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 the same. How fast is the height of the pile increasing when the pile is 25 feet high? Recall that the volume of a right circular cone with height h and radius of the base r is given by V=(π/3)(r^2)h
To find the rate at which the height of the pile is increasing, we need to use related rates. Let's call the height of the pile h and the radius of the base r.
We are given that the gravel is being dumped at a rate of 10 cubic feet per minute. This means that the volume of the pile, V, is increasing at a rate of 10 cubic feet per minute. We can express this as dV/dt = 10 ft^3/min.
Using the formula for the volume of a cone, V = (π/3)(r^2)h, we can differentiate both sides of the equation with respect to time t to find the rate of change of volume:
dV/dt = (π/3)(2r)(dr/dt)h + (π/3)(r^2)(dh/dt)
Since we want to find the rate at which the height of the pile is increasing, dh/dt, we need to find dr/dt, the rate at which the radius is changing.
We know that the base diameter and height of the cone are always the same, so the radius, r, is equal to the height, h. Therefore, we can substitute h for r in the equation:
dV/dt = (π/3)(2h)(dr/dt)h + (π/3)(h^2)(dh/dt)
Simplifying the equation:
10 = (2π/3)(h)(dh/dt) + (π/3)(h^2)(dh/dt)
Now, we can solve for dh/dt by isolating it on one side of the equation:
dh/dt = 10 / [(2π/3)(h) + (π/3)(h^2)]
Now, we can substitute the given height of the pile, h = 25 feet, into the formula to find the rate at which the height is increasing when the pile is 25 feet high:
dh/dt = 10 / [(2π/3)(25) + (π/3)(25^2)]
Calculating this expression will give us the answer, which corresponds to the rate at which the height of the pile is increasing when the pile is 25 feet high.
To find how fast the height of the pile is increasing, we need to use related rates. We know that the rate at which gravel is being dumped is 10 cubic feet per minute.
Let's assume that at a given time, the height of the pile is h(t) and the radius of the base is r(t). We are given that the base diameter and height of the pile are always the same, which means the radius is half the height, so r(t) = h(t)/2.
We are also given the formula for the volume of a right circular cone:
V = (π/3)(r^2)h
Substituting the values for r and simplifying, we get:
V = (π/3)((h/2)^2)h
V = (π/3)(h^3/4)
V = (πh^3)/12
Next, we differentiate both sides of the equation with respect to time t:
dV/dt = d((πh^3)/12)/dt
The left-hand side represents the rate at which the volume is changing, which is 10 cubic feet per minute. The right-hand side represents how the volume changes with respect to time, which we need to find.
So, we have:
10 = d((πh^3)/12)/dt
Next, we differentiate the equation with respect to h using the chain rule:
10 = (π/12) * (3h^2) * dh/dt
Now, we need to solve for dh/dt, which represents the rate at which the height is changing.
dh/dt = (10 * 12)/(π * 3h^2)
Simplifying further, we get:
dh/dt = 40/(π * h^2)
Given that the height of the pile is 25 feet, we can calculate the rate at which the height is increasing:
dh/dt = 40/(π * (25^2))
= 40/(625π)
≈ 0.0255 ft/min
Therefore, when the pile is 25 feet high, the height of the pile is increasing at a rate of approximately 0.0255 feet per minute.
V=(π/3)(r^2)h
but we are told that h = 2r or r = h/2
V = (1/3)π (h/2)^2 (h) = (1/3)π h^3/4
=(1/12)π h^3
dV/dt = (1/4)π h^2 dh/dt
10 = (1/4)π(625) dh/dt
dh/dt = ....
you finish it.