Answer:Therefore, the rate at which the height of the pile is increasing when the pile is 25 feet high is approximately 0.766 feet per minute.
Please note that the exact value depends on the approximation of π used.
Step-by-step explanation: To solve this problem, we can use the concept of related rates. We are given that gravel is being dumped from a conveyor belt at a rate of 20 cubic feet per minute. The pile of gravel forms a right circular cone, and we want to find how fast the height of the pile is increasing when the pile is 25 feet high.
Let's define the variables:
- V: Volume of the cone (cubic feet)
- r: Radius of the base of the cone (feet)
- h: Height of the cone (feet)
We know that the volume of a cone is given by the formula V = (1/3)πr^2h.
Since the base diameter and height of the pile are always equal, we can say that the radius (r) is equal to half of the height (h/2).
Now, let's differentiate the volume equation with respect to time (t):
dV/dt = (1/3)π(2r)(dh/dt)
We want to find dh/dt, which represents the rate at which the height of the pile is changing. We are given dV/dt, which is 20 cubic feet per minute.
We are also given that the height of the pile is 25 feet (h = 25).
Now we can plug in the values into the equation and solve for dh/dt:
20 = (1/3)π(2(h/2))(dh/dt)
Simplifying the equation, we have:
20 = (1/3)πh(dh/dt)
To solve for dh/dt, we can isolate it:
dh/dt = (20 * 3) / (πh)
Substituting h = 25 into the equation, we get:
dh/dt = (20 * 3) / (π * 25)
Simplifying further, we have:
dh/dt = 60 / (25π)