Rate of Change of Volume in a Sphere Assume that the radius $r$ of a sphere is expanding at a rate of $7$ in. /min. The volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Determine the rate at which the volume is changing with respect to time when $r = 16$ in.
I know I need to find the derivative of volume, and I think solve for $dr/dV$ and then plug in when $r= 16$. I've tried and I keep getting all kinds of wrong answers. Help would be greatly appreciated.
 A: Notice, volume of the sphere is given as $$V=\frac{4\pi}{3}r^3$$
differentiating volume $V$ w.r.t. time $t$ as follows 
$$\frac{dV}{dt}=\frac{4}{3}\pi\left(3r^2\frac{dr}{dt}\right)=4\pi r^2\frac{dr}{dt}$$
Now, setting expanding rate $\frac{dr}{dt}=7\ \mathrm{inch/min}$ & $r=16\ \mathrm{inch}$, one should get volume flow rate $$\frac{dV}{dt}=4\pi (16)^2(7)$$
$$=\color{red}{7168\pi\ \mathrm{inch^3/min}}$$
A: The rate at which Volume changes with respect to radius is the Area. So we can calculate volume change rate using:
$$ \dot V = \dot r 4 \pi r^2 $$
A: The volume of the sphere is: $$V=\frac{4\pi}{3}r^3$$
Differentiating volume with respect to radius gives:$$\frac{dV}{dr}=4\pi r^2$$
However, we want the differential of volume with respect to time. To do this we can multiply both sides by $\frac{dr}{dt}$:$$\frac{dr}{dt}\frac{dV}{dr}=4\pi r^2\frac{dr}{dt}$$
Cancelling out the $dr$ on the lefthand side gives us the differential of volume with respect to time:
$$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$$
Now, plugging in the rate of $7\frac{in.}{min.}$ for $\frac{dr}{dt}$ and $16in.$ for the radius, the rate at which the volume is changing is:$$\frac{dV}{dt}=4\pi (16in.)^2(7\frac{in.}{min.})=4\pi (256in.^2)(7\frac{in.}{min.})=$$
$${7168\pi\ \frac{in.^3}{min.}}$$
