Water is poured into a conical tank at 2 m$^3$/min. The tank's volume is $V=\dfrac{1}{27}\pi h^3$, where $h$ is the depth of the water in meters. How fast is the depth increasing when $h=6$ m?

Core Concept

Differentiating gives $\dfrac{dV}{dt}=\dfrac{\pi h^2}{9}\dfrac{dh}{dt}$. With $h=6$, $\dfrac{dV}{dh}=4\pi$, so $\dfrac{dh}{dt}=\dfrac{2}{4\pi}=\dfrac{1}{2\pi}$ m/min.