u = 3--2- v ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ WORKED

Example

4

6 A rectangular prism (box) CDEFGHIJ can be defined by three vectors r , s and t as shown ˜ ˜ ˜ at right. Express in terms of r , s and t : a the vector joining˜ C˜to H ˜r + s ˜ ˜ b the vector joining C to J s + t ˜ ˜ c the vector joining G to D r – s ˜ ˜ d the vector joining F to I r + s ˜ ˜ e the vector joining H to E t – s ˜ ˜ f the vector joining D to J s + t – r ˜ ˜ ˜ g the vector joining C to I r + s + t ˜ ˜ ˜ h the vector joining J to C. – s – t ˜

WORKED

Example

5

˜

J

I F

G

E

~t

H

~s C

~r

D

a, b Flight path

7 A pilot plans to fly 300 km north then 400 km east. a Make a vector drawing of her flight plan. b Show the resulting net displacement vector. c Calculate the length (magnitude) of this net displacement vector. 500 km d Calculate the bearing (from true north) of this net displacement vector.

eBook plus Digital doc: EXCEL Spreadsheet Position vector

53.1° clockwise from N

Chapter 7 Introduction to vectors

SLE 3: Use addition and subtraction in life-related situations.

307

8 Another pilot plans to travel 300 km east, then 300 km north-east. Show that the resultant bearing is 67.5 degrees. How far east of its starting point has the plane travelled? 512.1 km; find bearing using trigonometry

9 An aeroplane travels 400 km west, then 600 km north. How far is the aeroplane from its starting point? What is the bearing of the resultant displacement?

10 Each part of answer 721.1 km, 326.3° (clockwise from N) has coordinate labelled a, b, . . . j. The original 10 On a piece of graph paper draw a vector, a , that is 3 units east and 5 units north of the ˜ vectors a and b are origin. Draw another vector, b , that is 5 units east and 3 units north of the origin. ˜ ˜ ˜ also drawn. On the same graph paper, draw the following vectors. h

15 i a 5 ~

a f

c ~b –15 g –5 d 5 –5 j e –15

b

15

a a+b b a + 3b c a–b d b–a ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ e 3b – 4a f 0.5a + 2.5b g a – 2.5b h 4a ˜ ˜ ˜ ˜ ˜ ˜ ˜ i 2.5a – 1.5b j b – 2.5a ˜ ˜ ˜ ˜ 11 Find the direction and magnitude of a vector joining point A to point B, where B is 10 m east and 4 m north of A. Magnitude = 10.77, direction 68.2° True. 12 Consider a parallelogram defined by the vectors a and b , ˜ ˜ and its associated diagonals, as shown at right. Show that the vector sum of the diagonal vectors is 2a . ˜ 13 Show, by construction, that for any vectors u and v : ˜ ˜ 3 ( u + v ) = 3u + 3v ˜ ˜ ˜ ˜ (This is called the Distributive Law.)

12–18 Check with your teacher.

~b ~a

14 Show, by construction, that for any three vectors a , b and c : ˜ ˜ ˜ (a + b) + c = a + (b + c) ˜ ˜ ˜ ˜ ˜ ˜ (This is called the Associative Law.) 15 Show, by construction, that for any two vectors r and s : ˜ ˜ 3r – s = – ( s – 3r ) ˜ ˜ ˜ ˜ 16 As you will learn shortly, vectors can be represented by two values: the horizontal (or x) component and the vertical (or y) component. Consider the vector w , defined by joining the origin to the point (4, 5), and the ˜ vector v , defined by joining the origin to (2, 3). Find the horizontal and vertical com˜ ponents of each vector. Demonstrate, graphically, that the sum w + v has an x-component of 6 (that is, ˜ ˜ 4 + 2), and a y-component of 8 (that is, 5 + 3). 17 Using the same vectors, w and v , as in question 16, demonstrate graphically that the ˜ ˜ difference vector, w – v , has an x-component of 2 and a y-component of 2. ˜ ˜ 18 Using the same vectors, w and v , as in question 16, demonstrate graphically that: ˜ ˜ a the vector 4w has an x-component of 16 and a y-component of 20 ˜ b the vector – 2v has an x-component of −4 and a y-component of −6. ˜

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19 Using the results from questions 16, 17 and 18, what can you deduce about an algebraic method (as opposed to a graphical method) of addition, subtraction and multiplication of vectors? One can deduce that x and y components can be added/subtracted/ multiplied separately.

20 multiple choice

D ~a O b ~

In terms of vectors a and b in the figure above, the vector joining O to D is given by: ˜ ˜ E none of these A 3a + 3b B 2a + 4b C 3b – 2a D 2a – 3b ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ 21 multiple choice

E

~a O b ~

In terms of vectors a and b , the vector joining E to O above is: ˜ ˜ A 3a + 4b B 4b – 3a C 3a – 4b D – 3 a – 4b E none of these ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ 22 A girl walks the following route: 400 m north — 300 m east — 200 m north — 500 m west — 600 m south — 200 m east Make a vector drawing of these six paths. What is the net displacement vector? 0 ˜

23 Which of the following are vector quantities? speed velocity displacement force volume angle

Displacement, velocity, force

24 Which of the following are scalar quantities? speed time acceleration velocity length displacement

Speed, time, length

25 A 2-dimensional vector can be determined by its length and its angle with respect to (say) true north. What quantities could be used to represent a 3-dimensional vector? 1 magnitude and 2 angles

Position vectors in two and three dimensions Introduction As a vector has both magnitude and direction, it can be represented in 2-dimensional planes or 3-dimensional regions in space. (It is easier to discuss 2-dimensional vectors as they fit the page nicely!)

Position vectors in two dimensions In the figure at right, the vector u joins the point A to ˜ point B. An identical vector can be considered to join the origin with the point C. It is easy to see that u is made up of two components: one along the x-axis and ˜one parallel to the y-axis. Let i be a vector along the x-axis, with magnitude 1. Similarly,˜ let j be a vector along the y-axis, with magnitude 1. ˜

y B

~u A

x y ~u ~i

C j ~ D x

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Chapter 7 Introduction to vectors

We can say the vector u is the position vector of point C ˜ relative to the origin. With vectors, it is equivalent to travel along u from the ˜ to D and origin directly to C, or to travel first along the x-axis then along the y-axis to C. In either case we started at the origin and ended up at C. Clearly then, u is made up of some multiple of i in the x-direction and some˜ multiple of j in the ˜ y-direction. ˜ For example, if the point C has coordinates (6, 3) then u = 6i + 3 j . ˜ ˜ ˜

y 3

C ~u

O

6i~

D x

z

Position vectors in three dimensions

In 3 dimensions, a point in space has 3 coordinates, so a third component, along the z-axis, is needed. Let k be a vector 0 ˜ along the z-axis, with magnitude 1. The orientation is now such that the x-axis is coming directly out from the page as x shown at right. z For example, if the point C has coordinates (6, –2, 4) C (6, −2, 4), then its position vector would be ~v denoted by v = 6i – 2 j + 4k . 0 ˜ ˜ ˜ ˜ 4k 6i ~

Relationship between vectors and matrices

3j ~

–2j ~

~

y

y

x

From our earlier work on matrices, we know that a matrix is an array that can store numbers. These numbers could be the components of a vector, so for the vector a v = ai + b j + ck , the components could be expressed as b . ˜ ˜ ˜ ˜ c So as a matrix, the components of u = 6i + 3 j could be expressed as 6 and the ˜ ˜ ˜ 3 6 components of v = 6i – 2 j + 4k could be expressed as – 2 . ˜ ˜ ˜ ˜ 4 We will look further at the relationship between vectors and matrices later in the chapter.

The magnitude of a vector By using Pythagoras’ theorem on a position vector, we can find its length, or magnitude. Consider the vector u at right. y ˜ The magnitude of u , denoted as u or u, is given by: C (6, 3) ˜ ˜ ~u 2 2 3j u = 6 +3 ~ ˜ x 6i u = 45 ~ ˜ =3 5

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Consider, now, the general position vector relative to the origin, for the point with coordinates (x, y): u = xi + y j ˜ ˜ ˜ The magnitude of a vector, u = xi + y j , is given by u = x 2 + y 2 . ˜ ˜ ˜ ˜

The direction of a vector From what we already know about trigonometry, we can work out the angle (θ ) that u ˜ makes with the positive x-axis (that is, anticlockwise from the positive x-axis). This gives us the direction of u . ˜ y This angle can be calculated as: C (6, 3)

θ = tan−1 ( 3--6- )

~u

3j

~ θ = tan−1 0.5 x 6i ~ = 0.464 radians = 26.6° The result obtained by this method needs to be adjusted if the angle is in the 2nd, 3rd, or 4th quadrants. y The direction of a vector, u = xi + y j , is given by θ = tan−1 --- with appropriate x ˜ ˜ involved. adjustment depending on ˜the quadrant

WORKED Example 6 Using the vector shown at right, find: a the magnitude of u b the direction of u ˜(express the angle with respect to the positive ˜ x-axis) c the true bearing of u . ˜ THINK WRITE a

b

1

Use Pythagoras’ theorem or the rule for magnitude of a vector with the x- and y-components 3 and −5 respectively.

2

Simplify the surd.

1 2 3

c

1

2

The angle is in the 4th quadrant since x = 3 and y = −5. Use trigonometry to find the angle θ, from the x- and y-component values. Use a calculator to simplify. The negative sign implies that the direction is 59° clockwise from the x-axis. The true bearing from north is the angle measurement from the positive y-axis to the vector u . ˜

a u = ˜

u = ˜ =

y

θ

x ~u (3, –5)

3 2 + ( –5 )2

9 + 25 34

(= 5.831 to 3 decimal places)

b ˙

5 θ = tan−1  –----3

θ = −59° c

true bearing = 90° + 59° = 149°.

Chapter 7 Introduction to vectors

311

the magnitude and direction Graphics Calculator tip! Finding of a vector in two dimensions Consider the vector shown in worked example 6. The vector can be expressed in component form (or rectangular form) as 3i – 5 j . One way of finding the magnitude and ˜ the ˜ positive x-axis) is to convert the vector direction (the angle the vector makes with from rectangular form to polar form using a graphics calculator.

For the Casio fx-9860G AU 1. Press MENU and select RUN-MAT. Decide whether you want the angle displayed in degrees or radians. In this example we want degrees. Press SHIFT [SET UP] and alter the setting for Angle if necessary. Press F1 (Deg) for degrees.

s

s

2. Press EXE to accept this setting. Next press OPTN then F6 ( ) to show more options.

3. Press F5 (ANGL) and then F6 ( ) for more options. There are three options shown. The one we need is Pol(, which converts vectors in rectangular form to polar form. 4. Press F1 (Pol() and enter the components 3 and −5, separated by a comma (press , ). Press ) to close the set of brackets.

5. Press EXE to perform the conversion. The first number (which is highlighted) is the magnitude of the vector and the second number is the angle the vector makes with the positive x-axis.

For the TI-Nspire CAS 1. Decide whether you want the angle displayed in degrees or radians. In this example we want degrees. Press c to access the home screen and select 8: System Information followed by 2: System Settings. Press e until you reach Angle and then select Degree by using the arrow keys. Press · to accept this setting. 2. Continue pressing e until you highlight OK. Press · to select OK.

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3. We will first define 3i – 5 j as the vector u and then ˜ ˜Press / N and select convert this to polar form. 1: Add Calculator to open a new calculator document. Press b and select 1: Actions.

4. Select 1: Define. Press U (to name the vector as u) then = followed by / ( to set up square brackets. Enter the components 3 and −5, separated by a comma (press ,), within the square brackets. Use the right arrow key to move the cursor to the right of the second square bracket and press ·. 5. Press U to nominate vector u. Then press b and select 7: Matrix & Vector followed by C: Vector.

6. Select 4: Convert to Polar and press · to perform the conversion. The exact values for the magnitude and the angle are shown.

7. To obtain the approximate values, press / ·.

Note that if no other operations are required, you can enter the vector directly by using square brackets. You then continue in the same way to convert to polar form.

Unit vectors As we have seen, any vector u is composed of x and y (and z, in 3 dimensions) components denoted by xi , y j˜ (and zk ). The vectors, i , j and k are called unit ˜ vectors, as they each have ˜a magnitude of˜ 1. This allows us˜ to˜ resolve˜ a vector into its components.

Chapter 7 Introduction to vectors

313

If a 2-dimensional vector u makes an angle of θ with the positive x-axis and it has ˜ can find its x- and y-components using the formulas: a magnitude of u then we ˜ x = u cos q ˜ y = u sin q ˜

WORKED Example 7 Consider the vector u , whose magnitude is 30 and whose ˜ bearing (from N) is 310°. Find its x- and y-components and write u in terms of i and j . ˜ ˜ ˜ THINK 1 2

3

4

~u

W x

WRITE

Change the bearing into an angle with respect to the positive x-axis (θ). The angle between u and the positive ˜ y-axis is 360° − 310°.

N

y

E S

310°

y ~u

50°

θ

x

Calculate θ.

θ = 90° + 50° = 140°

Find the x- and y-components using trigonometry.

x = u cos θ ˜ = 30 cos 140° = –22.98 y = u sin θ ˜ = 30 sin 140° = 19.28 u = – 22.98i + 19.28 j ˜ ˜ ˜

Express u as a vector. ˜

the x- and y-components Graphics Calculator tip! Finding of a vector Vectors can be expressed in different forms. In the graphics calculator tip on page 311, we converted a vector in rectangular form to polar form so we could obtain the magnitude and direction of the vector. The reverse process can also be performed. Consider the vector in Worked example 7 where the magnitude is 30 and the angle to the positive x-axis is 140°. To find the x- and y-components, we convert the vector to rectangular form. Check that the calculator is set to degrees.

For the Casio fx-9860G AU 1. Press MENU and select RUN-MAT. Press OPTN then F6 ( ) to show more options. Press F5 (ANGL) and then F6 ( ) for more options. There are three options shown. The one we need is Rec(, which converts vectors in polar form to rectangular form. s

s

5

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2. Press F2 (Rec() and enter 30 for the magnitude followed by a comma. Then enter 140 for the angle and close the set of brackets.

3. Press EXE to perform the conversion. The first number (which is highlighted) is the x-component of the vector and the second number is the y-component. Hence u = – 22.98 i + 19.28 j . ˜ ˜ ˜

For the TI-Nspire CAS 1. Open a new Calculator document (press / N and select 1: Add Calculator). To enter the vector in polar form, first press / ( to set up square brackets. Then enter 30 for the magnitude followed by a comma. Press / k and highlight the angle symbol (∠). 2. Press · to insert the angle symbol on the calculator page and enter 140 for the angle. Move the cursor to the right of the second square bracket. The vector is now entered in polar form.

3. Press b and select 7: Matrix & Vector followed by C: Vector. Select 5: Convert to Rectangular and press · to perform the conversion. The exact values for the x- and y-components are shown. To obtain the approximate values, press / ·. Hence u = – 22.98 i + 19.28 j . ˜ ˜ ˜

WORKED Example 8 A bushwalker walks 16 km in a direction of bearing 050°, then walks 12 km in a direction of bearing 210°. Find the resulting position of the hiker giving magnitude and direction from the starting point.

1

Draw a clear diagram to represent the situation.

WRITE y 50°

240° m 6k a~1

~b 12 k m

THINK

210° 30°

x

Chapter 7 Introduction to vectors

THINK 2 3 4

315

WRITE

Express position vectors as angles from the direction of the x-axis. Simplify position vectors. Use the Triangle Law of addition of vectors.

a ~

θ

b ~

a = 16 cos 40° i + 16 sin 40° j ˜ j b˜ = 12 cos 240°˜ i + 12 sin 240° ˜a = 12.2567 i + ˜10.2846 j ˜ ˜ ˜ b˜ = −6 i − 10.3923 j ˜ a + b˜ = (12.2567 ˜− 6) i + (10.2846 − 10.3923) j ˜ ˜ = 6.2567 i − 0.1077 ˜ ˜ j ˜ ˜

a b ~+~ 5

Find the angle θ.

6

Find the magnitude.

7

State the resultant vector in terms of magnitude (distance) and direction (bearing).

y θ = tan–1 -x – 0.1077 = tan–1 ------------------6.2567 = −0.986° |a + b | = ˜ ˜ =

2

x +y

2

2

6.26 + ( – 0.11 )

2

= 39.2 = 6.26 Final position is 6.26 km from the starting point in a direction of bearing 091°.

Clearly, in 3 dimensions, this is much more difficult as you need two angles (for instance, an angle with respect to the x-axis and another with respect to the z-axis). Unit vectors can also be found in the direction of any vector. This is merely the original vector divided by its magnitude. The unit vector of any vector u , in the direction of u denoted by uˆ , is: ˜ ˜ ˜ u uˆ = ----˜ u ˜ ˜

WORKED Example 9

C (6, 3)

Find the unit vector in the direction of u . ˜

~u

3j ~

6i~

THINK 1

Express the vector in component form.

2

Compute the magnitude of the vector u . ˜

WRITE u = 6i + 3 j ˜ ˜ ˜ 2 u = 6 + 32 ˜ = 45 = 3 5 Continued over page

316

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THINK

WRITE

3

Divide each component of the original vector by the magnitude to get uˆ . ˜

4

Comfirm that uˆ has a magnitude of 1. ˜

6 3 uˆ = ---------- i + ---------- j ˜ 3 5˜ 3 5˜ 2 5 5 = ---------- i + ------- j 5 ˜ 5˜ 2

2

uˆ = x + y ˜ 20 5 = ------ + -----25 25 =

25 -----25

=1

the unit vector in the Graphics Calculator tip! Finding direction of the vector As seen above, the unit vector is obtained by dividing each component by the magnitude of the vector. We can also use a graphics calculator to achieve this. Consider the vector u in Worked example 9 which can be expressed in component form as 6i + 3 j . ˜ ˜ ˜

For the Casio fx-9860G AU 1. First find the magnitude of u . Repeat the steps ˜ the vector into shown on page 311 to change polar form. The magnitude is 6.7082. (Alternatively, calculate

2

2

6 + 3 .)

▼

2. Consider the components of u as the matrix 6 . ˜ the MAT 3 Press EXIT until you return to screen. Enter the matrix using square brackets (each row is listed within a set of square brackets) as shown in the screen at right. 3. Press Æ and then SHIFT [Mat] followed by ALPHA [U] to store this matrix as matrix U. 4. Press EXE to display the matrix and then press EXE again to return to your calculation screen. Next we need to divide each component of the vector by the magnitude of the vector to obtain the unit vector. Since we are using matrices, we need to change this to multiplying matrix U by the reciprocal of the magnitude.

317

Chapter 7 Introduction to vectors

Enter 6.7082 and press SHIFT [x -1] to obtain the reciprocal of the magnitude. Press ¥ and then enter matrix U by pressing SHIFT [Mat] followed by ALPHA [U]. 5. Press EXE . The approximate values for the components of the unit vector are shown. Hence uˆ = 0.8944i + 0.4472 j . ˜ ˜ ˜

For the TI-Nspire CAS 1. Begin with a new calculator page. To define the vector as u, first press b and select 1: Actions followed by 1: Define. Press U (to name the vector as u) then = followed by / ( to set up square brackets. Enter the components, separated by a comma, within the square brackets. Move the cursor to the right of the second square bracket and press ·. 2. Press b and select 7: Matrix & Vector followed by C: Vector.

3. Select 1: Unit Vector and press U to enter the required vector. Close the set of brackets by pressing ) and then press ·. The exact values for the components of the unit vector are shown. 2 5 5 Hence uˆ = ---------- i + ------- j . 5 ˜ 5˜ ˜ These steps also apply to finding the unit vector for a three-dimensional vector.

Locating vectors

y

A

In the figure at right, a is the position vector of point A ( OA ) a ~ ˜ B and b is the position vector of point B ( OB ) relative to the origin. b ~ ˜ x O The vector describing the location of A relative to B ( BA ) is easily found using vector addition as –b + a or a – b . ˜ ˜ ˜ ˜ Similarly, the vector describing the location of B relative to A ( AB ) is b – a . This ˜ ˜ result also applies in 3 dimensions and can be formalised as follows. If A and B are points defined by position vectors a and b respectively, then ˜ ˜ AB = b – a ˜ ˜

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WORKED Example 10

a Find the position vector locating point B (3, −3) from point A (2, 5). b Find the length of this vector. THINK

WRITE

a

1

Express the point A as a position vector a . ˜

a Let

OA = a = 2i + 5 j ˜ ˜ ˜

2

Express the point B as a position vector b . ˜

Let

OB = b = 3i – 3 j ˜ ˜ ˜

3

The location of B relative to A ( AB ) is b – a . ˜ ˜

AB = b – a ˜ ˜ = 3i – 3 j – ( 2i + 5 j ) ˜ ˜ ˜ ˜ = i – 8j ˜ ˜ b AB = b – a ˜ ˜ = 1 2 + ( –8 )2

b The length of AB is b – a . ˜ ˜

=

65 (or 8.06)

Magnitudes in three dimensions Pythagoras’ theorem also applies in the case of a 3-dimensional line or vector. Let x, y and z be the components of a vector, u , in 3-dimensional space, that is ˜ u = xi + y j + zk . The magnitude of u is ˜ ˜ ˜ ˜ ˜ u = x2 + y2 + z2 ˜

WORKED Example 11 Consider the point in 3-dimensional space given by the coordinates (2, 4, 3). Find the magnitude of the position vector, u , joining this point to ˜ the origin.

z (2, 4, 3) ~u y x

THINK 1 2

3

WRITE

Express u as a position vector. ˜ Since the vector is in 3-dimensional space, use the 3-D version of Pythagoras’ theorem to find the magnitude.

u = 2i + 4 j + 3k ˜ ˜ ˜ ˜ 2 2 u = x + y + z2 ˜

Substitute the components for each direction and compute the magnitude.

u = ˜ =

22 + 42 + 32 29

(= 5.39 to 2 decimal places)

Chapter 7 Introduction to vectors

319

We are now in a position to resolve the problem of finding, accurately, the resultant force acting on the television remote control when Daniel and Anna are pulling on it.

First redraw the diagram to show the addition of vectors. Taking the direction of Anna’s force as the ~i direction, Anna’s force, a = 50i~ + 0j ~ ˜ Daniel’s force, b = 40 cos 150°i~ + 40 sin 150°j ~ ˜ = −34.6410i~ + 20j

Resultant force

Daniel’s force 40 N

30° Anna’s 50 N force

150°

~

Resultant force, a + b = (50 − 34.6410)i~ + 20j ~ ˜ ˜ = 15.3590i~ + 20j Magnitude = | a + b | ˜ ˜ Magnitude = 25.2 N

~

20 Direction θ = tan–1 ------------------15.3590 Direction θ = 52.48°

remember remember

1. Magnitude of a vector: If u = xi + y j + zk , the magnitude is given by ˜ ˜ ˜ ˜ u = x2 + y2 + z2 ˜ Speed is the magnitude of velocity which is a vector quantity. 2. Direction of a vector (2-D only): If u = xi + y j , the direction is given by ˜ ˜ ˜ θ = tan−1 --y x

3. The x- and y-components of a vector: Given magnitude and direction, the xand y-components are given by: x = u cos θ ˜ y = u sin θ ˜ 4. Unit vector: The unit vector of a vector u , in the direction of u , is denoted by ˜ ˜ uˆ and is: ˜ u uˆ = ----˜ u ˜ ˜ 5. Locating vectors: If A and B are points with position vectors a and b ˜ ˜ respectively then AB = b – a . ˜ ˜

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Position vectors in two and three dimensions

7B WORKED

Example

6a, b

1 State the x, y and z components of the following vectors: a 3i + 4 j – 2k 3, 4, −2 b 6i – 3k 6, 0, −3 c 3.4i + 2 j + 1--2- k ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ 2 For each of the following find: i the magnitude of the vector ii the direction of each vector. (Express the direction with respect to the positive x-axis.) a b (–4, 7) y y (6, 6) ~v

c

d

y

x

3 a b c d

3 Find the true bearing of each vector in question 2.

6c Example

7

--12

eBook plus Digital doc: EXCEL Spreadsheet Position vector

4 Consider the vector w shown at right. Its ˜ bearing is 210° True. magnitude is 100 and its Find the x- and y-components of w , and express them as exact values (surds). ˜ State the answer in the form w = xi + y j . ˜ ˜ ˜ –50i – 50 3 j ˜ ˜ 5 multiple choice

i 72 i 65 i 4.88 i 320.16

045° 330.3° 224.2° 091.8°

B x-component =

3 1 ------- , y-component = --2 2 1 3 --- , y-component = ------2 2

C x-component = 5 3 , y-component = 5 D x-component = 5, y-component = 5 3 E none of the above 6 An aeroplane travels on a bearing of 147 degrees for 457 km. Express its position as a vector in terms of i and j . 248.9i – 383.3 j ˜ ˜ ˜ ˜ 7 A ship travels on a bearing of 331 degrees for 125 km. Express its position as a vector in terms of i and j . –60.6i + 109.3 j ˜ ˜ ˜ ˜

ii 225.8° ii 358.2°

eBook plus Digital docs:

y

N W

100

210°

SkillSHEET 7.1 Bearings

E

x S

w ~

A vector with a bearing of 60 degrees from N and a magnitude of 10 has: A x-component =

ii 45° ii 119.7°

x ~b (320, –10)

(–3.4, –3.5)

WORKED

2 a b c d

x

y

~a

Example

2,

w ~ x

WORKED

3.4,

Chapter 7 Introduction to vectors

WORKED

Example

8 SLE 3: Use addition and subtraction in life-related situations.

321

8 A pilot flies 420 km in a direction 45° south of east and then 200 km in a direction 60° south of east. Calculate the resultant displacement from the starting position giving both magnitude and direction. 615 km at 49.8° south of east 9 The instructions to ‘Black-eye the Pirate’s hidden treasure’ say: Take 20 steps in a north-easterly direction and then 30 steps in a south-easterly direction. However, a rockfall blocks the first part of the route in the north-easterly direction. How could you head directly to the treasure? 36 steps 11.3° south of east 10 Two scouts are in contact with home base. Scout A is 15 km from home base in a direction 30° north of east. Scout B is 12 km from home base in a direction 40° west of north. How far is scout B from scout A? 20.8 km

WORKED

Example

9 3--i 5˜

11 Find unit vectors in the direction of the given vector for the following: a y b y (3, 4)

+ 4--5- j ˜

--3- i 5˜

~a 0

c e

b = 4i + 3 j ˜ ˜ ˜ c = i + 2j ˜ ˜ ˜

– --45- j ˜

x

0 ~d

x + 3--5- j ˜ 1 2 ------- i + ------- j 3˜ 3 ˜ 4--i 5˜

(3, –4)

d e = – 4i + 3 j – 4--5- i + 3--5- j ˜ ˜ ˜ ˜ ˜ f f = – 3.5i + 2.7 j –0.792i + 0.611 j ˜ ˜ ˜ ˜

12 multiple choice A unit vector in the direction of 3i – 4 j is: ˜ ˜ C i– j A 3--5- i + 4--5- j B 3--5- i – 4--5- j ˜ ˜ ˜ ˜ ˜ ˜

D

3----i 25 ˜

4– ----j 25 ˜

E none of these

13 Not all unit vectors are smaller than the original vectors. Consider the vector v = 0.3i + 0.4 j . Show that the unit vector in the direction of v is twice as long as v . ˜ ˜ ˜ ˜ Check with ˜ your teacher. 14 Find the unit vector in the direction of w = – 0.1i – 0.02 j . –0.98i – 0.20 j ˜ ˜ ˜ ˜ ˜ 15 Find a unit vector in the direction of w for the vector of question 4. – 1--2- i – ------23- j ˜ ˜ ˜ 16 Consider the points A (0, 1) and B (4, 5) in the figure at B y (4, 5) right. A vector joining A to B can be drawn. Check with your teacher.

WORKED

Example

10 18 a –4i + 7 j ˜ ˜ b –3i – j ˜ ˜ c 4i – 7 j ˜ ˜ d 3i + j ˜ ˜ e – 2i ˜ f 4i ˜

a Show that an equivalent position vector is given by: 4i + 4 j . ˜ ˜ b Similarly, show that an equivalent position vector joining B to A is given by: –4i – 4 j . ˜ ˜ 17 For each of the following pairs of points find:

(0, 1) A x 17

a i 4i – 7 j ˜ ˜ b i 3i + j ˜ ˜ c i –4i + 7 j ˜ ˜ i the position vectors locating the second point from the first point d i –3i – j ˜ ˜ ii the length of this vector. e i 2i ˜ a (0, 2), (4, −5) b (2, 3), (5, 4) c (4, −5), (0, 2) f i – 4i ˜

d (5, 4), (2, 3)

e (3, 7), (5, 7)

f

(7, −3), (3, −3)

18 Find the position vectors from question 17, by going from the second point to the first.

ii

65

ii

10

ii

65

ii

10

ii 2 ii 4

322 19 a b c d e f

4 ---------- i 65 ˜

7 -j – --------65 ˜ 3 1 ---------- i + ---------- j 10 ˜ 10 ˜ 4 7 -j – ---------- i + --------65 ˜ 65 ˜ 3 1 - i – ---------- j – --------10 ˜ 10 ˜ i ˜ –i ˜

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

19 Find unit vectors in the direction of the position vectors for each of the vectors of question 17. 20 Let u = 5i – 2 j and e = –2i + 3 j . 5 2 ˜ ˜ ˜ ˜ - i – ---------- j ˜ ˜ --------29 ˜ 29 a Find: ˜

3 2 - i + ---------- j – --------13 13 ˜ ˜

3i + j ˜ ˜

13 iii uˆ 29 ii e 10 iv eˆ v u+e vi u + e u ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ b Confirm or reject the statement that u + e = u + e Reject, because magnitudes different. ˜ ˜ ˜ ˜ 21 Let u = –3i + 4 j and e = 5i – j . 5 1 ˜ ˜ ˜ 26 ˜ ˜ – 3--- i + 4--- j ---------- i – ---------- j ˜ 2i + 3 j 5 5˜ 26 ˜ 26 a Find: ˜ ˜ ˜ ˜

i

13 u 5 ii e iii uˆ iv eˆ v u+e vi u + e ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ b Confirm or reject the statement that u + e = u + e . Reject, because magnitudes different. ˜ ˜ ˜ ˜ 22 To find the distance between two vectors, a and b , simply find a – b . ˜ ˜ ˜ ˜ Find the distance between these pairs of vectors: 2 58 a 3i + 2 j and 2i + 3 j b 5i – 2 j and 2i + 5 j ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ 23 A river flows through the jungle from west to east at a speed of 3 km/h. An explorer wishes to cross the river by boat, and attempts this by travelling at 5 km/h due north. Find: a the vector representing the velocity of the river 3i ˜ b the vector representing the velocity of the boat 5 j ˜ c the resultant (net) vector of the boat’s journey 3i + 5 j ˜ ˜ d the bearing of the boat’s journey 031.0° 34 km/h e the magnitude of the net vector.

i

24 Consider the data from question 23. At what bearing should the boat travel so that it arrives at the opposite bank of the river due north of the starting position? 329.0°

Chapter 7 Introduction to vectors

WORKED

Example

11

323

25 Find the magnitude of the following 3-dimensional vectors. a b z z (–3, –4, 5)

5 2

5 2

0

y 0

y

(3, 4, –5)

x

x

11 0.5i – 2k + 3 j 3.64 2i – 2 2 j + k d ˜ ˜ ˜ ˜ ˜ ˜ 3 e –7i + 14 j – 21k 7 14 f i+ j+k ˜ ˜ ˜ ˜ ˜ ˜ 26 By calculating the difference between two position vectors, a vector representing the separation of the two vectors can be defined. Find the distance between the following 3-dimensional vectors. 35 13 a 4i + 3 j – 2k and 5i – 2 j + k b 2i + j – k and 5i + j + k ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ 62 c –i + 2 j + 3k and 3i + k 2 6 d i + 3 j – k and 8i + 5 j + 2k ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ 27 If four points C, D, E and F in 3-dimensional space are located as follows: C = (2, 6, 0), D = (3, −1, –2), E = (−4, 8, 10), F = (−2, −6, 6), show that CD is parallel to EF. CD = i – 7 j – 2k , EF = 2i – 14 j – 4k

c

˜

˜

˜

˜

˜

˜

28 A boat travels east at 20 km/h, while another boat travels south at 15 km/h. Find: a a vector representing each boat and the difference between the boats 20i , – 15 j , 20i + 15 j ˜ ˜ ˜ ˜ b the magnitude of the difference vector 25 c the bearing of the difference vector. 053.1° 29 Consider the vector u = 3i + 4 j and the vector v = 4i – 3 j . Find the angles of each ˜ respect ˜ ˜ to the x-axis. ˜Show˜ that˜ these two vectors are of these vectors with perpendicular to each other. Also show that the products of each vector’s corresponding x- and y-components add up to 0. Can you confirm that this is a pattern for all perpendicular vectors? 53.1°, −36.9° Difference = 90° SLE 3: Use addition and subtraction in life-related situations such as the effect of current flow on a boat.

30 A river has a current of 4 km/h westward. A boat which is capable of travelling at 12 km/h is attempting to cross the river by travelling due north. Find: 0.0417 h or 2.5 minutes a a vector representing the net velocity of the boat –4i + 12 j ˜ ˜ b the bearing of the actual motion of the boat 341.6° c how long it takes to cross the river, if the river is 500 m wide (from north to south). (Hint: The maximum ‘speed’ of the boat is still 12 km/h.)

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Multiplying two vectors — the dot product Introduction In a previous section we studied the result of multiplying a vector by a scalar. What happens if a vector is multiplied by another vector? There are two possibilities: either the result is a scalar (called the scalar product or dot product) or the result is a vector (called the cross, or vector product). In this course we will study only the former. The scalar or dot product of two vectors, u and v , is denoted by u ˜ ˜ ˜

⋅ v˜ .

Calculating the dot product There are two ways of calculating the dot product. The first method follows from its definition. (The second method is shown later.) Consider the two vectors u and v below. ˜ ˜ By definition, the dot product u v is given by: ˜ ˜ ~u u • v = u v cos θ [1] ˜ ˜ ˜ ˜ θ where θ is the angle between (the positive directions of) u and v . ~v ˜ ˜ Note: The vectors are not aligned as for addition or subtraction, but their two tails are joined.

◊

Properties of the dot product 1. The dot product is a scalar. It is the result of multiplying three scalar quantities: the magnitudes of the two vectors and the cosine of the angle between them. 2. The order of multiplication is unimportant (commutative property), thus u•v = v•u ˜ ˜ ˜ ˜ 3. The dot product is distributive, thus a • (u + v) = a • u + a • v ˜ ˜ ˜ ˜ ˜ ˜ ˜ 4. Since the angle between u and itself is 0° ˜ u•u = u 2 ˜ ˜ ˜

WORKED Example 12

Let u = 3i + 4 j and v = 6i . Find u • v . ˜ ˜ ˜ ˜ ˜ ˜ ˜ THINK 1

Find the magnitudes of u and v . ˜ ˜

2

Draw a right-angled triangle showing the angle that u makes with the positive x-axis ˜since v is along the ˜ x-axis.

WRITE u = 32 + 42 ˜ = 5 v = 62 ˜ = 6 y ~u

θ

5

3

x

Chapter 7 Introduction to vectors

THINK 3 Find cos θ, knowing that u = 5 and the x-component of u is 3. ˜ 4 Find u • v using equation 1. ˜ ˜ 5 Simplify.

325

WRITE cos θ = 3--5

u • v = u × v × cos θ ˜ ˜ ˜ ˜ = 5 × 6 × 3--5 = 18

Note: An easier method for finding the dot product will now be shown.

Unit vectors and the dot product Consider the dot product of the unit vectors i , j and k . Firstly, consider i • i in ˜ ˜ between ˜ them is 0°, cos θ =˜ 1,˜ thus detail. By definition, i = 1 and, since the angle ˜ i • i = 1 . To summarise these results: ˜ ˜ i • i = 1 (since θ = 0°) ˜ ˜ j • j = 1 (since θ = 0°) ˜k • ˜k = 1 (since θ = 0°) ˜ ˜ i • j = 0 (since θ = 90°) ˜ ˜ i • k = 0 (since θ = 90°) ˜ ˜ j • k = 0 (since θ = 90°) ˜ ˜ Using this information, we can develop another way to calculate the dot product of any vector. Let u = x 1 i + y 1 j + z 1 k and v = x 2 i + y 2 j + z 2 k , where x1, y1, z1, x2, y2, z2 ˜ we˜can write ˜ ˜ ˜ ˜ are constants. Then u˜ • v as:˜ ˜ ˜ u • v = ( x1 i + y1 j + z1 k ) • ( x2 i + y2 j + z2 k ) ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ = x1 x2 ( i • i ) + x1 y2 ( i • j ) + x1 z2 ( i • k ) + y1 x2 ( j • i ) + y1 y2 ( j • j ) ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ + y1 z2 ( j • k ) + z1 x2 ( k • i ) + z1 y2 ( k • j ) + z1 z2 ( k • k ) ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ Considering the various unit vector dot products (in brackets), the ‘like’ products ( i • i , j • j and k • k , shown underlined) are 1; the rest are 0. Therefore: ˜ ˜ ˜ ˜ ˜ ˜ [2] u • v = x1 x2 + y1 y2 + z1 z2 ˜ ˜ This is a very important result. We only need to multiply the corresponding x, y and z components of two vectors to find their dot product.

WORKED Example 13

Let u = 3i + 4 j + 2k and v = 6i – 4 j + k . Find u • v . ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ THINK WRITE u • v = ( 3i + 4 j + 2k ) • ( 6i – 4 j + k ) 1 Write down u • v using equation 2. ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ u•v = 3×6+4×–4+2×1 2 Multiply the corresponding ˜ ˜ components. 3

Simplify.

= 18 – 16 + 2 = 4

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the dot product Graphics Calculator tip! Finding of two vectors The following steps show how a graphics calculator can be used to find the dot product of two vectors. Consider the vectors u = 3i + 4 j + 2k and v = 6i – 4 j + k in Worked ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ example 13.

For the Casio fx-9860G AU 1. Press MENU and select RUN-MAT. Enter the components of vector u as a matrix. Then press Æ ˜ by ALPHA [U] to store and SHIFT [Mat] followed as matrix U. Press EXE to display the matrix and EXE again to return to the calculation screen. Repeat this procedure to enter the components of vector v as a matrix and store as matrix V. ˜ 2. To calculate the dot product using matrices, we first need to transpose the elements of matrix U before we multiply by matrix V. Press OPTN then F2 (MAT) followed by F4 (Trn) to access the transpose function. 3. Press SHIFT [Mat] then ALPHA [U] to enter matrix U. Press ¥ followed by SHIFT [Mat] then ALPHA [V] to enter matrix V.

4. Press EXE to display the value for the dot product.

For the TI-Nspire CAS 1. Open a new Calculator document. Define the vector u and the vector v by entering each set of ˜ ˜ brackets. components within square

2. Press b and select 7: Matrix & Vector then C: Vector followed by 3: Dot Product. Press U, then the comma key (,) followed by V and close the set of brackets by pressing ). Press · to obtain the value of the dot product.

Chapter 7 Introduction to vectors

327

Finding the angle between two vectors Now that we have two formulas (equations 1 and 2) for calculating the dot product, we can combine them to find the angle between the vectors: u • v = x1 x2 + y1 y2 + z1 z2 ˜ ˜ = u v cos θ ˜ ˜ Rearranging the final two equations, we obtain the result that: x1 x2 + y1 y2 + z1 z2 cos q = --------------------------------------------[3] u v ˜ ˜ Note: The angle will always be between 0° and 180° as 180° is the maximum angle between two vectors.

WORKED Example 14

Let u = 4i + 3 j + k and v = 2i – 3 j – 2k . Find the angle between them to the ˜ ˜ ˜ ˜ ˜ ˜ ˜ nearest degree. ˜ THINK

WRITE

1

Find the dot product using equation 2.

2

Simplify.

3

Find the magnitude of each vector.

⋅

u • v = ( 4i + 3 j + k ) ( 2i – 3 j – 2k ) ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ = 4 × 2 + 3 × −3 + 1 × −2 = −3 u = ˜ =

42 + 32 + 12

v = ˜ =

2 + ( –3 ) + ( –2 )

26 2

2

2

17

–3 cos θ = -------------------26 17

4

Substitute results into equation 3.

5

Simplify the result for cos θ.

6

Take cos−1 of both sides to obtain θ and round the answer to the nearest degree.

3 = – ------------442 = −0.142 695

θ = cos−1 (−0.142 695) = 98°

Special results of the dot product Perpendicular vectors If two vectors are perpendicular then the angle between them is 90° and equation 1 (page 324) becomes: u • v = uv cos 90° ˜ ˜ (since cos 90° = 0) = uv × 0 = 0 If u • v = 0 , then u and v are perpendicular. ˜ ˜ ˜ ˜

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WORKED Example 15

Find the constant a if the vectors u = 4i + 3 j and v = – 3 i + a j are perpendicular. ˜ ˜ ˜ ˜ ˜ ˜ THINK WRITE 1

Find the dot product using equation 2.

2

Simplify. Set u • v equal to zero since u and v are perpendicular. ˜ ˜ ˜ ˜ Solve the equation for a.

3 4

u • v = ( 4i + 3 j ) • ( – 3i + a j ) ˜ ˜ ˜ ˜ ˜ ˜ = – 12 + 3a u • v = –12 + 3a = 0 ˜ ˜ a=4

Parallel vectors If vector u is parallel to vector v then u = k v where k ∈ R. ˜ applying the dot product ˜ ˜ Note: When to˜ parallel vectors, θ (the angle between them) may be either 0° or 180° depending on whether the vectors are in the same or opposite directions.

WORKED Example 16

Let u = 5i + 2 j . Find a vector parallel to u such that the dot product is 87. ˜ ˜ ˜ ˜ THINK WRITE 1

2 3 4 5 6

Let the required vector v = ku . ˜ ˜ Find the dot product of u • v . ˜ ˜ Simplify. Equate the result to the given dot product 87. Solve for k. Substitute k = 3 into vector v . ˜

Let v = k ( 5i + 2 j ) ˜ ˜ ˜ = 5k i + 2k j ˜ ˜ u • v = ( 5i + 2 j ) • ( 5k i + 2k j ) ˜ ˜ ˜ ˜ ˜ ˜ = 25k + 4k = 29k 29k = 87 k=3 v = 15i + 6 j ˜ ˜ ˜

remember remember 1. Scalar (dot) product: The scalar or dot product of two vectors u and v is ˜ ˜ denoted by u • v . ˜ ˜ 2. Calculation of dot product: u • v = u v cos θ (where θ is the angle between the two vectors). ˜ ˜ ˜ ˜ 3. Algebraic calculation of dot product: Let u = x 1 i + y 1 j + z 1 k and v = x 2 i + y 2 j + z 2 k . ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ Then u • v = x 1 x 2 + y 1 y 2 + z 1 z 2. ˜ ˜ 4. Special results: (a) If u • v = 0 , then u and v are perpendicular. ˜ ˜ ˜ ˜ (b) If u = kv , k ∈ R , then u and v are parallel. ˜ ˜ ˜ ˜

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7C WORKED

Example

12

Multiplying two vectors — the dot product

1 Find the dot product of the vectors 3i + 3 j and 6i + 2 j using equation 1. ˜ ˜ ˜ ˜

23.99

2 Compare the result from question 1 with that obtained by finding the dot product using equation 2. Which is probably the most accurate? Dot product = 24; more accurate, since no angle needed WORKED

Example

13

3 Find u • v in each of the following cases. ˜ ˜ a u = 2i + 3 j + 5k , v = 3i + 3 j + 6k 45 ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ b u = 4i – 2 j + 3k , v = 5i + j – 2k 12 ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ c u = –i + 4 j – 5k , v = 3i – 7 j + k −36 ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ d u = 5i + 9 j , v = 2i – 4 j −26 ˜ ˜ ˜ ˜ ˜ ˜ e u = –3i + j , v = j + 4k 1 ˜ ˜ ˜ ˜ ˜ ˜ f u = 10i , v = – 2i −20 ˜ ˜ ˜ ˜ g u = 3 j + 5k , v = i 0 ˜ ˜ ˜ ˜ ˜ h u = 6i – 2 j + 2k , v = –i – 4 j – k 0 ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ 4 multiple choice The dot product of u = 3i – 3 j + 3k and v = i – 2 j + 6k is: ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ A 0 B 3 C 12 D 21

E 27

5 multiple choice Consider the two vectors shown at right. Their dot product is: A 30 B 21.2 C −21.2 D 0 E There is insufficient data to determine the dot product. 6 Consider the vectors u and v at right. Their magnitudes are ˜ u •˜ v . −36 7 and 8 respectively. Find ˜ ˜ 7 Let u = xi + y j . Show that u • u = x 2 + y 2 . ˜ ˜ ˜ ˜ ˜

45° v ~u 5 ~ 6 ~v 50° ~u

8 Let u = 2i – 5 j + k and let v = –i – 2 j + 4k . Find their dot product. 12 ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜

9 and 10 Check with your teacher.

9 Let u = 3i + 2 j , v = i – 2 j and w = 5i – 2 j . Demonstrate, using these vectors, ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ the property: w • (u – v) = w • u – w • v ˜ ˜ ˜ ˜ ˜ ˜ ˜ Formally, this means that vectors are distributive over subtraction. 10 Repeat question 9 for the property: w • (u + v) = w • u + w • v ˜ ˜ ˜ ˜ ˜ ˜ ˜ Formally, this means that vectors are distributive over addition.

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11 multiple choice If u = 5i + 4 j + 3k , which of the following is perpendicular to u ? ˜ ˜ ˜ ˜ ˜ A –5i – 4 j – 3k B 3i + 4 j + 5k C – 5i ˜ ˜ ˜ ˜ ˜ ˜ ˜ D –3i + 5k E –5i + 3k ˜ ˜ ˜ ˜ 12 multiple choice If ( u – v ) • ( u + v ) = 0 then: ˜ ˜ ˜ ˜ A u is parallel to v ˜ ˜ C u is perpendicular to v ˜ ˜ E None is true.

B u and v have equal magnitudes ˜ ˜ D u is a multiple of v ˜ ˜

13 multiple choice If ( u – v ) • ( u + v ) = v 2 then: ˜ ˜ ˜ ˜ ˜ A u = v ˜ ˜ C u is perpendicular to v ˜ ˜ E None is true.

B u must be equal to the zero vector, 0 ˜ ˜ D u must be equal to 2 v ˜ ˜

14 Find the dot product of the following pairs of vectors.

WORKED

Example

a 4i – 3k and 7 j + 4k −12 b i + 2 j – 3k and –9i + 4 j – k 2 ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ c 8i + 3 j and 2i – 3 j + 4k 7 d 5i – 5 j + 5k and 5i + 5 j – 5k −25 ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ 15 Find the angle between each pair of vectors in question 14 to the nearest degree. 15 a b c d

14

16 multiple choice The angle between the vectors 2i + 3 j and 2i – 3 j is closest to: ˜ ˜ ˜ ˜ A 0° B 67° C 90° D 113° E 180°

107° 87° 81° 109°

17 multiple choice The angle between the vectors 2i – 3 j and – 4 i + 6 j is closest to: ˜ ˜ ˜ ˜ A 0° B 69° C 90° D 111° E 180° WORKED

Example

15

WORKED

Example

16

18 Find the constant a, if the vectors v = ai + 3 j and u = 6i – 2 j are perpendicular. a = 1 ˜ ˜ ˜ ˜ ˜ ˜ 19 Find the constant a, such that v = ai – 2a j + 3k is perpendicular to ˜ ˜ ˜ ˜ u = 4i – 3 j + 2k . a = − 3--5˜ ˜ ˜ ˜ 20 Let u = 2i + 4 j . Find a vector parallel to u such that their dot product is 40. 4i + 8 j ˜ ˜ ˜ ˜ ˜ ˜ 21 Let u = 4i – 3 j . Find a vector parallel to u such that their dot product is 80. ˜ ˜ ˜ ˜

64 ------ i 5˜

------ j – 48 5 ˜

Chapter 7 Introduction to vectors

331

History of mathematics CHARLES LUTWIDGE DODGSON (1832 – 1898)

During his lifetime ... The Braille reading system is invented by Louis Braille. Morse code is developed. Famine devastates Ireland. Edison develops the light bulb. Charles Dodgson was an English writer and

mathematician. He wrote several mathematics books but is best known for his fictional works produced under the pen name Lewis Carroll. Dodgson was the son of a clergyman and was the third of eleven children. His education included a period at Rugby, a school where the game of Rugby originated in 1823. Dodgson went on to study at Christ Church College at Oxford University. After completing his studies he lectured in mathematics at Oxford from 1855 until 1881. He also became a clergyman in 1861 but did not take up a position in the church, possibly because he had a stutter.

Dodgson’s most famous achievements were two books which have become classics of children’s literature: Alice’s Adventures in Wonderland and Through the Looking-Glass. His main character was inspired by Alice Liddell who was the second daughter of the Dean of his college. Alice’s Adventures in Wonderland was published in 1865 and became a huge success. Through the Looking-Glass, published in 1872, was equally successful. Dodgson also wrote humorous poetry. His poem Jabberwocky is a masterpiece, although it is often thought of as being just a piece of nonsense verse. The word ‘chortle’ which is used for the first time in this poem was invented by Dodgson and is derived from the two words ‘chuckle’ and ‘snort’. Dodgson’s mathematical works include Euclid and his Modern Rivals, A Syllabus of Plane Algebraical Geometry, and An Elementary Treatise on Determinants. These were well written but have not survived the test of time.

Dodgson was interested in logic and loved puzzles. His books contain many mathematics and logic puzzles including this example from Through the Looking-Glass: ‘Contrariwise’, said Tweedledee, ‘if it was so, it might be; and if it were so, it would be; but as it isn’t, it ain’t. That’s logic.’ One of the puzzles he invented was ‘If 6 cats kill 6 rats in 6 minutes, how many will be needed to kill 100 rats in 50 minutes?’ Questions 1. What was Dodgson’s job at Oxford Alice’s Adventures University? Mathematics lecturer in Wonderland and 2. Name his two most famous works. Through the Looking-Glass 3. Who was the inspiration for the character of Alice? The daughter of the Dean of his college 4. If 6 cats kill 6 rats in 6 minutes, how many will be needed to kill 100 rats in 50 minutes? 12 cats

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Resolving vectors — scalar and vector resolutes Introduction Consider the two vectors, u and v , shown at right. The angle between ˜ by θ. It can be shown that v is them, as for a dot product,˜ is given ˜ made up of a component acting in the direction of u and another ˜ component acting perpendicular to u . ˜ Firstly we wish to find the component in the direction of u . ˜

~v

θ

~u

The scalar resolute To obtain the component of v in the direction of u , we perform the following construction: ˜ ˜ 1. Drop a perpendicular from the head of v to u (this is perpendicular to u ). This line ˜ ˜ ˜ joins u at point A. ˜ 2. We wish to find the length of the line OA. This construction is shown at right. ~v Let the length of v (its magnitude) be denoted by v . Then, ˜ ˜ from trigonometry: θ OA = v cos θ O A ˜ ~u But from the definition of the dot product: u • v = u v cos θ ˜ u˜ • v˜ = u˜ OA (from the first equation) ˜ ˜ ˜ Therefore, solving for OA: u•v OA = -------˜ ˜u ˜ u  = ----v  u˜  ˜ ˜ u But we know that ----˜ = uˆ , the unit vector in the direction of u , and therefore u ˜ ˜ OA = uˆ • v ˜ ˜ ˜ This quantity, the length OA, is called the scalar resolute of v on u . It effectively ˜ ˜ indicates ‘how much’ of v is in the direction of u . ˜ ˜ The scalar resolute of v on u is given by uˆ • v , where uˆ is the unit vector in the ˜ ˜ ˜ ˜ ˜ direction of u . ˜

WORKED Example 17

Let u = 3i + 4 j and a = 6i – 2 j . Find: ˜ scalar ˜ resolute ˜ u˜ ˜of a on ˜ a the b the scalar resolute of u on a . ˜ ˜ ˜ ˜ THINK WRITE a 1 Find the magnitude of u . a u = 32 + 42 ˜ ˜ =5 u ˆ by dividing u by u . uˆ = ----2 Find u ˜ u ˜ ˜ ˜ ˜ u = --˜5

333

Chapter 7 Introduction to vectors

THINK 3

WRITE

Simplify.

= =

4 5

Find the scalar resolute of a on u using uˆ • a . ˜ ˜ ˜ ˜ Simplify.

1 --- ( 3i + 5 ˜ 3 --- i + 4 --- j 5˜ 5

4 j) ˜

˜ uˆ • a = ( 3--5- i + 4--5- j ) • ( 6i – 2 j ) ˜ ˜ ˜ ˜ ˜ ˜ 18 8 = ------ − --5

=

5

10 -----5

=2 b

1

Find the magnitude of a . ˜

2

Find aˆ by dividing a by a . ˜ ˜ ˜

3

Find the scalar resolute of a on u using aˆ • u . ˜ ˜ ˜ ˜ Simplify.

4

b a = ˜ =

6 2 + ( –2 )2 40

a aˆ = ----˜ a ˜ ˜1 - ( 6i – 2 j ) = --------40 ˜ ˜ 1( 6i – 2 j ) • ( 3i + 4 j ) aˆ • u = --------40 ˜ ˜ ˜ ˜ ˜ ˜ 1 = ---------- ( 18 – 8 ) =

40 10--------40

=

10--------2

Notes: 1. The two scalar resolutes are not equal. v.u 2. The scalar resolute of v on u can easily be evaluated as ---------˜ ˜. u ˜ ˜ ˜

Vector resolutes

Consider, now, the vector joining O to A at right. Its magnitude ~v is just the scalar resolute ( uˆ • v ), while its direction is the same ~v ˜ ˜is called the vector resolute of v as u , that is uˆ . This quantity ˜ ˜ θ parallel to u ˜and is denoted by the symbol v || . O A ~u ˜ ˜ ~v The vector resolute of v parallel to u is given by: ˜ ˜ v || = ( uˆ • v )uˆ [4] ˜ ˜ ˜ ˜ Consider the geometry of the above figure. The original vector v can be seen to be the sum of two other vectors, namely v || and v ⊥ . This second vector˜ is called the vector ˜ be ˜computed simply as follows: resolute of v perpendicular to u and can ˜ ˜ v = v || + v ⊥ (by addition of vectors) ˜ ˜ ˜ v ⊥ = v – v || (by rearranging the vector equation) ˜ ˜ ˜ By substitution for v || from equation 4, the vector resolute of v perpendicular to u ˜ ˜ ˜ is given by: v ^ = v – ( uˆ • v )uˆ [5] ˜ ˜ ˜ ˜ ˜ In practice, once v || has been calculated, simply subtract it from v to get v ⊥ . ˜ ˜ ˜ ⊥

||

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WORKED Example 18

Let u = – 2 i + 3 j + k and v = 3i + 2 j – k . Find: ˜ scalar ˜ resolute ˜ ˜ of v ˜on u ˜ ˜ ˜ a the ˜ ˜ b the vector resolute of v parallel to u, namely v || ˜ to u , namely ˜ v^ . c the vector resolute of v˜ perpendicular ˜ ˜ ˜ THINK WRITE a

1

Find the magnitude of u . ˜

2

Find uˆ .

3

Find the scalar resolute using uˆ • v . ˜ ˜

4

Simplify.

a u ˜

=

( –2 )2 + 3 2 + 1 2

=

14

u = ----˜ u ˜ 1( –2i + 3 j + k ) = --------14 ˜ ˜ ˜ 1uˆ • v = --------( –2i + 3 j + k ) • ( 3i + 2 j – k ) 14 ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ uˆ ˜

=

1--------( –6 14

+ 6 – 1)

1= – ---------

14

b

c

1

Find v || using equation 4. ˜

b v || = ( uˆ • v )uˆ ˜ ˜ ˜ ˜ 1 - --------= ( – --------) [ 1 - ( –2i + 3 j + k ) ] 14 14 ˜ ˜ ˜

2

Simplify.

1

Find v ⊥ by subtraction of v || from v ˜ ˜ ˜ as in equation 5.

1= – ----( –2i + 3 j + k ) 14 ˜ ˜ ˜ 131---------= 7 i – 14 j – 14 k ˜ ˜ ˜ c v ⊥ = v – v || ˜ ˜ ˜ 31= 3i + 2 j – k – ( 1--7- i – ----j – ----k) ˜ ˜ 14 ˜ 14 ˜ ˜ ˜

2

Simplify by subtracting i , j , and k ˜ ˜ ˜ components.

v⊥ = ˜

20 ------ i 7˜

------ j – 13 ------ k + 31 14 ˜ 14 ˜

scalar and vector Graphics Calculator tip! Finding resolutes The dot product and the unit vector can be found on a graphics calculator, and then used to find a scalar resolute or a vector resolute. Consider the vectors u = – 2 i + 3 j + k and v = 3i + 2 j – k in Worked example 18. ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜

For the Casio fx-9860G AU 1. Press MENU and select RUN-MAT. Enter the components of both vectors as matrices and store as matrix U and matrix V.

Chapter 7 Introduction to vectors

2. To find the scalar resolute of v on u , we need to ˜ vector of u ) calculate the dot product of uˆ ˜ (the unit ˜ ˜ and the vector v . For a three-dimensional vector, we ˜ cannot calculate the magnitude of the vector using polar form as we did previously for the twodimensional vector. So calculate the value of 2

2

2

( –2 ) + 3 + 1 . 3. Press Æ followed by ALPHA [M] and then EXE to store this answer for future use.

4. As in the previous graphics calculator tip on page 326, the dot product is performed by multiplying the transpose of the first matrix by the second matrix. The first matrix in this case is uˆ , which is obtained by multiplying matrix U by ˜the reciprocal of magnitude M. (To access the transpose function, press OPTN then F2 (MAT) and F4 (Trn).) 5. Press EXE to display the scalar resolute.

6. To find the vector resolute of v parallel to u , we need to calculate ( uˆ • v )uˆ . This˜ is the scalar ˜resolute ˜ ˜ ˜vector uˆ . multiplied by the unit ˜

7. Press EXE to display the components for the required vector resolute.

8. To find the vector resolute of v perpendicular to u , ˜ we need to subtract the vector˜resolute of v parallel ˜ to u (calculated in Step 7) from v . Subtract the ˜ ˜ appropriate matrices.

9. Press EXE to display the components of the required vector resolute.

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For the TI-Nspire CAS 1. Open a new Calculator document. Define the vector u and the vector v by entering each set of ˜ ˜ brackets. components within square 2. To find the scalar resolute of v on u , we need to ˜ vector of u ) calculate the dot product of uˆ ˜ (the unit ˜ and the vector v . Access the Dot Product function˜ ˜ 7: Matrix & Vector then (press b and select C: Vector followed by 3: Dot Product). To enter uˆ , ˜ first access the Unit Vector function (press b and select 7: Matrix & Vector then C: Vector followed by 1: Unit Vector). Press U and then ) to close the set of inner brackets. Press the comma key (,) followed by V and then close the set of outer brackets by pressing ). Press · to display the value of the scalar resolute. 3. To find the vector resolute of v parallel to u , we ˜ / v to ˜show the need to calculate ( uˆ • v )uˆ . Press ˜ ˜ ˜ previous answer then press the multiplication key r. Repeat the steps above to access the Unit Vector function and press U. Close the set of brackets by pressing ). Press · to display the components of the required vector resolute. 4. The vector resolute can be found straight after defining the vectors u and v (without finding the ˜ seen˜in the screen at right. scalar resolute first) as

5. To find the vector resolute of v perpendicular to u , ˜ we need to subtract the vector˜resolute of v parallel to u (calculated in Step 4) from v . Press V˜ then ˜ followed by / v (the answer˜ from the previous line of working). Press · to display the components of the required vector resolute.

remember remember 1. Scalar resolute of v on u is given by: ˜ ˜ uˆ • v ˜ ˜ 2. Vector resolute of v , parallel to u is: ˜ v˜ || = ( uˆ • v )uˆ ˜ ˜ ˜ ˜ 3. Vector resolute of v , perpendicular to u : ˜ v ⊥ = v –˜ ( uˆ • v )uˆ or v˜ ⊥ = v˜ – v ˜|| ˜ ˜ ˜ ˜ ˜

Chapter 7 Introduction to vectors

7D

337

Resolving vectors — scalar and 1 a i vector resolutes b i

WORKED

Example

17

WORKED

Example

29

5 ---------11

2 ------3

2 d i

f i

21 e i – ---------

ii v || = --23- i + --23- j + --23- k iii v ⊥ = --43- i + --13- j + --53- k ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ 4 21 ------ ( 25i – 6 j – 8k ) - ( 2i + 3 j + 4k ) iii v ii v || = – ----= 29 29 ˜ ˜ ˜ ˜ ˜ ˜⊥ ˜ ˜ 15 5 5 15 - i + ------ j – ------ k ------ i + 17 ------ j – 28 ------ k v = – ii v || = ----iii 11 ˜ 11 11 ˜ 11 ˜ 11 11 ˜ ˜ ˜⊥ ˜ ˜

18

eBook plus Digital doc: WorkSHEET 7.1

1 For each of the following pairs of vectors, find: c i the scalar resolute of a on u . d ii the scalar resolute of u˜ on a˜ . ˜ ˜ e a u = 2i + 3 j and a = 4i + 5 j b u = 5i – 2 j and a = 3i – j ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ c u = – 2 i + 6 j and a = i – 4 j d u = 3i – 2 j and a = – 4 i – 3 j ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜1 ˜ 31i – ----j 2 a i --------ii v || = ----e u = 8i – 6 j and a = – 5 i + j 10 ˜ 10 10 ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ 82 b i --------ii v || = 8i + 10 j 2 For each pair of vectors u and v , find: 41 ˜ ˜ ˜ i the scalar resolute of ˜v on ˜u . c i 0 ii v || = 0 ˜ ˜ ˜ ii the vector resolute of ˜v , parallel to u , namely v || . ˜ ˜ ˜ iii the vector resolute of v , perpendicular to u , namely v ⊥ . ˜ ˜ ˜ a u = 3i – j ; v = 2i + 5 j b u = 4i + 5 j ; v = 8i + 10 j ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ c u = 4i + 3 j ; v = –3i + 4 j d u = i + j + k ; v = 2i + j – k ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ e u = 2i + 3 j + 4k ; v = 2i – 3 j – 4k f u = 3i + j – k ; v = 2 j – 3k ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ 3 An injured bushwalker is located at a position relative to a camp given by the vector 2i + 3 j . A searcher heads off from the camp ˜ in˜ a direction parallel to the vector 3i + 4 j . ˜ All measurements are in kilometres. ˜ a How far is the searcher from the camp when closest to the bushwalker? 3.6 km b What is the minimum distance between the searcher and the bushwalker?

i i i

23 13--------------13 17 29--------------29 10--------------– 13 10 13 – 6-----------13 23 – -----5-

ii ii ii ii ii

23 41--------------41 17 10--------------10 17--------------– 26 17 – 6--526--------------– 23 13

------ i + 51 ------ j iii v ⊥ = 17 10 ˜ 10 ˜ ˜ iii v ⊥ = 0 ˜ ˜ iii v ⊥ = – 3 i + 4 j ˜ ˜ ˜

0.2 km or 200 metres

4 A distressed yacht is located at a position given by the vector 5i – 2 j relative to a ˜ sent ˜ off from the cruiser. A rescue boat is cruiser and travels in a direction parallel to the vector 3i – j . If all measurements are in ˜ ˜ to the nearest metre, how kilometres find, close the rescue boat gets to the yacht.

316 metres

Vectors and matrices As we have seen earlier in the chapter, there is a relationship between vectors and matrices. If you are using a Casio fx-9860G AU graphics calculator, you will have used this relationship to perform certain functions or calculations. If we consider the vector v = ai + b j + ck , the components could be expressed ˜ ˜ ˜ ˜ a as the matrix b . c

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Similarly, for the vector w = pi + q j + rk , the components could be expressed as ˜ ˜ ˜ ˜ p the matrix q . r The properties of vector addition and subtraction, scalar multiplication and the dot product can be formulated in terms of the properties of matrices. Addition of v and w : ˜ ˜

Scalar multiplication of v : ˜

Dot product of v and w : ˜ ˜

a+ p p a b + q = b+q c+r r c 2a a 2 × b = 2b 2c c a b c

T

p q r

p = [a b c] q r = [ap + bq + cr] 1 Consider the vectors u = 2i + 3 j + 4k and v = 2i – 3 j – 4k . ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ Use matrix operations to calculate: a u • v −21 ˜ ˜ 21 b the scalar resolute of v on u – ---------29 ˜ ˜ 21 c the vector resolute of v parallel to u – ------ ( 2i + 3 j + 4k ) 29 ˜ ˜ ˜ ˜ ˜ 4 d the vector resolute of v perpendicular to u . ----- ( 25i – 6 j – 8k ) ˜ ˜ 29 ˜ ˜ ˜ In mathematics there is often more than one method to represent an idea. For example, a fraction of a number can be written using a proper fraction 3--- or a 4 decimal fraction 0.75. Each method has its advantages and disadvantages. A vector in two dimensions can be represented in three ways: 1 in component form using i and j ˜ ˜ 2 as a column matrix 3 by giving magnitude and direction. That is, 3i + 4 j , 3 or (5, 53.1˚) can all be used to represent the same vector. ˜ ˜ 4 2 Suppose that your class is going to use one of these forms of notation. By considering the strengths and weaknesses of each, decide which one you will use.

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Time-varying vectors Introduction Consider a vector where the x- and y-components are not constants, but vary with time. So, instead of writing u = xi + y j , we write u = x ( t )i + y ( t ) j . ˜ can be ˜ any ˜ often ˜there is ˜a limitation in that t ≥ 0. ˜ functions, but In theory, x(t) and y(t) 2 For example, suppose x(t) = t and y(t) = t. Thus, u = t 2 i + t j . ˜ We can tabulate values for u at various values of time t. ˜ ˜ ˜ t x(t) y(t) u ˜ 0 0 0 0 ˜ i+ j 1 1 1 ˜ ˜ 4i + 2 j 2 4 2 ˜ ˜ 9i + 3 j 3 9 3 ˜ ˜ 16i + 4 j 4 16 4 ˜ ˜ So at each point of time we get a ‘new’ vector; thus y we can talk about a ‘vector function of time’. 4 3 We can join the heads of each of these vectors, as 2 shown at right. The dashed line indicates the path that 1 the head of the vector takes. Note that the tail of all 1 2 3 4 5 6 7 8 9 10 x these vectors is the origin. How can we find the equation of this path?

Finding the equation of the path Consider a vector function of time u = x ( t )i + y ( t ) j . ˜ ˜ as parametric ˜ The expressions x(t) and y(t) are referred to equations. If we can solve these simultaneous equations, by eliminating t, we can get an expression in terms of x and y only. This is best seen by example.

WORKED Example 19

Let a particle’s position, as a function of time, be given by u = t 2 i + t j . Find the equation ˜ ˜ ˜ of the path, assuming t ≥ 0. THINK WRITE

3

Express the i and j components of u ˜ time ˜ ˜ functions. in terms of their Express each equation in terms of an identical function of t. Equate the two expressions.

4

Make y the subject.

1 2

x = t2 y=t x = t2 y2 = t2 t2 = x = y2 So x = y 2, or y=

x

The most difficult work is often in step 2, finding equivalent functions of t for both x and y. Sometimes squaring (or taking the square root of) one or more of the x(t) and y(t) functions will yield successful results. Otherwise the simultaneous equations can be solved using substitution as demonstrated in the following example.

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WORKED Example 20

Find the equation of the path of a particle whose position is given by: v = 2ti + ( t 2 – t ) j , t ≥ 0 ˜the graph˜ of its path. ˜ Sketch THINK 1 Express the i and j components in terms ˜ of their time˜functions. 2

Express t as a function of x.

3

Substitute for t in the equation y = t2 − t.

4

Simplify.

5 6 7

x Since t ≥ 0, --- ≥ 0. 2 State the domain of the equation. Use a graphics calculator to sketch the graph over the domain [0, ∞).

WRITE/DRAW x = 2t y = t2 − t x t = --2 x 2 x y =  --- – -- 2 2 2 x x = ----- – --4 2 x Since t ≥ 0, --- ≥ 0 2 x≥0 y

0

(1, – 1–)

2 3

x

4

WORKED Example 21

Let a particle’s position as a function of time be given by a Find the equation of the path. b Sketch the graph of the motion of the particle. c State the period of the motion. THINK a 1 Express the i and j components of u in ˜ terms of their˜ time ˜functions. 2 Square both sides of each equation so that a trigonometric identity can be used to eliminate t. Add the two equations. 3 2 2 4 Use the trigonometric identity cos + sin = 1 to simplify the equation. b The equation represents a circle of radius 1 and centre (0, 0).

u = cos ti + sin t j . ˜ ˜ ˜

WRITE/DRAW a x = cos t y = sin t x2 = cos2 t y2 = sin2t x2 + y2 = cos2t + sin2t x2 + y2 = 1 y 1

b

–1

c The period of cos t and sin t is 2π (the path makes one revolution every 2π).

c Period = 2π.

0 –1

1 x

Chapter 7 Introduction to vectors

341

WORKED Example 22

Let a particle’s position as a function of time be given by u = 2 cos ti + 3 sin t j . Find the ˜ ˜ ˜ equation of the path and sketch its graph. THINK 1 Express the i and j components of u in terms of ˜ ˜ ˜ their functions. 2

In this case, first eliminate the constants in front of the trigonometric functions.

3

Square both sides of the equation.

WRITE/DRAW x = 2 cos t y = 3 sin t x --- = cos t 2 y --- = sin t 3 x2 ----- = cos2 t 4 y2 ----- = sin2 t 9

4

Add the 2 equations.

5

Use the trigonometric identity cos2 + sin2 = 1.

6

This is the equation of an ellipse.

x2 y2 ----- + ---- = cos2 t + sin2 t 4 9 x2 y2 ----- + ---- =1 4 9 y 3

–2

0

2

x

–3

functions Graphics Calculator tip! Vector of time To draw the graph of a time-varying vector, we need to express the components in terms of parametric equations. Consider drawing the graph of the position vector u = 2cos t i + 3sin t j from Worked example 22 using a graphics calculator. ˜ ˜ ˜

For the Casio fx-9860G AU 1. Press MENU and select GRAPH. Ensure that the Angle setting is shown as radians. Press SHIFT [SET UP] and alter the setting for Angle if necessary. Press F2 (Rad) for radians. 2. Press EXE to accept this setting. Press F3 (TYPE) followed by F3 (Parm) to select the parametric equations option.

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

3. Enter the x-component by completing the entry line for Xt1= with 2 cos t. (Press X,q,T to enter t.) Similarly, enter the y-component by completing the entry line for Yt1= with 3 sin t. Press EXE after each entry. 4. Press F6 (DRAW) to display the graph. Note that with the standard window setting used, there are not equal scale values on the two axes so it doesn’t give a true representation of the shape of the graph.

s

342

5. Press SHIFT F2 (ZOOM) followed by F6 ( ) for more options. Press F2 (SQR) to correct the V-Window settings so that the x-axis and y-axis values are identical. (You can also experiment with other Zoom options to enhance the display.) 6. To obtain a clearer view of the graph, you can adjust the Window settings. Press SHIFT F3 (V-WIN) and adjust the values for Xmin, Xmax, Ymin and Ymax. Alternatively, press SHIFT F3 (V-WIN) followed by F1 (INIT) to initialise the V-Window. Press EXE and then press F6 (DRAW) to display the graph with this new setting.

For the TI-Nspire CAS 1. First ensure that the Angle setting is shown as radians. Press c to access the home screen and select 8: System Information followed by 2: System Settings. Press e until you reach Angle and then select Radian by using the arrow keys. Press · to accept this setting. 2. Continue pressing e until you highlight OK. Press · to select OK. Open a new Graphs & Geometry document (press / N and select 2: Add Graphs & Geometry). Press b and select 3: Graph Type followed by 2: Parametric.

3. Enter the x-component by completing the entry line for x1(t) with 2 cos t. Similarly, enter the y-component by completing the entry line for y1(t) with 3 sin t. (Use the arrow keys to move between entry lines.)

Chapter 7 Introduction to vectors

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4. Press · to display the graph. If the entry line is covering too much of the graph and you wish to hide it, press / G. (To bring the entry line back, press / G again.) 5. To obtain a clearer view of the graph, you can adjust the window settings. Press b and select 4: Window followed by 1: Window Settings. Adjust the values for XMin, XMax, YMin and YMax. Alternatively, press b and select 4: Window followed by 3: Zoom - In. Use the arrow keys to define the centre point of the zoom and press ·. This will increase the size of the graph by a factor of approximately 2. (Continue to press · to increase the size of the graph as required.) These techniques work well for 2-dimensional vectors, but 3-dimensional vectors usually are more difficult as the paths are much more complicated. Three-dimensional vectors will not be covered in this part of the course.

WORKED Example 23

Let u = (2t + 1) i + (3t – 4) j be the position vector of Ship A. ˜ ˜ ˜ Let v = (3t – 2) i + (2t + 3) j be the position vector of Ship B. ˜ ˜ Find where the ships’ paths˜ cross. THINK 1

Express the i and j components for the path of Ship A in˜terms˜of their time functions.

2

Make t the subject of each equation.

3

Equate the two expressions.

4

Simplify the equation and make y the subject. Call this equation A.

5 6

Express the i and j components of Ship B ˜ time ˜ functions. in terms of their Make t the subject of each equation.

WRITE Ship A: x = 2t + 1 y = 3t − 4 x–1 ----------=t 2 y+4 ------------ = t 3 x – 1 ----------y + 4----------= 3 2 3x − 3 = 2y + 8 3x – 11 y = -----------------2 Ship B: x = 3t − 2 y = 2t + 3

[A]

x + 2----------=t 3 y–3 ----------- = t 2 Continued over page

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THINK

WRITE x+2 y–3 ------------ = ----------3 2

7

Equate the two expressions.

8

Simplify the equation and make y the subject. Call this equation B.

2x + 4 = 3y − 9 2x + 13 y = -----------------3

[B]

9

The ships’ paths will cross when the two equations (equations A and B) are equal.

Equating equations A and B: 3x – 11 2x + 13 ------------------ = -----------------2 3

10

Solve for x.

11

Substitute x = 11.8 into equation B to find y.

9x − 33 = 4x + 26 5x = 59 x = 11.8 Substituting into equation B: 2 ( 11.8 ) + 13 y = ------------------------------3 = 12.2 The ships’ paths cross at the point with coordinate (11.8, 12.2).

12

State the solution. Note: Although the paths cross, the ships might not be there at the same time!

Note: Time-varying vectors will be considered in more detail in Chapter 8.

remember remember

If u = x ( t )i + y ( t ) j , t ≥ 0, then the equation of the path of a particle can be found ˜ equations ˜ by ˜solving the x(t) and y(t) simultaneously.

7E

Time-varying vectors

Use a graphics calculator to assist where appropriate in the following exercise.

WORKED

Example

19

1 For each of the following, find the equation of the path, assuming t ≥ 0. x a u = 2t i – t j y = – --2 ˜ ˜ ˜ b u = ( t – 1 )i – 3t j y = –3x – 3 ˜ ˜ ˜ c u = ( t + 3 )i + 4t 2 j y = 4 ( x – 3 ) 2 ˜ ˜ ˜ 3 x d u = 2t i + t 3 j y = ---˜ ˜ 8 ˜ 2 multiple choice The value of u when t = 0, for the vector u = ( t + 3 )i + 4t 2 j is: ˜ ˜ ˜ ˜ A 0 B 3i C 3i + 4 j D 4i + 4 j E none of these ˜ ˜ ˜ ˜

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Chapter 7 Introduction to vectors

WORKED

Example

20 b

y Note: x ≥ 1 0 3 x –1 1 (2, –1)

WORKED

Example

21

b

3 a Find the equation of the path of a particle whose position is given by: v = --2t- i + ( t 2 + t ) j y = 4x2 + 2x ˜ ˜ ˜ b Sketch the graph of its path.

y

x

0

4 a Find the equation of the path of a particle whose position is given by: v = ( t + 1 )i + ( t 2 – 2t ) j y = x2 − 4x + 3 ˜ ˜ ˜ b Sketch the graph of its path.

b

y 1

–1

0

5 Let a particle’s position as a function of time be given by u = cos 2t i + sin 2t j . ˜ ˜ ˜ a Find the equation of the path. x2 + y2 = 1 b Sketch the graph of the motion of the particle. c State the period of the motion of this particle. Period = π 6 A particle’s position as a function of time is given by u = 3cos 2t i + 3sin 2t j . ˜ ˜ ˜ a Find the equation of the path. x2 + y2 = 9 b Sketch its graph. c State the period of the motion of this particle. Period = π

1 x

–1

b

y 3

–3

0

3 x

–3

b

7 A particle’s position as a function of time is given by u = ( 1 + cos t )i + ( –2 + sin t ) j . ˜ ˜ ˜ a Find the equation of the path. (x − 1)2 + (y + 2)2 = 1 b Sketch its graph. c State the period of the motion of this particle. Period = 2π

y 0 –1

1

2 x

–2 –3

WORKED

Example

22

x2 8 ----- + y 2 = 1 9 y 1

8 Let a particle’s position, as a function of time, be given by: u = 3cos 2t i + sin 2t j ˜ ˜ ˜ Find the equation of the path and sketch its graph.

–3

x2 y2 9 ----- + ------ = 1 4 16

9 Let a particle’s position, as a function of time, be given by: u = 2cos t i – 4sin t j ˜ ˜ ˜ Find the equation of the path and sketch its graph.

WORKED

Example

23

eBook plus Digital doc: WorkSHEET 7.2

3 x

0 –1

y 4

10 Find the equation of the paths described by each of the following vector functions: –2 0 4t t2 + 4 –4 a u = ------------------ i + ------------------ j (Hint: Add x -and y-components and factorise.) 2 ˜ ( t + 2 ) ˜ ( t + 2 )2˜ What is the initial position of u ? ˜ When does x = y? What is the position at this time? y = 1 – x ; u ( 0 ) = i ; t = 2, u = 1--2- i + 1--2- j ˜ ˜ ˜ ˜ ˜ b u = ( t + 2 )i + ( t + 1 ) 2 j y = (x – 1)2, x ≥ 2 ˜ ˜ ˜ x – 3 )2 ( y + 1 )2 c u = ( 2 cos t + 3 )i + ( 3 sin t – 1 ) j (------------------ + ------------------- = 1 ˜ ˜ 4 9 ˜ 11 Let u = ( 3t + 1 )i + ( 4t – 2 ) j be the position vector of Ship A. ˜ ˜ Let v = ( 2t + 3 )i + ( 5t + 1 ) ˜j be the position vector of Ship B. ˜ ˜ 19˜ x = ----, y = --27Find where the ships’ paths cross. 7

2 x

12 Let u = t i + t 2 j . Find the equation of the path. Consider vector v = e t i + e 2t j . ˜ path ˜ both ˜vectors’ ˜ of this vector is the same as the path of u . Assuming ˜ Show˜ that the ˜ 2 equations start when t = 0, do these vectors ever coincide? y = x ; No, since v is always ‘ahead’ of u . ˜

˜

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summary Vectors and scalars • Definition: A vector is a quantity that has magnitude and direction. • Equality of vectors: Two vectors are equal if both magnitude and direction are equal. • Addition of vectors: To add two vectors, take the tail of one vector and join it to the head of the other. The result of addition is the vector from the tail of the first vector to the head of the second. • Subtraction of vectors: Subtract vectors by adding the negative of the second vector to the first vector. • Multiplication of vectors (by a scalar): Multiply the magnitude of the vector by the scalar, maintaining the direction of the original vector.

Position vectors in 2 and 3 dimensions • Magnitude of a vector: If u = xi + y j + zk , magnitude = u = ˜ ˜ ˜ ˜ ˜

x2 + y2 + z2

y • Direction of a vector (2-D only): If u = xi + y j , direction = θ = tan−1 -x ˜ ˜ ˜ • x- and y-components of a vector: Given magnitude and direction, the x- and y- components are given by: x = u cos θ ˜ y = u sin θ ˜ • Unit vector: The unit vector of a vector u , in the direction of u , denoted by uˆ is: ˜ ˜ ˜ u uˆ = ----˜ u ˜ ˜ • Locating vectors: If A and B are points with position a and b respectively then ˜ ˜ AB = b – a ˜ ˜

Multiplying two vectors — the dot product • Scalar (dot) product: The scalar or dot product of two vectors u and v is denoted ˜ ˜ by u • v . ˜ ˜ • Calculation of dot product: Where θ is the angle between two vectors: u • v = u v cos θ ˜ ˜ ˜ ˜ • Algebraic calculation of dot product: Let u = x 1 i + y 1 j + z 1 k and ˜ ˜ ˜ ˜ v = x2 i + y2 j + z2 k . ˜ ˜ ˜ Then u • v = x 1 x 2 + y 1 y 2 + z 1 z 2 . ˜ ˜ If u • v = 0 , then u and v are perpendicular. ˜ ˜ ˜ ˜ If u = kv, k ∈ R , then u and v are parallel. ˜ ˜ ˜ ˜

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Resolving vectors — scalar and vector resolutes • Scalar resolute of v on u : Let uˆ be a unit vector in the direction of u . The scalar ˜ ˜ ˜ ˜ resolute of v on u is given by uˆ • v . ˜ ˜ ˜ ˜ • Vector resolute of v parallel to u : Let uˆ be a unit vector in the direction ˜ ˜ ˜ of u . ˜ v || = ( uˆ • v )uˆ ˜ ˜ ˜ ˜ • Vector resolute of v perpendicular to u : ˜ ˜ v ⊥ = v – ( uˆ • v )uˆ or ˜ ˜ ˜ ˜ ˜ v ⊥ = v – v || ˜ ˜ ˜

Time-varying vectors • If u = x ( t )i + y ( t ) j , t ≥ 0, then the equation of the path of a particle can be found ˜ equations ˜ by ˜solving the x(t) and y(t) simultaneously.

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CHAPTER review 7A

7A,B

1 multiple choice If u = 4i – 3 j + 0.2k and v = 2i + 4 j – k , then 4u – 2.5v is equal to: ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ A 11i – 22 j + 3.3k B 6i + j – 0.8k C 8i – 12 j – 0.2k ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ D 21i – 13 j – 2.3k E 11i – 13 j ˜ ˜ ˜ ˜ ˜ 2 A fire observation tower reaches 40 m above the ground. Susan is 400 m from the tower, which is at a bearing of 60° (N 60° E) from her. State the position vector from Susan’s current position to the top of the tower. 200 3i + 200 j + 40k ˜

˜

˜

40 m N

400 m 60° E

Susan

7B

3 multiple choice The vector with a magnitude of 5 is: A 3i + j + k ˜ ˜ ˜ D 6i + 3 j – 4k ˜ ˜ ˜

7B

C 5i + 5 j + 5k ˜ ˜ ˜

B 2i + 5 j + 4k ˜ ˜ ˜ E 25i ˜

4 multiple choice Let the position vector for point P be 3i + 4 j – 5k and for point Q be – i + 3 j – 4k . Then ˜ ˜ ˜ ˜ ˜ ˜ the magnitude of the vector PQ is given by: A 3 2

B

134

C

553

D 4

E 1

7B

5 A boat sails 5 km due east from H, turns northward at a bearing of 45° (N 45° E) for a distance of 10 km and then travels due north for a further 5 km to point X. a Find the position vector from H to X. ( 5 + 5 2 )i + ( 5 + 5 2 ) j ˜ b Find the distance from H to X (correct to 2 decimal places).˜ 17.07 km

7C

6 multiple choice If u = 3i + 2 j – 4k and v = –4i + 5 j + k , then u • v is equal to: ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ A 10 B 0 C 480 D −6

E 33

Chapter 7 Introduction to vectors

349

7 multiple choice

7C

The angle between u = 3i – 2 j and v = 16i + 24 j is equal to: ˜ ˜ ˜ ˜ ˜ ˜ A 0° B 12.3° C 45° D 60°

E 90°

8 multiple choice A unit vector perpendicular to 3i – 4 j is: ˜ ˜ A 4i + 3 j B 0i + 0 j C 0.8i + 0.6 j ˜ ˜ ˜ ˜ ˜ ˜

7C E –3i + 4 j ˜ ˜

D 4i – 3 j ˜ ˜

9 multiple choice If u = 3i + a j , v = 2ai – a j and it is known that u is perpendicular to v , then a is: ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ A 0 B 1 C −1 D 6 E cannot be determined with the given information 10 multiple choice The angle between u = 4i – 2 j + 3k and v = 2i – 2 j is: ˜ ˜ ˜ ˜ ˜ ˜ ˜ 1−1 −1 --------A cos 0 B cos 58 1

7C 3

C cos−1 --------10

6

E cos−1 --------58

D cos−1 --------10

11 Let u = 4i + 3 j and v = –i + 2 j . Find the angle between the two vectors, in radians, to ˜ ˜ ˜ ˜ ˜ ˜ 4 decimal places. 1.3909 12 Let u = 3i – 5 j and v = –4i + j . Find: ˜ ˜ ˜ ˜ ˜ ˜ 3 5 - i – ---------- j a u + v – i – 4j b u – v 7i – 6 j c u • v −17 d uˆ --------34 ˜ 34 ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ e the angle between u and v . 135° ˜ 13 Find the angles that the vector v = 3i – 2 j + 4k makes with the x, y and z axes. ˜ ˜ ˜ ˜ 56.1°, 111.8°, 42.0° 14 Find value(s) of p, such that pi + 2 ( 1 – 3 p ) j is perpendicular to 2 pi + 3 j . ˜ ˜ ˜ ˜ 9 ± 69 ------------------Using the vectors a = i – 2 j and b = 2i + 3 j answer questions 15 and 16. 2 ˜ ˜ ˜ ˜ ˜ ˜ 15 multiple choice The scalar resolute of a on b is given by: ˜ ˜ 4 4 B – --------A – ----C – 4--13

7C

13

5

7A–C 7C 7C 7D

D

4 – ------5

E −4

16 multiple choice The vector resolute of b parallel to a is: ˜ ˜ 4A – -----B – 4--5- ( i – 2 j ) 5 ˜ ˜ 4 4 -( i – 2 j) --- ( 2i + 3 j ) – D – ----E 13 ˜ 5 ˜ ˜ ˜

7C

7D C

4 – ------ ( 2i 13 ˜

+ 3 j) ˜

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7D

17 Let u = 2i + 3 j – k and v = i + j – 2k . ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ 2 3 1 - i + ---------- j – ---------- k a Find a unit vector parallel to u . --------14 ˜ 14 14 ˜ ˜ ˜ b Resolve v into components parallel and perpendicular to u . ˜ ˜

7E

18 multiple choice

7E

v || = 1--2- ( 2i + 3 j – k ) , v ⊥ = – 1--2- ( j + 3k ) ˜ ˜ ˜ ˜ ˜ ˜ ˜

Let u ( t ) be a position vector of an object, whose position varies with time. If ˜ u = 3 sin t i + 2 cos t j then the path this object takes is: 2 ˜ ˜ 19 y = ----2- – 2 , hyperbolic ˜ x A a straight line y B a parabola C a circle 1 D an ellipse x 0 –2 E unable to be determined with the given information 1 19 Find the equation of the path of the time-varying position vector u = --- i + 2 ( t – 1 ) j . State t˜ ˜ ˜ the type of path (linear, parabolic, etc.). Hence, sketch its graph.

Modelling and problem solving 1 A radar station tracks a jet fighter flying with constant speed. If the radar station is considered to be at the origin, the fighter’s starting position is 2i + 8 j + k and 1 minute later it is at ˜ ˜ ˜ 8i – 4 j + 13k . The units are in kilometres. ˜ ˜ ˜ the vector which indicates the path of the fighter. 6i – 12 j + 12k a State ˜ ˜ ˜ b State a unit vector in the direction of this path. --13- ( ˜i – 2 j + 2k˜ ) ˜ c Find a vector, in terms of a parameter m, which represents the position of the fighter at any m time along the path. 2i˜ + 8 j + k˜ + ---3- ( ˜i – 2 j + 2k˜ ) ˜ ˜ d Find the point along the path where the fighter is closest to the station. 1--3- ( 10i + 16 j + 11k ) ˜ ˜ ˜ e Find the distance from the station at this point. 7.28 km f Find the speed of the plane in km/h. 1080 km/h g How fast is the plane rising (or falling)? 720 km/h

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Chapter 7 Introduction to vectors

2 Consider the box shown at right. z The coordinates (in cm) of point D are (3, 0, 4), while the coordinates of E are (0, 5.5, 4). D a Find the coordinates of point C. (3, 5.5, 0) b Express the line joining C to E as a vector. – 3i + 4k ˜ ˜ c Show that the two diagonals in the same plane as CE intersect with an angle of 73.7°. C d Find the volume of the box in litres. 66 cm3 = 0.066 litres x e Express the longest diagonal, from the origin, as a vector. 3i + 5.5 j + 4k ˜ ˜ ˜ f Find the length of this diagonal. 7.43 cm g Find the angle that this diagonal makes with the other long diagonal. 84.5°

E

y

3 The parallelogram OXYZ has O at the origin. The vector joining O to Z is given by 5i while a ˜ y the vector joining O to X is given by 2i + 7 j . ˜ ˜ a Sketch the parallelogram, labelling all vertices. b State the vectors joining Z to Y and Y to X. 2i + 7 j, – 5i ˜ ˜ ˜ c State the vectors which represent the diagonals of the parallelogram. 7i + 7 j, –3i + 7 j ˜ ˜ ˜ ˜ d Find the cosine of the angle between the diagonals. Express your answer in simplest surd O 2 29form. -----------29 e Find the angle that OX makes with the x-axis. 74.1° f State the vector resolute of the vector joining O to X in the direction of OZ. 2i ˜ g Let P be a point on the extended line of XY, such that the vector joining P to Z is perpendicular to OY. Find the coordinates of P. (−2, 7) h Find the area of the parallelogram. 35 square units 4 A river flows west–east at 5 m/s. A swimmer, in still water, can swim 3 m/s and tries to swim directly across the river from south to north. N a 5 a Draw a vector diagram to illustrate this situation. 3 b Find the resultant speed of the swimmer. 5.83 m/s c Find the bearing of the swimmer. 059° d If it took the swimmer 2 minutes to reach the opposite bank, how wide is the river? 360 m e How far downstream would the swimmer be carried? 600 m f Repeat parts b to e if the swimmer had started on the north bank.

Same result, except bearing = 180° − 059° = 121°

5 The top of an 8-m diving board lies over the swimming pool as illustrated at right. Kate Sally sits 25 metres away in the corner of the swimming pool and takes a bearing of 35° (N 35° E) to the feet of her friend, Kate, who is about to do a 8m N belly flop. X a State the position vector from Sally’s current 35° 25 m Sally location to her friend’s feet. 14.34 i + 20.48 j + 8 k ˜ ˜ ˜ E b Find the distance between Sally and her friend’s feet. 26.25 m c Determine the angle of elevation from Sally (at ground level) to the top of her friend’s head if her friend is 1.75 metres tall. (Give your answer to the nearest tenth of a degree.) 21.3°

X

Y(7, 7)

2i~ + 7j ~

5i~

Z

x

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6 Because of air resistance, a skydiver is eventually falling at a constant speed. If his target point on the ground is considered to be at the origin, the skydiver’s starting position is 6i + 4 j + 12k , while 1 minute later he is at 3i – 5 j + 4k . (The units are in kilometres.) ˜ ˜ ˜ ˜ ˜ a Indicate the skydiver’s path with a vector. −3 i ˜− 9 j − 8 k 1 ˜ ˜ ˜ -----------(−3 i − 9 j − 8 k ) eBook plus b State a unit vector in the path’s direction. ˜ ˜ ˜ 154 c Find the distance from the target to the skydiver at this point (that is, after 1 minute). 5 2 km Digital doc: Test Yourself d Find the speed of the skydiver in kilometres per hour. 744.6 km/h Chapter 7

Vector applications

8 syllabus reference Core topic: Vectors and applications

In this chapter 8A Force diagrams and the triangle of forces 8B Newton’s First Law of Motion 8C Momentum 8D Relative velocity 8E Using vectors in geometry

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Introduction

• two- and three-dimensional vectors and their algebraic and geometric representation • resolution of vectors into components acting at right angles to each other • applications of vectors in both life-related and purely mathematical situations

In the previous chapter, we discussed the theory of vectors. Vectors are the ideal mathematical tool for dealing with the motion of, and forces on, objects in two and three dimensions. In this chapter we consider these applications of vectors — forces in equilibrium, momentum and relative velocity.

Force diagrams and the triangle of forces Introduction An understanding of the causes of motion is fundamental to our understanding of the world. All around us objects are in motion. In the seventeenth century, Sir Isaac Newton succeeded in producing three simple laws which accurately described the motion of large objects. His mathematical model was based on the effect that forces have on the acceleration of objects treated as point particles. These effects are embodied in his First and Second Laws of Motion.

What is a force? We all know from experience what a force is. It is a ‘push’ or a ‘pull’. The force due to gravity acts on us all the time. A bar magnet repels a second bar magnet: a magnetic force acts here. The force of friction slows down the wheel of a bike when the brakes are applied. Air resistance retards the motion of athletes. In both these cases there is relative motion between two objects. The strings of a tennis racquet when stretched exert a force on a tennis ball while the strings and ball are in contact. In all these examples, objects which have unbalanced forces acting on them tend to undergo a change in their motion; objects which have a balanced set of forces acting on them maintain their motion. Objects under balanced forces are said to be in equilibrium. The forces that we will discuss in this chapter can be classified as one of three types: field forces, applied forces, and resistive forces.

Field forces Field forces act without physical contact between objects. The weight force which acts on an object due to the presence of an external gravitational field or the electric force which influences the motion of a charged particle are examples of field forces.

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Applied forces Applied forces are the pushes or pulls exerted on objects due to contact. They are forces with which we have daily experience. The normal contact force acting upwards on a book resting on a table or on us as we stand on the floor or sit on a chair are examples of applied forces. Other examples of applied forces include tensile forces in taut strings and cables (as in the cable used by a crane or rescue helicopter), and compressive forces acting on weightbearing rods.

Resistive forces Air drag and friction are examples of resistive forces. This type of force occurs when two objects move or attempt to move relative to one another. Air drag has been put to good use in the design of hang glider; it is found also, in the resistance between a moving body like a car and the air. An example of friction is seen in a bicycle that is slowing down on level ground, even without the brakes being applied.

What is a particle? We are all familiar with the notion of a particle, but in Newtonian dynamics a particle is used to model an object and is taken to be a point. That is, the size of the object is not relevant and any internal movements such as spin and change in shape are not included in the model. A particle can, in principle, model the movement of a bullet, a car, a diver or a planet, provided that we accept the model as an approximation to the motion of the real objects they represent. Any force that acts on a particle is said to act through the point that defines the position of the particle. For example, if we were to describe the various forces acting on a car, we would include the upwardly directed contact forces exerted on the car by the road at each of the four tyres in addition to frictional forces exerted on the tyres by the road. If the car was moving, we would add the drag forces due to the movement of the car through the air. There would be the weight force acting on the car due to the gravitational field of the earth. If we were to treat the car as a particle, we would describe all these forces as acting through a single point. The word particle serves to define the position of an object and sets it apart from the rest of the immediate environment.

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What assumptions do we make in Newtonian dynamics? In modelling motion we make the following assumptions so that problems can be solved to give reasonably accurate predictions. Term

Meaning

Light (body or string)

Object has no mass.

Smooth

No frictional forces are exerted.

Inextensible

Strings or ropes do not stretch.

Rigid

Objects do not change shape when forces are applied to them.

Perfectly elastic

Applied forces do not permanently deform an object (for example, a spring).

In many cases we ignore the presence of forces which would be insignificant, such as air drag on slowly moving objects.

The resultant force R

Crucial to a good understanding˜of Newton’s laws of motion is the concept of a net or resultant force acting on a particle. Force is a vector quantity because a force has not only a magnitude but also a direction. The unit of force is the newton (N). (This is a derived unit, which is simply one defined in terms of the standard units, namely distance, time and mass. This will be discussed later.) Forces can be described in two ways: 1. Using i – j – k notation ˜ ˜ ˜ For example, F = 2i – 4 j N (Only coplanar forces will be considered.) ˜ ˜ ˜ 2. Using size and direction notation For example, F = 200 N, N45°E is a force of magnitude 200 N directed at an angle ˜ 45° from north towards east. The net or resultant force is simply the vector sum of all real physical forces acting on the particle. It represents the sum or total force acting on a particle representing an object. It is not in itself a real force, only the sum of real forces. The net or resultant force acting on a particle is the vector sum of all real forces acting on that particle.

Force diagrams Individual forces are one of three types — field, contact or resistive — and are drawn as vectors which indicate their direction and magnitude.

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357

WORKED Example 1 Draw ‘vector diagrams’ to represent the forces involved in the situations shown below as a set of vectors acting on a point particle. Indicate the relative size of the force by the length of the vector arrows. Further indicate the nature of each of the forces acting by labelling them W for weight, N for normal contact, F for friction, A for applied force and D for ˜ ˜ ˜ ˜ ˜ or air resistance. air drag Constant velocity Accelerating A stationary person c b a

d

Velocity

f

e

Constant velocity Cricket ball through the air

Ball rolling down a slope

THINK

WRITE

In the force diagrams for a to f, treat each object as a particle. a There are two equal, opposing forces: the weight force down, which is a field force, and the normal contact force up.

a

~N

W ~

b

1

A cart moving at constant velocity has balanced horizontal forces.

2

Any resistive forces (D ) must be balanced by an applied force ( A˜) to keep the cart ˜ moving at constant velocity.

3

b

~N ~D

A ~ W ~

The cart will have balanced vertical forces arising from the weight force of the cart and the normal contact force.

c An accelerating cart will have balanced vertical forces as in a and b but unbalanced horizontal forces giving rise to an acceleration.

c

N ~ ~D

A ~ W ~ Continued over page

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THINK

WRITE

d

d

1

2

A ball moving through the air will have the vertical force of weight and the resistive force of air drag.

~D

W ~

The drag force will be in the opposite direction to that of the ball.

e A ball rolling down a slope will have a weight force directed vertically down, a normal contact force perpendicular to the slope and a resistive or frictional force.

e

f The parachutist will have two balanced vertical forces: one down due to the weight force and the second upwards due to air drag.

f

N ~

~F W ~ ~D

W ~

WORKED Example 2 Three forces — F 1 , F 2 and F 3 — act on a ball as shown in the force vector diagram at ˜ are described ˜ ˜ right. The three forces by the vectors: ~F3 j ~

~F2 = 4i – 5 j ˜ i~ ˜ = 10i + 2 j ˜ ˜ ~F1 = –6i + 7 j . ˜ ˜ Find the resultant force R , the sum of the three forces F 1 , F 2 and F 3 . ˜ resultant force R . ˜ ˜ ˜ Find the magnitude of the ˜ Find the angle that the vector R makes with the i vector. ˜ ˜ The force F 1 is changed so that R = 0; that is, the resultant force equals zero. Find the ˜ ˜ force F 1 . ˜ THINK WRITE

F1 ˜ F2 ˜ F3 ˜ a b c d

a

1

The resultant force forces F 1 , F 2 and ˜ ˜

R is the sum of the F˜ 3 . ˜

a

j ~ ~F1

2

Evaluate R . ˜

R ~ ~i

~F3 ~F2

R = ΣF ˜ ˜ = F1 + F2 + F3 ˜ ˜ ˜ = 4i – 5 j + 10i + 2 j – 6i + 7 j ˜ ˜ ˜ ˜ ˜ ˜ = 8i + 4 j ˜ ˜

C h a p t e r 8 Ve c t o r a p p l i c a t i o n s

THINK b

1 2

3

359

WRITE

The magnitude of the vector R is its length. b ˜ The symbol R or R is used to represent R = the magnitude˜ of a vector R . ˜ ˜ Recall that i • i = j • j = 1 and = i • j = j • i ˜ =˜ 0 , ˜and˜ evaluate R . ˜ ˜ ˜ ˜ ˜ =

R•R ˜ ˜ ( 8i + 4 j ) • ( 8i + 4 j ) ˜ ˜ ˜ ˜ 2 2 8 +4

=

80

= 4 5 c

1 2

3

d

Define θ. Then use the result for the dot product of two vectors a • b = a b cos θ. ˜ ˜ ˜ ˜

Evaluate θ to the nearest tenth of a degree.

1

The vector sum of all forces is now equal to zero. Set R = 0. ˜

2

Make F 1 the subject of the equation. ˜ Substitute F 2 and F 3 into the equation ˜ to find˜ F . and simplify ˜1

3

c Let the angle that R makes with i be θ. ˜ ˜ Let R • i = R i cos θ ˜ ˜ ˜ ˜ ( 8i + 4 j ) • i ˜ cos θ = --------------------------˜ ˜4 5×1 8 cos θ = ---------4 5 cos θ ≈ 26.6° d R = F1 + F2 + F3 ˜ ˜ ˜ ˜ = 0 F 1 = –( F 2 + F 3 ) ˜ ˜ ˜ F 1 = – ( 10i + 2 j – 6i + 7 j ) ˜ ˜ ˜ ˜ ˜ = –4i – 9 j ˜ ˜

The triangle of forces If three non-parallel forces acting on a particle have a resultant force of zero then the three forces can be represented in a triangle since the vector sum of the forces is zero. From the figure at right, if R = X + Y + Z = 0, then ˜ represented ˜ ˜ in a trithe three forces X , Y and Z ˜can be ˜ ˜ ˜ angle representing the magnitude and direction of the three forces. The advantage of representing three forces in a triangle is that the sine rule and/or cosine rule can be used to solve some problems involving three forces whose vector sum is zero. That is: Sine rule:

a b c -----------= ------------ = ------------- . sin A sin B sin C

Cosine rule: a2 = b2 + c2 − 2bc cos A.

c

B

a

A

C b

X ~

~Y

Z ~ X ~ Z ~ ~Y

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WORKED Example 3 Three forces — A, B and C — act on an object such that the resultant force is zero. The ˜ angle ˜ force A acts at an of ˜150° to the force B and they have the same magnitude of 20 N. ˜ ˜ a Determine the magnitude of C . ˜ b Find the angle that the force C makes with B to the nearest degree. ˜ ˜ THINK WRITE a

1

Draw a force vector diagram of three forces acting through a point with an angle of 150° between A and B and an ˜ C .˜ angle of θ between B and ˜ ˜

a

A ~

150°

θ C ~

~B 2

Since the resultant of the three forces is zero, place them in a triangle of forces.

3

Mark the angle between A and B as 180° − 150° = 30° and the˜ angle ˜ between B and C as 180° − θ. ˜ ˜

4

5 6

b

1

Substitute a = C, b = 20, c = 20 and A = 30° into the cosine rule. Solve for C to find the magnitude of C . ˜ State the solution.

Use the sine rule with a = 20, A = 180° − θ, b = 10.35 and B = 30.

2

Invert both sides of the equation.

3

Make sin(180 − θ)° the subject of the equation.

4

Find the value of 180° − θ.

5

Solve for θ.

6

State the solution to the nearest degree.

A ~ (180 – θ )°

C ~

30° B ~

C 2 = 20 2 + 20 2 – 2 × 20 × 20 cos 30° = 107.18 C = 10.35 The magnitude of force C is ˜ approximately 10.35 newtons. 20 10.35 b --------------------------------- = ----------------sin ( 180 – θ )° sin 30° sin ( 180 – θ )° sin 30° --------------------------------- = ----------------10.35 20 20 sin 30° sin ( 180 – θ )° = ------------------------10.35 = 0.9662 180° − θ = sin−1 (0.9662) = 75.06°

θ = 104.94° The angle between forces B and C is ˜ ˜ approximately 105°.

Note: The angle between two forces in the ‘real’ situation and the angle between them in a triangle of three forces are supplementary; that is, they sum to 180°.

1a

~N C h a p t e r 8 Ve c t o r a p p l i c a t i o n s

361

Book

remember remember

W ~

b

D ~

1. Force is a vector quantity. Its derived unit is the newton. 1 N = 1 kg m/s2 2. Types of force: (a) Field forces occur without physical contact. A common field force is the gravitational force often referred to as ‘weight’. (b) Normal contact forces occur between objects in contact or via strings, where they are known as tensile forces. (c) Friction forces occur when there is actual or attempted relative movement between two objects in contact. 3. The resultant force R is the vector sum of all physical ~F1 ˜ forces acting on an object. A force vector diagram ~F1 illustrates this with two forces F 1 and F 2 : ˜ ˜ non-parallel 4. If the resultant of three coplanar, forces F2 acting on a particle is zero then the three forces can be R = ~F1 + ~F2 ~ ~ represented in a triangle of forces. F ~2 a b c 5. Sine rule: ------------ = ------------ = ------------- . sin A sin B sin C 6. Cosine rule: a2 = b2 + c2 − 2bc cos A.

Ball W ~

c

~N ~F

Car

~A

W ~

d

~N F ~

e

A Boat ~ W ~

N ~

Sliding object W ~

8A WORKED

Example

1 1 f

N ~ ~F

~A Accelerating car W ~

g

~N

~F

Body at rest W ~ N ~

h

~F Sliding body W ~ Ball moving up

i ~D

W ~

j

~D Ball moving down W ~

Force diagrams and the triangle of forces

1 Draw vector diagrams to represent forces which act on the following objects. Indicate the relative size of the force by the length of the vector arrows. Further indicate the nature of each of the forces acting by labelling them: N for normal contact forces, W for gravitational forces (that is, weight), A for applied˜ forces, D for air resistance ˜ ˜ ˜ (drag) forces, and F for friction forces. ˜ a A book sitting on a table. b A ball falling vertically through the air at constant speed. c A car driving on a horizontal road at a constant speed. d A boat drifting through the water at constant speed. e A body sliding across a smooth horizontal surface at constant velocity. f A car accelerating on a horizontal road. g A body at rest on an inclined plane. h A body sliding down an inclined plane at constant speed. i A ball travelling vertically up (include air resistance). j A ball travelling vertically down (include air resistance). y 2 multiple choice 4 3 Refer to the diagram at right to answer the following questions. a Using i and j notation, the exact value of the resultant force is: ˜ ˜ + 7j 3 A ( 4 + 3 3 )i B 7 3i + ( 3 + 4 3 ) j ˜ ˜ ˜ ˜ C ( 4 + 3 3 )i + ( 3 + 4 3 ) j D ( 4 – 3 3 )i + 7 j ˜ ˜ ˜ ˜ E ( 4 + 3 3 )i + ( 4 3 – 3 ) j ˜ ˜

4 3 3

x

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b The magnitude and direction, anticlockwise from the i direction, respectively of a ˜ third force for the resultant force to be zero is nearest to: A 182 N, 49° B 15.3 N, 41° C 13.5 N, 41° D 182 N, 229° E 13.5 N, 221° 3 multiple choice A (5 newtons) ~ The three coplanar forces A , B and C act in such a way that ˜ of forces to assist in the resultant force is zero. ˜Use˜ a triangle C ~ answering the following questions. a The magnitude of C is: B (4 newtons) A 15 N B 21 N˜ C 3 N D −21 N E 0 N ~ b The angle between the forces A and C , to the nearest degree, is: ˜ C 143° ˜ A 37° B 53° D 137° E 127° WORKED

Example

2 4 a i 9i + 4 j ˜ ˜ ii –6i – 3 j ˜ ˜ iii ( 2 2 – 3 5 ) i – j ˜ ˜ b i 97 ii 3 5 iii

54 – 12 10

c i –9 i – 4 j ˜ ˜ ii 6i + 3 j ˜ ˜ iii ( 3 5 – 2 2 ) i + j ˜ ˜

6a i ii b i ii c i ii d i ii

61.4 N at N9.4°W 61.4 N at S9.4°E 50i – 5 j ˜ – 50 i + ˜5 j ˜ – 4 i + 12˜j ˜ 4i – 12 j˜ ˜ ˜ 2.9 N at N26.9°W 2.9 N at S26.9°E WORKED

Example

3

4 Two forces — F 1 and F 2 — act on an object. They are described by the vectors: ˜ ˜ i F 1 = 13i – 5 j, F 2 = –4i + 9 j ˜ ˜ ˜ ˜ ˜ ˜ eBook plus ii F 1 = 8i + 6 j , F 2 = –14i – 9 j ˜ ˜ ˜ ˜ ˜ ˜ Digital doc: iii F 1 = 2 2i – 3 j, F 2 = – 3 5i + 2 j . EXCEL ˜ ˜ ˜ ˜ ˜ ˜ Spreadsheet For each of the above find: Vector addition a the resultant force, R , acting on the object b the magnitude of the˜ resultant force, R ˜ that the resultant force is equal to zero. c a third force, F 3 , applied to the body so ˜ 5 Three forces — F 1 , F 2 and F 3 — act on an object. They are described by the vectors: ˜ ˜ ˜ i F 1 = 3i – 5 j, F 2 = 4i + 9 j and F 3 = –6i – 2 j ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ 5 a i i + 2 j ii ( 2 – 1 ) i ii F 1 = i – j, F 2 = – 2 i + 3 j and F 3 = 2i – 2 j . ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ b i 5 ii 2 – 1 For each of the above find: c i 63.4° ii 0° a the resultant force, R , acting on the object – i – 2 j d i ii ( 1 – 2 ) i ˜ ˜ b the magnitude of the˜ resultant force, R ˜ ˜ i vector c the angle that the vector R makes to the ˜ to the body so˜ that the resultant force is equal to zero. d a fourth force, F 4 , applied ˜ 6 In each of the following cases: i find the resultant force, R , for the specified forces acting on a body ˜ ii describe the additional force needed to attain equilibrium; that is, R = 0. ˜ a F = 25 N at N30°E and G = 45 N at N30°W ˜ ˜ b W = – 80 j and F = 50i + 75 j ˜ ˜ ˜ ˜ ˜ c A = 26i – 80 j and B = –30i + 92 j ˜ ˜ ˜ ˜ ˜ ˜ d F = 10 N at S50°W and G = 7.0 N at N45°W and H = 12 N at N70°E ˜ ˜ ˜ 7 Three forces — X , Y and Z — act on an object such that the resultant ˜ Y acts ˜ at an angle of 120° to the force X and has eBook plus force R = 0. The ˜force ˜ ˜ ˜ the same magnitude as the force X which is 10 N. Digital docs: a Determine the magnitude of Z˜ . 10 N SkillSHEET 8.1 Sine rule b Find the angle that the force Z˜ makes to X . 120° ˜ ˜ SkillSHEET 8.2 8 Three coplanar forces — A , B and C — have magnitudes of 1000 N, Cosine rule ˜ ˜ ˜ 1200 N and 1700 N respectively. They act on a body such that the resultant force is zero. Find the angle between A and B . 79.2° ˜ ˜

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9 Three forces — X , Y and Z — add to give a resultant force ˜ equal to zero. ˜ ˜ a Find β if α = 135°, Z = 200 N and X = 150 N. 77° ˜ ˜ b Find Z if α = 100°, β = 135° and Y = 27 N. 37.6 N ˜ ˜ c If X = Y and α = 60°, show that Z = 3 X . ˜ ˜ ˜ ˜

X ~

Z = 2 sin 60° × X ˜ ˜

~Y

β

α

θ

Z ~

History of mathematics S I R I S A AC N E W T O N ( 1 6 4 2 – 1 7 2 7 ) During his lifetime ... The Great Fire destroys much of London. The Great Plague of London causes the death of more than 12000 people. Huygens develops the pendulum clock. Lloyds of London starts its insurance society. Sir Isaac Newton was a great English mathematician and physicist. He developed the theory of differential calculus, discovered the law of gravitation and is regarded as the founder of modern physics. Newton was the son of a wealthy man but his father died just three months before his son was born. He was not happy as a child and his school reports describe him as idle and inattentive. It is thought that he suffered from bouts of depression throughout his life. In 1661 Newton was admitted to Trinity College, Cambridge. When bubonic plague broke out in London and closed the university in 1665 he went home to Lincolnshire. During the next two years he began the work which would lead to his revolutionary advances in mathematics, optics, physics and astronomy. He returned to Cambridge when it reopened after the plague and was elected to a major fellowship in 1668 at the age of only 26. He published his theory of differential calculus in 1671. However, the German philosopher Leibniz had been working in the same area and a bitter quarrel developed as to who had made

the discovery first. It became one of the great controversies of the times. In 1687 Newton published his major work, the Philosophiae Naturalis Principia Mathematica. The Principia is considered by many to be the greatest scientific book ever written. In it Newton gives a mathematical description of the laws of mechanics and gravitation and applies this theory to explain planetary and lunar motion. His three laws of motion can be expressed in simple terms as: 1. A body at rest or in constant motion will stay in that state until a force is applied; 2. Force = mass multiplied by acceleration; 3. For any action there is an equal and opposite reaction. In 1696 Newton took up a new direction when he was appointed Warden of the Royal Mint and then Master in 1699. He was responsible for an urgently needed reform of the coinage and also introduced measures to prevent counterfeiting. He was knighted in 1705. In 1703 he was elected President of the Royal Society and was re-elected each year until he died in 1727 at the age of 85.

Questions 1. Why did Newton leave Cambridge in 1665? Bubonic plague broke out and the university closed. 2. Who else claimed to have discovered differential calculus? Leibniz 3. What did Newton describe and apply in the Principia? The laws of mechanics and gravitation 4. Name two of Newton’s achievements as Master of the Royal Mint. He reformed the coinage and introduced measures to prevent counterfeiting.

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Newton’s First Law of Motion Historically, Newton’s First Law of Motion had its origins in some of the work done by Galileo. The law concerns itself with the motion of a particle when the resultant force R ˜ acting on it is zero. It was a commonly held belief that in order to have motion, an unbalanced force was required. This belief seemed quite natural: clearly, carts need to be pulled by horses in order to move and that balls rolling across a lawn come to rest. However, experiments showed that the force of friction could be used in both cases to explain the apparent necessity of an applied force to maintain constant motion. It was Newton who stated that in the absence of an unbalanced force (that is, R = 0) acting on ˜ a body, the body would continue in a state of uniform motion. The phrase ‘uniform motion’ is used to describe the motion of a body with zero acceleration; that is, the velocity of the body is a constant. Today we would write this as: If R = 0 (the resultant force acting on the body is zero) ˜ then v = c (the velocity of the body is a constant) and thus a = 0 (the acceleration of the body is zero). If the resultant force acting on a body is zero, then its acceleration is zero. It is important to recognise that the inference above is equally valid in reverse. That is, If a = 0 (the acceleration of the body is zero) then v = c (the velocity of the body is a constant) and thus R = 0 (the resultant force acting on the body is zero). ˜ When the net or resultant force acting on an object is equal to zero, it is commonly said that the body is ‘in equilibrium’. Such an object moves at a constant velocity. These types of situation belong to a class of problems called statics. To solve dynamics and related kinematics problems there is a clear strategy. 1. Read the question carefully. 2. Draw a clear diagram that contains all the information. 3. Superimpose arrows depicting vectors which act on the body in question. This is called a force vector diagram. 4. Find the resultant force vector that is the sum of all forces acting on the body.

Mass and weight We know from experience that some objects are harder to push than others; that is, it is more difficult to modify their motion. We call this property of matter inertial mass. The SI (Système International d’Unités) unit for mass is the kilogram. It is harder to stop or accelerate a large truck than a small car, the truck having the greater mass. Weight is a vector quantity. It is a force, equal to the product of the acceleration due to gravity and the mass on which it is acting. We often refer to ‘the weight of the object’ but the object does not possess weight as an intrinsic quantity; it does, however, possess mass. The SI unit for weight, since weight is a force, is the newton. Suppose that an astronaut is somewhere in space where there is a zero gravitational field (g = 0 N/kg); he may be 70 kg, but he is said to be ‘weightless’. The gravitational field strength g is a measure of the magnitude of the force of gravity acting on a unit mass. Consequently it has the dimensions of force per unit mass. In SI units this is the newton per kilogram (N/kg). The same 70 kg man on the surface of the Earth where the vector g has a magnitude of 9.8 N/kg down would have a weight of ˜

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686 N or 70 g N downward. That is, the weight W of a mass m in a gravitational field ˜ g is given by˜ the equation: ˜ W = mg 2 ˜ universally in examples and problems Note: The value for g of 9.8 m/s down˜ is used ˜ in this textbook. Sometimes the unit for g is quoted as N/kg. This is the acceleration ˜ due to gravity. For a mass of 1 kg, it becomes equal to the weight force when all other forces are ignored.

Resolving a force into its components In many cases the size of a force acting in a particular direction needs to be calculated. For example, when a jet aircraft is taking off from a runway the engines provide a thrust. Part of the thrust assists in lifting the plane up into the air (the vertical component) and the other component exerts a horizontal push making the plane move forward. In the diagram at right, the combined thrust of the jet’s engines is repj resented by a single vector F acting on a particle representing ~ ˜ the aircraft. ~i If F is the applied force acting on the plane due to the action ~F of the˜ engines then we can write the force vector as the sum of ~Fy θ two vectors F x and F y which are parallel to the unit vectors i ~Fx Aircraft ˜ ˜ ˜ and j respectively. ˜ F = Fx + Fy Thus ˜ ˜ ˜ F = Fxi + Fy j ˜ ˜ ˜ ˜ ˜ F x = F cos θ where ˜ ˜ and F y = F sin θ. ˜ ˜ The quantities Fx and Fy are referred to as the components of the vector force F in ˜ the i and j directions respectively. ˜ ˜ A force F can be resolved into perpendicular components: ˜ F = F cos θ i + F sin θ j ˜ ˜ ˜ where F = F and θ is the angle between F and i . ˜ ˜ ˜ For example, if the combined thrust of the engines is 2 × 106 N and the plane during take-off had a 15° elevation angle then the vertical component would be: Fy = F sin θ ˜ = 2 × 106 × sin 15° ≈ 5.18 × 105 N while the horizontal component would be: Fx = F cos θ ˜ = 2 × 106 × cos 15° ≈ 1.93 × 106 N or F ≈ (1.93 × 106) i + (5.18 × 105) j ˜ ˜ ˜ Note: The i and j components ˜ ˜ are usually horizontal and vertical components respectively but they can be rotated to suit a particular problem.

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WORKED Example 4 A water skier is being pulled along by rope attached to a speed boat across a horizontal lake. The rope makes an angle of 5° to the horizontal and exerts a force of 6000 N on the skier. a Calculate the horizontal component of the force exerted on the skier by the rope. b Calculate the vertical component of the force exerted on the skier by the rope. The skier is moving with a constant velocity. c Calculate the size of the horizontal resistance forces on the skier.

THINK The skier has four forces acting on him: the weight force, W , acting vertically downwards; the normal ˜ N , of the water on the skier, acting reaction, ˜ vertically upwards; the tension force, T , in the ˜ resistance rope, acting 5° to the horizontal; and the forces, F , acting horizontally against the direction ˜ of motion. 2 Draw a force vector diagram showing the forces acting on the skier. a Evaluate the magnitude of the horizontal component of T , TH. ˜ b Evaluate the magnitude of the vertical component of T , TV. ˜ c 1 Constant velocity means the acceleration is zero and in turn the resultant of the horizontal forces is zero. 2 The magnitude of the horizontal resistance forces, F, is equal to the horizontal component of the tension force, T . ˜

WRITE

1

~N ~T ~F

5°

W ~

a TH = 6000 cos 5° ≈ 5997 N b TV = 6000 sin 5° ≈ 523 N c Since acceleration = 0,

F = TH = 5997 N

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WORKED Example 5 In a science laboratory a 1.0 kg mass is suspended by two taut strings as shown at right. The tension forces in string 1 and string 2 are T 1 and T 2 respectively. ˜ ˜ diagram showing all three forces a Draw a force vector which act on the 1.0 kg mass. b By resolving vectors into i and j components find the ˜ magnitudes of T 1 and T 2˜ respectively. ˜ ˜ THINK WRITE a 1 The mass has three forces acting on it: a T ~2 the weight force vertically downwards and the two tension forces. One tension 42° j ~ force acts horizontally; the second acts at an angle of 42° to the horizontal. ~i 2

b

1 2 3 4

5

6 7

8

Draw the force vector diagram. The weight vector can be written as −g j . b ˜ The first tensile force can be written as T1 i . ˜ The second tensile force can be resolved as T 2 = −T2 cos 42° i + T2 sin 42° j . ˜ ˜ force, R , as˜ the Express the resultant ˜ sum of the three forces. Set the sum of the three vectors to zero in accordance with Newton’s First Law of Motion. Set the i component of R to zero and ˜ ˜ call it equation [1]. Set the j component of R to zero and ˜ ˜ call it equation [2]. Solve equation [2] for T2.

Solve for T1 by substituting T2 into equation [1]. Note: Part b could also be solved by drawing a triangle of forces and solving using trigonometry. 9

String 2 String 1

j ~

42°

~i

1.0-kg mass

T1 ~ W ~ = ~g

W = –g j ˜ ˜ T 1 = T1 i ˜ ˜ T 2 = –T2 cos 42° i + T2 sin 42° j ˜ ˜ ˜ R = (T1 − T2 cos 42°) i + ˜ ˜ (T2 sin 42° − g) j ˜ But R = 0 i + 0 j . ˜ ˜ ˜ T1 − T2 cos 42° = 0

[1]

T2 sin 42° − g = 0

[2]

g T2 = ----------------sin 42° ≈ 14.6 N T1 = T2 cos 42° ≈ 10.9 N W = g = T1 tan 42°

j ~

g = ----------------tan 42° ≈ 10.9 N and W = g = T2 sin 42°

T1

T2

W ~ = ~g ~i

~T2 42° ~T1

g = ----------------sin 42° ≈ 14.6 N (as in part b above).

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WORKED Example 6 A car of mass 800 kg is parked in a street which has an angle of j ~ elevation of 15°. The i direction is parallel down the street and the j ˜ ˜ direction is perpendicular to the street. ~i The car is subject to three forces, namely its weight, W , the normal 15° ˜ contact force, N , of the road acting on the car and the applied force ˜ is actually a static friction force) F . of the brake (this ˜ W , N and F , acting on the car, a Draw a vector diagram indicating the three forces, ˜ ˜ ˜ taking the car as a particle. b What is the magnitude of the resultant force R ? c Resolve the weight, W , into its components and˜ express it as a vector using i – j notation. ˜ ˜ ˜ d Calculate the magnitude of N , the normal contact force. ˜ e Calculate the magnitude of the applied force of the brake F . ˜

THINK WRITE a a 1 A stationary car parked on a street will have a vertical j weight force, a normal contact force and a static frictional ~ force resisting its sliding or rolling down the street. Draw the force vector diagram. 2 ~i b

1 2 3

c

1

c

~F 15° W ~

The car is in equilibrium since it is stationary. b Apply Newton’s First Law of Motion: the resultant R =0 ˜ force, R , must be zero. ˜ Therefore, the magnitude of the resultant force, R, is zero. R=0N Draw a diagram showing the resolution of the weight, W , into components parallel to i and j . ˜ ˜ ˜

N ~

j ~ 15° ~i

15° Wy W ~

2

The component of W parallel to i , Wx , is W sin 15°. ˜ ˜

Wx

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THINK 3 Substitute W = 800g and evaluate.

4 5

d

e

WRITE Wx = 800g sin 15° ≈ 207g W y = W cos 15°

The component of W parallel to j , W y , is W cos 15°. ˜ Substitute W = 800g and evaluate.

6

Express W in vector notation. ˜

1

The component of the net force parallel to the j ˜ vector is zero.

2

Solve for the magnitude of the normal N.

1

The component of the net force parallel to the i ˜ vector is zero.

2

Solve for the magnitude of the applied force of the brake, F.

= −800g cos 15° ≈ –773g W = Wx i + Wy j ˜ ˜ ˜ j W = 207g i − 773g ˜ ˜ ˜ d (N − 773g) j = 0 j ˜ ˜ N = 773g e (207g − F) i = 0 i ˜ ˜ F = 207g

Note: In general, when an object of mass m and hence weight mg is on an inclined ˜ components: plane with incline angle θ, the weight vector can be resolved into two 1. one, of magnitude mg sin θ , acting down the plane 2. one, of magnitude mg cos θ, acting perpendicular to the plane. This is opposed by the normal contact force, N . ˜

Friction A body resting on a table is acted on by a number of forces. As we have discussed the weight W acts downwards and is given ˜ by the formula

N

˜ Friction

G

˜

W = mass × g . W ˜ ˜ ˜ The normal reaction, N , is the force exerted by the table on the body which opposes, and just balances, the ˜weight. Suppose the body is further acted on by a horizontal force G . If the table is smooth, there is no friction and the mass will move to the right. ˜ If the table is not smooth a frictional force will oppose the motion. The frictional force depends on the roughness of the surface and the normal reaction N . ˜ Friction = µ × normal reaction, where µ is the coefficient of friction. F =µ× N ˜ ˜ Note that this formula gives the maximum value for friction. If, in the diagram, G is ˜ less than µ N , then the frictional force will just balance the force G . ˜ ˜

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WORKED Example 7

N

A body of mass 4 kg, at rest on a table, is acted on by a horizontal force, P , as shown in the figure at right. ˜ If the body just begins to move when P is 12 N, calculate ˜ the coefficient of friction between the body and the table. THINK 1

2 3

The weight of the body is m × g.

The normal force, N, must balance the weight. The friction will oppose P and its maximum value is µ × N.

˜ Friction

P

˜ W

WRITE

˜

W=m×g = 4 × 9.8 = 39.2 N N=W = 39.2 N F=µ×N 12 = µ × 39.2 µ = 0.3

WORKED Example 8 A body of mass 4 kg rests on a plane inclined at an angle of 30° to the horizontal. It is just prevented from moving by friction. Calculate the: a weight W ˜ N b normal reaction c frictional force F ˜ ˜ µ. d coefficient of friction

N

Friction

˜

W

˜

THINK

WRITE

a W=m×g m = 4 kg, g = 9.8 m/s2. The units are newtons.

a

W =m×g ˜ = 4 × 9.8 = 39.2 N

b First resolve each force into its components. Take the i direction to be down the plane and the j direction to be perpendicular to the plane. If a body is at rest the vector sum of forces is 0. If ai + b j + ci + d j = 0 then ˜ b =˜ −d. ˜ a = ˜−c and

b

N = Nj. ˜ cos 60 i + −39.2 sin 60 j W˜ = 39.2 ˜ j ˜ = 19.6 i + −33.9 ˜ ˜ ˜ F = −µN i ˜ ˜

c

c F = 19.6 N

d Coefficient of friction µ can be calculated using F = µN.

d

F+N+W=0 −F i + N j + 19.6 i + −33.9 j = 0 ˜ ˜ The i and˜ j components add˜to 0 N = 33.9 N F = µN 19.6 = µ × 33.9 µ = 0.58

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remember remember

371

j

~ 1. Components of a force: ~i A force F can be resolved into perpendicular components: ˜ ~F F = Fx + Fy ~Fy ˜ ˜ ˜ θ = F cos θ i + F sin θ j ˜ ˜ between F and i . Aircraft ~Fx where F = | F | and θ is the angle ˜ ˜ ˜ 2. Newton’s First Law of Motion is used for equilibrium or statics problems: R = 0 (the resultant force is zero) ˜ a = 0 (the acceleration is zero) v = c (the velocity is a constant). 3. The weight W (in N) of a mass m (in kg) in a gravitational field g is given by ˜ the value commonly used for g is 9.8 m/s2 (or 9.8˜ N/kg). W = m g where ˜ ˜ 4. Friction = µ × normal reaction, where µ is the coefficient of friction F =µ× N ˜ ˜ SLE 4: Use resolution of vectors to consider the equilibrium of a body subject to a number of coplanar forces acting at a point.

8B WORKED

Example

4

Newton’s First Law of Motion

Note: Use g = 9.8 m/s2 down for all problems involving weight. ˜ 1 A girl is pulling along her baby brother in a cart attached to a rope. The cart is on a horizontal path. The rope makes an angle of 20° to the horizontal and exerts a force on the cart of magnitude 25 N. a Calculate the horizontal component of the force exerted on the cart by the rope. 23.5 N b Calculate the vertical component of the force exerted on the cart by the rope. 8.6 N The cart moves along at a constant velocity; that is, the acceleration of the cart is zero. c Calculate the size of the horizontal friction force acting on the cart. 23.5 N 2 multiple choice A child of mass 40 kg is held on a ‘swinging rope’ at an angle of 25° to the vertical by a horizontal force of 300 N. If T is the tension force of the rope acting on the child, then: ˜ a the force vector diagram which best represents this situation is: A B C 25° 65°

~T

25°

300 N

~T

40g N

D 25°

40g N

E

~T

25°

~T

~T

40g N

40g N

300 N

40g N

300 N

300 N

300 N

b the horizontal component of T is: A 40 g N B 300 N ˜ C 127 N

D 272 N

E 200 N

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c

the vertical component of T is: ˜ A 300 N B 127 N C 40 g N d the magnitude of T is nearest to: ˜ A 494 N B 440 N C 477 N

a F ~

Drag force due to water

Tugboat T ~

Ship

T ~ Tugboat

WORKED

Example

5 a ~T2

60°

30° ~T1

j ~ ~i

a

8g N ~

~Tright

~Tleft

Speaker W ~

d –278 i + 245 j N; ˜ ˜ 278 i + 245 j N ˜ ˜

a

~N ~A = 120 N at 40° to vertical

~F W ~ = 40g ~

D 200 N

E 272 N

D 92 N

E 300 N

3 A ship of mass m is being pulled by two tugboats. It glides Tugboat 10° Ship through the water at constant velocity. The angle between the two ropes connecting the two tugboats to the front of the ship 10° Tugboat is 20° and they each support a tension of magnitude 20 000 N. a Draw a force vector diagram showing all three coplanar forces which act on the ship. b What is the magnitude of the resultant force acting on the ship? (Be careful; use 39 392.3 N Newton’s First Law of Motion.) 0 N c Calculate the magnitude of the force of friction due to the water acting on the ship. 4 A swing chair of mass 8 kg is suspended by two taut ropes as shown at right. The tension forces in rope 1 and rope 2 are T 1 and T 2 respectively. 60° 30° a˜ Draw˜ a force vector diagram showing Rope 1 Rope 2 all three forces which act on the swing. b By resolving vectors into i – j com˜ ˜ ponents find the exact magnitudes of T 1 and T 2 . T1 = 4g N; T2 = 4 3 g N ˜ ˜ 5 A speaker in an auditorium has a mass of 50 kg and is suspended from the ceiling by two 4.0-m ropes. The ropes are attached to the ceiling at points A and B whose separation is 6.0 m as shown at right. a Draw a force vector diagram showing all forces which act on the speaker. 6.0 m b Calculate the angle that each rope makes with A B the ceiling. 41.4° c Determine the vertical component of the ten4.0 m sion force in each rope. 245 N j ~ d Give the vectors for the tensions in the left and right rope respectively using i – j notation. ~i ˜ ˜ tension in e Calculate the magnitude of the each rope. 370.6 N f The speaker is to be raised by increasing the separation between the points A and B, but the ropes will break if the tension exceeds 4000 N. Find the maximum possible separation between A and B; that is, when the tensions in the ropes are equal to 4000 N. 8.0 m 6 Sam earns some extra pocket money by mowing his neighbour’s front lawn. When he pushes the lawnmower at a constant velocity he applies a force of 120 N down the shaft of the mower which is angled at 40° to the vertical. The lawnmower has a mass of 40 kg. a Draw a force vector diagram illustrating all four forces acting on the lawnmower. (Treat the lawnmower as a particle.)

120 N 40°

c R = ( F friction – 77.1 )i + ( N – 483.9 ) j = 0 ˜ ˜ ˜ with all forces in N.

Av = 91.9 N down AH = 77.1 N left

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b Calculate both the vertical and horizontal components of the force that Sam applies to the lawnmower. c Using i – j notation, write a vector equation for the resultant force, R , acting on ˜ ˜ the lawnmower in terms of the four forces acting on the lawnmower. ˜ d Determine the magnitude of the force of friction acting on the lawnmower as it moves across the lawn at constant velocity. 77.1 N e Determine the magnitude of the normal contact force. 483.9 N WORKED

Example

6 a

~N ~T Mass W ~

a

N ~ Mass

H ~

W ~

WORKED

Example

7

7 A 1.5 kg mass is placed on a smooth inclined plane angled at j ~ 30° to the horizontal. To stop it from sliding down the plane ~i a string is attached to the upper side as shown at right. 30° The unit vectors i and j are also shown. ˜ ˜ a Draw a force vector diagram showing the forces which act on the mass. Label the forces N for the normal contact force, T for the tension force and W for the ˜ ˜ force arising from the effect of gravity. ˜ b What is the magnitude of the resultant force, R ? 0i˜ + 0 j ˜ ˜ c Determine the weight vector, W , using i – j notation. –7.4 i – 12.7 j N ˜ ˜ ˜ ˜ d Determine the tension force, T , using i – j ˜ notation. 7.4 N ˜ ˜ ˜ e Determine the magnitude of the normal contact force, N. 12.7 N 8 A 1.5 kg mass is placed on a smooth inclined plane angled at j ~ 30° to the horizontal. To stop it from sliding down the plane ~i Horizontal a horizontal force is applied to the mass as shown at right. force, ~ H The unit vectors i and j are also shown. ˜ 30° a Draw a force vector ˜diagram showing the forces which act on the mass. Label the forces N for the normal contact ˜ and W for the force due to gravity. force, H for the horizontal force ˜ ˜ b Determine the weight vector, W , using i – j notation. – 7.4i˜ – 12.7 j Ν ˜ ˜ ˜ c Determine the horizontal force, H , using i˜– j notation. 7.4i – 4.3 j N ˜ ˜ ˜ ˜ d Determine the normal contact force, N , using˜ i – j notation. 17 j N ˜ ˜ ˜ ˜ 3W W --------------= e Show that H = ------. ˜ ˜ 3 ˜ 3 9 A box containing books has a mass of 40 kg. It requires a force of 300 N to move the box across the floor. Calculate the coefficient of friction between the box and the floor. 0.77 10 The maximum deceleration the BMV Tycoon can achieve under braking is 4 ms−2. If this vehicle has a mass of 1900 kg, calculate: a the frictional force on the car 7600 N b the normal reaction 18 620 N c the coefficient of friction between the tyres and the road. 0.41 11 A force of 200 N is needed to keep a log of mass 300 kg sliding along a horizontal path. Calculate the coefficient of friction between the log and the path. 0.07

WORKED

Example

8

12 A body of mass 6 kg rests on a plane inclined at an angle of 40° to the horizontal. It is just prevented from moving by friction. Calculate the: a weight 58.8 N b normal reaction 45.04 N c frictional force 37.8 N d coefficient of friction. 0.84

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13 A book lies on a horizontal table. One end of the table is raised until, at an angle of 35°, the book starts to slide. Calculate the coefficient of friction between the book and the table. 0.7 14 An object of mass 2 kg rests on a plane inclined at an angle of 40° to the horizontal. If the coefficient of friction between the object and the plane is 0.2, calculate the resultant force down the plane. 9.6 N

Momentum Kirsten is an investigator with the Traffic Accident Squad and she is called to investigate the collision between a large four-wheel-drive vehicle and a small hatchback. Luckily no one was hurt in the accident but there was considerable damage to the vehicles. An eyewitness report suggests that the small vehicle was speeding. The diagram below indicates the direction of travel of the vehicles. A indicates the direction of motion of the small car before the accident. B indicates the direction of motion of the A large car. C indicates the direction of motion of both 45º vehicles after impact. B 25º C What conclusion can Kirsten make concerning the speeds of the vehicles before impact? By investigating the accident scene Kirsten will be able to determine whether any of the cars were speeding. In particular she will look for any skid marks and look at the motion of the vehicles after collision. A key mathematical concept that will help her to understand the events leading up to the accident is momentum.

Definition of momentum One of the properties of a moving body is momentum. The momentum of a moving body is given by: Momentum = mass ¥ velocity Because velocity is a vector quantity, momentum is also a vector quantity; that is, it has both magnitude and direction. The standard units of momentum are N s (newton seconds) where mass is in kg and velocity is in m/s. (Alternative units for momentum are kg m/s.)

WORKED Example 9

Calculate the momentum of a car of mass 2.5 tonnes travelling north at 20 m/s. THINK 1 Use standard units. Convert tonnes to kilograms. 2 Use the formula for calculating the magnitude of momentum. 3

Momentum is a vector quantity so include direction as well as magnitude.

WRITE Mass = 2.5 t = 2500 kg Momentum = mass × velocity = 2500 × 20 = 50 000 N s Momentum = 50 000 N s in a northerly direction.

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WORKED Example 10 A tennis ball of mass 80 g is hit at a wall with a speed of 50 m/s. It rebounds with a speed of 40 m/s. What is the change in momentum? THINK

WRITE

1

State the required quantities in standard units. Mass should be in kg and velocity in m/s.

Mass = 80 g = 0.08 kg Initial velocity = 50 m/s towards the wall Final velocity = 40 m/s away from the wall

2

Calculate the initial momentum.

Initial momentum = mass × initial velocity = 0.08 × 50 = 4 N s towards the wall

3

Calculate the final momentum.

Final momentum = mass × final velocity = 0.08 × 40 = 3.2 N s away from the wall

4

Calculate the change in momentum by subtracting the initial momentum from the final momentum.

Change in momentum = final momentum − initial momentum = (3.2 N s away from the wall) − (4 N s towards the wall) = (3.2 N s away from the wall) − (−4 N s away from the wall) = 7.2 N s away from the wall

Conservation of momentum Momentum is a useful concept because it is conserved in collisions. That is, if two bodies, A and B, collide their total momentum before collision is equal to their total momentum after the collision. Total momentum before collision = total momentum after collision

WORKED Example 11 A car weighing 1.9 tonnes is travelling east at 22 m/s. A second car travelling north and weighing 2.4 tonnes passes through a stop sign at 14 m/s. It collides with the first car. If after the collision the cars move off together, calculate their final speed and direction. THINK 1 Momentum is conserved: the total momentum before collision is equal to the total momentum after collision. 2

First calculate the momentum of each vehicle before collision.

WRITE

Momentum of first car = 1900 × 22 Momentum of first car = 41 800 N s east Momentum of second car = 2400 × 14 Momentum of second car = 33 600 N s north Continued over page

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THINK 3 Find the total momentum before the collision by adding vectors.

WRITE

x 33 600 N s

θ 41 800 N s 2

2

x = 33 600 + 41 800 = 53 630 33 600 tan θ = ---------------41 800 θ = 38.8° Total momentum before collision = 53 630 N s at 38.8° north of east 4

Since momentum is conserved, state the total momentum after the collision.

5

Calculate the final speed (magnitude of the final velocity) by rearranging the formula for momentum. The mass of the cars as they move off together is the sum of the individual masses.

6

Write the final speed and direction of the cars after the collision.

Hence, total momentum after collision = 53 630 N s at 38.8° north of east momentum Final speed = --------------------------mass 53 630 Final speed = ----------------------------------( 1900 + 2400 ) Final speed = 12.5 m/s The final speed and direction of the vehicles as they move off together is 12.5 m/s at 38.8° north of east.

WORKED Example 12 Mass A, 2 kg, collides with mass B, 3 kg, moving as shown. a Calculate the momentum of each body before the collision. b Calculate the total momentum of the system before the collision.

A

6 m/s B

If, after the collision, mass A moves as shown, c calculate the final speed and direction of mass B.

30º

8 m/s A

40º

B x m/s

THINK a Calculate the momentum of A and the momentum of B before the collision, using the formula momentum = mass × velocity.

10 m/s

θ

WRITE a Before the collision: Momentum of A = 2 × 10 = 20 N s, to the right Momentum of B = 3 × 6 = 18 N s, 30° left of the initial = direction of A

C h a p t e r 8 Ve c t o r a p p l i c a t i o n s

THINK b 1 Draw a vector diagram showing the situation before the collision. Add the vectors for the momentum of A and momentum of B (head to tail) to produce the vector for the total momentum T . ˜ 2 Use the cosine and sine rules to calculate T and α, the magnitude and direction of the total momentum before the collision.

c

1

2

3

4

5

6

7

377

WRITE b

T

18 N s

˜

α 20 N s

30º

T 2 = 182 + 202 − 2 × 18 × 20 cos 150° = 1347.5 T = 36.7 N s sin α sin 150° ------------ = ------------------18 36.7 sin α = 0.24523 α = 14.2° Total momentum before the collision is 36.7 N s, 14.2° left of the initial direction of A. Calculate the momentum of A after c After the collision: the collision. Let B represent the Momentum of A = 2 × 8 momentum of B after the collision. = 16 N s, 40° left of the initial direction of A Let B represent the momentum of B after the collision. Since the momentum is conserved, state Total momentum after the collision = the total momentum after the collision. 36.7 N s, 14.2° left of the initial direction of A That is, momentum before the collision = momentum after the collision. Draw a vector diagram showing the θ B Ns situation after the collision. Add the β 16 N s vectors for the momentum of A and 40º 36.7 N s ω momentum of B (head to tail) to produce 14.2º the vector for the total momentum. Calculate B, the momentum of B2 = 162 + 36.72 − 2 × 16 × 36.7 cos ω mass B, using the cosine rule. = 162 + 36.72 − 2 × 16 × 36.7 cos 25.8° Note that the angle marked ω is B = 23.4 N s 40° − 14.2° = 25.8°. The speed of B can be calculated Momentum = mass × velocity from the momentum by using 23.4 = 3 × x momentum = mass × velocity. x = 7.8 m/s To calculate θ, note that as marked in the diagram β = θ +14.2°. First calculate β using the sine rule. State the final speed and direction of B after the collision.

sin β sin 25.8° ------------ = --------------------16 23.4 sin β = 0.2976 β = 17.3° Therefore θ = 3.1°. The final speed and direction of B is 7.8 m/s in a direction 3° right of the initial path of A as shown.

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Conservation of momentum using i and j notation ˜ ˜ Consider the problem posed in Worked example 12. Recalculate the answers by expressing the momentum vectors in i and j notation. ˜ ˜ Using our understanding of momentum we return to the scene of the accident with Kirsten (see page 374). The driver of the four-wheel-drive B vehicle, B, says she was travelling at 25º 45º 70 km/h (19.4 m/s). Further, the j A ~ masses of both vehicles are known: the C ~i mass of vehicle A is 1640 kg, the mass of vehicle B is 2060 kg. Momentum is conserved. That is, in vector terms, momentum of A + momentum of B = momentum of C. Momentum of B = mass × velocity = 2060 kg × (19.4 i + 0 j ) ˜ ˜ = 39 964 i + 0 j . ˜ ˜ Let the momentum of A before impact be A and let the momentum of both vehicles after collision be T. Then since momentum before = momentum after 39 964 i + A cos 45° i − A sin 45° j = T cos 25° i − T sin 25° j ˜ ˜ ˜ ˜ ˜ Therefore 39 964 + A cos 45° = T cos 25° [1] and A sin 45° = T sin 25° [2] Substituting [2] in [1] (noting that cos 45° = sin 45°) 39 964 + T sin 25° = T cos 25° T cos 25° − T sin 25° = 39 964 T(cos 25° − sin 25°) = 39 964 39 964 T = ---------------------------------------------( cos 25° – sin 25° ) = 82 623 N s Substituting in [2] gives A sin 45° = 82 623 sin 25° A = 49 381.5 N s Thus the speed of A before collision = momentum of A ÷ mass of A = 49 381.5 ÷ 1640 = 30.1 m/s = 108.4 km/h Kirsten concludes that car A was speeding.

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379

remember remember 1. The momentum of a moving body is given by the equation momentum = mass × velocity 2. Momentum is a vector quantity — it has both magnitude and direction. The units of momentum are N s or kg m/s. 3. Momentum is conserved in collisions: total momentum before collision = total momentum after collision.

8C WORKED

Example

9

Momentum

1 Calculate the momentum of the following objects whose mass and velocity are given: a mass = 2.5 kg, velocity = 16 m/s east 40 N s east b mass = 4 kg, velocity = 150 m/s south 600 N s south c mass = 250 g, velocity = 30 m/s north 7.5 N s north d mass = 2.5 tonnes, velocity = 13 m/s west 32 500 N s west e mass = 88 kg, velocity = 40 km/h north 978 N s north f mass = 3.4 tonnes, velocity = 120 km/h north 113 333 N s north 2 Calculate the speed of the following objects with momentum and mass given: a momentum = 2000 N s, mass = 25 kg 80 m/s b momentum = 3000 N s, mass = 80 g 37 500 m/s c momentum = 42 000 N s, mass = 2.1 tonnes 20 m/s 3 Calculate the mass of the following objects with momentum and speed given: a momentum = 2500 N s, speed = 50 m/s 50 kg b momentum = 22 000 N s, speed = 40 km/h 1980 kg 4 Give the total momentum of the following systems. a object A: mass = 3 kg, velocity = 10 m/s east object B: mass = 2 kg, velocity = 4 m/s east 38 N s east b object A: mass = 3 kg, velocity = 10 m/s east object B: mass = 2 kg, velocity = 4 m/s west 22 N s east c object A: mass = 3 kg, velocity = 10 m/s east object B: mass = 2 kg, velocity = 4 m/s north 31 N s 14.9° N of E d object A: mass = 3 kg, velocity = 10 m/s east object B: mass = 2 kg, velocity = 4 m/s 30° north of east 37.1 N s 6.2° N of E e object A: mass = 3 kg, velocity = 10 m/s east object B: mass = 2 kg, velocity = 4 m/s north of west 25 N s 13° N of E

WORKED

Example

10

5 At the carnival, a dodgem car with a mass of 300 kg with its driver travels at 6 m/s and crashes into a wall. If it rebounds at 3 m/s calculate the change of momentum. 2700 N s 6 A cyclist (plus bicycle, pack, clothes, etc.) has a combined mass of 100 kg. He is travelling at 10 m/s. a Find his momentum. 1000 kg m/s b If the cyclist slows down to 5 m/s, what is the change in his momentum? −500 kg m/s

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7 Find the change of momentum for: a a mass of 10 kg which slows from 12 m/s to 8 m/s −40 kg m/s b a mass of 8 kg which accelerates from 7 m/s to 13 m/s 48 kg m/s c a car of mass 0.95 tonne which increases speed from 44 km/h to 80 km/h. 9500 kg m/s d a horse of mass 400 kg which slows from a gallop of 50 km/h to a trotting speed of 18 km/h. −3556 kg m/s WORKED

Example xample

11

8 Laurie is about to collide, at 90°, with Alan. After the collision the two players move as one unit before coming to rest. Laurie has a mass of 85 kg and Alan has a mass of 100 kg. Calculate the common speed and direction of the two players immediately after the collision. 5.3 m/s at 44° to Alan’s current direction.

Laurie 8 m/s

Alan 7 m/s

WORKED

Example xample

12

9 Two dodgem cars collide. The mass of car A and its driver is 320 kg while the mass of car B and its driver is 280 kg. Calculate the final speed and direction of car B in each of the following situations. Before collision

9 a 6.6 m/s N b 6 m/s N c 3.4 m/s S

1

8

After collision

Car A

Car B

Car A

Car B

a

8 m/s north

2 m/s north

4 m/s north

?

b

8 m/s north

2 m/s south

1 m/s north

?

c

2 m/s north

8 m/s south

2 m/s south

?

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10 Using the data from question 8, assume that Laurie does not tackle Alan at right angles but rather at 60° as shown. Calculate the common velocity of the two players after the collision. 3.7 m/s at 58° to Alan’s current direction. Laurie 8 m/s Alan

60º 7 m/s

11 Two billiard balls, A and B, each of mass 100 g, collide and then move off. Use the principle of conservation of momentum to calculate the unknown values in the table below. θ

(The direction of motion is specified by giving the angle with the positive x-direction.) Before collision A

After collision B

A

B

v m/s

θ

v m/s

θ

v m/s

θ

2.2 at 47.9°

a

4

30°

0

—

2

10°

5.9 at 71.4°

b

4

60°

2

0°

3

−45°

1.67 at 65.4°

c

3

30°

2

150°

1

80°

2.4 at 234°

d

3

30°

4

−120°

2

0°

v m/s

θ

12 The principle of momentum can be used to calculate the amount of ‘recoil’ experienced when a weapon is fired. The momentum of a bullet and rifle before firing = the momentum of the bullet and rifle after firing. If a bullet has a mass of 25 g and is fired from a rifle at 180 m/s: a calculate the momentum of the rifle immediately after firing 4.5 N s b calculate the speed of the rifle immediately after firing if the rifle has a mass of 1.5 kg. 3 m/s

Collision momentum See Solutions Manual.

In the analysis of the accident scene, Kirsten concluded that the smaller vehicle was speeding. Her analysis is based on the evidence of the person driving the larger vehicle who claimed her speed was 70 km/h. What if this evidence was inaccurate? Imagine you are the accident investigator and have to write a report of the accident. In your report consider alternative scenarios in which the driver of car B was travelling at 70 km/h, at greater than 70 km/h and at less than 70 km/h.

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Relative velocity

SLE 3: Use addition and subtraction in life-related situations such as the effect of current flow on a boat; consider the concept of relative velocity.

As you sit at your desk reading this paragraph it appears that you are not moving significantly. While this may be apparent to someone in the room watching you, an observer with a different point of reference may disagree entirely. As you sit in the room you are moving around the Sun at about 100 000 km/h. Measurement of velocity depends on the frame of reference. Most people have experienced the sensation of movement while sitting in a stationary train and observing a nearby train moving. The visual messages tell our brain we are moving and, for a moment, we become disoriented. Consider the situation of a boy rowing a boat across a swiftly flowing river. The boy thinks he is rowing directly towards the bank opposite but to an observer on the shore his velocity is different. The situation can be described using vectors. Velocity of boy relative to river

Velocity of river Velocity of boy

In general the relationship between relative velocities is

where v a rel b ˜

v a = v a rel b + v b ˜ ˜ ˜ is the velocity of a ( v a ) relative to the velocity of b ( v b ). ˜ ˜

WORKED Example 13 The pilot of a boat heads due north at a speed of 12 km/h with respect to the water. The water moves at a velocity of 5 km/h in an easterly direction. Calculate the velocity of the boat as seen by an observer on the shore. THINK 1

2

3

The velocity of the boat relative to the water (vb rel w) is 12 m/s north. The velocity of the water (vw) is 5 km/h east. We want to find the velocity of the boat, vb. Use the relative velocity formula.

Write the answer as a speed and direction.

WRITE v b rel w = 12 j ˜ vw = 5 i ˜ ˜ ˜

vb

12

θ 5

v b = v b rel w + v w ˜ ˜ ˜ = 12 j + 5 i ˜ ˜ 2 2 | vb | = 12 + 5 = 13 12 tan θ = -----5 θ = 67.4° An observer on the shore sees the boat moving at 13 km/h at 67.4° north of east.

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383

WORKED Example 14 A pilot is to fly a plane to a destination which is 450 km from his present position in a direction N20°W. There is a wind from the east at 22 km/h and the plane has an airspeed of 300 km/h. In what direction should the plane head to reach the destination?

THINK 1 Use the relative velocity formula and draw a diagram. The wind velocity is v b and the ˜ v airspeed of the plane is a rel b . ˜ The speed of the plane, relative to the ground is v a . ˜

2

Use the sine rule: a b c -----------= ------------ = ------------sin A sin B sin C

WRITE v a = v a rel b + v b ˜ ˜ ˜ 22

300

va

20º

22 70º

300 x

sin 70° ----------x- = sin ---------------22 300 sin x = 0.0689 x = 3.95° The plane should head 16.05° west of north or N16.05°W.

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remember remember The equation for determining the velocity of object a relative to object b is v a = v a rel b + v b where v a rel b is the velocity of a ( v a ) relative to the velocity of ˜ ˜ ˜ b˜ ( v b ).˜ ˜

8D

Relative velocity

1 The pilot of a boat heads due east at a speed of 16 km/h with respect to the water. The water moves south at a velocity of 4 km/h. Calculate the velocity of the boat as seen by 13 an observer on the shore. 16.5 km/h at 14° S of E

WORKED

Example xample

2 A plane with an airspeed of 300 km/h heads due north. A wind blows from the west at 30 km/h. What is the velocity of the plane relative to the ground? 301.5 km/h at 5.7° W of N 3 A rowing crew heads across river at 15 m/s. The crew heads in a direction 45° north of west. At the same time the current flows at 3 m/s due north. What is the speed and direction of the boat on the river? 17.3 m/s at 52.1° N of W 4 The driver of a cross-river ferry wants to head to a pier directly across the stream, at right angles to the bank. His boat can travel at 20 km/h and there is a current flowing 14 downstream at 6 km/h. At what angle to the bank should the driver head in order to travel directly across the river? 72.5°

WORKED

Example xample

5 A passenger on a cruise ship walks at 6 km/h towards the stern (the back) of the ship. The ship travels at 19 km/h in a north-westerly direction. What is the speed of the passenger with respect to the water? 13 km/h 6 Copy and complete the following table. va ˜

vb ˜

v a rel b ˜

a

40 m/s N

20 m/s E

44.7 m/s at 26.6° W of N

b

5 m/s N

15 m/s S

20 m/s N

c

25 m/s NE

20 m/s E

d

4 m/s SE

32.95 m/s at 4.9° S of E

e

3 m/s 30° S of E

5 m/s 25° W of N

17.8 m/s at 7.6° W of N

30 m/s W 7.7 m/s at 52° S of E

7 A cyclist rides north at 15 km/h and observes that the wind appears to come to him from the north-east. On the return journey he rides at the same speed in the opposite direction. Now the wind appears to be coming from the south-east. Calculate the true speed and direction of the wind. 15 km/h W

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8 A cyclist rides north at 15 km/h and observes the wind appears to come to him from the north-east. On the return journey he rides at the same speed in the opposite direction. Now the wind appears to be coming from 30° south of east. Calculate the true speed and direction of the wind. 19.4 km/h 78° W of S eBook plus Digital doc: WorkSHEET 8.2

9 As a jogger runs along level ground at 9 km/h the rain appears to be heading directly towards her at an angle of 10° with the vertical. When she turns around and travels at the same speed in the opposite direction the rain still appears to be coming directly towards her but now makes an angle of 5° with the vertical. Calculate the speed and direction of the rain. 68.3 m/s at 2.5° with the vertical.

Using vectors in geometry

SLE 6: Solve problems from geometry using vectors.

The rules for addition, subtraction and dot product of vectors can be used to prove theorems and other general statements in geometry.

WORKED Example 15 In the drawing at right, ABC is a triangle. Point D is along the line BC such that BD = 1--3- BC. The vectors q , r and t are as shown ˜ ˜ in the diagram. Prove that: t = 1--3- (2 q + r ).˜ ˜ ˜ ˜

C

~r A

~t q ~

D B

THINK 1 2

WRITE

Express the line joining B to D as a vector v and the ˜ line joining B to C as a vector s . ˜ Redraw the diagram with v and s indicated. ˜ ˜ A

t ~ q ~

8

Express s as the vector sum of – q and r . ˜ ˜ ˜ Since D is one-third of the way from B to C, v = ˜ Express t as the vector sum of q and v . ˜ ˜ ˜ Substitute v = 1--- s into the expression for t . ˜ ˜ 3˜ Substitute s = –q + r into t = q + 1--- s . 3 ˜ ˜ ˜ ˜ ˜ ˜ Simplify by expanding the bracket.

9

Collect ‘like’ vectors.

10

‘Factorise’ the vectors, by removing common scalar factors. The required expression for t is obtained. ˜

3 4 5 6 7

1 --3

s. ˜

C

r ~

s ˜ v ˜ t ˜ t ˜ t ˜ t ˜ t ˜ t ˜

s ~ vD ~

= –q+r ˜ ˜ = 1--3- s ˜ = q+v ˜ ˜ = q + 1--3- s ˜ ˜ = q + 1--3- ( –q + r ) ˜ ˜ ˜ = q – 1--3- q + 1--3- r ˜ ˜ ˜ = 2--3- q + 1--3- r ˜ ˜ = 1--3- ( 2q + r ) ˜ ˜

B

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WORKED Example 16

Consider the isosceles triangle at right, where AB = BC. Show, using the properties of vectors only, that line BD, drawn such that BD is

B

perpendicular to AC, divides AC in half, that is: AD = DC .

A

THINK 1 Express AD as a vector, in terms of other

2

AD = AB + BD

AD • AD = ( AB + BD ) • AD

with AD itself. 2

2

3

Simplify, and recall that AD • AD = AD .

4

Since BD is perpendicular to

But BD • AD = 0 .

AD , BD • AD = 0 .

so

Express AB • AD using the definition of dot product.

6

Divide both sides by AD .

C

WRITE

vectors. (In this example, AD is the vector joining A to D.) Take the dot product of this expression

5

D

AD

= AB • AD + BD • AD

AD

2

= AB • AD = AB AD cos θ

AD = AB cos θ where θ is the angle between AB and AD .

7 8

Express CD as a vector in terms of other vectors. Take the dot product of this expression with CD itself.

9

Simplify.

10

Since BD and CD are perpendicular, BC • CD = 0 .

11

Express CB • CD in dot product form.

CD = CB + BD CD • CD = ( CB + BD ) • CD CD

2

= CB • CD + BD • CD

But BD • CD = 0 . CD

2

= CB • CD = CB CD cos α

12

Divide both sides by CD .

CD = CB cos α where α is the angle between CB and CD .

13

Write down the respective expressions for AD and CD .

14

Since the triangle is isosceles,

AD = AB cos θ CD = CB cos α But AB = CB and θ = α.

AB = CB and θ = α. 15

It can be concluded that AD = CD .

Therefore AD = CD .

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387

WORKED Example 17

Use a vector method to show that the diagonals of a rectangle bisect each other. THINK 1 Construct a rectangle ABCD.

WRITE B ~b A

C E a ~

D

Let the vector from A to D be a and ˜ the vector from A to B be b . ˜ Let E be the midpoint of diagonal AC.

Let AD = a ˜ Let AB = b ˜ Let E be the midpoint of AC.

4

Find AC in terms of a and b . ˜ ˜

AC = a + b ˜ ˜

5

Find AE which is half of AC in terms

AE = --12- ( a + b ) ˜ ˜

2

3

of a and b . ˜ ˜ 6

Determine BE in terms of a and b . ˜ ˜

BE = –b + 1--2- ( a + b ) ˜ ˜ ˜ 1 = --2- ( a – b ) ˜ ˜

7

Determine ED in terms of a and b . ˜ ˜

ED = – 1--2- ( a + b ) + a ˜ ˜ ˜ = 1--2- ( a – b ) ˜ ˜

8

Note that ED = BE .

= BE

9

This means that BE and ED are collinear and of equal length since they are equal vectors.

So E is the midpoint of BD.

10

State your conclusion.

The diagonals of a rectangle bisect each other.

Three-dimensional non-zero vectors See Solutions Manual.

If a , b and c are 3-dimensional non-zero vectors: ˜ ˜ ˜ 1 Show geometrically that: a+b+c ≤ a + b + c ˜ ˜ ˜ ˜ ˜ ˜ 2 Prove algebraically that: a + b ≥ a+b ˜ ˜ ˜ ˜ 3 Prove algebraically that: a•b ≤ a b ˜ ˜ ˜ ˜ Under what circumstances are all of the above statements equal?

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Vector geometry See Solutions Manual.

In any triangle ABC, the points D, E and F divide the lines BC, CA and AB respectively in the ratio 1:2. A

F

R E

Q P

C B

D

1 Show that the vector joining A to P is

6 --7

of the vector joining A to D.

2 Show that the vector joining R to P is

3 --7

of the vector joining A to D.

3 Can you calculate the ratio of the area of triangle QRP to the area of triangle ABC?

remember remember In vector proofs, it is simpler to use pronumerals to represent vectors, but not essential. Use as few vectors as necessary in completing a proof and apply the rules for vector addition, subtraction and dot products as appropriate.

8E WORKED

Example

15 Check with your teacher.

SLE 6: Solve problems from geometry using vectors: for example; prove that the angle of a semicircle is a right angle; Pythagoras’ theorem; the concurrency of (a) the medians and (b) the bisectors of the internal angles of a triangle.

Using vectors in geometry

1 In the triangle ABC, point D is the midpoint of BC. The vectors q , r and t are defined as indicated. ˜ Show that: ˜ ˜ t = ˜

1 --- r 2˜

+

q ~

˜

1 (n – 1) t = --- r + ----------------q n˜ n ˜ ˜

~t

A

1 --- q 2

2 ABC is a triangle with point D along the line BC such that: 1 BD = --- BC n The vectors q , r and t are as shown in the diagram. ˜ Prove that: ˜ ˜

C

~r

B

C

r ~ A

D

t ~ q ~

D B

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C h a p t e r 8 Ve c t o r a p p l i c a t i o n s

3 Consider the triangle ABC. Point D is the midpoint of line AB and E is the midpoint of line BC. Show, using vectors to represent various lines, that line DE is parallel to AC and DE = 1--- AC. Check with your teacher. 2

4 In the figure at right, D is the midpoint of CB and AD is perpendicular to CB. Let u be the vector joining D to A and v ˜ ˜ the vector joining D to B. Find, in terms of u and v only: ˜ ˜ a the vector, a , joining D to C − v˜ ˜ b the vector, b , joining C to A u + v ˜ ˜ ˜ c the vector, c , joining B to A u˜ − v˜ ˜ b.b = u2 + v2 d the dot products, b • b and c • c . ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ Hence, show that AC = AB. c .c = u2+ v2 ˜ ˜

5–12 Check with your teacher.

˜

B D

Example

16

C

A C

A

D

B

B

˜

5 In the figure at right, a is the vector joining point A to point ˜ B, b joins B to C and c joins C to A. ˜ ˜ Prove that a + b + c = 0 . ˜ ˜ ˜

~a

A WORKED

E

6 Consider the equilateral triangle at right. Show, using the properties of vectors only, that line BD, drawn such that BD is perpendicular to AC, divides AC in half, that is:

~b

~c

C

B

AD = DC . A

D

C

7 Show that the diagonals of a non-square rhombus intersect at right angles. (Hint: Make a drawing; a rhombus is a parallelogram with all 4 sides of equal length.) 8 If the length of a vector, a , is given by a , show geometrically that for any two ˜ ˜ vectors a and b : ˜ ˜ a + b ≥ a+b ˜ ˜ ˜ ˜ This is known as the triangle inequality. WORKED

Example

17

9 Use vectors to show that the diagonals of a parallelogram bisect each other. 10 Use vectors to show that the angle subtended by a diameter of a circle is a right angle. 11 Use a vector method to show that by joining the midpoints of a parallelogram, the figure formed is a parallelogram. 12 Consider any two major diagonals of a cube. Use a vector method to show that: a the diagonals bisect each other b the acute angle between the diagonals is 70.53°. 13 Use vectors to demonstrate Pythagoras’ theorem.

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summary Force diagrams and the triangle of forces • Force is a vector quantity. Its derived unit is the Newton. 1 N = 1 kg m/s2. • Types of force: Field forces occur without physical contact. A common field force is the gravitational force often referred to as ‘weight’: W = mg down. The magnitude of ˜ ˜ g is 9.8 m/s2. ˜ Normal contact forces occur between objects in contact or via strings where they are known as tensile forces. Friction forces occur when there is actual or attempted ~F1 relative movement between two objects in contact. ~F1 • The resultant force R is the vector sum of all physical forces acting on an object. ˜A force vector diagram illustrates this with two forces F 1 and F 2 : ~F2 R ˜ ˜ ~ = ~F1 + ~F2 F • A force F can be resolved into horizontal ˜ components. and vertical F = Fx + Fy ˜ ˜ ˜ = F cos θ i + F sin θ j ˜ ˜

~2

j ~ ~i ~F

θ

~Fy

~Fx

Newton’s First Law of Motion • Newton’s First Law of Motion is used for equilibrium or statics problems: (the resultant force is zero) R =0 (the magnitude of acceleration is zero) a˜ = 0 (the magnitude of velocity is a constant). v =c • Newton’s First Law of Motion: In the absence of an unbalanced force (that is, R = 0) a moving body moves in a state of uniform motion (or constant velocity). ˜ • Friction = µ × normal reaction, where µ is the coefficient of friction F =µ× N ˜ ˜

Momentum • The momentum of an object is the product of its mass and velocity. Momentum = mass × velocity • In a collision the momentum of the system is conserved: momentum of system before collision = momentum of system after collision • Momentum is a vector quantity. Its units are N s or kg m/s.

Relative velocity • The velocity of an object measured by an observer depends on the frame of reference of the observer. In general: v a = v a rel b + v b ˜ ˜ ˜ where v a rel b is the velocity of a relative to b. ˜

Using vectors in geometry • In proofs using vectors it is simpler to use pronumerals to represent vectors. • Use as few vectors as necessary in completing a proof and apply the rules for vector addition, subtraction and dot products as appropriate.

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CHAPTER review 1 multiple choice

8A

A body is in equilibrium under the action of three forces f 1 , f 2 and f 3 . ˜ ˜ ˜ f 1 = −3 i + 7 j and f 2 = 5 i + 2 j ˜ ˜ ˜ ˜ ˜ ˜ The magnitude of the vector f 3 is ˜ A 2i − 9 j B C 77 D − 85 85 ˜ ˜

17

E

2 multiple choice

8A

If the three forces shown are in equilibrium, the magnitude of the force F is A 34.6 N D 54.3 N

B 40 N E 80 N

20 N F

60°

C 20 2 N

20 N

3 multiple choice

8A

Two forces, equal in magnitude, are sufficient to keep the mass shown at right in equilibrium against the force of gravity. The magnitude of each force is A 8.9 N

B 10.8 N

D 19.6 N

E 22.8 N

50°

C 17.8 N

2 kg

4 An object is in equilibrium under the action of three forces f 1 , f 2 and f 3 . ˜ ˜ ˜ Calculate the magnitude and direction of f 3 . 89.4 N at 153.4° to f 1 ˜ ˜

˜

f3

˜ 138.6°

8A

f2 = 40 N f1 = 80 N

˜

5 Three coplanar forces have magnitudes of 400 N, 500 N and 600 N. They act on a body such that the resultant force is zero. Find the angle between the 500 N and 600 N forces.

8A

6 A body of mass 6.0 kg is suspended from a ceiling by two ropes. The angle between the two ropes is 90° and they are connected to the ceiling at points A and B respectively. The tension T2, in rope 2, is twice the tension T1, in rope 1. a Draw a force vector diagram for the 6.0 kg mass. b Calculate the magnitudes of the tensions T1 and T2. 26.3 N, 52.6 N c Find the angles that rope 1 and rope 2 make with the horizontal. 26.6°, 63.4°

8B

5x d If rope 1 has a length x, show that the distance AB is ---------- . 2

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8B

7 A mass of 4 kg rests on a plane which has a coefficient of friction of 0.25. Calculate the minimum force P , acting ˜ slipping parallel to the plane, needed to prevent the mass down the plane. 17.7 N

8C

8 multiple choice

P

˜

40°

The momentum of an 85 kg sprinter who runs 100 metres in 10 seconds is A 0.85 N s B 10 N s C 85 N s D 850 N s

8C

9 multiple choice

8C

10 multiple choice

8C

11 multiple choice

8C

12 The diagram shows the view from above as a ball, travelling horizontally, rebounds at a wall. The mass of the ball is 150 g and its speed, before and after the collision, is 20 m/s. Calculate the change in the momentum of the ball. 3 N s away from the wall

E 8500 N s

A tennis ball of mass 85 g hits a wall at 40 m/s and rebounds at 30 m/s. The change in momentum of the tennis ball is A 10 N s B 5.95 N s C 3.4 N s D 2.55 N s E 0.85 N s

How fast would an object of mass 45 g have to move to have the same momentum as a car of mass 1170 kg moving at 100 km /h? A approximately 500 000 m/s B approximately 600 000 m/s C approximately 700 000 m/s D approximately 800 000 m/s E approximately 900 000 m/s

A car of mass 1250 kg moves at 70 km/h in a northerly direction. It collides with a car of mass 1300 kg moving at 80 km/h in an easterly direction. The magnitude of the momentum of the system before the collision is closest to A 23 457 N s B 24 543 N s C 29 992 N s D 37 747 N s E 43 450 N s 30°

30°

8C

13 A 100 kg rugby player, A, tackles an opposition player, B, at right A B v = 8 m/s angles as shown. The opposition player has a mass of 85 kg. After the tackle they fall together. Calculate the direction and speed of their joint motion after the tackle. 6.0 m/s at 43.7° with A’s original motion v = 9 m/s

8D

14 multiple choice

8D

If v a = 2 i + 3 j and v b = −2 i + 2 j , the value of v a rel b is ˜ ˜ ˜ ˜ ˜ ˜ ˜ A 5j B 4i + j C −4 i − j D j ˜ ˜ ˜ ˜ ˜ ˜ 15 multiple choice

E

5

A traveller in a car is moving in a northerly direction. He observes a second vehicle and to him it appears to be travelling in a south-easterly direction. Which of the following could be possible values for the true direction of the second vehicle? A North B South C East D West E South-west

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16 A river flows from north to south at 5 km/h. A boat heads directly across the river from the river bank to the other side at 11 km/h. Find the true velocity of the boat. 12.1 km/h at 65.6°

8D

17 A cyclist travels at 15 km/h on a road heading east. When she is travelling in this direction the wind appears to be coming from a direction 60° E of N. When she turns around and travels west on the road at the same speed the wind appears to come from 60° W of N. Find the true speed and direction of the wind. 8.7 km/h from the north

8D

18 multiple choice

8E

to the bank

Which of the statements regarding p is true? A p =b −a B˜ p = 2 b − 4 a ˜ ˜ ˜ ˜ ˜ ˜ C p = 1--- b + 1--- a D p = 1--- b + 2--- a 2 ˜ 2 ˜ 3 ˜ 3 ˜ ˜ ˜ 2--1 E p = b + --- a 3 ˜ 3 ˜ ˜ 19 Show that the diagonals of a square bisect each other at 90°. 20 For any vectors (u + v ) • (u − ˜ ˜ ˜

A a

˜

2 cm

p

4 cm

˜ b

B

˜

8E 8E

u and v show that ˜ ˜ v ) = | u | 2 − | v |2 ˜ ˜ ˜

Modelling and problem solving 1 Arnie is pushing against a trailer, preventing it from rolling down a hill. The trailer has a mass of 200 kg and the hill is on an incline of 15° to the horizontal. At the moment there is no problem because Arnie is capable of pushing with a force of 1000 N parallel to the plane. However, it is raining and the trailer is filling with water at a rate of 25 litres per minute. How long will Arnie be able to hold the trailer and stop it from running down the hill? 7 minutes 27 seconds

2 When the surf gets big at Kirra there is always a sweep running from south to north.

Take-off area Sweep = 4 km/h 120 m Water Shore

Jodie is heading out at 8 km/h. If she heads directly out to sea and she wants to get to the take-off area, how far up the beach should she walk before paddling out? 60 m

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3 A ball of mass 0.20 kg is shot vertically in the air. It decelerates under the action of two forces: the weight force and the force of air resistance. When the ball moves with a speed of 40 m/s, it has an air resistance of 1.0 N. When the ball is stationary, the air resistance force is equal to zero. a Determine the magnitude and direction of the resultant force acting on the ball when it is moving upwards at a speed of 40 m/s. 2.96 N down b Determine the magnitude and direction of the resultant force acting on the ball when it is at its maximum height above the ground. 1.96 N down Later, the ball is travelling toward the ground at 40 m/s. c Determine the magnitude and direction of the resultant force acting on the ball when it is moving downwards at a speed of 40 m/s. 0.96 N down 4 In a James Bond movie scene, a car of mass 1500 kg rolls in a straight line from rest down a road inclined at 10° from the horizontal. It takes 10 seconds for the car to reach the bottom of the incline at a speed of 43.2 km/h where the road becomes level. The handbrake of the car was on, providing a constant retardation force of 20g newtons. a Find the value of the coefficient of friction correct to 3 decimal places. 0.038 At the bottom of the incline the car jolts, and the handbrake is rendered inactive. b How much further does the car travel? (Give your answer to the nearest tenth of a metre.) 193.5 m D (3, 4, − 1_3) 5 A pyramid, OABCD, is shown in the figure at right. The height of the pyramid is the length of DE, where E is the point on the C B (−2, 5, −2) base OABC such that DE is perpendicular to the base. (2, 6, 2) E a Show that the base, OABC, is a rhombus. Check with your teacher. O (0, 0, 0) A (4, 1, 4) b Use a vector method to find ∠AOC correct to the nearest tenth of a degree. 109.5° c The unit vector p i + q j + r k , p > 0, is perpendicular to both OA and OC. 1 1 - ; r = − ------p = -----i Show that q =˜0 and˜ find ˜the exact values of p and r. 2 2 5 2 ii Hence find the exact height of the pyramid. ---------- units 3

6 A parallelogram, OXYZ, has O at the origin. The vector joining O to Z is given by 6 i while ˜ the vector joining O to X is given by 3 i + 5 j . i ZY i j YX = − 6 ; = 3 + 5 ˜ ˜ a Sketch the parallelogram, labelling all vertices. ˜ ˜ ˜ b State the vectors joining Z to Y and Y to X. OY = 9 i + 5 j ; ZX = −3 i + 5 j c What vectors represent the diagonals of the parallelogram? ˜ ˜ ˜ ˜ d Find the angle between the diagonals to the nearest tenth of a degree. 91.9° e Find the angle that OY makes with the x-axis. 29.1° eBook plus f State the vector resolute of the vector joining O to X, perpendicular to OZ. 5 ˜j Digital doc: g Let P be a point on the line ZY, such that the vector joining P to X is perpendicular to ZX. Test Yourself Find the exact coordinates of P. (12 3--8- , 10 5--8- ) Chapter 8 h Find the exact area of the triangle XPZ. 31 7--8- square units a y X (3, 5)

Y (9, 5)

Z (6, 0) O

x

Sequences and series

9 syllabus reference Core topic: Structures and patterns

In this chapter 9A Arithmetic sequences 9B Geometric sequences 9C Applications of geometric sequences 9D Finding the sum of an infinite geometric sequence 9E Contrasting arithmetic and geometric sequences through graphs Fibonacci Sequence The Mandelbrot Set

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Introduction

• • • •

sum to infinity of a geometric progression purely mathematical and life-related applications of arithmetic and geometric progressions sequences and series other than arithmetic and geometric recognition of patterns in well-known structures including Pascal’s Triangle and Fibonacci sequence • applications of patterns

Discovered in 1980 by Benoit Mandelbrot, the Mandelbrot Set (see the image on page 446) is one of the most intricate and beautiful geometrical patterns in mathematics. The Mandelbrot Set is an image that captures many of the qualities people find fascinating about mathematics. Although it is generated by repeating a simple formula, its patterns are infinitely complex. If you select any portion of the Mandelbrot Set and magnify it you will see that no detail is lost — the magnified shape is as intricate and even contains parts that look like copies of the original. This notion of ‘worlds within worlds’ appeals to the philosopher in all of us. How is the Mandelbrot Set created? Using some of the concepts developed in the following sections we will see how a computer or a graphics calculator can produce this fascinating image.

Arithmetic sequences A sequence in mathematics is an ordered set of numbers. An arithmetic sequence is one in which: 1. the difference between any two successive terms is the same 2. the next term in the sequence is found by adding the same number. Consider the arithmetic sequence: 4, 7, 10, 13, 16, 19, 22. The difference between each successive term is +3, or similarly, the next term is found by adding 3 to the previous term. We can see that a positive common difference gives a sequence that is increasing. We say that the common difference is +3, stated as d = +3. +3 4

+3 7

+3 10

+3 13

+3 16

+3 19

22

The first term of the sequence is 4. We refer to the first term of a sequence as ‘a’. So in this example, a = 4. In the arithmetic sequence above, the first term is 4, the second term is 7, the third term is 10 and so on. Another way of writing this is: t1 = 4, t2 = 7 and t3 = 10. There are 7 terms in this sequence. Because there are a countable number of terms in the sequence, it is referred to as a finite sequence. The arithmetic sequence below –7 37,

–7 30,

–7 23,

–7 16,

9...

is an infinite sequence since it continues endlessly as indicated by the dots. The first term, a, is 37 and the common difference, d, is −7. We can see that a negative common difference gives a sequence that is decreasing. 1. An arithmetic sequence is a sequence of numbers for which the difference between successive terms is the same. 2. The first term of an arithmetic sequence is referred to as ‘a’. 3. The common difference between successive terms is referred to as ‘d’. 4. tn is the term number, for example, t6 refers to the 6th term in the sequence.

Chapter 9 Sequences and series

397

WORKED Example 1 Which of the following are arithmetic sequences? a 7, 13, 19, 25, 31, . . . b 1.3, 2.5, 3.7, 4.9, 6.3, . . . c −1 1--2- , −1, − 1--2- , 0, 1--2- , . . . THINK

WRITE

a

a 7, 13, 19, 25, 31, . . .

b

c

1

Write the sequence.

2

Calculate the difference between the first term, t1, and the second term, t2.

t2 − t1 = 13 − 7 =6

3

Calculate the difference between the second term, t2, and the third term, t3.

t3 − t2 = 19 − 13 =6

4

Calculate the difference between the third term, t3, and the fourth term, t4.

t4 − t3 = 25 − 19 =6

5

Calculate the difference between t5 and t4.

t5 − t4 = 31 − 25 =6

6

Check that the differences are the same and write your answer.

There is a common difference of 6, therefore d = 6. This is an arithmetic sequence.

1

Write the sequence.

2

Calculate the difference between t2 and t1.

2.5 − 1.3 = 1.2

3

Calculate the difference between t3 and t2.

3.7 − 2.5 = 1.2

4

Calculate the difference between t4 and t3.

4.9 − 3.7 = 1.2

5

Calculate the difference between t5 and t4.

6.3 − 4.9 = 1.4

6

Check that the differences are the same.

There is no common difference. This is not an arithmetic sequence.

1

Write the sequence.

b 1.3, 2.5, 3.7, 4.9, 6.3, . . .

c −1 1--- , −1, − 1--- , 0, 1--- , . . . 2

2

2

2

Calculate the difference between t2 and t1.

−1 −

3

Calculate the difference between t3 and t2.

− 1--- − (−1) =

4

Calculate the difference between t4 and t3.

0 − (− 1--- ) =

5

Calculate the difference between t5 and t4.

1 --2

6

Check that the differences are the same.

There is a common difference of --1- . 2 This is an arithmetic sequence.

(−1 1--- ) 2

=

2

2

−0=

1 --2

1 --2

1 --2

1 --2

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WORKED Example 2

Write down the value of a and d for each of the following arithmetic sequences. a 1.2, 3.6, 6, 8.4, 10.8, . . . b −1 2--5- , − 2--5- , 3--5- , 1 3--5- , 2 3--5- , . . . THINK WRITE a 1 Write the sequence. a 1.2, 3.6, 6, 8.4, 10.8, . . . a = 1.2 2 What is the first term? t2 − t1 = 3.6 − 1.2 3 What is the difference between t2 and t1? You need check only once as the question = +2.4 states that this is an arithmetic sequence. d = +2.4 The arithmetic sequence has a first term, 4 Write your answer. a, of 1.2 and a common difference, d, of +2.4. b 1 Write the sequence. b −1 2--- , − 2--- , 3--- , 1 3--- , 2 3--- , . . . 5

3

What is the first term? What is the difference between t2 and t1?

4

Write your answer.

2

5

5

5

5

a = −1 2--5 t2 − t1 = − 2--- − (−1 2--- ) 5 5 = +1 d = +1 The arithmetic sequence has a first term, a, of 1 2--- and a common difference, d, 5 of +1.

Finding the terms of an arithmetic sequence Consider the arithmetic sequence for which a = 8 and d = 10. +10

+10 8

18

+10 28

+10 38

48

t1 = 8 t1 = a t2 = 8 + 10 t2 = a + d t2 = a + 1d t3 = 8 + 10 + 10 t3 = a + d + d t3 = a + 2d t4 = 8 + 10 + 10 + 10 t4 = a + d + d + d t4 = a + 3d t5 = 8 + 10 + 10 + 10 + 10 t5 = a + d + d + d + d t5 = a + 4d We notice a pattern emerging. That pattern can be described by the equation: tn = 8 + (n − 1) × 10 where n represents the number of the term. For example, if n = 4, then the fourth term is: t4 = 8 + (4 − 1) × 10 t4 = 8 + 3 × 10 t4 = 38 Therefore, the 4th term is 38. We can generalise this rule for all arithmetic sequences. Now,

where

tn = a + (n − 1)d tn is the nth term a is the first term d is the common difference.

Chapter 9 Sequences and series

399

This rule enables us to find any term of an arithmetic sequence provided we know the value of a and d.

WORKED Example 3 Find the 20th term of the following arithmetic sequence. 5, 40, 75, 110, 145, . . . THINK 1 Find the value of a. 2 Find the value of d. You need to calculate only one difference as the question states that it is an arithmetic sequence. 3 Use the rule tn = a + (n − 1)d where n is 20 for the 20th term. 4

Write the answer.

WRITE a=5 d = t2 − t1 = 40 − 5 = 35 t20 = 5 + (20 − 1) × 35 = 5 + 19 × 35 = 670 The 20th term is 670.

If we are given only two terms of an arithmetic sequence, we are able to use the rule tn = a + (n − 1)d to set up two simultaneous equations to find the value of a and d and hence write down the rule for the arithmetic sequence.

WORKED Example 4

The third term of an arithmetic sequence is −1 and the fifth term is 11. a Write down the rule for the arithmetic sequence. b Find the 50th term of the sequence. THINK WRITE a 1 We know that t3 = −1 and that a t3 = a + 2d = −1 tn = a + (n − 1)d. 2 We know that t5 = 11 and that t5 = a + 4d = 11 tn = a + (n − 1)d. a + 2d = −1 3 Solve the 2 equations simultaneously using the elimination technique. a + 4d = 11 Eliminate a, by subtracting equation [1] 2d = 12 from equation [2]. d=6 Substituting d = 6 into [1]: 4 Evaluate a by substituting d = 6 into either of the two equations. a + 12 = −1 a = −13 tn = −13 + (n − 1) × 6 5 To find the rule, substitute values for a and d into tn = a + (n − 1) d. tn = −13 + 6n − 6 tn = −19 + 6n b tn = −19 + 6n b 1 To find the 50th term or t50, substitute n = 50 into the rule. t50 = −19 + 6 × 50 = −19 + 300 = 281 The 50th term is 281. 2 Write your answer.

[1] [2] [2] − [1]

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Graphics Calculator tip!

Listing the terms of an arithmetic sequence

If you know the rule for an arithmetic sequence, successive terms can be listed using a graphics calculator. The steps for generating these terms are shown below. Consider an example where the first term is 3 and the difference is 2 so the rule is tn = 3 + (n − 1) × 2.

For the Casio fx-9860G AU 1. Press MENU and select RUN-MAT. Press OPTN then F1 (LIST) followed by F5 (Seq). 2. Enter the rule (3 + (n − 1) × 2) followed by the variable name (n), the start value for n (1), the end value of n (say, 10) and the increment (1), each separated by a comma ( , ). Press ) to close the set of brackets. (To enter the variable n, press ALPHA [N].) 3. Press EXE to display the list of the first 10 terms. Use the down arrow key to scroll down the list.

For the TI-Nspire CAS 1. Open a new Lists & Spreadsheet document. Press b, then 3: Data.

2. Press 1: Generate Sequence and complete at least the first two entry boxes, pressing e to move to the next line. Enter 3 + (n − 1) × 2 for the Formula and 3 for the Initial Terms. Completing the boxes for Max No. Terms (the default value is 255) and Ceiling Value is optional. Press e until OK is highlighted. 3. Press · to display the list of terms. Use the NavPad to scroll down the list.

4. If you wish to widen Column A, use the up arrow on the NavPad to highlight the column and then press b followed by 1: Actions, 2: Resize and 1: Resize Column Width. Press the right arrow until you reach the required width and press ·.

Chapter 9 Sequences and series

401

The sum of a given number of terms of an arithmetic sequence When the terms of an arithmetic sequence are added together, an arithmetic series is formed. So, 5, 9, 13, 17, 21, . . . is an arithmetic sequence whereas 5 + 9 + 13 + 17 + 21 + . . . is an arithmetic series. The sum of n terms of an arithmetic sequence is given by Sn. Consider the finite arithmetic sequence below. 5, 10, 15, 20, 25, 30, 35, 40, 45, 50 The sum of this arithmetic sequence is given by S10 since there are 10 terms in the sequence. So, S10 = 5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50 = 275 Note that the sum of the first and last terms is 55. Also, the sum of the second and second last terms is 55. Similarly, the sum of the third and third last term is 55. This pattern continues with the fourth and fourth last terms as well as with the fifth and fifth last terms. There are in fact five lots of 55. We can formalise this pattern to obtain a rule which applies to all arithmetic sequences. Let Sn = a + (a + d) + (a + 2d) + . . . + (l − 2d) + (l − d) + l where l is the last term of the sequence. By reversing the order of the series above, we obtain Sn = l + (l − d) + (l − 2d) + . . . + (a + 2d) + (a + d) + a. By adding these two equations, we obtain 2Sn = (a + l) + (a + d + l − d) + (a + 2d + l − 2d) + . . . (a + l) + (a + l) + (a + l) 2Sn = (a + l) + (a + l) + (a + l) + . . . (a + l) + (a + l) + (a + l) 2Sn = n(a + l) where n represents the number of terms in the sequence. So, S n = 1--2- n ( a + l ) . The sum of n terms of an arithmetic sequence with a as its first term and l as its last term is given by n S n = --- ( a + l ) . 2 Now we know that the nth term of an arithmetic sequence is given by t n = a + ( n – 1 )d . So, for the sum of n terms, l is the last term; that is, tn = l. So, the last term is l = a + (n − 1)d. n n Substituting this into Sn = --- ( a + l ) we obtain Sn = --- { a + [ a + ( n – 1 )d ] } 2 2 n = --- [ 2a + ( n – 1 )d ] . 2 An alternative formula for the sum of n terms of an arithmetic sequence when the value of a and d are known is given by n S n = --- [ 2a + ( n – 1 )d ] . 2

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WORKED Example 5 Find the sum of the first ten given terms of the arithmetic sequence 4, 10, 16, 22, 28, 34, 40, 46, 52, 58. THINK Method 1 1 We know the values of the first and last term and that there are ten terms in the series. n 2 Use the series formula Sn = --- ( a + l ) . 2

WRITE a=4 l = 58 n = 10 n Sn = --- ( a + l ) 2 10 = ------ ( 4 + 58 ) 2 = 5 × 62 = 310

Method 2 (alternative) 1 We know the value of a and d and n.

2

3

n Use the formula Sn = --- [ 2a + ( n – 1 )d ] . 2

a=4 d = 10 − 4 = 6 n = 10 n Sn = --- [ 2a + ( n – 1 )d ] 2

Write the answer.

10 S10 = ------ [ 2 × 4 + ( 10 – 1 )6 ] 2 S20 = 5[8 + 9 × 6] = 5[8 + 54] = 5 × 62 = 310 The sum of the first 10 terms is 310.

the sum of an Graphics Calculator tip! Finding arithmetic sequence For the arithmetic sequence 4, 10, 16, 22, …, 58 considered in Worked example 5, we can use a graphics calculator to find the sum of the first 10 terms. The rule for this sequence is 4 + (n − 1) × 6.

For the Casio fx-9860G AU 1. Press MENU and select RUN-MAT. Press OPTN then F1 (LIST) followed by F5 (Seq). Enter the rule (4 + (n − 1) × 6) followed by the variable name (n), the start value (1), the end value (10) and the increment (1), each separated by a comma ( , ). Press ) to close the set of brackets.

Chapter 9 Sequences and series

s

2. To store this sequence of 10 terms as List 1, press Æ then F1 (List) and 1 followed by EXE .

s

3. Press F6 ( ) twice to display the Sum option and then press F1 (Sum).

4. Press F6 ( ) again for more options and then F1 (List) and 1 to indicate List 1. Press EXE to display the sum of the first 10 terms. The sum of the first 10 terms is 310.

For the TI-Nspire CAS 1. Open a new Lists & Spreadsheet document. Press b, then 3: Data followed by 1: Generate Sequence. Enter the rule 4 + (n − 1) × 6 with an initial term of 4. Press e until OK is highlighted.

2. Press · to display the list of terms. Use the NavPad to scroll across to cell B1.

3. To calculate the sum of the first 10 terms, we can add the numbers in cells A1 to A10. Enter the formula = sum(A1:A10) in cell B1. (Press =SUM(A1:A10).)

4. Press · to display the sum in cell B1. The sum of the first 10 terms is 310.

403

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

remember remember 1. In an arithmetic sequence: (a) the first term is a (b) the common difference is d. 2. Given an unspecified sequence, establish whether it is arithmetic by testing all terms for a common difference: d = t2 − t1 = t3 − t2 = t4 − t3 = . . . 3. To find the term of an arithmetic sequence use the following formula: tn = a + (n − 1) d where tn is the nth term a is the first term d is the common difference. 4. A series is the sum of terms in a sequence. 5. Sn is the sum of the first n terms in a series; for example, S25 represents the sum of the first 25 terms. 6. Given the number of terms in a series, n, the first term, a, and the last term, l, n use Sn = --- ( a + l ) . 2 7. Given the number of terms in a series, n, the first term, a, and the common n difference, d, use Sn = --- [ 2a + ( n – 1 )d ] . 2

9A WORKED

Example xample

1

1 State which of the following are arithmetic sequences. a 2, 7, 12, 17, 22, . . . b 3, 7, 11, 15, 20, . . . c 0, 100, 200, 300, 400, . . . d −123, −23, 77, 177, 277, . . . e 1, 0, −1, −3, −5, . . . f 6.2, 9.3, 12.4, 15.5, 16.6, . . . g 1--- , 1 1--- , 2 1--- , 3 1--- , 4 1--- , . . . h 1--- , 3--- , 1 1--- , 1 3--- , 2 1--- , . . . 2

WORKED

Example xample

Arithmetic sequences

2

2

2

2

4

4

4

4

1 a, c, d, g, h

4

2 For those arithmetic sequences found in question 1, write down the values of a and d.

2 WORKED

Example xample

3

WORKED

Example xample

4

3 For each of the arithmetic sequences given, find: a the 25th term of the sequence 2, 7, 12, 17, 22, 122 ... b the 30th term of the sequence 0, 100, 200, 300, 400, 2900 ... c the 33rd term of the sequence 5, −2, −9, −16, −23,−.219 ..

2 a a = 2, d = 5 c a = 0, d = 100 d a = −123, d = 100 g a = 1--2- , d = 1 h a = --14- , d =

4 Evaluate the following. a The 2nd term of an arithmetic sequence is 13 and the 5th term is 31. What is the 17th term of this sequence? 103 b The 2nd term of an arithmetic sequence is −23 and the 5th term is 277. What is the 20th term of this sequence? 1777 c The 2nd term of an arithmetic sequence is 0 and the 6th term is −8. What is the 32nd term of this sequence? −60 d The 3rd term of an arithmetic sequence is 5 and the 7th term is −19. What is the 40th term of this sequence? −217 e The 4th term of an arithmetic sequence is 2 and the 9th term is −33. What is the 26th term of this sequence? −152

--24

Chapter 9 Sequences and series

WORKED

Example

5

405

5 For each of the given series, find: a the sum of the first 20 terms of the sequence 3, 7, 11, 15, 19, 820 ... 270 b the sum of the first 15 terms of the sequence 4, 6, 8, 10, 12, . . . c the sum of the first 23 terms of the sequence −4, −1, 2, 5, 8, 667 ... 6 multiple choice The 41st term of the arithmetic sequence −4.3, −2.1, 0.1, 2.3, 4.5, . . . is: A 83.7 B 85.9 C 92.3 D 172.4 E 178.5 7 multiple choice The 2nd term of an arithmetic sequence is −2 and the 5th term is 2.5. The 27th term of this sequence is: A 32.5 B 35.5 C 42.5 D 89.5 E 96 8 multiple choice The sum of the first 21 terms of the sequence, 0, −3 1--- , −7, −10 1--- , −14, . . . is: 2 2 A −1470 B −735 C −700 D 36.75 E 735 9 multiple choice The first term of an arithmetic sequence is −5.2 and the 2nd is −6. The sum of the first 22 terms of the sequence is: A −598.4 B −299.2 C −242 D −70.4 E 70.4

10 State which of the following situations are arithmetic sequences. a A teacher hands out 2 lollies to the first student, 4 lollies to the second student, 6 lollies to the third student and 8 lollies to the fourth student. b The sequence of numbers after rolling a die 8 times. c The number of layers of paper after each folding in half of a large sheet of paper. d The house numbers on the same side of a street on a newspaper delivery route. 11 e The cumulative total of the number of seats in the first ten rows in a regular a a = 2, d = 2 d a = not specified, d = 2 cinema (for example, with 8 seats in each row, so there are 8 seats after the first e a = 8, d = 8 row, 16 seats after the first 2 rows, and so on).

10 a, d, e

11 For those arithmetic sequences found in question 10, where appropriate information is given, write down the value of a and d. 7th

12 For the following arithmetic sequences: a 4, 13, 22, 31, . . . which term, tn, will be equal to 58? b 9, 4.5, 0, . . . which term, tn, will be equal to −18? 7th c −60, −49, −38, . . . which term, tn, will be the first to be greater than 10? 8th d 100, 87, 74, . . . which term, tn, will be the first to be less than 58? 5th 13 A batsman made 23 runs in his first innings, 33 in his second and 43 in his third. If he continued to add 10 runs each innings, write down a rule for the number of runs he would have made in his nth innings. tn = 13 + 10n

406 tn = 37 + 3n

SLE 12: Investigate the use of an arithmetic progression in real-life situations.

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

14 In a vineyard, rows of wire fences are built to support the vines. The length of the fence in row 1 is 40 m, the length of the fence in row 2 is 43 m, and the length of the fence in row 3 is 46 m. If the lengths of the fences continue in this pattern, write down a rule for the length of a fence in row number n.

15 The first fence post in a fence is 12 m from the road, the next is 15.5 m from the road and the next is 19 m from the road. The rest of the fence posts are spaced in this pattern. a Write down a rule for the distance of fence post n from the road. tn = 8.5 + 3.5n b If 100 posts are to be erected, how far will the last post be from the road? 358.5 metres 16 The 1st term of an arithmetic sequence is 2 and the sum of the first 19 terms of the sequence is 551. Find: a the 19th term 56 b the first 3 terms of the sequence. 2, 5, 8 17 The 1st term of an arithmetic sequence is −4 and the sum of the first 30 terms of the sequence is 2490. Find: a the 30th term 170 b the first 3 terms of the sequence. −4, 2, 8 18 Sam makes $100 profit in his first week of business. If his profit increases by $75 each week, what would his profit be in total by the end of week 15? $9375 19 George’s salary is to start at $36 000 a year and increase by $1200 each year after that. How much will George have earned in total after 10 years? $414 000 20 A staircase is designed so that the height of each step increases by 0.8 cm for each step. If the height of the first step is 15 cm, what is the total height of the first 17 steps? 363.8 cm 21 Paula collects stamps. She bought 250 in the first month to start her collection and added 15 stamps to the collection each month thereafter. How many stamps will she have collected after 5 years? 1135 22 Proceeds from the church fete were $3000 in 1990. In 1991 the proceeds were $3400 and in 1992 they were $3800. If they continued in this pattern: a what were the proceeds from the year 2009 fete? $10 600 b how much in total would the proceeds from church fetes since 1990 have amounted to in the year 2009? $136 000

Geometric sequences A sequence in mathematics is an ordered set of numbers. A geometric sequence is one in which the first term is multiplied by a number, known as the common ratio, to create the second term which is multiplied by the common ratio to create the third term, and so on. The first term in a geometric sequence is referred to as a and the common ratio is referred to as r. Consider the geometric sequence where a = 1 and r = 3. The terms in the sequence are ×3

×3 1

3

×3 9

×3 27

81...

Chapter 9 Sequences and series

407

To discover the common ratio, r, of a geometric sequence you need to calculate the ratio t t t of successive terms, namely ---2- . You could alternatively calculate ---3- or ---4- and so on. t1 t2 t3 A geometric sequence is a sequence of numbers for which the ratio of successive terms is the same. t t t ---2- = ---3- = ---4- = º = common ratio t1 t2 t3 The first term of a geometric sequence is referred to as a. The common ratio between a term and its preceding term is referred to as r.

WORKED Example 6 Which of the following are geometric sequences? a 2, 10, 50, 250, 1250, . . . b −2, −6, 18, 54, −162, . . . THINK

WRITE

a

a 2, 10, 50, 250, 1250, . . . t ---2- = 10 -----2 t1

1

Write the sequence.

2

t Calculate the ratio of ---2- . t1

= 5 3

t Calculate the ratio of ---3- . t2

t ---3- = 50 -----10 t2 = 5

4

t Calculate the ratio of ---4- . t3

t ---4- = 250 --------50 t3 = 5

t Calculate the ratio of ---5- . t4

t ---5- = 1250 -----------250 t4

6

Check that all ratios are the same.

= 5 There is a common ratio of 5. This is a geometric sequence.

1

Write the sequence.

2

t Calculate the ratio of ---2- . t1

3

t Calculate the ratio of ---3- . t2

5

b

4

There is no need to check any further as the two ratios are not the same.

b −2, −6, 18, 54, −162, . . . t ---2- = –-----6–2 t1 = +3 t 18 ---3- = ----t2 –6 = –3 There is no common ratio. This is not a geometric sequence.

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WORKED Example 7 Write down the value of a and r for each of the following geometric sequences. a 1.2, −4.8, 19.2, −76.8, 307.2, . . . 9b −1, 3--- , − ----, 4

16

27 ------ , 64

81− -------, ... 256

THINK

WRITE

a

a 1.2, −4.8, 19.2, −76.8, 307.2, . . . a = 1.2

1 2 3

b

Write the sequence. a is the first term. t To find r use the ratio ---2- or any other t1 ratio.

t 2 – 4.8 ---- = ---------1.2 t1 = −4 or t 3 19.2 ---- = ---------t 2 – 4.8 = −4 This is a geometric sequence with the first term 1.2 and a common ratio of −4.

4

Write your answer.

1

Write the sequence.

2

a is the first term.

a = −1

t To find r use the ratio ---2- or any other t1 ratio.

--t2 4 ---- = ----t 1 –1

3

9b −1, 3--- , − ----, 4

16

27 ------ , 64

81− -------, ... 256

3

= − 3--4

or

9t 3 – ----16 ---- = ------3 t2 --4

9= – ----× 16

--43

= − 3--4

4

This is a geometric sequence with the first

Write your answer.

term −1 and a common ratio of − 3--- . 4

Finding the terms of a geometric sequence Consider the finite geometric sequence of seven terms for which a = 3 and r = 4. ×4

×4 3

Now,

t1 = 3 t2 = 3 × 4 t3 = 3 × 4 × 4 t4 = 3 × 4 × 4 × 4 t5 = 3 × 4 × 4 × 4 × 4

12

×4 48

×4 192

384

t1 = a t2 = a × r t3 = a × r × r t4 = a × r × r × r t5 = a × r × r × r × r

t2 = a × r1 t3 = a × r 2 t4 = a × r 3 t5 = a × r4

and so on . . .

409

Chapter 9 Sequences and series

We notice a pattern emerging. That pattern can be described by the equation: tn = 3 × 4n − 1. For example, if n = 5,

t5 = 3 × 44. We can generalise this rule for all geometric sequences. tn = ar n − 1 where tn is the nth term a is the first term r is the common ratio. This rule enables us to find any term of a geometric sequence provided we know the value of a and r.

WORKED Example 8 Find the 12th term of the geometric sequence: 2, 10, 50, 250, 1250, . . . THINK

WRITE

1

Find the value of a.

a=2

2

It has been stated that it is a geometric sequence, so find the value of r.

10 r = -----2 =5

3

Use the rule tn = a × r n − 1 to find the 12th term.

4

Write your answer.

t12 = 2 × 512 − 1 = 97 656 250 The value of the 12th term is 97 656 250.

WORKED Example 9

The 2nd term of a geometric sequence is 8 and the 5th is 512. Find the 10th term of this sequence. THINK 1 2 3

WRITE

We know that t2 = 8 and that tn = a × r n − 1.

t2 = a × r1 =8

We know that t5 = 512 and that tn = a × r n − 1.

t5 = a × r4 = 512

Solve the 2 equations simultaneously by eliminating a, to find r. Divide equation 2 by equation 1.

a × r1 = 8 a × r4 = 512 a------------× r 4- 512 = --------a×r 8 3 r = 64 r=4

[1] [2] [2] ÷ [1]

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THINK 4 To find a, substitute the value of r.

5

Write down the rule.

6

To find the 10th term, let n = 10.

7

Write your answer.

WRITE Substituting r = 4 into equation [1]: a×4=8 a=2 tn = 2 × 4n − 1 t10 = 2 × 49 = 524 288 The 10th term in the sequence is 524 288.

WORKED Example 10

The first three terms of a geometric sequence are 2, 6, and 18. Which numbered term would be the first to exceed 1 000 000 in this sequence? THINK

WRITE/DISPLAY

Method 1: Using logarithms 1

To find the rule for the sequence, find a and r and substitute them into tn = a × r n − 1.

a=2 t r = ---2 t1 = 6--2

=3 tn = 2 × 3n − 1 2

Set up the equation to be solved.

3

Express 3n − 1 and 500 000 in terms of logarithms to the base 10.

4

The next whole number term is the 13th.

2 × 3n − 1 = 1 000 000 3n − 1 = 500 000 log 3n − 1 = log 500 000 (n − 1) log 3 = log 500 000 log 500 000 (n − 1) = ---------------------------log 3 n − 1 = 11.9445 n = 12.9445 The 13th term would be the first to exceed 1 000 000.

Method 2: Trial and error 1

Find the rule for the sequence. See Method 1.

2

Set up the equation to be solved.

3

Try various values of n. With n = 8, value is less than 1 000 000. With n = 15, value is greater than 1 000 000. With n = 12, value is too small. With n = 14, value is too large. With n = 13, value just exceeds 500 000.

a = 2 and r = 3 tn = 2 × 3n − 1 2 × 3n − 1 = 1 000 000 3n − 1 = 500 000 Let n = 8, 377 = 2187 (too small) Let n = 15, 314 = 4 782 969 (too large) Let n = 12, 311 = 177 147 (too small) Let n = 14, 313 = 1 594 323 (too large) Let n = 13, 312 = 531 441 The 13th term is the required term.

Chapter 9 Sequences and series

THINK

411

WRITE/DISPLAY

Method 3: Using a graphics calculator 1

Find the rule for the sequence. See Method 1.

2

Locate the term that equals or exceeds 1 000 000. Two methods are shown for this example.

a = 2 and r = 3 tn = 2 × 3n−1

Method A: Generating the terms of the sequence For the Casio fx-9860G AU (a) Press MENU and select RUN-MAT. Press OPTN then F1 (LIST) followed by F5 (Seq). Enter the rule (2 × 3n − 1) followed by the variable name (n), the start value (1), the end value (say, 20) and the increment (1), each separated by a comma. Press ) to close the set of brackets. (b) Press EXE to display the list of terms.

(c) Scroll down until you reach the term that exceeds 1 000 000.

For the TI-Nspire CAS (a) Open a new Lists & Spreadsheet document. Press b, then 3: Data followed by 1: Generate Sequence. Enter the rule 2 × 3n − 1 with an initial term of 2. Press e until OK is highlighted.

(b) Press · to display the list of terms. Increase the width of Column A to view the terms more easily. (Press b then select 1: Actions, 2: Resize and 1: Resize Column Width.)

Continued over page

412

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THINK

WRITE/DISPLAY

(c) Scroll down until you reach the term that exceeds 1 000 000.

Method B: Solving an equation Set up an equation to solve.

2 × 3n − 1 = 1 000 000

For the Casio fx-9860G AU (a) Press MENU and select EQUA. Press F3 (SOLV) to select F3: Solver. A previous equation and its solution will be displayed.

(b) Enter the required equation in the first line and press EXE . Notice that even though the variable in the second line has changed to N, the value shown is not the calculated solution.

(c) Press F6 (SOLV) to display the solution. (If you wish to use the Solver again, press F1 (REPT).)

For the TI-Nspire CAS Open a new Calculator document. Use the solve function to find n when 2 × 3n − 1 = 1 00 000. Press b and select 3: Algebra followed by 1: Solve (or type the word solve and press (). Enter the equation to be solved and then the variable, separated by a comma. Press ) to close the set of brackets and press ·.

3

Write your answer. Note that the value of the 13th term may appear as 1.06E6 or 1.06288E6 which is approximately 1.06 × 106 or 1 060 000. It is written in exponential form since it is too large a number to fit on the screen.

The first term to exceed 1 000 000 is the 13th.

Chapter 9 Sequences and series

413

The sum of a given number of terms of a geometric sequence SLE 1: Establish the formula for the sum to n terms of a geometric progression.

When the terms of a geometric sequence are added, a geometric series is formed. So 3, 6, 12, 24, 48, . . . is a geometric sequence whereas 3 + 6 + 12 + 24 + 48 + . . . is a geometric series. The sum of n terms of a geometric sequence is given by Sn. Consider the general geometric sequence a, ar, ar 2, ar 3, . . . ar n − 1. Now, Sn = a + ar + ar 2 + ar 3 + . . . + ar n − 1. Also, multiplying each term by r, rSn = ar + ar 2 + ar 3 + ar 4 + . . . + ar n. So, rSn − Sn = −a + ar n since all the other terms cancel out. So, Sn (r − 1) = a(r n − 1) a(rn – 1) Sn = ---------------------r–1 This formula is useful if r < −1 or r > 1, for example, if r is 2, 10, 3.3, −4, −1.2. By calculating Sn − rSn instead of rSn − Sn, as we did earlier, we obtain an alternative form of the formula. That is, S n – rS n = a – ar n Sn( 1 – r ) = a( 1 – r n ) a(1 – rn) S n = ---------------------1–r This formula is useful if r is in between −1 and 1 (shown as −1 < r < 1). The sum of n terms, Sn, of a geometric sequence may be calculated using a( rn – 1 ) Sn = ---------------------- if r < −1 or r > 1 r–1 or a( 1 – rn ) Sn = ---------------------- if −1 < r < 1. 1–r

WORKED Example 11 Find the sum of the first 9 terms of the sequence 0.25, 0.5, 1, 2, 4, . . . THINK 1 Find the value of a. 2

Find the value of r by testing ratios of the given terms.

3

a(r n – 1) Since r > 1, use S n = ---------------------- . r–1

4

Write the answer.

WRITE/DISPLAY a = 0.25 t2 t 0.51---- = -----------3- = -----t 1 0.25 t 2 0.5 =2 =2 t5 4 t ---4- = 2------ = --t3 1 t4 2 =2 =2 9 0.25 ( 2 – 1 ) S9 = -----------------------------1 = 127.75 The sum of the first 9 terms is 127.75.

414

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remember remember 1. In a geometric sequence: (a) the first term is a,

2.

3.

4.

5.

t (b) the common ratio is r = ---2- . t1 Given an unspecified sequence, establish whether it is geometric by testing all t t t terms for a common ratio, r = ---2- = ---3- = ---4- = … t1 t2 t3 To find the terms of a geometric sequence use the following formula: tn = ar n − 1 where tn is the nth term a is the first term r is the common ratio. When the terms of a geometric sequence are added, a geometric series is formed. 2, 4, 8, 16, 32, 64 is a finite geometric sequence. 2 + 4 + 8 + 16 + 32 + 64 is a finite geometric series. The sum of n terms, Sn, of a geometric sequence may be calculated using a(rn – 1) S n = ---------------------- if r < –1 or r > 1, for example, r = –2, – 3--- , + 2 or + 1.5 2 r–1 or a(1 – rn) S n = ---------------------- if −1 < r < 1, for example, r = 0.2, 1–r

9B

1 --8

or −0.25.

Geometric sequences

2 b c e f h

a = 1, r = 4 a = −1, r = 2 a = 4, r = −3 a = −6, r = −10 a = 1.2, r = 2

1 State which of the following are geometric sequences. i a = 7, r = 1--2a 2, 6, 12, 24, 48, . . . b 1, 4, 16, 64, 256, . . . 6 j a = --12- , r = --12c −1, −2, −4, −8, −16, . . . d 1, 3, −9, 27, −81, . . . e 4, −12, 36, −108, 324, . . . f −6, 60, −600, 6000, −60 000, . . . 1 b, c, e, f, h, i, j g −5, 15, −25, 45, −85, . . . h 1.2, 2.4, 4.8, 9.6, 19.2, . . . 1- ----i 7, 3.5, 1.75, 0.875, 0.4375, . . . j 1--- , 1--- , 1--- , ----, 1- , . . . WORKED

Example xample

2

WORKED

Example xample

7 WORKED

Example xample

8

4

8

16

32

2 Look again at the geometric sequences found in question 1. Write down the values of a and r. 3 Find the value of the term specified for the given geometric sequences. a Find the 10th term of the geometric sequence 2, 12, 72, 432, 2592, . . . 20 155 392 b Find the 18th term of the geometric sequence 8, 16, 32, 64, 128, . . . 1 048 576 c Find the 11th term of the geometric sequence 5, 15, 45, 135, 405, . . . 295 245 d Find the 10th term of the geometric sequence 2.3, 2.76, 3.312, 3.9744, . . . 11.867 494 81 e Find the 9th term of the geometric sequence −2, −8, −32, −128, −512, . . . −131 072

Chapter 9 Sequences and series

WORKED

Example

9

WORKED

Example

10 eBook plus Digital doc: SkillSHEET 9.1 Solving indicial equations

WORKED

Example

11

415

4 Find the value of the term specified in each of the following geometric sequences. a The 2nd term of a geometric sequence is 6 and the 5th term is 162. Find the 10th term. 39 366 b The 2nd term of a geometric sequence is 6 and the 5th term is 48. Find the 12th term. 6144 c The 2nd term of a geometric sequence is 2 and the 5th term is 16. Find the 16th term. 32 768 d The 4th term of a geometric sequence is −32 and the 7th term is −256. Find the 14th term. −32 768 e The 4th term of a geometric sequence is −192 and the 7th term is −12 288. Find the 12th term. −12 582 912 f The 3rd term of a geometric sequence is 36 and the 6th term is −972. Find the 12th term. −708 588 5 Evaluate the following. a The first three terms of a geometric sequence are 5, 12.5 and 31.25. Which term would be the first to exceed 50 000? 12th b The first three terms of a geometric sequence are 3.2, 9.6 and 28.8. Which term would be the first to exceed 1 000 000? 13th c The first three terms of a geometric sequence are 5.1, 20.4 and 81.6. Which term would be the first to exceed 100 000? 9th d The first three terms of a geometric sequence are 4.3, 9.46 and 20.812. Which term would be the first to exceed 500 000? 16th 6 a Find the sum of the first 12 terms of the geometric sequence 2, 6, 18, 54, 162, . . . 531 440 b Find the sum of the first 7 terms of the geometric sequence 5, 35, 245, 1715, 12 005, . . . 686 285 c Find the sum of the first 15 terms of the geometric sequence 1.1, 2.2, 4.4, 8.8, 17.6, . . . 36 043.7 d Find the sum of the first 11 terms of the geometric sequence 3.1, 9.3, 27.9, 83.7, 251.1, . . . 274 576.3 7 multiple choice There is a geometric sequence for which a is positive and r = −2. It is true to say that: A only one term of the sequence is a positive number B the 3rd term will be a negative number C the 3rd term will be less than the 2nd term D the 5th term will be greater than the 6th term E the 4th term will be greater than the 3rd term. 8 multiple choice The 12th term of the geometric sequence 21, 63, 189, 567, . . . is: A 6804 B 413 343 C 1 240 029 D 3 720 087

E 5 931 980 229

9 multiple choice The sum of the first 10 terms of the geometric sequence 2.25, 4.5, 9, 18, 36, . . . is closest to: A 1149.75 B 2301.75 C 5318.81 D 6648.51 E 8342.65 10 multiple choice The 2nd term of a geometric sequence is −20 and the 5th is −1280. The sum of the first 12 terms of the sequence is: A −27 962 025 B −1 062 880 C −15 360 D 1 062 880 E 16 777 215

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11 On the first day Jenny hears a rumour. On the second day, she tells two friends. On the third day, each of these two friends tell two of their own friends, and so on. a Write the geometric sequence for the first five days of the above real-life situation. 1, 2, 4, 8, 16 b Find the value of r. 2 c How many people are told of the rumour on the 12th day? 2048

20 million, 10 million, 5 million, 2 1--2- million, 1 1--4- million, 625 000, 312 500

12 Decay of radioactive material is modelled as a geometric sequence where r = there are 20 million radioactive atoms, write the first 7 terms of the sequence.

1 --- . 2

If

13 The number of cells of a micro-organism, after each process of cell division, can be summarised as follows. 1, 2, 4, 8, 16 If the number of cells after each division continues to follow a geometric sequence, find: a a rule for the number of cells after n divisions tn = 2n − 1 b the number of cells after 12 divisions. 2048 14 The takings at a new cinema each month are recorded. If the takings each month continue to follow a geometric sequence, find: tn = 10 000 × 0.85n − 1 a a rule for the takings in month n b the takings in week 9. $2724.91

Month number

Takings

1

$10 000

2

$8500

3

$7225

Chapter 9 Sequences and series

417

15 A small town is renowned for spreading Number of citizens rumours. All of its citizens are aware in in the know Day number a short time of any new rumours. The spread of the rumour can be summarised 1 1 in the table given at right. 2 6 If the number of citizens who have been told the rumour each day con3 36 tinues to follow a geometric sequence, find: a a rule for the number of citizens in day n tn = 6n − 1 b the number of citizens told of the rumour by day 5 1296 c on which day all 4230 citizens will know of the rumour. 6 16 A ranger records the distance from a sand dune to the water’s edge at low tide over a number of years.

Year number

Distance from water’s edge (m)

1

60

2

64.8

3

69.984

If the distance from the water’s edge each year continues to follow a geometric sequence, find: a a rule for the distance from the water’s edge in year n tn = 60 × 1.08n − 1 b the distance from the water’s edge in year 6 (correct to 2 decimal places) 88.16 m c in which year the distance from the water’s edge will exceed 100 m. 8th year eBook plus Digital doc: WorkSHEET 9.1

17 How many terms of the geometric sequence 600, 180, 54, 16.2, 4.86, . . . are required for the sum to be greater than 855? Check with your teacher.

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Applications of geometric sequences Growth and decay of discrete variables is constantly found in real-life situations. Some examples are increasing or decreasing populations and increase or decrease in financial investments. Some of these geometric models are presented here.

WORKED Example 12 A city produced 100 tonnes of rubbish in the year 2008. Forecasts suggest that this may increase by 2% each year. If these forecasts are true, a what will be the city’s rubbish output in 2012? b in which year will the amount of rubbish reach 120 tonnes? c what was the total amount of rubbish produced by the city in the years 2008, 2009 and 2010? THINK

WRITE/DISPLAY

a

a a = 100

1

This is an example of a geometric sequence. Find the first term, a.

2

Determine the common ratio, r. The amount of rubbish increases by 2%, that is, the original amount plus an extra 2%. Note that r ≠ 0.02.

Increase by 2% 1 + 2% = 1 + 0.02 r = 1.02

3

Determine which term is represented by the amount of rubbish for the year 2012.

Year 2008 is the first term, so n = 1. Year 2009 is the second term, so n = 2. Year 2012 is the fifth term, so n = 5.

Use tn = ar n − 1 to find the amount of rubbish collected in the fifth year.

t5 = 100 × 1.025 − 1

Write your response.

The amount of rubbish produced in the fifth year, or 2012, will be 108.24 tonnes.

4

5

b Method 1: Using logarithms 1

Use tn = ar n − 1 and tn = 120.

2

Express 1.02 and 1.2 in terms of logarithms with base 10.

3

Write your answer.

= 100 × 1.0824 = 108.24

b 100(1.02)n − 1 = 120 (1.02)n − 1 = 1.2 log 1.02n − 1 = log 1.2 (n − 1) log 1.02 = log 1.2 log 1.2 n − 1 = ------------------log 1.02 n − 1 = 9.207 n = 10.207 During the 11th year, that is, during 2018, the rubbish will have exceeded 120 tonnes.

Chapter 9 Sequences and series

THINK

Method 2: Using trial and error 1

Use tn = ar n − 1 and tn = 120.

2

Try various values of n.

3

Write your answer.

419

WRITE/DISPLAY 100(1.02)n − 1 = 120 (1.02)n − 1 = 1.2 Let n = 10, (1.02)9 = 1.195 Let n = 11, (1.02)10 = 1.21 During the 11th year, that is, during 2018, the rubbish will have exceeded 120 tonnes.

Method 3: Using a graphics calculator 1 2

Use tn = ar n − 1 to write the rule.

tn = 100(1.02)n − 1

Locate the term that equals or exceeds 120. Two methods are shown for this example. Method A: Generating the terms of the sequence For the Casio fx-9860G AU (a) Press MENU and select RUN-MAT. Press OPTN then F1 (LIST) followed by F5 (Seq). Enter the rule (100 × 1.02n − 1) followed by the variable name (n), the start value (1), the end value (say, 20) and the increment (1), each separated by a comma. Press ) to close the set of brackets. (b) Press EXE to display the list of the first 20 terms. Scroll down until you reach the term that exceeds 120. (This screen is also useful for part c.)

For the TI-Nspire CAS (a) Open a new Lists & Spreadsheet document. Press b, then 3: Data followed by 1: Generate Sequence. Enter the rule 100 × 1.02n − 1 with an initial term of 100. Press e until OK is highlighted. (b) Press · to display the list of terms. Scroll down until you reach the term that exceeds 120. (You may like to save this file for part c.)

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THINK Method B: Solving an equation Set up an equation to solve.

WRITE/DISPLAY 100(1.02)n − 1 = 120

For the Casio fx-9860G AU Press MENU and select EQUA. Press F3 (SOLV) to select F3: Solver. (A previous equation and its solution will be displayed.) Enter the required equation in the first line and press EXE . Press F6 (SOLV) to display the solution. For the TI-Nspire CAS Open a new Calculator document. Press b and select 3: Algebra followed by 1: Solve. Enter the equation to be solved and then the variable, separated by a comma. Press ) to close the set of brackets and press ·.

c

3

Write your answer.

1

We need to find the sum of the first 3 terms.

The first term to exceed 120 tonnes is the 11th term or year 2018. c

Method 1: Using the formula a(rn – 1) Use S n = ---------------------- where n = 3. r–1

100 ( 1.02 3 – 1 ) S3 = -----------------------------------1.02 – 1 S3 = 306.04

Method 2: Using a graphics calculator Use the display generated by a graphics calculator from Method A of part b to add the first three terms. (For the TI-Nspire CAS calculator, find the sum of the numbers in cells A1 to A3.)

100 + 102 + 104.04 = 306.04 2

Write your answer.

The total output of rubbish for the years 2008, 2009 and 2010 will be 306.04 tonnes.

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WORKED Example 13

A computer system decreases in value each year by 15% of the previous year’s value. Find an expression for the value of the computer, which shall be referred to as Vn, after n years. Its initial purchase price is given as V1 = $12 000. THINK 1

2

3

WRITE

This is a geometric sequence since there is a 15% decrease on the previous year’s value. Find a and r. Note: Since this is a decreasing value, r is a value less than 1. We want an expression for the value after n years. Use tn = ar n − 1 which gives the value of the nth term. Use Vn instead of tn. Write your answer.

a = 12 000 r = 1 − 15% = 1 − 0.15 = 0.85 Vn = 12 000 × (0.85)n − 1

The value of the computer is given by the expression Vn = 12 000(0.85)n − 1.

Compound interest Consider the case where a bank pays compound interest of 5% per annum on an amount of $20 000. The amount is invested for 4 years and interest is calculated yearly. Compound interest receives its name because the interest which is earned is paid back into the account so that the next time interest is calculated, it is calculated on an increased amount. There is a compounding effect on the money in the account. If we calculated the amount in the account mentioned above each year, we would have the following amounts. Start After 1 year After 2 years After 3 years After 4 years

$20 000 $20 000 × 1.05 = $21 000 $20 000 × 1.05 × 1.05 = $22 050 $20 000 × 1.05 × 1.05 × 1.05 = $23 152.50 $20 000 × 1.05 × 1.05 × 1.05 × 1.05 = $24 310.13

The amounts 20 000, 21 000, 22 050, 23 152.50, 24 310.13, . . . form a geometric sequence where a = 20 000 and r = 1.05. We need to be a little careful, however, in using the formula tn = ar n − 1 in calculating compound interest. This is because the original amount in the account, that is, $20 000, in terms of the geometric sequence would be referred to as t1 or a. In banking terms, t1 would represent the amount in the account after the first lot of interest has been calculated and added in. To be clear and to be safe, it is best to use the following formula for compound interest. A = PR n r where R = 1 + --------100 A = amount in the account, $ P = principal, $ r = interest rate per period (that is, per year or quarter etc.), % n = the number of periods during the investment.

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WORKED Example 14

Helen inherits $60 000 and invests it for 3 years in an account which pays compound interest of 8% per annum compounding every 6 months. a What will be the amount in Helen’s account at the end of 3 years? b How much will Helen receive in interest over the 3-year period? THINK a 1 This is an example of compound interest. r Use A = PRn, where R = 1 + --------- . Interest 100 is calculated each 6 months so, over 3 years, there are 6 periods: n = 6. Interest is 8% per year or 4% per 6 months. So, r = 4%.

b

2

Write your answer.

1

Interest equals the amount in the account at the end of 3 years, less the amount in the account at the start of the investment. Write your answer.

2

WRITE a P = 60 000 n = 6 half years r = 4% per half year 4 So, R = 1 + --------100 = 1.04 A = PRn = 60 000(1.04)6 = 75 919.14 At the end of 3 years, Helen will have a total amount of $75 919.14. b Interest = Total amount − Principal = $75 919.14 − $60 000 = $15 919.14 Amount of interest earned over 3 years is $15 919.14.

WORKED Example 15

Jim invests $16 000 in a bank account which earns compound interest at the rate of 12% per annum compounding every quarter. At the end of the investment, there is $25 616.52 in the account. For how many years did Jim have his money invested? THINK Method 1: Using logarithms 1 We know the value of A, P, r and R. We need to find n using the compound interest formula. Note: There are 4 quarters per annum.

WRITE/DISPLAY A = 25 616.52 P = 16 000 -----r = 12 4 r = 3% per quarter 3and so R = 1 + -------= 1.03 100

2

3

Now substitute into A = PR . n

Express 1.601 and 1.04 in terms of logarithms with base 10.

A = PR 25 616.52 = 16 000(1.03)n 1.601 = 1.03n log 1.601 = log 1.03n So, log 1.601 = n × log 1.03 log 1.601 n = ---------------------log 1.03 0.2044 n = ---------------0.0128 n

Chapter 9 Sequences and series

THINK 4 5

Round up the number of periods to 16 to ensure the amount is reached. Write your answer.

Method 2: Trial and error 1 We know the value of A, P, r and R. We need to find n using the compound interest formula.

2

Try some different values of n.

3

Write your answer.

Method 3: Using a graphics calculator n 1 Use tn = PR to write the rule for the sequence.

2

Locate the term that equals or exceeds 25 616.52. You may wish to generate the sequence and scroll down to the required term (see page 400). Alternatively, an equation can be set up to be solved. This method is shown here. For the Casio fx-9860G AU Press MENU and select EQUA. Press F3 (SOLV) to select F3: Solver. (A previous equation and its solution will be displayed.) Enter the required equation in the first line and press EXE . Press F6 (SOLV) to display the solution.

423

WRITE/DISPLAY n = 15.92 It will take 16 periods where a period is 3 months. So, it will take 48 months or 4 years. A = 25 616.52 P = 16 000 -----r = 12 4 r = 3% per quarter 3and so R = 1 + -------100 = 1.03 Now, A = PRn So, 25 616.52 = 16 000(1.03)n 1.601 = 1.03n Let n = 5 1.035 = 1.159 Let n = 10 1.0310 = 1.344 Let n = 15 1.0315 = 1.558 Let n = 16 1.0316 = 1.605 It will take 16 periods where a period is 3 months. So, it will take 48 months or 4 years.

P = 16 000 -----r = 12 4 = 3% per quarter 3and so R = 1 + -------100 = 1.03 tn = 16 000(1.03)n For tn = 25 616.52, 16 000(1.03)n = 25 616.52

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THINK

WRITE/DISPLAY

For the TI-Nspire CAS Open a new Calculator document. Press b and select 3: Algebra followed by 1: Solve. Enter the equation to be solved and then the variable, separated by a comma. Press ) to close the set of brackets and press ·. 3

Write your answer.

Jim has money invested for 16 periods where a period is 3 months. So it will take 48 months or 4 years.

remember remember Geometric growth 1. Geometric growth or increase is expressed as a percentage increase. 2. Common ratio, r = 1 + % increase 3. r values are greater than 1, for example, an 8% increase gives r = 1.08. Geometric decay 4. Geometric decay or decrease is expressed as a percentage decrease. 5. Common ratio, r = 1 − % decrease 6. r values are less than 1, for example, an 8% decrease gives r = 0.92. Compound interest 7. The formula for compound interest is A = PRn r where R = 1 + --------100 A = amount in the account, $ P = principal, $ r = interest rate per period (that is, per year or quarter etc.) as a percentage n = the number of periods during the investment.

9C WORKED

Example xample

12

Applications of geometric sequences

1 A farmer harvests 4 tonnes of lucerne in his first year of production. In his business plan, he has estimated an annual increase of 6% on his lucerne harvest. a According to this plan, how many tonnes of lucerne should he harvest in his 7th year of production? 5.67 b In which year will his harvest reach 10 tonnes? 17th year c How much will he expect to harvest in the first 3 years? 12.7 tonnes

eBook plus Digital doc: EXCEL Spreadsheet Sequences and series

Chapter 9 Sequences and series

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2 A taxi driver estimates that the cost of keeping her taxi on the road increases by 4.5% each year. If the cost of keeping her taxi on the road in her first year of owning a taxi was $1800: a what was the cost in the 5th year? $2146.53 b during which year did costs exceed $2500? Year 9 c what were the total costs of keeping her taxi on the road in the first 3 years? $5646.65 WORKED

Example

13 WORKED

Example

14

3 The population of a town is decreasing by 10% each year. Find an expression for the population of the town, which will be referred to as Pn. The population in the first year, P1, was 10 000. Pn = 10 000 × (0.9)n − 1 4 $13 000 is invested in an account which earns compound interest of 8%, compounding quarterly. a After 5 years, how much is in the account? $19 317.32 b How much interest was earned in that period? $6317.32 5 The population of the newly established town of Alansford in its first year was 6000. It is predicted that the town’s population will increase by 10% each year. If this were to be the case, find: a the population of the town in its 10th year 14 147 b in which year the population of Alansford would reach 25 000. 16th year 6 The promoters of ‘Fleago’ flea powder assert that continued application of the powder will reduce the number of fleas on a dog by 15% each week. At the end of week 1, Fido the dog has 200 fleas left on him and his owner continues to apply the powder. a How many fleas would Fido be expected to have on him at the end of the 4th week? 123 b How many weeks would Fido have to wait before the number of fleas on him had dropped to less than 50? 10 weeks

7 Young saplings should increase in height by 9% each year under optimum conditions. If a batch of saplings which have been planted out measure 2.2 metres in their first year: a how high should they be in their 4th year? 2.85 m Year 11 b in which year should they exceed 5 metres in height? 8 A number of timber beams support a ramp. The first of the beams is 0.8 metres long and each successive beam is 3% longer than the previous one. a How long will the 7th beam in the line be? 0.96 m b Which beam will be the first to exceed 2 metres in length? 32nd beam

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9 multiple choice A colony of ants is studied and the population of the colony in week 1 of the study is 800. If the population of the colony is expected to increase at the rate of 2% each week, then the week in which the number of ants would exceed 1000 would be closest to: A 6 B 10 C 13 D 26 E 32 10 A company exported $300 000 worth of manufactured goods in its first year of production. According to the business plan of the company, this amount should increase each year by 7.5%. a How much would the company be expected to export in its 5th year? $400 640.74 b In which year would exports exceed $500 000? Year 9 c What is the total amount exported by the company in its first 7 years of operation? 2 636 196.56 11 Country football crowds have been decreasing by 3% each year since records of crowd attendance were kept. If the number of people attending in the first year that records were kept was 63 000 in a season: a how many people attended in the 5th year? 55 773 b when did the number of people attending in a year drop below 50 000? 9th year 12 $10 000 is invested in an account which earns compound interest of 10% per annum. Find the amount in the account after 5 years if the interest is compounded: a yearly $16 105.10 b every 6 months $16 288.95 c quarterly $16 386.16 d monthly. $16 453.09 13 $20 000 is invested in an account earning compound interest of 10% per annum compounding quarterly. What is the amount in the account after: a 1 year? $22 076.26 b 3 years? $26 897.78 c 5 years? $32 772.33 d 10 years? $53 701.28 14 $7000 is invested in an account which earns compound interest of 6% per annum compounding monthly. After 3 years, how much is in the account? $8376.76 15 In an account earning compound interest of 8% per annum compounding quarterly, an amount of $6000 is invested. When the account is closed, there is $7609.45 in the 15 account. For how many years was the account open? 3 years

WORKED

Example xample

16 Sue earns 12% interest per annum compounding quarterly on her investment of $40 000. For how many years would this investment need to operate for the amount to rise to $50 670.80? 2 years 17 Helena receives $15 627.12 after closing an investment account which earned compound interest of 9% per annum compounding every 6 months. If Helena originally deposited $12 000 in the account, for how long was it in the account? 3 years 18 Todd receives $66 277.33 after having invested an inheritance of $60 000 in an account earning compound interest of 12% per annum compounding monthly. For how long did Todd have the money invested? 10 months 19 An amount of $14 500 is invested in an account attracting compound interest of 6% per annum compounding quarterly. After a certain time the interest earned in the account is $1834.14. Find out for how long the amount had been invested. 2 years

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Finding the sum of an infinite geometric sequence SLE 1: Establish the formula for the sum to n terms of a geometric progression, and hence the formula for the sum to infinity of a geometric progression; verify the formula by mathematical induction.

If you are 2 metres away from a wall and you move 1 metre (or half-way) towards the wall and then move 1--- metre 2 (or half-way again) towards the wall and continue to do this, will you reach the wall? When will you reach the wall? Consider the following geometric sequence: 11, 1--- , 1--- , 1--- , ----, . . . This is an infinite geometric sequence 2

4

8

1m

16

since it continues on with an infinite number of terms. 2m Each term in the sequence is less than the previous term by a factor of 1--- , that is, r = 0.5. 2 If we were to add n terms of this sequence together, we would have:

1– 2

m

1– 4

1– 8

1m

n

1 × ( 1 – 0.5 ) Sn = --------------------------------1 – 0.5 n – 0.5 1-----------------= 0.5 n 1 0.5 = ------- − --------0.5 0.5 = 2 − 0.5n − 1 Consider 0.5n − 1 in the above equation. As n becomes very large, the term 0.5n − 1 becomes very small. Try this with your calculator. Let n = 5, 0.5n − 1 = 0.54 = 0.0625; therefore, S5 = 2 − 0.0625 = 1.9375 Let n = 10, 0.5n − 1 = 0.59 = 0.001 95; therefore, S10 = 2 − 0.001 95 = 1.998 05 Let n = 20, 0.5n − 1 = 0.519 = 0.000 001 9; therefore, S20 = 2 − 0.000 001 9 = 1.999 998 1 We can see that as n becomes larger, 0.5n − 1 becomes smaller. If n were to approach infinity (note that you can never reach infinity, you can only approach it), then the value of 0.5n − 1 would approach zero. So, Sn = 2 − 0.5n − 1 would become S∞ = 2. It is possible to generalise this in order to find the sum of an infinite geometric sequence. We use the symbol S∞ which is referred to as the sum to infinity of a geometric sequence. The sum to infinity of a geometric sequence for which −1 < r < 1 is given by a S • = ----------- . 1–r

WORKED Example 16 Find the sum to infinity of the geometric sequence 2, 0.4, 0.08, 0.016, 0.0032, . . . THINK 1

Find a and r.

WRITE a=2 t r = ---2t1 0.4 = ------2 = 0.2

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THINK 2

As r = 0.2 satisfies the condition a −1 < r < 1, use the formula S ∞ = ----------- . 1–r

3

Write your answer.

WRITE a S∞ = ----------1–r 2 = ---------------1 – 0.2 2 = ------0.8 = 2.5 The sum to infinity of the given sequence is 2.5.

WORKED Example 17 The sum to infinity of the geometric sequence is 6.25 and the value of r is 0.2. Write down the first 4 terms of the sequence. THINK 1

2

a Use the formula S ∞ = ----------- to find the 1–r value of a.

a = 5 and r = 0.2. Use these to generate the terms.

WRITE a S∞ = ----------1–r a 6.25 = ---------------1 – 0.2 6.25 × 0.8 = a a=5 The first 4 terms of the sequence are 5, 1, 0.2, 0.04.

WORKED Example 18 The sum to infinity of a geometric sequence is 15 and the value of a is 10. Write down the first 4 terms of the sequence. THINK 1

a Use the formula S ∞ = ----------- to find the 1–r value of r. Transpose the equation to make r the subject.

WRITE a S ∞ = ----------1–r 10 15 = ----------1–r 1–r =

10----15

r = 1 – 2--3r = 2

a = 10 and r = 1--- . Use these to generate 3 the terms in the sequence.

1 --3

The first 4 terms of the sequence are 10,

10 ------ , 10 ------ , 10 -----3 9 27

or 10, 3 1--- , 1 1--- , 3

9

10 ------ . 27

Chapter 9 Sequences and series

429

Converting recurring decimals to fractions We can use the sum to infinity formula to convert recurring decimals to fractions.

WORKED Example 19

. Express 1.2 as a fraction. THINK 1

2

WRITE

. We need to express 1.2 as the sum of a geometric sequence.

. 1.2 = 1.222 222 222 222 2 … = 1 + 0.2 + 0.02 + 0.002 + 0.0002 + … = 1 + (0.2 + 0.02 + 0.002 + 0.0002 + …) a = 0.2 0.02 r = ---------0.2 = 0.1 a S∞ = ----------1–r . 0.2 0.2 = ---------------1 – 0.1 . 0.2 0.2 = ------0.9 . 2 0.2 = --9

The terms in the bracket form an infinite geometric sequence where a = 0.2 and r = 0.1. Use the formula a S ∞ = ----------- . 1–r

Multiply both the numerator and 0.2 denominator of ------- by 10 to eliminate 0.9 the decimal. 3

Express the final answer.

. So 1.2 = 1 + = 1 2---

2 --9

9

WORKED Example 20 Express 0.645 . . . as a fraction. THINK 1

We need to express 0.645 as the sum of a geometric sequence.

2

The terms in the bracket form an infinite geometric sequence where a = 0.045 and r = 0.01.

WRITE 0.645 = 0.645 454 5… = 0.6 + 0.045 + 0.000 45 + 0.000 004 5 + … = 0.6 + ( 0.045 + 0.000 45 + 0.000 004 5 + … ) a = 0.045 0.00045 r = ------------------0.045 = 0.01 Continued over page

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THINK 3

4

5

WRITE a S ∞ = ----------1–r

a Use the formula S ∞ = ----------- . 1–r

0.045 0.045 = ------------------1 – 0.01 Multiply both the numerator and the 45 0.045 = --------0.045 990 denominator of ------------- by 1000 to eliminate 0.99 45 0.645 = 0.6 + --------the decimal. 990 6 -------45= ------ + Express the final answer. 10 990 594 45 = --------- + --------990 990 639 So 0.645 = --------Write your answer. 990

WORKED Example 21 An injured rabbit attempts to crawl back to its hole. It moves 30 metres in the first hour, 21 metres in the second hour and 14.7 metres in the third hour and so on. If the rabbit hole is 200 metres away, will the rabbit make it back to its hole? THINK 1 Determine what sort of sequence we have.

WRITE t1 = 30, t2 = 21 and t3 = 14.7

and

2

a Find the value of S ∞ = ----------- . 1–r

3

Write your answer.

21 -----30

= 0.7

14.7 ---------21

= 0.7

Now

So, we have a geometric sequence where a = 30, r = 0.7. 30 S∞ = ---------------1 – 0.7 S∞ = 100 The rabbit will cover a total of 100 metres. Since the rabbit hole is 200 metres away, the rabbit won’t make it.

remember remember Finding the sum of an infinite geometric sequence: 1. For decreasing or decaying geometric series, the sum of an infinite number of terms approaches a finite sum. 2. The sum to infinity of a geometric sequence for which −1 < r < 1 is given by a S ∞ = ----------1–r

Chapter 9 Sequences and series

431

SLE 13: Use geometric progressions in situations involving the sum to infinity.

9D WORKED

Example

16

Finding the sum of an infinite geometric sequence

1 Find the sum to infinity of the following geometric sequences. a 50, 25, 12.5, 6.25, 3.125, . . 100 . b 20, 16, 12.8, 10.24, 9.192, . . .100 1- ----10 d 1, 1--- , 1--- , ----, 1- , . . . --32c 4, 2.4, 1.44, 0.864, 0.5184, . . . 3 9 27 81 e 1, 1--- , 5

1- -------1----, 1 - , -------, 25 125 625

...

f

g 3, −0.6, 0.12, −0.024, 0.0048, .2.5 .. i 120, 48, 19.2, 7.68, 3.072, . . .200

2,

8- -------128- ------------------, 32- , ----------, 512 - , 10 100 1000 10 000

. . . 3 1--3h −12, 7.2, −4.32, 2.592, −1.5552, −. .7.5 . 5 j −50, 5, −0.5, 0.5, −0.05, . . . –45 ----11

5--4

2 Write down the first 3 terms of the geometric sequence for which: a r = 0.6 and S∞ = 25 10, 6, 3.6 b r = 0.2 and S∞ = 50 40, 8, 1.6 17 c r = 0.25 and S∞ = 8 6, 1.5, 0.375 d r = 0.9 and S∞ = 120 12, 10.8, 9.72 3 a 12.5, 6.25, 3.125 e r = −0.2 and S∞ = 3 1--- 4, −0.8, 0.16 f r = −0.5 and S∞ = 4 6, −3, 1.5 3 b 12.5, 9.375, 7.031 25 g r = −0.8 and S∞ = 5 9, −7.2, 5.76 h r = 0.2 and S∞ = −7.5 −6, −1.2, −0.24 d 48, 28.8, 17.28 7i r = 0.6 and S∞ = −60 −24, −14.4, −8.64 j r = −0.3 and S∞ = −11 ----−15, 4.5, −1.35 WORKED

Example

13

WORKED

Example

18

3 Write down the first 3 terms of the geometric sequence for which: b a = 12.5 and S∞ = 50 c a = 6 and S∞ = 8 6, 1.5, 0.375 a a = 12.5 and S∞ = 25 4 22- 22 -----, d a = 48 and S∞ = 120 e a = 2 4--- and S∞ = 3 2--- 2 --9- , ----27 81 9

WORKED

Example

19, 20 WORKED

Example

21

3

4 Express each of the following recurring decimals as fractions. . 8 . . 1 5 a 0.5 --9b 0.8. . --9c 0.4. . 4--9d 1.3 1 --314 55f 8.666 666 666 . . . 8 2--3- g 0.1 4 ----h 0.5 7 ----i 0.529 99 99

. 7 e 3.7 3 --9--------j 1.321 1 321 999

5 A defiant child walks 10 metres towards his mother in the first minute, 4 metres in the second minute and 1.6 metres in the third minute. If the child continues to approach in this same pattern, and if his mother is standing stationary, 20 metres from the 1 child’s initial position, will the child ever reach the mother? No — falls short by 3 --3- metres

262 --------495

6 A failing machine produces 35 metres of spouting in the first hour, 21 metres in the second hour and 12.6 in the third hour. If this pattern continues and 280 metres of spouting is required, how far short of the quota will the machine fall? 192.5 m 7 A nail penetrates 20 mm with the first hit of a hammer, 12 mm with the 2nd hit and 7.2 mm with the 3rd. If this pattern continues, will the 50 mm long nail ever be completely hammered in? Yes 8 A woman establishes a committee to raise money for a hospital. It raises $40 000 in the 1st year, $36 800 in the 2nd year and $33 856 in the 3rd year. If the fundraising continues in this pattern, how far short will they fall in raising $1 000 000? $500 000 9 An irrigation system sprays 25 mm of water over a crop in its 1st month, 20 mm in the 2nd month and 16 mm in the 3rd month. If the crop requires 100 mm of water during its lifetime, how far short or how far over is the irrigation system in supplying the correct amount? 25 mm too much 10 A will of a recently deceased woman specifies how her money is to be donated to a charity. Her total wealth of $12.5 million is to be donated for eternity with the first donation of $1 million in the first year. a What fraction of this first donation should be donated for the second year and sub-----sequent years? 23 25 $1 000 000, $920 000, $846 400, b Write the value of the donations for each of the first 5 years. $778 688, $716 392.96 c How much will be donated after 10 years? $7 070 144.32

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Contrasting arithmetic and geometric sequences through graphs When discrete variables are presented graphically some distinct features may be evident. This is especially so for discrete variables that have an arithmetic or geometric pattern.

Arithmetic patterns

Value of term tn

Value of term tn

Arithmetic patterns are distinguished by a straight line or a constant increase or decrease.

1

2 3 4 Term n

1

5

d is positive

2 3 4 Term n

5

d is negative

An increasing pattern or a positive common difference gives an upward straight line.

A decreasing pattern or a negative common difference gives a downward straight line.

Geometric patterns

Value of term tn

Value of term tn

Geometric patterns are distinguished by a curved line or a saw form.

1

2 3 4 Term n

5

An increasing pattern or a positive common ratio greater than 1 (r > 1) gives an upward curved line.

1

+

1

2

3

4

5

Term n

An increasing saw pattern occurs when the common ratio is a negative value less than −1 (r < −1).

Value of term tn

Value of term tn

5

A decreasing pattern or a positive fractional common ratio (0 < r < 1) gives a downward curved line.

+

–

2 3 4 Term n

1

–

2 3

4

5

Term n

A decreasing saw pattern occurs when the common ratio is a negative fraction (–1 < r < 0).

Chapter 9 Sequences and series

433

On the graph at right, the first 5 terms of a sequence are plotted. State whether the sequence could be arithmetic or geometric and give the value of a and the value of either d or r.

THINK

Value of term

WORKED Example 22 50 40 30 20 10 0

0

1 2 3 4 Term number

5

WRITE

2

Examine the difference between the value of each of the terms. In each case, they are the same, that is, 10. Find a.

There is a constant difference between each successive term so the graph shows an arithmetic sequence. Now, t1 = 40, so a = 40.

3

Find d.

Now, t2 = 30 and t1 = 40. So, d = t2 − t1 = −10.

1

WORKED Example 23 An amount of $10 000 is invested for 5 years and earns interest. Consider the following two cases: ii simple interest of 10% per annum ii compound interest of 10% per annum compounding yearly. a If the investment is earning simple interest, calculate the amount in the account at the end of each of the 5 years. b If the investment is earning compound interest, calculate the amount in the account at the end of each of the 5 years. c For each of the above cases, graph, on the same set of axes, the total assets over the 5 years. Use your graph or calculations to calculate the difference between the accounts after 4 years. THINK a Calculate how much is in the account earning simple interest at the end of each of the 5 years.

WRITE a After 1 year, amount in account = 10 000 + 10% of 10 000 = 10 000 + 1000 = $11 000 After 2 years, amount in account = 10 000 + 2 × 10% of 10 000 = 10 000 + 2000 = $12 000 After 3 years, amount in account = $13 000 After 4 years, amount in account = $14 000 After 5 years, amount in account = $15 000

Continued over page

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WRITE

b Calculate the amount in the account earning compound interest at the end of each of the 5 years using A = PRn where r R = 1 + --------- (that is, 1 + 10% = 1.1). 100

b After 1 year, amount in account = 10 000 × 1.11 = $11 000 After 2 years, amount in account = 10 000 × 1.12 = $12 100 After 3 years, amount in account = 10 000 × 1.13 = $13 310 After 4 years, amount in account = 10 000 × 1.14 = $14 641 After 5 years, amount in account = 10 000 × 1.15 = $16 105 c

c

1

Draw the graphs of the amount in the account earning simple interest and the amount in the account earning compound interest on the same set of axes.

Amount in account ($)

THINK

17 000 16 000 15 000 14 000 13 000 12 000 11 000 10 000

Compound interest

Simple interest

0 1 2 3 4 5 Number of years invested (n)

2

Use the values calculated for the end of the fourth year.

Difference in amounts = 14 641 − 14 000 = $641

3

Write your answer.

The compound interest account earned an extra $641 in interest after 4 years.

Graphics Calculator tip!

Comparison of simple and compound interest

A graphics calculator can be used to compare the yearly amounts in an account earning both simple interest and compound interest. Consider the scenario in Worked example 23 where the amounts are compared over the first 5 years. The steps are shown below. Then investigate this scenario further by drawing the graphs for both cases over the first 10 years.

For the Casio fx-9860G AU 1. Press MENU and select STAT. Enter 1, 2, 3, 4 and 5 into List 1 to indicate the 5 years to be considered. Enter a title for List 1 if you wish.

Chapter 9 Sequences and series

2. Press MENU and select RUN-MAT. Use the Seq function to enter the yearly amounts for the simple interest case. The rule is tn = 10 000 + n × 1000. Press Æ then F1 (List) and 2 followed by EXE to store these 5 terms as List 2. 3. Use the Seq function again to enter the yearly amounts for the compound interest case. The rule is tn = 10 000 × 1.1n. Press Æ then F1 (List) and 3 followed by EXE to store these 5 terms as List 3. 4. Press MENU and select STAT. The amounts at the end of each of the first 5 years are displayed. Enter a title for each list if you wish.

5. Press F1 (GRPH) then F6 (SET). Press F1 (GPH1) to display the properties of Statgraph1. This graph is to display the yearly amounts for the simple interest case, so set Graph Type as a scatterplot (press F1 (Scat)), XList as List 1 and YList as List 2. Keep Frequency as 1 and adjust the Mark Type if you wish. 6. Press EXE to accept the changes you have made to StatGraph1. Press F6 (SET) again and then press F2 (GPH2) to display the properties of Statgraph2. This graph is to display the yearly amounts for the compound interest case, so set Graph Type as a scatterplot, XList as List 1 and YList as List 3. Keep Frequency as 1 and adjust the Mark Type so it is different from the first graph. 7. Press EXE to accept the changes you have made to StatGraph2. Press F4 (SEL) and ensure that both StatGraph1 and StatGraph2 show DrawOn (and StatGraph3 shows DrawOff).

8. Press F6 (DRAW) to display both graphs on the same set of axes. Use the Trace function to investigate the graphs. Press SHIFT F1 (TRCE) and use the arrow keys to move to different points on the graphs. The coordinates of the point and the title of the graph are displayed.

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For the TI-Nspire CAS 1. Open a new Lists & Spreadsheet document. Enter 1, 2, 3, 4 and 5 into Column A to indicate the 5 years to be considered. Enter a title or variable name (say, n) for this column.

2. We wish to generate terms of the sequence for the simple interest case in Column B and the compound interest case in Column C. Increase the width of Columns B and C, and enter a title for each. Then return to Column B.

3. Press b, then 3: Data followed by 1: Generate Sequence. Enter the rule 10 000 + n × 1000 with a first term of 11 000 and a maximum number of terms of 5 to generate the yearly amounts for the simple interest case. Press e until OK is highlighted and then press ·. Move to Column C and again press b, then 3: Data and 1: Generate Sequence. Enter the rule 10 000 × 1.1n with a first term of 11 000 and a maximum number of terms of 5 to generate the yearly amounts for the compound interest case. Press e until OK is highlighted and then press ·. 4. Press / I to insert a new document and select 2: Add Graphs & Geometry. Press / b and select 4: Scatter Plot.

5. Click the a button to bring up an option screen and highlight the variable n for the x entry.

6. Press · and then press e to move to the y entry. Again click the a button, highlight the variable simple and press ·. This has entered the information for Scatterplot 1. Press d. Now press b and select 4: Window followed by A: Zoom - Fit to best display the graph for the simple interest case.

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7. Now press e to highlight the x entry for Scatterplot 2. Enter the information for the compound interest case in the same way. That is, select the variable n for x and the variable compound for y. Press / G to hide the function entry line. (Press / G again to bring it back.) Use the Trace function to investigate the graphs. Press b and select 5: Trace followed by 1: Graph Trace and use the NavPad to move to different points on the graphs to display the coordinates.

remember remember Value of term tn

Value of term tn

1. Arithmetic patterns

1

2 3 4 Term n

1

5

2 3 4 Term n

5

d is negative

d is positive

An increasing pattern or a positive common difference gives an upward straight line.

A decreasing pattern or a negative common difference gives a downward straight line.

Value of term tn

Value of term tn

2. Geometric patterns

1

2 3 4 Term n

5

An increasing pattern or a positive common ratio greater than 1 (r > 1) gives an upward curved line.

1

+

1

2

3

4

5

Term n

An increasing saw pattern occurs when the common ratio is a negative value less than −1 (r < −1).

Value of term tn

Value of term tn

5

A decreasing pattern or a positive fractional common ratio (0 < r < 1) gives a downward curved line.

+

–

2 3 4 Term n

1

–

2 3

4

5

Term n

An decreasing saw pattern occurs when the common ratio is a negative fraction (–1 < r < 0).

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

9E WORKED

Example xample

22

eBook plus

Contrasting arithmetic and geometric sequences through graphs

1 On the graph at right, the first five terms of a sequence are plotted. State whether the sequence could be arithmetic or geometric and give the value of a and the value of either d or r. Arithmetic sequence with a = 0 and d = 1

Value of term

438

5 4 3 2 1 0

Digital doc:

0

1

2 3 4 5 Term number

0

1

2 3 4 5 Term number

0

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2 3 4 5 Term number

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1

2 3 4 5 Term number

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2 3 4 5 Term number

2 On the graph at right, the first five terms of a sequence are plotted. State whether the sequence could be arithmetic or geometric and give the value of a and the value of either d or r. Arithmetic sequence with a = 10 and d = −2

Value of term

EXCEL Spreadsheet Sequences and series

12 10 8 6 4 2

3 On the graph at right, the first five terms of a sequence are plotted. State whether the sequence could be arithmetic or geometric and give the value of a and the value of either d or r. Geometric sequence with a = 10 and r = 1.5

Value of term

0

60 50 40 30 20 10

5 On the graph at right, the first five terms of a sequence are plotted. State whether the sequence could be arithmetic or geometric and give the value of a and the value of either d or r. Arithmetic sequence with a = 4 and d = −0.5

20 15 10 5 0

Value of term

4 On the graph at right, the first five terms of a sequence are plotted. State whether the sequence could be arithmetic or geometric and give the value of a and the value of either d or r. Geometric sequence with a = 20 and r = 0.5

Value of term

0

5 4 3 2 1 0

4 3 2

–15

–10

1 0

–5

5

10

tn 15

7c

439

100 80 60 40 20 0

5

8

Value of term

Geometric sequence with a = 100 and r = 0.8

n 7

6 On the graph at right, the first five terms of a sequence are plotted. State whether the sequence could be arithmetic or geometric and give the value of a and the value of either d or r.

6

Chapter 9 Sequences and series

0

1

2 3 4 5 Term number

7 Draw a graph showing the first 8 terms of each of the following sequences: a arithmetic, a = 7, d = 2 b geometric, a = 5, r = 1--2 c arithmetic, a = –14, d = 3.5 d arithmetic, a = 32, d = –5 10 e geometric, a = 12, r = -----f geometric, a = 0.02, r = 6 3

Value of term

180 150 120 90 60 30 0

0

1

2 3 4 5 Term number

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1

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5 Term number

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10

15

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25

7a

tn 30

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8 n

9 multiple choice 10 5 On the graph at right, the first five 0 terms of a sequence are plotted. –5 The sequence could be described –10 by which one of the following? A Arithmetic sequence with a = 10 and d = 20 B Arithmetic sequence with a = 10 and d = −20 C Geometric sequence with a = 10 and r = 10 D Geometric sequence with a = 10 and r = −20 E Geometric sequence with a = 10 and r = −1 Value of term

1

2

3

4

5

7b

tn 6

1

2

3

4

5

6

7

8 n

8 multiple choice On the graph at right, the first five terms of a sequence are plotted. The sequence could be described by which one of the following? A Arithmetic sequence with a = 10 and d = 10 B Arithmetic sequence with a = 10 and d = 0.5 C Geometric sequence with a = 10 and r = 0.5 D Geometric sequence with a = 10 and r = 2 E Geometric sequence with a = 10 and r = 1.5

10 An amount of $5000 is invested for 3 years and earns interest. Consider the following two cases: 23 i simple interest of 10% per annum ii compound interest of 10% per annum compounding yearly. a If the investment is earning simple interest, calculate the amount in the account at the end of each of the 3 years. $5500, $6000, $6500 b If the investment is earning compound interest, calculate the amount in the account at the end of each of the 3 years. $5500, $6050, $6655 c On the same set of axes, plot points showing the amount in each account at the end of each of the 3 years.

WORKED

5000

5500

6000

6500

1

2 Year

3

Legend Un Vn

Example

Amount ($)

10 c

11 An amount of $100 000 is invested for 3 years and earns: a simple interest of 15% per annum b compound interest of 15% per annum compounding yearly. On the same set of axes, plot points showing the amount in each account at the end of each of the 3 years.

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12 An amount of $10 000 is invested for 3 years and earns: a simple interest of 20% per annum b compound interest of 20% per annum compounding yearly. On the same set of axes, plot points showing the amount in each account at the end of each of the 3 years. 13 On the same set of axes, sketch the graphs of the sequences with the rule un = 10n and vn = 10 × 1.5n − 1. Use your graph to decide for how many of the first five terms un is greater than vn. 3 terms 14 On the same set of axes, sketch the graphs of the sequences with the rule un = 120 − 20n and vn = 100 × 0.8n − 1. Use your graph to decide for how many of the first five terms un is greater than vn. 0 terms 15 On the same set of axes, sketch the graphs of the sequences with the rule un = 120n and vn = 120 × 1.5n − 1. Use your graph to decide for how many of the first five terms un is greater than vn. 3 terms

eBook plus

16 On the same set of axes, sketch the graphs of the sequences with the rule un = 150 − 30n and vn = 100 × 0.5n − 1. Use your graph to decide for how many of the first five terms un is greater than vn. 4 terms

Digital doc: WorkSHEET 9.2

Reward time A teacher decides to give students lollies at an end-of-year party. He wishes to give different amounts to each student based on worst to best attendance record. 1 He decides to give 1 lolly to the student with the highest absence, 2 lollies to the student with the next highest absence, 4 to the next, then 8 and so on. a If there are 30 students in the class, show how much each of the first 6 students will receive. 1, 2, 4, 8, 16, 32 b How many lollies will the 30th student (with the best attendance record) expect to receive? 229 Legend

12 18 000

Amount ($)

17 000 16 000 15 000 14 000

Un Vn

13 000 12 000 11 000 1

2 Year

3

2 The teacher realises this is not practical. He investigates a distribution summarised as follows: 1, 3, 5, 7, . . . a What type of pattern is this? Explain why. Arithmetic sequence b How much will the 30th student expect to receive? 59 c How many lollies will the teacher need to give to the whole class? 900 d The value in c was still unacceptable to the teacher. What is the least total possible number of lollies required if each student is to receive a whole lolly using an arithmetic pattern. Show the pattern clearly. 465

3 The teacher decides against giving lollies. He now prefers to share one single cake. He decides to give out the pieces using the rule: 1--- cake to first student, 2 1 --- cake to second student, 1--- to third student and so on. 4 8 a How many cakes will he need if he continues the pattern for all 30 students? b How much of the cake will go to: 1 1i the 4th student? ----16 1----ii the 6th student? 1 64 iii the last student? -----30 2 There is some cake left. c Using this method the teacher realises there is something in it for him. Why?

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Chapter 9 Sequences and series

Changing shape E

In this investigation you will study how the area and perimeter of a closed shape change as you systematically modify the shape.

D

Consider a square of side length 1 unit as shown in diagram (a) below. Consider now making one change to each of the four sides as shown in the diagram where the side length, AB (b), is transformed into the articulated side length, AB (c). The new side length can be thought of as being composed of five sections all of equal length. B

A

B

A

B

C

A

B

(a)

1 unit

1 unit

(b)

(c)

1 Draw all the different shapes which can be made if all side lengths of the

1

A

original square undergo one transformation. (Hint: The new articulated edge length can be oriented two ways on each of the four sides. One way is to have the shape point out.) 2 For each of the five different shapes, calculate the area and perimeter of the

2 Shape Area Perimeter 20 -----3

E

7 --9

20 -----3

C, D

9 --9

20 -----3

B

11 -----9

20 -----3

A

13 -----9

20 -----3

3 Take the shape with the greatest area and now to each of the 20 sides apply

the transformation a second time to make a new shape of maximum area. Draw a diagram depicting the new shape. Calculate the perimeter and the area of this new shape.

4 We are now going to look for a pattern in the value of the perimeter and area The area increases in an for successive applications of the transformation when applied to all sides. arithmetic progression. The perimeter stays constant. Let n be the number of times the transformation is applied to the original

… + … 9--4- ( 9--5- )x – 1

5 Can you find a method for calculating the area and perimeter for n = x?

--49

square. For example, when n = 0 the area is 1 unit2 and the perimeter is 4 units. Investigate the value of the area and perimeter for n = 1 (question 2), 2 (question 3), 3, 4, 5, . . .

3

6 What happens to the value of the perimeter as n gets very large? Perimeter tends to infinity. Perimeter = = Area = 1 + =

137 --------81

20 -----3

×

5--3

7 What happens to the value of the area as n gets very large? Area tends to 2.

100-------9 4--9

+5×4×

1 ----2 9

8 If time permits you could extend this investigation to a new shape, for

example, an equilateral triangle or a cube. In the case of the cube you would investigate the surface area and volume.

Perimeter = 4 × ( 3--5- )x

5 --9

Area = 1 +

F

shape. Construct a table which details the area and perimeter of each of the five shapes, arranging the values in ascending order. What pattern do you notice, if any?

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Fibonacci Sequence SLE 8: Investigate the way in which a sequence can be defined recursively.

Consider the two sequences: 2, 4, 8, 16, 32, . . . and 2, 5, 11, 23, . . . Both are sequences whose terms can be predicted using a simple formula. However, the first sequence is considered fundamental and worthy of detailed study, whereas the second is not. The first sequence has numerous applications ranging from population growth to the study of investments. The second sequence has no such applications. Another sequence considered worthy of detailed examination is the Fibonacci Sequence: 1, 1, 2, 3, 5, 8, . . . This sequence takes the name of a mathematician, Fibonacci (better known as Leonardo of Pisa) who lived in Italy around AD 1200. The sequence arose as a result of the problem Fibonacci posed: If you had a pair of rabbits and it took a month for them to mature, and then produce a new pair after that, how many pairs would you have in twelve months? At the start of the first month

No. of pairs = 1

At the start of the second month

No. of pairs = 1

At the start of the third month

No. of pairs = 2

At the start of the fourth month

No. of pairs = 3

At the start of the fifth month

No. of pairs = 5

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This process can be continued and will reveal that the number of pairs at any stage can be obtained by adding the number of pairs at the two preceding stages. The resulting sequence is: 1, 1, 2, 3, 5, 8, 13, . . . In general then, if Fn stands for the number of pairs at the nth stage then Fn + 1 = Fn + Fn − 1 with F0 = 1 and F1 = 1. A sequence that specifies the general term by referring to the preceding terms is called recursive. An explict formula for the nth term is n

n

1 1+ 5 1 1– 5 Fn = -------  ---------------- − -------  ----------------   2 5 5 2  The Fibonacci Sequence has been used to describe patterns in nature from shells to pine cones.

Graphics Calculator tip!

Generating terms in the Fibonacci Sequence

A graphics calculator can be used to generate Fibonacci-type sequences in a number of ways.

For the Casio fx-9860G AU A spreadsheet approach is shown in the steps below. 1. Press MENU and select S-SHT to open a spreadsheet. Consider the first column, A, as Fn − 1, the second column, B, as Fn and the third column, C, as Fn + 1. Enter the two starting values in cells A1 and B1. That is, enter 1 in cell A1 and 1 in cell B1. 2. In cell C1, type the formula = A1 + B1. This will generate the third term.

3. Press EXE to display the 3rd term in cell C1. Next we need to repeat the last 2 terms (2nd and 3rd terms) in row 2 so they can be added to produce the next term. In cell A2, type the formula = B1 and press EXE . In cell B2, type the formula = C1 and press EXE . 4. Highlight cell C1 and press F2 (EDIT) followed by F2 (COPY) to copy the formula. Move to cell C2 and press F1 (PASTE) to paste the formula into this cell. This generates the 4th term of the sequence. 5. We can use this copy and paste technique to complete each column down to, say, the 10th row. For Column A, highlight cell A2, press EXIT to display the Copy option and then F2 (COPY) to copy the formula. Move to cell A3 and press F1 (PASTE) to paste the formula into this cell.

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Now move to cell A4 and press F1 (PASTE) to paste the formula into this cell. Continue repeating this paste step as you move down the cells of this column to row 10. Repeat the same process for Column B and then for Column C. Note that some cells may show a value of 0 until all the pasting has been completed. The first 10 terms of the Fibonacci sequence can be seen in Column A.

For the TI-Nspire CAS On this calculator, we can use the Generate Sequence function to display the terms of any Fibonacci-style sequence. The rule for a Fibonacci Sequence is Fn + 1 = Fn + Fn − 1 with F0 = 1 and F1 = 1 (or any two initial values). An equivalent rule to this is Fn = Fn − 1 + Fn − 2 with F1 = 1 and F2 = 1. The second version of the rule suits this calculator as the rule needs to be expressed as u(n) =. So the rule becomes u(n) = u(n−1) + u(n−2) with u(1) = 1 and u(2) = 1. 1. Open a new Lists & Spreadsheet document. Press b, then 3: Data and 1: Generate Sequence. Enter the rule as u(n−1) + u(n−2) with initial terms of 1 and 1. Press e until OK is highlighted. 2. Press · to display the list of terms.

Note that you can also use a spreadsheet approach to generate a Fibonacci Sequence. Follow the steps given for the Casio fx-9860G AU calculator above and use /C to copy a cell and /V to paste the formula into another cell.

The golden ratio An interesting, if mysterious, product of the Fibonacci Sequence is the ‘golden ratio’ (or golden mean). For centuries, artists have recognised in painting and architecture that some shapes are, of themselves, more appealing than others.

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The ratio of height to width (or width to height) of rectangles that appeal to the eye is called the ‘golden ratio’ and its value is 1.6 : 1. More exactly the ‘golden ratio’ is taken to be 1+ 5 ---------------- : 1 2 How is this number connected with the Fibonacci Sequence? We can write the Fibonacci Sequence as follows: F1, F2, F3, . . . Fn − 1, Fn, Fn + 1 where Fn is the nth term in the Fibonacci Sequence. Fn + 1 Consider the ratio of successive terms, -----------. As n increases, this fraction tends to a Fn fixed value, the ‘golden ratio’. For proof of this: Fn + 1 Let -----------= r. Fn By definition of the Fibonacci Sequence this means Fn + Fn – 1 ------------------------- = r Fn Fn Fn – 1 ------ + ------------ = r Fn Fn 1 1 + --- = r r r + 1 = r2 r2 − r − 1 = 0 1+ 5 r = ---------------2

(We reject the negative solution.)

Thus the ratio of terms in the Fibonacci Sequence tends to the ‘golden ratio’.

Fibonacci numbers 1 Use your graphics calculator to develop a table of the Fibonacci Sequence. a Find the ratio of the 20th and 21st terms. 1+ 5 ---------------- = 1.618 034 b How close is this ratio to the ‘golden ratio’? 2

10 946 ---------------- = 1.618 034 6765

2 Verify that the formula for the nth Fibonacci number is: n

n

1 1+ 5 1 1– 5 Fn = -------  ---------------- − -------  ----------------   2 5 5 2  for the first 5 terms of the sequence. Only use your calculator as a last resort See Solutions Manual and use techniques such as difference of squares to do the calculations. SLE 8: Investigate the way in which a sequence can be defined recursively.

3 Consider a population of rabbits where each pair produces another pair each month. However, after a pair of rabbits produces a sixth pair, both rabbits die. a Produce a spreadsheet to model the number of rabbits over time. b Produce a formula (either explicit or recursive) to model the number of rabbits in the population over time.

See Solutions Manual

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The Mandelbrot Set Now that we have developed some of the key concepts of sequences let us return to the development of the Mandelbrot Set. SLE 8: Investigate the way in which a sequence can be defined recursively. SLE 12: (Real and complex number systems): Investigate fractals; for example, the Mandelbrot Set, using a CAS calculator.

First recall that complex numbers can be represented as points on the complex plane. For each point c in the complex plane, consider the sequence formed by zn + 1 ← z2n + c where zn and c are complex numbers and z0 = 0. If the sequence converges to a point for a particular c then c belongs to the Mandelbrot Set. If the sequence diverges, then c does not belong to the Mandelbrot Set. If the sequence, for a particular c, does not diverge, this point in the complex plane is coloured black. Otherwise this point is assigned a colour. A two-colour (black and red) version of the Mandelbrot Set is arranged thus:

No — colour the point black. Does the sequence diverge? Yes — colour the point red. Colour versions of the Mandelbrot Set are arranged thus:

No — colour the point black. Does the sequence diverge? Yes — assign a colour depending on how quickly the divergence occurs. Consider, for example, the complex number: c = 0.2 + 0.3i

Chapter 9 Sequences and series

eBook plus Digital doc: EXCEL Spreadsheet Mandelbrot Set

447

Using your graphics calculator or a spreadsheet it is possible to determine whether or not this value of c belongs to the Mandelbrot Set. (That is, whether the sequence formed by zn + 1 ← z2n + c converges with a given value of c.) The following spreadsheet shows that 0.2 + 0.3i converges and hence belongs to the Mandelbrot Set.

For the complex number c = 0.8 + 0.1i we obtain the following spreadsheet screen. The number diverges and hence does not belong to the Mandelbrot Set.

This spreadsheet approach can also be used with a graphics calculator. Use the formulas shown in the spreadsheet above. Alternatively, a CAS graphics calculator can be used to generate a sequence of the type zn + 1 = (zn)2 + c and enable you to check for the convergence of the sequence for given values of c.

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For the TI-Nspire CAS 1. Open a new Calculator document (press /N and select 1: Add Calculator). So that we can use different values for c, we shall first define c. Press b and select 1: Actions and 1: Define. Enter c = 0.2 + 0.3i and press ·.

2. Insert a new Lists & Spreadsheet document (press /I and select 3: Add Lists & Spreadsheet). To generate a sequence, press b and select 3: Data and 1: Generate Sequence. Notice that the formula needs to be of the form u(n) =.

3. The formula zn + 1 = (zn)2 + c can also be written as zn = (zn − 1)2 + c. So, in terms of u rather than z, the formula to be entered into the calculator needs to be u(n) = [u(n – 1)]2 + c. Enter the formula in the first line and then press e to move to the next line. Enter 0 and c as the initial terms. Continue to press e until OK is highlighted. 4. Press · to generate the sequence. Resize column A to clearly view the terms.

5. Use the NavPad to scroll down the column. You will see that the terms converge to a value of approximately 0.0792 + 0.3565i.

6. To try other values of c, return to the Calculator document (press / and the left arrow on the NavPad). Again, press b and select 1: Actions and 1: Define. This time define c as 0.8 + 0.1i.

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7. Return to the Lists & Spreadsheet document (press / and the right arrow on the NavPad). The new terms of the sequence will be automatically displayed. Use the NavPad to scroll down the column to check whether the terms converge or diverge.

Draw the Mandelbrot Set In groups of four, use four sheets of 1-centimetre grid paper and proceed as follows. Assign each member of the group a portion of the Mandelbrot Set for which he or she is responsible. Ensure each member uses the same scale so that the four individual sheets can be collated to form a coherent picture. For each square on the grid paper: 1 Calculate the coordinates of the midpoint. 2 Use a graphics calculator or a spreadsheet to determine if the sequence zn + 1 ← z2n + c diverges with this value of c. 3 If it diverges, colour the square white, otherwise colour the square black. 4 Repeat until all squares are coloured.

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summary Arithmetic sequences Recognising arithmetic sequences • An arithmetic sequence is a sequence of numbers for which the difference between successive terms is the same. • Given an arithmetic sequence, identify: the first term, a and the common difference, d = t2 − t1 . • Given an unspecified sequence, establish whether it is arithmetic by testing all terms for a common difference: d = t2 − t1 = t3 − t2 = t4 − t3 = . . . Finding the terms of an arithmetic sequence tn = a + (n − 1) d where tn is the nth term. The sum of a given number of terms of an arithmetic sequence • A series is the sum of terms in a sequence. • Sn is the sum of the first n terms in series, for example, S25 represents the sum of the first 25 terms. • Given a number of terms in a series, n, the first term, a, and the last term, l, use n S n = --- ( a + l ) . 2 • Given a number of terms in a series, n, the first term, a, and the common difference, n d, use S n = --- [ 2a + ( n – 1 )d ] . 2

Geometric sequences Recognising geometric sequences • A geometric sequence is a sequence of numbers for which the ratio of successive terms is the same. • Given a geometric sequence, identify: the first term, a t and the common ratio, r = ---2- . t1 • Given an unspecified sequence, establish whether it is geometric by testing all t t t terms for a common ratio, r = ---2- = ---3- = ---4- = … t1 t2 t3 Finding the terms of a geometric sequence • tn = ar n − 1 where tn is the nth term a is the first term r is the common ratio. The sum of a given number of terms of a geometric sequence • 2, 4, 8, 6, 32, 64 is a finite geometric sequence. • 2 + 4 + 8 + 16 + 32 + 64 is a finite geometric series.

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• The sum of n terms, Sn, of a geometric sequence may be calculated using a(rn – 1) S n = ---------------------- if r > 1 or r < –1 for example, r = –2, – 3--- , +2 or +4.5 2 r–1 or a(1 – rn) S n = ---------------------- if −1 < r < 1, for example, r = 0.2, 1--- or −0.25. 8 1–r

Geometric growth • Growth or increase is expressed as a percentage increase. • Common ratio, r = 1 + percentage increase • r values are greater than 1, for example, an 8% increase gives r = 1.08.

Geometric decay • Decay or decrease is expressed as a percentage decrease. • Common ratio, r = 1 − percentage decrease • r values are less than 1, for example, an 8% decrease gives r = 0.92.

Compound interest • A = PRn r where R = 1 + --------100 A = amount in the account, $ P = principal, $ r = interest rate per period (that is, per year or quarter etc.) as a percentage n = the number of periods during the investment.

Finding the sum of an infinite geometric sequence • For decreasing or decaying geometric series, the sum of an infinite number of terms approaches a finite sum. • The sum to infinity of a geometric sequence for which −1 < r < 1 is given by a S ∞ = ----------- . 1–r

Contrasting arithmetic and geometric sequences through graphs

Value of term tn

Value of term tn

• Arithmetic patterns are distinguished by a straight line.

1

2 3 4 Term n

5

d is positive

An increasing pattern or a positive common difference gives an upward straight line.

1

2 3 4 Term n

5

d is negative

A decreasing pattern or a negative common difference gives a downward straight line.

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Value of term tn

Value of term tn

• Geometric patterns are distinguished by a curved line or a saw form.

1

2 3 4 Term n

5

An increasing pattern or a positive common ratio greater than 1 (r > 1) gives an upward curved line.

1

5

2

3

4

5

Term n

An increasing saw pattern occurs when the common ratio is a negative value less than −1 (r < −1).

Value of term tn

+

1

–

2 3 4 Term n

A decreasing pattern or a positive fractional common ratio (0 < r < 1) gives a downward curved line.

+ Value of term tn

452

1

–

2 3

4

5

Term n

A decreasing saw pattern occurs when the common ratio is a negative fraction (–1 < r < −0).

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CHAPTER review 1 multiple choice Which of the following could be the first 5 terms of an arithmetic sequence? A 1, 2, 4, 8, 12, . . . B 3, 3, 6, 6, 9, . . . C 56, 57, 58, 59, 60, . . . D −5, 5, 10, 15, 20, . . . E 1, 4, 9, 16, 25, . . . 2 multiple choice For the following sequence, −3.6, −2.1, −0.6, 0.9, 2.4, . . ., it is true to say that it is: A an infinite sequence with a = −3.6 and d = −0.15 B an infinite sequence with a = −3.6 and d = 1.5 C an infinite sequence with a = −0.15 and d = −3.6 D a finite sequence with a = −0.15 and d = −3.6 E a finite sequence with a = −3.6 and d = 0.15. 3 For the sequences below, state whether or not they are an arithmetic sequence. If they are, give the value of a and d. a −123, −23, 77, 177, 277, . . . Yes, a = −123, d = 100 b −5 1--- , −2 1--- , 3--- , 3 3--- , 6 3--- , . . . 4

4

4

4

4

9A

9A

9A

Yes, a = −5 --14- , d = 3

4 multiple choice For the arithmetic sequence, −1, 1, 3, 5, 7, . . . the value of a, the value of d and the rule for the sequence are given respectively by: A a = −1, d = 2, tn = −3 + 2n B a = −1, d = 2, tn = −3 − n D a = 2, d = −1, tn = 3 − n C a = 1, d = −1, tn = 2 − n E a = 2, d = −1, tn = 3 − n 5 multiple choice The 43rd term of the arithmetic sequence −7, 2, 11, 20, 29, . . . is: A −327 B −243 C 371 D 380

9A

9A E 387

6 multiple choice The 3rd term of an arithmetic sequence is 3.1 and the 7th term is −1.3. The value of the 31st term is: A −153.7 B −27.7 C 28.9 D 38.3 E 157.9

9A

7 If the second term of an arithmetic sequence is −5 and the 5th term is 16, which term in the sequence is equal to 226? Term number 35

9A

8 Blood donations at a suburban location increased by 40 each year. If there were 520 donations in the first year: a how many donations were made in the 15th year? 1080 b what was the total number of donations made over those 15 years? 12 000

9A

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9A

9 multiple choice

9A

10 multiple choice

9B

11 multiple choice

9B

12 multiple choice

9B

The sum of the first 24 terms of the sequence −16, −12, −8, –4, 0, . . . is: A 720 B 912 C 1344 D 1440 E 1488

The first term of an arithmetic sequence is 14 and the 3rd is 8. The sum of the first 30 terms of the sequence is: A −1770 B −1095 C −885 D 1725 E 2190

There is a geometric sequence for which a = 3 and r is a negative number. We can be certain that: A r is a fraction less than 1 B the 3rd term will be a positive number C the 3rd term will be greater than the 1st term D only one number in the sequence is positive E the 4th term will be greater than the 3rd term.

Which of the following is a geometric sequence? A 2, −2, 2, −2, 2, . . . B 2, 4, 6, 8, 10, . . . C 1, 1--- , − 1--- , 3 9 D 4, −4, 2, −2, 1, . . . E 100, 10, 0.1, 0.01, 0.001, . . .

1− ----, ... 81

13 For each of the sequences below, state whether or not they are a geometric sequence. If they are, state the value of a and r. a 5, 5--- , 5--- , 5--- , b

5----, 16

Yes, a = 5, r = −700, −70, −7, 7, 70, .No .. 2

9B

1----, 27

4

8

...

1--2

14 multiple choice The 19th term of the geometric sequence 3.25, 6.5, 13, 26, 52, . . . is: A 425 984 B 851 968 C 1 703 936 D 41 978 243

9B

15 multiple choice

9B

16 multiple choice

E 3 272 883 098

The 3rd term of a geometric sequence is 19.35 and the 6th is 522.45. The 12th term of the sequence is: A 16 539.15 B 417 629.75 C 126 955.35 D 380 866.05 E 1 142 598.15

The first 3 terms of a geometric sequence are 2.25, 4.5, 9. The first term which would exceed 1000 is: A t9 B t10 C t11 D t12 E t13

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17 The amount of garbage (in tonnes) collected in a particular area by the local council each year is recorded over 3 successive years.

Year number

9B

Amount of garbage (tonnes)

1

7.2

2

8.28

3

9.522

If the amount collected each year were to continue to follow a geometric sequence: a Write down a rule for the amount of garbage, tn, which would be collected in the area in year n. tn = 7.2 × 1.15n − 1 b How much garbage would be collected in the 8th year? (Answer correct to 2 decimal places.) 19.2 tonnes c In which year would the amount of garbage collected exceed 30 tonnes? Year 12 18 multiple choice The sum of the first 10 terms of the geometric sequence 8, 4, 2, 1, 1--- , . . . is closest to:

9B

2

A 15

B 16

C 17

D 18

E 20

19 multiple choice The 3rd term of a geometric sequence is 0.9 and the 6th is 7.2. The sum of the first 12 terms of the sequence is closest to: A −2

B 122

C 921

D 4122

9B

E 8190

20 How many terms of the geometric sequence 164, 131.2, 104.96, 83.968, 67.1744, . . . are required for the sum to exceed 800? 17

9B

21 multiple choice A tree increases in height each year by 5%. If it was 1.2 m high in its 1st year, in its 6th year its height would be closest to: A 1.53 m B 1.61 m C 5.5 m D 9.11 m E 3750 m

9C

22 multiple choice Profits in a company are projected to increase by 8% each year. If the profit in the first year was $60 000, in which year could a profit in excess of $100 000 be expected?

9C

A year 6

B year 7

C year 8

D year 9

E year 10

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9C

23 Anya invests $25 000 in an account earning compound interest of 10% per annum $33 622.22 compounding quarterly. a Find the amount in the account after 3 years. b Find how long it would take to have $40 965.41 in her account. 5 years

9D

24 multiple choice The sum to infinity of the geometric sequence 64- 256 ------ , -------1, 4--- , 16 , --------- , . . . is: 5

A

9D

25

1 --5

125

625 B 5--4

C 1 4--5

D 4

E 5

25 multiple choice The first term of the geometric sequence for which r = −0.5 and S∞ = 5 1--- is: 3

A −1

B −2 1--3

C 2 2---

9D 9D

26 Express 3.7 as a fraction.

9E

28 multiple choice

3

3 7--9-

The first five terms of a sequence are plotted on the graph at right. The sequence could be described by which of the following? A Arithmetic sequence with a = 50 and d = 25 B Arithmetic sequence with a = 50 and d = 0.5 C Geometric sequence with a = 50 and r = 0.5 D Geometric sequence with a = 50 and r = 1.5 E Geometric sequence with a = 50 and d = 2

Value of term

27 The batteries in a toy soldier are running down. The toy soldier marches 50 cm in the first minute, 30 cm in the second minute, 18 cm in the next and so on. By how much does the toy soldier fall short of marching 1.5 m? 25 cm

300 250 200 150 100 50 0

0

1

2 3 4 5 Term number

29 multiple choice The first five terms of a sequence are plotted on the graph at right. The sequence could be described by which of the following? A Arithmetic sequence with a = 10 and d = −5 B Arithmetic sequence with a = 10 and d = −0.5 C Geometric sequence with a = 10 and r = 5 D Geometric sequence with a = 10 and r = −5 E Geometric sequence with a = 10 and d = 5

15 10 5 0 –5 –10 –15

1

2

30 On the same set of axes, sketch the graph of the sequence with the rule: a un = 10n b vn = 10 × 2n − 1

3

4

30

5 Term number

80 70 Amount ($)

9E

E 10 2---

Value of term

9E

3

D 8

60 50 40 30

Legend

20

Vn = 10 × 2n – 1

10

Un = 10n 1

4 2 3 Term number (n)

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Modelling and problem solving 1 A newly established quarry produces crushed rock for the building of roads and freeways. The amount of crushed rock, in tonnes, it produces increases by 3 1--- tonnes each month and its 2 production for the first 3 months of operation is shown below. Month

Crushed rock produced (tonnes)

1

8

2

11.5

3

15

a Write down the amount of crushed rock produced in the 4th month. 18.5 tonnes b Write down a rule for tn, the amount of crushed rock produced in month n, expressed in terms of n, the nth month. tn = 4.5 + 3.5n c Write down the amount of crushed rock produced in the 60th month. 214.5 tonnes d During which month will the amount of crushed rock coming from the quarry exceed 100 tonnes? 28th month e The local council has ordered that after a total of 3050 tonnes of crushed rock has been extracted from the quarry, an environmental impact survey must be completed. After how many months will that happen? 40 months

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2 The amount of crushed rock produced each month at a second quarry is shown below. Month

Crushed rock produced (tonnes)

1

10

2

11

3

12.1

Given that production at this quarry increases geometrically, find: a the common ratio, r 1.1 b a rule for the amount of crushed rock produced, tn, in tonnes, expressed in terms of the number of months, n tn = 10 × 1.1n − 1 c the amount of crushed rock produced in the 5th month 14.641 tonnes d in which month the amount of crushed rock produced exceeds 30 tonnes 13th month e the total amount of crushed rock produced by the quarry in its first year of operation. 213.84 tonnes 3 During its first month of production, the second quarry produced more crushed rock than the first quarry. In the months after that, however, the first quarry produced more crushed rock than the second quarry. 24th month After how many months does the second quarry produce more than the first quarry again? 4 One bank offers a simple interest rate of 5% per annum on an investment of $100. For the same investment, another bank offers 5% interest compounded annually. When will the value of the investment earning compound interest be twice as much as the value of the investment eBook plus earning simple interest? 35 years Digital doc: Test Yourself Chapter 9

Permutations and combinations

10 syllabus reference Core topic: Structures and patterns

In this chapter 10A The addition and multiplication principles 10B Factorials and permutations 10C Arrangements involving restrictions and like objects 10D Combinations 10E Applications of permutations and combinations 10F Pascal’s Triangle, the binomial theorem and the pigeonhole principle

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• permutations and combinations and their use in purely mathematical and life-related situations

Introduction Combinatorics deals with determining the number of ways in which activities or events may occur. The study of combinatorics provides ways of answering questions such as: 1. How many doubles teams can be selected from a group of 6 volleyball players? 2. From a group of 4 candidates, in how many ways can a class captain and deputy class captain be selected? 3. How many different outfits can be chosen from 3 skirts and 5 tops? 4. If a Lotto ticket consists of a choice of 6 numbers from 45, how many different tickets are there? 5. How many different car number plates of 3 digits and 3 letters can be made using the digits 0 to 9 and the letters A, B and C?

The addition and multiplication principles To count the number of ways in which an activity can occur, first make a list. Let each outcome be represented by a letter and then systematically list all the possibilities. Consider the following question: In driving from Melbourne to Sydney I can take any one of 4 different roads and in driving from Sydney to Brisbane there are 3 different roads I can take. How many different routes can I take in driving from Melbourne to Brisbane? To answer this, let M1, M2, M3, M4 stand for the 4 roads from Melbourne to Sydney and B1, B2, B3 stand for the 3 roads from Sydney to Brisbane. Use the figure to systematically list the roads: M1B1, M1B2, M1B3 M1 B1 M2 M2B1, M2B2, M2B3 B M B2 M3 S M3B1, M3B2, M3B3 B 3 M4 M4B1, M4B2, M4B3 Hence, there are 12 different ways I can drive from Melbourne to Brisbane. In the above example it can be argued logically that if there are 4 ways of getting from Melbourne to Sydney and 3 ways of getting from Sydney to Brisbane then there are 4 × 3 ways of getting from Melbourne to Brisbane. This idea is formalised in the multiplication principle. The multiplication principle should be used when there are two operations or events (say, A and B), where one event is followed by the other. It states: If there are n ways of performing operation A and m ways of performing operation B, then there are n × m ways of performing A and B. Note: In this case ‘and’ means to multiply.

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A useful technique for solving problems based on the multiplication principle is to use boxes. In the example above we would write 1st

2nd

4

3

The value in the ‘1st’ column represents the number of ways the first operation — the trip from Melbourne to Sydney — can be performed. The value in the ‘2nd’ column stands for the number of ways the second operation — the trip from Sydney to Brisbane — can be performed. To apply the multiplication principle you multiply the numbers in the lower row of boxes. Now consider a different situation, one in which the two operations do not occur one after the other. I am going to travel from Melbourne to either Sydney or Adelaide. There are 4 ways of travelling from Melbourne to Sydney and 3 ways of travelling from Melbourne to Adelaide. How many different ways can I travel to either Sydney or Adelaide? It can be seen from the figure that there 1 1 are 4 + 3 = 7 ways of completing the 2 2 A S 3 journey. This idea is summarised in the M 3 4 addition principle. The addition principle should be used when two distinct operations or events occur in which one event is not followed by another. It states: If there are n ways of performing operation A and m ways of performing operation B then there are n + m ways of performing A or B. Note: In this case ‘or’ means to add.

WORKED Example 1 Two letters are to be chosen from A, B, C, D and E, where order is important. a List all the different ways that this may be done. b State the number of ways that this may be done. THINK a 1 Begin with A in first place and make a list of each of the possible pairs. 2 Make a list of each of the possible pairs with B in the first position. 3 Make a list of each of the possible pairs with C in the first position. 4 Make a list of each of the possible pairs with D in the first position. 5 Make a list of each of the possible pairs with E in the first position. Note: AB and BA need to be listed separately as order is important.

WRITE a AB AC AD AE BA BC BD BE CA CB CD CE DA DB DC DE EA EB EC ED

Continued over page

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THINK WRITE b There are 20 ordered pairs. b Method 1 Count the number of ordered pairs and answer the question. Alternatively, the multiplication principle could have been used to determine the number of ordered pairs. b Method 2 b 5 4 1 Rule up two boxes which represent the pair. 2 Write down the number of letters which may be selected for the first box. That is, in first place any of the 5 letters may be used. 3 Write down the number of letters which may be selected for the second box. That is, in second place, any of the 4 letters may be used. Note: One less letter is used to avoid repetition. 5 × 4 = 20 ways 4 Evaluate. Answer the question. There are 20 ways in which 2 letters 5 may be selected from a group of 5 where order is important. A selection where order is important is called an arrangement.

WORKED Example 2

How many ways could an arrangement of 5 letters be chosen from A, B, C, D, E and F? THINK 1

2

3 4

WRITE

Instead of listing all possibilities, draw 5 boxes to represent the 5 letters chosen. Label each box on the top row as 1st, 2nd, 3rd, 4th and 5th. Note: The word arrangement implies order is important. Fill in each of the boxes showing the number of ways a letter may be chosen. (a) In the 1st box there are 6 choices for the first letter. (b) In the 2nd box there are 5 choices for the second letter as 1 letter has already been used. 1st 2nd 3rd 4th 5th (c) In the 3rd box there are 4 choices for the third letter as 2 letters have already been used. 6 5 4 3 2 (d) Continue this process until each of the 5 boxes is filled. Use the multiplication principle as this is an No. of ways = 6 × 5 × 4 × 3 × 2 ‘and’ situation. = 720 Answer the question. An arrangement of 5 letters may be chosen 720 ways.

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WORKED Example 3

One or two letters are to be chosen from 6 letters A, B, C, D, E and F. In how many ways can this be done? THINK WRITE Determine the number of ways of choosing 1 letter. No. of ways of choosing 1 letter = 6. 1 2 Rule up two boxes for the first and second letters. 3 Determine the number of ways of choosing 2 letters 1st 2nd from 6. (a) In the 1st box there are 6 choices for the first letter. 6 5 (b) In the 2nd box there are 5 choices for the second letter as 1 letter has already been used. No. of ways of choosing 2 letters Use the multiplication principle (as this is an 4 =6×5 ‘and’ situation) to evaluate the number of ways of = 30 choosing 2 letters from 6. The number of ways of choosing 5 Determine the number of ways of choosing 1 or 2 1 or 2 letters is 6 + 30 = 36. letters from 6 letters. Use the addition principle as this is an ‘or’ situation. There are 36 ways of choosing 1 or 2 6 Answer the question. letters from 6.

WORKED Example 4

Jeannine’s restaurant offers its patrons a choice of 3 entrees, 9 main courses and 4 desserts. a How many choices of 3-course meals (entree, main, dessert) are available? b How many choices of entree and main course are offered? c How many choices of main course and dessert are offered? d How many choices of 2- or 3-course meals are available (assuming that a main course is always ordered)? THINK a 1 Rule up 3 boxes to represent each course — entree, main, dessert. Label each box on the top row as E, M and D. 2 3

4

Determine the number of ways of choosing each meal: entree = 3, main = 9, dessert = 4. Use the multiplication principle (as this is an ‘and’ situation) to evaluate the number of choices of 3-course meals. Answer the question.

WRITE a

E

M

D

3

9

4

No. of choices = 3 × 9 × 4 = 108 There are 108 choices of 3-course meals. Continued over page

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THINK

WRITE

b

b

1

2 3

4

c

1

2 3

4

d

1

2

Rule up 2 boxes to represent each course — entree, main. Label each box on the top row as E and M. Determine the number of ways of choosing each meal: entree = 3, main = 9. Use the multiplication principle (as this is an ‘and’ situation) to evaluate the number of choices of entree and main courses. Answer the question.

Rule up 2 boxes to represent each course — main and dessert. Label each box on the top row as M and D. Determine the number of ways of choosing each meal: main = 9, dessert = 4. Use the multiplication principle (as this is an ‘and’ situation) to evaluate the number of choices of main course and dessert. Answer the question.

Determine the number of ways of choosing 2- or 3-course meals, assuming that a main course is always ordered. Use the addition principle as this is an ‘or’ situation. Answer the question.

E

M

3

9

No. of choices = 3 × 9 = 27 There are 27 choices of entree and main course. c

M

D

9

4

No. of choices = 9 × 4 = 36 There are 36 choices of main course and dessert. d The number of ways of choosing 2- or 3-course meals, assuming that a main course is always ordered, is: 108 + 27 + 36 = 171 There are 171 ways of choosing 2- or 3-course meals, assuming that a main course is always ordered.

remember remember 1. The multiplication principle should be used when there are two operations or events (say, A and B) where one event is followed by the other. It states: If there are n ways of performing operation A and m ways of performing operation B, then there are n × m ways of performing A and B. 2. The addition principle should be used when two distinct operations or events occur in which one event is not followed by another. It states: If there are n ways of performing operation A and m ways of performing operation B, then there are n + m ways of performing A or B. 3. A selection where order is important is called an arrangement.

Chapter 10 Permutations and combinations

10A WORKED

Example

1 2 BG GB YB RB BY GY YG RG BR GR YR RY 3 ACB BAC CAB ABC BCA CBA WORKED

Example

2

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The addition and multiplication principles

1 Two letters are to be chosen from A, B and C, where order is important. a List all the different ways that this may be done. a AB BA CA AC BC CB b State the number of ways that this may be done. 6 2 List all the different arrangements possible for a group of 2 colours to be chosen from B (blue), G (green), Y (yellow) and R (red). 3 List all the different arrangements possible for a group of 3 letters to be chosen from A, B and C. 4 a In how many ways can an arrangement of 2 letters be chosen from A, B, C, D, E, F and G? 42 b In how many ways can an arrangement of 3 letters be chosen from 7 different letters? 210 c In how many ways can an arrangement of 4 letters be chosen from 7 different letters? 840 d How many different arrangements of 5 letters can be made from 7 letters? 2520 5 a A teddy bear’s wardrobe consists of 3 different hats, 4 different shirts and 2 different trousers. How many different outfits can the teddy bear wear? 24 b A surfboard is to have 1 colour on its top and a different colour on its bottom. The 3 possible colours are red, blue and green. In how many different ways can the surfboard be coloured? 6 c A new computer system comes with a choice of 3 keyboards, 2 different monitors and 2 different mouse attachments. With these choices, how many different arrangements are possible? 12 d Messages can be sent by placing 3 different coloured flags in order on a pole. If the flags come in 4 colours, how many different messages can be sent? 24 e A yacht race has a field of 12 competitors. In how many different ways can first, second and third place be filled by these 12 yachts? 1320

WORKED

Example

3

6 a One or 2 letters are to be chosen from the letters A, B, C, D, E, F and G. In how many ways can this be done? 49 b Two or 3 letters are to be chosen from the letters A, B, C, D, E, F and G. In how many ways can this be done? 252 c How many 1- or 2-digit numbers can be made using the digits 1, 3, 5 and 7 if no digit can be used more than once? 16 7 Nadia is in a race with 10 other girls. a If we are only concerned with the first, second and third placings, in how many ways can: i Nadia finish first? 90 ii Nadia finish second? 90 b In how many ways can Nadia finish first or second? 180 8 White Wolf is a horse in a race with 7 other runners. If we are concerned only with the first, second and third placings, in how many ways can White Wolf finish first or second or third? 126

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9 multiple choice There are 12 people on the committee at the local softball club. In how many ways can a president and a secretary be chosen from this committee? A 2 B 23 C 132 D 144 E 66 10 multiple choice Phone numbers consist of 8 digits. The first must be a 9. The second digit can be a 3, 4, 5 or 8. There are no restrictions on the remaining digits.

How many different telephone numbers are possible? A 4320 B 499 999 C 4 000 000 D 4 999 999

E 10 000 000

11 multiple choice A TV station runs a cricket competition called Classic Catches. Six catches, A to F, are chosen and viewers are asked to rank them in the same order as the judges. The number of ways in which the six catches can be ranked: A 1 B 6 C 30 D 720 E 128 12 The local soccer team sells ‘doubles’ at each of their games to raise money. A ‘double’ is a card with 2 digits on it representing the score at full time. The card with the actual full time score on it wins a prize. If the digits on the cards run from 00 to 99, how many different tickets are there? 100 13 Marcus has a briefcase that has a 4-digit security code. He remembers that the first number in the code was 9 and that the others were 3, 4 and 7 but forgets the order of the last 3 digits. How many different trials must he make to be sure of unlocking the briefcase? 6 14 Julia has a briefcase that has two 4-digit locks. She remembers that she used the digits 1, 3, 5 and 7 on the left lock and 2, 4, 6 and 8 on the right lock, but can not remember the order. What is the maximum number of trials she would need to make before she has opened both the left lock and the right lock? 48

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Chapter 10 Permutations and combinations

15 How many different 4-digit numbers can be made from the numbers 1, 3, 5 and 7 if the numbers can be repeated (that is 3355 and 7777 are valid)? 256 16 How many 4-digit numbers can be made from the numbers 1, 3, 5, 7, 9 and 2 if the numbers can be repeated? 1296 17 How many 4-digit numbers can be made from the numbers 1, 3, 5, 7, 9 and 0 if the numbers can be repeated? (Remember — a number cannot start with 0.) 1080 18 How many numbers less than 5000 can be made using the digits 2, 3, 5, 7 and 9 if repetition is not permitted? 133 19 A combination lock has 3 digits each from 0 to 9. a How many combinations are possible? 1000

9

9 0 1

3

7

6

2 3

0

7

4 5 6

6

8 5

4

4

0

3

2

Example

1

5

WORKED

The lock mechanism becomes loose and will open if the digits are within one either side of the correct digit. For example if the true combination is 382 then the lock will open on 271, 272, 371, 493 and so on. b How many combinations would unlock the safe? 27 c List the possible combinations that would open the lock if the true combination is 382.

1

471 472 473 481 482 483 491 492 493

7 8

371 372 373 381 382 383 391 392 393

2

4

c 271 272 273 281 282 283 291 292 293

9

8

20 Hani and Mary’s restaurant offers its patrons a choice of 4 entrees, 10 main courses and 5 desserts. a How many choices of 3-course meals (entree, main, dessert) are available? 200 b How many choices of entree and main course are offered? 40 c How many choices of main course and dessert are offered? 50 d How many choices of 2- or 3-course meals are available (assuming that a main course is always ordered)? 290 21 Jake is able to choose his work outfits from the following items of clothing: 3 jackets, 7 shirts, 6 ties, 5 pairs of trousers, 7 pairs of socks and 3 pairs of shoes. a How many different outfits are possible if he wears one of each of the above items? (He wears matching socks and shoes.) 13 230 b If Jake has the option of wearing a jacket and each of the above items, how many different outfits are possible? Explain your answer. 17 640 Jack may wear 13 230 outfits with a jacket or 4410 outfits without a jacket. Therefore he has a total of 17 640 outfits to choose from. The assumption made with this problem is that no item of clothing is exactly the same; that is, none of the 7 shirts is exactly the same.

Factorials and permutations Factorials

The Physical Education department is to display 5 new trophies along a shelf in the school foyer and wishes to know how many ways this can be done. Using the multiplication principle from the previous section, the display may be done in the following way: Position 1

Position 2

Position 3

Position 4

Position 5

5

4

3

2

1

That is, there are 5 × 4 × 3 × 2 × 1 = 120 ways.

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Depending on the number of items we have, this method could become quite time consuming. In general when we need to multiply each of the integers from a particular number, n, down to 1, we write n!, which is read as n factorial. Hence: 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40 320 n! = n × (n − 1) × (n − 2) × (n − 3) × . . . × 3 × 2 × 1 1. The number of ways n distinct objects may be arranged is n! (n factorial) where: n! = n × (n − 1) × (n − 2) × (n − 3) × . . . × 3 × 2 × 1 That is, n! is the product of each of the integers from n down to 1. 2. A special case of the factorial function is: 0! = 1.

Graphics Calculator tip! Calculating factorials The following steps show how to calculate 12! using a graphics calculator.

For the Casio fx-9860G AU 1. From the MENU select RUN-MAT. Press OPTN and then F6 ( ) for more options. s

468

2. Press F3 (PROB) and you will be able to see the function x!. Enter 12 and press F1 (x!). Press EXE to display the value of 12!.

For the TI-Nspire CAS 1. Open a new Calculator document. Enter 12 and then press / and k to access the symbol palette. Use the arrow keys to highlight the factorial symbol (exclamation mark).

2. Press · to insert the factorial symbol into the entry line. Press · to display the value.

Chapter 10 Permutations and combinations

469

WORKED Example 5 Evaluate the following factorials. a 7!

b 13!

18! c -------5!

9! d ----3!

( n – 1 )! e ------------------( n – 3 )!

THINK

WRITE

a

a 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

b

c

d

e

1

Write 7! in its expanded form and evaluate.

2

Verify the answer obtained using the factorial function on the calculator.

1

Write 13! in its expanded form and evaluate.

2

Verify the answer obtained using the factorial function on the calculator.

1

Write each factorial term in its expanded form.

2

Cancel down like terms.

=8×7×6

3

Evaluate.

= 336

4

Verify the answer obtained using the factorial function on the calculator.

1

Write each factorial term in its expanded form.

2

Cancel down like terms.

=9×8×7×6×5×4

3

Evaluate.

= 60 480

4

Verify the answer obtained using the factorial function on the calculator.

1

Write each factorial term in its expanded form.

2

Cancel down like terms.

b 13! = 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 ×3×2×1 = 6 227 020 800

c 8! × 7 × 6 × 5 × 4 × 3 × 2 × 1----- = 8------------------------------------------------------------------5! 5×4×3×2×1

d 9! × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1----- = 9---------------------------------------------------------------------------3×2×1 3!

e ( n – 1 )! -----------------( n – 3 )! ( n – 1 ) ( n – 2 ) ( n – 3 ) ( n – 4 ) ×… × 3 × 2 × 1 = ----------------------------------------------------------------------------------------------------------(n – 3)(n – 4) × … × 3 × 2 × 1 = (n − 1)(n − 2)

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In parts c, d and e of Worked example 5, there was no need to fully expand each factorial term. 8! 8 × 7 × 6 × 5! The factorial ----- could have first been simplified to -------------------------------- and then the 5! 5! 5! terms cancelled. 9! 9 × 8 × 7 × 6 × 5 × 4 × 3! The factorial ----- could have first been simplified to ------------------------------------------------------------ and 3! 3! then the 3! terms cancelled. ( n – 1 )! ( n – 1 ) ( n – 2 ) ( n – 3 )! The factorial ------------------ could have first been simplified to ----------------------------------------------------- and ( n – 3 )! ( n – 3 )! then the (n − 3)! terms cancelled.

Permutations The term permutation is often used instead of the term arrangement and in this section we begin by giving a formal definition of permutation. Previously, we learned that if you select 3 letters from 7 where order is important, the number of possible arrangements is: 1st

2nd

3rd

7

6

5

The number of arrangements = 7 × 6 × 5 = 210 7 × 6 × 5 × 4! 7! This value may also be expressed in factorial form: 7 × 6 × 5 = -------------------------------- = ----4! 4! Using more formal terminology we say that in choosing 3 things from 7 things where order is important the number of permutations is 7P3 = 7 × 6 × 5. The letter P is used to remind us that we are finding permutations. The number of ways of choosing r things from n distinct things is given by the rule: n Pr = n × (n − 1) × . . . × (n − r + 1) n ¥ ( n – 1 ) ¥ º ¥ ( n – r + 1 ) ( n – r )! = -----------------------------------------------------------------------------------------( n – r )! n! = -----------------( n – r )! The definition of nPr may be extended to the cases of nPn and nP0. Pn represents the number of ways of choosing n objects from n distinct things. n Pn = n × (n − 1) × (n − 2) × . . . × (n − n + 1) = n × (n − 1) × (n − 2) × . . . × 1 = n! From the definition: n! n Pn = -----------------( n – n )! n! = ----0! n! Therefore, equating both sides, we obtain: n! = ----- . 0! This can occur only if 0! = 1. n

Chapter 10 Permutations and combinations

471

n! P0 = -----------------( n – 0 )! n! = ----n! =1 In summary, the two special cases are: n

1. nPn = n! 2. nP0 = 1

WORKED Example 6

a Calculate the number of permutations for 6P4 by expressing it in expanded form. b Write 8P3 as a quotient of factorials and hence evaluate. THINK

WRITE

a

a 6P4 = 6 × 5 × 4 × 3 = 360 n! b nPr = -----------------( n – r )! 8! 8 P3 = -----------------( 8 – 3 )! 8! = ----5! 40 320 = ---------------120 = 336

b

2

Write down the first 4 terms beginning with 6. Evaluate.

1

Write down the rule for permutations.

2

Substitute the given values of n and r into the permutation formula.

3

Use a calculator to evaluate 8! and 5!

4

Evaluate.

1

WORKED Example 7 The netball club needs to appoint a president, secretary and treasurer. From the committee 7 people have volunteered for these positions. Each of the 7 nominees is happy to fill any one of the 3 positions. In how many different ways can these positions be filled? THINK

WRITE n! Pr = -----------------( n – r )!

1

Write down the rule for permutations. Note: Order is important, so use permutations.

n

2

Substitute the given values of n and r into the permutation formula.

7

3

Use a calculator to evaluate 7! and 4!

4

Evaluate. Answer the question.

5

7! P3 = -----------------( 7 – 3 )! 7! = ----4! 5040 = -----------24

= 210 There are 210 different ways of filling the positions of president, secretary and treasurer.

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Graphics Calculator tip! Calculating permutations To find the number of permutations of n objects taken r at a time, we need to calculate n Pr . The following steps show how to calculate 5P3 using a graphics calculator.

For the Casio fx-9860G AU 1. From the MENU select RUN-MAT. Press OPTN and then F6 ( ) for more options. s

472

2. Press F3 (PROB) and you will be able to see the function nPr. Enter 5 (for n), then press F2 (nPr) and enter 3 (for r). Press EXE to display the value.

For the TI-Nspire CAS 1. Open a new Calculator document. Press k to access the catalog and press 1 for the list of functions. Scroll down to select nPr(. You can do this more quickly by first pressing N.

2. Press · and then complete the entry line to obtain nPr(5,3). Press · to display the value. (Alternatively, you can use the letter keys to type npr followed by ( directly into the calculator screen instead of accessing it from the catalog. Press NPR followed by (, then complete the entry line and press ·.)

Arrangements in a circle Consider this problem: In how many different ways can 7 people be seated, 4 at a time, on a bench? By now you should quickly see the answer: 7P4 = 840. Let us change the problem slightly: In how many different ways can 7 people be seated, 4 at a time, at a circular table? A D The solution must recognise that when people are seated A B on a bench, each of the following represents a different D C arrangement: C B ABCD BCDA CDAB DABC C B However, when sitting in a circle, each represents the D C A B same arrangement. A D In each case B has A on the left and C on the right.

Chapter 10 Permutations and combinations

473

We conclude that the number 7P4 gives 4 times the number of arrangements of 7 people 7P in a circle 4 at a time. Therefore, the number of arrangements is --------4- = 210 . 4

In general, the number of different ways n people can be seated, r at a time, in a circle is: nP ---------r r

WORKED Example 8 How many different arrangements are possible if, from a group of 8 people, 5 are to be seated at a round table? THINK

WRITE

1

Write down the rule for the number of arrangements in a circle.

nP --------r r

2

Substitute the given values of n and r into the formula.

8P = --------55

3

Use a calculator to evaluate 8P5.

6720 = -----------5

4

Evaluate.

= 1344

5

Answer the question.

The number of ways of seating 5 from a group of 8 people at a round table is 1344.

remember remember 1. (a) The number of ways n distinct objects may be arranged is n! (n factorial) where: n! = n × (n − 1) × (n − 2) × (n − 3) × . . . × 3 × 2 × 1 (b) 0! = 1 (c) 1! = 1 2. (a) The number of different arrangements (permutations) when r things are chosen from n things and order is important is given by the rule nPr , where: n! n Pr = -----------------( n – r )! n (b) P = n! n

(c) nP0 = 1 3. The number of different ways n people can be seated, r at a time, in a circle is: nP --------r r

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10B

Factorials and permutations

1 Write each of the following in expanded form. a 4! 4 × 3 × 2 × 1 b 5! c 6! 5×4×3×2×1

2 Evaluate the following factorials. a 4! 24 b 5! 120 e 14! f 9! 362 880

WORKED

Example

5a, b 8.717 829 12 × 1010

3 Evaluate the following factorials. 9! 10! a ----- 3024 b -------- 151 200 5! 4!

WORKED

Example

5c, d

WORKED

Example

5e n(n − 1)(n − 2)(n − 3)(n − 4) WORKED

Example

6a Example

6b

Example

7

c 6! 720 g 7! 5040

c

c

d 10! 3 628 800 h 3! 6

7! ----- 840 3!

6! d ----- 720 0! 1 ------------------------------------------------------( n + 2 ) ( n + 1 )n ( n – 1 )

(-----------------n – 3 )! n!

( n – 2 )! d ------------------( n + 2 )!

(n + 3)(n + 2)

5 Calculate each of the following by expressing it in expanded form. a 8P2 8 × 7 = 56 b 7P5 c 8P7 8 × 7 × 6 × 5 × 4 × 3 × 2 = 40 320

6 Write each of the following as a quotient of factorials and hence evaluate. 5! 18! 9! - = 1 028 160 - = 60 480 b 5P2 ----- = 20 a 9P6 ---c 18P5 ------3! 13! 3! 7 Use your calculator to find the value of: a 20P6 27 907 200 b 800P2 639 200

WORKED

6×5×4×3×2×1

1 ------------------------------------n(n – 1)(n – 2)

4 Evaluate the following factorials. n! ( n + 3 )! a -----------------b ------------------( n – 5 )! ( n + 1 )!

7 × 6 × 5 × 4 × 3 = 2520

WORKED

d 7! 7 × 6 × 5 × 4 × 3 × 2 × 1

c

eBook plus Digital docs: SkillSHEET 10.1 Calculating nPr EXCEL Spreadsheet Permutations

18

P5 1 028 160

8 A soccer club will appoint a president and a vice-president. Eight people have volunteered for either of the two positions. In how many different ways can these positions be filled? 56 9 The school musical needs a producer, director, musical director and script coach. Nine people have volunteered for any of these positions. In how many different ways can the positions be filled? (Note: One person cannot take on more than 1 position.) 3024 Producer

Director

Coach

Musical Director

10 There are 14 horses in a race. In how many different ways can the 1st, 2nd and 3rd positions be filled? 2184 11 There are 26 horses in a race. How many different results for 1st, 2nd, 3rd and 4th can occur? 358 800 12 A rowing crew consists of 4 rowers who sit in a definite order. How many different crews are possible if 5 people try out for selection? 120

Chapter 10 Permutations and combinations

WORKED

Example

8

475

13 How many different arrangements are possible if, from a group of 15 people, 4 are to be seated in a circle? 8190 14 A round table seats 6 people. From a group of 8 people, in how many ways can 6 people be seated at the table? 3360 15 At a dinner party for 10 people all the guests were seated at a circular table. How many different arrangements were possible? 362 880 16 At one stage in the court of Camelot, King Arthur and 12 knights would sit at the round table. If each person could sit anywhere how many different arrangements were possible? 479 001 600

17 multiple choice Which one of the following permutations cannot be calculated? B 1P0 C 8P8 D 4P8 A 1000P100

E

5

P4

18 multiple choice The result of 100! is greater than 94!. Which of the following gives the best comparison between these two numbers? A 100! is 6 more than 94! B 100! is 6 times bigger than 94! C 100! is about 6! times bigger than 94! D 100! is about 10 000 more than 94! E 100! is 100P6 times bigger than 94! For questions 19 to 21 show your answers in the form nPr and then evaluate. 19 In how many ways can the letters of the word TODAY be arranged if they are used 5 P4 = 120 once only and taken: a 3 at a time? 5P3 = 60 b 4 at a time? c 5 at a time? 5P5 = 120 20 In how many ways can the letters of the word TUESDAY be arranged if they are used 7 P4 = 840 once only and taken: 7 P3 = 210 a 3 at a time? b 4 at a time? c 7 at a time? 7P7 = 5040 21 In how many ways can the letters of the word NEWTON be arranged if they are used once only and taken 6 at a time, assuming: a the first N is distinct from the second N? 6P6 = 720 b there is no distinction between the two Ns? 6P --------6- = 360 2

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Arrangements involving restrictions and like objects A 5-letter word is to be made from 3 As and 2 Bs. How many different arrangements can be made? If the 5 letters were all different, it would be easy to calculate the number of arrangements. It would be 5! = 120. Perhaps you can see that when letters are repeated, the number of different arrangements will be less than 120. To analyse the situation let us imagine that we can distinguish one A from another. We will write A1, A2, A3, B1 and B2 to represent the 5 letters. As we list some of the possible arrangements we notice that some are actually the same, as shown in the table.

A1A2B1A3B2

A1A2B2A3B1

A1A3B1A2B2

A1A3B2A2B1

A2A1B1A3B2

A2A1B2A3B1

A2A3B1A1B2

A2A3B2A1B1

A3A1B1A2B2

A3A1B2A2B1

A3A2B1A1B2

A3A2B2A1B1

B2A1A2B1A3

B1A1A2B2A3

B2A1A3B1A2

B1A1A3B2A2

B2A2A1B1A3

B1A2A1B2A3

B2A2A3B1A1

B1A2A3B2A1

B2A3A1B1A2

B1A3A1B2A2

B2A3A2B1A1

B1A3A2B2A1

Each of these 12 arrangements is the same — AABAB.

Each of these 12 arrangements is the same — BAABA.

The number of repetitions is 3! for the As and 2! for the Bs. Thus, the number of 5! different arrangements is ---------------- . 3! × 2! The number of different ways of arranging n things made up of groups of indistinguishable things, n1 in the first group, n2 in the second group and so on is: n! -------------------------------------- . n 1!n 2!n 3!ºn r! Note: If there are elements of the group which are not duplicated, then they can be considered as a group of 1. It is not usual to divide by 1!; it is more common to show only those groups which have duplications.

Chapter 10 Permutations and combinations

477

WORKED Example 9 How many different arrangements of 8 letters can be made from the word PARALLEL? THINK

WRITE

1

Write down the number of letters in the given word.

The word PARALLEL contains 8 letters; therefore n = 8.

2

Write down the number of times any of the letters are repeated.

The letter A is repeated twice; therefore n1 = 2. The letter L is repeated 3 times; therefore n2 = 3.

3

Write down the rule for arranging groups of like things.

n! -----------------------------------n 1!n 2!n 3!…n r!

4

Substitute the values of n, n1 and n2 into the rule.

8! = ---------------2! × 3!

5

Evaluate each of the factorials.

40 320 = ---------------2×6

6

Simplify the fraction.

40 320 = ---------------12

7

Evaluate.

= 3360

8

Answer the question.

3360 arrangements of 8 letters can be made from the word PARALLEL.

WORKED Example 10

How many different arrangements of 7 counters can be made from 4 black and 3 white counters? THINK 1 Write down the total number of counters. Write down the number of times any of 2 the coloured counters are repeated.

WRITE There are 7 counters in all; therefore n = 7. There are 3 white counters; therefore n1 = 3. There are 4 black counters; therefore n2 = 4.

3

Write down the rule for arranging groups of like things.

n! -----------------------------------n 1!n 2!n 3!…n r!

4 5

Substitute the values of n, n1 and n2 into the rule. Evaluate each of the factorials.

6

Simplify the fraction.

7

Evaluate. Answer the question.

7! = ---------------3! × 4! 5040 = --------------6 × 24 5040 = -----------144 = 35 Thirty-five different arrangements can be made from 7 counters, of which 3 are white and 4 are black.

8

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WORKED Example 11

A rowing crew of 4 rowers is to be selected, in order from the first seat to the fourth seat, from 8 candidates. How many different arrangements are possible if: a there are no restrictions? b Jason or Kris must row in the first seat? c Jason must be in the crew, but he can row anywhere in the boat? THINK a 1 Write down the permutation formula. Note: 4 rowers are to be selected from 8 and the order is important. 2 Substitute the given values of n and r into the permutation formula.

3 4 5

b

1

Use a calculator to evaluate 8! and 4!. Evaluate. Answer the question. Apply the multiplication principle since two events will follow each other; that is, Jason will fill the first seat and the remaining 3 seats will be filled in 7 × 6 × 5 ways or Kris will fill the first seat and the remaining 3 seats will be filled in 7 × 6 × 5 ways. J

c

7

6

5

or K

7

6

8! = -----------------( 8 – 4 )! 8! = ----4! 40 320 = ---------------24 = 1680 There are 1680 ways of arranging 4 rowers from a group of 8. b No. of arrangements = no. of ways of filling the first seat × no. of ways of filling the remaining 3 seats. = 2 × nPr 8

P4

5

2

Substitute the values of n and r into the formula and evaluate.

3

Answer the question.

1

Apply the addition principle, since Jason must be in either the first, second, third or fourth seat. The remaining 3 seats will be filled in 7 × 6 × 5 ways each time. J

7

6

5 + 7

J

6

5 +

7

6

J

5 + 7

6

5

J

2

Substitute the values of n and r into the formula.

3

Evaluate. Answer the question.

4

WRITE n! a n P r = -----------------( n – r )!

= 2 × 7P3 = 2 × 210 = 420 There are 420 ways of arranging the 4 rowers if Jason or Kris must row in the first seat. c No. of arrangements = No. of arrangements with Jason in seat 1 + No. of arrangements with Jason in seat 2 + No. of arrangements with Jason in seat 3 + No. of arrangements with Jason in seat 4.

No. of arrangements = 1 × 7P3 + 1 × 7P3 + 1 × 7P3 + 1 × 7P3 = 4 × 7P3 = 4 × 210 = 840 There are 840 ways of arranging the 4 rowers if Jason must be in the crew of 4.

Chapter 10 Permutations and combinations

479

WORKED Example 12

a How many permutations of the letters in the word COUNTER are there? b In how many of these do the letters C and N appear side by side? c In how many permutations do the letters C and N appear apart? THINK a 1 Count the number of letters in the given word. Determine the number of ways the 2 7 letters may be arranged. 3 Answer the question. b

4

Imagine the C and N are ‘tied’ together and are therefore considered as 1 unit. Determine the number of ways C and N may be arranged: CN and NC. Determine the number of ways 6 things can be arranged. Note: There are now 6 letters: the ‘CN’ unit along with O, U, T, E and R. Determine the number of permutations in which the letters C and N appear together. Evaluate.

5

Answer the question.

1

Determine the total number of arrangements of the 7 letters. Write down the number of arrangements in which the letters C and N appear together, as obtained in a. Determine the difference between the values obtained in steps 1 and 2. Note: The number of arrangements in which C and N are apart is the total number of arrangements less the number of times they are together. Answer the question.

1

2

3

c

2

3

4

WRITE a There are 7 letters in the word COUNTER. The 7 letters may be arranged 7! = 5040 ways. There are 5040 permutations of letters in the word COUNTER. b Let C and N represent 1 unit. They may be arranged 2! = 2 ways. Six things may be arranged 6! = 720 ways.

The number of permutations = 2 × 6! = 2 × 720 = 1440 There are 1440 permutations in which the letters C and N appear together. c Total number of arrangements = 7! = 5040 Arrangements with C and N together = 1440 The number of arrangements = 5040 − 1440 = 3600

The letters C and N appear apart 3600 times.

remember remember 1. The number of different ways of arranging n things made up of groups of indistinguishable things, n1 in the first group, n2 in the second group and so n! on is: ------------------------------------ . n 1!n 2!n 3!…n r! 2. When restrictions apply to arrangements, use the multiplication and addition principles as well as nPr .

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10C WORKED

Example

9

Arrangements involving restrictions and like objects

1 How many different arrangements can be made using the 6 letters of the word NEWTON? 360 2 How many different arrangements can be made using the 11 letters of the word ABRACADABRA? 83 160

WORKED

Example

10

eBook plus Digital doc:

3 How many different arrangements of 5 counters can be made using 3 red and 2 blue counters? 10

EXCEL Spreadsheet Permutations

4 How many different arrangements of 9 counters can be made using 4 black, 3 red and 2 blue counters? 1260 5 A collection of 12 books is to be arranged on a shelf. The books consist of 3 copies of Great Expectations, 5 copies of Catcher in the Rye and 4 copies of Huntin’, Fishin’ and Shootin’. How many different arrangements of these books are possible? 27 720

6 A shelf holding 24 cans of dog food is to be stacked using 9 cans of Yummy and 15 cans of Ruff for Dogs. In how many different ways can the shelf be stocked? 1 307 504 WORKED

Example

11

7 A cricket team of 11 players is to be selected, in batting order, from 15. How many different arrangements are possible if: a there are no restrictions? 5.4 × 1010 b Mark must be in the team at number 1? 3.6 × 109 c Mark must be in the team but he can be anywhere from 1 to 11? 4.0 × 1010 8 The Student Council needs to fill the positions of president, secretary and treasurer from 6 candidates. Each candidate can fill only one of the positions. In how many ways can this be done if: a there are no restrictions? 120 b Jocelyn must be secretary? 20 c Jocelyn must have one of the 3 positions? 60 9 The starting 5 in a basketball team is to be picked, in order, from the 10 players in the squad. In how many ways can this be done if: a there are no restrictions? 30 240 b Jamahl needs to play at number 5? 3024 c Jamahl and Anfernee must be in the starting 5? 6720

WORKED

Example

12

10 a How many permutations of the letters in the word MATHS are there? 120 b In how many of these do the letters M and A appear together? 48 c In how many permutations do the letters M and A appear apart? 72

Chapter 10 Permutations and combinations

481

11 A rowing team of 4 rowers is to be selected in order from 8 rowers.

a In how many different ways can this be done? 1680 b In how many of these ways do 2 rowers, Jane and Lee, sit together in the boat? 180 c In how many ways can the crew be formed without using Jane or Lee? 360 12 A horse race has 12 runners. a In how many ways can 1st, 2nd and 3rd be filled? 1320 b In how many ways can 1st, 2nd and 3rd be filled if Najim finishes first? 110 13 multiple choice If the answer is 10, which of the following statements best matches this answer? A The number of ways 1st and 2nd can occur in a race with 5 entrants. B The number of distinct arrangements of the letters in NANNA. C The number of permutations of the letters in POCKET where P and O are together. D The number of permutations of the letters in POCKET where P and O are apart. E 10P2 ÷ 4P2 14 multiple choice If the answer is 480, which of the following statements best matches this answer? A The number of ways 1st and 2nd can occur in a race with 5 entrants. B The number of distinct arrangements of the letters in NANNA. C The number of permutations of the letters in POCKET where P and O are together. D The number of permutations of the letters in POCKET where P and O are apart. E 10P2 ÷ 4P2 15 The clue in a crossword puzzle says that a particular answer is an anagram of STOREY. An anagram is another word that can be obtained by rearranging the letters of the given word. a How many possible arrangements of the letters of STOREY are there? 720 b The other words in the crossword puzzle indicate that the correct answer is O--T-How many arrangements are now possible? 24 c Can you see the answer? OYSTER eBook plus Digital doc: WorkSHEET 10.1

16 There are 30 students in a class. The students are arranged in order and asked to give the month and date of their birthday. a How many different arrangements of these dates are possible? 36530 b How many arrangements of these dates are possible if no 2 students have the same birthday? 365P30

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Combinations A group of things chosen from a larger group where order is not important is called a combination. In previous sections we performed calculations of the number of ways a task could be done where order is important — permutations or arrangements. We now examine situations where order does not matter. Suppose 5 people have nominated for a committee consisting of 3 members. It does not matter in what order the candidates are placed on the committee, it matters only whether they are there or not. If order was important we know there would be 5P3, or 60, ways in which this could be done. Here are the possibilities: ABC BDE EAB EBC CEA CBA EDB BAE CBE AEC

ABD BCD DAC ECD DEA DBA DCB CAD DCE AED

ABE BCE EAC BCA DEB EBA ECB CAE ACB BED

ACD CDE EAD BDA CDB DCA EDC DAE ADB CDB

ACE CAB EBD BEA CEB ECA BAC DBE AEB BEC

ADE DAB DBC CDA DEC EDA BAD CBD ADC CED

The 60 arrangements are different only if we take order into account; that is, ABC is different from CAB and so on. You will notice in this table that there are 10 distinct committees corresponding to the 10 distinct rows. Each column merely repeats, in a different order, the committee in the first row. This result (10 distinct committees) can be arrived at logically: 1. There are 5P3 ways of choosing or selecting 3 from 5 in order. 2. Each choice of 3 is repeated 3! times. 3. The number of distinct selections or combinations is 5P3 ÷ 3! = 10. This leads to the general rule of selecting r things from n things: 1. The number of ways of choosing or selecting r things from n distinct things where order is not important is given by the rule nCr , where: nP r n Cr = --------r! 2. The letter C is used to represent combinations.

WORKED Example 13

Write these combinations as statements involving permutations, then calculate them. a 7C2 b 20C3 THINK a

1

Write down the rule for nCr .

2

Substitute the given values of n and r into the combination formula.

WRITE a

nP Cr = --------r r! 7P 7 C2 = --------22! n

Chapter 10 Permutations and combinations

THINK

3

WRITE  7! -----  5! = ---------2!

Simplify the fraction.

7! = ----- ÷ 2! 5! 7! 1 = ----- × ----5! 2! 7! = ---------5!2! 7 × 6 × 5! = ----------------------5! × 2 × 1 4

7×6 = -----------2×1

Evaluate.

42 = -----2 = 21

b

1

Write down the rule for nCr .

2

Substitute the values of n and r into the formula.

3

Simplify the fraction.

b

nP Cr = --------r r!

n

20 P C3 = ----------33!

20

 20! --------  17! = ------------3! 20! = -------- ÷ 3! 17! 20! 1 = -------- × ----17! 3! 20! = ------------17!3! 20 × 19 × 18 × 17! = -------------------------------------------17! × 3 × 2 × 1

4

Evaluate.

20 × 19 × 18 = -----------------------------3×2×1 6840 = -----------6 = 1140

483

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

WORKED Example 14 In how many ways can a basketball team of 5 players be selected from a squad of 9 if the order in which they are selected does not matter? THINK

WRITE

1

Write down the rule for nCr . Note: Since order does not matter use the n Cr rule.

2

Substitute the values of n and r into the formula.

3

Simplify the fraction.

nP Cr = --------r r!

n

9P C5 = --------55!

9

 9! -----  4! = ---------5! 9! = ----- ÷ 5! 4! 9! 1 = ----- × ----4! 5! 9! = ---------4!5! 9 × 8 × 7 × 6 × 5! = -----------------------------------------4 × 3 × 2 × 1 × 5!

4

9×8×7×6 = -----------------------------4×3×2×1

Evaluate.

3024 = -----------24 = 126

The formula we use to determine the number of ways of selecting r things from n distinct things, where order is not important, is useful but needs to be simplified. nP n Cr = --------r r! n! ----------------( n – r )! = ----------------------r! n! = ----------------------( n – r )!r! n! Cr = ----------------------( n – r )!r!

n

Note: nCr may also be written as  n .  r

Chapter 10 Permutations and combinations

WORKED Example 15

Determine the value of the following. THINK a

b

1

Write down the rule for nCr .

2

Substitute the given values of n and r into the combination formula.

3

Simplify the fraction.

4

Evaluate.

1

Write down the rule for  n .  r

2

Substitute the given values of n and r into the combination formula.

3

Simplify the fraction.

4

Evaluate.

a

12

C5

485

b  10  2

WRITE n! Cr = ----------------------( n – r )!r! 12! 12 C5 = --------------------------( 12 – 5 )!5! 12! = ---------7!5! 12 × 11 × 10 × 9 × 8 × 7! = -----------------------------------------------------------7! × 5 × 4 × 3 × 2 × 1 12 × 11 × 10 × 9 × 8 = ------------------------------------------------5×4×3×2×1 95 040 = ---------------120 = 792 b  n = nCr  r n! = ----------------------( n – r )!r! 10! 10   = ------------------------- 2  ( 10 – 2 )!2! 10! = ---------8!2! 10 × 9 × 8! = -------------------------8! × 2 × 1 10 × 9 = --------------2×1 90 = -----2 = 45 a

n

WORKED Example 16

A committee consisting of 3 men and 4 women is to be chosen from 7 men and 9 women. In how many ways can this be done? THINK 1

Write down the rule for nCr . Note: Since order does not matter, use the nCr rule.

2

Write down the number of ways of choosing 3 men from 7.

WRITE n! Cr = ----------------------( n – r )!r!

n

Number of ways of choosing 3 men = 7C3. Continued over page

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

THINK 3 Write down the number of ways of choosing 4 women from 7. 4

Evaluate each of the combinations obtained in steps 2 and 3.

5

Use the multiplication principle to find the number of ways of choosing men and women.

6

Answer the question.

WRITE Number of ways of choosing 4 women = 9C4. 7! 9! 9 C3 = -----------------------C4 = -----------------------( 7 – 3 )!3! ( 9 – 4 )!4! 7! 9! = ---------= ---------4!3! 5!4! 7 × 6 × 5 × 4! 9 × 8 × 7 × 6 × 5! = -------------------------------= -----------------------------------------4!3 × 2 × 1 5! × 4 × 3 × 2 × 1 7×6×5 9×8×7×6 = --------------------= -----------------------------3×2×1 4×3×2×1 210 3024 = --------= -----------6 24 = 35 = 126 The number of ways of choosing 3 men and 4 women = 7C3 × 9C4 = 35 × 126 = 4410 There are 4410 ways of choosing 3 men and 4 women.

7

Graphics Calculator tip! Calculating combinations To find the number of combinations of n objects taken r at a time, we need to calculate n Cr . The following steps show how to calculate 5C3 using a graphics calculator.

For the Casio fx-9860G AU s

1. From the MENU select RUN-MAT. Press OPTN and then F6 ( ) for more options.

2. Press F3 (PROB) and you will be able to see the function nCr. Enter 5 (for n), then press F3 (nCr) and enter 3 (for r). Press EXE to display the value.

For the TI-Nspire CAS 1. Open a new Calculator document. Press k to access the catalog and press 1 for the list of functions. Scroll down to select nCr(. You can do this more quickly by first pressing N.

Chapter 10 Permutations and combinations

487

2. Press · and then complete the entry line to obtain nCr(5,3). Press · to display the value. (Alternatively, you can use the letter keys to type ncr followed by ( directly into the calculator screen instead of accessing it from the catalog. Press NCR followed by (, then complete the entry line and press ·.)

WORKED Example 17 Evaluate the following using your calculator and comment on your results. b 9C6 c 15C5 d 15C10 e 12C7 f 12C5 a 9C3 THINK a-f Use a graphics calculator (see above for steps) to evaluate the listed combinations.

Comment on your results.

WRITE a 9C3 = 84 b 9C6 = 84 c 15C5 = 3003 d 15C10 = 3003 e 12C7 = 792 f 12C5 = 792 So 9C3 = 9C6, 15C5 = 15C10 and 12 C7 = 12C5.

For each of the preceding examples, it can be seen that nCr = nCn − r . This may be derived algebraically: nP n–r Cn − r = ----------------( n – r )!

n

n!  -------------------------------  [ n – ( n – r ) ]! = ------------------------------------( n – r )!  n! -----  r!  = -----------------( n – r )! n! = ----- ÷ (n − r)! r! n! 1 = ----- × -----------------r! ( n – r )! n! = ----------------------r! ( n – r )! n! = ----------------------( n – r )!r! = nCr

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remember remember 1. The number of ways of selecting r things from n things when order is important is nPr . 2. The number of ways of selecting r things from n things when order is not important is nCr . nP 3. nCr = --------r r! n! = ----------------------( n – r! )r! 4. nCr may also be written as  n .  r 5. nCr = nCn − r

10D 8P 1 a --------33!

WORKED

Example

13

b ----------22! 1

c --------1! 5P

0

d --------0!

1 Write each of the following as statements in terms of permutations. a 8C3 b 19C2 c 1C1 d 5C0 2 Write each of the following using the notation nCr . 8P 9P 8P a --------2- 8C2 b --------3- 9C3 c --------02! 3! 0!

19 P

1P

Combinations

WORKED

Example

14

SLE 9: Investigate permutations and combinations which arise in games of chance such as, in poker, the number of hands containing two aces, the number of hands that are a full house.

eBook plus 10P

8

4

- 10C4 Digital doc: d --------4! EXCEL Spreadsheet 3 A committee of 3 is to be chosen from a group of people. In how many Combinations ways can this be done if the group contains: a 3 people? 1 b 6 people? 20 c 10 people? 120 d 12 people? 220 C0

4 A cricket team of 11 players is to be chosen from a squad of 15 players. In how many ways can this be done? 1365 5 A basketball team of 5 players is to be chosen from a squad of 10 players. In how many ways can this be done? 252 WORKED

Example

15

6 Determine the value of the following: b 11C1 11 a 12C4 495 e  21 54 264 f  10 120  7  15

c

15

C0 1 g  100 100  1 

d

12

C12 1 h  17 680  14

7 From a pack of 52 cards, a hand of 5 cards is dealt. a How many different hands are there? 2 598 960 b How many of these hands contain only red cards? 65 780 c How many of these hands contain only black cards? 65 780 d How many of these hands contain at least one red and one black card? WORKED

Example

16

eBook plus Digital doc: SkillSHEET 10.2 Listing possibilities

2 467 400

8 A committee of 3 men and 3 women is to be chosen from 5 men and 8 women. In how many ways can this be done? 560 9 A mixed netball team must have 3 women and 4 men in the side. If the squad has 6 women and 5 men wanting to play, how many different teams are possible? 100 10 A rugby union squad has 12 forwards and 10 backs in training. A team consists of 8 forwards and 7 backs. How many different teams can be chosen from the squad? 59 400

Chapter 10 Permutations and combinations

489

11 A quinella is a bet made on a horse race which pays a win if the punter selects the first 2 horses in any order. How many different quinellas are possible in a race that has: a 8 horses? 28 b 16 horses? 120 12 A CD collection contains 32 CDs. In how many ways can 5 CDs be chosen from the collection? 201 376 13 multiple choice At a party there are 40 guests and they decide to have a toast. Each guest ‘clinks’ glasses with every other guest. How many clinks are there in all? A 39 B 40 C 40! D 780 E 1560 14 multiple choice On a bookshelf there are 15 books — 7 geography books and 8 law books. Jane selects 5 books from the shelf — 2 geography books and 3 law books. How many different ways can she make this selection? A 3003 B 360 360 C 1176 D 366 E 15 Questions 15, 16 and 17 refer to the following information. The Maryborough Tennis Championships involve 16 players. The organisers plan to use 3 courts and assume that each match will last on average 2 hours and that no more than 4 matches will be played on any court per day. 15 In a ‘round robin’ each player plays every other player once. If the organisers use a round robin format: a How many games will be played in all? 120 b For how many days does the tournament last? 10 days 16 The organisers split the 16 players into two pools of 8 players each. After a ‘round robin’ within each pool, a final is played between the winners of each pool. a How many matches are played in the tournament? 57 b How long does the tournament last? 4 days 6 hours 17 A ‘knock out’ format is one in which the loser of every match drops out and the winners proceed to the next round until there is only one winner left. a If the game starts with 16 players how many matches are needed before a winner is obtained? 15 b How long would the tournament last? 1 day 4 hours 18 Lotto is a gambling game played by choosing 6 numbers from 45. Gamblers try to match their choice with those numbers chosen at the official draw. No number can be drawn more than once and the order in which the numbers are selected does not matter. a How many different selections of 6 numbers can be made from 45? 8 145060 b Suppose the first numbers drawn at the official draw are 42, 3 and 18. How many selections of 6 numbers will contain these 3 numbers? 11 480 c Suppose the first numbers drawn at the official draw are 42, 3, 18 and 41. How many selections of 6 numbers will contain these 4 numbers? 820 Note: This question ignores supplementary numbers. Lotto is discussed further in the next section. 6435, 6435 19 a Calculate the value of: 12 12 15 15 i C3 and C9 220, 220 ii C8 and C7 iii 10C1 and 10C9 10, 10 17 8 8 10 10 iv C3 and C5 56, 56 v C0 and C10 1, 1 SLE 6: Search for patterns in b What do you notice? Give your answer as a general statement such as ‘The value Pascal’s Triangle and verify n n n any claims algebraically or The value of C is the same as Cn − r . of Cr is . . .’. r WORKED

Example

otherwise.

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Applications of permutations and combinations

SLE 9: Investigate permutations and combinations which arise in games of chance.

Counting techniques, particularly those involving permutations and combinations, can be applied in gambling, logistics and various forms of market research. In this section we investigate when to use permutations and when to use combinations as well as examining problems associated with these techniques. Permutations are used to count when order is important. Some examples are: 1. the number of ways the positions of president, secretary and treasurer can be filled 2. the number of ways a team can be chosen from a squad in distinctly different positions 3. the number of ways the first three positions of a horse race can be filled. Combinations are used to count when order is not important. Some examples are: 1. the number of ways a committee can be chosen 2. the number of ways a team can be chosen from a squad 3. the number of ways a hand of 5 cards can be dealt from a deck. An interesting application of combinations as a technique of counting is a game that Australians spend many millions of dollars on each week — Lotto. To play, a player selects 6 numbers from 45 numbers. The official draw chooses 6 numbers and 2 supplementary numbers. Depending on how the player’s choice of 6 numbers matches the official draw, prizes are awarded in different divisions. Division 1: 6 winning numbers Division 2: 5 winning numbers and one of the supplementary numbers Division 3: 5 winning numbers Division 4: 4 winning numbers Division 5: 3 winning numbers and one of the supplementary numbers If the official draw was: Winning numbers 13

42

6

8

3 8

Supplementaries 20

12

2

11

34

A player who chose: 8 34 13 12 20 45 would win a Division 4 prize and a player who chose: 8 34 13 12 22 45 would win a Division 5 prize.

Lotto systems

40 25

A player may have 7 lucky numbers 4, 7, 12, 21, 30, 38 and 45, and may wish to include all possible combinations of these 7 numbers in a Lotto entry. This can be done as follows: 4 7 12 21 30 38 4 7 12 21 30 45 4 7 12 21 38 45 4 7 12 30 38 45 4 7 21 30 38 45 4 12 21 30 38 45 7 12 21 30 38 45

Chapter 10 Permutations and combinations

491

The player does not have to fill out 7 separate entries to enter all combinations of these 7 numbers 6 at a time but rather can complete a ‘System 7’ entry by marking 7 numbers on the entry form. A System 9 consists of all entries of 6 numbers from the chosen 9 numbers.

WORKED Example 18

a Ten points are marked on a page and no three of these points are in a straight line. How many triangles can be drawn joining these points? b How many different 3-digit numbers can be made using the digits 1, 3, 5, 7 and 9 without repetition? THINK a

b

WRITE a

n! Cr = ----------------------( n – r )!r!

n

1

Write down the rule for nCr . Note: A triangle is made by choosing 3 points. It does not matter in what order the points are chosen, so nCr is used.

2

Substitute the given values of n and r into the combination formula.

3

Simplify the fraction.

4

Evaluate.

5

Answer the question.

6

Verify the answer obtained by using the combination function on the calculator.

1

Write down the rule for nPr . Note: Order is important here.

2

Substitute the given values of n and r into the permutation formula.

5

3

Evaluate.

4

Answer the question.

5 × 4 × 3 × 2! = -------------------------------2! =5×4×3 = 60 Sixty 3-digit numbers can be made without repetition from a group of 5 numbers.

5

Verify the answer obtained by using the permutation function on the calculator.

10! C3 = --------------------------( 10 – 3 )!3! 10! = ---------7!3! 10 × 9 × 8 × 7! = ----------------------------------7! × 3 × 2 × 1 10 × 9 × 8 = -----------------------3×2×1 720 = --------6 = 120 120 triangles may be drawn by joining 3 points. 10

b

n! Pr = -----------------( n – r )!

n

5! P3 = -----------------( 5 – 3 )! 5! = ----2!

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

WORKED Example 19 Jade and Kelly are 2 of the 10 members of a basketball squad. In how many ways can a team of 5 be chosen if: a both Jade and Kelly are in the 5? b neither Jade nor Kelly is in the 5? c Jade is in the 5 but Kelly is not? THINK a

b

1

Write down the rule for nCr . Note: Order is not important, so nCr is used.

2

3

Substitute the given values of n and r into the combination formula. Note: If Jade and Kelly are included then there are 3 positions to be filled from the remaining 8 players. Simplify the fraction.

4

Evaluate.

5

Answer the question.

1

Write down the rule for nCr . Note: Order is not important, so nCr is used.

2

3

Substitute the given values of n and r into the combination formula. Note: If Jade and Kelly are not included then there are 5 positions to be filled from 8 players. Simplify the fraction.

4

Evaluate.

5

Answer the question.

WRITE a

n! Cr = ----------------------( n – r )!r!

n

8! C3 = -----------------------( 8 – 3 )!3! 8! = ---------5!3!

8

8 × 7 × 6 × 5! = -------------------------------5! × 3 × 2 × 1 8×7×6 = --------------------3×2×1 336 = --------6 = 56 If Jade and Kelly are included, then there are 56 ways to fill the remaining 3 positions. b

n! Cr = ----------------------( n – r )!r!

n

8! C5 = -----------------------( 8 – 5 )!5! 8! = ---------3!5!

8

8 × 7 × 6 × 5! = -------------------------------3 × 2 × 1 × 5! 8×7×6 = --------------------3×2×1 336 = --------6 = 56 If Jade and Kelly are not included, then there are 56 ways to fill the 5 positions.

Chapter 10 Permutations and combinations

THINK c

493

WRITE

1

c Write down the rule for nCr . Note: Order is not important, so nCr is used.

2

3

Substitute the given values of n and r into the combination formula. Note: If Jade is included and Kelly is not then there are 4 positions to be filled from 8 players. Simplify the fraction.

4

Evaluate.

5

Answer the question.

6

Verify each of the answers obtained by using the combination function on the calculator.

n! Cr = ----------------------( n – r )!r!

n

8! C4 = -----------------------( 8 – 4 )!4! 8! = ---------4!4!

8

8 × 7 × 6 × 5 × 4! = -----------------------------------------4 × 3 × 2 × 1 × 4! 8×7×6×5 = -----------------------------4×3×2×1 1680 = -----------24 = 70 If Jade is included and Kelly is not, then there are 70 ways to fill the 4 positions.

WORKED Example 20

Use the information on Lotto systems given on page 490. A player uses a System 8 entry with the numbers 4, 7, 9, 12, 22, 29, 32 and 36. The official draw for this game was 4, 8, 12, 15, 22, 36 with supplementaries 20 and 29. a To how many single entries is a System 8 equivalent? b List 3 of the player’s entries that would have won Division 4. c How many of the player’s entries would have won Division 4? THINK a

1

Write down the rule for nCr . Note: Order is not important, so nCr is used.

2

Substitute the given values of n and r into the combination formula. Note: A System 8 consists of all entries consisting of 6 numbers chosen from 8. Simplify the fraction.

3

WRITE a

n! Cr = ----------------------( n – r )!r!

n

8! C6 = -----------------------( 8 – 6 )!6!

8

8! = ---------2!6! 8 × 7 × 6! = ----------------------2 × 1 × 6! 8×7 = -----------2×1 Continued over page

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

THINK

WRITE

4

Evaluate.

5

Answer the question.

6

Verify each of the answers obtained by using the combination function on the calculator.

b List 3 of the player’s entries that would have won Division 4. Note: Division 4 requires 4 winning numbers. The player’s winning numbers are 4, 12, 22 and 36. Any of the other 4 numbers can fill the remaining 2 places. c

56 = -----2 = 28 A System 8 is equivalent to 28 single entries.

b Some of the possibilities are: 4 12 22 36 7 9 4 12 22 36 7 29 4 12 22 36 7 32

c

n! Cr = ----------------------( n – r )!r!

n

1

Write down the rule for nCr . Note: Order is not important, so nCr is used.

2

4

3

Substitute the given values of n and r into the combination formula. Note: To win Division 4 the numbers 4, 12, 22 and 36 must be included in the entry. The other 2 spaces can be filled with any of the other 4 numbers in any order. Simplify the fraction.

4

Evaluate.

5

Answer the question.

12 = -----2 =6 Six of the player’s entries would have won Division 4.

6

Verify each of the answers obtained by using the combination function on the calculator.

4! C2 = -----------------------( 4 – 2 )!2! 4! = ---------2!2!

4 × 3 × 2! = ----------------------2 × 1 × 2! 4×3 = -----------2×1

remember remember 1. Permutations are used to count when order is important. 2. Combinations are used to count when order is not important.

Chapter 10 Permutations and combinations

10E WORKED

Example

18

495

Applications of permutations and combinations

1 How many ways are there: a to draw a line segment between 2 points on a page with 10 points on it? 45 b to make a 4-digit number using the digits 2, 4, 6, 8 and 1 without repetition? 120 c to allocate 5 numbered singlets to 5 players? 120 d to choose a committee of 4 people from 10 people? 210 e for a party of 15 people to shake hands with one another? 105 2 How many ways are there: a for 10 horses to fill 1st, 2nd and 3rd positions? 720 b to give only 5 players in a group of 10 an unnumbered singlet? 252 c to choose a team of 3 cyclists from a squad of 5? 10 d to choose 1st, 2nd and 3rd speakers for a debating team from 6 candidates? 120 e for 20 students to seat themselves at 20 desks arranged in rows? 2.4 × 1018 3 The French flag is known as a tricolour flag because it is composed of the 3 bands of colour. How many different tricolour flags can be made from the colours red, white, blue and green, if each colour can be used only once in one of the 3 bands? 24 4 In a taste test a market research company has asked people to taste 4 samples of coffee and try to identify each as one of four brands. Subjects are told that no 2 samples are the same brand. How many different ways can the samples be matched to the brands? 24

376 992

5 In the gambling game roulette, if a gambler puts $1 on the winning number he will win $35. Suppose a gambler wishes to place five $1 bets on 5 different numbers in one spin of the roulette wheel. If there are 36 numbers in all, in how many ways can the five bets be placed?

WORKED

6 A volleyball team of 6 players is to be chosen from a squad of 10 players. In how many ways can this be done if: a there are no restrictions? 210 b Stephanie is to be in the team? 126 c Stephanie is not in the team? 84 d two players, Stephanie and Alison, are not both in the team together? 140

Example

19

7 A cross-country team of 4 runners is to be chosen from a squad of 9 runners. In how many ways can this be done if: a there are no restrictions? 126 b Tony is to be one of the 4? 56 c Tony and Michael are in the team? 21 d either Tony or Michael but not both are in the team? 70

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8 A soccer team of 11 players is to be chosen from a squad of 17. If one of the squad is selected as goalkeeper and any of the remainder can be selected in any of the positions, in how many ways can this be done if: a there are no restrictions? 8008 b Karl is to be chosen? 5005 c Karl and Andrew refuse to play in the same team? 5005 d Karl and Andrew are either both in or both out? 4004 9 multiple choice A netball team consists of 7 different positions: goal defence, goal keeper, wing defence, centre, wing attack, goal attack and goal shooter. The number of ways a squad of 10 players can be allocated to these positions is: 10! A 10! B 7! D 10P7 E 10C7 C -------7! 10 multiple choice A secret chemical formula requires the mixing of 3 chemicals. A researcher does not remember the 3 chemicals but has a shortlist of 10 from which to choose. Each time she mixes 3 chemicals and tests the result, she takes 15 minutes.

How long does the researcher need, to be absolutely sure of getting the right combination? A 1 hour B 7.5 hours C 15 hours D 30 hours E 120 hours WORKED

Example

20

11 Use the information on Lotto given on page 490. A player uses a System 8 entry with the numbers 9, 12, 14, 17, 27, 34, 37 and 41. The official draw for this game was 9, 13, 17, 20, 27, 41 with supplementaries 25 and 34. a To how many single entries is a System 8 equivalent? 28 b 9 17 27 41 12 14, 9 17 27 41 12 37, b List 3 of the player’s entries that would have won Division 4. 9 17 27 41 12 34 6 c How many of the player’s entries would have won Division 4?

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12 Use the information on Lotto given on page 490. A player uses a System 9 entry with the numbers 7, 10, 12, 15, 25, 32, 35, 37 and 41. The official draw for this game was 7, 11, 15, 18, 25, 39 with supplementaries 23 and 32. b 7 15 25 32 10 12, a To how many single entries is a System 9 equivalent? 84 7 15 25 32 10 35, b List 3 of the player’s entries that would have won Division 5. 7 15 25 32 10 37 c How many of the player’s entries would have won Division 5? 10 Questions 13 and 14 refer to the following information: Keno is a popular game in clubs and pubs around Australia. In each round a machine randomly generates 20 numbers from 1 to 80. In one entry a player can select up to 15 numbers. 13 Suppose a player selects an entry of 6 numbers. The payout for a $1 bet on 6 numbers is: Match 6 $1800 Match 5 $80 Match 4 $5 Match 3 $1 a In how many ways can an entry of 6 numbers contain 6 winning numbers? 38 760 Suppose an entry of 6 numbers has exactly 3 winning numbers in it. b In how many ways can the 3 winning numbers be chosen? 1140 c In how many ways can the 3 losing numbers be chosen? 34 220 d How many entries of 6 numbers contain 3 winning numbers and 3 losing numbers?

39 010 800

14 Suppose a player selects an entry of 20 numbers. The payout for a $1 bet on 20 numbers is: Match 20 $250 000 Match 16 $50 000 Match 14 $15 000 Match 12 $450 Match 10 $20 Match 8 $2 Match 2 $2 Match 0 $100 a In how many ways can an entry of 20 numbers contain 20 winning numbers? 1 b Suppose an entry of 20 numbers has exactly 14 winning numbers in it. eBook plus i In how many ways can the 14 winning numbers be chosen? 38 760 ii In how many ways can the 6 losing numbers be chosen? 50 063 860 Digital doc: WorkSHEET 10.2 iii How many entries of 20 numbers contain 14 winning numbers and 6 losing numbers? 1 940 475 213 600 iv How many entries of 20 numbers contain no winning numbers? 4 191 844 505 805 495

Pascal’s triangle, the binomial theorem and the pigeonhole principle Combinations are useful in other areas of mathematics, such as probability and binomial expansions. If we analyse the nCr values closely, we notice that they produce the elements of any row in Pascal’s triangle or each of the coefficients of a particular binomial expansion.

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Counting paths Dorothy and Toto enter a maze, and they have a compass. To prevent themselves from going round in circles they decide that they will only travel south or east and never north or west. The maze is shown below left and each intersection is labelled.

A B C D E F G 1 2 3 4 5 6 7

Dorothy and Toto enter the maze at A1. There are 2 ways to get to B2: A1 → A2 → B2 or A1 → B1 → B2 Are you able to list the 6 ways of going from A1 to C3? (Remember that they can only travel south or east.) If you answered: A1 → A2 → A3 → B3 → C3 A1 → A2 → B2 → B3 → C3 A1 → A2 → B2 → C2 → C3 A1 → B1 → B2 → B3 → C3 A1 → B1 → B2 → C2 → C3 A1 → B1 → C1 → C2 → C3 then you are correct. How many different ways can they A B C D E F G travel to get to D6, E5 and E6? 1 If you answered 56, 70 and 126 respectively, then you are correct. 2 2 Can you devise a method of counting the number of ways of 3 6 getting to each intersection and so show that there are 924 ways of 4 getting to G7? Use the grid at right to help record your results. 5 70 This activity provides a link between two apparently separate 6 ideas — combinations and Pascal’s 56 126 Triangle. Let us consider Pascal’s Triangle and then look at its 7 connection with combinations.

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Pascal’s Triangle The triangle below was named after the French mathematician Blaise Pascal. He was honoured for his application of the triangle to his studies in the area of probability. 0th row 1st row 1

2nd row

0th position 1 1

This 5 is in the 1st position in the 5th row.

1 1

1 1

2 3

4 5

6

1

3 6

10 15

1 1 4 10

20

1 5

15

1 6

1

The next entry here is 21. It is the sum of the 2 numbers above it — 6 and 15. Each element in Pascal’s Triangle can be calculated using combinations. For example, 10 is the 2nd element in the 5th row of Pascal’s Triangle; that is, 5C2 = 10 (this assumes 1 is the zeroth (0th) element). Compare the values from Pascal’s Triangle with the number of ways of getting to each intersection in the investigation ‘Counting paths’. What do you notice? Pascal’s Triangle shows that the rth element of the nth row of Pascal’s Triangle is given by nCr . It is assumed that the 1 at the beginning of each row is the 0th element. Another application of combinations is the binomial theorem. This theorem gives a rule for expanding an expression such as (a + b)n. Expanding expressions such as this may become quite difficult and time consuming using the usual methods of algebra. (a + b)1 = a + b (a + b)2 = a2 + 2ab + b2 (a + b)3 = a3 + 3a2b + 3ab2 + b3 (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 To use the binomial theorem you will need to recall the following conventions and terminology of algebra: 1. In the term 3a2b, ‘3’ is called the coefficient of the term. 2. In the term a2b, the coefficient of the term is 1 even though it is not written. 3. In the term 3a2b, the power of ‘a’ is 2 and the power of ‘b’ is 1. The binomial theorem is defined by the rule: (a + b)n = nC0 an + nC1an − 1b + nC2 an − 2b2 + . . . + nCr an − rbr + . . . + nCnbn = an + nC1an − 1b + nC2 an − 2b2 + . . . nCr an − rbr + . . . + bn

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When expanding brackets which are in the form (a + b)n using the binomial theorem, recall: 1. The power of a in the first term of the expansion corresponds to the power of n and in each successive term decreases by 1 until it corresponds to the power of 0. 2. The power of b starts at 0 and in each successive term increases by 1 until it corresponds to the power of n. 3. The coefficient of the rth term is nCr . 4. The rth term is obtained by using nCr an − r br. Again, this assumes that the initial term of the expansion is the 0th element. The binomial theorem is particularly useful in probability calculations.

WORKED Example 21

Refer to Pascal’s Triangle on page 499 and answer the following questions. a What number is in the 4th position in the 6th row? b Complete the 7th row in Pascal’s Triangle. c The numbers 7 and 21 occur side by side in the 7th row. What element in the 8th row occurs below and in between these numbers? THINK WRITE a 1 Locate the 6th row and the 4th position. a 6th row ⇒ 1 6 15 20 15 6 1 Note: Remember the 0th row is 1 and the first row is 1 1. In the 6th row the 1 on the left is in the 0th position. The number in the 4th position in the 6th 2 Answer the question. row is 15. b 1 Write down the elements of the 6th row. b 6th row ⇒ 1 6 15 20 15 6 1 2 Obtain the 7th row. (a) Place the number 1 at the beginning of 7th row ⇒ 1 7 21 35 35 21 7 1 the row. (b) Add the first 2 adjacent numbers from the 6th row (1 and 6). (c) Place this value next to the 1 on the new row and align the value so that it is in the middle of the 2 numbers (directly above) which created it. (d) Repeat this process with the next 2 adjacent numbers from the 6th row (6 and 15). (e) Once the sums of all adjacent pairs from the sixth row have been added, place a 1 at the end of the row. Answer the question. The 7th row is 3 1 7 21 35 35 21 7 1. c 7 21 c 1 Add the numbers 7 and 21 in order to obtain the element in the 8th row which occurs below and in between these 28 numbers. The element in the 8th row which occurs 2 Answer the question. below and in between 7 and 21 is 28.

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WORKED Example 22 Use combinations to calculate the number in the 5th position in the 9th row of Pascal’s Triangle. THINK 1 Write down the combination rule. 2

3 4

WRITE n

Cr

C5 = 126

Substitute the values for n and r into the rule. Note: The row is represented by n = 9. The position is represented by r = 5. Evaluate using a calculator. Answer the question.

9

The value of the number in the 5th position in the 9th row is 126.

WORKED Example 23

Use the binomial theorem to expand (a + 2)4. THINK 1 2 3

Write down the rule for the binomial theorem. Substitute the values for a, b and n into the rule: a = a, b = 2 and n = 4. Simplify.

WRITE (a + b)n = an + nC1an − 1b1 + . . . nCr an − rbr + . . . bn (a + 2)4 = a4 + 4C1a321 + 4C2a222 + 4C3a123 + 24 = a4 + 4 × a3 × 2 + 6 × a2 × 4 + 4 × a × 8 + 16 = a4 + 8a3 + 24a2 + 32a + 16

WORKED Example 24

What is the 4th term in the expansion of (x + y)7? THINK 1

2 3

4

Write down the rule for the rth term. Note: The rule for the 4th term is obtained from the binomial theorem: (a + b)n = an + nC1an − 1b1 + . . . nCr an − rbr + . . . bn Substitute the values for a, b, n and r into the rule: a = x, b = y, n = 7 and r = 4. Simplify. Note: The 0th term corresponds to the first element of the expansion. Answer the question.

WRITE rth term = nCr an − rbr

Cr an − rbr = 7C4x7 − 4y4

n

= 35x3y4 The 4th term is equal to 35x3y4.

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Pigeonhole principle

SLE 17: Apply the pigeonhole principle to solve problems in situations such as showing there are at least 2 in a group of 8 people whose birthdays fall on the same day of the week in any given year.

Henri Poincaré, a famous mathematician, once described mathematics as ‘the art of giving the same name to different things’. Consider three phenomena, which on the surface appear different — population growth, the value of investments and radioactive decay. Each can be described by one mathematical concept: exponential change. The mathematician gives three seemingly different things the same name. The pigeonhole principle is a good example of how mathematics gives the same name to different things. The pigeonhole principle states that: If there are (n + 1) pigeons to be placed in n pigeonholes then there is at least one pigeonhole with at least two pigeons in it. In this statement: 1. Note the precise use of language; in particular the importance of the phrase ‘at least’. 2. Some may view the pigeonhole principle as an obvious statement, but used cleverly it is a powerful problem-solving tool.

WORKED Example 25 In a group of 13 people show that there are at least 2 whose birthday falls in the same month. THINK 1 Think of each person as a pigeon and each month as a pigeonhole. 2 If there are 13 pigeons to be placed in 12 holes at least one hole must contain at least two pigeons.

WRITE There are 12 months and 13 people. Using the pigeonhole principle: 13 people to be assigned to 12 months. At least one month must contain at least two people. That is, at least two people have birthdays falling in the same month.

Generalised pigeonhole principle: If there are (nk + 1) pigeons to be placed in n pigeonholes then there is at least one pigeonhole with at least (k + 1) pigeons in it.

WORKED Example 26 In a group of 37 people show that there are at least 4 whose birthdays lie in the same month. THINK 1 Think of each person as a pigeon and each month as a pigeonhole. 2 Use the generalised pigeonhole principle. 3 (nk + 1) pigeons to be allocated to n holes; n = 12 → k = 3

WRITE There are 12 months and 37 people. Using the generalised pigeonhole principle: 37 people to be assigned to 12 months. The value of n is 12 and k is 3. So at least one month has at least (k + 1) or 4 people in it. That is, at least 4 people have birthdays falling in the same month.

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WORKED Example 27 On resuming school after the Christmas vacation, many of the 22 teachers of Eastern High School exchanged handshakes. Mr Yisit, the social science teacher said ‘Isn’t that unusual — with all the handshaking, no two people shook hands the same number of times’. Not wanting to spoil the fun, the mathematics teacher, Mrs Pigeon said respectfully, ‘I am afraid you must have counted incorrectly. What you say is not possible.’ How can Mrs Pigeon make this statement?

THINK

WRITE

1

Think of the possible number of handshakes by a person as a pigeonhole.

For each person there are 22 possible numbers of handshakes; that is 0 to 21.

2

If two or more people have 0 handshakes, the problem is solved. Consider the cases where there is 1 person with 0 handshakes or 0 persons with 0 handshakes.

1 person with 0 handshakes: If there is 1 person with 0 handshakes there can be no person with 21 handshakes. Thus there are 21 people to be assigned to 20 pigeonholes. Therefore there must be at least one pigeonhole with at least two people in it. 0 people with 0 handshakes: If there is no person with 0 handshakes there are 22 people to be assigned to 21 pigeonholes. Therefore there must be at least one pigeonhole with at least two people in it (at least two people have made the same number of handshakes).

3

Conclude using a sentence.

Thus, there are at least two people who have made the same number of handshakes.

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70 56

35 28 1

1

8

1

1

7

6

1

5

21

15

4 1

56

21 35

15 20

10 10

6 3

1

28

1 6

1 5

1 4

1 3

1 2

1 1

0th term

rth term

6. When expanding brackets which are in the form (a + b)n using the binomial theorem, recall: (a) The power of a in the first term of the expansion corresponds to the power of n and in each successive term decreases by 1 until it corresponds to the power of 0. (b) The power of b starts at 0 and in each successive term increases by 1 until it corresponds to the power of n. (c) The coefficient of the rth term is nCr . (d) The rth term is obtained by using nCr an − rbr. Again, this assumes that the initial term of the expansion is the 0th element. 7. The pigeonhole principle: If there are (n + 1) pigeons to be placed in n pigeonholes then there is at least one pigeonhole with at least two pigeons in it.

0 1 2 3 4 5 6 7 8

1 Row

1. Pascal’s Triangle shows that the rth element of the nth row of Pascal’s Triangle is given by nCr . 2. Each new row in Pascal’s Triangle is obtained by first placing a 1 at the beginning and end of the row and then adding adjacent entries from the previous row. 3. Row 1 is the row containing the elements ‘1 and 1’. 4. The ‘1’ on the left-hand side of each row is in the 0th position of that row. 5. The binomial theorem is defined by the rule: (a + b)n = an + nC1an − 1b + nC2an − 2b2 + . . . + nCr an − rbr + . . . + bn

7

1

8

1

1

remember remember

10F

Pascal’s Triangle, the binomial theorem and the pigeonhole principle

1 Write the first 8 rows in Pascal’s Triangle. WORKED

Example

21

WORKED

Example

22

WORKED

Example

23

2 Refer to Pascal’s Triangle on page 499 and answer the following eBook plus questions. a What number is in the 4th position in the 8th row? 70 Digital doc: EXCEL b Complete the 9th row in Pascal’s Triangle. 1 9 36 84 126 126 84 36 9 1 Spreadsheet c If 9 and 36 occur side by side in the 9th row, what element in the Pascal’s Triangle 10th row occurs below and in between these numbers? 45 3 Use combinations to: a calculate the number in the 7th position of the 8th row of Pascal’s Triangle 8 b calculate the number in the 9th position of the 12th row of Pascal’s Triangle 220 c generate the 10th row of Pascal’s Triangle. 4 Use the binomial theorem to expand: a (x + y)2 x2 + 2xy + y2 b (n + m)3 n3 + 3n2m + 3nm2 + m3 c (a + 3)4 a4 + 12a3 + 54a2 + 108a + 81

10

C0 10C1 10C2 10C3 10C4 10C5 10C6 10C7 10C8 10C9 10C10 1 10 45 120 210 252 210 120 45 10 1

Chapter 10 Permutations and combinations

WORKED

Example

24

505

5 a What is the 4th term in the expansion of (x + 2)5? 80x b What is the 3rd term in the expansion of (p + q)8? 56p5q3 2 c What is the 7th term in the expansion of (x + 2)9? 4608x 6 multiple choice x 1 9 36 84 126 126

84

36

9

1

7 multiple choice 16x3 is definitely a term in the binomial expansion of: B (x + 4)3 C (x + 2)4 D (x + 4)4 A (x + 2)3

E (x + 2)5

b i 8 a In Pascal’s Triangle, calculate the sum of all elements in the: The sum of the elements in each row i 0th row 1 ii 1st row 2 iii 2nd row 4 of Pascal’s triangle 8 16 iv 3rd row v 4th row vi 5th row 32 is a power of 2: Row Sum b i What do you notice? 0 20 = 1 ii Complete the statement: ‘The sum of the elements in the nth row of Pascal’s 1 21 = 2 n Triangle is . . .’ ‘The sum of the elements in the nth row of Pascal’s triangle is 2 .’ 2 22 = 4 3 23 = 8 9 Use the result from question 8 to deduce a simple way of calculating: 26 = 64 4 24 = 16 5 5 2 = 32 6 6 6 6 6 6 6

C 0 + C 1 + C 2 + C3 + C 4 + C 5 + C 6

10 In a cricket team consisting of 11 players, show that there are at least 2 whose phone numbers have the same last digit. 25

WORKED

Example

11 Two whole numbers add to give 21. Show that at least one of the numbers is greater than 10. 12 A squad of 10 netballers is asked to nominate when they can attend training. They can choose Tuesday only, Thursday only or Tuesday and Thursday. Show that there is at 26 least one group of at least 3 players who agree with one of these options.

WORKED

Example

13 S&M lollies come in five great colours — green, red, brown, yellow and blue. How many S&Ms do I need to select to be sure I have 6 of the same colour? 26 14 The new model WBM roadster comes in burgundy, blue or yellow with white or black trim. That is, the vehicle can be burgundy with white or burgundy with black and so on. How many vehicles need to be chosen to ensure at least 3 have the same colour combination? 13 15 Is it possible to show that in a group of 13 people, there are at least 2 whose birthdays fall in February? 16 Nineteen netball teams entered the annual state championships. However, it rained frequently and not all games were completed. No team played the same team more 27 than once. Mrs Organisit complained that the carnival was ruined and that no two teams had played the same number of games. Show that she is incorrect in at least part of her statement and that at least two teams played the same number of games.

WORKED

Example

SLE 6: Search for patterns in Pascal’s Triangle and verify any claims algebraically or otherwise.

A row of Pascal’s Triangle is given above. What number is located at position x? A 8 B 28 C 45 D 120 E 136

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History of mathematics B L A I S E PA S C A L — ( 1 6 2 3 – 1 6 6 2 ) During his life . . . Construction of the Taj Mahal is started. Rembrandt completes many of his famous paintings. Oliver Cromwell governs England. Blaise Pascal was a French mathematician and physicist who studied combinatorics and developed the theory of probability. He was born in the town of Clermont in France. His father was a taxation officer. His mother died when he was only 4. Pascal was a sickly child and so was not sent to school initially but was educated at home by his father. Because he was not healthy his father forbad him from studying mathematics. It took about 5 years before Blaise could convince his father to let him try. When Blaise was 16, his father was in trouble with the courts because he would not set any more taxes. He had to leave Paris, and the family moved to Rouen. Blaise Pascal discovered and proved a major theorem of geometry when he was only 16 years old. This theorem was about the intersections of points on a conic plane. When he was 18 he became very ill. He eventually recovered, after being temporarily paralysed and close to death. After this scare he became very religious and started to study philosophy and religion. His research into mathematics and science often conflicted with his religious beliefs. At age 19, Pascal invented a calculating machine that could do simple addition and subtraction. He sold many of these machines and they were so well made that some still exist today.

He demonstrated that air pressure decreases with height by taking accurate measurements at various levels on the side of the Puy de Dôme mountain. He persuaded his brother to climb the mountain and take measurements using a heavy barometer. Like many mathematicians, Blaise Pascal had arguments with other mathematicians, including René Descartes, who came to visit him. Descartes did not believe that Pascal was capable of such difficult mathematics and claimed that Pascal had stolen some of his ideas from Descartes himself. Blaise Pascal developed the pattern of numbers now known as Pascal’s Triangle that is used in probability, permutations and combinations. When Blaise Pascal’s father died, his unmarried sister went into a convent and he was left to live free of family and spiritual conflicts. His health improved and he took up an active social life including gambling and driving a fast, horse-drawn carriage! In late 1654 he was involved in an accident. His horses went over the edge of a bridge and were killed, but he survived. Pascal was shaken up by this and again saw the event as a message from God. In 1655 he moved in with his married sister. Later that year, Pascal became ill and eventually died from the effects of a brain tumour and stomach ulcer in 1662. The computer language ‘Pascal’ is named after him. Questions 1. How old was Pascal when he proved his theorem on conics? 16 2. What did he develop at age 19 that earned him a lot of money? Calculating machine 3. Upon which mountain was his work on air pressure done and who did the real work? Puy de Dôme; his brother 4. What is ‘Pascal’s Triangle’ used for? Probability 5. What did he die from? Brain tumour

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summary The addition and multiplication principles • Combinatorics is often called ‘counting’ and deals with counting the number of ways in which activities or events can happen. • The multiplication principle should be used when there are two operations or events (say, A and B) where one event is followed by the other. It states that: If there are n ways of performing operation A and m ways of performing operation B, then there are n × m ways of performing A and B. • The addition principle should be used when two distinct operations or events occur in which one event is not followed by another. It states that: If there are n ways of performing operation A and m ways of performing operation B then there are n + m ways of performing A or B. • A selection where order is important is called an arrangement.

Factorials and permutations • The number of ways in which n distinct objects may be arranged is n! (n factorial) where: n! = n × (n − 1) × (n − 2) × (n − 3) × . . . × 3 × 2 × 1 • 0! = 1 • 1! = 1 • The number of different arrangements or permutations when r things are chosen from n things and order is important is given by the rule nPr , where: n! • nPr = -----------------( n – r )! • nPn = n! • nP0 = 1 • The number of different ways in which n people can be seated, r at a time, in a circle is: nP --------r . r

Arrangements involving restrictions and like objects • The number of different ways of arranging n things made up of groups of indistinguishable things, n1 in the first group, n2 in the second group and so on is: n! -----------------------------------. n 1!n 2!n 3!…n r! • When restrictions apply to arrangements use the multiplication and addition principles as well as nPr .

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Combinations • The number of ways of selecting r things from n things when order is not important is nCr . nP • nCr = ---------r r! n! = ----------------------( n – r )!r! • nCr may also be written as  n .  r • nCr = nCn − r

Applications of permutations and combinations • Permutations are used to count when order is important. • Combinations are used to count when order is not important.

Pascal’s Triangle and the binomial theorem • Pascal’s Triangle shows that the rth element of the nth row of Pascal’s Triangle is given by nCr . • Each new row in Pascal’s Triangle is obtained by first placing a 1 at the beginning and end of the row and then adding adjacent entries from the previous row. • The top row is row 0. • Row 1 is the row containing the elements ‘1 1’. • The ‘1’ on the left-hand side of each row is in the 0th position of that row. • The binomial theorem is defined by the rule: • (a + b)n = an + nC1an − 1b + nC2 an − 2b2 + . . . + nCr an − rbr + . . . + bn 0th term

rth term

• When expanding brackets which are in the form (a + b)n using the binomial theorem, recall: 1. The power of a in the first term of the expansion corresponds to the power of n and in each successive term decreases by 1 until it corresponds to the power of 0. 2. The power of b starts at 0 and in each successive term increases by 1 until it corresponds to the power of n. 3. The coefficient of the rth term is nCr. 4. The rth term is obtained by using nCr an − rbr. • Points 3 and 4 both assume that the initial term of the expansion is the 0th element.

The pigeonhole principle • If there are (n + 1) pigeons to be placed in n pigeonholes then there is at least one pigeonhole with at least two pigeons in it. • The generalised pigeonhole principle: If there are (nk + 1) pigeons to be placed in n pigeonholes then there is at least one pigeonhole with at least (k + 1) pigeons in it.

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CHAPTER review 1 multiple choice Barbie’s wardrobe consists of 5 different tops, 4 different skirts and 3 different pairs of shoes. The number of different outfits Barbie can wear is: A 5 B 12 C 60 D 80 E 120

10A

2 multiple choice How many different 3-digit numbers can be made from the numbers 1, 3, 5, 7 and 9 if the numbers can be repeated? A 60 B 125 C 243 D 729 E 999

10A

3 multiple choice There are 7 candidates seeking election to the positions of either president or secretary of the Soccer Club Committee. If one of these candidates, George, is to be either president or secretary, in how many ways can positions be filled? A 12 B 21 C 42 D 49 E 56

10A

4 How many 4-digit numbers less than 4000 can be made using the digits 1, 2, 3, 5, 7 and 9 if: a repetition is not permitted? 180 b repetition is permitted? 648

10A

5 multiple choice The permutation 9P6 is equal to:

10B

A 9×8×7

B 9×8×7×6×5×4×3

9! C ----6!

9! D ----3!

9! E ----4!

6 multiple choice There are 12 horses in a race. In how many different ways can the 1st, 2nd and 3rd positions be filled? A 12P3 B 123 C 312 D 12C3 E 12C12

10B

7 multiple choice A round table seats 5 people. From a group of 8 people, in how many ways can 5 people be seated at the table: 8P 8P 9! B 8P5 E 8C5 A ----C --------5D --------56! 5 5!

10B

8 Use your calculator to place these in ascending order: 19P6, 12P9, 2000P2.

10B

2000

P2, 19P6, 12P9

510 10C

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9 multiple choice How many different arrangements can be made using the 8 letters of the word NONSENSE? A 1680 B 2520 C 3360 D 5040 E 40 320

10C

10 How many different arrangements of 4 letters can be made from the letters of the word PILL? 12

10D

11 multiple choice Which of the following is equivalent to 8C2? 6P A --------22!

8P B --------62!

8P C ---------26!2!

8P D --------26!

8P E --------22!

10D

12 multiple choice

10D 10D

13 Use your calculator to place these in ascending order: 19C6, 22C15, 2000C2.

10D

15 Two cards are dealt from a pack of 52. What is the number of ways that: a both are black? 325 b both are aces? 6 c the cards are of different colours? 676

10E

16 multiple choice

10E

17 A ward in a city hospital has 15 nurses due to work on Friday. There are 3 shifts that need to be staffed by 5 nurses on each shift. How many different arrangements for staffing these 3 shifts are possible, assuming that each nurse only works 1 shift? 756 756

A committee of 4 men and 3 women is to be formed from 5 men and 8 women. In how many ways can this be done? A 61 B 280 C 1320 D 20 160 E 40 320 19

C6, 22C15, 2000C2

14 A committee of 3 men and 4 women is to be formed from 7 men and 5 women. In how many ways can this be done? 175

A cycling team of 3 riders is to be chosen from a squad of 8 riders. In how many ways can this be done if one particular rider, Jorge, must be in the team? A 56 B 336 C 21 D 210 E 420

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18 multiple choice

10F

What is the 4th term in the expansion of (p + 1)7? A p4 B 35p3 C 35p4

D 21p3

E 21p4

19 multiple choice

10F

A row of Pascal’s Triangle is given below. What number is located at position x? 1 9 36 84 126 126 84 36 9 1 x A 48

B 120

C 56

D 210

E 252

20 a In the 10th row of Pascal’s Triangle, what is the 6th entry? 210 b Write the 10th row of Pascal’s Triangle using combinations. 10C 10C 10C 10C 10C . . . 10C 0 1 2 3 4 10 c What is the sum of the elements of the 10th row? 1024

10F

21 In the expansion of (x + 3)10, what is the: a 2nd term? 405x8 b 3rd term? 3240x7

10F

c

9th term? 196 830x

Modelling and problem solving 1 Assume that car number plates are sequenced as follows: DLV334 → DLV335 → ... DLV339 → DLV340 → ... DLV999 → DLW000 and so on. Using this sequence, how many number plates are there between DLV334 and DNU211 inclusive? 50 878 2 How many paths are there from A to B if you are only allowed to move either down or to the right on the lines of the grid? 2 944 656 A

B

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3 Poker is a card game in which initially each person is dealt 5 cards. a How many different hands are possible? (Order is not important.) 2 598 960 b How many hands contain only diamonds? 1287 c How many hands contain only red cards? 65 780 d When Wild Bill Hickock died at Deadwood, Dakota, he was holding in his hand 2 pairs — aces and eights. This is called the dead man’s hand. In how many ways can you be dealt the dead man’s hand? 1584 4 From a group of 20 female students, 2 female staff, 18 male students and 3 male staff, a committee of 6 is to be formed. Find the number of different committees if: a there are no restrictions 6 096 454 b all committee members must be students 2 760 681 c one female staff member, one male staff member and 4 students must be on the committee 442 890 d there is an equal number of males and females on the committee 2 048 200 e one particular student must be on the committee 850 668 f one particular student must not be on the committee 5 245 786 g the committee must comprise 2 male staff members, 2 male students, 1 female staff member and 1 female student. 18 360 5 Two women and three men approach an ATM at the same time. a How many different queues are possible if the position of each person in the queue is taken into account? 120 b How many queues of at least two people are possible if the position of each person in the queue is not taken into account? 26 6 A school is using identification cards (ID cards) that consist of 2 letters selected from A to D inclusive followed by 3 digits chosen from 0 to 9 inclusive. a How many different ID cards can be issued to students if a digit may be used more than once but all 2 letters of each ID are different? 12 000 b New ID cards are issued to all students each year and the old cards discarded. However, the old ID numbers are not used again. If, on average, the school’s population increases by eBook plus 10% each year and was 2000 during the year when the ID cards were first used, how many Digital doc: years will elapse before cards with numbers already used will have to be issued? 5 years Test Yourself Chapter 10

Dynamics

11 syllabus reference Option topic: Dynamics

In this chapter 11A Displacement, velocity and acceleration 11B Projectile motion 11C Motion under constant acceleration

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Displacement, velocity and acceleration

y

To explore the motion of a particle in a plane, we must consider both its magnitude and direction. As this motion may not be in a straight line, it is convenient to use a vector approach to calculate the displacement, velocity and acceleration at time t. To illustrate this vector approach, consider the point P in the following diagram. Relative to the origin, O, the position vector of the displacement of P from P (x, y)

O

• derivatives and integrals of vectors • Newton's laws of motion in vector form applied to objects of constant mass • application to: – straight line motion in a horizontal plane with variable force – vertical motion under gravity without air resistance – projectile motion without air resistance

the origin, at time t, may be written as OP = x i + y j where x and y denote ˜ magnitude of the displacement along the Ox and Oy˜ axes respectively. The x

position of P varies with time so it is appropriate to think of OP as a vector which is time dependent. Displacement at time t is the change in position of the point P, and as the reference point is the origin, the displacement of P can be represented by the position vector.

If we write r (t) to represent OP at time t, then we may say that r (t) = x(t) i + y(t) j ˜ ˜ ˜ ˜ where x(t) and y(t) are the respective horizontal and vertical components of this displacement at time t. As a general rule, when we write the magnitude of the displacement or position vector r (t), we will write r where it is understood that r = r ( t ) . ˜ ˜

Vector expressions for velocity and acceleration If the position vector r (t) = x(t) i + y(t) j represents the displacement of P at time t, ˜ ˜ ˜ 2 dr d r then -----˜ represents the instantaneous velocity and -------2˜- represents the acceleration at dt dt time t. (It is often convenient to make use of the time-derivative notation of r˙ (t) for ˜ velocity and r˙˙(t) for the acceleration at time t.) We calculate the speed of a particle at ˜ time t as r˙ = r˙( t ) and the magnitude of the acceleration at time t as r˙˙ = r˙˙( t ) . In our ˜ ˜ work, the distance or magnitude of the displacement will be measured in metres (m), the speed or magnitude of the velocity in metres per second (m/s) and the magnitude of the acceleration in metres per second per second (m/s2).

WORKED Example 1 Relative to the origin, O, the displacement (in metres) of a particle at time t seconds is given by r (t) = 4t i + (10t - 5t2) j . ˜ ˜ ˜ a Find expressions for the velocity and acceleration vectors at time t seconds. b Determine the initial position and speed of the particle. c When t = 4 seconds, calculate i the distance of the particle from the origin ii the velocity (both magnitude and direction) of the particle iii the measure of the angle between the velocity and displacement vectors. d Is the acceleration constant for this motion?

Chapter 11 Dynamics

THINK a 1 Write the position vector. 2

3

b

1

2

. dr Find r (t) = -----˜ for the velocity dt ˜ vector. . 2 .. .. dr d r Find r (t) = -----˜ or r (t) = -------˜- for the 2 dt ˜ ˜ dt acceleration vector. The initial position occurs when t = 0. Substitute t = 0 into the displacement vector expression.

The initial speed occurs when t = 0. Substitute t = 0 into the velocity vector expression. Remember that speed is the magnitude of velocity.

WRITE a r (t) = 4t i + (10t − 5t2) j ˜ ˜ ˜ . dr r (t) = -----˜ dt ˜ = 4 i + (10 − 10t) j ˜ ˜ So v (t) = 4 i + (10 − 10t) j m/s ˜ ˜ . ˜ .. dr r (t) = -----˜ dt ˜ = −10 j ˜ So a (t) = −10 j m/s2 ˜ ˜ b When t = 0, r (0) = (4 × 0) i + (10 × 0 − 5 × 02) j ˜ ˜ ˜ = 0i + 0 j ˜ ˜ the initial position of the This means that particle is at the origin. When t = 0, v (0) = 4 i + (10 − 10 × 0) j ˜ ˜ ˜ = 4 i + 10 j ˜ ˜ The initial speed of the particle is the magnitude of the initial velocity. 2

v ( 0 ) = 4 + 10 ˜ = 2 29 c i

1

2

3

Distance is the magnitude of displacement, so first calculate r (4) ˜ to find the displacement at t = 4. Find the magnitude of the resulting vector expression to calculate the distance.

State the answer.

515

2

The initial speed is 2 29 m/s. c i r (t) = 4t i + (10t − 5t2) j ˜ ˜ ˜ r (4) = 16 i − 40 j ˜ ˜ ˜ The required distance is the magnitude of the displacement. 2

r ( 4 ) = 16 + ( – 40 ) ˜ = 8 29

2

The distance of the particle from the origin at t = 4 s is 8 29 m.

ii

1

Calculate v (4) to find the velocity vector at t ˜= 4.

2

Find the magnitude of the velocity.

ii v (t) = 4 i + (10 − 10t) j ˜ ˜ ˜ v (4) = 4 i − 30 j ˜ ˜ ˜ 2 2 v ( 4 ) = 4 + ( – 30 ) ˜ = 2 229 ≈ 30.27 m/s Continued over page

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THINK 3

To find the direction, draw a diagram to represent the velocity vector v (4) = 4 i − 30 j . ˜ ˜ ˜

WRITE Let θ be the angle the velocity vector makes with the horizontal. y O

4

2 229

x 30

~v (4) = 4i~ – 30j ~ 4

Use trigonometry to obtain an equation to solve for θ.

30 tan θ = -----4 θ ≈ 82°24′

5

Write the answer.

When t = 4 s, the velocity of the particle is 2 229 m/s (or approx. 30.27 m/s) downwards at an angle of 82˚24′ to the horizontal.

iii

1

Write the two vectors for velocity and displacement at t = 4 s.

2

To calculate the angle between the velocity and displacement vectors we can use the dot product of the two vectors. The dot product of two vectors a and b is given as a • b = a˜ b cos˜ θ where θ is the ˜ ˜ ˜ ˜ included angle by a and b . ˜ ˜ Calculate the product on the left of the equation by multiplying the corresponding components of i ˜ and j , and substitute the ˜ magnitudes of r (4) and v (4) on the right of the ˜equation. ˜

3

d

1 2

iii When t = 4, r (4) = 16 i − 40 j and v (4) = 4 i − 30 j ˜ ˜ ˜ ˜ ˜ ˜ Let θ be the angle between the two vectors. Using the dot product: r (4) • v (4) = r ( 4 ) v ( 4 ) cos θ ˜ ˜ ˜ ˜

(16)(4) + (−40)(−30) = (8 29 )(2 229) cos θ

4

Solve the equation to find θ.

1264 cos θ = ---------------------16 6641 θ ≈ 14°12′

5

Write the answer.

The angle between the velocity and displacement vectors is 14˚12′.

Consider the vector expression for acceleration found in part a. Since −10 j does not contain t, it is a ˜ constant. State the answer.

d a (t) = −10 j ˜ ˜ Since −10 j is a constant, the acceleration is ˜ constant for this motion.

Chapter 11 Dynamics

517

WORKED Example 2 At time t seconds, a particle has a position vector given by the expression r (t) = 2t i + (25 − t2) j metres. ˜ ˜ ˜ a Use a graphics calculator to plot the trajectory of this particle across the interval 0 ≤ t ≤ 5 seconds. b Repeat part a above using an Excel spreadsheet. c Determine the equation of this trajectory in the form y = f (x). THINK

WRITE/DISPLAY

a

a r (t) = 2t i + (25 − t2) j ˜ ˜ ˜ Let x = 2t and y = 25 − t2

1

Consider the components of the position vector. Assign x to the horizontal component and y to the vertical. This produces two parametric equations that we can use to graph the trajectory.

2

Use a graphics calculator to generate a graph of the particle’s motion.

SLE 9: Use the parametric facility of a graphing calculator to model the flight of a projectile.

For the Casio fx-9860G AU (a) Press MENU and select GRAPH. Press F3 (TYPE) followed by F3 (Parm) to select the parametric equations option. (b) Enter the x-component by completing the entry line for Xt1 with 2t. (Press X,q,T to enter t.) Similarly, enter the y-component by completing the entry line for Yt1 with 25 − t2. Press EXE after each entry. (c) Press F6 (DRAW) to display the graph. To obtain a clearer view of the graph, you can adjust the View Window settings. Press SHIFT F3 (V-WIN) and adjust the values for Xmin, Xmax, Ymin and Ymax. (d) Press EXE until you return to the equations screen and then press F6 (DRAW) to display the graph with this new setting. To display the t, x and y values at different points on the graph, press SHIFT F1 (TRCE) and use the arrow keys to move along the line. Continued over page

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THINK

WRITE/DISPLAY

For the TI-Nspire CAS (a) Open a new Graphs & Geometry document. Press b and select 3: Graph Type followed by 2: Parametric. In the entry panel, enter x = 2t and y = 25 − t2. Change the domain for t to 0 ≤ t ≤ 5. (b) Press · to display the graph of the particle’s trajectory. To adjust the viewing window, press b, then select 4: Window and 1: Window Settings. Enter the settings as shown.

(c) Highlight OK and press ·. To see more of the screen, press /G to hide the entry panel. (Press /G again to bring the entry panel back.) To investigate the coordinates of the points along the graph, press b, select 5: Trace and 1: Graph Trace. b

1

Set up a spreadsheet with 3 columns titled t, x and y. In the first column, enter values from 0 to 5 for t. Enter the appropriate formula for both x and y and generate corresponding values in columns 2 and 3.

b t

x

y

0

0

25

1

2

24

2

4

21

Digital doc:

3

6

16

EXCEL Spreadsheet Plotting the trajectory of a particle

4

8

9

5

10

0

eBook plus SLE 16: Use spreadsheets to investigate problems.

Generate a graph of the particle’s trajectory.

y Metres

2

c

1

Write the parametric equations for the position vector.

30 25 20 15 10 5 0

Trajectory

0

2

c x = 2t y = 25 − t2

4

6

8

[1] [2]

10

12

x

Chapter 11 Dynamics

THINK 2

519

WRITE/DISPLAY

Eliminate the shared parameter t between x = 2t and y = 25 − t2. Simplify to obtain an equation in the form y = f (x).

x Rearranging equation [1]: t = --2 Substituting for t in equation [2]: x y = 25 −  ---  2

2

2

x = 25 − ----4 = 1--- (100 − x2) 4

3

Write the answer.

The equation of the trajectory is y = 1--- (100 − x2). 4

WORKED Example 3 At time t seconds, the displacement (in metres) of a particle A is given by r A(t) = (8 − t) i + (8 − 4t + t2) j and the displacement (in metres) of a particle B is given by ˜ ˜ ˜ r B(t) = (t + 2) i + (2 − 2t + t2) j . ˜ ˜ ˜ a If these particles collide, determine when they collide. b What are the coordinates of the impact point? c Find the Cartesian equations of the trajectories of the particles. d Use a graphics calculator to verify the coordinates of the impact point. THINK

WRITE/DISPLAY

a

a At impact point, horizontal components are equal: 8−t=t+2 [1] At impact point, vertical components are equal: 8 − 4t + t2 = 2 − 2t + t2 [2]

1

If the particles collide, they meet at a common point. So the horizontal components of the two vectors will be equal and the vertical components will be equal. Produce two equations to reflect this.

2

Solve equation [1] to find t.

Rearranging equation [1]: 2t = 6 2t = 3

3

Check the solution by substituting t = 3 into the left and right sides of equation [2].

Check by substituting t = 3 into equation [2]: LHS = 8 − 4t + t2 = 8 − 12 + 9 = 5 RHS = 2 − 2t + t2 = 2 − 6 + 9 = 5 LHS = RHS so solution is correct.

4

Write the answer.

Particles A and B collide after three seconds. Continued over page

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THINK

WRITE/DISPLAY

b

b r A(t) = (8 − t) i + (8 − 4t + t2) j ˜ ˜ ˜ At the point of impact, t = 3. So r A(3) = (8 − 3) i + (8 − 12 + 9) j ˜ ˜ ˜ So r A(3) = 5 i + 5 j ˜ ˜ ˜ The coordinates of the impact point are (5, 5).

1

Substitute for t in the expression for the displacement of either particle A or B. (In this case, select particle A.)

2

Write the answer.

c In both expressions, assign the variables x and y respectively to the horizontal and vertical components. Eliminate the shared parameter t to obtain each Cartesian equation.

c Let x represent the horizontal component and y the vertical component of the displacement. For particle A: x = 8 − t and y = 8 − 4t + t2 Hence, t = 8 − x and so y = 8 − 4(8 − x) + (8 − x)2 = x2 − 12x + 40 yA = (x − 6)2 + 4 For particle B: x = t + 2 and y = 2 − 2t + t2 Hence, t = x − 2 and so y = 2 − 2(x − 2) + (x − 2)2 = x2 − 6x + 10 yB = (x − 3)2 + 1

d Verify that the coordinates of the impact point are (5, 5) by drawing the graphs of the displacement for each particle and locating the point of intersection.

d

For the Casio fx-9860G AU (a) Press MENU and select GRAPH. Press F3 (TYPE) followed by F1 (Y=) to select the function option. Complete the entry line for Y1 with the equation for particle A. Similarly, complete the entry line for Y2 with the equation for particle B. (b) Press F6 (DRAW) to display the graph. (Adjust the View Window settings if necessary.) Press SHIFT F5 (G-SLV) followed by F5 (ISCT) to display the coordinates of the intersection point.

Point of intersection is (5, 5) as required.

Chapter 11 Dynamics

THINK

521

WRITE/DISPLAY

For the TI-Nspire CAS (a) Open a new Graphs & Geometry document. Complete the entry line for f1(x) with the equation for particle A. Press ·. Similarly, complete the entry line for f2(x) with the equation for particle B and press ·. Adjust the Window Settings if necessary. (b) To display the point of intersection, press b, select 6: Points & Lines and 3: Intersection Point(s). Navigate the pointer to each line and press ·. (Press d to fix any errors.) The coordinates of the point of intersection will be displayed. Point of intersection is (5, 5) as required.

remember remember 1. If the position vector r (t) = x(t) i + y(t) j represents the displacement of P at ˜ ˜ ˜ 2 dr d r time t, then -----˜ represents the instantaneous velocity and -------2˜- represents the dt dt instantaneous acceleration at time t.

SLE 1: Given the position vector of a point as a function of time such as r (t ) = t i + t 2 j + sin t k ~ ~ ~ ~ ` ` ` ` determine the velocity and acceleration vectors.

2. For the magnitude of a vector x i ˜ 3. The dot product of two vectors a ˜ θ is the angle included by a and ˜

SLE 9: Use the parametric facility of a graphing calculator to model the flight of a projectile.

2

2

+ y j : xi + y j = x + y ˜ ˜ ˜ and b is given as a • b = a b cos θ where ˜ ˜ ˜ ˜ ˜ b. ˜

SLE 16: Use spreadsheets to investigate problems.

11A WORKED

Example

1

Displacement, velocity and acceleration

1 Relative to the origin, O, the displacement (in metres) of a particle at time t seconds is v (t) = 6 i + (3 − 14 t) j , a (t) = −14 j given by r (t) = 6t i + (3t − 7t2) j . ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ a Find expressions for the velocity and acceleration vectors at time t seconds. b Determine the initial displacement and speed of the particle. r (0) = 0 m, v(0) = 3 5 m/s ˜ ˜ c When t = 2 seconds, calculate: i the distance of the particle from the origin r˜ (2) = 12 ˜i − 22 j m 25.7 m/s downwards at an ˜ ii the velocity (both magnitude and direction) of the particle angle of 13°30′ to the vertical iii the measure of the angle between the velocity and displacement vectors. 15°7′ d Is the acceleration constant for this motion? Yes, 14 m/s2 downwards

522 WORKED

Example

3 e i v (t) = i + 6t j , a (t) = 6 j g i v (t) = 2t i − 4 j , a (t) = 2 i ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ii r (0) = 0 m, v(0) = 4 m/s ii r (0) = 4 i m, v(0) = 1 m/s ˜ ˜ ˜ ˜ iii y2 = 16x iii y = 3(x − 4)2 -------h i v (t) = 1 - i − 6t j , t ≠ 0; 2 t ˜ ˜ f i v (t) = 2t i − j , a (t) = 2 i ˜ 1 ˜ ˜ ˜ ˜ ˜ - i − 6j, t ≠ 0 a (t) = − ---------4t t ˜ ii r (0) = 0 m, v(0) = 1 m/s ˜ ˜ ˜ ˜ ii r (0) = 0 m, v(0) is undefined iii y2 = x ˜ ˜ iii y = −3x4

2

WORKED

Example

3

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

2 At time t seconds, a particle has a position vector given by the expression r (t) = 5t i + (49 − t2) j metres. ˜ ˜ ˜ a Use a graphics calculator to plot the trajectory of this particle across the interval 0 ≤ t ≤ 7 seconds. Check with your teacher. b Repeat part a above using an Excel spreadsheet. Check with your teacher. 1(1225 − x2) c Determine the equation of this trajectory in the form y = f(x). y = ----25 3 For each of the following position vectors (a to h) for a particle, where the displacement is measured in metres and time in seconds, i calculate the corresponding vector expressions for the velocity and acceleration ii find the initial position and speed iii determine the Cartesian equation of the trajectory of the particle. 3 a i v (t) = 2 i − j , a (t) = 0 c i v (t) = − i − 6 t j , a (t) = −6 j a r (t) = 2t i + (5 − t) j ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ 2˜ ii r (0) = 10 j m, v(0) = 1 m/s b r (t) = t i + (6t − t ) j ii r (0) = 5 j m, v(0) = 5 m/s ˜ ˜ ˜ ˜ ˜ ˜ ˜ iii y = 10 − 3x2 iii y = 5 − 1--2- x c r (t) = −t i + (10 − 3t2) j ˜ ˜ d i v (t) = 3 i + (5 − 4t) j , ˜ b i v (t) = i + (6 − 2 t) j , ˜ ˜ d r (t) = 3t i + (5t − 2t2) j ˜ ˜ ˜ ˜ ˜ ˜ ( t ) = − 4j a 2 ˜ a (t) = −2 j ˜ e r (t) = (4 + t) i + 3t j ˜ ˜ ˜ ˜ ˜ ii r (0) = 0 m, v(0) = 34 m/s ˜ 2 r ii (0) = m, v (0) = m/s 37 0 f r (t) = t i − t j ˜ ˜ ˜ ˜ 2 ˜ ˜ iii y = --9x- (15 − 2x) ˜ 2 iii y = 6 x − x g r (t) = t i − 4t j ˜ ˜ ˜ h r (t) = t i − 3t2 j ˜ ˜ ˜ 4 At time t seconds, the displacement (in metres) of a particle A is given by r A(t) = (2 − t) i + (3 − 2t + t2) j and the displacement (in metres) of a particle B ˜ ˜ ˜ is given by r B(t) = (t − 8) i + (t2 + t − 12) j . yA = (x − 1)2 + 2, ˜ ˜ ˜ yB = (x + 8.5)2 − 12.25 a If these particles collide, determine when they collide. t = 5 seconds b What are the coordinates of the impact point? (−3, 18) c Find the Cartesian equations of the trajectories of the particles. d Use a graphics calculator to verify the coordinates of the impact point.

5 The displacement (in metres) of a particle at time t seconds is given by v (t) = 2 i + 2(1 − t) j , a (t) = −2 j r (t) = 2t i + (4 + 2t − t2) j . ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ a Find expressions for the velocity and acceleration vectors at time t seconds. b Determine the initial velocity of the particle. 2 2 m/s upwards at an angle of 45° to the vertical c When t = 3 seconds, calculate 2 5 m/s downwards at 37 m i the distance of the particle from the origin an angle of 26°34′ to the vertical ii the velocity (both magnitude and direction) of the particle iii the measure of the angle between the velocity and acceleration vectors. 26°34′ .. . d Show that r (t) + r (t) = 2 i − 2t j . ˜ ˜ ˜ ˜ 6 At time t seconds, a particle has a position vector given by the expression r (t) = 2t i + (5 − t)2 j metres. ˜ ˜ ˜ a Use a graphics calculator to plot the trajectory of this particle across the interval 0 ≤ t ≤ 10 seconds. Check with your teacher. b Repeat part a using an Excel spreadsheet. Check with your teacher. c For what values of t are the velocity vectors and acceleration vectors perpendicular? t = 5 seconds

Chapter 11 Dynamics

523

7 P and Q are particles with displacements (in metres) at time t seconds given by 12 r P(t) = (t + 2) i + (5 − t)3 j and r Q(t) = ---------- i + (t − 3) j . t –2˜ ˜ ˜ ˜ ˜ ˜ a If these particles collide, determine when and where they collide. P and Q do not collide. b Verify your response to part a by using suitable technology. ------ − 1 c Express the displacement of Q as an equation written in Cartesian form. y = 12 x 8 The displacement (in metres) of a particle at time t seconds is given by 2 r (t) = --- i + 3t j . t˜ ˜ ˜ a Use a graphics calculator to plot the path of this particle across the interval 1 ≤ t ≤ 10 seconds. Check with your teacher. b Calculate the average velocity across the interval 1 ≤ t ≤ 10 seconds. − --15- i + 3 j m/s ˜ ˜ c Calculate the instantaneous velocity and acceleration of the particle when 2 4 - i + 3 j , a (5) = --------- i v (5) = − ----t = 5 seconds. 25 ˜ 125 ˜ ˜ ˜ ˜ d What can be said about the acceleration as t increases? Approaches zero 9 The displacement (in metres) of an object at time t seconds is given by r (t) = (2 + t) i + (3 + 12t − 5t2) j . ˜ ˜ ˜ a From what position is this object projected? 2 i + 3 j ˜ ˜ 145 m/s b Calculate the speed of projection. c At what angle to the horizontal is this object projected? 85°14′ d When t = 2 seconds, i calculate the velocity (both magnitude and direction) of the object ii determine the position of the object relative to the origin. 4 i + 7 j ˜

65 m/s downwards at an angle of 7°08′ to the vertical

˜

10 Two particles are projected from the origin, O, such that at time t seconds the displacement (in metres) of the first particle is given by r 1(t) = 20t i + (15t − 5t2) j and ˜ ˜ ˜ i + the displacement (in metres) of the second particle is given by r 2(t) = (80 − 12t) ˜ ˜ 2 (a + 9t − 5t ) j where a is a real number. ˜ a What are the initial positions of the particles? r 1(0) = 0 , r 2(0) = 80 i + a j ˜ ˜ ˜ ˜ ˜ b If these two particles meet after 2 1--2- seconds of flight, what is the value of a in the expression for r 2(t)? a = 15 c Locate the point˜ of collision. (50, 6.25) d Calculate the angle between the velocity vectors at this time. 100°18′ 11 A particle is projected from the origin so that at time t seconds its displacement (in metres) is given by r (t) = 20t i + (c + bt − 5t2) j where both b and c are real ˜ ˜ ˜ numbers. y

1y = − ----(x − 50)2 + 26 --1480

The point Q is a point on an inclined plane OQ where Q has coordinates (80, 15). When the particle reaches Q, its . velocity vector is given by r (t) = 20 i − 15 j m/s. ˜ ˜ a What are the values of b and c? b = 25, c˜ = −5 b Find the equation of the trajectory of the particle. c How far up the incline does the projectile land? 81.4 m 12 A particle is projected from the origin up a plane that is inclined at an angle of θ to the horizontal such that at time t seconds, its displacement (in metres) is given by r (t) = 35 cos α t i + (35 sin α t − 5t 2) j where α is the ˜ ˜ angle of projection of the particle. ˜

Q O

x

y

Q α O

x

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9Rmax = 76 ----m 16 when α ≈ 63°26′

a If the point Q is located a distance R metres along the incline, show that R cos θ = 35 cos α t and that R sin θ = 35 sin α t − 5t2. 4R b Deduce that if tan θ = 3--- , then t = ------------------------ . 4 175 cos α 16R c Now show that (4 sin α − 3 cos α) cos α = ------------ . 1225 d By making suitable use of a graphics calculator, find the greatest value that R may have and the value of α when this occurs if 0 < α < 90°. Give your answer correct to the nearest minute. e Use an Excel spreadsheet to validate your solution obtained in part d above.

Projectile motion eBook plus Digital doc: Introduction to Integral Calculus

SLE 8: Model the path of a projectile without air resistance, using the vector form of the equations of motion starting with a = −g j where upwards ~ ~ ` is positive. `

A projectile is a body that is projected at or near the Earth’s surface, and which has no mechanical means of propulsion after leaving a firing mechanism. The only external force acting on such a body is its weight force. For speeds that are relatively low, we will assume that air-resistance is negligible and that the projectile behaves like a point mass. (In reality, shape, size, mass and speed are indeed significant, particularly so for high-speed motion.) It should be understood that the motion of a projectile occurs in the vertical plane containing an initial velocity vector, and that the motion is parabolic. When the projectile is set in motion, the only external force is the particle’s weight force (by Newton’s Second Law of Motion, force = mass × acceleration). If we write g to represent gravitational acceleration, then weight = mass × g; that is, W = mg. In the ideal situation where there are no resisting forces acting on the projectile, the net force of the system is zero. F Hence, if F represents the upwards force of motion, then the equation F + W = 0 describes the sum of the forces acting vermg tically on the projectile. Since we can write F = ma, then F + W = 0 becomes ma + mg = 0 so a = −g m/s2. Note: Various approximations for g exist, and in our work we shall use g = 9.8 m/s2 unless stated otherwise. The unit of force is the newton (N). In the vertical plane, we may express the sum of the forces acting using vector notation. F + mg j = 0 ˜ ˜ m a + mg j = 0 ˜ ˜ m a = −mg j ˜ ˜ a = −g j . ˜ ˜ Since a = ax i + ay j , the horizontal component of acceleration (ax) is zero and hence ˜ ˜ ˜ the horizontal component of the velocity is always a constant. . Now if r (t) represents the displacement of a particle at time t, then r (t) represents the ˜ ˜ .. associated velocity vector and r (t) represents the acceleration vector. ..˜ Since a = −g j and a = r (t) ˜ ˜ .. ˜ ˜ r (t) = −g j ˜. ˜ r (t) = (−g j ) dt = −gt j + c 1 ˜ ˜ ˜ ˜ so v (t) = −gt j + c 1 ˜ ˜ ˜ This is the velocity vector at time t where c 1 is a vector constant of integration. ˜

∫

Chapter 11 Dynamics

525

If we integrate this expression for velocity, we obtain an expression for the displacement of the particle. r (t) = − 1--- gt2 j + c 1t + c 2 2 ˜ ˜ ˜ ˜ The constant vectors c 1 and c 2 can be determined by any initial data. ˜ ˜

WORKED Example 4

A particle moves so that its acceleration at time t is given by a = 2 i + (3 + 4t) j . Find ˜given that the ˜ ˜ initial vector expressions for the velocity and displacement of the particle velocity is 3 i + 4 j and the initial position is 0 i + 0 j . ˜ ˜ ˜ ˜ THINK WRITE v = a dt 1 Integrate the expression for a to ˜ ˜ ˜ find v . = (2 i + (3 + 4t) j ) dt ˜ ˜ ˜ = 2t i + (3t + 2t2) j + c 1 ˜ ˜ ˜ When t = 0, v = 3 i + 4 j and so 2 Apply the initial condition (initial ˜ ˜ ˜ velocity) to determine the constant 3i + 4 j = 0i + 0 j + c 1 ˜ ˜ ˜ ˜ ˜ vector of integration. c 1 = 3i + 4 j ˜ ˜ ˜ v = 2t i + (3t + 2t2) j + 3 i + 4 j 3 Substitute the expression for c 1 into v . ˜ ˜ ˜ ˜ ˜ ˜ ˜ Hence the velocity vector is 4 Simplify to write the vector expression for the velocity of the particle. v = (2t + 3) i + (4 + 3t + 2t2) j ˜ ˜ ˜ r = v dt 5 Integrate the expression for v to ˜ ˜ ˜ find r . ˜ = [(2t + 3) i + (4 + 3t + 2t2) j ] dt ˜ 3 2 ˜ = (t2 + 3t) i + (4t + --- t2 + --- t3) j + c 2 2 3 ˜ ˜ ˜ r i j When t = 0, = 0 + 0 and so Apply the initial condition (initial 6 ˜ ˜ ˜ position) to determine the constant 0i + 0 j = 0i + 0 j + c 2 ˜ ˜ ˜ ˜ ˜ vector of integration. c 2 = 0i + 0 j ˜ ˜ ˜ Hence the displacement vector is 7 Substitute the expression for c 2 into r ˜ ˜ 3 2 and write the vector expression for the r = (t2 + 3t) i + (4t + --- t2 + --- t3) j 2 3 ˜ ˜ ˜ displacement of the particle.

∫ ∫

∫ ∫

Components of velocity and displacement For projectile motion where resistance to motion is ignored, we have established that a = −g j and hence v = −gt j + c . The initial velocity data will determine the vector ˜ ˜ ˜ ˜ c. ˜ constant, ˜ y

v

O

~v

x

vy

vx

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

For an object projected at an angle of θ to the horizontal, the velocity vector can be expressed as v = vx i + vy j . ˜ ˜ ˜ Using trigonometric ratios, the initial velocity (at t = 0) can be written as v = v cos θ i + v sin θ j . ˜ ˜ ˜ To find c , we substitute t = 0 and the initial velocity into v = −gt j + c . ˜ ˜ ˜ ˜ So v cos θ i + v sin θ j = 0 j + c ˜ ˜ ˜ ˜ c = v cos θ i + v sin θ j ˜ ˜ ˜ Hence v = −gt j + c becomes ˜ ˜ ˜ v = −gt j + v cos θ i + v sin θ j ˜ ˜ ˜ ˜ v = v cos θ i + (v sin θ − gt) j ˜ ˜ ˜ The horizontal component of the velocity, vx, is given by vx = v cos θ and the vertical component of the velocity, vy , is given by vy = v sin θ − gt. In a similar way, the components of the displacement can be established by integrating v = v cos θ i + (v sin θ − gt) j with respect to t. ˜ ˜ ˜ That is,

∫

r = [v cos θ i + (v sin θ − gt) j ] dt ˜ ˜ ˜ = vt cos θ i + (vt sin θ − 1--- gt2) j + c 2 ˜ ˜ ˜ If the origin is the launch point of the projectile, then r (0) = 0 i + 0 j and so c = 0 i + 0 j . ˜ ˜ ˜ ˜ ˜ ˜ Hence r = vt cos θ i + (vt sin θ − 1--- gt2) j . 2 ˜ ˜ ˜ The horizontal component of the displacement, rx, is given by rx = vt cos θ and the vertical component of the displacement, ry , is given by ry = vt sin θ − 1--- gt2. 2

What do we do if the launch point is not at the origin? Now if the launch point is at (a, b), then r (0) = a i + b j ˜ ˜ ˜ and so c = a i + b j . ˜ ˜ ˜ r = vt cos θ i + (vt sin θ − 1--- gt2) j + c becomes 2 ˜ ˜ ˜ ˜ 1--- 2 r = (a + vt cos θ) i + (b + vt sin θ − gt ) j 2 ˜ ˜ ˜

y

v P b

O

a

x

Characteristics of flight The launch velocity and angle of projection determine such things as the time of flight, greatest height reached above the point of projection and the range. The period of time for which the projectile is moving under the influence of gravity alone is called the time of flight of the projectile. The range is the horizontal displacement of the projectile from its launch point. To determine the greatest height reached, we must find when the vertical component of velocity is zero; that is, solve v y = 0 for t. The vertical component of the displacement using this value for t determines the greatest height reached.

Chapter 11 Dynamics

527

WORKED Example 5 A projectile is fired from the foot of a cliff and its displacement (in metres) at time t seconds is given by r (t) = 10t i + (24t - 1--- gt2) j . 2 ˜ ˜ ˜ Use g = 9.8 m/s2 in your calculations. a Determine an expression for the velocity vector of the projectile at time t seconds. b Calculate the launch speed and angle of projection of the projectile. c Find the greatest height reached and the time of flight of the projectile. d Calculate the impact velocity of the projectile when it returns to the ground (as a horizontal plane through the origin). THINK a Find the derivative of the displacement vector to obtain an expression for the velocity vector. b

1

To find the launch speed, substitute t = 0 into the velocity vector and then calculate the magnitude.

2

Draw a diagram of the velocity vector at t = 0 to assist in calculating the angle of projection or launch angle.

WRITE a r (t) = 10t i + (24t − 1--- gt2) j 2 ˜ . ˜ ˜ v (t) = r (t) ˜ ˜ v (t) = 10 i + (24 − gt) j ˜ ˜ ˜ b At launch, t = 0. Therefore, v (0) = 10 i + 24 j ˜ ˜ ˜ 2 2 v (0) = 10 + 24 ˜ = 26 Let θ be the angle of projection. ~v (0) = 10i~ + 24j ~

y

26

10

O

x

Use trigonometry to obtain an equation to solve for θ.

tan θ =

4

Write the answer.

The projectile is launched with a speed of 26 m/s upwards at an angle of 67°23′ to the horizontal.

1

The greatest height occurs when the vertical component of the velocity is 0.

3

c

24

24 -----10

θ ≈ 67°23′

c If the vertical component of v is vy then ˜ vy = 24 − gt At the greatest height, vy = 0. 0 = 24 − gt t=

24 -----g

≈ 2.45 s Continued over page

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THINK 2

d

WRITE

To calculate the greatest height, substitute for t in the expression for the vertical component of the displacement vector.

3

For the time of flight, calculate when the vertical component of the displacement is 0; that is, the time when the projectile returns to the ground. Note that there are two solutions for t when the displacement is 0.

1

Since we know that the projectile returns ------ s, substitute for t in to the ground at t = 48 g the velocity vector.

2

Calculate the magnitude of the velocity at this time. (This is the speed.)

3

Find the direction of the projectile at impact. Use a diagram showing the velocity vector at t =

48 ------ s g

and

trigonometry to find the required angle.

When t =

24----, g

the vertical component of

displacement, ry, becomes: ------ = 24 × ry  24  g

=

288 --------g

=

288 --------9.8

24 -----g

−

1--2

×g×

2  24 ------  g

≈ 29.4 m The greatest height reached is 29.4 m. Projectile returns to the ground when ry(t) = 0. 24t − 1--- gt2 = 0 2 t(24 − 1--- gt) = 0 2 -----t = 0 or t = 48 g The solution t = 0 gives the time when the projectile is launched from the ground so the projectile returns to the ground at ------ s (or ≈ 4.90 s). t = 48 g Hence, the time of flight of the projectile is 4.90 s. d v (t) = 10 i + (24 − gt) j ˜ ˜ ˜ ------ = 10 i + (24 − g × 48 ------ ) j v  48  g ˜ ˜ g ˜ = 10 i − 24 j ˜ ˜ velocity The magnitude of the 2

2

= 10 + ( – 24 ) = 26 m/s Let θ be the angle the velocity vector at t= y

48 ------ s g

makes with the horizontal. x

10

O 24

26

( 48 ) ~v —g = 10i~ – 24j ~

tan θ = 4

Write the answer.

24 -----10

θ ≈ 67°23′ At impact, the velocity of the projectile is 26 m/s downwards at an angle of 67°23′ to the horizontal.

Chapter 11 Dynamics

529

Consider the calculations in Worked example 5. What conclusions can you draw when you compare the launch velocity and the impact velocity of the projectile? What can you conclude about the time taken for the projectile to reach its greatest height compared with the total time of the flight? What condition must you have for these observations to be true in other situations?

WORKED Example 6 A particle is fired from the top of a 125 m high cliff with a velocity of 50 m/s inclined at an angle of 60∞ to the horizontal. Use g = 9.8 m/s2 in your calculations. y 50 m/s 60º

150 m

125 m O

A

200 m

B

x

a Show that the velocity vector of the projectile at time t seconds is given by v (t) = 25 i + (25 3 − gt) j m/s. ˜ ˜ ˜ b Develop a vector expression for the displacement, r (t), of the particle at time t seconds. ˜ c Determine if this projectile will clear a 150 m high building situated 200 m from the base of the cliff and in the same horizontal plane. THINK

WRITE

a

a For projectile motion, a = −g j ˜ ˜

1

Write an expression for the acceleration vector. In projectile motion, the horizontal component of a is 0 and the ˜ vertical component is −g. (The negative signifies a downward direction.)

2

Integrate the expression for a to find v . ˜ ˜

3

Determine the initial velocity. Draw a diagram to represent the launch velocity and its horizontal and vertical components.

v = ˜ =

∫ a˜ dt ∫ (−g ˜j ) dt

= −gt j + c 1 ˜ ˜

50

vy

60º vx 4

Write expressions for the horizontal component, vx, and the vertical component, vy.

vx = 50 cos 60° vy = 50 sin 60° Continued over page

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THINK 5

Use the components to write the expression for the velocity vector at t = 0. This is the initial or launch velocity of the particle.

6

Apply the initial condition (launch velocity) to determine the constant vector of integration.

7

Substitute the expression for c 1 into v . ˜ ˜ Simplify to write the vector expression for the velocity of the particle.

8

b

c

1

Integrate the expression for v to find r . ˜ ˜

2

Apply the initial condition to determine the constant vector of integration. The initial position of the particle is at the point (0, 125).

3

Substitute the expression for c 2 into r ˜ ˜ and write the vector expression for the displacement of the particle.

1

Write an expression for the range of the projectile. The horizontal component of r gives the range. ˜ Calculate when the range is 200 m.

2

3

4

5

Write an expression for the height of the projectile. The vertical component of r gives the height. ˜ Substitute t = 8 to calculate the height of the projectile at this time. Write the answer.

WRITE Since v = vx i + vy j ˜ ˜ ˜ v (0) = 50 cos 60° i + 50 sin 60° j ˜ ˜ ˜ = 25 i + 25 3 j ˜ ˜ When t = 0, v = 25 i + 25 3 j and so ˜ ˜ ˜ 25 i + 25 3 j = 0 j + c 1 ˜ ˜ ˜ ˜ c 1 = 25 i + 25 3 j ˜ ˜ ˜ v = −gt j + 25 i + 25 3 j ˜ ˜ ˜ ˜ Hence the velocity vector is v = 25 i + (25 3 − gt) j ˜ ˜ ˜ b r = ˜ =

∫ v˜ dt ∫ [25 ˜i + (25

3 − gt) j ] dt ˜ = 25t i + (25 3 t − 1--- gt2) j + c 2 2 ˜ ˜ ˜ When t = 0, r = 0 i + 125 j and so ˜ ˜ ˜ 0 i + 125 j = 0 i + 0 j + c 2 ˜ ˜ ˜ ˜ c 2 = 0 i +˜ 125 j ˜ ˜ ˜ Hence the displacement vector is r = 25t i + (125 + 25 3 t − 1--- gt2) j 2 ˜ ˜ ˜

c Range = rx = 25t When the range is 200 m, 25t = 200 t=8s Height of projectile = rx = 125 + 25 3 t − 1--- gt2 2

When t = 8, ry = 125 + 25 3 × 8 − ≈ 157.8 m

1 --2

× g × 82

Since the height of the projectile is about 157.8 m when it reaches a range of 200 m, the projectile easily clears the 150 m high building.

Chapter 11 Dynamics

531

In the next worked example, we examine how to find the angle of projection of a particle, given prescribed data. Even though the principles and techniques are the same as those used in solving purely numeric problems, we need to employ some trigonometry in order to reach the solution. Note: The angle of projection that gives a projected particle its greatest range is often intuitively accepted as an angle of elevation of 45° … but this is true only when the point of projection and point of impact are on the same horizontal plane (or more accurately, at the same elevation).

WORKED Example 7

y A missile is fired from a point on level ground with a Target velocity V m/s and an angle of elevation of q to the horizontal. The target is positioned on top of an 80 m high V 80 m tower which is located 100 m away. The base of the tower is in the same horizontal plane as the point of projection O x 100 m of the missile. a If g is the gravitational acceleration of the missile, then show that the motion of this missile satisfies V2 cos q (5 sin q − 4 cos q) = 250g. b Use a graphics calculator to show those values of q in the domain 0 < q < 90° for which the expression cos q (5 sin q − 4 cos q) is positive. State the greatest possible value this expression has and the value of q which produces it. Provide supporting argument for your solutions. c Extend from your answer to part b above to find the least value of V for which the missile can reach the target. (Use g = 9.8 m/s2.) d Generate a spreadsheet to validate the solution obtained using a graphics calculator.

THINK a 1 Use a first principles approach to develop an expression for the velocity of the missile at time t seconds after launch.

WRITE a a = −g j ˜ ˜ v = a dt ˜ ˜

∫ = (−g j ) dt ∫ ˜

= −gt j + c 1 ˜ ˜

2

Determine the initial velocity. Draw a diagram to represent the launch velocity and write expressions for its horizontal and vertical components.

V

vy

vx

vx = V cos θ and vy = V sin θ Since v = vx i + vy j ˜ ˜ ˜ v (0) = V cos θ i + V sin θ j ˜ ˜ ˜

Continued over page

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THINK 3

Apply the initial condition (launch velocity) to determine the constant vector of integration in v = −gt j + c 1. ˜ ˜ ˜

4

Substitute the expression for c 1 ˜ into v = −gt j + c 1. ˜ ˜ ˜ Simplify to write the vector expression for the velocity of the missile at time t seconds.

5

6

Integrate the expression for v to ˜ find r . ˜

WRITE When t = 0, v = V cos θ i + V sin θ j and ˜ ˜ ˜ v = −(g × 0) j + c 1 so ˜ ˜ ˜ V cos θ i + V sin θ j = 0 j + c 1 ˜ ˜ ˜ ˜ c 1 = V cos θ i + V sin θ j ˜ ˜ ˜ v = −gt j + V cos θ i + V sin θ j ˜ ˜ ˜ ˜ Hence the velocity vector is v (t) = V cos θ i + (V sin θ − gt) j ˜ ˜ ˜

∫ v˜ dt = [V cos θ i + (V sin θ − gt) j ] dt ∫ ˜ ˜

r = ˜

= Vt cos θ i + (Vt sin θ − 1--- gt2) j + c 2 2 ˜ ˜ ˜ When t = 0, r = 0 i + 0 j and so ˜ ˜ ˜ 0i + 0 j = 0i + 0 j + c 2 ˜ ˜ ˜ ˜ ˜ c 2 = 0i + 0 j ˜ ˜ ˜

7

Apply the initial condition to determine the constant vector of integration. The initial position of the missile is at the point (0, 0).

8

Substitute for c 2 and write the ˜ for the vector expression displacement of the missile at time t seconds.

Hence the displacement vector is

9

Write an expression for the range of the missile at time t. Use the given information that the range to the target is 100 m to obtain an expression for t.

Range of missile is given by rx = Vt cos θ When rx = 100, Vt cos θ = 100 100 t = ---------------V cos θ

10

Write an expression for the height of the missile at time t. Use the information that the height of the target at the given range is 80 m 100 and that this occurs at t = ---------------V cos θ seconds.

Height of missile is given by

r (t) = Vt cos θ i + (Vt sin θ − 1--- gt2) j 2 ˜ ˜ ˜

ry = Vt sin θ − 1--- gt2 2

100 When ry = 80, t = ---------------- and so V cos θ Vt sin θ − 1--- gt2 = 80 becomes 2

100 V × ---------------- × sin θ − 1--- g × 2 V cos θ

100 2  --------------- = 80  V cos θ

533

Chapter 11 Dynamics

THINK 11 Continue simplifying. Multiply both sides of the equation by V2 cos2 θ.

WRITE 100 sin θ 5000g -------------------- − ------------------= 80 2 2 cos θ V cos θ 2

2

2

2

100 sin θ × V cos θ 5000g × V cos θ ----------------------------------------------- − -----------------------------------------2 2 cos θ V cos θ = 80 V 2 cos2 θ 100V 2 sin θ cos θ − 5000g = 80 V 2 cos2 θ 20V 2 cos θ (5 sin θ − 4 cos θ) = 5000g V 2 cos θ (5 sin θ − 4 cos θ) = 250g (as required) b

1

b Use a graphics calculator to graph the function f(θ) = cos θ (5 sin θ − 4 cos θ). You will need to use x in place of the variable θ.

For the Casio fx-9860G AU (a) Press MENU and select GRAPH . Ensure that your calculator is set to degrees. (Press SHIFT [SET UP] to change the Angle setting.) Press F3 (TYPE) followed by F1 (Y=) to select the function option. Complete the entry line for Y1 with the expression (cos x) (5 sin x − 4 cos x) and press EXE . (b) Press F6 (DRAW) to display the graph. To obtain a clearer view of the graph, you can adjust the View Window settings. Press SHIFT F3 (V-WIN) and adjust the values for Xmin, Xmax, Ymin and Ymax. (c) Press EXE until you return to the function screen. Press F6 (DRAW) to display the graph with the new setting. The graph shows where the expression is positive for 0 < x < 90°. Use the Trace function (press SHIFT F1 (TRCE)) to investigate points along the line. (You can also find the maximum value for the expression and the value of x which produces it in this way.) (d) To find the maximum value (other than using the Trace function), press SHIFT F5 (G-SLV) followed by F2 (MAX). The coordinates of the required point are displayed. Continued over page

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THINK

WRITE

For the TI-Nspire CAS (a) Ensure that your calculator is set to degrees. (Access the home screen and select 8: System Info to change the Angle setting.) Open a new Graphs & Geometry document. Complete the entry line for f1(x) with cos (x)(5 sin(x) − 4 cos(x)). (b) Press · to display the graph of the function. To adjust the viewing window, press b, select 4: Window and 1: Window Settings. Enter the settings as shown. (c) Highlight OK and press ·. The graph shows where the expression is positive for 0 < x < 90˚.

(d) From the home screen (press c), select 1: Calculator. Press b and select 4: Calculus followed by 7: Function Maximum. Press F1(X),X), then press ·. This gives the value of x (or θ) for the maximum value of f(x) (or f(θ)). (e) To find the function value for this x value, press F1( followed by /v to copy the answer value from above and then ) to close the set of brackets. Press · to display the maximum value.

2

Write a conclusion from your observations and calculations.

The graph shows where the expression cos θ (5 sin θ − 4 cos θ) is positive in the domain 0 < θ < 90°. There is one local maximum in this domain. The greatest possible value of this expression is approx. 1.202 when the angle is approx. 64.3°.

Chapter 11 Dynamics

THINK

WRITE

c

c V 2 cos θ (5 sin θ − 4 cos θ) = 250g 250g V 2 = ------------------------------------------------------cos θ ( 5 sin θ – 4 cos θ )

2

The least value of V 2 occurs when the denominator is greatest. Use the maximum value for cos θ (5 sin θ − 4 cos θ) found in part b.

3

Substitute the maximum value for the denominator and the value for g to calculate the least value for V for which the missile can reach the target.

4

Write the answer.

1

Set up a spreadsheet with 3 columns titled t, x and y. In the first column, enter values for t. Enter the appropriate parametric equations of the displacement as x = Vt cos θ and y = Vt sin θ − 1--- gt2 using the calculated 2 value for V and θ and substituting 9.8 for g. Generate corresponding values in columns 2 and 3.

The least value of V 2 occurs when cos θ (5 sin θ − 4 cos θ) is greatest. The maximum value of cos θ (5 sin θ − 4 cos θ) is 1.202 (from part b). 250 × 9.8 V 2 = ---------------------1.202 Since V > 0, then V ≈ 45.1. The required least velocity of the missile to reach the target is approx. 45.1 m/s at an angle of elevation of approx. 64.3°. d

Theta

64.3299

Velocity

45.1000

Delta t

x

y

0

0

0

0.25

4.884

9.856

0.50

9.768

19.099

0.75

14.653

27.730

Metres

Generate the graph to provide a visual validation of the results.

.......

Digital doc: EXCEL Spreadsheet Missile flight

2

t

.......

eBook plus

0.2500

y 90 80 70 60 50 40 30 20 10 0

.......

Rearrange the equation from part a to make V 2 the subject of the equation.

.......

d

1

535

4.75

92.800

82.525

5.00

97.684

80.744

5.25

102.568

78.350

Missile flight

0

20

40

60 80 Metres

100 120

x

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M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

remember remember

dv 1. Finding the velocity given the acceleration: a = -----˜- → v = a dt. dt ˜ ˜ ˜ dr 2. Finding the displacement given the velocity: v = -----˜ → r = v dt. dt ˜ ˜ ˜ Remember that the constant of integration is always a vector constant. 3. The period of time for which the projectile is moving under the influence of gravity alone is called the time of flight of the projectile. 4. The range is the horizontal displacement of the projectile from its launch point.

11B SLE 3: Given the force on an object as a function of time and suitable prescribed conditions, such as velocity and displacement at certain times, use integration to find the position vector of the object.

WORKED

Example xample

2 e

f

g

h

1 Given the following information about the motion of a particle at time t, determine a vector expression for the particle’s displacement. a v (t) = i and r (0) = 3 i + 2 j r (t) = (t + 3) i + 2 j ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ b v (t) = 2t i − 3 j and r (0) = i + j r (t) = (t2 + 1) i + (1 − 3t) j ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ dr c -----˜ = 4 i + 7t j and r (1) = 2 i − j r (t) = 2(2t − 1) i + --12- (7t2 − 9) j ˜ dt ˜ ˜ ˜ ˜ ˜ ˜ ˜ dr 1--d -----˜ = (3 + t) i − 2 j and r (2) = i + 5 j r (t) = (3t + 2 t − 7) i + (−2t + 9) j ˜ ˜ dt ˜ ˜ ˜ ˜ ˜ ˜ . 3--- 2 1--- 2 e r (t) = 3t i + (4 + t) j and r (0) = − i + j r (t) = ( 2 t − 1) i + (4t + 2 t + 1) j ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ . 1--1--- 2 2 r ( t ) = − t ( t − 4) − t ( t + 18) j i f r (t) = (2 − t) i + (6 + t ) j and r (0) = 0 i + 0 j ˜ 2 ˜ 3 ˜ ˜ ˜ ˜ ˜ ˜ ˜ g v (t) = 15 i + (10 − 3t) j and r (0) = 20 j r (t) = (15t) i + (10t − 3--2- t2 + 20) j ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ dr 1 2   1--h -----˜ = −2t i + 5 + ---- j and r (1) = 2 i − 7 j r (t) = (−t + 3) i + (5t − t − 11) j 2 ˜ dt ˜ ˜ ˜  ˜ ˜ ˜ t ˜ 2 Determine vector expressions at time t for the velocity, v , and displacement, r , ˜ of a particle which moves according to the following ˜ relationships involving acceleration, a . ˜ a a (t) = 2 i and when t = 0, v = i − 3 j and r = − i + 5 j ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ dv b -----˜- = 3 i − 5 j and when t = 0, v = 2 i − j and r = 0 i + 0 j dt ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ .. c r = − i + t j , v (0) = 3 i − 4 j and r (0) = 2 i − 3 j ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ .. . d r = 2t i − 3 j , r (1) = − 4 j and r (0) = i + j ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ dv e -----˜- = −5 i + (t + 1) j , r (0) = 3 i + 4 j and r (2) = 8 i + 3 j dt ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ f a (t) = i + (6 − t) j , r (0) = −5 i + 12 j and r (1) = − i + 4 j ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ dv g -----˜- = (2t + 1) i − (3 − 5t) j , r (0) = 8 j and r (1) = −3 i + 9 j dt ˜ ˜ ˜ ˜ ˜ ˜ ˜ .. h r = 3 i + (12 − 5t) j , v (3) = 3 i − 4 j and r (5) = 2 i − 3 j ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜

1513v (t) = (−5t + ----) i + ( 2--1- t2 + t − ----)j, 2 ˜ 6 ˜ 1513r (t) = (− --5- t2 + ----t + 3) i + ( --1- t3 + 2--1-˜t2 − ----t + 4) j 2 2 6 6 ˜ ˜ ˜ 65)j, v (t) = (t + 2--7- ) i + (− 2--1- t2 + 6t − ----6 ˜ ˜ 65t + 12) j r (t) = ( 2--1- t2 + 2--7- t − 5) i + (− 6--1- t3 + 3t˜ 2 − ----6 ˜ ˜ ˜ 23) i + ( 2--5- t2 − 3t + 3--5- ) j , v (t) = (t2 + t − ----6 ˜ ˜ 23t) i + ( --5- t3 − --3- t2˜+ 3--5- t + 8) j r (t) = ( --1- t3 + --1- t2 − ----3 2 6 6 2 ˜ ˜ ˜ 35)j, v (t) = (3t − 6) i + (− 2--5- t2 + 12t − ----2 ˜ ˜ ˜ 3 2 11 35 116 - ) i + (− --5- t + 6t − ----- t + --------) j r (t) = ( 2--3- t2 − 6t − ----2 ˜ 6 2 3 ˜ ˜

2

Projectile motion

Unless otherwise stated, use g = 9.8 m/s2.

a v (t) = (2t + 1) i − 3 j , ˜ ˜ r (t) = (t2 + t − 1) i +˜ (−3t + 5) j ˜ ˜ ˜ b v (t) = (3t + 2) i + (−5t − 1) j , ˜ ˜ r (t) = ( 3--2- t2 + 2t) i + (− 5--2- t2 − ˜t) j ˜ ˜ ˜ c v (t) = (−t + 3) i + ( 1--2- t2 − 4) j , ˜ ˜ ˜ r (t) = (− --12- t2 + 3t + 2) i + ( --16- t3 − 4t − 3) j ˜ ˜ ˜ d v (t) = (t2 − 1) i + (−3t − 1) j , ˜ ˜ 1--- 3 3--- 2˜ r (t) = ( 3 t − t + 1) i + (− 2 t − t + 1) j ˜ ˜ ˜

4

∫ ∫

Chapter 11 Dynamics

WORKED

Example

5 19.2 m/s at an angle of 51°20′ above the horizontal

537

3 A projectile is fired from the foot of a cliff and its displacement (in metres) at time t seconds is given by r (t) = 12t i + (15t − 1--- gt2) j . v (t) = 12 i + (15 − gt) j 2 ˜ ˜ ˜ ˜ ˜ ˜ a Determine an expression for the velocity vector of the projectile at time t seconds. b Calculate the launch speed and angle of projection of the projectile. c Find the greatest height reached and the time of flight of the projectile. 11.5 m 3.06 seconds d Calculate the impact velocity of the projectile when it returns to the ground in a horizontal plane through the origin. 19.2 m/s downwards at an angle of 51°20′ to the horizontal

4 A projectile is fired from the origin, O, and its displacement (in metres) at time t seconds is given by r (t) = 15t i + (30t − 1--- gt2) j . 2 ˜ ˜ v (t) = 15 i + (30 − gt) j a Determine an expression for the velocity˜ vector of the projectile at time t seconds. ˜ ˜ ˜ 33.5 m/s at an angle of b Calculate the launch speed and angle of projection of the projectile. 63°26′ above the c Determine the greatest height reached by the projectile. 45.9 m horizontal d Will this projectile clear a 4 m high wall located 90 m from the launch point in the same horizontal plane? No, as it falls short by about 40 cm. WORKED

Example

6

5 A particle is fired from the top of a 50 m high cliff with a velocity of 36 m/s inclined at an angle of 30° to the horizontal.

y 30º 50 m O

47.7 m/s downwards at an angle of 49°12′ to the horizontal

x

a Show that the velocity vector of the projectile at time t seconds is given by v (t) = 18 3 i + (18 − gt) j m/s. ˜ expression ˜ ˜ for the displacement, r (t), of the particle at time b Develop a vector ˜ t seconds. r (t) = 18 3 t i + (50 + 18t − --12- gt2) j Maximum height is 66.5 m which is ˜ ˜ ˜ 16.5 m above the launch platform. c Calculate the greatest height reached by the projectile. d What is the impact velocity of the projectile when it reaches a point in the horizontal plane through the origin, O? e At this impact point, what is the range of the projectile? 172.1 m

Particles don’t collide but they do 6 At time t, the velocities of two particles A and B are given by pass through two dr B common points at v A(t) = 4 i + (5 − 2t) j and -------˜ = −3 i + (4t − 7) j . different times. At r A(t) = 4(t − 1) i + (3 + 5t − t2) j , dt ˜ ˜ ˜ ˜ ˜ ˜ ˜ (13.70, 5.55), one r B(t) = 3(8 − t) i + (6 − 7t + 2t2)˜ j When t = 0, r A = −4 i + 3 j and r B = 24 i + 6 j . particle arrives when ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ t = 4.43 s and the a Determine vector expressions for the displacement of both particles at time t. other when t = 3.43 s. b Do these particles ever collide? If so, calculate when and where. If not, explain At the point what happens. (18.20, −0.06), one 1yA = 3 + ----(x + 4)(16 − x), 16 particle arrives when c Find the Cartesian equations of the trajectories of the particles. 1-= 6 + (24 − x)(27 − 2x) y B t = 5.55 s and the 9 d Use a graphics calculator to verify your response to part b above. other when t = 1.93 s.

7 The acceleration of a particle λ, at time t, is given by a λ(t) = −8 j . For this particle, ˜ r λ(2) = −4 i + 3 j and r λ(5) = 11 i + 15 j . The displacement of a˜ second particle, ξ, ˜ ˜ ˜ ˜ ˜ − 10 − 2t2) j . r λ(t) = (5t − 14) i + (32t − 45 − 4t2) j at time t, is given˜ by r ξ (t) = (6 + t) i + (15t ˜ ˜ ˜ ˜ ˜ a Determine the vector expression for the displacement˜ of the particle, λ, at time, t. b Do these particles ever collide? If so, calculate when and where. c Validate your response to part b above by using suitable technology. When t = 5 s, the particles do collide at (11, 15). Particles appear to collide at (9.15, 17.4) but actually pass through this point at different times; so they don’t collide. Particle λ at t = 4.63 s and particle ξ at time t = 3.15 s.

538 WORKED

Example xample

7

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

8 A missile is fired from a point on level ground with a velocity V m/s and an angle of elevation of θ to the horizontal. The target is positioned on top of a 40 m high tower which is located 80 m away. The base of the tower is in the same horizontal plane as the point of projection of the missile. y V 40 m

O

SLE 16: Use spreadsheets to investigate problems.

x

80 m

a If g is the gravitational acceleration of the missile, then show that the motion of this missile satisfies V 2 cos θ (2 sin θ − cos θ) = 80g. b Use a graphics calculator to show those values of θ in the domain 0 < θ < 90° for which the expression cos θ (2 sin θ − cos θ) is positive. State the greatest possible value this expression has, and the value of θ which produces it. Provide supporting argument for your solutions. 0.618 when θ ≈ 58°17′ c Extend from your answer to part b above to find the least value of V for which the missile can reach the target. (Use g = 9.8 m/s2.) 35.6 m/s d Generate a spreadsheet to validate the solutions obtained using a graphics calculator. 9 A particle is projected with an initial speed of 35 m/s in such a direction that it just passes over the top of a 5 m high wall. The foot of the wall is situated 50 m from the point of projection of the particle and, as shown at right, is in the same horizontal plane through the point of projection. Determine all possible angles of projection, θ. θ ≈ 17°46′ and θ ≈ 77°56′ 10 A particle is projected horizontally from the top of a tower 10 m high with an initial speed of 15 m/s. 21.429 m

y 35 m/s 5m

O

x

50 m

y 15 m/s

If the surrounding ground is horizontal, how far from the base of the tower does the particle land?

10 m

O

x

7 - , from the top 11 A projectile is fired at an angle θ below the horizontal, where tan θ = ----24 of a cliff, 75.6 m high, which overlooks the sea. If the projectile reaches the sea at the instant 2 s after projection, find its initial speed and the distance from the base of the cliff to the point of impact of the projectile with the sea. 100 m/s, 192 m

539

Chapter 11 Dynamics

12 Two particles, P and Q, are fired simultaneously from points A and B respectively which are located 60 m apart on horizontal ground, as shown below. The particle P has an initial speed of 24 m/s and an angle of projection θ, where θ = tan−1  3--- . The particle 4 Q has an initial speed of 18 m/s with an angle of projection φ where φ = tan−1  4--- . 3

y 24 m/s

Do these particles collide? If so, when and where do they collide? Yes, they meet at (38.4, 9.2)

18 m/s

when t = 2 s.

ø A

60 m

B

x

13 Two walls are situated 16 m apart on y level ground. One wall is 7.1 m high V and the other is 4.4 m high. A projectile is to be fired from the point on the ground which is 16 m from the 7.1 m high wall and on the side remote from 7.1 m 4.4 m the second wall as seen in the figure at x O 16 m 16 m right. Determine the initial speed and angle of projection required if this projectile will just clear both walls. 20 m/s at an angle of approx. 36°52′ 14 A particle is projected with an initial speed of 35 m/s from the base of a plane inclined at an angle φ to the horizontal where φ = tan−1  3--4- . If the projectile is initially angled at θ to the horizontal, where tan θ = 5--- , find how far along the plane the projec2 tile lands, and the time of flight. 75.431 m, approx. 4.64 s 15 A particle is projected up a plane inclined at an angle θ to the horizontal, the angle of projection being φ above the horizontal. If the initial speed of projection is V, a show that the range, R, on this inclined plane is given by 2

y

V

ø

sin ( φ – θ ) cos φ R = 2V ----------------------------------------------2 x O g cos θ b determine the maximum range (Rmax) on the plane and the value of φ for which 2 V this occurs. Rmax = ------------------------- when φ = 45° + --θ(1+sin θ )g 2

eBook plus Digital doc: WorkSHEET 11.1

16 A sprinkler sprays water symmetrically about its vertical axis at a constant speed of V. The initial direction of the spray varies continuously between angles of 15° and 60° to the horizontal. a Prove that from a fixed position, O, on level ground, the sprinkler will wet the sur2 2 V V face of an annular region with respective internal and external radii ------ and ------ . 2g g b Now show that if the sprinkler is located appropriately to a rectangular garden bed 2 V of size 6 m by 3 m, the entire garden will be watered provided ------ ≥ 1 + 7 . 2g

540

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Motion under constant acceleration Even though our study in this course is restricted to motion involving constant acceleration, this study does include many situations of practical significance. The common example of bodies in free fall towards the Earth’s surface, regardless of their size, mass or composition, involves motion under constant acceleration — that of acceleration due to gravity. Galileo illustrated this property by dropping various sized cannon balls from the top of the Leaning Tower of Pisa. Provided the effects of air resistance are minimised (that is, the object should not be too large in area or fall too far) all bodies will fall with constant acceleration. Other variations will be considered later in your studies. Since the acceleration due to gravity depends on the distance from the centre of the Earth, it varies slightly at different places on the Earth’s surface — being greater at the poles than at the equator and less at high altitudes. It is usual to state the constant of acceleration due to gravity as 9.8 m/s2 unless you are told otherwise. Since acceleration is constant, the displacement of a falling object can be modelled by the general quadratic expression lt 2 + mt + n This can be used to model the motion of a body in free fall or one tossed upwards. The important difference is that the velocity and acceleration are in different directions. When vector notation is used the vertical displacement vector is written as r = (lt 2 + mt + n) j ˜ ˜ Because j does not depend on t the derivative of this function is as follows: ˜ v = (2lt + m) j ˜ ˜ a = 2l j and ˜ ˜ The following analysis examines the motion of a ball as it rests at the top of a cliff. As soil falls away the ball begins to move. (a) At t = 0, find v. At t = 0 the ball is at rest, therefore v = 0. (b) If the bottom of the cliff is the origin and the cliff is h metres high, find r when t = 0. At t = 0, r = h j ˜ ˜ (c) Given the constant acceleration towards the origin equals −9.8 m/s2 (a negative value is used to indicate that the direction of the acceleration is downwards), find m and l.

Chapter 11 Dynamics

541

Use the original equation for displacement and differentiate: r = (lt2 + mt + n) j [1] ˜ ˜ v = (2lt + m) j [2] ˜ ˜ a = 2l j [3] ˜ ˜ when t = 0 From part (a) above v = [2l(0) + m] j in [2] ˜ ˜ but v = 0 (initially at rest) ˜ Therefore, m = 0 From part (b) above, if at t = 0, r = h j ˜ ˜ [1] r = (lt 2 + mt + n) j ˜ 2 ˜ r = [l(0) + m(0) + n] j ˜ ˜ at t = 0) =hj (as given ˜ Therefore, n=h so a = −9.8 j (as a = g j ) From (c) g = −9.8 m/s2 ˜ ˜ ˜ ˜ a = 2l j Also ˜ ˜ So 2l j = −9.8 j ˜l = −4.9 ˜ Therefore, Therefore the general equation for this motion can be written: r = (−4.9t 2 + h) j ˜ Do not attempt to memorise ˜these equations. Equations [1], [2] and [3] are the basic equations and all others are specific rearrangements of these. In general, 1. l can be readily found if the acceleration is given: a = 2l. 2. m can be readily found if the initial velocity is given: v = 2l(0) + m. 3. n can be readily found if the initial displacement is given: r = l(0)2 + m(0) + n

WORKED Example 8 Examine the motion of a stone thrown upwards with a velocity of 6 m/s from the top of a cliff 60 m high. Find the time taken for it to reach the ground and its velocity on impact. THINK 1 As the acceleration is constant, the vertical displacement is quadratic. Write the three equations of motion. 2 The position of the origin needs to be stated. Generally where the motion begins or finishes can be thought of as the origin. 3 Use the given information to find each value of n, m and l. First use the initial position to find n. 4

Rewrite the equation for r with the ˜ new information.

WRITE r = (lt2 + mt + n) j v˜ = (2lt + m) j ˜ ˜ a˜ = 2l j ˜ ˜ Let the origin be the top of the cliff.

At t = 0, r = 0 j and ˜ 2 + m(0) + n] j r˜ = [l(0) ˜ ˜ So 0=0+0+n therefore n = 0 r = (lt 2 + mt) j ˜ ˜

Continued over page

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THINK 5 Use the equations for a to find l. ˜ Constant downwards acceleration away from the origin so the acceleration is negative. 6

Use the initial velocity to find m. Initial velocity at t = 0 is 6 m/s up (positive sign).

7

Rewrite r with all current information. ˜ Find the time when the stone reaches the ground at r = −60 j . ˜ notation. ˜ Drop the vector Solve for t using the quadratic formula where a = −4.9, b = 6 and c = 60.

8

WRITE a = 2l j and ˜ j a˜ = −9.8 ˜ = −9.8 ˜ So 2l l = −4.9 r = (−4.9t 2 + mt) j ˜ ˜ At t = 0, v = 6 or v = 6 j ˜ j ˜ 6 j = [2l(0) + m] ˜ =6 ˜ m r = (−4.9t 2 + 6t) j ˜ ˜ At r = −60 j ˜ −60 j ˜ = (−4.9t 2 + 6t) j ˜0 = −4.9t 2 + 6t +˜ 60 2

–b ± b – 4ac t = ------------------------------------2a –6 ± 36 – 4 ( – 4.9 ) ( 60 ) = ----------------------------------------------------------2 ( – 4.9 ) t = 4.16 s or –2.94 s Reject the negative solution. Therefore the stone reaches the ground 4.16 s after it was thrown upwards.

9

Find the velocity on impact by finding v when t = 4.16. Note that the negative sign indicates a vector moving downwards, in the same direction as the negative acceleration.

v = (−9.8t + 6) j ˜ At ˜t = 4.16 s v = [−9.8(4.16) + 6] j ˜ = −34.81 j ˜ ˜

10

State the solution.

Therefore, the stone reaches the ground with a velocity of 34.81 m/s downwards 4.16 s after launching.

WORKED Example 9 A stone (A) is thrown upwards from a cliff with a velocity of 30 m/s. After stone A has been in motion for 4 s another stone (B) is dropped from the same point. Find when and where the two stones will meet. THINK 1 Draw a diagram and state the origin position and direction of positive motion. Write the 3 equations of motion.

WRITE/DRAW 30 m/s

r = (lt2 + mt + n) j ˜ O v = (2lt + m) j ˜ B ˜ ˜ a = 2l j ˜ ˜ The origin is at the top of the cliff and there is positive motion up. A

Chapter 11 Dynamics

THINK 2

Stone A: Substitute the given information to find r . ˜

3

Use information about v to find m. ˜

4

Rewrite r . ˜ Use information about a to find l.

5

543

WRITE/DRAW Stone A: At t = 0, r = 0 j and ˜ ˜ 2 + m(0) + n] j r = [l(0) ˜ ˜ So 0=0+0+n therefore n = 0 r = (lt 2 + mt) j ˜ ˜ At t = 0, v = 30 j and ˜ ˜ + m] j v = [2l(0) ˜ ˜ So 30 = 0 + m m = 30 r = (lt 2 + 30t) j ˜ ˜ At t = 0, a = −9.8 j and ˜ a = 2l j ˜ ˜ So −9.8 = 2l ˜ l = −4.9 r = (−4.9t 2 + 30t) j ˜ ˜ Stone B: At t = 0, a = −9.8 and l = −4.9 At t = 0, v = 0 j and ˜ ˜ + m] j v = [2l(0) ˜ ˜ therefore m = 0 At t = 0, r = 0 j and ˜ ˜ 2 + m(0) + n] j r = [l(0) ˜ ˜ therefore n = 0

6

For stone B: Repeat the method used for stone A. For many of your problems l = –4.9.

7

Rewrite the equation for r using all ˜ current information.

8

Find where the two stones meet: when r A = r B; but tB = tA – 4. ˜ all information ˜ Express in terms of A.

The two stones meet where r A = r B ˜ ˜ (−4.9tA2 + 30tA) j = −4.9tB2 j −4.9tA2 + 30t˜A = −4.9(tA ˜– 4)2 = −4.9(tA2 − 8tA + 16) 0 = 39.2tA − 30tA – 78.4 = 9.2tA − 78.4 tA = 8.52 s

9

Find the displacement of B at tA = 8.52 s and tB = tA – 4 = 4.52 s. The information for stone B is easier to use to find the displacement because A has had upwards and downwards motion (both unknown) whereas B has had downwards motion only.

rB = −4.9 (4.52)2 = −100.1 m The stones meet about 100 m below the top of the cliff 8.52 seconds after stone A is thrown.

r = −4.9t 2 j ˜ ˜

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In general this procedure can be shortened. If an object starts at the origin, n = 0 and m has the value of the initial velocity (upwards is positive, downwards is negative). If the body is influenced by gravity then l = −4.9. If an object is initially at rest then m = 0, but if an object is part of a system that is itself moving, and the object is released from that moving system, then the initial velocity (m) of the object will be the velocity of the whole system at the instant prior to release.

remember remember 1. 2. 3. 4. 5.

If the origin is the starting point, then n = 0. If the body is initially at rest, then m = 0. In vertical motion, displacement up is positive; displacement down is negative. Velocity up is positive; velocity down is negative. Acceleration up is positive; acceleration down is negative. SLE 5: Model vertical motion under gravity alone.

11C

Motion under constant acceleration

1 A particle moving from rest with constant acceleration reaches a speed of 16 m/s in 4 seconds. Find: a the acceleration 4 m/s2 b the distance travelled. 32 m 2 An object travelling at 8 m/s accelerates uniformly over a distance of 20 metres until it reaches a speed of 18 m/s. Find: a the acceleration 6.5 m/s2 b the time taken. Approx. 1.54 s 3 a A racing car accelerates constantly from rest and covers a distance of 400 metres in 10 seconds. Find its velocity at the end of the 400 metres. 80 m/s Approx. 8.94 s b Another car travels the 400 metres with a constant acceleration of 10 m/s2. Find its time for the 400 metres. 4 A train travelling at a constant speed decelerates uniformly for 30 seconds over a distance of 270 metres, coming to a stop. Find: a the initial speed 18 m/s b the acceleration. −0.6 m/s2

Chapter 11 Dynamics

545

5 A parachutist free-falls from an aircraft for 6 seconds. If the acceleration due to gravity is 9.8 m/s2, find: a the speed of the parachutist after 6 seconds 58.8 m/s b the distance travelled after 6 seconds. 176.4 m 6 A ball is thrown up from the ground with an initial velocity of 19.6 m/s. The acceleration due to gravity is 9.8 m/s2 downwards, that is, −9.8 m/s2. Find: a the maximum height attained by the ball 19.6 m b the total time taken for the ball to return to the ground. 4 s WORKED

Example

8

7 A stone is dropped from a bridge which is 39.2 metres above a river. a How long does it take the stone to reach the water? 2.83 s b What is its speed on impact? 27.72 m/s 8 A ball is dropped from a tower and reaches the ground in 4 seconds. Find: a the height of the tower 78.4 m b the velocity of the ball when it hits the ground. 39.2 m/s

eBook plus Digital docs: SkillSHEET 11.1 Vertical motion

9 A particle is projected vertically up from the top of a building that is 50 metres above the ground. If the initial speed of the particle is 28 m/s, find: a the maximum height, above the ground, that it reaches 90 m b total time taken to reach the ground 7.14 s c the speed of the particle when it reaches the ground. 42 m/s 10 A train travels a distance of 1800 metres in 90 seconds while accelerating uniformly 10from rest. What is its velocity at the end of 500 metres? 21.08 m/s or 20 --------------3 11 A car accelerates uniformly from rest, increasing its speed from 5 m/s to 25 m/s in 10 seconds. Find: a the acceleration 2 m/s2 b the distance travelled, from rest, in 12 seconds 144 m c the time taken to increase its speed from 15 m/s to 30 m/s. 7.5 s 12 A sprinter accelerates uniformly to his top speed after running 30 metres of a 100-metre race. He maintains this speed for the remainder of the race and takes 10.4 seconds to complete it. Find: a the top speed of the athlete 12.5 m/s b the time taken to reach the top speed. 4.8 s 13 A bus is travelling at 16 m/s when the brakes are applied, reducing the speed to 6 m/ s in 2 seconds. Assuming the retardation is constant, find: a the acceleration −5 m/s2 b the distance travelled 2 seconds after the brakes are applied 22 m c how long after applying the brakes the tram comes to a stop 3.2 s d the braking distance of the bus. 25.6 m 14 A car moving from rest with uniform acceleration takes 12 seconds to travel 144 metres. What is its speed after 6 seconds? 12 m/s 15 A bus travels 60 metres in 10 seconds and the next 60 metres in 15 seconds. If the acceleration is constant, find: a how much further it will travel before coming to rest 24.5 m b how many more seconds it takes before coming to rest. 17.5 s

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16 A juggler throws balls vertically into the air so that they rise to a height of 4.4 metres above the ground. He fails to catch one and it hits the ground with a speed of 1.155 times that of its initial speed. Find: a the speed of projection of the ball 8.04 m/s b the height from which the ball is thrown 1.1 m c the total time the ball is in the air. 1.77 s 17 An object is projected vertically up from a 14 metre tower and reaches the ground 4 seconds later. a What is the projection speed of the object? 16.1 m/s b What is the maximum height above the ground that is attained by the object? 27.23 m

b

18 a An object is dropped from the top of a building 39.2 m high. Calculate: i its velocity when it is halfway to the ground −19.6 m/s ii its velocity on striking the ground −27.7 m/s iii the time taken to each the ground. t = 8 s b Repeat the above questions for the case when the object is thrown to the ground with a velocity of 4.9 m/s.

i −20.2 m/s ii −28.15 m/s iii t = 2.4 s

19 a A stone rolls off the top of a cliff and is found to take 4 seconds to reach the sea below. What is the height of the cliff? 78.4 m b What is the difference in time to reach the bottom between part a and if the stone were launched vertically upwards from this cliff with a velocity of 20 m/s? 2.53 s longer 20 A vertical slit 1.5 m long is positioned in a stone wall 9.8 m below the top of the wall. A small object is dropped from the top of the wall so that it falls in line with the slit. For what length of time is the falling object visible through the slit? 0.1044 s 21 A body is dropped from the top of a 100 m high tower at the same time as a body is launched vertically upwards from the bottom of the tower with a velocity of 25 m/s. 9 Find when and where the two bodies are at the same height above the ground. 4 s, 78.4 m

WORKED

Example xample

22 A body is rising with a velocity of 20 m/s and it releases a small particle when it is 50 m above the ground. How long will it take for the small particle to reach the ground? 5.83 s 23 A cage is descending into a well at a constant velocity of 2 m/s when an object falls through the wire of the cage. If the object reaches the water at the bottom of the well 10 seconds before the cage find the height above the water level at which the object fell out of the cage. 195 m eBook plus Digital doc: WorkSHEET 11.2

24 A fireworks rocket is fired vertically upwards with uniform acceleration of 19.6 m/s2. After 2 seconds a small particle is released from the rocket. How long after release will the particle fall to the ground? 8.9 s

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summary Displacement, velocity and acceleration • Displacement gives the position of a particle relative to some reference point (usually the origin). • Instantaneous velocity is the rate of change of displacement with respect to time and is represented by the derivative of the displacement variable. • Average velocity during a time interval is the change in velocity across that interval. • Speed is the magnitude of velocity just as distance is the magnitude of displacement. • Instantaneous acceleration is the rate of change of velocity with respect to time and is represented by the derivative of the velocity variable.

Vector relationships • If the position vector r (t) = x(t) i + y(t) j represents the displacement of P at time t, ˜ ˜ ˜ 2 . .. dv dr d r then -----˜ (or r (t)) represents the instantaneous velocity and -------2˜- (or r (t) or -----˜- ) dt dt ˜ ˜ dt represents the instantaneous acceleration at time t. dv • Finding the velocity given the acceleration: a = -----˜- → v = a dt. dt ˜ ˜ ˜ dr • Finding the displacement given the velocity: v = -----˜ → r = v dt. dt ˜ ˜ ˜ Remember that the constant of integration is always a vector constant.

∫ ∫

2

2

• For the magnitude of a vector x i + y j : xi + y j = x + y . ˜ ˜ ˜ ˜ • The dot product of two vectors a and b is given as a • b = a b cos θ where θ is ˜ ˜ ˜ ˜ ˜ ˜ the angle included by a and b . ˜ ˜

Projectile motion • The period of time for which the projectile is moving under the influence of gravity alone is called the time of flight of the projectile. • The range is the horizontal displacement of the projectile from its launch point.

Motion under constant acceleration • r = (lt2 + mt + n) j , v = (2lt + m) j , a = 2l j ˜ ˜ ˜ ˜ ˜ ˜ • If the origin is the starting point, then n = 0. • If the body is initially at rest, then m = 0. • In vertical motion, displacement up is positive; displacement down is negative. • Velocity up is positive; velocity down is negative. • Acceleration up is positive; acceleration down is negative.

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CHAPTER review 11A 11A 11A 11A 11A

2 multiple choice The vector which represents the average displacement (in m) in the first two seconds is: A 4i + 6 j B 2 i + 3.5 j C 4i + 7 j D 8 i + 12 j E 4i + 4 j ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ 3 multiple choice The particle reaches its greatest height after how many seconds? A 0 B 2 C 3 D 4 E 5 4 multiple choice The acceleration of the particle in m/s2 is given by: A 2i B −2 j C 2i − 2 j D 0i + 0 j E 2j ˜ ˜ ˜ ˜ ˜ ˜ ˜ 5 multiple choice The Cartesian equation of the trajectory of the particle is: x x A y = 4x B y = 8x – x2 C y = ------ (32 – x) D y = --- (2 – x) E y = x2 – 16x 16 8 6 Relative to the origin, O, the displacement of a particle at time t seconds is given by r (t) = 6t i + 3t(4 − t) j metres. ˜ ˜ the velocity and acceleration vectors at time t seconds. a˜ Find expressions for b Determine the initial displacement and speed of the particle. Origin, 6 5 m/s c When t = 2 seconds, calculate: i the distance of the particle from the origin 12 2 m ii the velocity (both magnitude and direction) of the particle 6 m/s to the right iii the measure of the angle between the velocity and displacement vectors. 45°

11A

7 At time t, a particle A has a displacement given by r A(t) = t i + (4t − t2) j , while a second ˜ particle B has a displacement given by r B(t) = t i + ˜(t − 10) ˜j . ˜ ˜ collide. 5˜ seconds a If these particles collide, determine when they b What are the coordinates of the impact point? (5, −5) 2 c Find the Cartesian equations of the trajectories of the particles. yA = 4x − x , yB = x − 10 d Use a graphics calculator to verify the coordinates of the impact point.

11A

8 At time t seconds, a particle has a position vector given by the expression r (t) = (t + 1) i + (25 − t2) j metres. ˜ ˜ to plot the trajectory of this particle across the interval a˜ Use a graphics calculator 0 ≤ t ≤ 5 seconds. b Repeat part a using an Excel spreadsheet. c Determine the equation of this trajectory in the form y = f (x). y = 25 − (x − 1)2

v (t) = 6 i + (12 − 6t) j m/s, ˜ ˜ ˜ a (t) = −6 j m/s2 ˜ ˜

11A

Questions 1 to 5 refer to a particle that has a displacement r (t) at time t seconds, where ˜ r (t) = 4t i + (8t − t2) j metres. ˜ ˜ ˜ 1 multiple choice The initial speed of the particle in m/s is: A 4 B 4 5 C 0 D 2 13 E 2 10

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v (t) = t i + (3t − t2) j m/s, 2 ˜ ˜ r (t) = 1--2- t2 i + ---t6- (9 −˜ 2t) j m ˜ ˜ ˜

Questions 9 to 11 refer to a projectile that has velocity v (t) at time t seconds, where v (t) = 6 i + (10 − 2t) j m/s. The initial displacement of ˜this projectile is given by ˜ r˜ (0) = 4˜ j m. ˜ ˜ 9 multiple choice The angle of projection of this particle, correct to the nearest degree, is: A 39° B 49° C 51° D 59° E 61°

11B

10 multiple choice The range (in m) of the projectile through the foot of the launch platform is nearest in value to: A 45 B 50 C 60 D 70 E 75

11B

11 multiple choice The greatest height (in m) reached by the particle is closest in value to: A 29 B 32 C 35 D 39

11B E 95

12 A particle moves so that its acceleration at time t is given by a = i + (3 − 2t) j . Find vector ˜ ˜ that v = 0˜ and r = 0 expressions for the velocity and displacement of the particle given ˜ ˜ when t = 0.

13 A projectile is fired from the foot of a cliff and its displacement (in metres) at time t seconds is given by r (t) = 6t i + (45t − 1--- gt2) j . 2 v (t) = 6 i + (45 − gt) j m/s ˜ 2 in your ˜ calculations. ˜ Use g = 10 m/s ˜ ˜ ˜ a Determine an expression for the velocity vector of the projectile at time t seconds. b 3 229 m/s at an angle b Calculate the launch speed and angle of projection of the projectile. of elevation c Find the greatest height reached and the time of flight of the projectile. 101 1--- m, 9 s 4 of 82°24′ d Calculate the impact velocity of the projectile when it returns to the ground in a horizontal plane through the origin. 3 229 m/s downwards at an angle of 7˚36′ to the vertical

11B 11B

14 A particle is fired from the top of a 60 m high building with a velocity of 24 m/s inclined at an angle of 30° to the horizontal. a Show that the velocity vector of the projectile at time t seconds is given by v (t) = 12 3 i + (12 − gt) j m/s. ˜ expression ˜ ˜ for the displacement, r (t) metres, of the particle at time b Develop a vector ˜ t seconds. r (t) = 12t 3 i + (60 + 12t − 1--2- gt2) j ˜ ˜ ˜ c Determine if this projectile will clear a 14 m high tower located 94 m from the launch point and in the same horizontal plane. Clears by about 5 cm

11B

15 A missile is fired from a point on level ground with a velocity V m/s and with an angle of elevation of θ to the horizontal. The target is positioned on top of a cliff 30 m above the ground in the same horizontal plane as the point of projection of the missile, and with a range of 60 m from it. a If g is the gravitational acceleration of the missile, show that the motion of this missile satisfies V 2 cos θ (2 sin θ – cos θ) = 60g. b Use a graphics calculator to find those values of θ in the domain 0 < θ < 90° for which the expression cos θ (2 sin θ – cos θ) is positive. State the greatest possible value this expression has, and the value of θ that produces it. Provide supporting argument for your solutions. Expression is positive for 26°34′ ≤ θ < 90°. Maximum is 0.618 when θ ≈ 58°17′. c Extend from your answer to part b above to find the least value of V for which the missile can reach the target. (Use g = 9.8 m/s2.) 30.845 m/s d Generate a spreadsheet to validate the solution obtained using a graphics calculator.

11B

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11C

16 A stone is projected vertically up from the ground with an initial velocity of 24.5 m/s. Taking the acceleration due to gravity to be −9.8 m/s2, find: a the maximum height reached by the stone 30.625 m b the times at which its height is 20 metres above the ground. t ≈ 1 s and 4 s

11C

17 multiple choice

11C

18 multiple choice

11C

19 multiple choice

An object is dropped from the top of a 200 metre high building. If the acceleration due to gravity is 9.8 m/s2, what will be the height of the object after 5 seconds? A 151 m B 49 m C 122.5 m D 20 m E 77.5 m

A particle initially moving at 6 m/s is subject to a constant retardation of 2 m/s2. The distance, in metres, travelled before coming to rest is: A 27 B 8 C 9 D 10 E 12

A bus travels 500 metres in 25 seconds when accelerating uniformly from rest. The acceleration, in m/s2, is: A 0.4 B 1.6 C 1.25 D −1.2 E 0.625

11C

20 A parachutist drops from an aeroplane so that the constant acceleration during free fall due to gravity and air resistance is 8 m/s2. The parachute is released after 6 seconds, uniformly retarding the parachutist in 28 seconds to a constant speed of 2.5 m/s. This speed is maintained until the parachutist reaches the ground which is 1101 metres below the point of release. a How long is the parachutist in the air? b After how long has the parachutist fallen 2 min 14 s half the distance (answer to the nearest tenth of a second)? 16.2 s

11C

21 Jogger A is running in a straight line at a constant speed of 4 m/s when passing jogger B who has stopped to tie a lace. Jogger B heads off in the same direction as jogger A 6 seconds later, accelerating uniformly at 2 m/s2 until reaching a speed of 5 m/s. a Sketch a velocity–time graph showing the motion of both joggers. b How long is it after jogger A first passes B jogger B until B catches up to A? 36.25 s A c How far has jogger B travelled to catch t (s) jogger A? 145 m d How far ahead will jogger B be after jogger B has travelled 225 metres? 16 m

21 a v (m/s) 5 4 0

_6 8.5

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Modelling and problem solving

780 m

1 Car A is 600 metres from the centre of the intersection when it starts from rest and accelerates uniformly at 4 m/s2, reaching a speed of 24 m/s which it maintains. At the instant car A takes off, car B is 780 metres from the centre of the intersection and travelling at a constant speed of 28 m/s. When car B is 52 metres from the centre of the intersection it decelerates uniformly at 5 m/s2. a Which car gets to the centre of the intersection first? Car A b How far past the centre of the intersection is the first car Car B when the second car reaches it? 8.4 m c If all other conditions remain the same, what constant acceleration would: ii the second car need to have for a collision to occur? −2 m/s2 ii the first car need to have for a collision to occur? 3.58 m/s2 Car A d If all other conditions remain the same, at what constant speed would: Centre of intersection ii the first car need to travel for a collision to occur? 600 m (Use a graphics calculator or a numerical method to assist.) 23.62 m/s ii the second car need to travel for a collision to occur? 28.33 m/s (Give answers correct to 2 decimal places where appropriate.) 2 Martin is trying out his new spaceship by challenging a local alien to a race. He places his hand on the throttle and his spaceship starts to accelerate at a constant rate a for a time ∆t. During that time ∆t, he travels a distance s through space. Martin’s spaceship has an initial speed u at the beginning of the time interval and a final speed v after a time ∆t. The local space police are on to Martin and want to photograph him just as he breaks the speed limit. However, they need to set their camera up at the precise point, P, where Martin is travelling at his average speed. The police know when the average speed will occur; it will occur at a time T = ∆t ----- . They want to find out where the average speed will occur. Let x be 2 the distance travelled when the spaceship reaches its average velocity for the time interval ∆t. All distances and times are in metres (m) and seconds (s). It is clear that x is a fraction of s, the total distance covered during the time interval ∆t. Positive Average speed occurs here x

P

a The initial velocity (u) is 0, the acceleration (a) is 1.0 and the time interval (∆t) is 10. Find x the value of x and hence state the value of the fraction -- . --xs = 1--4s b Find equations for x and s in terms of the acceleration, a; the initial speed, u; and the time u∆t+ 1--8- a(∆ t)2 s = 1--2- a(∆ t)2 + u∆ t interval ∆t. x = -------2 s c Under what conditions will x = --- ? When a = 0 2

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3 When a projectile is fired from a body out into space, it decelerates due to the gravitational k pull of the body. The general equation for this deceleration is a = − ----2 where r is the distance r from the centre of the body and k is a positive constant. a On the surface of the moon (r = 1760 km), the deceleration is a = 1.6 m/s2. Find the value of k. k = 4.96 × 1012 b If a body is launched from the surface of the moon with a velocity of 500 m/s, find the velocity of the body at a distance r from the centre of the moon; that is, find v(r). c Find the distance from the centre of the moon when the body is momentarily stationary. 1.84 × 106 m 12

4.96 × 10 b v(r) = 2  --------------------------– 2.69 × 10 r

6 

Appendix

553

Instructions for the TI-89 Titanium graphics calculator Chapter 1 — Number systems: the Real Number System

Graphics Calculator tip: Square, cube and nth roots (page 7) ................................... 555 Graphics Calculator tip: Approximation mode ........................................................... 556 Chapter 2 — Number systems: complex numbers

Graphics Calculator tip: Simple algebra of complex numbers (page 93) .................. 557 Graphics Calculator tip: Modulus and Argument (page 101) .................................... 559 Graphics Calculator tip: Expressing complex numbers in polar form (page 105) .... 561 Graphics Calculator tip: Expressing complex numbers in Cartesian form (page 106) ............................................................................................................... 561 Graphics Calculator tip: Pascal’s Triangle coefficients (page 115) ............................ 562 Chapter 3 — Matrices

Graphics Calculator tip: Solving matrix equations (page 156) .................................. 563 Investigation: Matrix multiplication using a graphics calculator (page 162) ............ 564 Graphics Calculator tip: Alternative method for adding the elements in each row in a dominance matrix (page 167) ......................................................................... 565 Chapter 5 — Matrices and their applications

Investigation: Performing Gaussian elimination using a graphics calculator (page 217) ............................................................................................................... 566 Graphics Calculator tip: Matrix operations (page 237) .............................................. 568 Chapter 6 — Transformations using matrices

Graphics Calculator tip: Graphing the original and its image (page 256) ................. 571 Chapter 7 — Introduction to vectors

Graphics Calculator tip: Finding the magnitude and direction of a vector in two dimensions (page 311) ........................................................................................... 572 Graphics Calculator tip: Finding the x- and y-components of a vector (page 313) ............................................................................................................... 573 Graphics Calculator tip: Finding the unit vector in the direction of the vector (page 316) ............................................................................................................... 574 Graphics Calculator tip: Finding the dot product of two vectors (page 326) ............ 574 Graphics Calculator tip: Finding scalar and vector resolutes (page 334) .................. 575 Graphics Calculator tip: Vector functions of time (page 341) ................................... 576

554

Appendix

Chapter 9 — Sequences and series

Graphics Calculator tip: Listing the terms of an arithmetic sequence (page 400) .... 577 Graphics Calculator tip: Finding the sum of an arithmetic sequence (page 402) ..... 578 Worked example 10 (page 410) .................................................................................. 579 Worked example 12 (page 418) .................................................................................. 580 Worked example 15 (page 422) .................................................................................. 581 Graphics Calculator tip: Comparison of simple and compound interest (page 434) ............................................................................................................... 582 Graphics calculator tip: Generating terms in the Fibonacci Sequence (page 443) ............................................................................................................... 584 Chapter 10 — Permutations and combinations

Graphics Calculator tip: Calculating factorials (page 468) ........................................ 585 Graphics Calculator tip: Calculating permutations (page 472) .................................. 585 Graphics Calculator tip: Calculating combinations (page 486) ................................. 586 Chapter 11 — Dynamics

Worked example 2 (page 517) .................................................................................... 587 Worked example 3 (page 519) .................................................................................... 588 Worked example 7 (page 531) .................................................................................... 589

Appendix

555

Chapter 1 page 7 Graphics Calculator tip! Square, cube and nth roots A graphics calculator can be used to find the square root, cube root or higher root of a number. 1. From the MENU (press APPS or 2ND [QUIT]), highlight Home.

2. Press ENTER . To calculate the square root of a number (for example, 8 ), press 2ND [ ] followed by the number (8 in this case), close the brackets and press ENTER . Your calculator may give this answer as a simplified surd, in this case, 2 2 . If so, you will need to change the set up of your calculator to approximation mode. (See the graphics calculator tip that follows.) 3. To calculate higher order roots you will need to use a fractional index. That is, for the cube root use 1--- , 3 for the fourth root use 1--- and so on. To calculate 4 3 8 , enter 8 then press Ÿ and enter (1 ÷ 3) as the index.

556

Appendix

Graphics Calculator tip! Approximation mode The following steps show how to set your calculator in approximation mode. 1. From the MENU, highlight Home. (Press APPS to display the MENU.)

2. Press ENTER and then press MODE to display your set up options. Press F2 (Page 2) to display the second page of options.

3. Press the down arrow key until the entry next to Exact/Approx is highlighted and then press the right arrow to display your options.

4. Select 3: APPROXIMATE for approximation mode (press 3 or highlight the required mode and press ENTER ). Your selection will be flashing.

5. Press ENTER to save this setting. Notice that the bottom of the display shows APPROX.

Appendix

557

Chapter 2 page 93 Graphics Calculator tip! Simple algebra of complex numbers Operations with complex numbers, finding the real and imaginary parts of a complex number and finding the complex conjugate can be achieved with a graphics calculator. You may not need to use a graphics calculator with simple complex numbers but it can be useful in more complicated questions.

Operations with complex numbers 1. From the MENU, highlight Home. (Press APPS to display the MENU.)

2. Press ENTER . To perform simple algebra on complex numbers, use the standard keys to enter the expression. To enter i, press 2ND [i]. For example: (a) input (2 − 2i)(1 + 3i) and then press ENTER . (b) input (2 − 2i) ÷ (1 + 3i) and then press ENTER . (c) input (2 − 2i)^3 and then press ENTER . Notice that a multiplication symbol (a dot) appears on the screen for any multiplication operations in these examples. The calculator has assumed that 2i means 2 × i. 3. A complex number can be stored and then retrieved if a number of operations need to be performed on it.

▼

▼

Input 2 − 2i and then press STO followed by Z to store this expression as the variable z. Press ENTER . Input 1 + 3i and then press STO followed by ALPHA [W] to store this expression as the variable w. Press ENTER . We can now calculate expressions involving z and/or w, for example, z × w, z ÷ w and z3. Press ENTER to obtain each answer.

Appendix

Features of a complex number 1. To find the complex conjugate, the real part or the imaginary part of a complex number or expression, press CATALOG . You may then access conj( for the conjugate, real( for the real part and imag( for the imaginary part of a complex number as required. The screen at right shows conj( highlighted. 2. Press ENTER to display the selection in the entry line of the Home screen. Enter the complex number, close the brackets and press ENTER . 3. Alternatively, you can enter an expression, for example, (2 − 2i)(1 + 3i), and then press STO followed by ALPHA [A] to store the output as the variable a. Press ENTER . ▼

558

(a) To find ( 2 – 2i ) ( 1 + 3i ) , press CATALOG and select conj(. (To select conj(, highlight conj( and then press ENTER .) Key in the variable assigned to the stored data (a in this case, so press ALPHA [A]) and close the brackets. Press ENTER . (b) To find Re ((2 − 2i)(1 + 3i)), press CATALOG and select real(. Key in the variable assigned to the stored data and close the brackets. Press ENTER . (c) To find Im ((2 − 2i)(1 + 3i)), press CATALOG and select imag(. Key in the variable assigned to the stored data and close the brackets. Press ENTER .

Appendix

559

Chapter 2 page 101 Graphics Calculator tip! Modulus and Argument Your graphics calculator is also able to calculate the modulus and Argument of a complex number. Consider 3 + i and −1 − 2 i from Worked example 23.

Modulus (magnitude, or absolute value) of a complex number For this activity your calculator will need to be changed into AUTO or EXACT mode and must not be in APPROX mode. 1. From the Home screen, press CATALOG and access abs(. 2. (a) Press ENTER to display abs( in the entry line of the Home screen, then enter 3 + i and close the brackets. Press ENTER to obtain the answer. (b) The modulus of −1 − 2 i is also shown in the screen at right. (c) The modulus of a stored complex number can also be calculated. For example, using the expression from the previous graphics calculator tip where we assigned the variable a to (2 − 2i)(1 + 3i), we can obtain the magnitude as shown in the screen at right.

Argument of a complex number 1. Decide whether you want the angle shown in radians or degrees. See the instructions below for changing the system settings for Angle. 2. Press CATALOG and access angle(.

3. Press ENTER to display angle( in the entry line of the Home screen, then enter the required complex number 3 + i and close the brackets. Press ENTER to obtain the answer. If the calculator is set to radians, the answer will be shown as π--- . If the 6 calculator is set to degrees, the answer will be shown as 30 for 30°. 4. (a) The Argument of −1 − 2 i is shown as an exact answer in radians in the screen at right. (b) For an approximate answer, switch the calculator back to approximation mode as shown earlier on page 556.

Appendix

▼

(c) To convert this angle in radians to degrees, minutes and seconds, first press 2ND [ANS]. Then press CATALOG and select DMS. Now that you are back in the Home screen, press ENTER to obtain the answer. Note: You can convert degrees to radians by selecting Rad in the CATALOG.

▼

560

5. (a) The Argument of a stored complex number can also be calculated. See the screen at right for the Argument of the variable a where a is assigned to (2 − 2i)(1 + 3i). The exact answer is shown. (b) For the approximate answer, set the calculator back to approximation mode. The calculator is set to degrees for this example.

Changing the system settings for Angle The following steps show how to set your calculator in degrees mode. 1. From the MENU, highlight Home. (Press APPS to display the MENU.)

2. Press ENTER and then press MODE to display your set up options.

3. Press the down arrow until the entry next to Angle is highlighted and then press the right arrow to display your angle options.

4. Select 2: DEGREE for degree mode (press 2 or highlight the required mode and press ENTER ). Your selection will be flashing.

5. Press ENTER to save this setting. Notice that the bottom of the screen displays DEG.

The same method is used to change back to radians.

Appendix

561

Chapter 2 page 105 Graphics Calculator tip! Expressing complex numbers in polar form Complex numbers in Cartesian form (also known as rectangular form) can be written in polar form if we know the modulus and the Argument. Consider 1 + i and 1 − 3 i from Worked example 24. To begin this activity, have your calculator set in DEGREES and EXACT form.

▼

1. From the MENU, select Home. Enter the complex number in Cartesian form. Press CATALOG and select Polar. Press ENTER to show the modulus and Argument. For 1 + i, we can see that the modulus is 2 and the Argument is 45° (as the calculator is set to degrees). For 1 − 3 i, the modulus is 2 and the Argument is −60°. 2. If the calculator is set to radians, the answer is shown in a different format but we can still see the iπ modulus and Argument. Another way of ----4 π expressing 2 cis --- (the polar form of 1 + i) is 2 e . 4 (This is beyond the scope of this course.)

Chapter 2 page 106 Graphics Calculator tip! Expressing complex numbers in Cartesian form Complex numbers in polar form can be written in Cartesian form by entering both the modulus and the Argument into the calculator. Consider 3 cis --π- from Worked example 4 25. The modulus is 3 and the Argument is --π- or 45°. 4

1. From the MENU, select Home. To enter the complex number 3 cis π--- , first open a set of brackets and press 4 3 to enter the modulus of 3. Then press 2ND [– ] to enter the ∠ function. Complete the entry line by entering the angle in radians ( π--- in this case) or 4 degrees (45°) as appropriate to the calculator setting. The screen at right shows the angle in radians. Close the set of brackets and press ENTER . 2. If the calculator is set to degrees with the Argument entered as 45, the same result is obtained.

562

Appendix

Chapter 2 page 115 Graphics Calculator tip! Pascal’s Triangle coefficients The coefficients of each term of the expansion of (a + bi)n can be found using your graphics calculator. For example, the coefficients of the expansion of (a + bi)5 can also be written as: 5 C0 a5 + 5C1 a4(bi)1 + 5C2 a3(bi)2 . . . 5C5(bi)5 where 5C0, 5C1, . . ., 5C5 represent the coefficients. The following steps show how to calculate 5C3 using a graphics calculator. 1. From the Home screen, press CATALOG and access nCr(. You can find nCr( more quickly by first pressing ALPHA [N].

2. Press ENTER and then complete the entry line to obtain nCr(5, 3). Press ENTER to display the value. Note that with this calculator, we can obtain the actual expansion of (a + bi)5. 1. From the MENU, select Home. Press 2ND [F6] and press 1 to select 1: Clear a-z.... This sets the variables a-z to their default values and makes them ready for use. 2. Press ENTER to indicate YES. Enter (a + bi)^5 and press ENTER . You will see a small arrow at the end of the answer line indicating that there are more terms. Use the arrow keys to see more of the expansion. The full expansion is shown as a(a4 − 10a2b2 + 5b4) + (5a4 − 10a2b2 + b4)bi.

Appendix

Chapter 3 page 156 Graphics Calculator tip! Solving matrix equations Most graphics calculators provide a facility for calculating inverses of matrices. To solve the equations in Worked example 12, follow these steps. 1. From the MENU, highlight Data/Matrix Editor. 2. Press ENTER and select 3: New... to open a new matrix. 3. Ensure that Type is shown as Matrix (to change, press the right arrow to display the options and select 2: Matrix), then enter a for the variable (press ALPHA [A]), 2 for the Row dimension and 2 for Col dimension to define the matrix as a 2 × 2 matrix with variable name a. 4. Press ENTER to accept the settings. Fill in the 2 × 2 matrix. Press ENTER after each number to move the cursor from cell to cell. 5. Press F1 (Tools) and select 3: New... to define a second matrix.

6. Enter the settings shown at right to define a 2 × 1 matrix with variable name b.

7. Enter the values 16 and 5 for matrix b as shown.

8. Press APPS to return to the MENU and select Home. Enter the calculation a-1 ¥ b as shown to obtain the result.

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Appendix

Chapter 3 page 162

Matrix multiplication using a graphics calculator Worked example 13 may be solved using a graphics calculator as follows. 1. From the MENU, highlight Data/Matrix Editor. 2. Press ENTER and select 3: New... to open a new matrix.

3. Enter the settings shown at right to define the matrix as a 3 × 3 matrix with variable name a.

4. Press ENTER and complete the matrix.

5. Press F1 (Tools) and select 3: New... to define a second matrix.

6. Enter the settings shown at right to define a 3 × 1 matrix with variable name b.

7. Enter the values for matrix b as shown.

8. Press APPS to return to the MENU and select Home. Enter the calculation a ¥ b as shown to obtain the result.

Appendix

565

Chapter 3 page 167 Alternative method for adding the

Graphics Calculator tip! elements in each row in a dominance matrix

Multiplying a square matrix by a column vector with the same number of rows and all entries shown as 1 has the effect of adding the elements in each row of the matrix. In the example on page 165, the dominance vector V1 could have been obtained using the following steps. (The main advantage is if the dominance matrix is 5 × 5 or larger. You don’t need to arrow across the screen to see the elements when you are adding them.) 1. From the MENU, select Data/Matrix Editor and establish a 4 × 4 dominance matrix with a variable name m. (Refer to the graphics calculator tip on page 563 if you are unsure how to do this.) Note that you will need to use the arrow keys to display the fourth column of the matrix. 2. Press F1 (Tools) and select 3: New... to create a new 4 × 1 vector with variable name n. Enter 1 as shown for every element.

3. Press APPS to return to the MENU and select Home. Perform the multiplication m ¥ n as shown.

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Appendix

Chapter 5 page 217

Performing Gaussian elimination using a graphics calculator Matrix row operations can be performed on your graphics calculator. To demonstrate this, we will repeat the steps of Worked example 4 to find the inverse of 121 A = 101 . 013 1. Enter Matrix A. (a) From the MENU, select Data/Matrix Editor. Select 3: New... and enter the settings shown to create a 3 × 3 matrix with variable a (to represent matrix A). (b) Press ENTER and then enter the values for matrix A as shown.

2. Set up an augmented matrix B = [A|I]. Enter the augmented matrix [A|I] first and then save as matrix B.

▼

(a) Press APPS to return to the MENU and select Home. (b) Press CATALOG and select augment(. Press ALPHA [A] for matrix A and then the comma key , . To enter the identity matrix, press CATALOG and select identity(. Press 3 (for a 3 × 3 identity matrix) and press ) twice to close the two sets of brackets. Press STO and then press ALPHA [B] to save the augmented matrix as matrix B. Press ENTER to display the matrix. 3. Replace R2 with R2 − R1 (or − R1 + R2). (a) To perform the required row operation, press CATALOG and access mRowAdd(.

Appendix

▼

(b) Press ENTER to display mRowAdd( in the entry line of the Home screen. Input the scalar multiplier, the matrix name, the row number to be multiplied and then the row number for the result to be added to, each separated by a comma. In this case, enter −1, B, 1, 2. Press ) to close the set of brackets. Then press STO and ALPHA [C] to save the matrix as matrix C. Press ENTER to display the matrix. Compare this screen with the matrix obtained in Step 2 of Worked example 4. 4. Swap R2 and R3. (a) To perform the required row operation, press CATALOG and access rowSwap(.

▼

(b) Press ENTER to display rowSwap( in the entry line of the Home screen. Input the matrix name and the two row numbers to be swapped, each separated by a comma. In this case, enter C, 2, 3. Press ) to close the set of brackets. Then press STO and ALPHA [D] to save the matrix as matrix D. Press ENTER to display the matrix. Compare this screen with the matrix obtained in Step 3 of Worked example 4. 5. Replace R3 with R3 + 2R2.

▼

(a) As before, press CATALOG and select mRowAdd(. (b) Enter 2, D, 2, 3 to represent 2 × R2 + R3 in matrix D. Press ) to close the set of brackets. Then press STO and ALPHA [E] to save the matrix as matrix E. Press ENTER to display the matrix. Compare this screen with the matrix obtained in Step 4 of Worked example 4. 6. Multiply R3 by 1--- . 6

(a) To perform the required row operation, press CATALOG and access mRow(.

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Appendix

▼

(b) Press ENTER to display mRow( in the entry line of the Home screen. Enter 1--- , E, 3 to 6 represent 1--- × R3 in matrix E. Press ) to 6 close the set of brackets. Then press STO and ALPHA [F] to save the matrix as matrix F. Press ENTER to display the matrix. Compare this screen with the matrix obtained in Step 5 of Worked example 4. 7. Now that we have row-echelon form, continue performing operations until reduced row-echelon form is achieved.

Chapter 5 page 237 Graphics Calculator tip! Matrix operations The graphics calculator can perform a number of matrix operations and can provide quick and reliable answers to some of the problems that you have encountered in this chapter. A number of operations are shown below, some of which have already been covered in earlier graphics calculator tips. Consider the matrix A =

2 3 2 0 1 4 . 2 1 –1

Entering a matrix (a) From the MENU, select Data/Matrix Editor. Select 3: New... and enter the settings shown to create a 3 × 3 matrix with variable a (to represent matrix A).

(b) Press ENTER and then enter the values for matrix A as shown.

Appendix

Inverse of A Press APPS to return to the MENU and select Home. With matrix A entered in the calculator, press ALPHA [A] then the power key, Ÿ , and type in the index of −1 to specify the matrix A−1. Press ENTER to display the required matrix.

Powers of A (a) From the Home screen, and with matrix A entered in the calculator, press ALPHA [A] to recall matrix A to the entry line. To find A2, press the power key, Ÿ , and type in the required index of 2. Press ENTER to display the required matrix. (b) Similarly, press ALPHA [A] followed by Ÿ and then 3 to specify A3. Press ENTER to display the required matrix.

Determinant of A (a) From the Home screen, press CATALOG and access det(.

(b) Press ENTER to display det( in the entry line and then press ALPHA [A] to specify matrix A. Close the set of brackets and press ENTER to display the answer.

The identity matrix (a) From the Home screen, press CATALOG and access identity(.

(b) Press ENTER to display identity( in the entry line of the Home screen. For a 3 × 3 identity matrix, press 3 . Close the set of brackets and press ENTER to display the required matrix.

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Appendix

Calculate (I – A)−1 and store it as B (a) Press ( to open a set of brackets and specify the 3 × 3 matrix I as shown above. Press the subtraction key (–) and then press ALPHA [A] to specify the matrix A. Press ) to close the set of brackets and then press the power key ( Ÿ ) and type in −1 as the index to specify the inverse matrix. (b) Press STO and then ALPHA [B] to store the matrix as matrix B. Press ENTER to display the resulting matrix B. ▼

Alternatively, you can store the resulting matrix after the calculation has been performed. Input (I – A)−1 and press ENTER to display the resulting answer matrix. To store this as matrix B, press STO and then ALPHA [B]. Press ENTER . ▼

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12

12

Calculate the product A−1 3 4 by entering 3 4 as a list 56 56 From the MENU, select Data/Matrix Editor and create a new matrix of dimension 3 × 2 and store it as matrix B. Return to the Home screen and enter A−1 as previously shown, then multiply by B.

Fill cells of a matrix with a given value (a) From the MENU, select Data/Matrix Editor and set up a new matrix A with specified dimensions. (Let’s use 3 × 3.) Press APPS to return to the MENU and select Home. Press CATALOG and access Fill.

(b) Press ENTER to display Fill in the entry line of the Home screen. Enter the given value. (Let’s use 5.) Press the comma key ( , ) and then press ALPHA [A] to enter matrix A. Press ENTER . You will see a message on the screen indicating that this command has been done. To display matrix A, press ALPHA [A] and then ENTER .

Appendix

Chapter 6 page 256 Graphics Calculator tip! Graphing the original and its image We can use a graphics calculator to draw the original function and its image on the same axes. Consider the function y = x2 and its image y′ = x′ 2 − 2x′ − 3 found in Worked example 4. 1. From the MENU, highlight Graph.

2. Press ENTER . To draw the graph of y = x2, press ♦ [Y=] and enter x2 in the function entry line next to y1(x).

3. Press ♦ [GRAPH] to view the graph of this curve.

4. For the second equation, press ♦ [Y=] and enter x2 − 2x − 3 in the function entry line next to y2(x) and press ♦ [GRAPH] to view the graphs.

5. To see the two curves more clearly, alter the scale settings of the axes. Press ♦ [WINDOW] and enter the settings shown.

6. Press ♦ [GRAPH] to redraw the graph with the new scale.

7. The TI-89 Titanium does not show the function names on the screen as other calculator models may do. However, you can trace each function by pressing F3 (Trace). The coordinates of the cursor are shown and the number of the function (1 or 2) is displayed in the top right of the screen.

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Appendix

Chapter 7 page 311 Graphics Calculator tip! Finding the magnitude and direction of a vector in two dimensions Consider the vector shown in Worked example 6. The vector can be expressed in component form (or rectangular form) as 3 i − 5 j . One way of finding the magnitude and ˜ the ˜ positive x-axis) is to convert the vector direction (the angle the vector makes with from rectangular form to polar form using a graphics calculator. 1. From the MENU, select Home. Decide whether you want the angle displayed in degrees or radians. In this example we want degrees. Press MODE and select 2: DEGREE at the Angle line. 2. Press ENTER until you return to a blank Home screen. We will first define 3 i − 5 j as the vector u and then convert this to polar ˜form.˜ Press CATALOG to access Define. 3. Press ENTER to display Define in the entry line and then press ALPHA [U] to name the vector as u. Press = and then open a set of square brackets (press 2ND [[].) Enter the components 3 and −5, separated by a comma (press , ), then close the set of square brackets (press 2ND []]) and press ENTER . ▼

4. Press ALPHA [U] for vector u. Then press CATALOG to access Polar.

▼

5. Press ENTER to display Polar in the entry line and then press ENTER again to display the exact value of the vector in polar form.

6. To obtain the approximate values, change to approximation mode (press MODE and select 3: APPROXIMATE at the Exact/Approx line). Press ENTER to recalculate in APPROX mode. Note that if no other operations are required, you can enter the vector directly by using square brackets. You then continue in the same way to convert to polar form.

Appendix

573

Chapter 7 page 313 Graphics Calculator tip! Finding the x- and y-components of a vector Vectors can be expressed in different forms. In the previous graphics calculator tip on page 572, we converted a vector in rectangular form to polar form so we could obtain the magnitude and direction of the vector. The reverse process can also be performed. Consider the vector in Worked example 7 where the magnitude is 30 and the angle to the positive x-axis is 140°. To find the x- and y-components, we convert the vector to rectangular form. Before beginning this activity, check that the calculator is set to degrees and is in AUTO mode. 1. From the MENU, select Home. To enter the vector in polar form, open a set of square brackets, enter 30 for the magnitude followed by a comma. Press 2ND [– ] and enter 140 for the angle. Close the set of square brackets and press ENTER to display the vector in polar form. ▼

2. Move the cursor to the right of the expression in the entry line. Press CATALOG to access Rect and press ENTER .

3. Press ENTER to perform the conversion. The exact values for the x- and y-components are shown.

4. To obtain the approximate values, switch the calculator to APPROX mode and press ENTER to perform the calculation again. Hence u = −22.98 i + 19.28 j . ˜ ˜ ˜

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Appendix

Chapter 7 page 316 Graphics Calculator tip! Finding the unit vector in the direction of the vector As seen previously, the unit vector is obtained by dividing each component by the magnitude of the vector. We can also use a graphics calculator to achieve this. Consider the vector u in Worked example 9 which can be expressed in component form as 6 i + 3 j . ˜ ˜ ˜ To begin this exercise, set your calculator to EXACT mode. 1. From the MENU, select Home. To define the vector [6, 3] as u, first press CATALOG and select Define. Press ALPHA [U] (to name the vector as u) followed by = and then 2ND [[] to open a set of square brackets. Enter the digits 6 and 3 separated by a comma. Close the set of square brackets and press ENTER . 2. Press CATALOG to access unitV(. 3. Press ENTER to display unitV( in the entry line of the Home screen and then press ALPHA [U] to enter the required vector. Close the set of brackets and press ENTER . The exact values for the components of the unit vector are shown. 2 5 5 Hence u = ---------- i + ------- j . 5 ˜ 5˜ ˜ These steps also apply to finding the unit vector for a three-dimensional vector.

Chapter 7 page 326 Graphics Calculator tip! Finding the dot product of two vectors The following steps show how a graphics calculator can be used to find the dot product of two vectors. Consider the vectors u = 3 i + 4 j + 2 k and v = 6 i − 4 j + k in ˜ ˜13. ˜ ˜ ˜ ˜ ˜ ˜ Worked example 1. From the MENU, select Home. Define the vector u ˜ and the vector v by entering each set of components ˜ within square brackets. 2. Press CATALOG and select dotP(. Press ALPHA [U] then the comma key ( , ) followed by ALPHA [V]. Close the set of brackets and press ENTER to obtain the value of the dot product.

Appendix

575

Chapter 7 page 334 Graphics Calculator tip! Finding scalar and vector resolutes Finding the dot product and the unit vector on a graphics calculator can be used to find a scalar resolute or a vector resolute. Consider the vectors u = −2 i + 3 j + k and ˜ ˜ ˜ ˜ v = 3 i + 4 j − k in Worked example 18. ˜ ˜ ˜ ˜ 1. From the MENU, select Home. Define the vector u ˜ and the vector v by entering each set of components ˜ within square brackets.

2. To find the scalar resolute of v on u , we need to calculate the dot product of uˆ ˜ (the ˜unit vector of u ) and the vector v . Access the ˜dot product function ˜ ˜ and select dotP( ). To enter uˆ , first (press CATALOG ˜ and access the unit vector function (press CATALOG select unitV( ) and then press ALPHA [U]. Press ) to close the set of inner brackets and then press the comma key ( , ) followed by ALPHA [V]. Close the set of outer brackets and press ENTER to display the value of the scalar resolute. 3. To find the vector resolute of v parallel to u , we need ˜ [ANS] to˜ show the to calculate ( uˆ • v )uˆ . Press 2ND ˜ ˜ ˜ previous answer then press the multiplication key. Repeat the steps above to access the unit vector function and press ALPHA [U]. Close the set of brackets and press ENTER to display the components of the required vector resolute. 4. The vector resolute can be found straight after defining the vectors u and v (without finding the ˜ seen˜ in the screen at right. scalar resolute first) as

5. To find the vector resolute of v perpendicular to u , ˜ we need to subtract the vector˜resolute of v parallel ˜ to u (calculated in step 4) from v . Press ALPHA [V] ˜ – followed by 2ND [ANS] ˜ to show the then previous answer. Press ENTER to display the components of the required vector resolute.

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Appendix

Chapter 7 page 341 Graphics Calculator tip! Vector functions of time To draw the graph of a time-varying vector, we need to express the components in terms of parametric equations. Consider drawing the graph of the position vector u = 2cos t i + 3sin t j from Worked example 22 using a graphics calculator. ˜ ˜ ˜ 1. From the MENU, highlight Graph.

2. Press ENTER . First ensure that the angle setting is shown as radians. Press MODE and select 1: RADIAN at the Angle line.

3. Before leaving this screen, select 2: PARAMETRIC at the Graph line.

4. Press ENTER to save the settings. Press ♦ [Y=] and enter the x-component by completing the entry line for xt1(t) with 2cos t and then press ENTER . Similarly, enter the y-component by completing the entry line for yt1(t) with 3sin t and press ENTER . 5. Press ♦ [GRAPH] to display the graph.

To obtain a clearer view of the graph, you can adjust the window settings. Press ♦ [WINDOW]. Adjust the values for xmin, xmax, ymin and ymax. Alternately, press F2 (Zoom) and select 2: ZoomIn. Use the arrow keys to define the centre point of the zoom and press ENTER . This will increase the size of the graph.

Appendix

577

Chapter 9 page 400 Graphics Calculator tip! Listing the terms of an arithmetic sequence If you know the rule for an arithmetic sequence, successive terms can be listed using a graphics calculator. The steps for generating these terms are shown below. Consider an example where the first term is 3 and the difference is 2 so the rule is tn = 3 + (n − 1) 2. 1. From the MENU, select Data/Matrix Editor and then 3: New....

2. Adjust the settings with Type shown as Data and Variable as n.

3. Press ENTER to accept the settings. With the cursor in the cell containing c1, press CATALOG to access seq(. 4. Press ENTER to display seq( in the entry line. The sequence function has four parts each separated by a comma (,). The first part is the formula (enter 3 + (n − 1) × 2), the second part is to define the variable (enter n), while the third and fourth parts define the value of n for the initial term (enter 1) and the ceiling value of n (enter 255). Enter this information as shown in the entry line and press ENTER . 5. Use the arrow keys to scroll down the list.

6. If you wish to widen the column, press F1 (Tools) and select 9: Format. Press ENTER and select a cell width of 10.

7. Press ENTER to display the table with wider columns.

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Appendix

Chapter 9 page 402 Graphics Calculator tip! Finding the sum of an arithmetic sequence For the arithmetic sequence 4, 10, 16, 22, . . ., 58 considered in Worked example 5, we can use a graphics calculator to find the sum of the first 10 terms. The rule for this sequence is 4 + (n − 1) × 6. 1. From the MENU, select Data/Matrix Editor and then 3: New.... Enter the settings shown with variable n.

2. Press ENTER to accept the settings. With the cursor in the cell containing c1, press CATALOG and select seq(. Enter the sequence 4 + (n − 1) × 6 as shown in the entry line.

3. To calculate the sum of the first 10 terms, we can add the terms in the sequence. Move the cursor to an empty cell in column 2 (say, r1 c2). Press CATALOG and select sum(. Again press the CATALOG function and select seq(. 4. Re-enter the sequence, the variable, the starting value of n (enter 1) and the last value of n (enter 10) as shown in the entry line.

5. Press ENTER to display the sum of the term.

Note that the sum of the sequence could have been calculated without actually displaying the sequence first.

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579

Chapter 9 page 410

WORKED Example 10 The first three terms of a geometric sequence are 2, 6 and 18. Which numbered term would be the first to exceed 1 000 000 in this sequence? THINK

WRITE/DISPLAY

Method 3: Using a graphics calculator 1

Find the rule for the sequence. See Method 1 on page 410.

2

Locate the term that equals or exceeds 1 000 000. Two methods are shown for this example.

a = 2 and r = 3 tn = 2 × 3n−1

Method A: Generating the terms of the sequence (a) From the MENU, select Data/Matrix Editor and then 3: New... with variable n as shown.

(b) Press ENTER to accept the settings. Press CATALOG and select seq( to enter the sequence 2 × 3n−1 into column 1 (c1) as shown.

(c) Press ENTER and scroll down until you reach the term that exceeds 1 000 000. You will need to widen the column to see this number. (Refer to page 577 for instructions on widening a column.) Method B: Solving an equation Set up an equation to solve.

2 × 3n−1 = 1 000 000

From the Home screen, press CATALOG and select solve(. Enter the equation 2 × 3n−1 = 1 000 000 to be solved and then the variable, separated by a comma. Close the set of brackets and press ENTER . 3

Write your answer.

The first term to exceed 1 000 000 is the 13th.

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Chapter 9 page 418

WORKED Example 12 A city produced 100 tonnes of rubbish in the year 2008. Forecasts suggest that this may increase by 2% each year. If these forecasts are true, a what will be the city’s rubbish output in 2012? b in which year will the rubbish reach 120 tonnes? c what was the total amount of rubbish produced by the city in the years 2008, 2009, 2010? For the solution to part a, refer to page 418. THINK

WRITE/DISPLAY

b Method 3: Using a graphics calculator n−1 to write the rule. 1 Use tn = ar 2 Locate the term that equals or exceeds 120. Two methods are shown for this example.

b tn = 100(1.02)n−1

Method A: Generating the terms of the sequence (a) From the MENU, select Data/Matrix Editor and 3: New... with variable n as shown.

(b) Press ENTER to accept the settings. Press CATALOG and select seq( to enter the sequence 100 × 1.02n−1 into column 1 (c1) as shown. Press ENTER to display the terms. (c) Scroll down until you reach the term that exceeds 120. You may need to widen the column to see this number. (Refer to page 577 for instructions on widening a column.) Method B: Solving an equation From the Home screen, press CATALOG and select solve(. Enter the equation 100 × 1.02n−1 = 120 to be solved and then the variable, separated by a comma. Close the set of brackets and press ENTER . 3

Write your answer.

The first term to exceed 120 tonnes is the 11th term or year 2018.

Appendix

THINK c 1 We need to find the sum of the first 3 terms.

581

WRITE/DISPLAY c

Method 2: Using a graphics calculator Return to the Data/Matrix Editor used earlier. With the cursor in row 1 of column 2 (r1c2), press CATALOG and select sum( and then press CATALOG again and select seq( to find the sum the first three terms of the sequence. The entry line should display sum(seq(100 ¥ 1.02^(n–1), n, 1, 3)). 2

Write your answer.

The total output of rubbish for the years 2008, 2009 and 2010 will be 306.04 tonnes.

Chapter 9 page 422

WORKED Example 15 Jim invests $16 000 in a bank account which earns compound interest at the rate of 12% per annum compounding every quarter. At the end of the investment, there is $25 616.52 in the account. For how many years did Jim have his money invested? THINK Method 3: Using a graphics calculator n n 1 Use t = PR to write the rule for the sequence.

WRITE/DISPLAY P = 16 000 r =

12 -----4

= 3% per quarter

and so R = 1 +

3 --------100

= 1.03 tn = 16 000(1.03)n 2

3

Locate the term that equals or exceeds 25 616.52. You may wish to generate the sequence and scroll down to the required term (see page 577). Alternatively, an equation can be set up to be solved. This method is shown here. From the Home screen, press CATALOG and select solve(. Enter the equation 16 000 × 1.03n−1 = 25 616.52 to be solved and then the variable, separated by a comma. Close the set of brackets and press ENTER . Write your answer.

Jim has invested money for 16 periods where a period is 3 months. So it will take 48 months or 4 years.

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Chapter 9 page 434 Graphics Calculator tip! Comparison of simple and compound interest A graphics calculator can be used to compare the yearly amounts in an account from earning both simple interest and compound interest. Consider the scenario in Worked example 23 where the amounts are compared over the first 5 years. The steps are shown below. Then investigate this scenario further by drawing the graphs for both cases over the first 10 years. 1. From the MENU, select Data/Matrix Editor and then 3: New... with variable n. Each row of the table can be considered as a year of the investment.

2. We wish to generate terms of the sequence for the simple interest case in column 1 (c1) and the compound interest case in column 2 (c2). Increase the width of the columns to 10 (see page 577 for instructions) and then return to column 1. 3. With the cursor in the cell containing c1, press CATALOG and select seq(. Enter the rule 10 000 + n × 1000 with variable n, initial value for n of 1 and a maximum number of terms of 5 to generate the yearly amounts for the simple interest case. 4. Move to column 2 and repeat step 3 using the rule 10 000 × 1.1n to generate the yearly amounts for the compound interest case.

5. To see the graph of the two functions, first press APPS and select Graph from the MENU. Press ♦ [Y=] to display the function editor.

6. Enter the functions y1 = 10000 + 1000x and y2 = 10000 ¥ 1.1x. Press ENTER after each function is entered.

Appendix

7. Press ♦ [WINDOW] and enter the following settings.

8. Press ♦ [GRAPH] to display both graphs.

9. Press F3 (Trace) and use the arrow keys to compare the value of each investment at various points in time. The number displayed in the top right of the screen indicates which graph is being traced.

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Appendix

Chapter 9 page 443 Graphics Calculator tip! Generating terms in the Fibonacci Sequence On this calculator, we can use the Sequence form of entering a function to display the terms of any Fibonacci style sequence. The rule for a Fibonacci Sequence is Fn + 1 = Fn + Fn − 1 with F0 = 1 and F1 = 1 (or any two initial values). An equivalent rule to this is Fn = Fn − 1 + Fn − 2 with F1 = 1 and F2 = 1. The second version of the rule suits this calculator as the rule needs to be expressed as u(n) =. So the rule becomes u(n) = u(n−1) + u(n−2) with u(1) = 1 and u(2) = 1. 1. From the MENU, highlight Table.

2. Press ENTER and then press MODE . Select 4: SEQUENCE at the Graph line.

3. Press ENTER to return to the Table screen and then press ♦ [Y=]. Enter u1(n–1) + u1(n–2) in the entry line for u1(n) and press ENTER . Also set the initial values of 1 and 1 by entering {1, 1} for ui1 and press ENTER . 4. Press ♦ [TBLSET] and define the table beginning at 1 and with increments of 1. Press ENTER to save the settings.

5. Press ♦ [TABLE] to display the table. The required sequence is shown in the second column.

Appendix

585

Chapter 10 page 468 Graphics Calculator tip! Calculating factorials The following steps show how to calculate 12! using a graphics calculator. 1. From the MENU, select Home. Enter 12 and then press CATALOG to access ! (factorial symbol).

2. Press ENTER to insert the factorial symbol into the entry line of the Home screen and then press ENTER again to perform the calculation.

Chapter 10 page 472 Graphics Calculator tip! Calculating permutations To find the number of permutations of n objects taken r at a time, we need to calculate n Pr. The following steps show how to calculate 5P3 using a graphics calculator. 1. From the MENU, select Home. Press CATALOG to access nPr(.

2. Press ENTER and then complete the entry line to obtain nPr(5, 3). Press ENTER to display the value.

(Alternatively, you can use the letter keys to enter npr followed by ( directly into the entry line of the Home screen instead of accessing it from the CATALOG . Press ALPHA [N] then ALPHA [P] and ALPHA [R] followed by ( , then complete the entry line and press ENTER .)

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Appendix

Chapter 10 page 486 Graphics Calculator tip! Calculating combinations To find the number of combinations of n objects taken r at a time, we need to calculate Cr. The following steps show how to calculate 5C3 using a graphics calculator.

n

1. From the MENU, select Home. Press CATALOG to access nCr(.

2. Press ENTER and then complete the entry line to obtain nCr(5, 3). Press ENTER to display the value.

(Alternatively, you can use the letter keys to enter ncr followed by ( directly into the entry line of the Home screen instead of accessing it from the CATALOG . Press ALPHA [N] then ALPHA [C] and ALPHA [R] followed by ( , then complete the entry line and press ENTER .)

Appendix

Chapter 11 page 517

WORKED Example

2

At time t seconds, a particle has a position vector given by the expression r (t) = 2t i + (25 − t2) j metres. ˜ Use a ˜graphics calculator ˜ a to plot the trajectory of this particle across the interval 0 £ t £ 5 seconds. b Repeat part a above using an Excel spreadsheet. c Determine the equation of this trajectory in the form y = f (x). THINK

WRITE/DISPLAY

a

a r (t) = 2t i + (25 − t 2) j ˜ ˜ ˜ Let x = 2t and y = 25 − t 2

1

2

Consider the components of the position vector. Assign x to the horizontal component and y to the vertical. This produces two parametric equations that we can use to graph the trajectory. Use a graphics calculator to generate a graph of the particle’s motion. (a) From the MENU, select Graph. Press MODE and select 2: PARAMETRIC at the Graph line.

(b) Press ENTER to return to the graphing screen and then press ♦ [Y=]. In the entry line, enter 2t for xt1(t) and press ENTER . Similarly, enter 25 − t2 for yt1(t) and press ENTER . (c) Press ♦ [WINDOW] to adjust the viewing window. Enter the settings as shown.

(d) Press ♦ [GRAPH] to draw the graph of the function. Press F3 (Trace) to investigate points on the curve.

For the solution to parts b and c, refer to pages 518 and 519.

587

588

Appendix

Chapter 11 page 519

WORKED Example

3

At time t seconds, the displacement (in metres) of a particle A is given by r A(t) = (8 − t) i + (8 − 4t + t2) j and the displacement (in metres) of a particle B is given by r˜ B(t) = (t + 2)˜i + (2 − 2t + t2)˜ j . ˜ If these particles ˜ ˜ a collide, determine when they collide. b What are the coordinates of the impact point? c Find the Cartesian equations of the trajectories of the particles. d Use a graphics calculator to verify the coordinates of the impact point. For the solution to parts a, b and c, refer to pages 519 and 520. THINK

WRITE/DISPLAY

d Verify that the coordinates of the impact point are (5, 5) by drawing the graphs of the displacement for each particle and locating the point of intersection. (a) From the MENU, select Graph. Ensure that the Graph mode is set to FUNCTION (press MODE and select 1: FUNCTION at the Graph line). Complete the entry line for y1(x) with the equation for particle A. Press ENTER . Similarly, complete the entry line for y2(x) with the equation for particle B and press ENTER . (b) Adjust the WINDOW settings as shown at right.

d

(c) To display the graphs, press ♦ [GRAPH].

(d) To display the point of intersection, first press F5 (Math) and select 5: Intersection.

(e) Navigate the pointer to each line and press ENTER . Then navigate the pointer to set a lower and upper bound. This is done by pressing ENTER at any point either side of the intersection point. The coordinates of the point of intersection will be displayed.

Appendix

589

Chapter 11 page 531

WORKED Example

7

y A missile is fired from a point on level ground with a Target velocity V m/s and an angle of elevation of q to the V horizontal. The target is positioned on top of an 80 m high 80 m tower which is located 100 m away. The base of the tower is in the same horizontal plane as the point of projection O x 100 m of the missile. a If g is the gravitational acceleration of the missile, then show that the motion of this missile satisfies V2 cos q (5 sin q − 4 cos q) = 250g. b Use a graphics calculator to show those values of q in the domain 0 < q < 90° for which the expression cos q (5 sin q − 4 cos q) is positive. State the greatest possible value this expression has and the value of q which produces it. Provide supporting argument for your solutions. c Extend from your answer to part b above to find the least value of V for which the missile can reach the target. (Use g = 9.8 m/s2.) d Generate a spreadsheet to validate the solution obtained using a graphics calculator. For the solution to parts a, c and d, refer to pages 531 to 535.

THINK b Use a graphics calculator to graph the function f(θ) = cosθ (5sinθ − 4 cosθ). You will need to use x in place of the variable θ. (a) Ensure that your calculator is set to degrees. From the MENU, select Graph. Complete the entry line for y1(x) with cos(x)(5sin(x) − 4cos(x)).

WRITE/DISPLAY b

(b) Press ♦ [WINDOW] to enter the settings as shown.

(c) Press ♦ [GRAPH]. The graph shows where the expression is positive for 0 < x < 90°.

Continued over page

590

Appendix

THINK (d) To display the maximum point, first press F5 (Math) and select 4: Maximum.

(e) Navigate the pointer to set a lower and upper bound. This is done by pressing ENTER at any point either side of the maximum. The coordinates of the maximum point will be displayed.

WRITE/DISPLAY

CHAPTER 1 Number systems: the Real Number System Exercise 1A — Classification of numbers 1 a e i m q u y 2 a e i m q u y 3 B 7 a e i m q u y 8 a e i m q u y 9 C

Q I Q I Q Q I Q Q I I I I I Z+ Z+ Z+ Z− Z+ Z− Z− Z+ Z− Q Q Z− I Z+

b f j n r v

Q Q Q Q I I

b f j n r v

Q I Undefined Q I Q

4 b f j n r v

E Q Q Z− Z+ Z− Q

5 c g k o s w

C Q Z+ Z− Q Z+ Z+

6 d h l p t x

D Q Z+ Q Q Q Z+

b f j n r v

I I Q Q I Z+

c g k o s w

I Z− Z+ Z+ I Q

d h l p t x

Q I Z− I Z− I

10 C

c g k o s w

Q Q Q I I I c g k o s w

d h l p t x Q I I Q Q Q

11 B

I I Q Q I Q Q Q I Q Q Q

d h l p t x

12 D

Exercise 1C — Surds 1 b d f g h i l m o q s t w 2 A 4 E 5 B 7 2 (when a = 64) 8 m=4

1 a 2 3

b 3 2

c 2 6

d 2 14

e 3 3

f 5 3

g 5 5

h 3 11

j 2 15

k 4 7

l

m 2 17

n 5 6

o 6 5

p 13 2

q 2 22

r 3 15

s 9 2

t 10 2

u 7 5

v 8 5

w 8 7

x 7 15

b 15 2

c 24 10

d 24 7

f 10 17

g 21 6

h 40 2

j 18 7

k – 28 5

l

10

o 2 2

p 5

r 2 6

s 2 3

t 2 2

i

3 6

e 36 5 i

– 30 3

m 64 3

u

1 --3

15

e

c

d

4 --9

f

1 --6

g

17 -----45

h

19 -----45

i

31 -----45

j

32 -----45

28 k 2 ----45

l

53 -----99

m

4 -----33

-----n 1 34 99

--------o 3 367 495

p

361 --------999

q

427 --------999

r

868 -----------1665

s

323 --------999

--------t 3 152 333

u

13 -----18

v

157 --------300

w

1237 -----------1980

x

y

2 -----13

7 --2

11

c 6a 2

d 3ab 6

e 3a 10b

f 4a 3ab

g

13a 2

h

5a 2 b

6

i

13ab 2ab

j

2a 2 b 3

k

2ab 2

17ab

l

4x 3 5y

2 3ab

m 5x 3 y 2 5

n 24x y

p 14xy 7xy

q 54c 3 d 2 2cd r 9c 2 d 2 14d

o 20xy 5x

s 18c 3 d 4 5cd t 28c 5 d 5 6 e f 3 30

w 7e 5 f 5 2ef

xy 4 6xy

z

4 E

5 C

1 5 6 --3

x y

u x

22ef 3 6 --4

e f2 7f

3

6 D

7 C

Exercise 1E — Addition and subtraction of surds 1 a 7 5

b 17 2

c 8 3

d 19 7

e 15 5 + 5 3 f 15 2 + 7 6

g 4 11

h 5 13

i

13 2

10 7 – 11 5 k – 3 6

l

–7 2+5 6

j 6 E

x

m 17 3 – 18 7 n 5 xy p

x – 5 y + 7 xy

o 8 x+3 y

1A

➔

3 E 4 D 5 C 7 Irene. It can also be written as 0.02 .

5611 -----------9000

7

b 9ab

1 --9

b

3 --2

18 30

y −8 3

y

2 a

w

v 20 5

3 a 4a

1 c def g j k l n opq r t uvw x y 5 --9

n

2

q

v

8 --9

7 2

y 9 5 2 a 4 2

Exercise 1B — Recurring decimals 7 --9

6 C

Exercise 1D — Simplifying surds

2 2 --3

2 --9

answers

591

Answers

1E

answers

592

Answers

b 2 2

Exercise 1G — The Distributive Law

c 5( 5 + 6)

d –6 6+2 3

1 a

e 7 3

f 10 2

c 2 5 – 10

d

g 4 5

h 5 5

e 126 2 – 14 3

f 10 21 – 4 6

g 72 + 14 30

h – 30 15 + 80 6

2 a 10 ( 2 – 3 )

i

14 3 + 3 2

j 11 – 4 11

k 3 6+6 3

17 2

l

i

m 15 10 − 10 15 + 10 n 0 o – 8 11 + 22

b

q 12 30 – 16 15

7 --2

5 --8

x

3 a 34 a – 6 2a

f 3

d 32a + 2 6a + 8a 2

e a 2a

f

g 3a a + a 2 3a

h ( a 2 + a ) ab

4ab ab + 3a 2 b b

a + 2 2a

j 3 ab ( 2a + 1 )

k – 6ab 2a + 4a 2 b 3 3a 4 D 5 E 6 A

– 2a b 7 E 8 B b ( 6 6 + 8 3 ) cm

l

9 a 12 2 cm c ( 18 – 2 3 + 2 5 ) cm

d 3π 5 m

e ( 18 2 + 2 5 ) m

f

21 11 m

Exercise 1F — Multiplication of surds 14

1 a

b

55

42

c

d 2 6

f 6 2

g 10

3 7

j 4 10

k 27

l

m 10 33

n 96 6

o 180 5

p 126

q 120

r 144

s 120 3

t 360 3

e 4 3 i

u 2 6 y

2 --5

6

v

6

w

2 2--3-

h 5 15

x

3 a 27 + 10 2

b 16 + 4 15

c 18 + 6 5

d 53 + 10 6

e 35 + 12 6

f 53 + 12 10

g 104 + 60 3

h 14 – 6 5

i

10 – 2 21

4 a d g j m p s v y

−46 50 10 51 7 44 76 2x − 3y xy(49x − 9y)

5 A

6 C

j 37 – 8 10 b e h k n q t w z

18 6 −5 26 17 53 17 9x − 16y xy(81x − 25y)

c f i l o r u x

7 E

11 −2 7 −1 63 343 x−y 4x3 − 25y 8 D

b 59 – 12 15 + 9 5 – 6 3

9 a 57 – 12 15

c 2

d 2 3

6a 3 b 4

e

6

f

15

g 4

h

3 --5

h 15x 6 y 2 2

i

3 ------4

j

4 2 ---------3

5 k ------2

l

5 6

j

1 --2

a 3 b 2 2ab b 75π cm2 d 6 6 m2

e ( 45 π + 96 10 ) m2

f

6 D

4x + 2 5xy – 10y

7

c 20 11 m2 5 C

j

b

g 3x 2 y 2 10xy

3 a 98 cm2

15x + 26 xy + 8y

5

f

a 2 b 4 5ab

i

1 a

e 6a 5 b 2 2b

4 E

5

h 180 – 30 3 – 18 6 + 9 2

Exercise 1H — Division of surds

d 5abc 2 2abc

9 --2

4 --3

b x2 y3 x

c 3a 4 b 2 2ab

i

30 3

z 3 3

2 a x2 y y

2 55 – 2 22 – 4 15 + 4 6

g 10 35 + 14 14 – 15 10 – 42

b 52 a – 29 3a

c 6 6ab

i

– 35 – 11

e 112 – 140 3 + 24 6 – 90 2

v –3 2

w 15 2

– 24 2 + 12

d 24 3 – 18 30 – 8 10 + 60

t 0

2+2 3

6 + 10

c – 4 – 40 3

r – 2 5 – 5 2 – 2 30 + 2 15 u

b 3 10 – 7 5

2 a 3 10 + 9 2 – 5 5 – 15

p 39 3

s 12 ab + 7 3ab

21 + 6 3

7 A

72 15 m2 8 15 360 2

m 2 3

n 1 --15-

o 1

p 2 6

q 1 4--5-

r 3 3

s 2 17

t 2 2--5-

x u -y

1 v --------x3 y2

2 w --------x3 y4

x 6x xy

2x 2 2 a 2xy 3y b ----------3y 2 2 e ---------------------3m 3 n m 3 B 4 E

3b 2 2b d ------------------2a a

4 a c ---------3 f

15 -------------2m 2 n 2

d 3 7 m

6 C

b 4 6 cm

c 7 11 m

e 5 13 cm

f

5 ------3 b 126 L

8 a 4 5 9 a 35 2 cm

15 -----2

5 cm

c 2 2

b

2 21 e ------------7 i

5 6 ---------6

9 11 + 9 h ---------------------20 i

5 14 + 2 10 – 25 7 – 10 5 -----------------------------------------------------------------------155

j

12 2 – 17

19 – 4 21 k ------------------------5

Exercise 1I — Rationalising denominators 1 a 5---------22

2 21 – 35 ----------------------------14

15 15 – 20 6 g -----------------------------------13

5 A

7 a 4 13 m

f

l

9 2 + 154 -----------------------------4

7 3 b ---------3

4 11 c ------------11

4 6 d ---------3

– 20 2 + 9 10 + 4 30 – 9 6 m -------------------------------------------------------------------------2

f

10 ---------2

2 15 g ------------5

3 35 h ------------5

3 n ------12

j

4 15 ------------15

5 7 k ---------14

l

8 15 ------------15

3 3+2 6 o --------------------------18

10 o ---------3

– ( 10 3 + 15 6 + 9 2 + 27 ) p ---------------------------------------------------------------------42

3 10 – 2 33 b --------------------------------6

12 3 – 4 + 3 6 – 2 q ---------------------------------------------------52

12 5 – 5 6 c -----------------------------10

9 10 d ------------5

60 2 + 10 30 – 6 10 – 5 6 r -----------------------------------------------------------------------35

3 10 + 6 14 e --------------------------------4

f

3 22 – 4 10 g --------------------------------6

21 – 15 h -------------------------3

8 21 m ------------49 2 a

i

8 105 n ---------------7

2+2

14 – 5 2 ---------------------6

6 15 – 25 k ------------------------70 3 B 4 D 21 7 a ± ---------7

j

5 C 15 b ± ---------3

115 + 31 21 s ------------------------------148

5 6 ---------3

5–2

3+ 6 b ---------------3 2 2+ 5 c -----------------------3

71 – 12 33 ---------------------------17

102 + 48 6 v ---------------------------95 6 A

2 6 c ± ---------3

– 9 154 + 132 + 42 2 – 8 77 w --------------------------------------------------------------------------50 7 3+9 x ------------------3 21 5 – 6 14 – 5 70 – 20 y -----------------------------------------------------------------27 – 6 + 6 2 + 10 – 2 5 z --------------------------------------------------------2 9 2+8 2 a ------------------14 9 7 – 13 3 b -----------------------------120 16 210 – 12 14 c -----------------------------------------77 d 6–7 2

1E

➔

2 6+ 7 d -----------------------17 8 11 + 4 13 e --------------------------------31

t

u 18 2 + 10 6 – 9 3 – 15

12 – 10 ---------------------16

Exercise 1J — Rationalising denominators using conjugate surds 1 a

593

answers

Answers

1J

answers

594

Answers

6 a

– ( 45 + 15 14 + 9 10 + 6 35 ) e ---------------------------------------------------------------------------5 f

66 + 24 6 ------------------------5 b

g 5 – 4 14

x

–2

–1

–0

–1

2

3

4

5

6

y

–8

–6

–4

–2

0

2

4

6

8

|y|

–8

–6

–4

–2

0

2

4

6

8

i j

– ( 41 + 6 30 ) --------------------------------12

–4

Exercise 1L — Solving equations using absolute values

–6 + 2 15 + 3 10 – 5 6 b -------------------------------------------------------------6 4 A 6 35 14 5 a −2 ----b ------------19 19 210 2 – 120 6 a ------------------------------41

1 a x = ±5 c x = 1 --12- or x = − --12-

312 35 c – ------------------361

1

1 1--2-

<x<2

295 2 – 382 h ------------------------------49 7 Yes

–1

0

0

1

15

d 1 --2

e 8

f 2a

g 12

h

i 3.21

j 0

k − 2--3-

l 4

m 10 q 27

n 10 r 15

o a2b2 s −72

p −16 t −54

v −8

w −11

x 30

z −6cd 3 E

4 D

5 B

0

1

1 1–2 2 2

–1

1 3 4

1

2

3

c x < 3 or x >

4 1--2-

1

2

3

0

1

2 2–2 3

c 2<x<3 d −9 < x < 11

3

4

3

0

2

0

1

1 1–3

4

2

3 1

4 4–2 5 1

4

–5 –4 –3 –2 –1 0 1 2 3 4 5

b −5 < x < −3

Exercise 1K — Further properties of real numbers — modulus

4

b x < 1 or x >

3 a −4 < x < 4

42 5 + 28 7 b --------------------------------17

3

1 1--3-

d x > 2 1--2- or x < 2

11 + 5 8 a ----------------------6

2

–3 –2 –1

2 a x < 0 or x > 2

c 0.75

2 --5

d x = 1 1--2- or x = 3 1--4-

0

d 2<x<3

g 7 2+4

y −3a 2 C

c x = 3 or x = − 1--5-

c

103 – 90 2

1 --2

b No solutions d x=5 b x = 10 or x =

b x > 2 or x < −2

99 120 – 50 460 2 d --------------------------------------------1681 e 44

u

2 a x=2 c No solutions 3 a x = 6 or x = 1 1--3-

1 a 1<x<2

1 --4

x = ±9

f

Exercise 1M — Solving inequations

99 238 – 50 400 2 c --------------------------------------------1681

b

b x = 4 or x = −6 d x = --23- or x = −2

e x = − 1--2-

200 2 – 126 b ------------------------------41

1 a 19

x

2

230 + 257 3 – 137 5 – 80 15 3 a ----------------------------------------------------------------------------431

f

c R and y ≥ 0

y = 2x – 4

4

959 + 281 77 + 182 7 + 6 11 h ----------------------------------------------------------------------------629 3 + 7 65 – 16 11 --------------------------------------------28

y

y = |2x – 4|

–6 –5 –4 –3 –2 –1 0

1 –10

4 a x < −1 --13- or x > − --23b x < 1 1--4- or x > 1 3--4c x < 3 or x > 9 d x < −3 or x > − --13-

2 –5 –3

0

0

3

4

5

0

5

10

–2

–1 –11–3 – 2–3 1 11– 4

0

1

13–4 2

2

3

2 3 4 5 6 7 8 9 10 –4 –3 –2 –1 _ 1– 0 3

1

2

e 1 2--3- < x < 7, x ≠ 3 f

34 a −11 35 C 39 D

0 112– 2 3 4 5 6 7 8 3

3 1--5- < x < 5 1--3- , x ≠ 4

3 31–5

5 51–3

4

6

7

Chapter review 1 A 2 a Irrational, since equal to non-recurring and nonterminating decimal b Rational, since can be expressed as a whole number c Rational, since given in a rational form d Rational, since it is a recurring decimal e Irrational, since equal to non-recurring and nonterminating decimal 3 A 4 B b Z+ c Q d I 5 a Z− 6 D 7 E 8 a

62 -----99

b

337 --------900

c

157 --------165

9 E 10 a b

b −3 36 C 40 B

2m , 25m ,

3

m,

m ------ , 16

3

8m

10 1 a 5------------ amps 3

5 102 b ---------------- amps 6

135 38 c ------------------- amps 19 2 a 2 37 cm

5 34 d ------------- amps 2 b 18 π 37 cm3

c 54 π 3 cm3

d 18 π ( 37 + 3 3 ) cm3 3 10 π 2 15 π 3 a 360 10 cm3 b ----------------- cm c ----------------- cm π π

CHAPTER 2 Number systems: complex numbers Exercise 2A — Introduction to complex numbers

e

20 -----m

11 B 13 a 72x 3 y 4 2xy 14 A

12 C b – --14- x 2 y 5 xy

15 a 25 3

b 3ab ab

16 a

b ( 17 – 4 6 ) cm

5 m

d 22 cm c ( 26 – 4 2 ) m 17 D 18 a 27 b 720 2 19 A 20 D 21 B 23 E 25 C 5 7 26 ---------4 x 5y 27 a ------------2 28 E 29 C

22 A

2

2323 – 594 14 30 a ------------------------------------50

g

2 --3

3i

d

i

h

6 --5

i

c −3, −8 f 0, 2 1−i 1 + 0i

d 0 + 0i h 1 – 2i

0 −9

d −6 h 2

C

d E

c 3a

b

Im (z)

0

c

3 Re (z)

2

d

Im (z) –2 –1 –10 –2 –3 –4 –5 –2 – 6i –6

e

–1 –2 –3 –4 –5

3+i 1

Im (z) 1 2 3 4 Re (z)

Re (z)

7 + 3i 3 2 1 0 1 2 3 4 5 6 7 Re (z)

f

Im (z)

Im (z) –8 + i 3

0 –1 –2

1

2

3 Re (z) 5 – 2i

4 – 5i

Im (z)

–8

2 1 0 Re (z)

1J

➔

3 7– 3 31 ----------------------40 3 32 ------- m2 2 33 A

7i

f

1

2277 – 606 14 b ------------------------------------50 c 51 – 12 14 – 18 7 + 27 2

11 i

c 7i

2 a 9, 5 b 5, −4 d −6, 11 e 27, 0 g –5, 1 h 0, –17 3 a −1 + i b 1+i c e −1 + 2i f −1 + i g 4 z = −2 − 3i, w = 7 + 3i b 15 c 5 a −5 e 2 f 0 g 6 4−i 7 a E b C c 8 Check with your teacher.

1 a b

b 5i

Exercise 2B — Basic operations using complex numbers

24 ( 23 6 – 48) cm

x2 y

38 A

Modelling and problem solving

1 a 3i 20 ------ , m

37 C

answers

595

Answers

2B

answers

596

Answers

2 a 4−i d 9 − 13i 3 a −12 + 3i d −25 + 3i 4 a 7 − 23i d 63 − 37i 5 a 111 + 33i d 61 6 14 + 52i 8 a −8 e −30

b e b e b e b e

1 − 14i −12 + 4i −19 − 8i −50 − 48i 4 + 45i −85 − 132i 31 − 8i −53 7 −3

b −5 f −115

21 16 b x = ------ , y = – -----41 41 d x = −2, y = −3

0 –2

Re (z)

3

d

Im (z)

Re (z)

2

2 – 4i

–4

f

Im (z)

Im (z)

–10

h 32 Re (z)

Re (z)

Im (z) –88 + 16i

Im (z) 3 2 1

0 Re (z)

–88 32 – 24i

z, i4z, –i2z

–2 –3

9

d 2 a

7 + 3i

e 5 − 2i

Im (z) z=3+i Re (z) z= 3 – i

f

2 + 15 2 5 – 6 2--------------------------i ----------------------- + 7 7

−

23 -----29

i

e

43 -----53

+

18 -----53

i

3–i b ---------10 – 3 – 2i e -----------------13

23 -----10

17 -----2

b

9 -----10

+ --92- i

c

17 -----5

10 –29

17 a −12 + 11i

4 + 3i c -------------25 f

3 + 2i ---------------------5

-----d − 16 5

-----e − 14 5

11 −33 + 58i

b −30 − 19i

c 0

18 Check with your teacher. 19 a –4, 16, –64 20 x = −1, y = ± 2 21 a = − --12- , b = 22 x = 2, y = 1; x = −2, y = −1 23 a i 13 ii 5 2 3 - − ------ i c i ----ii --15- − --25- i 13 13

i3z, –iz

b 5 + 9i

14 -----29

i

1 --2

d −8 + 16i e −2 + 10i – 4 + 7i f -----------------65

Exercise 2C — Conjugates and division of complex numbers 1 a 7 − 10i

d

26 -----25

7 10 + 24i

–3 –2 –1–10 1 2 3 Re (z) i2z

7 - + c − ----25

b 0−i

12 a D b C c B 13, 14 and 15 Check with your teacher. 16 −16

–10i

16 –24

+ 1--2- i

2+i 6 a ----------5 5 – 4i d -------------41

10 Re (z)

Im (z)

1 --2

8 a

0

iz, i5z

z = – 4 –5i

11 Re (z)

Im (z)

6 – 2i

0

Im (z)

3 Check with your teacher.

11 – 2i

6 Re (z)

0

z = – 4 + 5i

5 a 0−i

Im (z)

0

12

z = –1 – 3i

c

3 + 4i

0 –2

g

Re (z)

4

c C b

0

Im (z)

z = –1 + 3i

d 35

b B

4

e

b

Re (z)

c x = 1, y = 5

c

−4 − 2i −9 − 5i 12 + 23i −41 − 28i −50 − 13i 176 − 61i 22 − 48i 32 − 126i

c −9

9 a x = 5, y = −2

10 a E 11 a Im (z)

c f c f c f c f

c 3 − 12i f

– 6 + 11 i

Exercise 2D — Radians and coterminal angles 1

Im (z) 3— π 4

π – 2

5— π π 6 7— π 6 5— π 4

3— π 2

π – 4 π – 6

7— π 4

0 2π

Re (z)

π 2 a --4

π b --3

c

3π -----4

3π d -----2

5π -----6

e

e i

0

3 a 210° b 225° c 240° d 300° 4

ii

Im (z)

130

9 Re (z)

Im (z) –7

(b)

(c)

f i

(a) (d)

z+w–u

ii 10

Im (z)

answers

597

Answers

z2

6 Re (z) (e) (f)

0

4 a

8 Re (z)

Im (z)

History of mathematics Probability He was a foreigner. Tutoring students and writing books Newton De Moivre predicted it.

1 2 3 4 5

z1

z2

6 4 2

z3 z4

2 4 6 8 10 Re (z)

–4 –2

b 42.5 square units

Exercise 2E — Complex numbers in polar form 1 a

z = 4 5

b

Im (z)

5 a

z = 4 + 8i

8

Im (z) 12 10 8 6 4 z 2 –4 –2 0

w

u 2 4 6 8 Re (z)

b 24 square units 0

4 Re (z)

6 a 0.588 2 a 13

b 3

d 3 5 3 a i

c f 5

5

e

ii

Im (z) z–w 4

17

65

2π e – -----3 π i – --2

π 7 a – --2 –1 0

b i

Im (z) 6

Re (z)

ii

0 1

c i

–8

d i

9 a

w–u

ii

53 c

0 –2

7 Re (z) w+z

d 2.034 h π

π b --6 f

6π -----7

149, – 2.182

π c – --8

3π d -----4

2π g -----5

11 π h --------12

3π b 5 2, -----4

π d 8, --6 f 2 10, 1.893

π g 4, --3

Re (z)

Im (z)

e

π – --4 π g --2 c

j 0

2π c 2, – -----3 ii 10

6

f −1.030

π 8 a 3 2, – --4

Re (z)

Im (z) 0

37

u+z

5π e – -----6

π b --6

3π 2 cis -----4

π b 2 2 cis --6

3π 10 cis  – ------  4

π d 2 5 cis  – ---  3 f

2 3π ------- cis -----4 4

2B ➔

2π e cis  – ------  3

2E

answers

598

Answers

3 2 b ---------- (1 + i) 2

10 a – 1 + 3i 5 c ------- ( − 3 + i ) 2

d 2–2 3 i

14 e ---------- (1 + i) 2

f 8i

11 C

12 B

13 D

Chapter review 1 B 4 C 7 a −48 g – 3

14 E

15 D

Exercise 2F — Basic operations on complex numbers in polar form 3π 1 a 6 cis ----4 d

2π 6 cis -----3

π b 20 cis --3 π e 2 7 cis  – ---  6

2 a –3 2+3 2 i c 3 10 – 3 10 i e

c

6 3 2 d – ------- + ---------- i 2 2

π 8 2 cis  – ------  12 11 π b 4 cis --------12

3π 2 cis  – ------  10 

3π d 2 2 cis  – ------  14 

3 2 7π e ---------- cis -----4 12 3 6 3 6 ii ---------- + ---------- i 2 2 ii −16

c i 9 cis π

ii −9

3π d i 32 cis -----4

ii – 16 2 + 16 2i

6 a − 1--43 1 d ------- – ------ i 64 64 7 – 64 3 – 64i 8 1 9 a B 8 π 10 ---, – --9 6 12 −64 + 64i

b − 1--8- i

c − 1--4- + 1--4- i

e 0.171 – 0.046i

f

b C

29 10 A

b 210° 16 D

17 B

19 E

20 A

117 44 b ---------------- + ---------------- i 15 625 15 625

c ±(1 − 2i) Modelling and problem solving π π 1 a cos ------ + sin ------ i 12 12 b z = 1 + 3 i, w = 2 + 2 i 2 + 6 + ( 6 – 2 )i c -----------------------------------------------------4 6+ 2 6– 2 d i -------------------ii -------------------iii 2 – 3 4 4 e Check with your teacher. 2 a i 2 ii 0 b x=3 3 a iz = −2 + 3i, i2z = −3 − 2i, i3z = 2 − 3i, i4z = 3 + 2i b i4z = z c d Im z Im z

16

4 3 2 1

11 16 − 16i

i z

d ±(2 − i)

i 4z

0 Re z –4 –3 –2 –1 –1 1 2 3 4 –2 2 i z –3 i 3z –4

0 –4 –3 –2 –1 –1 1 2 3 4 Re z –2 i 2z –3 i 3z –4

e 13 f i, ii and iii One-quarter turn (rotation by 90°) in an anticlockwise direction. g One-quarter turn in an anticlockwise direction h Circle with centre at the origin and radius 2

x +y

2

Im z z = x + yi zi r

2 π 13 ---, – --------5 120 b ±(3 − 2i)

4 3 2 1

iz 4

c E

±( 1 + 2 + – 1 + 2 i )

c 6 5 11 A

b π

3π 18 7 2 cis  – ------  4

r =

14 a ±(3 + 2i) c

9 C 1 12 ------ ( 12 – 14i ) 17 π 13 a --6 14 a 135° 15 B

iz

π 5 a i 3 3 cis --4 b i 16 cis π

b

22 a –527–336i

b 10 + 10 3 i

π b 8 3 cis  – ---  2

π 4 a 3 cis --2

8 a 11 − i

3 −1 − 2i 6 C b a = −549, b = 296

21 B

21 – 7 i

5π 3 a 4 2 cis -----12 c

π c 6 5 cis  – ---  4

2 B 5 E

y x

Re z zi

zi 2

3

CHAPTER 3 Matrices

Exercise 3B — Multiplying matrices 1 a A (2 × 2), B (2 × 2), C (3 × 2), D (1 × 2), E (2 × 3), I (2 × 2) b CA, DB, AE, AI, IA, IB, A2, EC c (3 × 2), (1 × 2), (2 × 3), (2 × 2), (2 × 2), (2 × 2), (2 × 2), (2 × 2)

Exercise 3A — Operations with matrices 1

Matrix

Order

2, 1 element

1, 3 element

A

2×2

8

—

B

3×1

5

—

C

1×4

—

10

D

2×3

4

4

E

3×3

1

2

2 a

33 09

3 a

–4 6 8 14 9 9 0 27

d

b

4 a C

–7 3 85

c

6 –3 –2 8

b

–9 6 12 12

c

e

7 18 – 12 – 4

b D

c E

d

11 6 – 4 20

d A

e B

2 0 14 40 0 6 0 18

b

0 8 0 0 10 16 0 12 0

c

2 8 14 4 10 16 6 12 18

d

3 4 21 65 8 9 6 27

e

4 0 28 80 0 12 0 36

f

–1 4 –7 –2 5 8 –3 6 –9

b –2 0 –6 2

8 a

9

10 a

11

1 0 2 2

3 2 0 1

1 2 1 0

b

0 0 1 2

1 1 0 3

b No

3 a

4 3 0 –9

b

2 –7 24 9

c

e

2 0 0 –3

f

10 – 4 g – 24 – 9

i

00 00

j

32 –8 5

6 a

3 1 0

b

32 31

00 00

d

32 –8 5

31 0 0 31

h

10 – 11 16 1

10 01

d B

Sharks have a total of 32 points. Dolphins have a total of 31 points.

18 12 7 a 14 15 b 6 10 14 1 9 16 c Southport 120, Broadbeach 99, Lions 74, Eagles 70

1 2 3 0

82 54 75 68 91 82

10 25 12 b Shop A = $820, Shop B = $345

8 a

15 14 104 7 10 52

b

13 7 5 1 31 18 26 12 4 4 4 17 15 16

14 8 5 1 35 19 29 13 4 4 5 18 19 16 b True

c False

d True

History of mathematics 1 2 3 4

Matrix theory and number theory Computer development Cross of Honour Caltech

2E

➔

12 a True

10 20 8 26 – 5 10 – 4 12

10 ii 1 0 iii 01 01 b All are I c Multiplicative inverses b C c D 5 a A

2 –6 c – 1 0 1 d – 1 – 12 2 3 –1 –2 –1

0 0 1 1

2 a

4 a i

7 Different orders 0 1 3 1

20 14 – 4 18 – 8 2 – 3 44 22 2 – 2 –8 –8 6 4 5 4 5 2 – 3 1 1 – 8 – 21 14 15 4 5 1 0 28 13 – 24 – 30

3 –6 –2 –1

5 a

2 0 6 a 3 –1

d

answers

599

Answers

3B

answers

600

Answers

Exercise 3C — Powers of a matrix 1 a

4 –2 0 0

8 –4 0 0

b

16 – 8 0 0

c

2 –1

13 a 2 a

1 0 0 0 1 0 0 0 1

b

1 0 0 0 1 0 0 0 1

3 a

1 0 0 8 9 0 2 4 1

b

1 00 26 27 0 10 13 1

4 a

1 0 0 0 1 0 0 0 1

c

2 3

b

14 a x = −2, y = 1 c x = −2, y = 3

b x = 1, y = 2 d x = 7, y = 4

Exercise 3E — The transpose of a matrix Check with your teacher.

Exercise 3F — Applications of matrices

–1 –1 –1 1 1 2 –1 –1 –1

Exercise 3D — Multiplicative inverse and solving matrix equations 1 AB = 6 1 0 01

1 1 h ------ – 132 – 114 or --- – 44 – 38 15 186 162 5 62 54

1 g ------ 1 – 5 15 2 5

1 a A –1 = --- B 6

1 a (5, −1) b (3, 0) d (0, 0) e (4, 4) 2 a and b Answers will vary. y d i 0 –1

1 b B –1 = --- A 6

4– 3

2

c (10, 2) f (−2, −3) c det = 0 ii

y

Both lines

0

2 x

x

3

–3

–2 –3

2 MN = – 2 1 0 , 01 3 a 5

b 12

1 4 a --- 10 – 3 5 –5 2 1 d --- – 1 3 8 4 –4 5 a C

b E

M –1 c −2

1 = – --- N , 2 d −8

1 b ------ 0 3 12 – 4 – 2 1 e --- – 5 1 7 3 –2 c D

N –1

1 = – --- M 2

e 7

f 14

1 c --- 1 6 2 0 –2 f

1 ------ 4 1 14 – 6 2

1 d --- 1 2 8 – 2 – 12

0 8 –1 –2

1 b --- 1 2 4 –2 0

c

1 e --- – 11 2 8 40

f

12 2 –2 –1 1 1 2 --8 – 2 – 12

1 b --- – 2 – 8 8 1 0

10, 11 Check with your teacher. 1 12 a --- – 31 – 22 2 24 18 1 d --- 18 23 2 – 12 – 16

C

1 a T

K B

8 a D − det = 0 b E − det = 0 c F − Not a square matrix 9 a

Exercise 3G — Dominance matrices

d A

6 Answers will vary. 1 7 a --- 1 6 2 0 –2

e In i there are parallel lines; in ii there is only one line. 3 a E b B 4 a C b D 5 16, 4 6 15, 10 7 Anh 8 $51 070

1 b --- – 5 5 2 14 – 8

1 c --- – 6 2 6 –6 4

1 e ------ 78 103 f 30 – 24 – 34

10 01

b Cameron, Breanna, Kayley, Teagan 2 Mair, Ann, Janine 3 a Hamilton, Leslie, Cunningham, Barnes b 20 points to Hamilton, 15 to Leslie, 10 to Cunningham, 5 to Barnes 0 0 1 1 0 1 0 1 1 0 4 a M= 0 0 0 0 1 0 0 1 0 1 1 1 0 0 0 b 5 points to Warwick, 4 points to Ipswich, 3 points to Stanthorpe, 2 points to Clifton, 1 point to Goondiwindi

Chapter review 1 D

2 B

6 a –3 2 –2 –4

3 C b

–7 –6 6 –4

4 B c

5 a, f, g, i 5 – 14 14 12

7 a

100 142 100

100 384 100

b

8 a A

c

1 00 7 16 8 1 00

b B

9 D

10 C

12 A

11 C

93 –6 0

13

1 15 a ------ 1 2 10 – 3 4

14 B

b (5, −1)

16 a [0.25 0.40 0.20 0.15] [800] = [a b c d] b A = 800B c [200 320 160 120] 17 a

˙ 0 1.12 0 0.95

b

601

CHAPTER 4 An introduction to groups Exercise 4A — Modulo arithmetic 1 a 4, 12, 20, 28, 36 … 2 a 0, 1, 2 c 0, 1, … 10 3 a + 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4 b

× 0 1 2 3

0 0 0 0 0

1 0 1 2 3

2 0 2 0 2

3 0 3 2 1

c

× 0 1 2 3 4

0 0 0 0 0 0

1 0 1 2 3 4

2 0 2 4 1 3

3 0 3 1 4 2

278.88 27.55 761.60 46.55

b 4, 10, 16, 22 b 0, 1, 2, … 8

answers

Answers

18 James, Cameron, Glen, William Modelling and problem solving 1 a 1: Despatch for Deluxe model takes 1 hour. b 14: Packaging at Plant 1 has a wage rate of $14 per hour. c 3 × 3, 3 × 2, 3 × 2 d

433.25 420.50 529.50 514.00 605.25 587.50

e The total costs for the Standard model at Plants 1 and 2 f The assembly costs for each model at Plant 1 g i $529.50 ii $514.00 2 a 145x + 103y + 121z = 20 260 130x + 110y + 90z = 18 400 142x + 115y + 80z = 19 200 b

145 103 121 x 20 260 130 110 90 y = 18 400 142 115 80 z 19 200

0.025 544 –0.093 523 0.066 579 c A−1 = – 0.039 222 0.091 991 – 0.044 166 0.011 042 0.033 767 – 0.042 189 d The cost of an adult’s ticket is $75, a child’s ticket is $50 and a pensioner’s ticket is $35. a 3 A=

b 2

a – ----- – – a b a

4 A=

b 2

History of mathematics 1 He tutored students. 2 Abelian groups are those that have the property of commutativity.

Exercise 4B — The terminology of groups 3+2 1 ------------ = 2 1--2- and 2 1--2- is not an element of the set of 2 whole numbers. 2 1 ° 3 = 1 + 9 = 10 Not a whole number, ∴ not closed. 3 a IE = 1 b b = 2 − a 4 IE = 0 a ° 0 = a + 0 − a × 0 5 No identity. 4a × ( 1--2- )2 = a but 1--2- ° a ≠ a 6 Assuming this operation has an identity then let a+b ------------ = a ab a + b = a2b a = a2b − b But a ≠ a2b − b therefore the operation has no identity. 7 Let (0, 1) = (a, b) = IE. Therefore, (0, 1) ° (c, d) = (0 × d + 1 × c, 0 × c + 1 × d) = (c, d) and (a, b) ° (0, 1) = (a × 1 + b × 0, a × 0 + b × 1) = (a, b). 8 Let (a + b)2 = a where b = IE Take the square root of both sides: a + b = ± a If a is negative then a ∉ R. Since an identity must be applicable to all elements of the set, there is no IE for a ° b.

3C

➔

a–a -------------- 1 – a b

4 0 4 3 2 1

4B

answers

602

Answers

Exercise 4C — Properties of groups 1 a [R, +] It is closed, associative, IE+ = 0, inverse = −a, therefore it is a group. b It is Abelian. 2 a Closed, associative, no IE since 0 ∉ {even numbers}, there is an inverse; therefore not a group. b Closed, associative, no IE since 1 ∉ {even numbers}, no inverse; therefore not a group. 3 a 12 + 13 is not closed; not a group. b 12 × 13 is closed, and associative, IE× = 1, there is an inverse; so it is a group. 4 Check with your teacher. 5 × 1 2 3 4 1 1 2 3 4 2 2 4 1 3 3 3 1 4 2 4 4 3 2 1 Closed, associative, IE× = 1 and there is an inverse; therefore it is a group. 6 a + 4 16 64 256 … 4 8 20 68 260 … 16 20 32 80 272 … 64 68 80 128 320 … 256 260 272 320 512 … . . . . . . . . . . . . . . . b Under addition: not closed, associative, no IE+ since 0 ∉ 22n, no inverse (always +ve); not a group. c Under multiplication: closed, associative, IE× = 1 is not present as no 20 (0 ∉ {even numbers}), inverse is 2–2n; not a group. 7

× 0 1 2 0 0 0 0 1 0 1 2 2 0 2 1 It is closed and associative, IE× = 1; inverse does not exist since there are no 1s in the first row or column. This is not a group; therefore, it is not Abelian, even though the commutative law does apply. 8 a Yes b No, not closed c No, no inverse for b d Yes 9 a 5 10 20 ° 5 5 10 20 10 10 10 20 20 20 20 20 b It is closed, associative, IE° = 5, no inverse; so not a group.

10

° N L R A

N N L R A

L L A N R

R R N A L

A A R L N

Closed, associative, IE° = N, there is an inverse, N appears in every row and column.

Exercise 4D — Cyclic groups and subgroups 1 Closed, IE+ = 0, inverses exist (3 + 9 ≡ 0 mod 12, 6 + 6 ≡ 0 mod 12), therefore subgroup. Generators are 3 and 9. 2 Closed, IE = 0 0 , the inverse of a 0 is – a 0 00 0b 0 –b which is a member of M, therefore a subgroup. 3 a <3> = [{0, 3, 6, 12}, + mod 12]; <6> = [{0, 6}, + mod 12]; <9> = [{0,3,6,12}, + mod 12] b Generators are 3 and 9. 4 a Cyclic. Generators are 1 and −1. b Not cyclic. c Cyclic. Generators are 6 and −6. d Cyclic. Generators are 3 and 1--3- .

History of mathematics 1 He worked towards having women accepted at Cambridge University. 2 Abstract algebra, group algebra, n-dimensional geometry, matrices and determinants

Exercise 4E — Further examples of groups — transformations 1

A

C

B

C

R0

B A

R120 C B

a

° R0 R120 R240

R0 R0 R120 R240

R120 R120 R240 R0

R240 A

R240 R240 R0 R120

b IE = R0, Inverse exists for all elements. It is an Abelian group because the table is symmetrical about the leading diagonal. 2 A RR

C a

° R0 RV RL RR

RL

RV R0 R0 RV RL RR

B RV RV R0 — —

RL RL — R0 —

b Does not form a group. 3 Not Abelian.

RR RR — — R0

4 a

3

2 1

Identity exists: 0 0 is the identity element. 00 Inverses exist: the inverse of A is −A.

3 1

4 2

4

3

ii 1 2 iii The set of 2 × 2, non12 singular matrices. 9 Identity = I. Inverse is present as I is present in each row and column. Closed and associative

4

RV 1 2

2

1

RH

3

R0 4

2

1

b

2

3

8 a Closed: addition of 2 × 2 matrices results in a 2 × 2 matrix. Associative: matrix addition is associative.

4 3

R180

12 34

b i

2

3

10 a IE+ = 0 0 (Remember 0 is a complex number.) 0 0 Inverse =

4

1

3

RR

RL 4

c

1

1

3

4 2

4 1 R H

3 2 4

2

3 R180

4

R0

3 4 2

1

3

1

RV

1

1

3 4

2 5 …H H H H… RH …H H H H…

RV R180 R0

…H H H H… …H H H H… …H H H H…

z –z 1 - 1 2 where the b IE× = I Inverse = -------------2 2 z1 + z2 z2 z1 determinant is real. The inverse exists if the determinant ≠ 0. 11 b Yes. 00 Closed, associative, IE, Inverse of 0 0 is 1 1 --- --z z z z i 0 – 1 0 – i 0 1 0 12 a , , , 0 i 0 –1 0 –i 0 1 b Yes.

Chapter review 1 a

+ 0 1 2 3 4

0 0 1 2 3 4

1 1 2 3 4 0

2 2 3 4 0 1

3 3 4 0 1 2

4 4 0 1 2 3

b

× 0 1 2 3 4 5

0 0 0 0 0 0 0

1 0 1 2 3 4 5

2 0 2 4 0 0 4

3 0 3 0 3 0 3

4 0 4 2 0 4 2

6

7 a

°

f1

f2

f3

f4

f1 f2 f3 f4

f1 f2 f3 f4

f2 f1 — —

f3 f4 — —

f4 f3 — —

5 0 5 4 3 2 1

2 a Yes b No, not closed 3 a Yes b No, 0 does not have an inverse c Yes d Yes 4 a No, no identity b No, not associative 5 Check with your teacher. 6 a Commutativity b There is only element x such that p ° x = q and x ° q = p. c Each element has a unique inverse. 7 a Yes b No, 0–1 doesn’t exist c No, inverses don’t always exist d Yes 8 Check with your teacher. 9 a No, there is no generator.

4C ➔

b Not a group

–z1 –z2 z2 –z1

2

R0 3

4

1

4

2

1

R180 2

603

answers

Answers

4E

answers

604

Answers

b No, the operation is not closed on these elements. c = {0, a}, = {0, b}, = {0, c} 10, 11 and 12 Check with your teacher.

Exercise 5E — Inverse of a 3 ¥ 3 matrix 1 a iii 1

Modelling and problem solving 1 and 2 Check with your teacher. iii

CHAPTER 5 Matrices and their applications

b iii −12

Exercise 5A — Inverse matrices and systems of linear equations 1

26 9

5

17 1

8 c

2

42 36

3

10.38 17.31

6

100.29 92.06

7

98.04 108.82

3.18 0.681

iii

400 370

4

Exercise 5B — Gaussian elimination 1 x = 1, y = 1 3 x = 2, y = 1, z = −3 5 x = −2, y = −1, z = 0 7 a

1 --2

– --14-

– 1--2-

3 --4

b

1 --2

2 1--3- – 2--3- – 1 1--3c

–1

0

1

1 --3

1 --3

– 1--3-

iii

2 a = −3, b = −1 4 x = 1, y = 2, z = 0 6 x = −3, y = 5 1--4- , z = −3 3--4-

–3 2 2 –1

d

1 --2

iii

0

– 1 – 1--2- 1

8 a x = 2.7, y = 0.6 b x = 2 1--5- , y = 1 1--5- , z = − 3--5-

2 a

c x = −2 --17- , y = −2 --37- , z = 1 --37-

History of mathematics 3 Science of determining the size and shape of the Earth.

c

Exercise 5C — Introducing determinants 1 2

2 13

5 a −5 e 61 i −24

b −33 f 0 j 122

3 − --12-

4

3 --4

c 81 g 0

d −3 h 24

Exercise 5D — Properties of determinants 1 2 3 4 5 6 7 8 9

a a a a a a a a a

−3 0 0 22 −18 2 8 1 −6 b

b b b b b b b b −4

−3 c Property 2 Property 3 −22 c −36 c 2 c Property 7 Property 8 c 3 d

4 a

Property 1 Property 4 Property 5 Property 6 −8

e 3

3

f 40

3 –2 –1 1

iv

3 –1 –2 1

iii

9 –6 –5 2

iv

iii

8 – 12 8 2 2 –2 –4 8 –4

d iii −33

– 1--4- 0

0

9 –5 –6 2

c iii 8

480.68 481.82

d

1.81 1.81

3 –1 –2 1

iii

–3 –1 5 1 0 –1 6 2 –9

iv

iv

– 0.16 0.08 0.44 0.52 0.24 – 0.68 d 0.28 0.36 – 0.52

2 3 –4

b

1 --2

– 1--6-

8 2 –4 – 12 2 8 8 –2 –4

1 --4

– 1--2-

b

– 1.5 1 1.25 0.5 0 – 0.25 – 1.5 1 0.75

5 -----12

1 – 1 1--2-

iii

13 – 12 – 4 –5 –3 –1 11 0 – 11

– --34-

1 --4

1 – --14-

1 – 1--2-

13 – 5 11 – 12 – 3 0 – 4 – 1 – 11 13 – ----33

12 -----33

4 -----33

5 -----33

1 -----11

1 -----33

– 1--3-

0

1 --3

1 –3 –2 0 0 0.5 – 1 4 2.5 – 1.5 0.5 0.5 – 0.85 0.45 0.25 0.6 – 0.2 0

19 X = –2 15

1 0 3

5 x=± 3 6 a (y − z)(v − u) b xy2 − xz − x2y + yz2 + x2 − y2z c (1 − x)(1 − y) + y + x − 1 d 2a3 − 3a2b + b3 7 a a = 1, 2 b a = ± 1--2- c a = 0, 3 d a = 0, 1, 2 e x = 3, 4 f x = −1, 3

8

1 --3

5 --6

– 1 2--3-

– --16-

7 -----12

– --23-

0 – 1--2-

1

8 x= 0.3 0.05 0.03 0.15 – 0.06 0.07 – 0.24 0.03 0.13

9 a −189 b

c

1 --2

0.25 – 0.5 0.75 0.5 0 – 0.5 – 0.25 0.5 0.25

12 a 1 + i b 0 c −4 − 7i 13 Check with your teacher.

Exercise 5F — Cramer’s Rule for solving linear equations x = 2, y = −3 2 x = −1, y = 4 x = −2, y = −5 4 x = 0, y = 3 x = 3, y = −1, z = 2 x = −11 --13- , y = −5 --13- , z = − --13-

Chapter review 1 G = 50, H = 10 2 X = 15.15 (15.15 tonnes of aluminium, 14.1 tonnes 14.1 of gold) 3 x = −1, y = 8, z = 9 4 a

0.2 – 0.4 0.2 0.6

5 a −2 6 a 0 b −30 c −52 d 13 e −24 f 0 g −15

1 –1 0 5 –7 1 –2 3 0

b 10 c −2 Property 3 Property 7 Property 4 Property 1 Property 5 (twice) Property 2 Property 6

7 a iii −2

iii

b

4 –2 –3 1

b iii −12

2 -, z = , y = −2 ----27

2 --9

Modelling and problem solving

0.25 0.5 – 0.25 a A′ = – 0.5 0 0.5 0.75 – 0.5 0.25

1 3 5 6

17 -----27

7 –5 –1 1 iv ------ – 4 8 4 12 –5 7 –1

u b a u v d c v 9 a y = ----------------------- and x = ----------------------a b a b c d c d where ax + by = u and cx + dy = v b x = 2, y = −1

1 0

10 Check with your teacher. 11 a ′A =

–7 5 1 4 –8 –4 5 –7 1

iii

answers

605

Answers

4 –3 –2 1

iv

–2 1 1.5 – 0.5

iii

–7 4 5 5 –8 –7 1 –4 1

0 4 0 0

1 0 0 2

0 0 0 1

0 2 C= 0 0 1 3 a = ±3

3 1 0 0

0 0 –2 0

1 0 0 0

2 413 1 692 56 524 3 313 8 382 4 a D = 8 844 7 433 25 989 13 159 16 487 7 195 c

(I − A) =

3713 -----------4532

0

0

3945 -----------4273 – 383 -----------4273 – 106 -----------4273 – 136 -----------4273 – 52 -----------4273 – 416 -----------4273 – 915 -----------4273 –9 -----------4273 – 18 -----------4273 – 42 -----------4273

– 821 -----------4532 – 24 -----------4532 – 48 -----------4532 – 100 -----------4532 – 130 -----------4532 – 209 -----------4532 –4 -----------4532 – 29 -----------4532 – 23 -----------4532

– 1160 ---------------81 609 – 1716 ---------------81 609 69 438 ---------------81 609 – 778 ---------------81 609 – 136 ---------------81 609 – 97 ---------------81 609 – 981 ---------------81 609 – 2 120 ---------------81 609 – 34 ---------------81 609 – 75 ---------------81 609 – 192 ---------------81 609

0 – 195 -----------8570 – 107 -----------8570 4733 -----------8570 – 209 -----------8570 – 64 -----------8570 – 103 -----------8570 – 522 -----------8570 –9 -----------8570 – 38 -----------8570 – 44 -----------8570

4 532 4 273 81 609 8 570 14 100 b X = 10 501 17 256 48 238 13 756 19 116 10 168 – 17 ---------------14 100 – 285 ---------------14 100 – 1 918 ---------------14 100 – 15 ---------------14 100 16 713 ---------------14 100 – 82 ---------------14 100 – 106 ---------------14 100 – 1 370 ---------------14 100 – 14 ---------------14 100 – 11 ---------------14 100 – 11 ---------------14 100

– 93 ---------------10 501 – 41 ---------------10 501 – 1 982 ---------------10 501 – 83 ---------------10 501 – 78 ---------------10 501 10 194 ---------------10 501 – 1 587 ---------------10 501 – 2 606 ---------------10 501 –7 ---------------10 501 – 39 ---------------10 501 – 88 ---------------10 501

0 –7 ---------------17 256 – 939 ---------------17 256 – 36 ---------------17 256 – 128 ---------------17 256 – 150 ---------------17 256 14 203 ---------------17 256 – 1 226 ---------------17 256 – 25 ---------------17 256 – 61 ---------------17 256 – 113 ---------------17 256

–2 ---------------48 238 –2 ---------------48 238 – 1 511 ---------------48 238 – 78 ---------------48 238 – 1 415 ---------------48 238 – 373 ---------------48 238 – 2 140 ---------------48 238 39 454 ---------------48 238 – 452 ---------------48 238 – 349 ---------------48 238 – 423 ---------------48 238

–2 ---------------13 756 –3 ---------------13 756 – 2 979 ---------------13 756 – 141 ---------------13 756 – 628 ---------------13 756 – 216 ---------------13 756 – 649 ---------------13 756 – 2 302 ---------------13 756 13 719 ---------------13 756 – 365 ---------------13 756 – 438 ---------------13 756

–3 –6 ---------------- ---------------19 116 10 168 –4 0 --------------10 168 – 1 601 – 672 ---------------- ---------------19 116 10 168 – 127 – 33 ---------------- ---------------19 116 10 168 – 286 – 41 ---------------- ---------------19 116 10 168 – 152 – 64 ---------------- ---------------19 116 10 168 – 384 – 274 ---------------- ---------------19 116 10 168 – 848 – 1 348 ---------------- ---------------19 116 10 168 –1 –5 ---------------- ---------------19 116 10 168 17 543 – 71 ---------------- ---------------19 116 10 168 – 129 8 699 ---------------- ---------------19 116 10 168

5A ➔

iii

0 1 B= 0 1 0

5F

answers

606

Answers

CHAPTER 6 Transformations using matrices Exercise 6A — Geometric transformations and matrix algebra 1 a (0, 0) b (0, −26) c (0, 4) 2 a (2, −5) b (4, −9) c (5, 0) 3 a A′(4, 2), B′(7, 7), C′(11, 4) b A′(4, 0), B′(7, 5), C′(11, 2) c A′(0, −2), B′(3, 3), C′(7, 0) d A′(0, 0), B′(3, 5), C′(7, 2) 4 Check with your teacher. 5 y′ = 2x′ + 7 y

d (51, 7) d (−2, −4)

y B' 5 4 3 2 –3 1 –2 B –2 –3 –4 –5 –6 –7

c C′(2, −7)

b B′(−1, 5)

C 2 4 A

6

8 x

A' C'

4 A′(7, −6), B′(−1, 2), C′(15, −10)

y = 2x + 3

7

3 a A′(8, −7)

y B'(–1, 2) 5 C(0, 5) B(–4, 1) A(4, 1) –4 –1 45 7 10

y' = 2x' + 7

15 x

–5 A'(7, –6) –10

3 –3.5 –1.5 x

0

5 a

6 a y′ = −(x′ − 3)2 − 1 b y′ = (x′ − 5)2 − 5 c y′ = x′2 + 3x′ − 4 d x′2 − 4x′ + y′2 − 4y′ + 4 = 0 e x′2 + 8x′ + y′2 + 2y′ + 8 = 0 7 a

–2 –3

–1 6

b

c

–4 –7

c

Exercise 6B — Linear transformations 1 2 2 a iii iii iii iv b iii 1 a

1 c 2 –3 1 3 2 A′(0, −1), B′(4, 2), C′(−5, −2) A′(−1, 2), B′(−2, 0), C′(3, −1) A′(2, −4), B′(4, 0), C′(−6, 2) A′(1, −2), B′(2, 0), C′(−3, 1) y

B'(4, 2) C(–3, 1) B(2, 0) x

0

C'(15, –10)

1 --3

– 1 2--3-

1 --13-

– --16-

b

1 1 1.75 2.5

d

– 1--3-

1.5

– 2 2--3-

2

Exercise 6C — Linear transformations and group theory 1 a d 2 a 3 a 4 a b d

0 1 P′(9, 5) b P′′(5, 1) c –2 1 Check with your teacher. P′′(1, 4) b Check with your teacher. Non-singular b Singular P′(31, 18) det A = 1 (non-singular) 2 ------ x′ + -----iii y′ = 4--7- x′ ii y′ = 10 17 17

iii 10x′2 − 34x′y′ + 29y′2 = 2 5 y′ = 1--2- x′ − 2 y y = 3x + 2

A' C'(–5, –2)

1.5 1 1.5 1

A(1, –2)

iii

y A'(–1, 2) C(–3, 1)

2

y' = _12 x' – 2

B(2, 0) x

0

C'(3, –1) A(1, –2)

B'(–2, 0)

iii

– _23 0

y

–2

C'(–6, 2) C

B 0

B'(4, 0) x

A A'(2, –4)

iv No change

6 a 17x′2 − 26x′y′ + 10y′2 = 9 b 10y′2 + x′2 − 2x′y′ = 81 c 13x′2 + 10x′y′ + 2y′2 = 9

4

x

7 a y′ = x′ b y′ = 3--2- x′ 8 Check with your teacher. 9 a, c i

c y′ = 1--2- x′

b iii, iii

y y' = _3x' + _13

y

1 y = 2x – 1

_1 3

_1

() y' = _3x' – _13 ()

3

y' = 0.3x' + 0.2

_1 0

–1

–3

x

1

0.2

y = –3x + 1

0

2 –_ 3

answers

607

Answers

x

_1 2

4 No change, rotation about the centre of the circle. –1

Exercise 6E — Reflections 1 a a, c ii

y y = –x + 4

d y' = 4

4

e 0

b i y′ =

3 -----10

x

4 1 --5

x′ +

–1 0 0 1

– 1--2-

3 ------2

3 ------2

1 --2

2 a (1 −

3 ------2

2 -, c ( 3--------2

–1 0 c 0 –1 1 --2

3 + )

,

9 2 ---------2

e (−3, 2)

0 1 d –1 0

x --3

+

1 --3

a iii y′ =

x --3

−

1 --3

b iii

x

1 0 0 1

(i)' (vi) (iii)

(vi)' (i) (iii)' mx = 0

b iii (3, 1) ii (4, −2) iv (−2, −4) v (3, 0)

3 3 ---------2

3 3 - − --- ) , − -----2 2 1 --2

+

3 ------2

iii (−1, 3) vi (−2, 1)

y (iv) (iii)'

)

(ii) (i)' (v)(v)' x (i) (ii)'

my = 0 (vi)'

2 x′ 3 a iii y′ = ---- + ------2 4 a iii y′ =

(ii) (v) 0

3 1 - − --- , ( -----2 2

f

(iv)'

(ii)' (v)'

d ( 1--2- −

iii (1, −3) vi (2, −1)

y (iv)

b (0, −4)

)

0 1 1 0

0 –1 –1 0

f

2 a iii (−3, −1) ii (−4, 2) iv (2, 4) v (−3, 0)

ii y′ = 4

Exercise 6D — Rotations 0 –1 b 1 0

c

1 0 and then 0 0 –1 4

10 a x′2 − 6x′y′ + 9y′2 − 2x′ + 5y′ = 0 b 11y′ = 4x′ − 5

1 a

1 0 0 –1

b

0

(vi) (iii) (iv)'

c iii (1, −3) iv (−4, 2)

y

ii (−2, −4) v (0, −3)

iii (3, 1) vi (1, 2)

y (iv)

− y = _x +√_2 2

0.35 –0.7

0

(vi)'

(iv)'

1

_1 3

4

1

(vi)

x

0

(ii) (iii)' (v) x (i)

(iii) (v)' (i)' (ii)' my = – x

y = –3x + 1 3 ------2

,

3 3 ---------2

+ 1--2- )

iii (2 +

3 , 2 3 − 1)

6A ➔

d iii ( 3--2- −

6E

answers

608

Answers

iii ( –-----21- −

3 3 ---------2

,

– 3 ---------2

2

+ 3--2- )

iv (−1 + 2 3 , − 3 − 2) iv ( 3--2- , vi (−1 −

3 ------2

3 3 ---------2

)

2

2

(a)

x y b ------ + ----- = 1 64 9

, − 3 + --12- ) y

y

2

x 4y 3 a ----- + -------- = 1 4 9

x (b)

my =√–3

x2 — 4

3

(iv)

(i)'

(iii)'

(v)' (ii)' (ii) (v)

0 (vi) (vi)'

(i)

(iii)

(iv)'

4 a y′ = --18- x′2

x

y +— 9 =1 2

b y′ = 8x′2 y

y

y = 2 x2

y' = 8x' 2 y = 2x2

y' = –18 x' 2

3 a iii y′ = −x′ iii y′ = −x′ iii y′ = x′ iv y′ = --12- ( 3 − 1)2x′ or y′ ≈ 0.268x′

x

iii y′ = −x′2 iii y′ = ± – x′ b iii y′ = x′2 2 c iii y′ = 2x′ + 1 iii y = −2x′2 − 1

2

d iii y′ = −x′

x

5 y′ = 3x′ −

–x′ – 1 iii y′ = ± ----------------2

1 --2

y 2

iii y′ = x′

y' = 3x' – 1–2

iii y′ = ± x′

y = 3x – 2 (c)

Exercise 6F — Dilations 1 a

i ii iii iv v vi

(4, 1) (−8, 3) (0, 3) (6, 0) (4, 5) (1, −3)

(ii)'

(v) (v)' (iii) (iii)'

(ii)

(i) 0

(iv)

(vi)

b

i ii iii iv v vi

(2, −3) (−4, −9) (0, −9) (3, 0) (2, −15) ( 1--2- , 9)

x

0

y

(i)' (iv)' x

6 a iii (1, 1--2- )

(vi)'

iv (1 --12- , 0)

ii (−2, 1 1--2- )

iii (0, 1 1--2- )

v (1, 2 --12- )

vi ( --14- , −1 --12- )

y

y

(v)

(vi)'

(iii) (ii) (v)' (ii)' (iii)' (i) (i)' 0 (iv)' (iv) x (vi)' (vi)

(v) (ii)

(iii) (i) (iv) (iv)' x

0

(ii)'

b

(i)'

(vi)

(iii)'

(−8, −4) (16, −12) (0, −12) (−12, 0) (−8, −20) (−2, 12)

i ii iii iv v vi

(vi)'

(ii) (iv)' (i)'

y (v)' (iii) (i) 0 (iv) (vi)

(ii)'

(iii)'

(v)' 3 --2

2 a y′ = x′ − 2 b y′ = −3x′ + 2

y

x

(v)' 2

y = 3x – 2

2

x y 7 a ------ + ------ = 1 16 36

(a)

y

2

x

0

16y b 4x2 + ----------- = 1 9

(b)

(a)

x x2 — 4

(b)

y +— 9 =1 2

Exercise 6G — Shears 1 ii a (3, 6) a y

2 a

b (2, 5) y

6

B'(10, 3) C(6, 3) C'(15, 3) B(1, 3)

P' 5

P'

b

y

y

C'(6, 21) D'(5, 15) B'(1, 6)

x

A = A' D(5, 0)

B

C

A = A' D(5, 0) 1

P 0

x

3

P

0

c

x

2

B'(–2, 3)

c (−4, −7) a

d (3, 4) y

B(1, 3) B'

C'(3, 3)

A = A' D(5, 0) x

4

C(6, 3) D(5, 0) x C'(6, –3)

A = A' D'(5, –5)

P'

3 ii a

x

0

0 –2

y

b

y' = 4x' y=x

y

y' = 2x'

x

3 P

x

–7

P'

y

y

P –4

d

y B(1, 3) C(6, 3)

x

answers

609

Answers

x y = –x

e (−2, −9) a y

f

c

y

d

y = x2

y y = 2x + 5

y 0

–2

(0, −5)

x

x

0

x

P –5

–5

y' = x' 2 + 3x'

P = P'

x y' = 5x' + 5

ii a

P' –9

y

b

y =x

y y' = 1–2 x'

y' =

b (2 1--4- , 1)

1 ii a (3, 0) a

y

1– x' 4

x x

y

y = –x 0

(3, 0)

P P'

1

P = P' x

0

1

c

x

2

d

y

y

y = x2

y = 2x + 5

y' = (x' –3y')2

c (−3 --34- , 1)

d (2 --12- , −2)

a

Chapter review

1

–4–3–2–1 0

x

0 –2

e (−3 1--4- , −5) a

x

x y

y P P'

y' = 2–7 x' + 5–7

f

y

3

x

P' P

(−1 1--4- , −5)

1 2 3 4 5 6

Translations, rotations and reflections a (−1, 3) b (2, 4) c (3, 1) Check with your teacher. y′ = 2x′2 − 8x′ + 7 A′(3, 0) C′(12, −3) – 2 1--3- 2 2--3– 2 2--3- 2 1--3-

y

7

y′ = −2x′ + 3

y y' = –2x' + 3

0

–2

x

–1 0

x x

P' P

–5

y = –x + 1

6E

➔

P' P –5

6G

answers

610 8

Answers

2 ( 5--------2

c

,

y′ =

x′ ---2

2 − 3--------2

5 --2

d (1 + 9

3 -) b ( – 3--2- , − 3--------2

a (−2, 1)

2 a d 3 C 4 a 5 D 6 a

)

3 , − 3 + 2 --12- )

+1

y

x

a (−3, −1) c (− 1--2- +

11

b (2, 2) 3 ------2

3,

y′ = − 1--3- x′

+ 1) d (−2, 3)

y

x y' = – 1–3 x' y = –3x

12 13 14

a (2, 2) c (−16, 4) y′ = −x′ a (3, 2 1--2- )

A to C

b s + t + u + v c –s – t ˜ ˜ ˜ ˜ ˜ ˜ e –u – t – s ˜ ˜ ˜ b D to B

c B to D

d A to C

r+s b s+t c r–s ˜ ˜ ˜ ˜ ˜ ˜ d r+s e t–s f s+t–r ˜ ˜ ˜ ˜ ˜ ˜ ˜ g r+s+t h –s–t ˜ ˜ 7 a, b˜ ˜ ˜ c 500 km Flight d 53.1° clockwise from N path 8 512.1 km; find bearing using trigonometry 9 721.1 km, 326.3° (clockwise from N) 10 Each part of answer has h 15 coordinate labelled a, b, . . . j. b a f i a The original vectors a and b ~ 5 ˜ ˜ c are also drawn. ~b 11 Magnitude = 10.77, direction –15 g –5–5 d 5 15 j e 68.2° True. –15 12–18 Check with your teacher. 19 One can deduce that x and y components can be added/subtracted/multiplied separately.

y' = 1–2 x' + 1

y = –2x + 2

10

s+t ˜ ˜ –v – u – t ˜ ˜ ˜

b (1, 2) d (0, 2) b (2,1)

B 21 D Displacement, velocity, force Speed, time, length 1 magnitude and 2 angles

22 0 ˜

Modelling and problem solving

20 23 24 25

1 Rotation of 90°: 0 – 1 1 0

Exercise 7B — Position vectors in two and three dimensions

15

2

y′ = 2x′ + 2x′

1 a 3, 4, −2

or reflection in y-axis: – 1 0 0 1

b 6, 0, −3

c 3.4,

2,

1 --2

CHAPTER 7 Introduction to vectors

2 a i 72 ii 45° ii 119.7° b i 65 c i 4.88 ii 225.8° d i 320.16 ii 358.2° 3 a 045° b 330.3° c 224.2° d 091.8° 4 –50i – 50 3 j 5 C ˜ ˜ 6 248.9i – 383.3 j 7 –60.6i + 109.3 j ˜ ˜ ˜ 8 615 km at 49.8°˜ south of east 9 36 steps 11.3° south of east 10 20.8 km

Exercise 7A — Vectors and scalars

11 a

or translation 4 left: – 4 0 2 Dilation of 3 about the origin followed by reflection in the y-axis: – 3 0 0 3

1 a i

~s ~r

iii ~s

ii

r +~s ~ ~s

–s ~

~r

~s

~r – ~s –r ~ ~s – ~r

b

i Same as 1 a i except scaled by a factor of 2. ii Same as 1 a ii except scaled by a factor of 2. iii –4r 3s~ ~s ~r

~

3s~ – 4r ~

12 14 16 17

3 --- i 5˜

+ 4--5- j b 3--5- i – 4--5- j c 4--5- i + 3--5- j ˜ ˜ ˜ ˜ ˜ 4 3 1 2 f –0.792i + 0.611 j d – --5- i + --5- j e ------- i + ------- j 3˜ 3 ˜ ˜ ˜ ˜ ˜ B 13 Check with your teacher. 3 -j –0.98i – 0.20 j 15 – 1--2- i – -----˜ ˜ 2˜ ˜ Check with your teacher. a i 4i – 7 j ii 65 ˜ ˜ b i 3i + j ii 10 ˜ ˜ c i –4i + 7 j ii 65 ˜ ˜ d i –3i – j ii 10 ˜ ˜ e i 2i ii 2 ˜ f i – 4i ii 4 ˜

18 a –4i + 7 j ˜ ˜ d 3i + j ˜ ˜ 4 7 - i – ---------- j 19 a --------65 ˜ 65 ˜ 4 7 -j c – ---------- i + --------65 ˜ 65 ˜ e i ˜ 20 a i 29

b –3i – j ˜ ˜ e – 2i ˜ b

c 4i – 7 j ˜ ˜ f 4i ˜ 3 1 ---------- i + ---------- j 10 ˜ 10 ˜ 3 1 -j d – ---------- i – --------10 ˜ 10 ˜ f –i ˜ ii 13

3 2 2 -j - i + ---------- j iv – --------– --------13 29 13 ˜ ˜ ˜ v 3i + j vi 10 ˜ ˜ b Reject, because magnitudes different. 5 ---------- i 29 ˜

iii

21 a

i 5

ii

26

5 1 - i – ---------- j iv --------iii – --35- i + --45- j 26 ˜ 26 ˜ ˜ ˜ v 2i + 3 j 13 vi ˜ ˜ b Reject, because magnitudes different.

22 a

2

b

58

b 5j ˜ 34 km /h e

23 a 3i ˜ d 031.0°

c 3i + 5 j ˜ ˜

25 a 5 2 d

11

26 a

35

b

b 5 2

c 3.64

1 Mathematics lecturer 2 Alice’s Adventures in Wonderland and Through the Looking-Glass 3 The daughter of the Dean of his college 4 12 cats

Exercise 7D — Resolving vectors — scalar and vector resolutes 1 a i

23 13 ---------------13

ii

23 41 ---------------41

b i

17 29 ---------------29

ii

17 10 ---------------10

10 ---------------c i – 13 10

17 ---------------ii – 26 17

13 d i – 6-----------13

ii – 6--5-

-----e i – 23 5

26 ---------------ii – 23 13

2 a i

1 ---------10

b i

82 ---------41

c i 0

e 7 14

f

13

c 2 6

3 d

62

30 a –4i + 12 j ˜ ˜ b 341.6° c 0.0417 h or 2.5 minutes

29

iii v ⊥ = ˜ 5 f i --------11

iii v ⊥ = ˜ 3 a 3.6 km

1 51 3 17 - i – ------ j iii v ⊥ = ------ i + ------ j ii v || = ----10 10 10 ˜ 10 ˜ ˜ ˜ ˜ ˜ ii v || = 8i + 10 j iii v ⊥ = 0 ˜ ˜ ˜ ˜ ˜ iii v ⊥ = – 3 i + 4 j ii v || = 0 ˜ ˜ ˜ ˜ ˜ 2 2 2 ---ii v || = 3 i + 3 j + 3 k ˜ ˜ ˜ ˜ 4 1 5 --- i + --- j – --- k 3˜ 3 3˜ ˜ ------ ( 2i + 3 j + 4k ) ii v || = – 21 29 ˜ ˜ ˜ ˜ 4 ------ ( 25i – 6 j – 8k ) 29 ˜ ˜ ˜ 15 5 5 - i + ------ j – ------ k ii v || = ----11 ˜ 11 11 ˜ ˜ ˜ 28 17 ------ i + ------ j – ------ k – 15 11 ˜ 11 ˜ 11 ˜ b 0.2 km or 200 metres

4 316 metres

Exercise 7E — Time-varying vectors

Exercise 7C — Multiplying two vectors — the dot product 1 23.99 2 Dot product = 24; more accurate, since no angle needed 3 a 45 b 12 c −36 e 1 f −20 g 0 4 E 5 C 6 −36 9 and 10 Check with your teacher. 11 D 12 B 13 D a −12 b 2 c 7 14

2 ------3

iii v ⊥ = ˜ 21 e i – ---------

27 CD = i – 7 j – 2k , EF = 2i – 14 j – 4k ˜ ˜ ˜ ˜ ˜ ˜ 28 a 20i , –15 j , 20i + 15 j b 25 c 053.1° ˜ ˜ ˜ ˜ 29 53.1°, −36.9° Difference = 90°

x 1 a y = – --2 c y = 4( x – 3 )2 2 B 3 a y = 4x2 + 2x

b y = –3x – 3 x3 d y = ----8 b

y

d −26 h 0 8 12

x

0

d −25

b 87° c 81° d 109° 17 E 18 a = 1 48 20 4i + 8 j ------ i – ------ j 21 64 5˜ 5 ˜ ˜ ˜

4 a y = x2 − 4x + 3

b

y Note: x ≥ 1 0 3 x –1 1 (2, –1)

7A

➔

15 a 107° 16 D 19 a = − 3--5-

History of mathematics

d i

24 329.0°

611

answers

Answers

7E

answers

612

Answers

5 a x2 + y2 = 1

b

c Period = π

y 1

2 19 y = ----2- – 2 , hyperbolic x

y

1 x

0

–1

18 D

1

6 a x2 + y2 = 9

y 3

b –3

c Period = π 3 x

0 –3

2

2

7 a (x − 1) + (y + 2) = 1

b

Modelling and problem solving 1 a 6i – 12 j + 12k b 1--3- ( i – 2 j + 2k ) ˜ ˜ ˜ ˜ ˜ ˜ m c 2i + 8 j + k + ---- ( i – 2 j + 2k ) ˜ ˜ ˜ ˜ 3 ˜ ˜ 1 --- ( 10i 3 ˜

+ 16 j + 11k ) ˜ ˜ e 7.28 km f 1080 km/h d

y 0 –1

1

2 x

–2 –3

c Period = 2π x2 8 ----- + y 2 = 1 9

y

0 –1

g 720 km/h

2 a (3, 5.5, 0) b – 3i + 4k ˜ ˜ d 66 cm3 = 0.066 litres e 3i + 5.5 j + 4k f 7.43 cm ˜ ˜ ˜ 3 a y X

1

–3

x

0 –2

–1

3 x

g 84.5°

Y(7, 7)

2i~ + 7j ~

y

x2 y2 9 ----- + ------ = 1 4 16

O

4

0

–2

2 x

–4

10 a y = 1 – x ; u ( 0 ) = i ; t = 2, u = 1--2- i + 1--2- j ˜ ˜ ˜ ˜ ˜ b y = (x – 1)2, x ≥ 2 ( x – 3 )2 ( y + 1 )2 c ------------------- + ------------------- = 1 4 9 2 ------ , y = --11 x = 19 7 7 12 y = x2; No, since v is always ‘ahead’ of u . ˜ ˜

Chapter review A 2 200 3i + 200 j + 40k ˜ ˜ ˜ B 4 A a ( 5 + 5 2 )i + ( 5 + 5 2 ) j b 17.07 km ˜ ˜ D 7 E 8 C 9 D E 1.3909 a – i – 4j b 7i – 6 j ˜ ˜ ˜ ˜ 3 5 - i – ---------- j d --------c −17 34 ˜ 34 ˜ e 135° 13 56.1°, 111.8°, 42.0° 1 3 5 6 10 11 12

9 ± 69 14 ------------------2 15 B 16 B 2 3 1 - i + ---------- j – ---------- k 17 a --------14 ˜ 14 14 ˜ ˜ 1 b v || = --2- ( 2i + 3 j – k ) , v ⊥ = – 1--2- ( j + 3k ) ˜ ˜ ˜ ˜ ˜ ˜ ˜

5i~

x

Z

b 2i + 7 j, – 5i c 7i + 7 j, –3i + 7 j ˜ ˜ ˜ ˜ ˜ ˜ ˜ 29 d 2-----------e 74.1° f 2i 29 ˜ g (−2, 7) h 35 square units N 4 a 5

3

b 5.83 m/s c 059° d 360 m e 600 m f Same result, except bearing = 180° − 059° = 121° 5 a 14.34 i + 20.48 j + 8 k b 26.25 m c 21.3° ˜ ˜ ˜ 1 6 a −3 i − 9 j − 8 k b ------------- (−3 i − 9 j − 8 k ) ˜ ˜ ˜ ˜ ˜ ˜ 154 c 5 2 km

d 744.6 km/h

CHAPTER 8 Vector applications Exercise 8A — Force diagrams and the triangle of forces 1 a

b

~N

D ~

Book

Ball W ~

W ~

c

d

~N ~F

Car W ~

~A

~N F ~

A Boat ~ W ~

e

~F

Sliding object

h

W ~

W ~

j

~D Ball moving down

~D W ~

W ~

2 a C 3 a C 4 a i 9i + 4 j ˜ ˜ b i 97

b E b E

ii –6i – 3 j iii ( 2 2 – 3 5 ) i – j ˜ ˜ ˜ ˜ ii 3 5 iii 54 – 12 10 c i – 9 i – 4 j ii 6i + 3 j iii ( 3 5 – 2 2 ) i + j ˜ ˜ ˜ ˜ ˜ ˜ a i i + 2j ii ( 2 – 1 ) i ˜ ˜ ˜ b i 5 ii 2 – 1 c i 63.4° ii 0° d i – i – 2j ii ( 1 – 2 ) i ˜ ˜ ˜ a i 61.4 N at N9.4°W ii 61.4 N at S9.4°E ii – 50 i + 5 j b i 50i – 5 j ˜ ˜ ˜ ˜ c i – 4 i + 12 j ii 4i – 12 j ˜ ˜ ˜ d i 2.9 N at N26.9°W ii 2.9 N at˜ S26.9°E a 10 N b 120° 79.2° a 77° b 37.6 N c Z = 2 sin 60° × X ˜ ˜

5

6

7 8 9

History of mathematics Bubonic plague broke out and the university closed. Leibniz The laws of mechanics and gravitation He reformed the coinage and introduced measures to prevent counterfeiting.

1 2 3 4

Exercise 8B — Newton’s First Law of Motion 1 a 23.5 N b 8.6 N 2 a D b B 3 a Drag force F ~

c 23.5 N c C

Tugboat

due to water

T ~

Ship

d A

b 0N c 39 392.3 N

T ~

Tugboat

4 a ~T2

60°

30° ~T1

j ~

~A = 120 N at 40° to vertical

8g N ~

b T1 = 4g N; T2 = 4 3 g N

~F W ~ = 40g ~

b Av = 91.9 N down AH = 77.1 N left c R = ( F friction – 77.1 )i + ( N – 483.9 ) j = 0 ˜ ˜ ˜ with all forces in N. d 77.1 N e 483.9 N 7 a ~N b 0i + 0 j ˜ ˜ T ~ c – 7.4 i – 12.7 j N Mass ˜ ˜ d 7.4 N W ~ e 12.7 N 8 a

N ~ Mass

H ~

W ~

9 10 11 12 13 14

b – 7.4i – 12.7 j Ν ˜ ˜ c 7.4i – 4.3 j N ˜ ˜ d 17 j N ˜

0.77 a 7600 N b 18 620 N c 0.41 0.07 a 58.8 N b 45.04 N c 37.8 N d 0.84 0.7 9.6 N

Exercise 8C — Momentum 1 a 40 N s east b 600 N s south c 7.5 N s north d 32 500 N s west e 978 N s north f 113 333 N s north 2 a 80 m/s b 37 500 m/s c 20 m/s 3 a 50 kg b 1980 kg 4 a 38 N s east b 22 N s east c 31 N s 14.9° N of E d 37.1 N s 6.2° N of E e 25 N s 13° N of E 5 2700 N s 6 a 1000 kg m/s b −500 kg m/s 7 a −40 kg m/s b 48 kg m/s c 9500 kg m/s d −3556 kg m/s 8 5.3 m/s at 44° to Alan’s current direction. 9 a 6.6 m/s N b 6 m/s N c 3.4 m/s S 10 3.7 m/s at 58.6° to Alan’s current direction. 11 a 2.2 at 47.9° b 5.9 at 71.4° c 1.67 at 65.4° d 2.4 at 234° 12 a 4.5 N s b 3 m/s

Exercise 8D — Relative velocity 1 16.5 km/h at 14° S of E 2 301.5 km/h at 5.7° E of N

7E

➔

~i

~N

~N ~F Sliding body

Ball moving up

b 41.4° c 245 N d –278 i + 245 j N; 278 i ˜+ 245 j ˜ N ˜ N f˜ 8.0 m e 370.6

W ~

6 a

Body at rest

i

Speaker

W ~

~N

~F

~Tright

~Tleft

~A Accelerating car

W ~

g

5 a

N ~

f

N ~

613

answers

Answers

8D

answers

614

Answers

CHAPTER 9 Sequences and series

17.3 m/s at 52.1° N of W 72.5° 13 km/h a 44.7 m/s at 26.6° W of N b 5 m/s N c 17.8 m/s at 7.5° W of N d 32.95 m/s at 4.9° S of E e 7.7 m/s at 52° S of E 7 15 km/h W 8 19.4 km/h 78° W of S 9 68.3 km/s at 2.5° with the vertical. 3 4 5 6

Exercise 9A — Arithmetic sequences 1 a, c, d, g, h 2 a a = 2, d = 5 d a = −123, d = 100 h a = 1--4- , d = 2--4-

Exercise 8E — Using vectors in geometry 1–3 Check with your teacher. 4 a −v ˜ b u +v ˜ ˜ c u −v ˜ ˜ d b .b = u2 + v2 ˜ ˜ ˜ ˜ c .c = u2 + v2 ˜ ˜ ˜ ˜ 5–12 Check with your teacher.

Chapter review 1 4 5 6 7 8 12 13 14 16 17 18

B 2 A 3 B 89.4 N at 153.4° to f 1 ˜ 138.6° b 26.3 N, 52.6 N c 26.6°, 63.4° 17.7 N D 9B 10 C 11 D 3 N s away from the wall 6.0 m/s at 43.7° with A’s original motion B 15 C 12.1 km/h at 65.6° to the bank 8.7 km/h from the north D

Modelling and problem solving 1 7 minutes 46 seconds 2 60 m 3 a 2.96 N down b 1.96 N down c 0.96 N down 4 a 0.038 b 193.5 m 5 a Check with your teacher. b 109.5° c i p=

1 ------2

1 ; r = − -----2

5 2 ii ---------- units 3

6 a y X (3, 5)

Y (9, 5)

Z (6, 0) O

x

b YX = −6 i ; ZY = 3 i + 5 j ˜ ˜ ˜ c OY = 9 i + 5 j ; ZX = −3 i + 5 j ˜ ˜ ˜ ˜ d 91.9° e 29.1° f g (12 3--8- , 10 5--8- )

h 31 7--8- square units

c a = 0, d = 100 g a = 1--2- , d = 1

3 a 122 b 2900 c −219 4 a 103 b 1777 c −60 d −217 e −152 5 a 820 b 270 c 667 6 A 7 B 8 B 9 B 10 a, d, e 11 a a = 2, d = 2 d a = not specified, d = 2 e a = 8, d = 8 12 a 7th b 7th c 8th d 5th 13 tn = 13 + 10n 14 tn = 37 + 3n 15 a tn = 8.5 + 3.5n b 358.5 metres 16 a 56 b 2, 5, 8 17 a 170 b −4, 2, 8 18 $9375 19 $414 000 20 363.8 cm 21 1135 22 a $10 600 b $136 000

Exercise 9B — Geometric sequences 1 b, c, e, f, h, i, j 2 b a = 1, r = 4 c a = −1, r = 2 e a = 4, r = −3 f a = −6, r = −10 h a = 1.2, r = 2 i a = 7, r = 1--21 1 j a = --2- , r = --23 a 20 155 392 b 1 048 576 c 295 245 d 11.867 494 81 e −131 072 4 a 39 366 b 6144 c 32 768 d −32 768 e −12 582 912 f −708 588 5 a 12th b 13th c 9th d 16th 6 a 531 440 b 686 285 c 36 043.7 d 274 576.3 7 D 8 D 9 B 10 A 11 a 1, 2, 4, 8, 16 b 2 c 2048 12 20 million, 10 million, 5 million, 2 1--2- million, 1 --14- million, 625 000, 312 500 13 a tn = 2n − 1 b 2048 14 a tn = 10 000 × 0.85n − 1 b $2724.91 15 a tn = 6n − 1 b 1296 c 6 n−1 16 a tn = 60 × 1.08 b 88.16 m c 8th year 17 Check with your teacher.

Exercise 9C — Applications of geometric sequences 5j ˜

1 a 5.67 b 17th year 2 a $2146.53 b Year 9 3 Pn = 10 000 × (0.9)n − 1

c 12.7 tonnes c $5646.65

4 5 6 7 8 9 10 11 12

a $19 317.32 b $6317.32 a 14 147 b 16th year a 123 b 10 weeks a 2.85 m b Year 11 a 0.96 m b 32nd beam C a $400 640.74 b Year 9 c 2 636 196.56 a 55 773 b 9th year a $16 105.10 b $16 288.95 c $16 386.16 d $16 453.09 13 a $22 076.26 b $26 897.78 c $32 772.33 d $53 701.28 14 $8376.76 15 3 years 16 2 years 17 3 years 18 10 months 19 2 years

7 a

tn 30 25 20 15 10

615

answers

Answers

5

b

0

1

2

3

4

5

6

7

8 n

0

1

2

3

4

5

6

7

8 n

tn 6 5 4 3 2

Exercise 9D — Finding the sum of an infinite geometric sequence 1 a 100 e

5 --4

i 200 2 a c e g i 3 a c e 4 a

5 7 8 9 10

c 10

d

f 3 1--3-

g 2.5

h −7.5

2 4--95 --9

c

4 --9

d 1 1--3-

g

14 -----99

h

57 -----99

6

192.5 m

22 -----27

,

5 0

8 --9

f 8 2--3-

d

5

6

7

8

1

2

3

4

5

6

7

8

e

0

1

2

3

4

5

6

7

n

tn 8 × 104 6 × 104 4 × 104 2 × 104

f

8 n

tn 6000 5000 4000 3000 2000 1000 0

1

2

3

4

5

6

7

8 n

8D

➔

Arithmetic sequence with a = 0 and d = 1 Arithmetic sequence with a = 10 and d = −2 Geometric sequence with a = 10 and r = 1.5 Geometric sequence with a = 20 and r = 0.5 Arithmetic sequence with a = 4 and d = −0.5 Geometric sequence with a = 100 and r = 0.8

4

tn 40 35 30 25 20 15 10 5 0 –5

c $7 070 144.32

1 2 3 4 5 6

3

–15

b $1 000 000, $920 000, $846 400, $778 688, $716 392.96

Exercise 9E — Contrasting arithmetic and geometric sequences through graphs

2

–10

--------j 1 321 999

No — falls short by 3 1--3- metres Yes $500 000 25 mm too much -----a 23 25

n

1

–5

22 -----81

b

tn 15 10

40, 8, 1.6 12, 10.8, 9.72 6, −3, 1.5 −6, −1.2, −0.24 −15, 4.5, −1.35 12.5, 9.375, 7.031 25 48, 28.8, 17.28

262 --------495

c

5 j –45 ----11

b d f h j b d

e 3 7--9i

3 --2

b 100

10, 6, 3.6 6, 1.5, 0.375 4, −0.8, 0.16 9, −7.2, 5.76 −24, −14.4, −8.64 12.5, 6.25, 3.125 6, 1.5, 0.375 ,

1

9E

Answers

30 80 70 Amount ($)

Amount ($)

8 D 9 E 10 a $5500, $6000, $6500 b $5500, $6050, $6655 c 6500 6000

Legend Un Vn

5500 5000 1

2 Year

60 50 40 30

Vn = 10 × 2n – 1

10

Un = 10n

3

1

Amount ($)

150 000 140 000 Legend Un Vn

120 000 110 000 1

2 Year

3

12 18 000 17 000 16 000 15 000 14 000

CHAPTER 10 Permutations and combinations

Legend Un Vn

13 000 12 000

Exercise 10A — The addition and multiplication principles

11 000 1

13 14 15 16

2 Year

3

3 terms 0 terms 3 terms 4 terms

Chapter review 1 3 4 7 8 9 13 14 17 18 21 23 24 25 26 27 28 29

2 3 4 Term number (n)

Modelling and problem solving 1 a 18.5 tonnes b tn = 4.5 + 3.5n c 214.5 tonnes d 28th month e 40 months 2 a 1.1 b tn = 10 × 1.1n − 1 c 14.641 tonnes d 13th month e 213.84 tonnes 3 24th month 4 35 years

160 000

130 000

Legend

20

11

Amount ($)

answers

616

C 2 B a Yes, a = −123, d = 100 A 5 C 6 Term number 35 a 1080 b A 10 C 11 a Yes, a = 5, r = 1--2b B 15 D 16 a tn = 7.2 × 1.15n − 1 b c Year 12 B 19 C 20 A 22 C a $33 622.22 b E D 3 7--925 cm D A

b Yes, a = −5 1--4- , d = 3 B 12 000 B 12 No B 19.2 tonnes 17 5 years

A

1 a AB BA CA AC BC CB b 6 2 BG GB YB RB BY GY YG RG BR GR YR RY 3 ACB BAC CAB ABC BCA CBA 4 a 42 b 210

c 840

d 2520

5 a 24

c 12

d 24

6 a 49 7 a i 72 b 144 8 126 9 C 10 C 11 D 12 100 13 6 14 48 15 256 16 1296 17 1080 18 48 19 a 1000

b 6

b 252 c 16 ii 72

b 27

e 1320

c 271 272 273 281 282 283 291 292 293

371 372 373 381 382 383 391 392 393

Exercise 10C — Arrangements involving restrictions and like objects

471 472 473 481 482 483 491 492 493

20 a 200 b 40 c 50 d 290 21 a 13 230 b 17 640 Jack may wear 13 230 outfits with a jacket or 4410 outfits without a jacket. Therefore he has a total of 17 640 outfits to choose from. The assumption made with this problem is that no item of clothing is exactly the same; that is, none of the 7 shirts is exactly the same.

Exercise 10B — Factorials and permutations 1 a b c d 2 a d g 3 a 4 a b c d 5 a b c 6 a 7 8 9 10 11 12 13 14 15 16 17 18 19

4×3×2×1 5×4×3×2×1 6×5×4×3×2×1 7×6×5×4×3×2×1 24 b 120 c 720 3 628 800 e 8.717 829 12 × 1010 f 362 880 5040 h 6 3024 b 151 200 c 840 d 720 n(n − 1)(n − 2)(n − 3)(n − 4) (n + 3)(n + 2) 1 ------------------------------------n(n – 1)(n – 2) 1 ------------------------------------------------------( n + 2 ) ( n + 1 )n ( n – 1 ) 8 × 7 = 56 7 × 6 × 5 × 4 × 3 = 2520 8 × 7 × 6 × 5 × 4 × 3 × 2 = 40 320 9! 5! 18! ----- = 60 480 b ----- = 20 c -------- = 1 028 160 3! 3! 13! 27 907 200 b 639 200 c 1 028 160

a 56 3024 2184 358 800 120 8190 3360 362 880 479 001 600 D E a 5P3 = 60

3.6 × 109 20 3024 48 180 110 14 D b 24 b 365P30 b b b b b b

c c c c c

4.0 × 1010 60 6720 72 360

c OYSTER

Exercise 10D — Combinations 8P 3 1 a --------3!

19 P b ----------22!

1P c --------11!

5P d --------00!

2 a 8C2 b 9C3 c 8C0 d 10C4 3 a 1 b 20 c 120 d 220 4 1365 5 252 6 a 495 b 11 c 1 d 1 e 54 264 f 120 g 100 h 680 7 a 2 598 960 b 65 780 c 65 780 d 2 467 400 8 560 9 100 10 59 400 b 120 11 a 28 12 201 376 13 D 14 C 15 a 120 b 10 days 16 a 57 b 4 days 6 hours 17 a 15 b 1 day 4 hours 18 a 8 145060 b 11 480 c 820 19 a i 220, 220 ii 6435, 6435 iii 10, 10 iv 56, 56 v 1, 1 b The value of nCr is the same as nCn − r .

20 a

7

b

7

21 a

6

b --------- = 360 2

P4 = 120

P4 = 6P 6

840

c

5

c

7

P5 = 120 P7 = 5040

1 2 3 4 5 6 7 8 9 10

a 45 b 120 c 120 d 210 e 105 a 720 b 252 c 10 d 120 e 2.4 × 1018 24 24 376 992 a 210 b 126 c 84 d 140 a 126 b 56 c 21 d 70 a 8008 b 5005 c 5005 d 4004 D D

9E

➔

5

P6 = 720

360 83 160 10 1260 27 720 1 307 504 a 5.4 × 1010 a 120 a 30 240 a 120 a 1680 a 1320 B a 720 a 36530

Exercise 10E — Applications of permutations and combinations

b

P3 = 210

1 2 3 4 5 6 7 8 9 10 11 12 13 15 16

answers

617

Answers

10E

answers

618

Answers

11 a 28 b 9 17 27 41 12 14, 9 17 27 41 12 37, 9 17 27 41 12 34 c 6 12 a 84 b 7 15 25 32 10 12, 7 15 25 32 10 35, 7 15 25 32 10 37 c 10 13 a 38 760 b 1140 c 34 220 d 39 010 800 14 a 1 b i 38 760 ii 50 063 860 iii 1 940 475 213 600 iv 4 191 844 505 805 495

Exercise 10F — Pascal’s triangle, the binomial theorem and the pigeonhole principle 1 Row 0 1 2 1 3 1 4 1 4 5 1 5 6 1 6 15 7 1 7 21 8 1 8 28 56

1 1

1 2

3

1 3

6 10

10 20

35

5 6 8

9

35 70

21 56

16 Calculating machine Puy de Dôme; his brother Probability Brain tumour

1 4 5 8 12 15 16 20 21

C 2 B 3 A a 180 b 648 D 6 A 7 C 2000 P2, 19P6, 12P9 9 A 10 12 11 E B 13 19C6, 22C15, 2000C2 14 175 a 325 b 6 c 676 C 17 756 756 18 B 19 D a 210 b 10C0 10C1 10C2 10C3 10C4 . . . 10C10 c 1024 a 405x8 b 3240x7 c 196 830x

Modelling and problem solving 1 50 878 2 2 944 656 3 a 2 598 960 b 1287 c 65 780 d 1584 4 a 6 096 454 b 2 760 681 c 442 890 d 2 048 200 e 850 668 f 5 245 786 g 18 360 5 a 120 b 26 6 a 12 000 b 5 years

1 6 28

CHAPTER 11 Dynamics

1 7

1 8

1

70 1 9 36 84 126 126 84 36 9 1 45 8 220 10 C0 10C1 10C2 10C3 10C4 10C5 10C6 10C7 10C8 10C9 10 C10 1 10 45 120 210 252 210 120 45 10 1 a x2 + 2xy + y2 b n3 + 3n2m + 3nm2 + m3 c a4 + 12a3 + 54a2 + 108a + 81 a 80x b 56p5q3 c 4608x2 B 7 D a i 1 ii 2 iii 4 iv 8 v 16 vi 32 b i The sum of the elements in each row of Pascal’s triangle is a power of 2: Row Sum 0 20 = 1 1 21 = 2 2 22 = 4 3 23 = 8 4 24 = 16 5 25 = 32 ii ‘The sum of the elements in the nth row of Pascal’s triangle is 2n.’ 26 = 64 13 26 14 13

History of mathematics 1 2 3 4 5

1 5

15

2 a b c 3 a b c 4

1 4

Chapter review

Exercise 11A — Displacement, velocity and acceleration 1 a v (t) = 6 i + (3 − 14t) j , a (t) = −14 j ˜ ˜ ˜ ˜ ˜ b r (0) = 0 m, v(0) = 3 5 m/s ˜ ˜ c i r (2) = 12 i − 22 j m ˜ ˜ ˜ at an angle of 13°30′ to ii 25.7 m/s downwards the vertical iii 15°7′ d Yes, 14 m/s2 downwards 2 a, b Check with your teacher. 2 1 - (1225 − x ) c y = ----25 3 a

i v (t) = 2 i − j , a (t) = 0 ˜ ˜ ˜ ˜ ˜ ii r (0) = 5 j m, v(0) = 5 m/s ˜ ˜ iii y = 5 − 1--2- x

i v (t) = i + (6 − 2t) j , a (t) = −2 j ˜ ˜ ˜ ˜ ˜ ii r (0) = 0 m, v(0) = 37 m/s ˜ ˜ iii y = 6x − x2 c i v (t) = − i − 6t j , a (t) = −6 j ˜ ˜ ˜ ˜ ˜ ii r (0) = 10 j m, v(0) = 1 m/s ˜ ˜ iii y = 10 − 3x2 d i v (t) = 3 i + (5 − 4t) j , a (t) = −4 j ˜ ˜ ˜ ˜ ˜ ii r (0) = 0 m, v(0) = 34 m/s ˜ ˜ iii y = --9x- (15 − 2x) b

e

i v (t) = i + 6t j , a (t) = 6 j ˜ ˜ ˜ ˜ ˜ ii r (0) = 4 i m, v(0) = 1 m/s ˜ ˜ 2 iii y = 3(x − 4)

i v (t) = 2t i − j , a (t) = 2 i ˜ ˜ ˜ ˜ ˜ ii r (0) = 0 m, v(0) = 1 m/s ˜ ˜ iii y2 = x g i v (t) = 2t i − 4 j , a (t) = 2 i ˜ ˜ ˜ ˜ ˜ ii r (0) = 0 m, v(0) = 4 m/s ˜ ˜ iii y2 = 16x 1 - i − 6t j , t ≠ 0; h i v (t) = -------2 t ˜ ˜ ˜ 1 - i − 6j, t ≠ 0 a (t) = − ---------4t t ˜ ˜ ˜ ii r (0) = 0 m, v(0) is undefined ˜ ˜ iii y = −3x4 a t = 5 seconds b (−3, 18) c yA = (x − 1)2 + 2, yB = (x + 8.5)2 − 12.25 a v (t) = 2 i + 2(1 − t) j , a (t) = −2 j ˜ ˜ ˜ ˜ ˜ b 2 2 m/s upwards at an angle of 45° to the vertical c i 37 m ii 2 5 m/s downwards at an angle of 26°34′ to the vertical iii 26°34′ a, b Check with your teacher. c t = 5 seconds 12 - − 1 a P and Q do not collide. c y = ----x 1 a Check with your teacher. b − --5- i + 3 j m/s ˜ ˜ 2 4 - i + 3 j , a (5) = --------- i c v (5) = − ----25 ˜ 125 ˜ ˜ ˜ ˜ d Approaches zero a 2i + 3 j b 145 m/s c 85°14′ ˜ ˜ d i 65 m/s downwards at an angle of 7°08′ to the vertical ii 4 i + 7 j ˜ a r 1(0) = 0 ,˜ r 2(0) = 80 i + a j b a = 15 ˜ ˜ ˜ ˜ ˜ c (50, 6.25) d 100°18′ 2 1 1 - (x − 50) + 26 --a b = 25, c = −5 b y = − ----80 4 c 81.4 m 9 - m when α ≈ 63°26′ d Rmax = 76 ----16 f

4 5

6 7 8

9

10 11 12

Exercise 11B — Projectile motion

d v (t) = (t2 − 1) i + (−3t − 1) j , ˜ ˜ ˜ r (t) = ( --13- t3 − t + 1) i + (− --32- t2 − t + 1) j ˜ ˜ ˜ 1 2 13 ------ ) i + ( --- t + t − ------ ) j , e v (t) = (−5t + 15 2 2 6 ˜ ˜ 15 1 3 1˜ 2 13 - t + 3) i + ( --- t + --- t − ------ t + 4) j r (t) = (− --52- t2 + ----2 6 2 6 ˜ ˜ ˜ ------ ) j , f v (t) = (t + 7--2- ) i + (− 1--2- t2 + 6t − 65 6 ˜ ˜ ˜ 65t + 12) j r (t) = ( --12- t2 + --72- t − 5) i + (− --16- t3 + 3t2 − ----6 ˜ ˜ ˜ 2 23 5 2 5 g v (t) = (t + t − -----6- ) i + ( --2- t − 3t + --3- ) j , ˜23 ˜ 5 3 3 2˜ 5 - t) i + ( --- t − --- t + --- t + 8) j r (t) = ( --13- t3 + --12- t2 − ----6 ˜ 6 2 3 ˜ ˜ ------ ) j , h v (t) = (3t − 6) i + (− 5--2- t2 + 12t − 35 2 ˜ 11 ˜ 2˜ 35 5 3 116 - ) i + (− --- t + 6t − ------ t + --------- ) j r (t) = ( --32- t2 − 6t − ----6 2 ˜ 2 3 ˜ ˜ 3 a v (t) = 12 i + (15 − gt) j ˜ ˜ ˜ b 19.2 m/s at an angle of 51°20′ above the horizontal. c 11.5 m 3.06 seconds d 19.2 m/s downwards at an angle of 51°20′ to the horizontal 4 a v (t) = 15 i + (30 − gt) j ˜ an angle of˜63°26′ above the ˜ b 33.5 m/s at horizontal c 45.9 m d No, as it falls short by about 40 cm. 5 b r (t) = 18 3 t i + (50 + 18t − --12- gt2) j ˜ ˜ ˜ c Maximum height is 66.5 m which is 16.5 m above the launch platform. d 47.7 m/s downwards at an angle of 49°12′ to the horizontal e 172.1 m 6 a r A(t) = 4(t − 1) i + (3 + 5t − t2) j , ˜ ˜ ˜ r B(t) = 3(8 − t) i + (6 − 7t + 2t2) j ˜ ˜ ˜ b Particles don’t collide but they do pass through two common points at different times. At (13.70, 5.55), one particle arrives when t = 4.43 s and the other when t = 3.43 s. At the point (18.20, −0.06), one particle arrives when t = 5.55 s and the other when t = 1.93 s. c yA = 3 + yB = 6 +

1 ------ (x + 4)(16 − x), 16 1 --- (24 − x)(27 − 2x) 9

7 a r λ(t) = (5t − 14) i + (32t − 45 − 4t2) j ˜ b When t = 5 s, the˜ particles do collide˜at (11, 15). Particles appear to collide at (9.15, 17.4) but actually pass through this point at different times; so they don’t collide. Particle λ at t = 4.63 s and particle ξ at time t = 3.15 s. 8 b 0.618 when θ ≈ 58°17′ c 35.6 m/s 9 θ ≈ 17°46′ and θ ≈ 77°56′ 10 21.429 m 11 100 m/s, 192 m 12 Yes, they meet at (38.4, 9.2) when t = 2 s. 13 20 m/s at an angle of approx. 36°52′ 14 75.431 m, approx. 4.64 s 15 b Rmax =

2

V -------------------------(1+sin θ )g

when φ = 45° +

θ --2

10E

➔

1 a r (t) = (t + 3) i + 2 j ˜ ˜ ˜ − 3t) j b r (t) = (t2 + 1) i + (1 ˜ ˜ ˜ c r (t) = 2(2t − 1) i + --12- (7t2 − 9) j ˜ ˜ ˜ d r (t) = (3t + 1--2- t − 7) i + (−2t+ 9) j ˜ ˜ ˜ e r (t) = ( 3--2- t2 − 1) i + (4t + 1--2- t2 + 1) j ˜ ˜ ˜ f r (t) = − --12- t(t − 4) i − --13- t(t2 + 18) j ˜ ˜ ˜ g r (t) = (15t) i + (10t − 3--2- t2 + 20) j ˜ ˜ ˜ h r (t) = (−t2 + 3) i + (5t − --1t- − 11) j ˜ ˜ ˜ 2 a v (t) = (2t + 1) i − 3 j , ˜ ˜ r (t) = (t2 + t − 1) i +˜ (−3t + 5) j ˜ ˜ ˜ b v (t) = (3t + 2) i + (−5t − 1) j , ˜ ˜ r (t) = ( 3--2- t2 + 2t) i + (− 5--2- t2 − ˜t) j ˜ ˜ ˜ c v (t) = (−t + 3) i + ( 1--2- t2 − 4) j , ˜ ˜ r (t) = (− 1--2- t2 + 3t + 2) i + ( 1--6-˜t3 − 4t − 3) j ˜ ˜ ˜

619

answers

Answers

11B

answers

620

Answers

Exercise 11C — Motion under constant acceleration 1 2 3 4 5 6 7 8 9

a a a a a a a a a

4 m/s2 6.5 m/s2 80 m/s 18 m/s 58.8 m/s 19.6 m 2.83 s 78.4 m 90 m

10 21.08 m/s or 11 12 13 14 15 16 17 18 19 20 21 22 23 24

b b b b b b b b b 7.14 s

32 m Approx. 1.54 s Approx. 8.94 s −0.6 m/s2 176.4 m 4s 27.72 m/s 39.2 m/s c 42 m/s

14 15

20 10 ---------------3

a 2 m/s2 b 144 m c a 12.5 m/s b 4.8 s a −5 m/s2 b 22 m c 3.2 s 12 m/s a 24.5 m b 17.5 s a 8.04 m/s b 1.1 m c a 16.1 m/s b 27.23 m a i −19.6 m/s ii −27.7 m/s b i −20.2 m/s ii −28.15 m/s a 78.4 m b 2.53 s longer 0.1044 s 4 s, 21.6 m above the ground 5.83 s 24.04 m 8.9 s

8 c y = 25 − (x − 1)2 9 D 10 C 11 A 12 v (t) = t i + (3t − t2) j m/s, 2 ˜ ˜ r (t) = --12- t2 i + t---6- (9 −˜2t) j m ˜ ˜ ˜ 13 a v (t) = 6 i + (45 − gt) j m/s ˜ ˜ b 3 229 m/s at an angle˜ of elevation of 82°24′ c 101.25, 4.5 s

7.5 s d 25.6 m

1.77 s iii t = 8 s iii t = 2.4 s

Chapter review 1 B 2 A 3 D 4 B 5 C 6 a v (t) = 6 i + (12 − 6t) j m/s, a (t) = −6 j m/s2 ˜ ˜ ˜ ˜ ˜ b Origin, 6 5 m/s c i 12 2 m ii 6 m/s to the right iii 45° 7 a 5 seconds b (5, −5) c yA = 4x − x2, yB = x − 10

16 17 18 19 20 21

d 3 229 m/s downwards at an angle of 7˚36′ to the vertical b r (t) = 12t 3 i + (60 + 12t − 1--2- gt2) j ˜ ˜ 5 cm ˜ c Clears by about b Expression is positive for 26°34′ ≤ θ < 90°. Maximum is 0.618 when θ ≈ 58°17′. c 30.845 m/s a 30.625 m b t ≈ 1 s and 4 s E C B a 2 min 14 s b 16.2 s a v (m/s) b 36.25 s c 145 m B 5 d 16 m A 4 0

_6 8.5

t (s)

Modelling and problem solving 1 a Car A b 8.4 m c i −2 m/s2 ii 3.58 m/s2 d i 23.62 m/s ii 28.33 m/s 2 u∆t 1 - + --- a(∆t) 2 a --xs = --14b x = -------s = --12- a(∆t)2 + u∆t 2 8 c When a = 0 3 a k = 4.96 × 1012 12

× 10 b v(r) = 2  4.96 --------------------------- – 2.69 × 10 r

c 1.84 × 106 m

6 

Index

Abel, Niels Henrik 183 Abelian groups 185, 265 absolute value of a number 49–50 solving equations using 51–4 acceleration motion under 540–4 vector expressions for 514–21 addition principle 461–4 adjoint matrix 229 algebraic structures 178 applied forces 355 Argand diagrams 80–1, 97 argument of complex number 95, 97–103 arithmetic sequences definition 396–8 finding terms rule 398–9 graphs of 432 listing terms of 400 simple interest 433–7 sum of terms 401–3 arrangements 462 in a circle 472–3 involving restrictions and like objects 476–9 see also permutations associativity of operations 181, 264 augmented matrices 210 axioms related to operations and whole numbers 188 binary number systems 17 binary operations 180 binomial theorem 499–501 Cartesian form of complex numbers 106 Cayley, Arthur 194 Cayley table 179–80, 186–8 circle, number of arrangements in 472–3 cis θ 103 closure of operations 181, 263 coefficient of friction µ 369 coefficient matrix 155 cofactor matrix 228 collision momentum 381 combinations 482–7 applications of 490–4 combinatorics 460 common differences 396 common ratio 407 commutativity of operations 182 complex conjugate numbers 87–8 complex number multiplication by constant (scalar) 82 by i 85

in polar form 111 of two numbers 83–4 complex numbers addition 81–2 applications 120–1 Argand diagrams 80–1 argument 95–9, 100 in Cartesian form 106–7 conjugates 87–8 definition 76–9 division 88–9 equality of two numbers 84–5 geometrical representation 95–6 imaginary part 77 modulus 96–7, 100 multiplicative inverses 89–90 Pascal’s Triangle 114–15 plotting 84 polar form 103–6 powers 114–17 in quadratic equations 80 real part 77 simple algebra 93–5 subtraction 81–2 composition of transformation 263 composition and transformations 192–3 compound interest 421–3 compared with simple interest 434–7 conformable matrix 137 congruent transformations 270, 276 conjugate pairs of surds product of pair 34 rationalising denominators using 44–7 conjugates of complex numbers 87–8 conservation of momentum 375–8 constant acceleration, motion under 540–4 coterminal angles 96 counting paths 498 Cramer’s rule 234–7 cryptography 199–200 cube roots 7 curves, images of 266–8 translation of 255 cyclic groups and subgroups 189–91 De Moivre, Abraham 108 decimals, recurring 11–13, 429–30 determinants of matrices 148 2 × 2 and 3 × 3 matrices 222–3 Cramer’s rule 234–7 expansion of 225–6 finding area of triangle 243

621

622

Index

determinants of matrices (continued) finding equation of line 243 properties 224–5 dilations 284–9 about origin 288–9 dilation factor 284 parallel to x- and y-axes 285–8 displacement, velocity and acceleration 514–21 Distributive Law 32–4 division of complex numbers 88–9 of surds 36–8 Dodgson, Charles Lutwidge 331 dominance matrices 164–8 dot product of two vectors angle between vectors 327 calculation 324 perpendicular and parallel vectors 327–8 properties 324–5 and unit vectors 325 using calculator 326 DOTS (difference of two squares) identity 34 equation solving using absolute values 51–4 using matrices 133–4 equilibrium state 364 equilibrium values in input–output analysis Escher, Maurits Cornelius 290 factorials 467–70 Fibonacci numbers 445 Fibonacci Sequence 442–3 generating terms in 443–4 golden ratio 444–5 field forces 354 FOIL method 32 force diagrams 356–9 force vector diagrams 364 forces resolving into components 365–9 resultant of 356 triangle of 359–60 types 354–5 friction force 369–70 functions in groups 193–4 Gauss, Carl Friedrich 216 Gaussian elimination to find inverse of matrix 213–15 to solve simultaneous equations 209–13 using calculator 217–21 geometric sequences common ratio 407 in compound interest 421–4, 433–7 definition 406–8

206–8

finding terms of 408–12 graphs of 432 in growth and decay 418–21 sum of terms 413 see also infinite geometric sequences geometric transformations, and matrix algebra 250–6 geometry, vectors in 385–8 golden ratio (golden mean) 444–5 graphics calculator use addition of elements in dominance matrices 167–8 combinations 486–7 comparison of simple and compound interest 434–7 complex numbers 93–5 complex numbers in Cartesian form 106–7 complex numbers in polar form 105–6 compound interest 434–7 dot product of two vectors 326 factorials 468 Gaussian elimination 217–21 graphing original and translated image 256–8 listing terms of arithmetic sequence 400 Mandelbrot Sets 447–9 matrix equations 156–8 matrix multiplication 162–4 matrix operations 237–42 modulus and argument of complex numbers 101–3 Pascal’s Triangle coefficients 115 permutations 472 plotting missile flight 533–5 plotting particle trajectory 517–19, 520–1 roots 7 scalar and vector resolutes 334–6 simultaneous equations 242 sum of arithmetic sequence 402–3 terms in Fibonacci Sequence 443–4 terms of geometric sequence 411–12, 419–20 unit vector in direction of vector 316–17 vector functions of time 341–3 vector magnitude and direction in two dimensions 311–12 x- and y-components of vector 313–14 gravity, acceleration under 540–4 group theory applications 197–8 and linear transformations 263–8 groups Abelian 185 cyclic 189–91 inner and outerfunctions 193–4 permutations 191 properties 184–8 terminology 180–3 transformations of shapes 192–3 Hamilton, William Rowan

118

Index

idempotent matrices 149–50 identity element (IE) 181–3, 264 identity matrix, multiplicative 139–40 images, graphing using calculator 256–8 images of curves non-singular transformations 266–7 singular transformations 267–8 images as result of translation 251 imaginary number i 76 multiplication by 85 imaginary part of complex numbers 77 inequations, solving 55–62 inertial mass 364 infinite geometric sequences converting recurring decimals to fractions 429–30 sum of 427–9 see also geometric sequences input–output analysis 206–8 integers 4 interest, simple and compound 421–4, 433–7 inverse of 3 × 3 matrix 228–32 of set elements 181–2 inverse of matrix 147–9 by Gaussian elimination 213–15 inverse transformations 264–5 irrational numbers 4–5 proof by contradiction 18–19 surds 17–18 Leontief matrix 207 linear equation solving using Cramer’s rule 234–7 using determinants 222–37 using inverse matrices 206–21 linear transformation matrix 261 linear transformations 259–62 and group theory 263–8 Lotto systems 490–1, 493–4 Mandelbrot Set 446–9 mass and weight 364–5 matrices addition 130 adjoint 229 applications 145–6 augmented 210 coefficient 155 cofactor 228 conformable 137 determinants of 148, 222–7 dilation 285 dominance 164–8 equality of 133 Gaussian elimination in 209–15 identity 139–40

623

information summarising using 158–9 introduction 128–9 inverse of 147–9, 228–32 linear transformation 261 multiplication by scalar 131–3 multiplication of 137–40, 162–4 multiplicative inverse 146–7, 150–1 powers of 143–4 reflection 276 simple equation solving using 133–4 simultaneous equation solving using 155–8 singular 149–50 subtraction of 131 transition 159 transpose of 154 and vectors 337–8 matrix algebra transformations 250–1 translations 251–5 matrix applications input–output analysis 206–8 modelling and problem solving 247–8 matrix equations 133–4, 150–2 matrix operations, using calculator 237–42 maze, paths through 498 mediator 276 missile flight 531–5 modular arithmetic 17 modulo arithmetic 179–80 modulus of complex number 96–7, 100–3 of a number 49–50 momentum conservation of 375–8 definition 374–5 motion Newton’s First Law of 364–70 Newton’s Second Law of 524 of projectiles 514–35 under constant accleration 540–4 multiplication of complex numbers 82–4, 85, 111–14 of matrices 137–40, 150–1, 162–4 of matrix by scalar 131–3 of surds 27–30 multiplication principle 460–4 multiplicative identity matrix 139–40 multiplicative inverse 146–7 of complex number 89–90 negative numbers, square root of 76 Newton, Isaac (Sir) 363 First Law of Motion 364–70 Second Law of Motion 524 Newtonian dynamics, assumptions made in nilpotent matrices 149–50

356

624

Index

non-singular transformations 266–7 nth roots 7 number systems, other 16–17 real 3–5 operations on numbers 180 operations and whole numbers, axioms relating to

188

particles, in Newtonian dynamics 355 Pascal, Blaize 506 Pascal’s triangle 114, 499–501 Pauli matrices 198 perfect square identities 33 permutations 191, 470–2 applications of 490–4 see also arrangements pigeonhole principle 502–3 π (pi) 5 approximations for 63 place value number system 16 polar form, of complex numbers 96–106 position vectors direction 310 locating 317–18 magnitude 309–10 magnitude and direction using calculator 311–12 magnitude in three dimensions 318–19 relationship with matrices 309 resultant 314–15 in two and three dimensions 308–9 unit vectors 312–13, 315–17 x- and y-components using calculator 313–14 powers of complex numbers 114–17 of a matrix 143–4 projectile motion 524–36 proof by contradiction 18 quadratic equations, complex numbers in quaternions 197–8 radian measure 95–6 rational numbers 3–4 recurring decimals 11–13 rationalising denominators of surds 40–2 using conjugate surds 44–7 Real Number System classification 3–5 relationships between subsets 5 real numbers application and modelling 64–5 investigations 15 modulus 49–50 real part of complex number 77 recurring decimals 11–13 conversion to fractions 429–30 reflection matrix 276

80

reflection transformation 276 reflections in line y + x tan θ 280–1 in line y = x 279–80 with mediator not through origin in x-axis where y = 0 276, 277 in y-axis where x = 0 277, 278 relative velocity 382–4 resistive forces 355 resolution of forces 365–9 resolution of vectors scalar resolute 332–3, 334 using calculator 334–6 vector resolutes 333, 334 resultant force 356, 364 rewards allotment 440 roots, square, cube and nth 7 rotations general 272–5 special 270–1

282–3

scalar quantities 300 scalar resolutes 332–3 sequences arithmetic 396–403 geometric 406–13 Mandelbrot Set 446–9 set notation 7–8 shapes, changes in area and perimeter 441 shears, definition 291–2 parallel to x- and y-axes 292–5 simultaneous equations in input–output analysis 206–8 simultaneous equations solving using calculator 156–8, 242 using Gaussian elimination 209–13 using inverse matrices 206 using matrices 155–6 singular matrices 149–50 singular transformation 267–8 square roots 7 of negative numbers 76 statics 364 subgroups 189–90 successive translations 253–5 surds, addition and subtraction 24–5 conjugate pairs 34 definition 17–18 Distributive Law applied to multiplication of 32–4 division 36–8 multiplication 27–30 proof of irrationality 18–19 rationalising when denominator of fraction 40–2, 44–7 simplifying 21–2 squaring 29–30

Index

Taussky-Todd, Olga 142 ternary operations 180 time-varying vectors equation of path 339–41, 343–4 using calculator 341–3 transformations, congruent 270 definition 250–1 dilations 284–9 linear 259–68 reflections 276–83 rotations 192–3, 270–5 shears 291–5 transition matrix 159 translations, of a curve 255 definition 251–3 successive 253–4 transpose of a matrix 154 triangle of forces 359–60 unary operations 180 unit vectors 312–13, 315–17 and dot product 325 vector (force) diagrams vector quantities 300 vectors addition of 301–5

356–9

definition 300 dot product of 324–8 equality of 301 expressions for velocity and acceleration finding angle between 327 in geometry 385–8 and matrices 309, 337–8 multiplication of 324–8 multiplying by a scalar 302 negative of 301 notation 301 properties 301 relationship with matrices 337–8 scalar and vector resolutes 332–4 three-dimensional non-zero 387 time-varying 339–44 see also position vectors velocity relative 382–4 vector expressions for 514–21

625 514–21

weight and mass 364–5 whole numbers and operations, axioms relating to 188 zero matrix

133