Water level in a conical tank

Text

Water flows out of the bottom of a cone-shaped tank at $2\ \text{ft}^3/\text{min}$. The height of the cone is twice its radius. How fast is the water level dropping when the depth is $h=4\text{ ft}$?

Solution

Since $h=2r$, $r=h/2$ and $V=\tfrac13\pi r^{2}h=\dfrac{\pi h^{3}}{12}$. Differentiating: $\dfrac{dV}{dt}=\dfrac{\pi h^{2}}{4}\dfrac{dh}{dt}$. With $\dfrac{dV}{dt}=-2$ and $h=4$: $-2=4\pi\dfrac{dh}{dt}$, so $\dfrac{dh}{dt}=-\dfrac{1}{2\pi}\approx-0.159\text{ ft/min}$. $\boxed{\dfrac{dh}{dt}=-\dfrac{1}{2\pi}\approx-0.159\ \text{ft/min}}$