Question

At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 6 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? (Hint: The formula for the volume of a cone is V = 1 3 πr2h.)

Answer 1

Answer:

2/(3π) ft/min ≈ 0.212 ft/min

Explanation:

Volume of a cone is:

V = ⅓ π r² h

The diameter is three times the altitude, so:

2r = 3h

r = 3/2 h

Substituting:

V = ⅓ π (3/2 h)² h

V = ⅓ π (9/4 h²) h

V = ¾ π h³

Taking derivative with respect to time:

dV/dt = 9/4 π h² dh/dt

Given dV/dt = 6 and h = 2:

6 = 9/4 π (2)² dh/dt

6 = 9π dh/dt

dh/dt = 2/(3π)

dh/dt ≈ 0.212 ft/min