Related rate problems

**MadHotThunder** · February 25th 2009, 01:30 PM

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. ><"

**skeeter** · February 25th 2009, 02:28 PM

you are given $\frac{dr}{dt}$ , and the problem wants you to find $\frac{dV}{dt}$ when the surface area = $100 \pi$

$V = \frac{4}{3}\pi r^3$

$\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}$

since surface area = $4\pi r^2 = 100\pi$

$\frac{dV}{dt} = 100 \pi \cdot \frac{dr}{dt}$

finish up

**MadHotThunder** · February 25th 2009, 02:36 PM

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

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