Paradoxes And Sophisms In Calculus

From Mahobe Resources (NZ) Ltd We hope you enjoy these sample pages of the book Paradoxes and Sophisms in Calculus. We have provided the first 42 pages of the book. The actual book can be purchased from our website www.mahobe.co.nz This book is a supplementary resource intended to enhance the teaching and learning of a first-year university Calculus course. It can also be used in upper secondary school. It consists of selected paradoxes and sophisms that can be used as a pedagogical strategy by creating surprise and interest in the subject. In this book the following major topics from a typical single-variable Calculus course are explored: Functions, Limits, Derivatives and Integrals. As with the author’s previous book “Counter-Examples in Calculus” (Maths Press, Auckland, New Zealand, 2004, ISBN 0-476-01215-5, 116 p.) the intention of this book is to encourage teachers and students to use it in the teaching and learning of Calculus, with these purposes: • for deeper conceptual understanding • to reduce or eliminate common misconceptions • to advance one’s mathematical thinking, that is neither algorithmic nor procedural • to enhance generic critical thinking skills – analysing, justifying, verifying, checking, proving which can benefit students in other areas of life • to expand the ‘example set’ - a number of examples of interesting functions for better communication of ideas in mathematics and in practical applications • to make learning more emotional, active and creative The book can be useful for: • upper secondary school teachers and university lecturers as a teaching resource • upper secondary school and first-year university students as a learning resource.

If you enjoy the book then why not purchase it! Go to the Mahobe Resources (NZ) Ltd website www.mahobe.co.nz Go ahead and enjoy the first pages of this very informative and thought provoking book.

Copyright © 2007 Sergiy Klymchuk

This e-book is copyright. All rights reserved. Apart from any fair dealing for the purpose of study, teaching, review, or otherwise permitted under the Copyright Act, no part of this publication may be reproduced by any process without written permission from the author.

ISBN 978-0-473-12798-5

The printed version first published December 2005 ISBN 0-473-10550-0 Maths Press PO Box 109-760 Newmarket Auckland 1031 New Zealand

Klymchuk, Sergiy. Paradoxes and Sophisms in Calculus

Contents Preface……………………………………………………………….7 Paradoxes 1. Functions and Limits………….………………………11 2. Derivatives and Integrals………………………........15 Sophisms 1. Functions and Limits………….………………………16 2. Derivatives and Integrals………………………........28 Solutions to Paradoxes 1. Functions and Limits………….………………………34 2. Derivatives and Integrals………………………........43 Solutions to Sophisms 1. Functions and Limits………….………………………45 2. Derivatives and Integrals………………………........53

Preface Dear God, If I have just one hour remaining to live, Please put me in a Calculus class So that it will seem to last forever. A bored student’s prayer There are many interesting non-routine problems, puzzles, paradoxes and sophisms. Some of them can change the life of students forever. They can enthuse, enlighten and inspire. They can open the mind and drive a person into the labyrinths of knowledge. They can encourage passion for invention and discovery. The teaching/learning process loses its effectiveness without emotional involvement. What we tend to remember most are the events, situations, news and facts that bring with them strong feelings and emotions. The purpose of this book is to create and use such feelings and emotions as a pedagogical strategy in a Calculus course. The book is a supplementary resource intended to enhance the teaching and learning of a first-year university Calculus course. It can also be used in upper secondary school. The following major topics from a typical single-variable Calculus course are explored in the book: Functions, Limits, Derivatives and Integrals. The book consists of two parts: paradoxes and sophisms. The word paradox comes from the Greek word paradoxon which means unexpected. There are several usages of this word, including those that deal with contradiction. In this book, the word means a surprising, unexpected, counter-intuitive statement that looks invalid but in fact is true. The word sophism comes from the Greek word sophos which means wisdom. In modern usage it denotes intentionally invalid reasoning that looks formally correct, but in fact contains a subtle mistake or flaw. In other words, it is a false proof of an incorrect statement. Each such ‘proof’ contains some sort of error in reasoning.

7

Most students are exposed to sophisms at school. Often finding and analyzing the mistake in a sophism can give a student deeper understanding than a recipe-based approach in solving standard problems. The former works on a psychological level whereas the latter on a procedural level. Some basic examples of the sophism include division by zero, taking only a non-negative square root and so on. These tricks are used to ‘prove’ statements like “1 = 2”. In this book the tricks from Calculus are used to ‘prove’ such statements. Some of the paradoxes presented in the book (such as ‘Cat on a Ladder’ and ‘Encircling the Earth’) are at a lower level than Calculus, but they can still be used in Calculus classes to demonstrate that sometimes our intuition fails, even when we are dealing with very familiar shapes (like circles for example). Much of the book’s content can be used as edutainment – both education and entertainment. As with my previous book, “Counter-Examples in Calculus” (Maths Press, Auckland, New Zealand, 2004, ISBN 0-476-01215-5) the intention of this book is to encourage teachers and students to use it in the teaching/learning of Calculus with these purposes: x x x x

x

x

For deeper conceptual understanding To reduce or eliminate common misconceptions To advance one’s mathematical thinking, that is neither algorithmic nor procedural To enhance generic critical thinking skills – analysing, justifying, verifying, checking, proving which can benefit students in other areas of life To expand the ‘example set’ - a number of examples of interesting functions for better communication of ideas in mathematics and in practical applications To make learning more emotional, active and creative

The book can be useful for: x Upper secondary school teachers and university lecturers as a x

teaching resource Upper secondary school and first-year university students as a learning resource

8

Acknowledgement My thanks to Dr Farida Kachapova and Brody Radford from the Auckland University of Technology for proofreading and formatting the text and putting the material on AUT Online, the University’s web-based learning system.

Contact Details for Feedback Please send your questions and comments about the book to the postal address: Dr Sergiy Klymchuk Associate Professor School of Mathematical Sciences Faculty of Design and Creative Technologies Auckland University of Technology Private Bag 92006 Auckland 1020 New Zealand or e-mail to: [email protected]

Sergiy Klymchuk June 2006

9

Paradoxes I see it but I don't believe it! Georg Cantor (1845 – 1918), in a letter to R. Dedekind (1877)

1. Paradoxes: Functions and Limits 1. Laying bricks Imagine you have an unlimited amount of the same ideal homogeneous bricks. You are constructing an arc by putting the bricks one on top of another without using any cementing solution between them. Each successive brick is further to the right than the previous (see the diagram below).

How far past the bottom brick can the top brick extend? 2. Spiral curves Construct two similar-looking spiral curves that both rotate infinitely many times around a point, with one curve being of a finite length and the other of an infinite length. 3. A tricky curve Construct a curve that is closed, not self-crossing, has an infinite length and is located between two other closed, not self-crossing curves of a finite length.

11

4. A tricky area A square with sides of 1 unit (and therefore area of 1 square unit) is divided into 9 equal squares, each with sides 1/3 unit and areas of 1/9 square unit, then the central square is removed. Each of the remaining 8 squares is divided into 9 equal squares and the central squares are then removed. The process is continued infinitely many times. The diagram below shows the first 4 steps. At every step 1/9 of the current area is removed and 8/9 is left, that is at every step the remaining area is 8 times bigger than the area removed. After infinitely many steps what would the remaining area be?

5. A tricky ‘next term’ What is the next term in the sequence 2, 4, 8, 16? 6. A tricky shape It looked like the cross-section of an object of circular shape. To determine whether it was a circle, a student suggested measuring its several diameters (the length of line segments passing through the centre of symmetry of the figure and connecting its two opposite boundary points). The student reasoned that if they all appeared the same, the object had a circular shape. Was the student’s reasoning correct? 7. Rolling a barrel A person holds one end of a wooden board 3 m long and the other end lies on a cylindrical barrel. The person walks towards the barrel, which is rolled by the board sitting on it. The barrel rolls 12

without sliding. This is shown in the following diagram:

What distance will the person cover before reaching the barrel? 8. A cat on a ladder Imagine a cat sitting half way up a ladder that is placed almost flush with a wall. If the base of the ladder is pushed fully up against the wall, the ladder and cat are most likely going to fall away from the wall (i.e. the top of the ladder falls away from the wall). Part 1: If the cat stays on the ladder (not likely perhaps?) what will the trajectory of the cat be? A, B or C?

A

B

C

Part 2: Which of the above options represents the cat’s trajectory if instead of the top of the ladder falling outwards, the base is pulled away? A, B or C? 9. Sailing A yacht returns from a trip around the world. Different parts of the yacht have covered different distances. Which part of the yacht has covered the longest distance?

13

10.

Encircling the Earth Imagine a rope lying around the Earth’s equator without any bends (ignore mountains and deep-sea trenches). The rope is lengthened by 20 metres and the circle is formed again. Estimate how high approximately the rope will be above the Earth: A) 3 mm B) 3 cm C) 3 m?

11. A tricky equation To check the number

of

solutions

to

the

equation

x

§1· ¨ ¸ one can sketch the graphs of two inverse functions © 16 ¹

log 1 x 16

x

y

log 1 x and y 16

§1· ¨ ¸ . © 16 ¹ 3 2 1

-2

0

0

2

4

-1 -2

From the graphs we can see that there is one intersection point and x

therefore one solution to the equation log 1 x 16

§1· ¨ ¸ . But it is easy © 16 ¹

1 1 and x satisfy the 2 4 equation. So how many solutions does the equation have? to check by substitution that both x

14

2. Paradoxes: Derivatives and Integrals 1.

An alternative product rule The derivative of the product of two differentiable functions is the product of their derivatives: (uv)c u cv c . In which cases is this ‘rule’ true?

2.

A ‘strange’ integral Evaluate the following integral

dx

³ dx .

3.

Missing information? At first glance it appears there is not enough information to solve the following problem: A circular hole 16 cm long is drilled through the centre of a metal sphere. Find the volume of the remaining part of the sphere.

4.

A paint shortage

1 , the x x-axis and the line x = 1 is impossible. There is not enough paint in f 1 the world, because the area is infinite: ³ dx lim(ln b  ln 1) f. b of x 1

To

paint

the

area

bounded

by

the

curve

y

However, one can rotate the area around the x-axis and the resulting solid of revolution would have a finite volume: f 1 1 1 S ³ 2 dx S lim(  ) S . This solid of revolution contains the area b of b 1 1 x which is a cross-section of the solid. One can fill the solid with S cubic units of paint and thus cover the area with paint. Can you explain this paradox?

15

Sophisms 1 + 1 = 3 for large values of 1. A student joke

1. Sophisms: Functions and Limits 1. 1 = 0

1

n

Let us find the limit lim ¦ n of

k 1

2

n k

using two different methods.

a) n

1

lim ¦ n of

lim

n2  k

k 1

n of

1 n2 1

1

 lim

n2  2

n of

 ...  lim n of

1 n2  n

b) The lower and upper boundaries for the sum

1

n

k 1

2

n n

Since

1

n

<¦

k 1 n

lim ¦

n of k 1

2

n k 1 2

n n

n

<¦

k 1

lim n of

1

n

¦

k 1

¦

0  0  ...  0

n2  k

0

are:

1 n2 n

1

2

n n n

then by the Squeeze Theorem lim ¦

n of k 1

and

1 n2  k

n

lim ¦

n of k 1

1 n

2

lim n of

n n2

1

1.

Comparing the results in a) and b) we conclude that 1 = 0. 2. 1 = 0

1 x

Let us find the limit lim( x sin ) using two different methods. xo0

1 1 d 1 then  x d x sin d x . Applying the Squeeze x x 1 Theorem we conclude that lim( x sin ) 0 . x o0 x sin x 1 . Rewriting the limit we receive b) It is known that lim x o0 x 1 sin(1 / x) lim( x sin ) lim 1. x o0 x o0 (1 / x) x

a) Since  1 d sin

Comparing the results in a) and b) we conclude that 1 = 0.

16

3. 1 = 0 Let us find the limit lim( x x ) using two different methods. x o0

a) First find the limit of the base and then the other limit: lim( x x ) (lim x) x lim 0 x 0 . x o0

x o0

x o0

b) First find the limit of the power and then the other limit:

lim( x x )

x

x o0

( lim x )

lim x 0

xo0

1.

x o0

Comparing the results in a) and b) we conclude that 1 = 0. 4. 1 = ’ Let us find the limit lim n n using two different methods. n of

a) First find the limit of the expression under the radical and then the other limit: lim n n n lim n lim n f f . n of

n of

nof

b) First find the limit of the nth root and then the other limit: 1

lim n n n of

lim n n

n

n of

lim

1

lim n 0

nof n

nof

lim1 1 . nof

Comparing the results in a) and b) we conclude that 1 = ’. 5. sin kx

k sin x

sin x x o0 x

It is known that lim

1.

Using this fact we can find the following two limits:

sin kx x o0 x

a) lim

b) lim x o0

k sin x x

Since lim x o0

sin kx x o0 kx

k lim

sin x x o0 x

k lim

sin kx x

lim x o0

sin u u o0 u

k lim

k.

k.

k sin x sin kx then x x

k sin x and sin kx x

k sin x .

6. 1 = 0 We know that the limit of the sum of two sequences equals the sum of their limits, provided both limits exist. We also know that this is true for any number k of sequences in the sum. Let us take n equal sequences 17

1 n

and find the limit of their sum

when nń’:

1 1 1 lim(   ...  ) n of n n n

1 1 1 lim  lim  ...  lim 0  0  ...  0 0 . n of n nof n n of n 1 1 1 1 On the other hand, the sum   ...  is equal to n u 1. n n n n  n

So we receive 1 = 0. 7. 1 = -1 Let us find two limits of the same function:

x 1 x y y a) lim lim lim lim lim(1) 1 . x of y of x  y x o f y of x x of 1 y y 1 x y x lim 1 1 . lim lim b) lim lim y of x of x  y y of x of y y of 1 x x y x y Since lim lim lim lim , the results from a) and b) must x of y of x  y y of x of x  y be equal. Therefore we conclude that 1 = -1. 8. a = 1/a Let a be any non-zero number. Let us find two limits of the same function:

ax 1 ax  y 1 1 y a) lim lim . lim lim lim x of y of x  ay x of y o f x x of a a a y y a ax  y x lim a a . b) lim lim lim lim y of x of x  ay y o f x of ay y of 1 x ax  y ax  y Since lim lim , the results from a) and b) must lim lim x of y of x  ay y of x of x  ay be equal. Therefore we conclude that a = 1/a.

18

9. 1 = 2 a) Take two line segments of length 1 unit and 2 units and establish a one-to-one correspondence between their points as shown on the diagram below:

The number of points on the line segment of length 1 unit is the same as the number of points on the line segment of length 2 units, meaning 1 = 2. b) Take two circles of radius 1 unit and 2 units and establish a one-to-one correspondence between their points as shown on the diagram below:

The number of points on the circumference of the inner circle is the same as the number of points on the circumference of the outer circle, so we can conclude that 1 = 2. 10. R = r Two wheels of different radius are attached to each other and put on the same axis. Both wheels are on a rail (see the diagrams on the next page). After one rotation the large wheel with radius R covers the distance AB which is equal to the length of its circumference 2›R. The small wheel with radius r covers the distance CD which is equal to the length of its circumference 2›r. It is clear that AB = CD, therefore 2›R = 2›r and R = r.

19

Wheel

Rail Cross-section showing the wheel and the rail.

C

D

A

B

11. 2 > 3 We start from the true inequality: 2

1 1 §1· §1· ! or ¨ ¸ ! ¨ ¸ 4 8 © 2¹ © 2¹

3

Taking natural logs of both sides: 2

§1· §1· ln¨ ¸ ! ln¨ ¸ © 2¹ © 2¹

3

Applying the power rule of logs:

§1· §1· 2 ln¨ ¸ ! 3 ln¨ ¸ © 2¹ © 2¹

§1· © 2¹

Dividing both sides by ln¨ ¸ : 2 > 3.

20

12. 2 > 3 We start from the true inequality: 2

1 1 §1· §1· ! or ¨ ¸ ! ¨ ¸ 4 8 © 2¹ © 2¹

3

Taking logs with the base 2

§1· §1· log 1 ¨ ¸ ! log 1 ¨ ¸ 2© 2¹ 2© 2¹

1 of both sides: 2

3

Applying the power rule of logs:

§1· §1· 2 log 1 ¨ ¸ ! 3 log 1 ¨ ¸ 2© 2¹ 2© 2¹ §1· Since log 1 ¨ ¸ 1 we obtain: 2 > 3. 2© 2¹ 1 1 ! 4 2

13.

We start from the true equality:

1 2

1 2

Taking natural logs of both sides:

ln

1 2

ln

1 2

Doubling the left hand side we obtain the inequality:

2 ln

1 1 ! ln 2 2

Applying the power rule of logs: 2

§1· §1· ln¨ ¸ ! ln¨ ¸ © 2¹ © 2¹ 2

1 1 1 §1· Since y = ln x is an increasing function ¨ ¸ ! or ! . 2 4 2 © 2¹

21

14. 2 = 1 Let us take an equilateral triangle with sides of 1 unit. Divide the upper sides by 2 and transform them into a zig-zagging, segmented line as shown on the diagram below:

1

1

1

1

1

1

a) The length of this segment line is 2 units because it is constructed from two sides of 1 unit each. We continue halving the triangle sides infinitely many times. At any step the length of the segment line equals 2 units. b) On the other hand from the diagram we can see that with more steps, the segment line gets closer and closer to the base of the triangle which has length 1 unit. That is lim S n 1 , where S n is the n of

length of the segment line at step n. Comparing a) and b) we conclude that 2 = 1. 15. › = 2 Let us take a semicircle with diameter d. We divide its diameter into n equal parts and on each part construct semicircles of diameter as shown on the following diagram:

22

d n

a) The arc length of each small semicircle is of n semicircles is Ln nń’ is: lim Ln n of

lim n of

Sd 2

Sd

un

Sd

2n Sd . 2

2

Sd 2n

. The total length Ln

. Therefore the limit of Ln when

b) From the diagram we can see that when n increases, the curve consisting of n small semicircles gets closer to the diameter, which has length d. That is lim Ln n of

d.

Comparing a) and b) we see that

Sd 2

d and conclude that › = 2.

16. › = 0 Let us find the lateral surface area of a cylinder with height 1 unit and radius 1 unit using the following approach. Divide the cylinder into n horizontal strips. Divide the circumference of each cross-section by points into m equal parts. Rotate all odd-numbered circumferences in such a way that the points on them are exactly midway between the points on the even-numbered circumferences. Form 2mn equal isosceles triangles by joining any two adjacent points on each circumference with a point midway between them on the circumferences above and below.

23

Using simple geometry of a right-angled triangle it can be shown that the area of the resulting polyhedral surface is:

S mn

2m sin

S m

1  4n 2 sin 4

S 2m

.

When both m and n tend to infinity this area tends to the lateral surface area of the cylinder. The limit of Smn is found using the well

sin x 1 . Let us consider 3 cases. x o0 x

known formula lim a) n = m

S · § sin ¸ 4 2 ¨ sin S m ¨ 2 m m ¸ lim 2S 1 m of S 4 ¨ S ¸ ¨ ¸ m © 2m ¹ S

lim S m m of

lim 2m sin

mof

S m

1  4m 6 sin 4

S 2m

4

f.

b) n = m2

S

lim S m m of

2S 1 

lim 2m sin m of

S4 4

S m

1  4m 4 sin 4

.

24

S 2m

S · § sin 4 ¨ sin S ¨ 2m ¸¸ m lim 2S 1 m of S 4 ¨ S ¸ ¸ ¨ m © 2m ¹

4

c) n = m3 4

S

S · § sin ¸ 4 2 ¨ sin S S m 1 S m ¨ 2m ¸ f. lim S m lim 2m sin 1  4m 6 sin 4 lim 2S m of S m 2m m of 4 ¨ S ¸ m of ¸ ¨ m © 2m ¹ In all three cases a), b) and c) when mń’ the polyhedral surface tends to the lateral surface of the cylinder. So the limit lim S m in all m of

three cases a), b) and c) must be the same. This is only possible if › = 0. 17. Achilles and the Tortoise This is one of the sophisms created by the Greek philosopher Zeno in the 5th century B.C. Sometimes they are called Zeno’s paradoxes, but in the sense of “paradox” and “sophism” accepted in this book they are considered sophisms. In a race between Achilles, the fastest of Greek warriors, and a tortoise that had a head start, Achilles will never pass the tortoise. Suppose the initial distance between them is 1 unit and Achilles is moving 100 times faster than the tortoise. When Achilles covers the 1 th distance of 1 unit the tortoise will have moved of a unit 100 further from its starting point. When Achilles has covered the 1 1 th distance of of a unit the tortoise will move of a unit 100 100 2 1 further, and so on. The tortoise is always ahead of Achilles by 100 n of a unit no matter how long the race is. This means that Achilles will never reach the tortoise. 18. A snail Imagine a snail moving at a speed of 1 cm/min along a rubber rope 1 m long. The snail starts its journey from one end of the rope. After each minute the rope is uniformly expanded by 1 m. Below is a ‘proof’ that at some stage the snail will eventually reach the other end of the rope. 1 th In the first minute the snail will cover the first of the rope. 100 1 th of the rope. In the second minute the snail will cover 200

25

1 th of the rope, and so on. 300 The distance covered by the snail after n minutes will be 1 § 1 1 1· ¨1    ...  ¸ . The sum in the brackets represents the first n 100 © 2 3 n¹

In the third minute the snail will cover

terms of the harmonic series, which is divergent. The sum in the brackets (and therefore the distance) can be made bigger than any number. So no matter how big the length of the rope is, the distance covered by the snail at some stage will be bigger than the length of the rope. This means that the snail will reach the other end of the rope. 19. 1,000,000 § 2,000,000 If we add 1 to a big number the result would be approximately equal to the original number. Let us take 1,000,000 and add 1 to it. That is 1,000,000 § 1,000,001. Similarly 1,000,001 § 1,000,002. And 1,000,002 § 1,000,003. And so on… 1,999,999 § 2,000,000. Multiplying the left-hand sides and the right-hand sides of the above equalities we receive:

1,000,000 u 1,000,001 u ... u 1,999,999 | 1,000,001 u 1,000,002 u ... u 2,000,000. Dividing both sides by 1,000,001 u ... u 1,999,999 we conclude that 1,000,000 | 2,000,000. 20. 1 = -1 Since a u b

1

1

a u b , it follows that (1) u ( 1) 1 u 1 i u i i2

1.

21. 2 = -2 Two students were discussing square-roots with their teacher. The first student said: “A square root of 4 is -2”. 26

The second student was sceptical, and wrote down

4

2.

Their teacher commented: “You are both right”. The teacher was correct, so 2 = -2. 22. 2 = 1 Let us find the equation of a slant (or oblique) asymptote of the x2  x  4 function y using two different methods. x 1 x2  x  4 6 a) By performing long division: . The last term, x2 x 1 x 1 6 tends to zero as x o f . Therefore as x o f the function x 1 approaches the straight line y x  2 , which is its slant asymptote. b) Dividing both numerator and denominator by x we receive: 4 x 1 2 x x4 x . Both 4 and 1 tend to zero as x o f . Therefore 1 x x x 1 1 x as x o f the function approaches the straight line y x  1 which is its slant asymptote.

x2  x  4 has only one slant asymptote. Therefore x 1 from a) and b) it follows that x  2 x  1 . The function y

Cancelling x we receive 2 = 1.

27

2. Sophisms: Derivatives and Integrals 1. 1 = C, where C is any real number Let us apply the substitution method to find the indefinite integral ³ sin x cos xdx using two different methods:

ªu sin x º u2 sin 2 x a) ³ sin x cos xdx «  C1 » ³ udu 2  C1 2 ¬du cos xdx ¼ ªu cos x º u2 cos 2 x b) ³ sin x cos xdx « »  ³ udu  2  C 2  2  C 2 , ¬du  sin xdx ¼ where C1 and C2 are arbitrary constants. Equating the right hand sides in a) and b) we obtain

sin 2 x cos 2 x  C1 =   C2 . 2 2 Multiplying the above equation by 2 and simplifying we receive sin 2 x  cos 2 x 2C 2  2C1 or sin 2 x  cos 2 x C since the difference of two arbitrary constants is an arbitrary constant. On the other hand we know the trigonometric identity sin 2 x  cos 2 x 1 . Therefore 1 = C. 2. 1 = 0 Let us find the indefinite integral integration by parts

³ udv

1 1º ª u du  « x x2 » « » ¬dv dx v x ¼ 1 1 That is, ³ dx 1  ³ dx . x x 1 ³ x dx

1

³ x dx

using the formula for

uv  ³ vdu :

1 §1· § 1· ¨ ¸ x  ³ x¨  2 ¸dx 1  ³ dx . x © x¹ © x ¹

Subtracting the same expression

1

³ x dx

from both sides we receive

0 = 1. 3. Division by zero Let us find the indefinite integral

f c( x) dx

³ f ( x) a)

dx

dx

³ 2x  1

by the formula

ln f ( x)  C using two different methods: 1

dx

³ 2x  1 2 ³ 1 x

1 1 ln x   C1 . 2 2

2 28

b)

dx

1

2dx

1

³ 2 x  1 2 ³ 2 x  1 2 ln 2 x  1  C 2 .

Equating the right hand sides in a) and b) we obtain

1 1 ln x   C1 2 2

1 ln 2 x  1  C 2 2

Since C1 and C2 are arbitrary constants that can take any values,

1 1 1 ln 2 x  1 . ln x  2 2 2 1 1 Solving for x we receive x  2 x  1, x  . 2 2

let C1 = C2 = 0. Then

Substituting this value of x into the original integral gives zero in the denominator, so division by zero is possible! 4. sin 2 x

1 for any value of x

Let us differentiate the function y = tan x twice:

yc

1 , cos 2 x

y cc

2 sin x . cos 3 x

The second derivative can be rewritten as

2 sin x cos 3 x

2 sin x cos x u cos 2 x

1 2 yy c ( y 2 )c . 2 cos x Integrating both sides of the equation y cc ( y 2 )c we receive 1 1 sin 2 x 2 2 y c y or tan x , or . cos 2 x cos 2 x cos 2 x From here sin 2 x 1 . y cc

ªS ¬

2 tan x u

º ¼

5. 0  « , S » 2 S

Let us estimate the integral

dx

³ 1  cos 0

2

x

.

1 1 d d 1 on [0, S ] then 2 1  cos 2 x S S dx d³ dS . 2 0 1  cos 2 x

a) Since

29

S S 1 dx dx d d ³2 ³ 1  cos 2 x ³ dx or 0 0 0

S

b) On the other hand

dx x 0 t 0º ªt tan x dx cos 2 x » dx ³ 1  cos 2 x ³ 1  tan 2 x  1 ««dt » x t 0 S 0 0 cos 2 x ¬ ¼ ªS º Comparing a) and b) we conclude that 0  « , S » . ¬2 ¼ S

S

0

dt

³ 2  t2 0. 0

6. ln 2 is not defined Let us find the area enclosed by the graph of the function y

1 , x

the x-axis and the straight lines x = -2 and x = -1 using two different methods (see the diagram on the next page). a) On one hand, the derivative of the function y = ln x is y c

1 and x

1 is F ( x) ln x . We can apply x 1 1 the Newton-Leibnitz formula to the integral ³ dx (the limits are 2 x therefore an antiderivative of f ( x)

finite and the function is continuous on [-2,-1]) to find the required area: A

1 1  ³ dx 2 x

(ln(1)  ln(2)) . The area is undefined since the

logarithm of a negative number does not exist. b) On the other hand, this area is the same as the area enclosed by

1 , the x-axis, and the straight lines x

the graph of the function y

x = 1 and x = 2 due to the symmetry of the graph about the origin. Therefore the area equals: A

2

1

³ xdx ln 2  ln1 ln 2. 1

Comparing a) and b) we conclude that ln 2 is not defined.

30

1.5

1

0.5

-4

0

-2

0

2

-0.5

-1

-1.5

-2

7. › is not defined Let us find the limit lim

Sx  sin x

x  sin x sin x S x lim S. x of sin x 1 x x of

a) lim x of

Sx  sin x x  sin x

using two different methods.

b) Since both numerator and denominator are differentiable we can

f ( x) ª f º f c( x) lim «¬ f »¼ xof g c( x) which gives us: x of g ( x ) S  cos x , which is undefined. lim x of 1  cos x

use the well known rule lim

lim x of

Sx  sin x x  sin x

ªf º «¬ f »¼

Comparing the results in a) and b) we conclude that › is not defined. 8. 0 = C, where C is any real number We know the property of an indefinite integral: ³ kf ( x)dx k ³ f ( x)dx where k is a constant. Let us apply this property for k = 0. a) The left hand side of the above equality is where C is an arbitrary constant. a) The right hand side is 0 ³ f ( x) dx

³ 0 f ( x)dx ³ 0dx C ,

0.

Comparing a) and b) we conclude that 0 = C, where C is any real number.

31

9. 1 = 2 Let us find the volume of the solid of revolution produced by rotating the hyperbola y 2 x 2  1 about the x-axis on the interval [-2, 2] using two different methods. 2 2 2 x3 a) V S ³ y 2 dx S ³ ( x 2  1)dx S (  x) 3 2 2 2

4 S (cubic units). 3

b) Since the hyperbola is symmetrical about the y-axis we can find the volume of a half of the solid of revolution, say on the right from the y-axis and then multiply it by 2. Obviously the point (1,0) is a vertex to the right of the origin and the right branch of the hyperbola is to the right of the vertex (1,0). Therefore the volume of 2 2 2 x3 4 2 2 the right half is V1 S ³ y dx S ³ ( x  1)dx S (  x) S (cubic 3 3 2 1 1 units) and the total volume V

2V1

Comparing a) and b) we obtain

4 S 3

8 S (cubic units). 3 8 S or 1 = 2. 3

10. An infinitely fast fall Imagine a cat sitting on the top of a ladder leaning against a wall. The bottom of the ladder is pulled away from the wall horizontally at a uniform rate. The cat speeds up, until it’s falling infinitely fast. The ‘proof’ is below.

l

y

x

32

l 2  x 2 , where x

By the Pythagoras Theorem y

x(t ) , y

y (t )

are the horizontal and vertical distances from the ends of the ladder to the corner at time t. Differentiation of both sides with respect to t gives us y c



xxc l 2  x2

. Since the ladder is pulled

uniformly x c is a constant. Let us find the limit of y c when x approaches l: lim y c x ol

§ xxc lim¨¨  2 x ol l  x2 ©

· ¸¸ ¹

f . When the bottom of the

ladder is pulled away by the distance l from the wall, the cat falls infinitely fast.

11.

A positive number equals a negative number sin x is continuous and non-negative on a) The function f ( x) 1  cos 2 x 3S the interval ª«0, º» . Therefore by the definition of the definite ¬ 4¼

integral the area enclosed by the function f(x) and the x-axis on the 3S interval ª«0, º» is a positive number. ¬ 4¼ b) On the other hand, since the function F ( x) tan 1 (sec x) is an antiderivative of the function f(x) (this is easy to check by 3S differentiation) calculating the area as the integral of f(x) on ª«0, º» ¬ 4¼ 3 4

we receive a negative number:

S

sin x

³ 1  cos 0

2

x

dx

 tan  1 2 

Hence a positive number equals a negative number.

33

S 4

.

Solutions to Paradoxes 1. Solutions to Paradoxes: Functions and Limits 1. Laying bricks The top brick can be infinitely far from the bottom brick! The x-coordinate of the position of the centre of mass of a system of n objects with masses m1, m2,…, mn is defined by the formula:

x0

m1 x1  m2 x 2  ...  mn xn . m1  m2  ...  mn

Let us consider two bricks. For the upper brick not to fall from the lower brick the perpendicular distance from the centre of mass of the upper brick should not be beyond the right edge of the lower brick. That is, the maximum value of the x coordinate of the centre of mass of the upper brick is l: x0 l . So the maximum shift is:

'x1

l . 2 y y

x

l

34

Let us consider three bricks.

y

x

l

For the top brick we have the maximum possible shift: 'x1

l . Let 2

us find the maximum possible shift for the middle brick. Again, the perpendicular distance from the centre of mass of the system of the middle and top bricks should not extend past the right edge of the lower brick. In other words, the maximum value of the x coordinate of the centre of mass of the system of the middle and the top bricks is l: x0 l . Expressing x0 for the system of the middle and the top bricks from (1) we obtain:

l l l m( 'x 2  )  m('x2   ) 2 2 2 2m From here 'x 2

l.

l . 4

l l . , ... , 'x n 8 2n l 1 1 Adding all shifts we receive: 'x1  'x 2  ...  'x n (1   ...  ) . n 2 2

In a similar way we can obtain: 'x3

l , 'x4 6

When nń’ the sum in the brackets tends to infinity. This means that the maximum possible shift of the top brick with respect to the bottom brick can be made as large as we want. Comment: In practice it is of course impossible. Starting from a certain value of n we will not be able to make shifts of the length

l as they will be too small to perform. 2n 35

2. Spiral curves a) Let us construct a spiral curve of a finite length. Draw a line segment of length d. Draw a semicircle with diameter d on one side of the line segment. Then on the other side of the line segment draw a semicircle of diameter d/2. Then on the other side draw a semicircle of diameter d/4, and so on.

The length of the curve is:

S

d d d 1 1 1  S  S  ... Sd (    ...) Sd . 2 4 8 2 4 8

b) Let us construct a spiral curve of infinite length. Draw a line segment AB of length d with midpoint P. Draw a circle with the centre C on the line segment at distance a from P. On one side of the line segment draw a semicircle of the diameter d. On the other side of the line segment draw a semicircle of the diameter AE, where point E is the midpoint of PB. Then on the other side of the line segment draw a semicircle of the diameter EF, where point F is the midpoint of AD and so on (see the following diagram).

36

The curve has infinitely many rotations around point C and each rotation has a length bigger than the circumference 2›a, so the length of the curve is infinite. 3. A tricky curve One example of such a curve is the famous Koch snowflake. We start with an equilateral triangle and build the line segments on each side according to a simple rule. At every step each line segment is divided into 3 equal parts, then the process is repeated infinitely many times. The resulting curve is called the Koch curve or Koch snowflake and is an example of a fractal. The first four iterations are shown below:

The initial triangle and all consecutive stars and snowflakes are located between the circumferences inscribed into the triangle and circumscribed around it. Both circumferences have finite lengths. If the perimeter of the initial triangle is 1 unit, then the perimeter 37

4 units. The perimeter of 3 2 1 16 § 4 · the snowflake in the second iteration is 48 u ¨ ¸ units. 27 9 © 3 ¹ n 4 § · The perimeter of the snowflake in the nth iteration is ¨ ¸ units. © 3¹

of the star in the first iteration is 12 u

1 9

As nń’ the perimeter of the snowflake tends to infinity. The Koch curve has an infinite length but bounds a finite area, which is between the area of the circle inscribed into the initial triangle and the area of the circle circumscribed around it. 4. A tricky area Although at every step the remaining area is 8 times bigger than the area removed, after infinitely many steps the remaining area will be zero and the area removed will be 1 square unit. Let us show this. After the first step the remaining area equals 1 

1 9

8 1  8u After the second step the remaining area is 9 81

§8· ¨ ¸ . ©9¹

64 81

8 . 9 2

n

Similarly, after the

nth

§8· step the remaining area is ¨ ¸ and if n ©9¹

tends to infinity this area tends to zero. This figure is called the Sierpinski carpet and is another example of a fractal. 5. A tricky ‘next term’ a) The expected answer is 32, but there are infinitely many other correct answers. Actually the next term in the sequence 2, 4, 8, 16 can be any number. Let the nth term be a n 2 n  (n  1)(n  2)(n  3)(n  4) x . The first 4 terms are 2, 4, 8, 16. One can make a5 equal to any number by determining x from the formula for a5 . This can be done to obtain the formula for the nth term. For example, let the 5th term be –4. From the equation  4 2 5  4 u 3 u 2 u 1u x we obtain x = -1.5 and the formula for the nth term is a n 2 n  (n  1)(n  2)(n  3)(n  4)(1.5) . b) Another interesting example has a geometrical flavour. Draw a circle, and put two dots on the circumference and connect them with a line segment. The circle is divided into 2 regions. Put a third 38

dot and connect all dots. The circle is now divided into 4 regions. Put a fourth dot and connect all dots. The circle is now divided into 8 regions. Put a fifth dot and connect all dots. The circle is now divided into 16 regions. It looks like we have a clear pattern. But when you put the sixth dot and connect all dots the circle is divided into 30 regions! 6. A tricky shape No, the student was not right. A figure can be of constant diameter yet not be a circle. As an example, consider the following curve. In an equilateral triangle draw circular arcs with the radius equal to the side of the triangle from each vertex. The resulting figure is a curved triangle, which is called the Reuleaux triangle (see the diagram below). One of its properties is that it has a constant diameter. When it rolls on a horizontal surface its center moves along a sine curve with ups and downs (unlike a circle whose center does not move up and down – only along a straight horizontal line). For this reason it is not practical to use it as a wheel, but it does have practical applications. It is used in the Wankel rotary engine, and in some countries manhole covers are shaped like the Reuleaux triangle.

7. Rolling a barrel The barrel rolls as long as the person continues walking. The velocity of the point on the top of the barrel equals the velocity of the walking person and is twice the velocity of the axis of the barrel, so the person will cover 6 m by the time he reaches the barrel. 8. A cat on a ladder Part 1. Most people are confident that C is the answer to Part 1. Without much difficulty, one can imagine the ladder rotating about a central point, i.e. where the base of the ladder touches the wall. 39

An arc is the result and in this case represents a quarter of the circumference of a circle.

Part 2. However, Part 2 isn’t so easy! Many people conclude that A is the correct answer. It sounds reasonable that as the ladder slides outwards away from the wall that it would appear to drop quickly, then level out as it approaches the horizontal. Surprisingly, the answer to this problem is also C. Try it out by making a model (see the sequential sketches below). With a paper ladder, with a point marking half way, slowly slide the ladder down and away from the wall. After each small amount of movement, put a dot on the page at the place where the centre of the ladder lies. Note that as the ladder approaches the horizontal, further lateral movement is minimal.

Here is a simple proof: 40

B

D C

O

A

Let AB be the ladder. Point C (where the cat sits) is always at the same distance (half the length of the ladder) from the point O regardless of the position of the ladder. This comes from the fact that diagonals in a rectangle are the same and are divided in half by the point of their intersection. You may well be very surprised to see that the trajectory is the same in both cases. Do not be alarmed however – in this case the intuition of many people fails. My colleague tried this test out with a class of 100 4th year engineering students in Australia, Germany, New Zealand and Norway. These young men and women, aged about 21, are expected to be able to quickly conceptualise shapes, dimensions, movements and forces. The students were given 40 seconds to find the answer. They were told that it was a mental exercise, with no calculations or drawings permitted. The results were startling, for although 74% of the students gave C, the correct answer to Part 1, 86% were wrong in Part 2. 14% and 34% gave B as the answer in Parts 1 and 2 respectively. 9. Sailing The top of the yacht has covered the longest distance. The shape of the Earth is approximately spherical, so the top of the yacht has the longest radius compared to lower parts and therefore has the longest circumference.

41

10.

Encircling the Earth Approximately 3 m high. This is a surprising answer for many people. Let r be the radius of the Earth and R be the radius of the circle after adding 20 metres to the rope. The difference between the two circumferences is 20 m: 2SR  2Sr 20 or 2S ( R  r ) 20. From here the difference between the two radii is R  r | 3m. The answer does not depend on the original length of the rope.

11.

A tricky equation The rough sketch of the graphs is “too rough”. Both functions are decreasing for all x in their domains but they are very close to both axes, and in fact have 3 intersection points. It can be shown that the equation has 3 solutions.

42