contoh folio addmath 2010 (project work 1)

Additional Mathematics Project Work 1

1

Additional mathematics

project work 1

Nama kt sini

480123-3912-1329

5sc1

Pn. Cikgu

Sekolah Menengah Kebangsaan ……….


2

Contents

Appreciation 3

Objectives 4

Introduction 5

Procedure and Findings 6-9

Further Exploration 10-14

Conclusion 15

Reflection 16

References 17

Appendix 18


3

Appreciation

Firstly, I would like to give a big thanks to my parents for providing

everything, such as money, to buy anything that are related to this project work,

their advice and support. Then, I want to thank my teacher; Pn. Cikgu for teaching

me Additional Mathematics form 5 and guiding me throughout this project.

Last but not least, my friends who were doing this project with me and

sharing our ideas and knowledge. They were helping each other so we can

complete our project without any problems.


4

Objectives The objectives of this project work are:

Apply and adapt a variety of problem-solving strategies to solve problems.

Develop mathematical knowledge through problem solving in a way that

increases students’ interest and confidence.

Develop positive attitude towards mathematics.

Improve thinking skills and creativity.

Promote efficiency of mathematical communication.

Provide learning environment that stimulates and enhances effective

learning.


5

Introduction

Around AD 1000, the Islamic mathematician Ibn al-Haytham (Alhacen) was

the first to derive the formula for the sum of the fourth powers of an arithmetic

progression, using a method that is readily generalizable to finding the formula

for the sum of any higher integral powers, which he used to perform integration.

In the 11th century, the Chinese polymath Shen Kuo developed 'packing'

equations that dealt with integration.

In the 12th century, the Indian mathematician, Bhāskara II, developed an

early derivative representing infinitesimal change, and he described an early form

of Rolle's Theorem. Also in the 12th century, the Persian mathematician Sharaf al-

Dīn al-Tūsī discovered the derivative of cubic polynomials, an important result in

differential calculus.

In the 14th century, Indian mathematician Madhava of Sangamagrama,

along with other mathematician-astronomers of the Kerala School of astronomy

and mathematics, described special cases of Taylor series, which are treated in

the text Yuktibhasa.

In the 19th century, calculus was put on a much more rigorous footing by

mathematicians such as Cauchy, Riemann, and Weierstrass. It was also during

this period that the ideas of calculus were generalized to Euclidean space and the

complex plane. Lebesgue generalized the notion of the integral so that virtually

any function has an integral, while Laurent Schwartz extended differentiation in

much the same way.

Calculus is a ubiquitous topic in most modern high schools and universities

around the world.


6

Procedure and Findings

(A) Function I

Maximum point (0,4.5) and pass through point (2,4) 𝑦 = 𝑎 𝑥 − 𝑏 2 + 𝑐 b=0, c=4.5 𝑦 = 𝑎 𝑥 − 𝑂 2 + 4.5 𝑦 = 𝑎𝑥2 + 4.5 ---------- (1) Substitute (2,4) into (1) 4 = 𝑎 2 2 + 4.5 4𝑎 + 4.5 = 4 4𝑎 = −0.5 𝒚 = −𝟎. 𝟏𝟐𝟓𝒙𝟐 + 𝟒. 𝟓


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Function II

Maximum point (0,0.5) and pass through point (2,0) 𝑦 = 𝑎 𝑥 − 𝑏 2 + 𝑐 b=0, c=0.5 𝑦 = 𝑎 𝑥 − 0 2 + 0.5 𝑦 = 𝑎𝑥2 + 0.5 ---------- (2) 0 = 𝑎 2 2 + 0.5 4𝑎 = −0.5 𝑎 = −0.125 𝒚 = −𝟎. 𝟏𝟐𝟓𝒙𝟐 + 𝟎. 𝟓


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Function III

Maximum point (2,4.5) and pass through point (0,4) 𝑦 = 𝑎 𝑥 − 𝑏 2 + 𝑐 b=2, c=4.5 𝑦 = 𝑎 𝑥 − 2 2 + 4.5 ---------- (3) Substitute (0,4) into (3) 4 = 𝑎 0 − 2 2 + 4.5 4 = 4𝑎 + 4.5 4𝑎 = −0.5 𝑎 = −0.125 𝒚 = −𝟎. 𝟏𝟐𝟓 𝒙 − 𝟐 𝟐 + 𝟒. 𝟓


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(b)

= 4 × 1 − 2 −0.125𝑥2 + 0.5 ⅆ𝑥

2

0

= 4 − 2 −0.125𝑥3

3+ 0.5𝑥

0

2

= 4 − 2 2

3− 0 = 4 −

4

3

= 𝟐𝟐

𝟑𝒎𝟐


10

Further Exploration

(a) (1) Structure 1

Area = 22

3𝑚2

Volume = 22

3× 0.4

= 16

15𝑚3

Cost = 16

15× 840

= RM896.00


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Structure 2

Area = 4 × 1 −1

2× 4 × 0.5

= 3𝑚2 Volume = 3 × 0.4 = 1.2𝑚3 Cost = 1.2 × 840 = RM1008.00


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Structure 3

Area = 4 × 1 −1

2× 1 + 4 × 0.5

= 2.75𝑚2

Volume = 2.75 × 0.4 = 1.1𝑚3 Cost = 1.1 × 840 = RM924.00


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Structure 4

Area = 4 × 1 −1

2× 2 + 4 × 0.5

= 2.5𝑚2 Volume = 2.5 × 0.4 = 1𝑚3 Cost = 1 × 840 = RM840 Conclusion The minimum cost to construct is Structure 4. (2) As the president of the Arts Club, I would like to choose structure 4 because it is the cheapest and I can minimize the construction cost.


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(B) (1)

k (m) Area to be painted (m2)

0 3

0.25 2.9375 0.5 2.875

0.75 2.8125 1 2.75

1.25 2.6875

1.5 2.625

1.75 2.5625

2 2.5

Area to be painted was calculated using this formula: 4 −1

2 4 + 𝑘 0.5

(B) (2) Areas to be painted are: 3, 2.9375, 2.875, 2.8125, 2.75, 2.6875, 2.625, 2.5625, and 2.5 This is an Arithmetic Progression (AP) with Common difference, d = 2.9375 - 3 = -0.0625 or d = 2.875 - 2.9375 = -0.0625 When k increases by 0.25m, the area to be painted decreases by 0.0625 m (C)

Area of concrete structure = 4 × 1 −1

2 4 + 𝑘 0.5

= 4 −1

4 4 + 𝑘

= 4 − 1 −1

4𝑘

= 3 −𝑘

4 𝑚2

When 𝑘 → 4 1

4𝑘 → 1

∴ Area → 2𝑚2


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CONCLUSION

I notice that quadratic function and integration so close in our daily life. Solving

problems will be easy by using calculus as well as quadratic function. As the result, we can

calculate and identify problems involving integration (calculus) and quadratic function.


16

Reflection

I found a lot of information while conducting this project. Moreover, this project

encourages the student to think critically to identify and solve problems. It is also encourage

student to gather information using the technologies such as the internet, improve thinking skills

and promote effective mathematical communication.

Lastly, I proposed this project should be continue because it brings a lot of advantages to

the student and also test the student’s understanding in Additional Mathematics.


17

Reference

Additional Mathematics Textbooks

http://en.wikipedia.org/wiki/Calculus

http://en.wikipedia.org/wiki/Integration_(mathematics)


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Appendix

contoh folio addmath 2010 (project work 1)

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