Gravel is being dumped from a conveyor belt at a rate of 20 {\rm ft}^3{\rm /min}. It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is 18 ft high?
v = 1/3 πr^2 h = π/3 (d/2)^2 h = π/12 h^3
dv/dt = π/4 h^2 dh/dt
Now just plug in your numbers.
To find the rate at which the height of the pile is increasing, we need to relate the variables in the problem. Let's denote the height of the pile as h and the radius of the base of the pile as r.
We can use the formula for the volume of a right circular cone:
V = (1/3)πr^2h
We are given that the rate at which gravel is being dumped is 20 ft^3/min, which means that the volume of the pile is increasing at a rate of dV/dt = 20 ft^3/min.
We are asked to find the rate at which the height of the pile is increasing, i.e., dh/dt. To relate this to the volume and the variables r and h, we can take the derivative of the volume equation with respect to time:
dV/dt = (1/3)π(2rh)(dh/dt)
Now, we can plug in the given values:
dV/dt = 20 ft^3/min
Since the base diameter and height of the pile are always the same, we have r = h/2. Substituting this into the equation above gives:
20 = (1/3)π(h/2)(dh/dt)
Simplifying, we get:
dh/dt = (6/π) ft/min
The rate at which the height of the pile is increasing is therefore dh/dt = (6/π) ft/min.
To solve this problem, we can use related rates. We are given that the gravel is being dumped at a rate of 20 ft^3/min. We need to find the rate at which the height of the pile is changing when the pile is 18 ft high.
Let's denote the height of the pile as h and the radius of the base as r. Since the base diameter and height are always the same, the base radius is equal to h/2.
The volume V of a right circular cone is given by the formula V = (1/3) * π * r^2 * h.
To find the rate of change of the height (dh/dt) with respect to time, we differentiate the volume equation with respect to time (t):
dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)
Since the base radius is equal to h/2, we substitute r = h/2 in the differentiation equation:
20 = (1/3) * π * (2 * (h/2) * dr/dt * h + (h/2)^2 * dh/dt)
Simplifying the equation, we get:
20 = (1/3) * π * (h * dr/dt * h + h^2/4 * dh/dt)
60 = π * (h * dr/dt * h + h^2/4 * dh/dt)
Now, we need to solve for dh/dt, the rate at which the height of the pile is changing. We can rearrange the equation to isolate dh/dt:
dh/dt = (60 - h * dr/dt * h) / (π * h^2/4)
We are given that the pile is 18 ft high, so we can substitute h = 18 into the equation:
dh/dt = (60 - 18 * dr/dt * 18) / (π * 18^2/4)
dh/dt = (60 - 18 * dr/dt * 18) / (81π)
To find dh/dt, we need the value of dr/dt. However, it is not given in the problem statement. We need more information to solve for the rate at which the height of the pile is changing.
Let me know if you have any additional information that could help us find dr/dt.