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 the same. How fast is the height of the pile increasing when the pile is 18 feet high?
To find the rate at which the height of the pile is changing, we need to use related rates and the properties of a right circular cone.
Let's start by assigning variables to the quantities involved:
- Let V represent the volume of the cone (in cubic feet).
- Let r represent the radius of the base of the cone (in feet).
- Let h represent the height of the cone (in feet).
Given:
- The rate at which the gravel is being dumped is 30 cubic feet per minute.
- The base diameter and height of the cone are always the same.
We know the formula for the volume of a cone:
V = (1/3) * π * r^2 * h
To relate the rate at which the height is changing to the rate at which the volume is changing, we can take the derivative of both sides with respect to time (t):
dV/dt = (1/3) * π * (2r * dr/dt) * h + (1/3) * π * r^2 * dh/dt
Where dr/dt represents the rate at which the radius is changing and dh/dt represents the rate at which the height is changing.
Since the base diameter and height are the same, the radius r will always be half the height:
r = h/2
Substituting this into the equation:
dV/dt = (1/3) * π * (2(h/2) * dr/dt) * h + (1/3) * π * (h/2)^2 * dh/dt
Simplifying:
dV/dt = (1/3) * π * h * (dh/dt + 2(h/2) * dr/dt)
We know that dV/dt, the rate at which the volume is changing, is 30 cubic feet per minute.
30 = (1/3) * π * h * (dh/dt + h * dr/dt)
We're interested in finding dh/dt, the rate at which the height is changing when the pile is 18 feet high.
Now, we need to find dr/dt, the rate at which the radius is changing. Since the base diameter and height are always the same, the radius r is always half of the height. Therefore, dr/dt is also half of dh/dt.
dr/dt = (1/2) * dh/dt
Substituting this back into the equation:
30 = (1/3) * π * h * (dh/dt + h * (1/2) * dh/dt)
To find dh/dt, we need to solve for it.
Multiplying through the parentheses:
30 = (1/3) * π * h * (dh/dt + (h/2) * dh/dt)
Combining like terms:
30 = (1/3) * π * h * ((3/2) * dh/dt)
Now, we can solve for dh/dt:
dh/dt = (30 * 3) / (π * h * (3/2))
dh/dt = 60 / (π * h)
When the pile is 18 feet high (h = 18), we can calculate the rate at which the height is changing:
dh/dt = 60 / (π * 18)
dh/dt ≈ 1.05 feet per minute
Therefore, when the pile is 18 feet high, the height of the pile is increasing at a rate of approximately 1.05 feet per minute.
dV/dt = d/dt[(pi/3)*R^2*H]
= 30 ft^3/min
= (pi/3)(1/4)*d/dt[H^3] since H= 2R
dV/dt = 30 = (pi/4)*3H^2*dH/dt
Solve for dH/dt when H = 18