Question

Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 3 cubic feet per minute. If the pool has radius 7 feet and height 8 feet, what is the rate of change of the height of the water in the pool when the depth of the water in the pool is 4 feet

Answer 1

Answer:

Therefore the height of the water in the pool changes at the rate of $\frac{1}{3\pi}$ feet per minute.

Step-by-step explanation:

Given that  the shape of swimming pool is right circular cylinder.

The  rate of water pouring in the pool = 3 cubic feet per minute.

It means the rate of change of volume is 3 cubic feet per minute.

$\frac{dv}{dt}=3$ cubic feet per minute.

When the volume of the swimming pool changed it means the height of the water level of the pool change and the radius of the swimming pool remains constant.

Let the height of the pool be h.

The volume of the pool is = $\pi 3^2 h$  cubic feet

                                          $=9\pi h$ cubic feet

Therefore,

v $=9\pi h$

Differentiating with respect to t

$\frac{dv}{dt}= 9\pi \frac{dh}{dt}$

Putting $\frac{dv}{dt}=3$

$3=9\pi \frac{dh}{dt}$

$\Rightarrow \frac{dh}{dt} =\frac{3}{9\pi}$

$\Rightarrow \frac{dh}{dt} =\frac{1}{3\pi}$

The change of height of the pool does not depend on the depth of the pool.

Therefore the rate of change of height of the water in the pool is $\frac{1}{3\pi}$ feet per minute.