Gravel is being dumped from a conveyor belt at a rate of 50 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=13πr^2h
so, what's the trouble? You have the formula, so crank it out.
h = 2r (diameter = base), so
v = π/3 r^2h = π/3 (h/2)^2h = π/12 h^3
dv/dt = π/4 h^2 dh/dt
50 = π/4 * 25^2 dh/dt
dh/dt = 8/(25π) ft/min
Well, this seems like a real conundrum. Let's tackle it with a smile, shall we?
We know that the rate of change of volume with respect to time is 50 cubic feet per minute. So, we have dV/dt = 50 ft³/min.
Now, we need to find how fast the height of the pile is increasing, dh/dt, when the pile is 25 feet high.
To do that, we need to express h in terms of r. Lucky for us, since the base diameter and height are always the same, the radius r is equal to h/2.
Now, the volume formula V = (1/3)πr^2h becomes V = (1/3)π(h/2)²h = (π/12)h³.
Differentiating both sides of the equation with respect to time, we get dV/dt = (π/4)h² * dh/dt.
Plugging in the values we know, we have 50 = (π/4)(25)² * dh/dt.
Now we just solve for dh/dt:
dh/dt = 50 / [(π/4)(25)²]
dh/dt = 50 / [(π/4)(625)]
dh/dt = (200 / π) ft/min.
So, my friend, when the pile is 25 feet high, the height of the pile is increasing at a rate of approximately 63.7 ft/min. Now it's time for you to elevate the joy with your newfound knowledge!
To solve this problem, we will need to use related rates. We are given that the rate at which gravel is being dumped from the conveyor belt is 50 cubic feet per minute and we need to find the rate at which the height of the pile is increasing when the pile is 25 feet high.
Let's assign variables to the relevant quantities:
- The rate at which the height is increasing is dh/dt (in ft/min).
- The height of the pile is h (in ft).
- The radius of the base of the pile is r (in ft).
- The rate at which gravel is being dumped is dV/dt = 50 ft^3/min.
From the given information, we can deduce that the volume of the cone is increasing with time. The volume can be expressed as:
V = 1/3 * π * r^2 * h
To relate the variables, we differentiate both sides of the equation with respect to time t:
dV/dt = 1/3 * π * (2r * dr/dt * h + r^2 * dh/dt)
We know that dV/dt = 50 ft^3/min and we need to find dh/dt when h = 25 ft.
We also know that the radius of the base and the height of the pile are always the same. Therefore, r = h/2.
Substituting this information into the equation, we have:
50 = 1/3 * π * (2(h/2) * dr/dt * h + (h/2)^2 * dh/dt)
Simplifying the equation:
50 = 1/3 * π * (h * dr/dt * h + 1/4 * h^2 * dh/dt)
50 = 1/3 * π * (h^2 * dr/dt + 1/4 * h^2 * dh/dt)
150 = π * (h^2 * dr/dt + 1/4 * h^2 * dh/dt)
150/π = h^2 * dr/dt + 1/4 * h^2 * dh/dt
Now, we can substitute the known values into the equation. Since r = h/2, dr/dt = dh/dt * 1/2.
150/π = h^2 * (dh/dt * 1/2) + 1/4 * h^2 * dh/dt
150/π = (h^2/2 + h^2/4) * dh/dt
150/π = (3h^2/4) * dh/dt
Simplifying further:
dh/dt = (150/π) / (3h^2/4)
dh/dt = (150/π) * (4/3h^2)
dh/dt = 200/πh^2
Now, we can calculate the rate at which the height of the pile is increasing when the pile is 25 feet high by substituting h = 25 into the equation:
dh/dt = 200/π(25^2)
dh/dt = 200/π(625)
dh/dt = 0.032 ft/min
Therefore, when the pile is 25 feet high, the height of the pile is increasing at a rate of 0.032 feet per minute.
To find the rate at which the height of the pile is changing, we need to use related rates and take the derivative of the formula for the volume of a cone with respect to time.
First, let's determine the variables we know and the one we're looking for:
Given:
- The rate at which gravel is being dumped onto the pile is 50 cubic feet per minute.
To find:
- The rate at which the height of the pile is changing when the pile is 25 feet high.
Formula for the volume of a cone:
V = (1/3)πr^2h
Given that the base diameter and height are always the same, we can infer that the radius (r) is equal to half of the base diameter, which is also equal to half the height of the cone.
Let's assign some variables:
- Volume of the cone: V
- Radius of the cone: r
- Height of the cone: h
Since r = h/2, we can rewrite the volume formula in terms of h only:
V = (1/3)π(h/2)^2h
V = (1/12)πh^3
Now, we need to take the derivative of both sides of the equation with respect to time (t):
dV/dt = (1/12)π * 3h^2 * dh/dt
The left side of the equation, dV/dt, represents the rate at which the volume is changing, which is given as 50 cubic feet per minute.
Substituting the known values into the equation:
50 = (1/12)π * 3(25^2) * dh/dt
Now, we need to solve for dh/dt, which represents the rate at which the height is changing when the height is 25 feet:
dh/dt = 50 / [(1/12)π * 3(25^2)]
dh/dt = 4 / (π * 3(25^2))
Simplifying further:
dh/dt = 4 / (75π)
dh/dt = 4 / 235.62
Therefore, the rate at which the height of the pile is increasing when the pile is 25 feet high is approximately 0.017 feet per minute.