Question

At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 10 cubic feet per minute. The diameter of the base of the cone is about three times the altitude. At what rate is the height of the pile changing at when the pile is 15 feet high.

Answer 1

0.0063 ft / min

Explanation:

From the problem statement we have that the variation in volume with respect to time is equal 10, like this:

dV / dt = 10 ft ^ 3 / min

We know that the volume of the cone is given by the equation:

V = (1/3) * Pi * r ^ 2 * h

Now, we are mentioned that the diameter of the base of the cone is approximately three times the altitude, therefore:

2r = 3h, solving for r we have:

r = (3/2) * h, replacing the volume equation:

V = (1/3) * Pi * [(3h / 2) ^ 2] * h, solving we have:

V = (3/4) * Pi * h ^ 3

They ask us to calculate the change in height with respect to time, that is dh / dt

Therefore we derive both sides with respect to time, we have to:

dV / dt = (3/4) * Pi [3 * (h ^ 2) dh / dt] = (9/4) * pi * (h ^ 2) * dh / dt

reorganizing for dh / dt, we have:

dh / dt = [4/9 * pi * (h ^ 2)] * dV / dt

Knowing that h is 15 and dV / dt is 10, we replace these values:

dh / dt = [4/9 * 3.14 * (15 ^ 2)] * 10 = 0.0063 ft / min

0.0063 ft / min would be the change in height with respect to time.