Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 8 cubic feet per minute. If the pool has radius 6 feet and height 11 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 8 feet?

dh/dt = 0,07 ft/min

Step-by-step explanation:

The swimming pool has the shape of right circular cylinder, therefore its volume is

V(c) = π*x²*h

Where x is the radius of the base and h the height

We take differentiation on both sides of the equation to get:

dV/dt  =  π*x²*dh/dt

The rate of change in height of water in the pool, is independent of the height of the water, since the pool is a right crcular cylinder, and dV/dt is constant at 8 ft³/min.

Then:

8  = π*x²*dh/dt

dh/dt = 8 /  π*x²

dh/dt = 8/113,04

dh/dt = 0,07 ft/min

answer: all the values are within because it has a pattern that repeats from -6 to 4.

good luck!