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.
Best Answer
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 $$\frac{dV}{dt}=4\pi (16)^2(7)$$ $$=\color{red}{7168\pi\ \mathrm{inch^3/min}}$$