## Calculus (3rd Edition)

$\frac{9}{8\pi}\,m/min$
We set up the fraction: $\frac{radius}{height}=\frac{r}{h}=\frac{2}{3} \implies r=\frac{2}{3}h$ The volume of the tank is $V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi(\frac{2}{3}h)^{2}h=\frac{4\pi}{27}h^{3}$ Now we find the rate of change of the volume with height: $\frac{dV}{dh}=\frac{d}{dh}(\frac{4\pi}{27}h^{3})=\frac{12\pi}{27}h^{2}$ $\frac{dV}{dh}|_{h=2\,m}=\frac{48\pi}{27}\,m^{2}$ Finally, we find the rate of change of the height with time: $\frac{dh}{dt}=\frac{\frac{dV}{dt}}{\frac{dV}{dh}}=\frac{2\,m^{3}/min}{\frac{48\pi}{27}m^{2}}=\frac{9}{8\pi}m/min$