Question

Water is flowing into a vertical cylindrical tank of radius 2 ft at the rate of 8 cubic/minute. The rate at which the water level is rising, is _______________.

Answer 1

Let h be the water level in the cylindrical tank at time t minutes.

Radius of the cylinder, r = 2 ft

∴ Volume of the water in the cylindrical tank at time t, V = $\mathrm{\pi }{r}^{2}h$ = $\mathrm{\pi }\times {\left(2\right)}^2\times h$

$V=4\mathrm{\pi }h$

Differentiating both sides with respect to t, we get

$\frac{dV}{dt}=4\mathrm{\pi }\times \frac{dh}{dt}$

Now,

$\frac{dV}{dt}$ = 8 cubic feet/minute (Given)

$\therefore 8=4\mathrm{\pi }\times \frac{dh}{dt}$

$\Rightarrow \frac{dh}{dt}=\frac{8}{4\mathrm{\pi }}=\frac{2}{\mathrm{\pi }}$ ft/min

Thus, the water level in the tank is rising at the rate of $\frac{2}{\mathrm{\pi }}$ feet/minute.


Water is flowing into a vertical cylindrical tank of radius 2 ft at the rate of 8 cubic/minute. The rate at which the water level is rising, is $\underline{\frac{2}{\mathrm{\pi }}\mathrm{feet}/\mathrm{minute}}$.