Calculus (3rd Edition)

$\frac{1000\pi}{3}cm^{3}/s$
The volume of a circular cone is $V=\frac{1}{3}\pi r^{2}h$ $\frac{dV}{dt}=\frac{1}{3}\pi \times\frac{d}{dt}(r^{2}h)$ Applying the product rule, we have $\frac{d}{dt}(r^{2}h)=2r\times\frac{dr}{dt}\times h+r^{2}\times\frac{dh}{dt}$ Given: $r=10\,cm$, $h=20\,cm$, $\frac{dr}{dt}=2\,cm/s$ and $\frac{dh}{dt}=2\,cm/s$ Substituting the given values, we get $\frac{d}{dt}(r^{2}h)=1000\,cm^{3}/s$ Then, $\frac{dV}{dt}=\frac{1}{3}\pi\times1000\,cm^{3}/s=\frac{1000\pi}{3}cm^{3}/s$