Question

The radius of a cylinder is decreasing at a constant rate of 7 feet per second. the volume remains a constant 72 cubic feet. at the instant when the height of the cylinder is 2 feet, what is the rate of change of the height?

Answer 1

The rate of change of the height of the cylinder is 3.5 feet per second.

Step-by-step explanation:

To find the rate of change of the height of the cylinder, we can use the equation for the volume of a cylinder: V = πr^2h. Since the volume remains constant at 72 cubic feet, we can differentiate the equation with respect to time to find the rate of change.

dV/dt = 0 = 2πrh(dr/dt) + πr^2(dh/dt)

Given that the radius is decreasing at a constant rate of 7 feet per second (dr/dt = -7), plugging in the values we have, 0 = 2π(2)(-7)(2) + π(2)^2(dh/dt)

Solving for dh/dt, we get dh/dt = 14π/π(4) = 14/4 = 3.5 feet per second.