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 22 feet high?
d = 2 r = 3 h ... r = 3/2 h
v = 1/3 * π * r^2 * h
v = π/3 * 9 h^3 / 4
dv = 9 π h^2 dh / 4
dh = 4 dv / (9 π h^2)
To find the rate at which the height of the pile is changing, we need to use related rates. Let's denote the height of the pile as 'h' (in feet) and the radius of the base of the cone as 'r' (in feet).
Given information:
- The sand is falling off at a rate of 6 cubic feet per minute. This means that the rate of change of the volume of the pile, dV/dt, is equal to 6. Since we are interested in the rate of change of the height, we need to find dh/dt.
- The diameter of the base of the cone is approximately three times the altitude. This means that the radius, r, is equal to h/3.
We can express the volume of a cone in terms of its height and radius using the formula:
V = (1/3) * π * r^2 * h
Let's differentiate the volume formula with respect to time (t) to find the rate of change:
dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)
Since dr/dt is not provided in the given information, we can solve for it using the relationship between r and h:
r = h/3
dr/dt = dh/dt/3
Substituting the values back into the volume equation:
6 = (1/3) * π * (2 * (h/3) * (dh/dt/3) * h + (h/3)^2 * dh/dt)
Simplifying the equation:
6 = (1/3) * π * (2h/3 * dh/dt/3 * h + h^2/9 * dh/dt)
18 = 2π * h * (dh/dt/9h + dh/dt/9)
18 = 2π * (2dh/dt/9)
dh/dt = 81/4π
Finally, we have the required rate of change of the height of the pile when the pile is 22 feet high:
dh/dt = 81/4π ≈ 6.48 cubic feet per minute
Therefore, the height of the pile is changing at a rate of approximately 6.48 cubic feet per minute when the pile is 22 feet high.