calculus 3 project 1

CALC 143-71: Project #1Due on Friday, Jan 18, 2013

Linda Patton 3:10

Brandon Kirklen, Katie Sutherlin, and Grace Cowell

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Brandon, Katie, and Grace CALC 143-71 (Linda Patton 3:10): Project #1

Contents

Problem Intro 3

Part (a) 3

Part (b) 4

Part (c) 6

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Brandon, Katie, and Grace CALC 143-71 (Linda Patton 3:10): Project #1

Problem Intro

An infinite cake is created as follows. The base of the cake is cylindrical. It is 4 inches tall and 12 inches in

diameter. There are infinitely many cylindrical layers in the cake, and each layer is half as tall and has 23

the diameter of the previous layer.

Part (a)

How tall is the cake?

In order to find the height of the cake, we must first set up the sequence that will model that height. We

are told that the N1 layer is 4 inches tall with each successive layer being one half the previous layer. The

created sequence must model the following pattern:

4, 2, 1,1

2,

1

4, ...

From that sequence we can create an equation driven form to find the height of any layer N.

F (n) = 8

(1

2

)n

This equation can then in turn be used to create a series that will approximate the total height of the cake.

∞∑n=1

8

(1

2

)n

= 8 ∗∞∑

n=1

(1

2

)n

(by the properties of convergant series

)This resulting series is geometric with a = 1/2 and r = 1/2, where a is the first term and r is the common

ratio. The sum of a geometric series is equal to first term1−common ratio thus:

∞∑n=1

(1

2

)n

=a

1 − rwith a =

1

2and r =

1

2

=12

1 − 12

= 1

Thus the final height is

Height =

∞∑n=1

8

(1

2

)n

= 8 ∗∞∑

n=1

(1

2

)n

= 8 ∗ 1

= 8 inches

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Brandon, Katie, and Grace CALC 143-71 (Linda Patton 3:10): Project #1 Part (a)

Part (b)

What is the total volume of the cake?

This problem begins with the basic equation of a volume. The shape we are finding here is a series of

cylinders which has the volume equation of V = πr2 ∗ h. The challenge is that r and h are not simple

variables but rather dynamic variables which have a set value at each iteration of cake layers. By finding

the equations for each of these series then multiplying them in accordance to the properties of series, we can

find the volume of each layer then add them all up, the function of a series.

The first step is to find the two series which would approximate the radius and height of each layer. We’ve

already found a series which accurately models the height of each layer in part (a) of this project which will

fit nicely into this equation but we have yet to find one which models the change in radius. By applying the

same logic we did to the first series, we know the change between each layer is 23 and n1 = 6 thus the two

series would be:

r =

∞∑n=1

9

(2

3

)n

and

h =

∞∑n=1

8

(1

2

)n

Using these two series as well as our initial equation we get:

V = π ∗

( ∞∑n=1

9

(2

3

)n)2

∗

( ∞∑n=1

8

(1

2

)n)

Using series properties, we can simplify the second term:( ∞∑n=1

9

(2

3

)n)2

=

∞∑n=1

92

((2

3

)n)2

=

∞∑n=1

81

(2

3

)2

n

=

∞∑n=1

81

(4

9

)n

Part (b) continued on next page. . . of 9

Brandon, Katie, and Grace CALC 143-71 (Linda Patton 3:10): Project #1 Part (b) (continued)

Then plug this into the volume equation:

V = π ∗∞∑

n=1

81

(4

9

)n

∗∞∑

n=1

8

(1

2

)n

= π ∗∞∑

n=1

81 ∗ 9

(4

9∗ 1

2

)n

= 648π

∞∑n=1

(2

9

)n

Geometric Series

= 648π ∗29

1 − 29

= 648π ∗ 2

7

Volume of Cake =1296

7π

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Brandon, Katie, and Grace CALC 143-71 (Linda Patton 3:10): Project #1 Part (b)

Part (c)

What is the surface area of the cake? This part is the trickiest, because you should compute exactly the

amount of cake that would need to be frosted. Don’t frost parts of the cake that are covered by other layers.

Also, don’t frost the bottom of the base.

The problem of finding the surface area has two components, the sides as well as the tops of each layer. This

gives us that the total surface area is equal to the sum of the area of the top and walls or S(T ) = S(U) +S(W ).

Because we are only frosting the parts of each layer that are exposed to the air, it at first seems like we

have yet another series to compute. This series would be the area of each cake top minus the area of the

bottom of the layer above. However, thinking topologically, we can easily see a simpler solution.

At first glance it appears this would be the visual representation of the areas we are attempting to compute.

Each layer is separate and the areas are distinct from each other.

Part (c) continued on next page. . . of 9

Brandon, Katie, and Grace CALC 143-71 (Linda Patton 3:10): Project #1 Part (c) (continued)

However, by stacking these layers on top of each other we get the following.

This, of course, is mathematically equal to a simple circle of diameter 12 inches.



The formula for the area of a circle is A = π ∗ r2 and r = 6 thus the surface area of the top of the cake is

S(U) = Surface Area of Upper

S(U) = π ∗ r2 and r = 6

S(U) = π ∗ 62

= 36π

The second part of the surface area is the area of the walls of the cake. This portion is a bit more difficult

mathematically. We begin with the formula for the surface area of the walls of a cylinder S(W ) = 2 ∗π ∗ r ∗hthen define our variables. First, the radius of each layer:

r =

∞∑n=1

9 ∗(

2

3

)n

= 9 ∗∞∑

n=1

(2

3

)n

Next, the height of each layer:

h =

∞∑n=1

8 ∗(

1

2

)n

= 8 ∗∞∑

n=1

(1

2

)n

Then, we combine these into the original equation:

S(W ) = 2 ∗ π

(9 ∗

∞∑n=1

(2

3

)n

∗ 8 ∗∞∑

n=1

(1

2

)n)

= 2 ∗ 9 ∗ 8 ∗ π ∗

( ∞∑n=1

(2

3

)n

∗∞∑

n=1

(1

2

)n)

= 144π ∗∞∑

n=1

(2

3∗ 1

2

)n

= 144π ∗∞∑

n=1

(1

3

)n

(Geometric Series)

= 144π ∗13

1 − 13

= 144π ∗ 1

2

=144

2π

= 72π



These two parts S(U) and S(W ) are then combined to form the final answer:

S(T ) = S(U) + S(W )

= 36π + 72π

= 108π

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calculus 3 project 1

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