Related Rates (One Formula)

Question

The radius of a cone is decreasing at a constant rate of 5 feet per second. The volume remains a constant 432 cubic feet. At the instant when the height of the cone is 66 feet, what is the rate of change of the height? The volume of a cone can be found with the equation V=1/3πr^2h. Round your answer to three decimal places.

Answer 1

Since the volume remains constant, we know that dV/dt has to be 0.

But by the product rule plus the chain rule, we also know dV/dt = pi/3 2rh*dr/dt + pi/3 r^2 dh/dt.

so we have

pi/3 2rh*dr/dt + pi/3 r^2 dh/dt = 0. (*)

We're given dr/dt = -5 ft/s, and h = 66 ft at this particular instant. This allows us to solve for r at that time

since V = 432 = pi/3 r^2 *66.

Once we solve for r, we can plug everything into the starred equation earlier to solve for dh/dt