Set 7 Problem number 2

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

When a certain amount of paint is spread over a sphere of radius 4.6 meters, there is a uniform coat with area density 8.5 gallons/m ^ 2.

• Use the technique of proportionality to determine the area density if the same amount of paint is spread over a sphere of radius 9.15 meters.

Solution

We know that the area over which the paint is spread is 4 `pi r ^ 2.

• The area is therefore proportional to the square of the radius.

We know also that the area density is inversely proportional to the area of the sphere.

• So the area density is inversely proportional to the square of the radius.

We therefore write y = k/x ^ 2, with y representing area density and x representing radius.

• Since area density is 8.5 gal/m ^ 2 when radius is 4.6 meters, it follows that 8.5 gal/m ^ 2 = k/( 4.6 meters) ^ 2.
• Solving for k, we obtain k = 179.85 gallons.
• The relation y = k / x^2 therefore becomes

• y = k/x ^ 2 = 179.85 gallon / x^2.

• The constant k = 179.85 gallons is called the proportionality constant.

• To find the area density for any radius, we simply substitute that radius for x, and y will give us the area density.
• In this case, we obtain for radius 9.15 meters,

• y = 179.85 gal / ( 9.15 meters) ^ 2 = 2.14 gal/meter ^ 2.

Generalized Solution

If we know that a quantity Q, when spread uniformly over a sphere of radius r1, has density `sigma1, we can find the density of the same amount spread uniformly over a sphere of any radius r.

The key is to understand that area is proportional to the square of radius, since A = 4 `pi r^2, and that area density being inversely proportional to area is inversely proportional to the square of the radius. 

By the inverse square proportionality

`sigma / `sigma1 = (r1 / r)^2,

we have

`sigma = `sigma1 * (r1 / r) ^ 2.

Explanation in terms of Figure(s), Extension

The figure below depicts two spheres with radii r1 and r2.

• Their areas are 4 `pi r1^2 and 4 `pi r2^2, so the area ratio is (r2 / r1) ^ 2, as indicated.
• Any quantity spread over the larger sphere will be spread more thinly than the same quantity over the smaller, with a area density ratio which is the inverse of the area ratio.
• The area density ratio will therefore be (r1 / r2) ^ 2.