A winery has a cylindrical fermentation tank 20 feet high with a radius of 6 feet. If the tank is being filled with grape juice at a rate of 90 cubic feet per minute, how fast is the surface of the juice rising at any time t ?

Question

anonymous · Answer

$$V=\pi r^2h$$
You're given that $\dfrac{dV}{dt}=90$ and you must find $\dfrac{dh}{dt}$.

Differentiating with respect to $t$, you have
$$\frac{dV}{dt}=2\pi r\frac{dr}{dt}h+\pi r^2\frac{dh}{dt}$$
Notice that as the level of the juice rises, the radius does not change, so $\dfrac{dr}{dt}=0$:
$$\frac{dV}{dt}=\pi r^2\frac{dh}{dt}$$
Plug in the known values:
$$90=\pi 6^2\frac{dh}{dt}~~\Rightarrow~~\frac{dh}{dt}=\frac{90}{36\pi}\approx0.796\text{ ft/min}$$