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smk puncak alam42300 kuala Selangor
Selangor DARUL EHSAN
ADDITIONAL MATHEMATICS PROJECT WORK (KPMT) 2014
NAME : NOR HIDAYAH BINTI WAHID
IC NO. : 970910385016
CLASS : 5 BESTARI
TEACHER’S NAME : PN ROHAYAH SANI
ADDITIONAL MATHEMATICS PROJECT WORK (KPMT) 2014 Page  1
CONTENT
CAPTION PAGE
ACKNOWLEDGEMENT 3
OBJECTIVE 4
INTRODUCTION 5
PART 1 6,7
PART 2 8,9,10,11,12
PART 3 13,14,15,16,17
FURTHER EXPLORATION 18
REFLECTION 19
CONCLUSION 20
REFERENCE 21
ADDITIONAL MATHEMATICS PROJECT WORK (KPMT) 2014 Page  2
ACKNOWLEDGEMENT
Grace be upon Allah, with his blessing, I completed my Additional Mathematics Project Work. Of course, it was very hard work but the project was most interesting. I have learned a lot from it. But then again, all of this would not have been possible without a few people in my life.
First of all, I would like to thank my Additional Mathematics teachers, Pn. Rohaya binti Sani for helping me and also my classmates. She never fails to teach and guide us so we can complete this project work with flying colours.
The other person who played an important role in helping me in helping me to complete my project is my parents, they have been very helpful and supportive of me. I would also like to thank them for the financial support they have given me.
Last but not least, I would like to thank to all my friends for giving and share the information that they have with me.
ADDITIONAL MATHEMATICS PROJECT WORK (KPMT) 2014 Page  3
OBJECTIVES
The objectives of carrying out this project work are:
To apply and adapt a variety of problemssolving strategies to solve problems.
To improve thinking skills.
To promote effective mathematical communication.
To develop mathematical knowledge through problem solving in a way that increases students’ interest and confidence.
To use the language of mathematics to express mathematical ideas precisely.
To provide learning environment that stimulates and enhances effective learning.
To develop positive attitude towards mathematics
ADDITIONAL MATHEMATICS PROJECT WORK (KPMT) 2014 Page  4
INTRODUCTION
The idea of calculus had been developed earlier in Egypt, Greece ,China, India,Iraq, Persia and Japan. The use of calculus began in Europe,during the 17 th century,when Isaac Newton and Gottfried Wilhelm Liebniz built on the work of earlier mathematics to introduce the basic principles. The development of calculus was built on earlier concepts of instantaneous motion and area under the curve .
Application of differential calculus include computations involving velocity and acceleration, the slope of curve and optimization. Application of integral calculus include computations involving are,volume,arc length,centre of mass , work and pressure. Calculus is also used to gain a more orecise understanding of nature of space,time and motion.
ADDITIONAL MATHEMATICS PROJECT WORK (KPMT) 2014 Page  5
PART 1GOTTFRIED WILHELM LEIBNIZ (PIONEER OF MODERN CALCULUS)
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ADDITIONAL MATHEMATICS PROJECT WORK (KPMT) 2014 Page  7
Born on First July 1646 in Leipzig Saxony
His parent name was Friedrich Leibniz and Catharina Schmuck
His father died when Leibniz was six years old,
and from that point on he was raised by his mother
which was his father's third wife.
Leibniz's father had been a Professor of Moral Philosophy at the University of Leipzig and the
boy inherited his father's personal library at the age of seven.
He enrolled in his father's former university at age
15, and completed his bachelor's degree in philosophy in December
1662
In 1666, at age 20, Leibniz wrote his first book, On the Art of Combinations, the
first part of which was also
his habilitation thesis in philosophy
ADDITIONAL MATHEMATICS PROJECT WORK (KPMT) 2014 Page  8
Leibniz is credited, along with Sir Isaac Newton, with the discovery of calculus (that
comprises differential and integral calculus). According to Leibniz's
notebooks, a critical breakthrough occurred on November 11, 1675,
when he employed integral calculus for the first time to find
the area under the graph of a function y = ƒ(x)
The product rule of differential calculus is still called "Leibniz's law". In addition, the theorem that
tells how and when to differentiate under the integral sign is called
the Leibniz integral rule.
He introduced several notations used to this day,
for instance the integral sign ∫ representing an
elongated S, from the Latin word summa and the d used
for differentials, from the Latin word differentia.
Leibniz exploited infinitesimals in developing the calculus,
manipulating them in ways suggesting that they
had paradoxical algebraic properties. George Berkeley, in a tract calledThe Analyst and also in De Motu, criticized these. A recent
study argues that Leibnizian calculus was free of
contradictions, and was better grounded than Berkeley's
empiricist criticisms.
From 1711 until his death, Leibniz was engaged in a dispute
with John Keill, Newton and others, over whether Leibniz had
invented the calculus independently of Newton. This
subject is treated at length in the article LeibnizNewton
controversy.
Infinitesimals were officially banned from mathematics by the
followers of Karl Weierstrass, but
survived in science and engineering, and even in rigorous mathematics,
via the fundamental computational device
known as the differentials.
Beginning in 1960, Abraham Robinson worked out a rigorous
foundation for Leibniz's infinitesimals, using model
theory, in the context of a field of hyperreal numbers. The
resulting nonstandard analysis can be seen as a belated
vindication of Leibniz's mathematical reasoning.
Robinson'stransfer principle is a mathematical implementation of
Leibniz's heuristic law of continuity, while the standard part function implements the
Leibnizian transcendental law of homogeneity.
part 2
A car travel along a road and its velocitytime graph function is illustrated in Diagram 1. The staright line PQ is parallel to the straight line RS.
a) From the graph,find
(i) the acceleration of the car in the first hour,
When t = 0
V = 60 (0) + 20
V = 20
When t = 1
V = 60(1) + 20
V = 80
= 80km /h−20km /h
1hour
=60kmh−2
(ii) the average speed of the car in the first two hours.
Area of A = 12× (20+80 )×1
= 50 km
Area of B = 12×(0.5+1)×80
= 60 km
2 hours =totaldistancetotal time
= 110km2hours
ADDITIONAL MATHEMATICS PROJECT WORK (KPMT) 2014 Page  9
¿55kmh−1
b) What is the significance of the position of the graph
(i) above the taxis
The position of the graph above the taxis shows that the car travel to its destination.
(ii) below the taxis
The position of the graph below the taxis means that the car is going in the opposite direction, it travel back where it comes from.
c) Using two different methods, find the total distance travelled by the car
1st method is area under the graph
o Area A = 50 kmo Area B = 60 km
o Area C = 12×(1.5+0.5)×80
¿ 12×(2hours)×80
= 80 km
Hence the total distance travelled by the car
50km+60km+80km=190 km
2nd method is integration
Since PQ is parallel to RS, the gradient of RS is also 160.
Instead of using y=mx+c
We use v=mt+c
Substitude R (2.5,0) into v=mt+c
0=−160 (2.5 )+c
ADDITIONAL MATHEMATICS PROJECT WORK (KPMT) 2014 Page  10
c=400
Hence, the equation of RS is v=−160 t+400
The gradient of TU is 160
v=mt+c
v=160 t+c
Substitute U (4,0) into v=160 t+c
0=160 (4 )+c
c=−640
Hence, the equation of TU is v=160 t−640
Area of A ¿∫0
1
V dt
¿∫0
1
(60 t+20 )dt
=[ 60 t 2
2+20 t ]10
¿ [30 t2+20 t ]10
=[30 (1 )2+20 (1 )−30 (0 )2+20(0)]
=50
Hence, the distance travelled is 50 km
Area of B ¿∫1.0
1.5
v dt
¿∫1.0
1.5
80dt
ADDITIONAL MATHEMATICS PROJECT WORK (KPMT) 2014 Page  11
¿ [ 80 t ] 1.51.0
¿ [80 (1.5 )−80(1.0)]
¿40
So the distance travelled by the car is 40 km
Area of C ¿∫1.5
2
(−160 t+320 t )dt
¿ [−160 t2
2+320 t ] 2
1.5
¿ [−80 t 2+320t ] 21.5
¿¿
= 20
So,the distance travelled by the car is 20 km
Area of D ¿∫2.5
3
(−160 t+400 )dt
¿ [−160 t2
2+400 t ] 3
2.5
¿ [−80 t 2=400 t ] 32.5
¿ [−80 (3 )2+400 (3 )−(−80 (2.5 )2+400(2.5))]
¿−20
Area ¿−20
¿20
ADDITIONAL MATHEMATICS PROJECT WORK (KPMT) 2014 Page  12
So, the distance travelled is 20 km
Area of E ¿∫3
3.5
(−80 )dt
¿ [−80 t ] 3.53
¿ [−80 (3.5 )−(−80 (3))]
¿−40
Area ¿−40
¿40
So, the distance travelled is 40 km.
Lastly,
Area of F ¿∫2.5
4
(160 t−640)dt
¿ [ 160 t2
2−640 t ] 4
2.5
¿ [ 80 t−640 t ] 42.5
¿−20
Area ¿−20
¿20
Total distance travelled by the car is
50+40+20+20+40+20=190 km
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d) Based on the above graph, write an interesting story of the journey in not more than 100 words.
Gideon with his friends went to the Port Dickson which is 110 km far away from their hometown that is Puncak Alam. Gideon drives his car from home at velocity of 60 km/h. After an hour, he drives with the velocity of 80 km/h constantly. Then, he decelerate for the last half an hour before destination. For 30 minutes, they have fun at the place and then, they back home. For the first 30 minutes, Gideon drives 80 km/h. Then, he drives with the velocity of constant and decelerate for the last half an hour until they arrived home.
PART 3
Diagram 2 shows a parabolic satellite disc which is symmetrical at the yaxis. Given that the diameter of the disc is 8 m and the depth is 1 m.
a) Find the equation of the curve y = ƒ(x) .
Since the parabolic satellite disc is symmetrical at the yaxis, the curve y = ƒ(x) can be written on y = α x2+ c . It can be seen that the curve y = ƒ(x) cuts the yaxis at the point (0,4). Substitute (0,4) into y = α x2+ c, and you will get y = α x2+ 4
y
4 m 4 m
ADDITIONAL MATHEMATICS PROJECT WORK (KPMT) 2014 Page  14
(4,5)1 m
4
x0 4
Substitude the point (4,5) into y=a x2+45=a (4 )2+4
¿ 116
y= f (x ) is now written as y=1
16x2+4
So , f ( x )= x2
16+4
b) To find the approximate area under a curve, we can divide the region into several vertical strips, then we add up the areas of all the strips. Using a scientific calculator or any suitable computer software, estimate the area bounded by the curve y = ƒ(x) at (a), the xaxis, x = 0 and x = 4.
y=f (x )= x2
16+4
f ( x )= x2
16+4
1st strip => When x=0,
f (0 )= 02
16+4
¿4
ADDITIONAL MATHEMATICS PROJECT WORK (KPMT) 2014 Page  15
2nd strip => When x=0.5 ,
f (0.5 )=0.52
16+4
¿4.0156
3rd strip => When x=1 ,
f (1 )= 12
16+4
¿4.0625
4th strip => When x=1.5 ,
f (1.5 )=1.52
16+4
¿4.1406
5th strip => When x=2 ,
f (2 )= 22
16+4
¿4.25
6th strip => When x=2.5
f (2.5 )=2.52
16+4
¿4.3906
7th strip => When x=3
f (3 )= 32
16+4
¿4.5625
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8th strip => When x=3.5
f (3.5 )=3.52
16+4
¿4.765
(i) Area of 1st strip ¿4.0000×0.5¿2.0000
2nd strip ¿4.0156×0.5¿2.0078
3rd strip ¿4.0625×0.5¿2.0313
4th strip ¿4.1406×0.5¿2.0703
5th strip¿4.2500×0.5 ¿2.1250
6th strip ¿4.3906×0.5¿2.1953
7th strip¿4.5625×0.5 ¿2.2813
8th strip¿4.7656×0.5 ¿2.3828
Total area ¿17.0938m2
(ii) Area of 1st ¿2.00782nd ¿2.03133rd ¿2.07034th¿2.12505th¿2.19536th¿2.28137th ¿2.38288th ¿5×0.5
ADDITIONAL MATHEMATICS PROJECT WORK (KPMT) 2014 Page  17
¿2.5
Total area ¿17.5938m2
(iii) Area of 1st and 2nd ¿4.01563rd and 4th ¿4.14065th and 6th ¿4.39067th and 8th ¿4.7656
Total area ¿17.3124m2
c) i) Calculate the area under the curve using integration.
=∫0
4
y dx
=∫0
4
( x2
16+4)dx
=[ x3
16 (3 )+4 x]4
0
=[[ 43
48+4 (4 )]−(0)]
=17 13
(ii) Compare your answer in c (i) with the values obtained in (b). Hence, discuss which diagram gives the best approximate area.
 The diagram 3 (iii) gives the best approximate area,which 17.3124 m2
(iii) Explain how you can improve the value in c (ii).
 We can improve the value in c (ii) by having more strips from x = 0 to x = 4
d) Calculate the volume of satellite disc.
y= x2
16+4
ADDITIONAL MATHEMATICS PROJECT WORK (KPMT) 2014 Page  18
x2=16( y−4 )
x2=16 y−64
volume=π∫4
5
x2dy
=π∫4
5
(16 y−64 )dy
¿ π [16 y2
2−64 y ] 5
4
¿ π [8 y2−64 y ] 54
¿8 πm2
ADDITIONAL MATHEMATICS PROJECT WORK (KPMT) 2014 Page  19
FURTHER EXPLORATIONA gold ring in diagram 4(a) has the same volume as the solid of revolution obtained when
the shaded region in diagram 4(b) is rotated about 360° about the xaxis.
f ( x )=1.2−5 x2
let , f ( x )= y
y=1.2−5 x2
When x=0.2
y=1.2−5 (0.2 )2
y=1
Find
(a) The volume of gold needed,
y=1.2−5 x2
y2=(1.2−5 x2 )2
y2=1.44+25 x4−12 x2
Volume of gold needed
¿ π ∫−0.2
0.2
(1.44+254−12x2 )dx−π ∫−0.2
0.2
1dx
¿ π [1.44 x=2.5x5
5−
12x3
3 ] 0.2−0.2
−π [x ] 0.2−0.2
¿ π [1.44 x−5 x5−4 x3 ] 0.2−0.2
−π [x ] 0.2−0.2
¿0.5125 π−0.4 π
ADDITIONAL MATHEMATICS PROJECT WORK (KPMT) 2014 Page  20
¿0.36191 cm3
(b) The cost of gold needed for the ring.(Gold density is 19.3 gcm−3. The price of the gold can be obtained from the goldsmith)
Gold density is 19.3 gcm−3
Let x be the weight of gold in g 1cm3=19.3
0.36191 cm3=x gSo x=6.9849Finally, 150×6.9849
¿ RM 1047.74
REFLECTION
What have you learnt while conducting the project? What moral values did you practise? Express your fellings and opinions creatively through the usage of symbols, drawings, lyrics of a song or a poem.
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CONCLUSION
I have done many researches throughout the internet and discussing with some friends who have helped me a lot in completing this project.Through the completion of this project work, i have also learned many skills and techniques in ICT. In addition, this project work helps me to understand more about the uses of calculus in our daily life. This project work also exposes me to the techniques of application of additional mathematics in real life situations. While conducting this project work, a lot of information that i found today. Apart from that, this project work encourages the students to work together and share their knowledge. It also encourages the student to gather information from the internet, improve thinking skills, and promote effective mathematical communication. Last but not least, I proposed this project should be continued to the next generation because it brings a lot of moral value to the students and also test the students’ comprehending additional mathematics.
ADDITIONAL MATHEMATICS PROJECT WORK (KPMT) 2014 Page  22
REFERENCE
http://www.bumigemilang.com/category/kerjaprojekmatematiktambahanadditionalmathematicsprojectwork/
http://www.physicalsclassroom/class/1dkin/u114a.cfm
http://en.wikipedia.org/wiki/GottfriedWelhelmLeibniz
ADDITIONAL MATHEMATICS PROJECT WORK (KPMT) 2014 Page  23