Water is flowing into a vertical cylindrical tank

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

Radius of the cylinder, r = 2 ft

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

$V=4 \pi h$

Differentiating both sides with respect to t, we get

$\frac{d V}{d t}=4 \pi \times \frac{d h}{d t}$

Now,

$\frac{d V}{d t}=8$ cubic feet $/$ minute   (Given)\

$\therefore 8=4 \pi \times \frac{d h}{d t}$

$\Rightarrow \frac{d h}{d t}=\frac{8}{4 \pi}=\frac{2}{\pi} \mathrm{ft} / \mathrm{min}$

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

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