Answer the following question.

Question

At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 16 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 22 feet high? The formula for the volume of a cone is V= 1/3𝜋r²h.

h'=

Answer 1

Hi Anikka,

We want to get the volume equation in one variable. Right now it's in terms of r and h. Since the question is asking for the rate of change of the height, we should convert r into h. The question states that the diameter of the cone is approximately 3x the altitude. That means:

d = 3h

or

2r = 3h

r = 3h/2

So we can rewrite the volume equation as:

V = 1/3π(3h/2)²*h

= 1/3π*9h²/4 * h

V(h) = 3π/4*h³

The goal is to find the rate of change of the height, which is dh/dt. We can implicitly differentiate V(h) with respect to time to get dh/dt.

d/dt( V(h) = 3π/4*h³ )

dV/dt = 9π/4*h²*dh/dt

dV/dt is the rate of change of the volume, which the question says is 16 cubic feet per minute.

h is the height of the pile, which the question says is 22 feet.

16 = 9π/4*(22)²*dh/dt

16 = 1089π*dh/dt

dh/dt = 16/(1089π)

dh/dt ~ 0.00468 ft/ min

Hope this helps!