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?
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.
