Gravel is being dumped from a conveyor belt at a rate of 15 ft^3/hr and its courseness is such that is forms a pile in the shape of an inverted right cone whose height is three times its base radius. How fast is the height of the pile increasing when the pile has a height of 10 feet?
h=3r, so r = h/3
v = π/3 r^2 h = π/3 (h/3)^2 h = π/27 h^3
dv/dt = π/9 h^2 dh/dt
So, plug in your numbers to get dh/dt.
To find the rate at which the height of the pile is increasing, we need to use related rates. We have the following information:
- The rate at which gravel is being dumped: 15 ft^3/hr
- The shape of the pile: an inverted right cone
- The relationship between the height and base radius of the cone: height = 3 * base radius
Let's represent the height of the pile as h (in feet) and the base radius as r (in feet).
We need to find dh/dt, the rate at which the height of the pile is changing when the pile has a height of 10 feet.
First, we need to find the relationship between h and r.
Since we have an inverted right cone, the volume V of the pile can be given by:
V = (1/3) * π * r^2 * h
We want to find dh/dt, which represents the rate at which the height is changing with respect to time. To do this, we can differentiate both sides of the equation V = (1/3) * π * r^2 * h with respect to time t.
dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)
We know that dV/dt is given as 15 ft^3/hr, and we need to find dh/dt when h = 10 ft.
Replacing V with the rate of gravel being dumped, we have:
15 = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)
Now, we can substitute the relationship between h and r, which is h = 3r. Therefore, we have:
15 = (1/3) * π * (2r * dr/dt * 3r + r^2 * dh/dt)
Simplifying further:
15 = (2/3) * π * (3r * dr/dt * r + r^2 * dh/dt)
15 = (2/3) * π * (3r^2 * dr/dt + r^3 * dh/dt)
We are given that h = 10 ft. Since h = 3r, we can substitute 10 for h in the equation:
15 = (2/3) * π * (3r^2 * dr/dt + r^3 * dh/dt) [Substituting h = 10]
Now, we can solve for dh/dt:
15 = (2/3) * π * (3r^2 * dr/dt + 10^3 * dh/dt)
We need to find dh/dt when h = 10 ft. At that point, we know that r = h/3 = 10/3 ft. Plugging in these values, the equation becomes:
15 = (2/3) * π * (3 * (10/3)^2 * dr/dt + (10/3)^3 * dh/dt)
Simplifying further:
15 = (2/3) * π * (100/3 * dr/dt + 1000/27 * dh/dt)
Now, we can solve for dh/dt:
dh/dt = (15 * 3 * 27) / (2 * π * 100) ft/hr
Calculating this expression gives:
dh/dt ≈ 0.204 ft/hr
Therefore, the height of the pile is increasing at a rate of approximately 0.204 feet per hour when the pile has a height of 10 feet.