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SEKOLAH MENENGAH KEBANGSAAN SRI TEBRAU JOHOR BAHRU
ADDITIONAL MATHEMATICS PROJECT WORK 2009
TITLE
Name: Class: Teacher’s Name:
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CONTENT 1.
Content………………………………………………………………………………………………….2
2.
Preface…………………………………………………………………………………………………..3
3.
Introduction…………………………………………………………………………………………..4
4.
Part 1 Pictures of circle or part of a circle……………………………………………….5 Definition of Pi (π)……………………………………………………………………6, 7 Brief History of π………………………………………………………………………7, 8
5.
Part 2 Relation between lengths of arcs PQR, PAB and BCR………..9, 10, 11 Relation between lengths of arcs PQR, PAB BCD and DER and its tabulate findings………………………………………………………11, 12, 13, 14 Generalization………………………………………………………………………….. 15 Showing of the generalization is true……………………….16, 17, 18, 19
6.
Part 3
Expression of y in terms of π and x……………………………………….20, 21 Diameters of two fish ponds………………………………………………..21, 22 Reduction of non-linear equation to simple linear form…………….20 Graph…………………………………………………………………………………………20 Two methods to determine the area of flower pot………………23, 24
Diameters of the flower beds……………………………………………….25, 26 7.
Conclusion……………………………………………………………………………………………27
8.
Reference…………………………………………………………………………………………….28
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PREFACE This project is prepared by the students of class 5 Science 1 (2009) and it is based on the Additional Mathematics textbooks, internet search and reference book. Additional Mathematics is an elective subject in secondary school. Each student who takes this subject has to carry out a project work on the given tasks. The project work for the year 2009 is about Circle by using the principle of Pi (π). The aim of doing this project is to improve the skills in using mathematics for students. Working on this also gives a chance for students to apply their skills on what they had learnt to solve an assigned project. Therefore, every student stands a chance to improve their thinking skills, usage of languages and grammar as well as mathematics skills throughout the project. After doing this project, the student will be able to master and understand more on the application of Additional Mathematics that they learnt in their school syllabus. The student can also learn some values during the completion of the project such as to learn how to work together or to be cooperative, improving their communication skills, responsibility and also not to give up easily on the task given.
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INTRODUCTION Additional Mathematics is one of the compulsory subjects for SPM science stream candidates. All of the students would have to carry out a project work based on a topic given and must be submitted in three weeks time. The objective of carrying out this project is to apply and adapt a variety of problem-solving strategies that we had learnt to solve the problems. Moreover, our thinking skills can be improved. It also promotes effective mathematical communication. Our confidence and interest towards mathematics will be increase through solving various types of problems. Besides, the aim of doing this project is to use the language of mathematics to express mathematical ideas precisely. This also stimulates and enhances effective earning. During the project, we are able to develop our positive attitude towards mathematics. This makes the lesson to be more fun, useful and meaningful.
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PART 1 There are a lot of things around us related to circles or parts of a circle. Thus, we need to understand the relations of circles in our environment in order to solve various problems involving circles.
FIND YOURSELF
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Definition of Pi (π) Pretty much everyone is familiar with what π is. Take a circle on a plane. Measure the distance around the outside of it, which is called the circumference. Divide that by the diameter of the circle. That's π.
It also shows up in almost anything else involving measurements of circles and angles, from things like the sin function to the area of a circle to the volume of a sphere. Where it gets interesting is when you start to ask about how to compute it. You get the relatively obvious things - like equations based on integrals to calculate the area of a circle. But then, you get the surprising ones. After all, π is a fundamental geometric number; it comes from the circumference of a circle. So why in the world is the radius of a circle related to an infinite sum of the reciprocals of odd numbers? It is. 𝜋
is the sum of the infinite series 1/1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + 1/13 ...., or more formally: 4
𝜋
Or how about this? = 2
What about a bit of probability? Pick any integer at random. What's the 6 probability that neither it nor any of its factors is a perfect square? . 𝜋²
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How about a connection between circles and prime numbers? Take the set of 1 6 all of the prime numbers P, then the product of all factors (1 − ) is . 𝑝²
𝜋²
What's the average number of ways to write an integer as the sum of two 𝜋 perfect squares? . 4
There's also a funny little trick for memorizing π, called a piem (argh!). A piem is a little poem where the length of each word is a digit of π.
Brief History of π The oldest value we know for π comes from the Babylonians. (Man, but those guys were impressive mathematicians; almost any time you look at the history of fundamental numbers and math, you find the Babylonians in the roots.) They tended to work in ratios, and the approximation that they used 25/8s (3.125), which is not a terribly bad approximation. Especially when you realize when they came up with this approximation: 1900BC! The next best approximation came from Egypt, around the time of Pharaoh Amenemhat in the mid 17th century BC, where it had been refined to 256/81 (3.1605). Which isn't such a great step forward; it's actually a hair farther from the true value of π than the Babylonian approximation. We don't see any real progress until we get to the Greek. Archimedes (yes, that Archimedes) worked out a better approximation. He used a really neat trick. He worked out how to compute the perimeter of a 96-sided polygon; and then worked out the perimeter of the largest 96-gon that could be drawn inside the circle; and the smallest 96-gon that the circle could be drawn inside. Here's a quick diagram using octagons to give you a clearer idea of what he did:
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That gives you an approximation of π as the average of 223/71 and 22/7 - or 3.14185. And next, we find progress in India, where the mathematician Madhava worked out a power series definition of π, which allowed him to compute π to 13 decimal places. 13 decimal places, computing a power series completely by hand! Astounding! Even better, during the same century, when this work made its way to the great Persian Arabic mathematicians, they worked it out to 9 digits in base-60 (base-60 was in inheritance from the Babylonians). 9 digits in base 60 is roughly 16 digits in decimal! And finally, we get back to Europe; in the 17th century, van Ceulen used the power series to work out 35 decimal places of π. Alas, the publication of it was on his tombstone. Then we get to the 19th century, when William Rutherford calculated 208 decimal places of π. The real pity of that is that he made an error in the 153rd digit, and so only the first 152 digits were correct. (Can you imagine the amount of time he wasted?) That was pretty much it until the first computers came along, and once that happened, the fun went out of trying to calculate it, since any bozo could write a program to do it. There's a website that will let you look at its computation of the first 2 hundred million digits of π. The name of π came from Euler (he of the great equation, eiπ + 1 = 0). It's an abbreviation for perimeter in Greek. There's also one bit of urban myth about π that is, alas, not true. The story goes that some state in the American Midwest (Indiana, Iowa, Ohio, Illinois in various versions) passed a law that π=3. Didn't happen.
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PART 2
Diagram 1 (a) Diagram 1 shows a semicircle PQR of diameter 10cm. Semicircles PAB and BCR of diameter d1 and d2 respectively are inscribed in PQR such that the sum of d1 and d2 is equal to 10 cm. By using various values of d1 and corresponding values of d2, I determine the relation between length of arc PQR, PAB, and BCR as shown below: 1
By using the formula of a semicircle, 𝜋𝑑 2
d1
d2
Length of arc PQR
Length of arc PAB
Length of arc BCR
(cm)
(cm)
in terms of π (cm)
in terms of π (cm)
in terms of π (cm)
1
9
5π
1 π 2
9 π 2
2
8
5π
π
4π
3
7
5π
3 π 2
7 π 2
4
6
5π
2π
3π
5
5
5π
5 π 2
5 π 2
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6
4
5π
3π
2π
7
3
5π
7 π 2
3 π 2
8
2
5π
4π
π
9
1
5π
9 π 2
1 π 2
10
0
5π
5π
0
Table 1
From the Table 1 we know that the length of arc PQR is not affected by the different in d1 and d2 in PAB and BCR respectively. The relation between the length of arcs PQR, PAB and BCR is that the length of arc PQR is equal to the sum of the length of arcs PAB and BCR, thus which can get this equation:
SPQR = SPAB + SBCR Examples below will show that the relation above is true: 1)
Let d1 = 1 and d2 = 9, Since
SPQR = SPAB + SBCR,
5𝜋 =
1 9 𝜋+ 𝜋 2 2
5𝜋 =
10 𝜋 2
5𝜋 = 5𝜋
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2)
Let d1 = 4 and d2 = 6,
SPQR = SPAB + SBCR,
Since
5𝜋 = 2𝜋 + 3𝜋 5𝜋 = 5𝜋 From two examples above, both shows that
SPQR = SPAB + SBCR (b)(i) Using various values of d1 and d2 and the corresponding values of d3, determine the relation between the lengths of arcs PQR, PAB, BCD and DER. Tabulate your findings.
Diagram 2 1
By using the formula of a semicircle, 𝜋𝑑 2
d1 (cm)
d2 (cm)
d3 (cm)
Length of arc PQR in terms of π (cm)
1
1
8
5π
1
2
7
5π
Length of arc PQR in terms of π (cm) 1 𝜋 2 1 𝜋 2
Length of arc PQR in terms of π (cm) 1 𝜋 2 1π
Length of arc PQR in terms of π (cm) 4π 7 𝜋 2
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1 𝜋 2 1 𝜋 2 1 𝜋 2 1 𝜋 2 1 𝜋 2 1 𝜋 2
3 𝜋 2
3π
2π
5 𝜋 2
5 𝜋 2
2π
3π
3 𝜋 2
7 𝜋 2
1π
1
3
6
5π
1
4
5
5π
1
5
4
5π
1
6
3
5π
1
7
2
5π
1
8
1
5π
2
1
7
5π
1π
1 𝜋 2
1 𝜋 2 7 𝜋 2
2
2
6
5π
1π
1π
3π
2
3
5
5π
1π
3 𝜋 2
5 𝜋 2
2
4
4
5π
1π
2π
2π
2
5
3
5π
1π
5 𝜋 2
3 𝜋 2
2
6
2
5π
1π
3π
1π
2
7
1
5π
1π
1 𝜋 2
3
1
6
5π
7 𝜋 2 1 𝜋 2
3
2
5
5π
3
3
4
5π
3
4
3
5π
3
5
2
5π
3 𝜋 2 3 𝜋 2 3 𝜋 2 3 𝜋 2 3 𝜋 2
4π
3π
1π
5 𝜋 2
3 𝜋 2
2π
2π
3 𝜋 2
5 𝜋 2
1π
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3
6
1
5π
3 𝜋 2
3π
4
1
5
5π
2π
1 𝜋 2
1 𝜋 2 5 𝜋 2
4
2
4
5π
2π
1π
2π
4
3
3
5π
2π
3 𝜋 2
3 𝜋 2
4
4
2
5π
2π
2π
1π
4
5
1
5π
2π
1 𝜋 2
5
1
4
5π
5 𝜋 2 1 𝜋 2
5
2
3
5π
5
3
2
5π
5
4
1
5π
6
1
3
5π
3π
1 𝜋 2
1 𝜋 2 3 𝜋 2
6
2
2
5π
3π
1π
1π
6
3
1
5π
3π
1 𝜋 2
7
1
2
5π
3 𝜋 2 1 𝜋 2
7
2
1
5π
8
1
1
5π
5 𝜋 2 5 𝜋 2 5 𝜋 2 5 𝜋 2
7 𝜋 2 7 𝜋 2 4π
2π
1π
3 𝜋 2
3 𝜋 2
1π
2π
1π 1 𝜋 2
1π 1 𝜋 2 1 𝜋 2
Table 2 From the Table 2 we know that the length of arc PQR is not affected by all the possible different in d1, d2, d3 in PAB, BCD and DER respectively. ©David Tan 2009 @ www.davidtn.net Please do not copy and paste it as it’s for the purpose of reference.
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Hence, the relation between the length of arcs PQR, PAB, BCD and DER is that the length of arc PQR is equal to the sum of the length of arcs PAB, BCD and DER, which can get this equation:
SPQR = SPAB + SBCD + SDER The examples below will prove that the relation stated above is true, 1)
Let d1 = 5, d2 = 1 and d3 = 4, Since
SPQR = SPAB + SBCD + SDER,
5𝜋 =
5 1 𝜋 + 𝜋 + 2𝜋 2 2
5𝜋 =
10 𝜋 2
5𝜋 = 5𝜋 2)
Let d1 = 2, d2 = 3 and d3 = 5, Since
SPQR = SPAB + SBCD + SDER,
3 5 5𝜋 = 1𝜋 + 𝜋 + 𝜋 2 2 5𝜋 =
10 𝜋 2
5𝜋 = 5𝜋
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(b)(ii) Based on your findings in (a) and (b)(i), make generalisations about the length of the arc of the outer semicircle and the lengths of arcs of the inner semicircles for n inner semicircles where n = 2, 3, 4.... Base on the findings in the table in (a) and (b) above, it can conclude that: The length of the arc of the outer semicircle is equal to the sum of the length of arcs of any number, n of the inner semicircles.
Souter = S2 + S3 + S4 + ……. + Sn, where n = 2,3,4,...
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(c) For different values of diameters of the outer semicircle, show that the generalisations stated in b (ii) is still true.
Diagram 3 Diagram 3 shows a big semicircle with n number of small inner circle. From the diagram, we can see that d = d1 + d2 + ............. + dn So the length of arc of the outer semicircle, 1
𝜋𝑑
2
2
SOUTER = 𝜋𝑑 =
The sum of the length of arcs of the inner semicircles, SINNER =
𝜋d 1 2
+
𝜋d 2 2
+ ........... +
𝜋d n 2
𝜋
SINNER = (𝑑1 + d2 + .......... + dn) 2
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Substitute, For Example: Assume the diameter of outer semicircle is 20cm and 4 semicircles are inscribed in the outer semicircle such that the sum of d1, d2, d3, d4 is equal to 20cm. Let d1=4, d2=6, d3=3 and d4=7, 𝜋
SINNER = (𝑑1 + d2 + .......... + dn) 2
𝜋
SINNER = (4 + 6 + 3 + 7) 2
𝜋
SINNER = (20) 2
1
𝜋𝑑
2
2
As mentioned above, SOUTER = 𝜋𝑑 =
,
𝜋
𝜋20
2
2
SINNER = (20) =
,
where 20 is the diameter, d as stated above. SOUTER =
𝜋(𝑑) 2
SINNER =
𝜋(20) 2
∴ SINNER = SOUTER
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Another example, Assume again the diameter of outer semicircle is 30cm and 4 semicircles are inscribed in the outer semicircle such that the sum of d1, d2, d3, d4 is equal to 30cm. Let d1=6, d2=7, d3=8 and d4=9, 𝜋
SINNER = (𝑑1 + d2 + .......... + dn) 2
𝜋
SINNER = (6 + 7 + 8 + 9) 2
𝜋
SINNER = (30) 2
1
𝜋𝑑
2
2
As mentioned above, SOUTER = 𝜋𝑑 = 𝜋
𝜋30
2
2
SINNER = (30) =
,
,
where 30 is the diameter, d as stated above. SOUTER =
𝜋(𝑑) 2
SINNER =
𝜋(30) 2
∴ SINNER = SOUTER
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From two different of diameters given in the examples above, we can see that both of them carried one conclusion that is:
SINNER = SOUTER As a result, we can conclude that the length of the arc of the outer semicircle is equal to the sum of the length of arcs of any number of the inner semicircles. This is true for any value of the diameter of the semicircle. In other words, for different values of diameters of the outer semicircle, it shows that the generalisation stated in b (ii) is still true.
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PART 3 The Mathematics Society is given a task to design a garden to beautify the school by using the design as shown in Diagram 4. The shaded region will be planted with flowers and the two inner semicircles are fish ponds.
Diagram 4 a)
The area of the flower plot is y m2 and the diameter of one of the fish ponds is x m. Express y in terms of π and x.
Area of ADC = 1 = 𝜋𝑟² 2 1 10 = 𝜋( )² 2 2 =
25 𝜋 2
Area of AEB = 1 = 𝜋𝑟² 2 =
1 𝑥 𝜋( )² 2 2
1 𝑥² = 𝜋( ) 2 4 𝑥² = 𝜋 8 Area of BFC = 1 = 𝜋𝑟² 2 1 𝑥 = 𝜋(5 − )² 2 2 1 𝑥² = 𝜋(25 − 5𝑥 + ) 2 4 =
25 5𝑥 𝑥² 𝜋− 𝜋+ 𝑥 2 2 8
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Area of shaded region = Area of ADC – (Area of AEB + Area of BFC) 25 𝑥2 25 5𝑥 𝑥2 = 𝜋−[ 𝜋+ 𝜋− 𝜋+ 𝜋 ] 2 8 2 2 8 25 𝑥2 25 5𝑥 𝑥2 = 𝜋−( 𝜋+ 𝜋− 𝜋+ 𝜋) 2 8 2 2 8 =
25 𝑥2 25 5𝑥 𝑥2 𝜋− 𝜋− 𝜋+ 𝜋− 𝜋 2 8 2 2 8
5𝑥 𝑥2 = 𝜋− 𝜋 2 4 =
10𝑥 − 𝑥² 𝜋 4
Therefore,
𝑦= (b)
10𝑥 − 𝑥² 𝜋 4
Find the diameters of the two fish ponds if the arc of the flower plot is 22 16.5 m2. (Use π = ). 7 Given y = 16.5 cm, 16.5 =
10𝑥 − 𝑥² 22 ( ) 4 7
16.5 =
220𝑥 − 22𝑥² 28
462 = 220x – 22x2 0 = 22x2 - 220x + 462 0 = x2 – 10x + 21
÷22
Factorise, 0 = (x - 7)(x – 3) ∴ x = 7 cm or x = 3 cm ©David Tan 2009 @ www.davidtn.net Please do not copy and paste it as it’s for the purpose of reference.
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∴From the diagram, we can assume that the diameter AB is 3 cm and the diameter BC is 7 cm. (c)
Reduce the non-linear equation obtained in (a) to simple linear form and hence, plot a straight line graph. Using the straight line graph, determine the area of the flower plot if the diameter of one of the fish ponds is 4.5 m. By using linear law, 5𝑥 𝑥2 𝑦= 𝜋− 𝜋 2 4 Change it to linear form of Y = mX + c, 𝑦 −𝑥 5 = + 𝜋 𝑥 4 2 x 𝑦 𝑥
0
1
2
3
4
5
6
7
7.9
7.1
6.3
5.5
4.7
3.9
3.1
2.4
GRAPH CONSTRUCT A GRAPH OF X/Y AGAINST X USING THE TABLE GIVEN ABOVE. Thus, plot a graph of y/x against x and draw the line of best fit. Find the value of y/x when x = 4.5 m. Then multiply y/x you get with 4.5 to get the actual value of y.
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From the graph, when x = 4.5, Area of flower pot =
𝑦 𝑥
𝑦 𝑥
= 4.3
×x
= 4.3 × 4.5 = 19.35 m2 (d)
The cost of constructing the fish ponds is higher than that of the flower plot. Use two methods to determine the area of the flower plot such that the cost of constructing the garden is minimum. We need to get the largest value of y so that the cost of constructing the garden is minimum, Method 1: Differentiation 5𝑥 𝑥2 𝑦= 𝜋− 𝜋, 2 4 𝑑𝑦 −𝜋𝑥 5𝜋 = + 𝑑𝑥 2 2 𝑑2 𝑦 −𝜋 = 𝑑𝑥 2 2 y has a maximum value.
At maximum point, 𝑑2 𝑦 −𝜋 = 𝑑𝑥 2 2 −𝜋𝑥 5𝜋 + =0 2 2 𝜋𝑥 5𝜋 = 2 2 ©David Tan 2009 @ www.davidtn.net Please do not copy and paste it as it’s for the purpose of reference.
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x=5m When x = 5 m, Maximum value of y = −𝜋(52 ) 5(5)𝜋 = + 4 2 = 6.25𝜋 m2 Method 2: Completing the Square −𝜋𝑥² 5𝜋𝑥 𝑦= + 4 2 −𝜋 2 = 𝑥 − 10𝑥 4 =
−𝜋 2 𝑥 − 10𝑥 + 25 − 25 4
=
−𝜋 [ 𝑥 − 5)² − 25 ] 4
=
−𝜋 25𝜋 (𝑥 − 5)² + 4 4 −𝜋
y is a ⋂ shape graph as a = 4 Hence, it has a maximum value. When x = 5 m, maximum value of the graph = 6.25π m2.
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(e)
The principal suggested an additional of 12 semicircular flower beds to the design submitted by the Mathematics Society as shown in Diagram 4. The sum of the diameters of the semicircular flower beds is 10 m.
Diagram 5 The diameter of the smallest flower bed is 30 cm and the diameter of the flower beds are increased by a constant value successively. Determine the diameter of the remaining flower beds. By using arithmetic progression, (i) The principal suggested an additional of 12 semicircular flower beds to the design submitted by the Mathematics Society. (n = 12) (ii) The sum of the diameters of the semicircular flower beds is 10 m. (S12 = 10 m = 1000 cm) (iii) The diameter of the smallest flower bed is 30 cm. (a = 30 cm) (iv) The diameter of the flower beds are increased by a constant value successively. (d =?) n S12 = [ 2𝑎 + 𝑛 − 1 𝑑] 2 12 2 30 + 12 − 1 𝑑 2 1000 = 6(60 + 11𝑑) 1000 = 360 + 66𝑑 66𝑑 = 640 𝑑 = 9.697 cm 1000 =
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Since the first flower bed is 30 cm, Hence the diameters of remaining 11 flower beds expressed in arithmetic progression are:
Tn (flower bed), given Tn = a + (n – 1)d, where a = 30 and d = 9.697
Diameter (cm)
T1
30.00
T2
39.70
T3
49.39
T4
59.09
T5
68.79
T6
78.49
T7
88.18
T8
97.88
T9
107.6
T10
117.3
T11
127.0
T12
136.7
©David Tan 2009 @ www.davidtn.net Please do not copy and paste it as it’s for the purpose of reference.
©David Tan 2009 @ www.davidtn.net
CONCLUSION Pi (π) is a very useful mathematics related to circle in which it helps the mankind to solve many problems easily involving circle. We are able to know how we can use this unit to solve various problems involving objects that are circular in shape of even part of a circle shape. Besides, in this project work we need to use a lot of mathematical concept in order to get the answer. This makes me understand more about other mathematical concept besides Pi (π). So, after doing this project, I am quite impressed with the usage of circle and its ways to help us in solving problems although there are some errors occur. Besides that, I also learnt many things for this which I can never find them in the textbook or reference book or even in our school syllabus. I am doing many researches to understand its usage and its principles when apply to solve the problem involving circles. Furthermore, I am able to interpret carefully when handling such mind twisting problem that is in Part 3. This experience that I gain from this project work can makes me apply to other subjects so that it will make me more careful when handling such question mentioned. I am really appreciating the government as they gave us this opportunity to do this project in the process of understanding and learning deeply into circles. I would like to thanks my additional mathematics teacher as without his help, I would not be able to accomplice this project.
©David Tan 2009 @ www.davidtn.net Please do not copy and paste it as it’s for the purpose of reference.
©David Tan 2009 @ www.davidtn.net
REFERENCE BOOK 1)
ACE Analysis Additional Mathematics SPM, Oxford Fajar Sdn. Bhd. Wong Pek Wei, Lee Beck Tun.
WEBSITES 1)
http://images.google.com/
2)
http://en.wikipedia.org/wiki/Circle
3)
http://www.gap-system.org/~history/HistTopics/Pi_through_the_ages.html
©David Tan 2009 @ www.davidtn.net Please do not copy and paste it as it’s for the purpose of reference.