Answer
$\frac{dr}{dt}=1\frac{ft}{min}$
$\frac{dS}{dt}=40\pi \frac{ft^2}{min}$
Work Step by Step
Given volume (V)=$\frac{4}{3}\pi r^3$ and $\frac{dV}{dt}=100\pi ft^3$ and radius r=5
on taking the derivative of the volume, we get:
$\frac{dV}{dt}=\frac{4}{3}\pi \frac{dr^3}{dt}$
$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$
$100\pi=4\pi(5)^2\frac{dr}{dt}$
$\frac{dr}{dt}=1\frac{ft}{min}$
the rate of surface area change is:
$\frac{dS}{dt}=8\pi r\frac{dr}{dt}$
$\frac{dS}{dt}=8\pi (5)\frac{dr}{dt}$
$\frac{dS}{dt}=40\pi \frac{ft^2}{min}$
