Question

Suppose a spherical snowball is melting and the radius is decreasing at a constant rate, changing from 12 in to 8 in in 45 minutes. How fast was the volume changing when the radius was 10 in?

Answer 1

To find how fast the volume is changing when the radius was 10 inches, we can use the formula for the volume of a sphere, which is V = (4/3)πr^3. We are given that the radius is decreasing at a constant rate, so we can find the derivative of the volume with respect to time. When the radius was 10 inches, the volume was changing at a rate of -800π cubic inches per minute.

Step-by-step explanation:

To find how fast the volume is changing when the radius was 10 inches, we can use the formula for the volume of a sphere, which is V = (4/3)πr^3. We are given that the radius is decreasing at a constant rate, so we can find the derivative of the volume with respect to time. Let's call the radius r and the time t. From the given information, we know that r is changing from 12 inches to 8 inches in 45 minutes.

Now, we can find the rate at which the volume is changing by taking the derivative of V with respect to t. This can be done using the chain rule. We have:

dV/dt = dV/dr × dr/dt = (4/3)π(3r^2) × (-dR/dt)

Substituting the given values, we have:

dV/dt = (4/3)π(3(10^2)) × (-4/45) = -800π

Therefore, when the radius was 10 inches, the volume was changing at a rate of -800π cubic inches per minute.