Water drains from an inverted conical tank 10 m tall with a top radius of 5 m at a rate of 2 m$^3$/min. How fast is the depth of the water decreasing when the depth is 4 m?

Core Concept

Since $\dfrac{r}{h}=\dfrac{5}{10}$, $r=\dfrac{h}{2}$ and $V=\dfrac{1}{3}\pi\left(\dfrac{h}{2}\right)^2h=\dfrac{\pi h^3}{12}$. Then $\dfrac{dV}{dt}=\dfrac{\pi h^2}{4}\dfrac{dh}{dt}$, so $-2=\dfrac{\pi(16)}{4}\dfrac{dh}{dt}$ and $\dfrac{dh}{dt}=-\dfrac{1}{2\pi}$ m/min.