# Estimate the length of rope in the spiral. The width of the boards is 8 cm. Explain how you get your result

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Introduction

03/09/03 11 AM 2 Assignment 1 Johanna Trautwein

A. Estimate the length of rope in the spiral. The width of the boards is 8 cm. Explain how you get your result.

In the picture, the width of the 2nd to last board = 1 cm

The width of each board in real size = 8 cm

Therefore, the ratio of length at the 2nd to last board: real size is 1: 8

The width of the rope on paper = 0.25 cm

Therefore, the width of the rope = 0.25 cm x 8 = 2 cm.

Let the diameter of the spiral = 6.5 board widths

Then the diameter of the spiral = 6.5 x 8 cm = 52 cm

The length of the rope in the spiral = π (diameter of most outer loop + diameter of 2nd most outer loop + diameter of 3rd most outer loop + … + diameter of 12th most outer loop)

Diameter of most outer loop = diameter of the spiral

Diameter of 2nd most outer loop = diameter of the spiral - 2 width of rope

Diameter of 3rd

Middle

n = no of terms

L = last term

Svn = Sum of n cylinders’ volume

ri = radius of the ith cylinder

Imagine approximating this cone with cylinders of equal height. The radius of each cylinder is the radius of the cone’s cross section, at the height of the cylinder’s base.

a. There are ten cylinders. What is the height of each cylinder?

The ten cylinders are equal in height

Sum of all the cylinders’ height = height of the cone

Therefore the height of each cylinder = h/10

b. Each cylinder has a different radius.

i. Find the radius of the top cylinder

The cylinders are all in height and the slope of the cone is constant.

Therefore the difference between the radiuses of each cone is constant.

So when the radius of the bottom cylinder = r

Then, the top cylinder’s radius = r/10

ii. Find the radius of the next cylinder down

If at the base, the radius = r

Conclusion

πr2h

Svn = π(n/2 (a + L))2 h

Sv100 = π(50 (r/100 + r))2 h

Sv100 = π(50r/100 + 50r)2 h

Sv100 = π(r/2 + 100r/2)2 h

Sv100 = π(101r/2)2 h

Sv100 = (10201πr2h)/4

e. Now imagine using n cylinders. Find the total volume of the n cylinders.

Svn = π(n/2 (a + L))2 h

Svn = π(n/2 (r/n + r))2 h

Svn = π(r/2 + nr/2)2 h

Svn = π(r2/4 + n2r2/4)h

Svn = πr2h/4 + πn2r2h/4

f. As n gets larger (going to infinity), what happens to the volume of the cylinders?Explain why this is the volume of the cone. Hint: Can you express the volume as πr2h(something)?

Volume of a cone = πr2h(1/3)

Svn = 1/4(πr2h/4 + πn2r2h)

Lim 1/4(πr2h + πn2r2h)

n→∞

Lim 1/4(πr2h/n2 + πr2h)

n→∞

As n→∞

πr2h/n2→0

Lim = πr2h/4

n→∞

Therefore as n→∞, Svn→ πr2h/4

The volume of a cone is πr2h/3

Therefore as n approaches infinity, the sum of the volumes of cylinders comes closer to the volume of the cone, but will never get there.

This student written piece of work is one of many that can be found in our GCSE Height and Weight of Pupils and other Mayfield High School investigations section.

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