AddMaths Project Work 2 2011

8/6/2019 AddMaths Project Work 2 2011

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Name : Yong Tian Kai

Class : 5A

School : SMJK TIONG HUA

I/C Num : 931024-12-6229

Angka Giliran: HF007A046


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Table of Content

Topic Page

Title 1

Table of Content 2

Questions 3-5

Introduction 6-8

Task Specification 9-11

Answers 12-26

Conclusion 27

Reflection 28


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INTRODUCTION

There are a lot of things around us related to circles or parts of a circle. A circle is

a simple shape of Euclidean geometry consisting of those points in a plane which is the

same distance from a given point called the centre. The common distance of the points

of a circle from its center is called its radius.

Circles are simple closed curves which divide the plane into two regions,

an interior and an exterior. In everyday use, the term "circle" may be used

interchangeably to refer to either the boundary of the figure (known as the

perimeter) or to the whole figure including its interior. However, in strict

technical usage, "ci rcle" refers to the perimeter while the interior of the circle is

called a disk. The circumference of a circle is the perimeter of the circle (especially when

referring to its length).

A circle is a special ellipse in which the two foci are coincident. Circles are conic

sections attained when a right circular cone is intersected with a plane perpendicular to

the axis of the cone.

The circle has been known since before the beginning of recorded history. It is

the basis for the wheel, which, with related inventions such as gears, makes much

of modern civilization possible. In mathematics, the study of the circle has helped

inspire the development of geometry and calculus. Circles had been used in daily lives

to help people in their living.


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Brief History of Geometry

Geometry was one of the two fields of pre-modern mathematics the other being the study ofnumbers.

Geometry began with a practical need to measure shapes. It is the science of shape and sizeof things. It is believed that geometry first became important when an Egyptian Pharaoh wanted totax farmers who raised crops along the Nile River. To compute the correct amount of tax thePharaohs agents had to be able to measure the amount of land being cultivated. Around 2 900 BC(Before Century) the first Egyptian pyramid was constructed. Knowledge of geometry was essentialfor building pyramids, which consisted of a square base and triangular faces. The earliest record of aformula for calculating the area of a triangle dates back to 2 000 BC. The Egyptians (5000 500 BC)and the Babylonians (4 000 500 BC) developed practical geometry to solve everyday problems. Itwas the early Greeks (600 BC 400 AD) that developed the principles of modern geometrybeginning with Thales of Miletus (624 547 BC). Thales is credited with bringing the science ofgeometry from Egypt to Greece. Thales studied similar triangles and wrote the proof thatcorresponding sides of similar triangles are in proportion. The next great Greek geometer wasPythagoras (569 475 BC). Pythagoras is regarded as the first pure mathematician to logicallydeduce geometric facts from basic principles. Pythagoras founded a brotherhood called thePythagoreans, who pursued knowledge in mathematics, science, and philosophy. Some peopleregard the Pythagorean School as the birthplace of reason and logical thought. The most famousand useful contribution of the Pythagoreans was the Pythagorean Theorem. The theory states thatthe sum of the squares of the legs of a right triangle equals the square of the hypotenuse. Euclid of

Alexandria (325 265 BC) was one of the greatest of all the Greek geometers and is considered bymany to be the father of modern geometry . Euclid is best known for his 13-book treatise TheElements.The Elementsis one of the most important works in history and had a profound impact on thedevelopment of Western civilization.

Euclid began The Elements with just a few basics, 23 definitions, 5 postulates, and 5 common notions

or general axioms. An axiom is a statement that is accepted as true. From these basics, he provedhis first proposition. Once proof was established for his first proposition, it could then be used as partof the proof of a second proposition, then a third, and on it went. This process is known as theaxiomatic approach. Euclids Elements form the basis of the modern geometry that is still taught inschools today. Archimedes of Syracuse (287 212 BC) is regarded as the greatest of the Greekmathematicians and was also the inventor of many mechanical devices including the screw, thepulley, and the lever. The Archimedean screw a device for raising water from a low level to a higherone is an invention that is still in use today. Archimedes works include his treatiseMeasurement of a Circle, which was an analysis of circular area, and his masterpiece On the Sphere andthe Cylinderin which he determined the volumes and surface areas of spheres and cylinders. Therewere no major developments in geometry until the appearance of Ren Descartes (1 596 1 650BC). In his famous treatise Discourse on the Method of Rightly Conducting the Reason in the Search forTruth in the Sciences, Descartes combined algebra and geometry to create analytic geometry. Analyticgeometry, also known as coordinate geometry, involves placing a geometric figure into a coordinatesystem to illustrate proofs and to obtain information using algebraic equations. The next greatdevelopment in geometry came with the development of non-Euclidean geometry. Carl Friedrich


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Gauss (1 777 1 855 BC) who along with Archimedes and Newton is considered to be one of thethree greatest mathematicians of all time, invented non-Euclidian geometry prior to the independentwork of Janos Bolyai (1 802 1 860 BC) and Nikolai Lobachevski (1792 1 856 BC). Non-Euclidiangeometry generally refers to any geometry not based on the postulates of Euclid, includinggeometries for which the parallel postulate is not satisfied. The parallel postulate states that through a

given point not on a line, there is one and only one line parallel to that line. Non-Euclidian geometryprovides the mathematical foundation for Einsteins Theory of Relativity. The most recentdevelopment in geometry is fractal geometry. Fractal geometry was developed and popularized byBenoit Mandelbrot in his 1982book The Fractal Geometry of Nature. A fractal is a geometric shape,which is self-similar (invariance under a change of scale) and has fractional (fractal) dimensions.Similar to chaos theory, which is the study of non-linear systems; fractals are highly sensitive to initialconditions where a small change in the initial conditions of a system can lead to dramatically differentoutputs for that system.


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TASK SPECIFICATION

Part I

Identifying the Problem: Question asks to provide information about how Mathematics is used

in baking and cake decorating.

Strategy: Apply knowledge obtained from Additional Mathematics and also consult resources

from the Internet.

Part II

Question 1

Identifying the Problem: Question asks to find the diameter of the baking tray to be used to fit

the cake.

Strategy: Find diameter, d using the formula of volume of cylinder, with the height, h andvolume, v of cake per kilogram and the value of.Question 2(a)

Identifying the Problem: Question asks to find and tabulate the different values heights and

diameters of the baking tray to be used, if the volume remains the same .

Strategy: Find diameter, d using the formula of volume of cylinder and volume, v of cake per

kilogram and the value of with varying values of height, h.Question 2(b)(i)

Identifying the Problem: To state the range of heights unsuitable for baking of cakes and givereasons.

Strategy: Obtain the range of heights not suitable for cakes by compare and contrast with logic

opinion.

Question 2(b)(ii)

Identifying the Problem: Question asks to suggest the dimension most suitable for the cake, and

provide reasons.

Strategy: Compare and contrast and the heights and diameters in table from Question 2(a) and

suggest logical reasons.


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Question 2(c)(i)

Identifying the Problem: Question asks to reduce the equation to linear form in order to display

the relation between h and d. After doing so, question requires us to plot a graph based on the

equation.

Strategy: Using the linear law, and the linear equation, reduce the equation in Question 2(a) by

using logarithm and plot a graph of h against d using software.

Question 2(c)(ii)(a)

Identifying the Problem: Question asks to determine the diameter of the round cake pan if the

height of the cake is 10.5 cm using the graph.

Strategy: Express the height of the cake using logarithms Base 10. Then plot the graph to obtain

the value of diameter, d in logarithmic terms. Finally, express the value without logarithms.

Question 2(c)(ii)(b)Identifying the Problem: Question asks to determine the height of the cake, h obtained if the

diameter of the round tray is 42 cm.

Strategy: Express the diameter of the baking tray, d in logarithms Base 10. Then plot the point

on the graph in order to obtain the logarithm value of the height. Finally, express the value in

normal terms.

Question 3(a)

Identifying the Problem: Question asks to calculate the amount of fresh cream required to

decorate the cake based on the dimensions suggested in Question 2(b)(ii).

Strategy: Using h = 11 cm and d = 46.89293 cm, and the given thickness of the cream = 1 cmuniformly, calculate the amount of fresh cream required.

Question 3(b)

Identifying the Problem: Question asks to give 3 other suggestions for cake shaped which has the

same height and volume as 2(b)(ii), and then calculate the amount of fresh cream to be used .

Strategy: Provide 3 other cake shape suggestions by logic thinking and calculate the dimensions,

given the volume and height are constant. After that, calculate the amount of fresh cream required for

each shape.


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Part III

Identifying the Problem: Question asks to find the dimensions of cake that requires the

minimum amount of fresh cream to decorate by using 2 different method, including calculus.

After that, to state whether if such dimensions are suitable for the baking of the cake and give

reasons.

Strategy: Find the dimension that requires the minimum amount of fresh cream by comparing

values of height against volume of cream used. The second method would be using

differentiation. Finally, apply logic thinking to come to a conclusion, whether the cake is suitable

to be baked or not.

Further Exploration (a)

Identifying the Problem: Question asks to obtain the volume of the first, second, third, and

fourth cakes, and compare the values to determine whether the volumes form a number pattern.

After that, provide explanation and elaboration.

Strategy: Given height of cake is 6.0 cm each and the radius of the largest cake is 31.0 cm, after

that the radius of the second cake is 10% less, the third radius is 10% less than the second cake

and so on. Use the information given to obtain the volumes of the cakes, and compare by

division to determine the existence of the number pattern.

Further Exploration (b)

Identifying the Problem: Question asks to calculate the maximum number of cakes to be baked

by the bakery if the total mass must not exceed 15 kg. After that, verify the answer using other

methods.

Strategy: Express the mass of the cake given 1kg of cake has a volume of 3800 cm3. Then findthe number of terms using the formula to given the sum is a geometric progression. After that,

verify the answer by trial and improvement to prove the number of cakes that the bakery needs to

bake.


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ANSWERS FOR PART I

Uses of Mathematics in Cake Baking and Decorating

One very importantaspect ofcake baking istheamount of ingredientsrequired. This is

becauseacakeshould bereasonable pricing. Atthesametime, beauty isalso a priority incake

baking, ingredientsshouldnot be waster unnecessarily.

Calculus can be applied in cake baking. Particularly, the second derivative is greatly

used. This is because the second derivative allows bakeries to calculate the maximum or

minimum amount of ingredients needed to increase profit and efficiency. This allows positive

growth in business. At the same time, the bakeries will not under-order, or over-order the

ingredients.

In the baking of more complex cakes, such as multi-storey cakes or multi-layered cakes,

progressions are applied. Progressions allow us to calculate the size or volume of a subsequent

layer. Also, it allows us to estimate the quantity of ingredients needed. Usually, geometric

progressions are used.

Lastly, ratios are used in cake baking. More often than not bakers need to estimate the

amount of ingredients used or substitute the ingredient with another if that ingredient is notavailable. For example, we often read cookbooks guiding us to use 3 parts of water to 1 part of

flour. This ratio of water to flour 3:1, allows us to bake a cake of different sizes. Although we

may bake a smaller or larger cake, the flour and water used still obeys the proportions set. We

are then allowed to creatively bake the cake.


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ANSWERS FOR PART II

Answer for Question 1

Given volume ofcake perkilogram = 3800cm3,and = 3.142

Total volume ofcake = 3800cm35kg

= 19000cm3

Givend = 2r,thusr =

Volume ofcake = Volume ofcylinder

Volume ofcake = Basearea HeightVolume ofcake =

19000 = (3.142)r27

=

= d = 58.78339783

d58.7834cmTherefore,thediameter ofthe bakingtraythatcan be used is58.7834cm.


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Answer for Question2(a)

Volume ofcake = Volume ofcylinder

Volume ofcake = Basearea HeightVolume ofcake = 19000 = (3.142)r2h

=

=

d =

d =

Usingtheequation,thetable is formedas below:

h/cm d/cm

1.0 155.52625

2.0 109.97366

3.0 89.79312

4.0 77.76313

5.0 69.55345

6.0 63.49333

7.0 58.78339

8.0 54.98683

9.0 51.54208

10.0 49.18172

11.0 46.89293

12.0 44.89656

13.0 43.13522

14.0 41.56613

15.0 40.15670


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Answer for Question2(b)(i)

Therange ofheightsnotsuitable forthecake ish .

This is becausethediameter ofthecakes would betoo largeto fit inthe oven. Also,thecake would

betoo wideand unattractive.

Answer for Question2(b)(ii)

Thedimensionthat I think is mostsuitable forthecake ish = 11 cm andd = 46.89293cm.

This is becausethecake isableto fit inthe oven. A cakethat istoo tall ortoo high may beeasily

burnt.

Answer for Question2(c)(i)

d =

= = + 2.1918

Y= mX + C

Theequation of linearrelation betweenhandd is = + 2.1918.


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0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 2.1918 1.9418 1.6918 1.4418 1.1918 0.9418 0.6918 0.4418 0.1918


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Answer for Question2(c)(ii)(a)

Whenh = 10.5cm, = = 1.0212

From thegraph, when = 1.0212,then = 1.6812 = 1.6812

d = 101.6812

d = 47.9954cm

Therefore, whentheheight ifthecake is 10.5cm,thediameter oftheroundcake panneeded

must be largerthan47.9954cm.

Answer for Question2(c)(ii)(b)

Whend = 42cm, = = 1.6232

From thegraph, when

= 1.6232,then

= 1.1372

= 1.1372h = 10

1.1372

h = 13.7151 cm

Therefore,theheight ifthecake whenthediameter is42cm is 13.7151 cm.


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Answer for Question3(a)

h = 11 cm,d = 46.89293cm

Thickness ofcream = Height ofcream

Amount ofcream = Volume ofcream

Volume ofcream = Volume ofcream atsides + Volume ofcream atthetop

Volume ofcream atthetop = Area oftop surface Height ofcream= 1 cm= (3.142)

(1)

= 1727.2727cm3

Volume ofcream atsides = Area ofsides Height ofcream= (circumference ofcakeheight ofcake) Height ofcream= (2 10) 1= 23.142 10 1= 1473.37586cm

3

Therefore,estimatedamount of freshcream requiredto decoratethecake

= 1727.2727 + 1473.37586

= 3200.65cm3


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Answer for Question3(b)

h = 11cm, volume ofcake = 19000 cm3

1st

shape- Cuboid

Volume ofcake = Basearea Height19000 cm

3= Basearea 11 cm

Basearea = 1727.2727cm2

Length Width = 1727.2727cm2By Trial and Improvement, 1727.2727cm

2= 43.1818cm40 cm

Volume ofcream required

= Volume of Cake with Cream Volume of Cake without Cream= (43.1818+2) (40+1) (11+1) (19000)= 22229.4456 19000= 3229.4456cm

3


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2nd

shape Triangular-basedshape

Volume ofcake = Basearea Height

19000 cm

3

= Basearea 11 cmBasearea = 1727.2727cm

2

Length Width = 1727.2727cm2By Trial and Improvement,3454.5454cm

2= 57.5757cm 60 cm

Volume ofcream required

= Volume of Cake with Cream Volume of Cake without Cream= (57.5757+1) (60+2) (11+1) 19000= 2790.1604cm

3


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3rd

shape Pentagonal-basedshape

Volume ofcake = Basearea Height19000 cm

3= Basearea 11 cm

Basearea = 1727.2727cm2

5 Area of Isoceles Triangle ofeachside = 1727.2727cm2Area ofeach isosceles Triangle = 345.4545cm

2

(

Length

Width) ofeach isoscelestriangle = 345.4545cm

2

By Trail and Improvement,690.909cm2

= 25cm 27.6363cmVolume ofcream required.

= Volume of Cake with Cream Volume of Cake without Cream= (5) (25+1) (27.6363) (11+1) (19000)= 2556.314cm

3


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Conclusion for Question3(c)

Thethreesuggestedshapearecuboidshapecake,triangular-basedcakeand pentagonal-based

cake.

The volume ofcream required forcuboidshapedcake = 3229.4456cm3.

The volume ofcream required foratriangular basedcake is2790.1604cm3.

The volume ofcream required fora pentagonal-basedcake is2556.314cm3.

Therefore, from the values,the pentagonal-basedcakerequiresthe leastamount of fresh

cream to be used, which is only2556.314cm3.


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ANSWERS FOR PART III

Method 1 By comparing values of height against volume of cream used

According to the table above, the minimum volume of cream used is 3303.66 cm3 when h = 18cm.

When h = 18cm, r = 18.3 cm


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Method2 Usingdifferentiation

Assuming that the surface area of the cake is proportionate to the amount of fresh cream needed to

decorate the cake.*

Formula for surface area of cake = h =

Surface area in contact with cream

= 2 + 2 = 2 + The values, when plotted into a graph will from a minimum value that can be obtained throughdifferentiation.

0 = 6.284r3 3800038000 = 6.284r36407.104 = r3

18.22 = rWhen r = 18.22 cm, h = 18.22 cm

The dimensions of the cake that requires the minimum amount of fresh cream to decorate is

approximately 18.2 cm in height and 18.2 cm in radius.

I would bake a cake of such dimensions because the cake would not be too large for the cutting or

eating of said cake, and it would not be too big to bake in a conventional oven.

* The above conjecture is proven by the following

When r = 10,

~ the total surface area of the cake is 4114.2 c~ the amount of fresh cream needed to decorate the cake is 4381.2 c~ the ratio of total surface area of cake to amount of fresh cream needed is 0 .94

When r = 20,

~ the total surface area of the cake is 3156.8 c~ the amount of fresh cream needed to decorate the cake is 3308.5 c~ the ratio of total surface area of cake to amount of fresh cream needed is 0 .94

Therefore, the above conjecture is proven to be true.


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ANSWERS FOR FURTHER EXPLORATION

Answer for (a)

Volume of cake = Given height of the cake = 6cm and radius of largest cake = 31cm which decreases by 10%,

Volume of FIRST cake = (3.142) (31)2 (6)

= 18116.772 cm3

Volume of SECOND cake = (3.142) (31 90%)2 (6)= 14674.585 cm

3

Volume of THIRD cake = (3.142) (27.9 90%)2 (6)= 11886.414 cm3

Volume of FOURTH cake = (3.142) (25.11 90%)2 (6)= 9627.995 cm

3

18116.772 cm3

, 14674.585 cm3

, 11886.414 cm3

, 9627.995 cm3

SECOND FIRST: 14674.585 18116.772 = 0.81THIRD SECOND: 11886.414 14674.585 = 0.81FOURTH

THIRD: 9627.995

11886.414 = 0.81

The numbers form a geometric progression with the first number, a = 18116.772 and acommon ratio, r = 0.81.

The conjecture is proven. Volume of cake form a geometric progression.


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Answer for (b)

Sn = =

15 kg = 57000 cm3

57000 >

11400 > 18116.772(1-0.8n)

0.629 > 1-0.8n

0.371 > 0.8n0.371 < 0.8

n

log 0.371 < n log 0.8 < n

4.444 < n

n = 4

Verification of answer

If n = 4

Total volume of 4 cakes

= 18116.772

+ 14676.585

+ 11886.414

+ 9627.995

= 54307.766 Total mass of cakes

= 14.29 kg

If n = 5

Total volume of 5 cakes

= 18116.772 + 14676.585 + 11886.414 + 9627.995 + 7798.676 = 62106.442 Total mass of cakes

= 16.34 kg

Total mass of cakes must not exceed 15 kg.

Therefore, maximum number of cakes needed to be made = 4


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CONCLUSION

After doing research, answering questions, drawing graphs, and some problem solving, I

saw that the usage of geometry, calculus and progressions are important in daily life.

Geometry is the study of angles and triangles, perimeter, area and volume. It differs fromalgebra in that one develops a logical structure where mathematical relationshipsare proved and applied.

A geometric progression (GP), also known as a geometric sequence, is a

sequence of numbers where each term after the first is found by multiplying theprevious one by a fixed non-zero number called the common ratio.

Differentiation is essentially the process of finding an equation which will giveyou the gradient (slope, "rise over run", etc.) at any point along the curve. Sayyou have y = x^2.The equation y' = 2x will give you the gradient of y at any pointalong that curve.

As the conclusion, geometry, calculus and progressions are part of our

necessities. Thus, we should be thankful of the people who contribute in the idea of

geometry, calculus and progressions because without them, we cant done the multi-

storey cake, and its hard to find out the volume of ingredients needed for the cake.


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REFLECTION

I have done many researches throughout the internet and discussing with friends who have helped me alot in completing this project. Through the completion of this project, I have learned many skills

and techniques. This project really helps me to understand more about the uses of progressions in ourdaily life.This project also helped expose the techniques of application of additional mathematics in real life

situations.

While conducting this project, I found a lot of information. I have learnt how to bake a wedding tieredcake stands with good quality and proper height.Apart from that, this project encourages the student to work

together and share their knowledge. It is also encourage student to gather information from the internet, improvethinking skills and promote effective mathematical communication.Last but not least, I proposed this project

should be continue because it brings a lot of moral values to the student and also test the students understanding inAdditional Mathematics.

AddMaths Project Work 2 2011

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