what rate is the height of the pile changing

Question

At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 4 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 12 feet high? (Hint: The formula for the volume of a cone is

V = 1/3πr2h.)

h' = _____ft/min

Answer 1

To solve this problem, we set up a simple differential equation:

1. We are told the rate of change of volume, dV/dt = 4m³/minute
2. We are also told that the radius r and the height h are related (D = 2r = 3h) so r = (3/2)h

Let us then being by rearranging our equation for volume

1. V = (1/3)πr²h .... substitute our relationship between r and h to get rid of r as a variable:
2. V = (1/3)π(9/4)h³π ..... simplifying:
3. V = (3/4)πh³ ..... note: this is true here because r = (3/2)h, but is not true for all cases.

We now have an expression for Volume in terms of height.

Let us derive both sides of this equation with respect to time:

1. d/dt (V) = d/dt ((3/4)πh³) .... recall that (3/4) and π are constants, not variables or functions of t
2. dV/dt = (3/4)(π)(d/dt(h³)) .... recall d/dt (f(h)) = h'(dh/dt)
3. dV/dt = (3/4)(π)(3h²)dh/dt .... simplifying:
4. dV/dt = (9/4)πh²(dh/dt)

We now have a formula that relates change in Volume to Change in height where

1. dV/dt = change volume per time
2. dh/dt = change in height per time

We are told that at dV/dt = 4m³/minute = 141ft³/minute and that h = 12ft at the moment in question so:

141ft³/minute = (9/4)π(12ft)²(dh/dt)

Solving for (dh/dt), we get:

(dh/dt) = .139 ft/min

Solved!