Gravel is being dumped from a conveyor belt at a rate of 30 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is 24 feet high?

Question

Gravel is being dumped from a conveyor belt at a rate of 30 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is 24 feet high?

Recall that the volume of a right circular cone with height h and radius of the base r is given by
v=1/3pir^2h

Answer 1

Let the radius be r, so the diameter is 2r

then the height is 2r

V = (1/3)π r^2 h
= (1/3) π r^2 (2r)
= (2/3)π r^3

dV/dt = 2π r^2 dr/dt

when h = 24
2r = 24
r = 12 and dV/dt = 30

30 = 2π (144) dr/dt
dr/dt = 30/(288π) = 5/(48π) ft/min
or appr .033 ft/min

check my arithmetic.

Answer 2

To find the rate at which the height of the pile is increasing, we need to determine how the volume of the pile changes with respect to time. Using the given information, we can write the equation for the volume of the cone pile:

V = (1/3)πr^2h

where V is the volume, r is the radius of the base, and h is the height of the pile.

To relate the height of the pile to the rate of change of the height with respect to time, we can differentiate both sides of the equation with respect to time (t) using implicit differentiation:

dV/dt = (1/3)π * 2r * (dh/dt) + (1/3)πr^2 * (dh/dt)

The first term on the right side represents the rate of change of the volume with respect to the radius, while the second term represents the rate of change of the volume with respect to the height. Since the radius and height of the cone pile are always the same, we can simplify the equation:

dV/dt = (2/3)πr * (dh/dt) + (1/3)πr^2 * (dh/dt)

Since we are given the rate at which the volume is changing (dV/dt = 30 ft^3/min), we can plug in the values and solve for the rate of change of the height (dh/dt):

30 = (2/3)πr * (dh/dt) + (1/3)πr^2 * (dh/dt)

Simplifying further:

30 = (dh/dt) * ((2/3)πr + (1/3)πr^2)

Now, we can substitute the known values:

30 = (dh/dt) * ((2/3)π(24/2) + (1/3)π(24/2)^2)

30 = (dh/dt) * ((2/3)π(12) + (1/3)π(12)^2)

Simplifying the expression:

30 = (dh/dt) * (8π + 48π)

30 = (dh/dt) * 56π

Now we can solve for (dh/dt):

(dh/dt) = 30 / (56π) ≈ 0.168 ft/min

Thus, when the pile is 24 feet high, the height of the pile is increasing at a rate of approximately 0.168 feet per minute.