
Related rate problems
The radius r of a sphere is increasing at the uniform rate of .3 inches per second. At the instant when the surface area S becomes 100pi square inches, what if the rate of increase, in cubic inches per second, in the volume V?
s=4pir^2 v=(4/3)pir^2
:/ All I know is a have to take the derivative of each. ><"

you are given $\displaystyle \frac{dr}{dt}$ , and the problem wants you to find $\displaystyle \frac{dV}{dt}$ when the surface area = $\displaystyle 100 \pi$
$\displaystyle V = \frac{4}{3}\pi r^3$
$\displaystyle \frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}$
since surface area = $\displaystyle 4\pi r^2 = 100\pi$
$\displaystyle \frac{dV}{dt} = 100 \pi \cdot \frac{dr}{dt}$
finish up

30 pi!
Thank you, I was halfway through this problem when you responded. :3