Random Proportionality Problems version 1



Introduction

The problems below offer a range of examples of the use of proportionality as applied to geometric situations.  The first 8 problems are accompanied by solutions (to be found at the end of the page). 

The basic ideas are that volumes and weights are related to a given linear dimension over a set of geometrically similar solids by a proportionality function of the form y = k x ^ 3, while areas and quantities proportional to areas are related to the given linear dimension by a proportionality function of the form y = k x ^ 2.  It follows that the ratio of volumes between geometrically similar objects is y2 / y1 = (x2 / x1) ^ 2, while the ratio of areas is (y2 / y1) = (x2 / x1) ^ 2.

These ideas are applied the problems of finding proportionalities of scale, areas and volumes of cubes (the fundamental shape from which the proportionalities are derived), sandpiles and balloons.

The problems of determining the factors by which the frequency and period of a pendulum change for a given change in pendulum length are also solved for the experimentally observed proportionalities between frequency and length, and between period and length.

Problems 1-8

Problem 1: By what factor does the volume of a cube change if its edge changes from 7 to 8?

Solutions

Problem 2: By what factor does the area of a square change if its diagonal changes from 8 to 9?

Solutions

Problem 3: By what factor does the length of an edge of a cube change if its volume changes from 85 to 8?

Solutions

Problem 4: By what factor does the length of an edge of a cube change if the total area of its faces changes from 69 to 2?

Solutions

Problem 5: By what factor does the volume of a cube change if its area changes from 71 to 5?

Solutions

Problem 6: By what factor does the area of the face of a cube change if its volume changes from 52 to 8?

Solutions

Problem 7: A pendulum has period T = k L^.5, where L is length. By what factor does the period change if the length changes from 9 to 7?

Solutions

Problem 8: A pendulum has frequency f = k L^-.5, where L is length. By what factor does frequency change if length changes from 6 to 4?

Solutions

Problems 9-40 (random order, random parameters)

Problem 9: By what factor does the surface area of a balloon change if its circumference changes from 6 to 8?

Problem 10: If a cube with weight 10 has surface area 45, what is the surface area a cube with weight 2?

Problem 11: If a sandpile with base diameter 10 requires 22 ounces of cement to cover its surface, how much cement is required for a sandpile with base diameter 9?

Problem 12: By what factor does the length of an edge of a cube change if the total area of its faces changes from 3 to 6?

Problem 13: By what factor does the altitude of a sandpile change if its volume changes from 5 to 2?

Problem 14: By what factor does the total surface area of a balloon change if its volume changes from 4 to 2?

Problem 15: By what factor does the volume of a cube change if its edge changes from 7 to 10?

Problem 16: By what factor does the length of an edge of a cube change if its volume changes from 8 to 9?

Problem 17: If a sandpile with base diameter 3 has weight 37, what is the weight of a sandpile with base diameter 9?

Problem 18: If a cube requiring 6 grams of paint has edge length 99, what is the edge length of a cube requiring 9 grams of paint?

Problem 19: By what factor does the volume of a sandpile change if its surface area changes from 3 to 7?

Problem 20: By what factor does the area of the face of a cube change if its volume changes from 6 to 3?

Problem 21: If a sandpile with weight 5 has base diameter 32, what is the base diameter of a sandpile of weight 2?

Problem 22: By what factor does the volume of a balloon change if its surface area changes from 3 to 6?

Problem 23: By what factor does the total surface area of a sandpile change if its volume changes from 6 to 10?

Problem 24: By what factor does the volume of a sandpile change if its altitude changes from 10 to 3?

Problem 25: By what factor does the altitude of a sandpile change if its total surface area changes from 5 to 7?

Problem 26: By what factor does the volume of a cube change if its area changes from 7 to 10?

Problem 27: By what factor does the circumference of a balloon change if its volume changes from 5 to 7?

Problem 28: If a sandpile requiring 5 ounces of cemet to cover its surface has base diameter 70, what is the base diameter of a sandpile requiring 3 ounces of cement?

Problem 29: If a sandpile with weight 3 has surface area 91, what is the surface area of a sandpile with weight 6?

Problem 30: If a cube with surface area 3 has weight 71, what is the weight of a cube with surface area 8?

Problem 31: A pendulum has frequency f = k L^-.5, where L is length. By what factor does frequency change if length changes from 6 to 10?

Problem 32: By what factor does the surface area of a sandpile change if its altitude changes from 10 to 2?

Problem 33: By what factor does the volume of a balloon change if its circumference changes from 10 to 5?

Problem 34: If a cube with weight 8 has edge length 78, what is the edge length of a cube of weight 3?

Problem 35: If a cube with edge length 7 requires 39 grams of paint, how much paint is required for a cube with edge length 9?

Problem 36: A pendulum has period T = k L^.5, where L is length. By what factor does the period change if the length changes from 6 to 2?

Problem 37: If a sandpile with surface area 6 has weight 16, what is the weight of a sandpile with surface area 7?

Problem 38: If a cube with edge length 2 has weight 30, what is the weight of a cube with edge length 8?

Problem 39: By what factor does the area of a square change if its diagonal changes from 3 to 8?

Problem 40: By what factor does the circumference of a balloon change if its total surface area changes from 5 to 10?

Solutions to Problems 1-8

**** Solution 1: When a cube is inflated to the scale of another cube all three of its dimensions change proportionally. Cube volume y and edge length x are therefore related by the p = 3 power proportionality y = k x^3. It follows that the ratio y2 / y1 is equal to (k x2^3) / (k x1^3) = (x2 / x1)^3.

Since x = edge length changes from 7 to 8, the edge length ratio is x2 / x1 =8 / 7 = 1.142.

The ratio y2 / y1 is therefore equal to (x2 / x1)^3 = 1.142^3 = 1.492. This is the factor by which volume will change.

**** Solution 2: When a square is inflated to the scale of another square both of its dimensions change proportionally. Square area y and diagonal length x are therefore related by the p = 2 power proportionality y = k x^2. It follows that the ratio y2 / y1 is equal to (k x2^2) / (k x1^2) = (x2 / x1)^2.

Since x = edge length changes from 8 to 9, the edge length ratio is x2 / x1 =9 / 8 = 1.125.

The ratio y2 / y1 is therefore equal to (x2 / x1)^2 = 1.125^2 = 1.265. This is the factor by which volume will change.

**** Solution 3: When a cube is inflated to the scale of another cube all three of its dimensions change proportionally. Cube volume y and edge length x are therefore related by the p = 3 power proportionality y = k x^3. It follows that the ratio y2 / y1 is equal to (k x2^3) / (k x1^3) = (x2 / x1)^3.

We can find the ratio y2 / y1 of volumes directly from the given information. We desire the ratio (x2 / x1) of lengths. Since y2 / y1 = (x2 / x1)^3, we can take the 1/3 power of both sides to obtain x2 / x1 = (y2 / y1) ^ (1/3).

Since y = volume changes from 85 to 8, the volume ratio is y2 / y1 = 8 /85 = .09411.

The ratio x2 / x1 is therefore equal to (y2 / y1) ^ (1/3) = .09411 ^ (1/3) = .4548.

This is the factor by which edge length will change.

**** Solution 4: The faces of a cube can be covered with tiny squares. When a cube is inflated to the scale of another cube all three of its dimensions change proportionally. However, each square changes only in its two dimensions. Face area y and edge length x are therefore related by the p = 2 power proportionality y = k x^2. It follows that the ratio y2 / y1 is equal to (k x2^2) / (k x1^2) = (x2 / x1)^2.

We can find the ratio y2 / y1 of areas directly from the given information. We desire the ratio (x2 / x1) of lengths. Since y2 / y1 = (x2 / x1)^2, we can take the 1/2 power of both sides to obtain x2 / x1 = (y2 / y1) ^ (1/2).

Since y = area changes from 69 to 2, the area ratio is y2 / y1 = 2 /69 = .02898.

The ratio x2 / x1 is therefore equal to (y2 / y1) ^ (1/2) = .02898 ^ (1/2) = .1702.

This is the factor by which edge length will change.

**** Solution 5: We can begin by finding the edge length ratio, which we can then use to find the ratio of volumes.

The faces of a cube can be covered with tiny squares. When a cube is inflated to the scale of another cube all three of its dimensions change proportionally. However, each square changes only in its two dimensions. Face area y and edge length x are therefore related by the p = 2 power proportionality y = k x^2. It follows that the ratio y2 / y1 is equal to (k x2^2) / (k x1^2) = (x2 / x1)^2.

We can find the ratio y2 / y1 of areas directly from the given information. We desire the ratio (x2 / x1) of lengths. Since y2 / y1 = (x2 / x1)^2, we can take the 1/2 power of both sides to obtain x2 / x1 = (y2 / y1) ^ (1/2).

Since y = area changes from 71 to 5, the area ratio is y2 / y1 = 5 /71 = .07042.

The ratio x2 / x1 is therefore equal to (y2 / y1) ^ (1/2) = .07042 ^ (1/2) = .2653.

This is the factor by which edge length will change.

Since the volume of a cube changes in three dimensions, the volume ratio will be V2 / V1 = (x2 / x1) ^ 3 = .2653 ^ 3 = .01868.

Note that we could have found an expression for the volume ratio in terms of the area ratio y2 / y1, using the fact that x2 / x1 = (y2 / y1) ^ (1/2).

We would have obtained V2 / V1 = (x2 / x1) ^ 3 = ( (y2 / y1)^(1/2) ) ^ 3 = (y2 / y1) ^ (3/2).

Substituting y2 = 5 AND y1 = 71 we obtain V2 / V1 = (5 / 71) ^ (3/2) = .01868.

**** Solution 6: We can begin by finding the edge length ratio, which we can then use to find the ratio of areas.

When a cube is inflated to the scale of another cube all three of its dimensions change proportionally.

Cube volume y and edge length x are therefore related by the p = 3 power proportionality y = k x^3. It follows that the ratio y2 / y1 is equal to (k x2^3) / (k x1^3) = (x2 / x1)^3.

We can find the ratio y2 / y1 of volumes directly from the given information. We desire the ratio (x2 / x1) of lengths. Since y2 / y1 = (x2 / x1)^3, we can take the 1/3 power of both sides to obtain x2 / x1 = (y2 / y1) ^ (1/3).

Since y = volume changes from 52 to 8, the volume ratio is y2 / y1 = 8 /52 = .1538.

The ratio x2 / x1 is therefore equal to (y2 / y1) ^ (1/3) = .1538 ^ (1/3) = .5358.

This is the factor by which edge length will change.

Since the face of a cube changes in only two dimensions, the area ratio will be A2 / A1 = (x2 / x1) ^ 2 =.5358 ^ 2 = .2871.

Note that we could have found an expression for the area ratio in terms of the volume ratio y2 / y1, using the fact that x2 / x1 = (y2 / y1) ^ (1/3).

We would have obtained A2 / A1 = (x2 / x1) ^ 2 = ( (y2 / y1)^(1/3) ) ^2 = (y2 / y1) ^ (2/3).

Substituting y2 = 8 AND y1 = 52 we obtain A2 / A1 = (8 / 52) ^ (2/3) = .2871.

**** Solution 7: Since pendulum period is a p = .5 power function T = k L^.5 of length, the ratio of periods is T2 / T1 = (k L2 ^ .5) / (k L1 ^ .5) = (L2 / L1) ^ .5.

Thus a change in length from 9 to 7 results in ratio T2 / T1 = (7 / 9) ^ .5 = .8819.

This ratio is the factor by which the period changes.

**** Solution 8: Since pendulum frequency is a p = -.5 power function f = k L^-.5 of length, the ratio of periods is f2 / f1 = (k L2 ^ -.5) / (k L1 ^ -.5) = (L2 / L1) ^ -.5.

Thus a change in length from 6 to 4 results in ratio f2 / f1 = (4 / 6) ^ -.5 = 1.224.

This ratio is the factor by which the frequency changes.