A cylindrical tank with radius 3 m is being filled with water at a rate of 2 m3/min. How fast is the height of the water increasing (in m/min)

.0707 m / min

Explanation:

The computation is shown below:

= Flow of the volume ÷ crossectional area

A = π × r^2

= 9 π

= 28.27 m^2

so, the fast of the height of the water rise is

=(2 m^3 / min) ÷ 28.27 m^2

= .0707 m / min

an obstruction

d eras and periods