Gravel is being dumped from a conveyor belt at a rate of 20 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 7 ft high?
To find the rate at which the height of the pile is increasing, we need to use related rates and apply the formula for the volume of a cone.
Let's denote the height of the pile as h, and the radius of the pile (and consequently the base diameter) as r. Since the base diameter and height are always equal, we have r = h.
The volume of a cone can be expressed as V = (1/3) * π * r^2 * h. Since r = h, we can simplify the formula to V = (1/3) * π * h^3.
We are given that the gravel is being dumped at a rate of 20 ft^3/min. This means that the rate of change of the volume of the pile with respect to time, dV/dt, is 20 ft^3/min.
Using the chain rule, we can find an expression for dV/dt in terms of dh/dt, the rate at which the height is changing:
dV/dt = dV/dh * dh/dt
To find dV/dh, we differentiate V = (1/3) * π * h^3 with respect to h:
dV/dh = π * h^2
Now we can substitute the given values and solve for dh/dt:
20 = π * h^2 * dh/dt
To find the height of the pile when it is 7 ft high, we substitute h = 7 into the equation:
20 = π * 7^2 * dh/dt
Simplifying further:
dh/dt = 20 / (49π)
Therefore, the height of the pile is increasing at a rate of (20 / (49π)) ft/min when the pile is 7 ft high.