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 equal to each other. How fast is the height of the pile increasing when the pile is 14 feet high?
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To find how fast the height of the pile is increasing, 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 pile as r (in feet). Since the base diameter and height are always equal, we can say that r = h/2.
We are given that the pile is formed in the shape of a right circular cone. The volume of a cone is given by the formula:
V = (1/3) * π * r^2 * h
We also know that the rate at which gravel is being dumped is 30 cubic feet per minute. Therefore, the rate of change of volume with respect to time (dh/dt) is 30 cubic feet per minute.
Now, let's differentiate the volume equation with respect to time:
dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)
Since we are trying to find how fast the height of the pile is increasing, we need to solve for dh/dt when h = 14 feet.
We already know the value of dV/dt (which is 30 cubic feet per minute), and we can substitute the values of r (h/2) and h (14 feet) into the equation.
30 = (1/3) * π * (2 * (h/2) * dr/dt * h + (h/2)^2 * dh/dt)
Simplifying the above equation, we have:
30 = (1/3) * π * (h^2 * dr/dt + (h^2/4) * dh/dt)
Substituting the known values, we have:
30 = (1/3) * π * (14^2 * dr/dt + (14^2/4) * dh/dt)
Now, we need to solve for dh/dt. Rearranging the equation:
(1/3) * π * (14^2/4) * dh/dt = 30 - (1/3) * π * 14^2 * dr/dt
Simplifying the equation further:
dh/dt = (4/14^2) * ((3/π) * (30 - (1/3) * π * 14^2 * dr/dt))
Substituting the value of dr/dt, which is given as 30 cubic feet per minute:
dh/dt = (4/14^2) * ((3/π) * (30 - (1/3) * π * 14^2 * 30))
Evaluating the above expression will give us the rate at which the height of the pile is increasing when the pile is 14 feet high.