How do you solve for related rates of a snowball increasing with a diameter and cm/min

Question

A spherical snowball is being rolled down a hill in such a way that its diameter is increasing at rate of 0.2 cm/min.

At what rate is the volume of the snowball increasing when the diameter is 18 cm?

Answer 1

Hi Joshua! Let me make this explanation as simple as possible!

Let's start simple. What is the formula for the volume of a sphere? V = (4/3) * π * r^3 , right?

We can relate that equation with rates by simply taking the derivative on both sides with respect to TIME, NOT volume or radius! (Why? Well, that's because the problem indicates that the rate is measured in minutes (time) )! Make sense?

The derivative is ⇒ dV/dt = (4π / 3) * 3 * r^2 * dr/dt 〈By Power Rule and Chain Rule〉

"diameter is increasing at rate of 0.2 cm/min" is equivalent to dr/dt = 0.1 cm/min because D / 2 = R, right?

Now the rest is simple. Just substitute all the values.

dV/dt = (4π / 3) * 3 * (18 / 2) ^2 * (0.1) = 101.8 cm^3 / min >>>>>>>>>>(ANSWER). Hope this helped!