1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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 _______________.

Open in App
Solution

## 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 = $\mathrm{\pi }{r}^{2}h$ = $\mathrm{\pi }×{\left(2\right)}^{2}×h$ $V=4\mathrm{\pi }h$ Differentiating both sides with respect to t, we get $\frac{dV}{dt}=4\mathrm{\pi }×\frac{dh}{dt}$ Now, $\frac{dV}{dt}$ = 8 cubic feet/minute (Given) $\therefore 8=4\mathrm{\pi }×\frac{dh}{dt}$ $⇒\frac{dh}{dt}=\frac{8}{4\mathrm{\pi }}=\frac{2}{\mathrm{\pi }}$ 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 $\overline{)\frac{2}{\mathrm{\pi }}\mathrm{feet}/\mathrm{minute}}$.

Suggest Corrections
1
Join BYJU'S Learning Program
Related Videos
Basics
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program