Gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 18 feet high? (Recall that the volume of a right circular cone with height h and radius of the base r is given by
V = (1/3) * pi * r^2 * h)
so, you know V, and
dV/dt = 1/3 pi (2r dr/dt + dh/dt)
You know h = 2r and that h = 18 and that dV/dt = 10.
Piece o' cake, no?
Ok, but i'm still a little confused on how I get dh/dt by itself
To find how fast the height of the pile is increasing when the pile is 18 feet high, we need to use related rates by differentiating the equation with respect to time.
Given:
- Gravel is being dumped at a rate of 10 cubic feet per minute.
- The pile forms a right circular cone with a base diameter and height that are always equal.
We are asked to find:
- The rate at which the height of the pile is increasing when the pile is 18 feet high.
Let's start by assigning variables:
- Let V represent the volume of the cone (in cubic feet).
- Let r represent the radius of the cone's base (in feet).
- Let h represent the height of the cone (in feet).
We know the formula for the volume of a right circular cone is:
V = (1/3) * pi * r^2 * h
To use this formula to find the rate of change of the height, we need to differentiate both sides of the equation with respect to time (t). This will involve finding derivatives of V, r, and h with respect to t.
Differentiating both sides of the equation with respect to t:
dV/dt = (1/3) * pi * (2r * dr/dt * h + r^2 * dh/dt)
The term dh/dt represents the rate at which the height is changing with respect to time.
We are given that the volume is increasing at a rate of 10 cubic feet per minute, so we can substitute dV/dt with 10:
10 = (1/3) * pi * (2r * dr/dt * h + r^2 * dh/dt)
Since the base diameter and height are always equal, we can express r in terms of h:
r = h/2
Substituting this into the equation:
10 = (1/3) * pi * (2 * (h/2) * dr/dt * h + (h/2)^2 * dh/dt)
10 = (1/3) * pi * (h * dr/dt * h + (h^2/4) * dh/dt)
10 = (1/3) * pi * (h^2 * dr/dt + h^2 * dh/dt/4)
10 = (1/3) * pi * h^2 * (dr/dt + dh/dt/4)
Now, we need to find dr/dt in terms of h. Since the base diameter and height are equal, we can express r in terms of h:
r = h/2
Differentiating both sides with respect to t:
dr/dt = (d(h/2)/dt)
dr/dt = (1/2) * (dh/dt)
Substituting this result back into the equation:
10 = (1/3) * pi * h^2 * ((1/2) * (dh/dt) + dh/dt/4)
10 = (1/3) * pi * h^2 * (3/4) * (dh/dt)
10 = (3/4) * (1/3) * pi * h^2 * (dh/dt)
10 = (1/4) * pi * h^2 * (dh/dt)
Simplifying the equation:
40 = pi * h^2 * (dh/dt)
Finally, we need to find the value of dh/dt when the pile is 18 feet high. Substituting h = 18 into the equation:
40 = pi * (18)^2 * (dh/dt)
40 = pi * 324 * (dh/dt)
dh/dt = 40 / (pi * 324)
dh/dt ≈ 0.03884 feet per minute
Therefore, the height of the pile is increasing at a rate of approximately 0.03884 feet per minute when the pile is 18 feet high.