The height of the pile is changing at a rate of 0.0024 feet per minute when the pile is 15 feet high.
The Breakdown
The formula for the volume of a cone is given by:
V = (1/3) × π × r² × h
where V is the volume,
π is a mathematical constant (approximately 3.14159),
r is the radius of the base of the cone,
h is the height of the cone.
The height of the pile as h(t), where t represents time. We are given that dh/dt = 10 cubic feet per minute, which represents the rate at which the height is changing.
We are also given that the diameter of the base of the cone is approximately three times the altitude. Since the diameter is twice the radius, we can say that the radius is (1/2) × (3h) = (3/2)h.
Now, we can substitute the values into the volume formula:
V = (1/3) × π × [(3/2)h]² × h
V = (1/3) × π × (9/4) × h³
V = (3/4) × π × h³
To find the rate at which the height is changing when the pile is 15 feet high, we need to differentiate the volume equation with respect to time:
dV/dt = (3/4) × π × 3h² × dh/dt
We know that dV/dt = 10 cubic feet per minute, and we want to find dh/dt when h = 15 feet. Substituting these values into the equation, we get:
10 = (3/4) × π × 3(15)² × dh/dt
Simplifying further:
10 = (3/4) × π × 3 × 225 × dh/dt
10 = (3/4) × π × 675 × dh/dt
Now, we can solve for dh/dt:
dh/dt = 10 / [(3/4) × π × 675]
dh/dt ≈ 0.0024 feet per minute
Therefore, the height of the pile is changing at a rate of 0.0024 feet per minute when the pile is 15 feet high.