Related Rates question

Question

At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 2 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate is the height of the pile changing when the pile is 2 feet high?

Answer 1

Let

h(t) be the height of the cone at time t,

r(t) = 3h(t)/2 be the radius of the cone at time t, (1)

V(t) = πr²h/3 be the volume of the cone at time t, (2)

dV/dt = 2ft²/min be the rate of increase of the volume V. (3)

(1) and (3) are given in the problem, and (2) is the formula for the volume of a cone.

We are to calculate the rate of change in height, i.e., dh/dt.

First express V in terms of h, and then take the derivative.

Substituting (1) into (2)

V = 3πh³/4

Taking derivative of both sides:

dV/dt = 9πh²/4 dh/dt

Therefore

dh/dt = 4 dV/dt / 9πh²

To get the actual rate we substitute the given dV/dt and the given h = 2ft:

dh/dt = 8 ft³/min / 36π ft² = 0.07 ft/min