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 24 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)
shoot, they've done the work for you
V = 1/3 pi r^2 h
but h = 2r (the diameter)
V = 2/3 pi r^3
dV/dt = 2pi r^2 dr/dt
50 = 2pi*144 dr/dt
dr/dt = 50/288pi = .055 ft/min
Well, well, well, looks like we have a conundrum involving gravel and a conical pile. Let's get to the bottom of this, shall we?
Given that the gravel is being dumped at a rate of 50 cubic feet per minute, we can determine that the volume of the pile is increasing at a rate of 50 cubic feet per minute. Using the volume formula for a right circular cone, which you so kindly provided, we can express the volume V as:
V = (1/3)πr^2h
Now, we want to find out how fast the height of the pile is increasing when the pile is 24 feet high. To solve this, we need to find a connection between the height h and the radius r.
Luckily, we know that the base diameter and the height are always the same in our situation. This means that the radius r is equal to half the height h. So, r = h/2.
Let's substitute this relationship for r into our volume formula:
V = (1/3)π(h/2)^2h
= (1/3)π(h^3/4)
Now, we can differentiate both sides of the equation with respect to time t:
dV/dt = (1/3)(3h^2/4)(dh/dt)
The left-hand side represents the rate at which the volume changes, which we already know is 50 cubic feet per minute. So, we have:
50 = (1/3)(3h^2/4)(dh/dt)
Simplifying this equation a bit, we get:
dh/dt = (4/3)(50/h^2)
Now, all that's left is to substitute the value of h = 24 into this equation to find the rate at which the height is increasing when the pile is 24 feet high:
dh/dt = (4/3)(50/24^2)
≈ (4/3)(50/576)
≈ (4/3)(25/288)
≈ 33/36
≈ 0.92 feet per minute
So, my friend, when the pile is 24 feet high, the height of the pile is increasing at a rate of approximately 0.92 feet per minute. Keep pouring that gravel and watch that pile grow!
To find the rate at which the height of the pile is increasing when the pile is 24 feet high, we need to find the derivative of the volume function with respect to time and then substitute the given values.
Given:
Rate of dumping gravel from conveyor belt = 50 cubic feet per minute
Height of the pile = 24 feet
First, we need to express the volume of the cone, V, as a function of height, h, and the radius of the base, r.
The volume of a cone is given by the formula:
V = (1/3)πr^2h
We know that the radius and height of the cone are always the same, so r = h.
Substituting r = h in the volume formula:
V = (1/3)πh^3
Next, we differentiate both sides of the equation with respect to time, t, using the chain rule:
dV/dt = d/dt [(1/3)πh^3]
dV/dt = (1/3)π * 3h^2 * dh/dt
dV/dt = πh^2 * dh/dt
Now we can substitute the given values:
dV/dt = π * (24 ft)^2 * dh/dt
dV/dt = 576π * dh/dt
We also know that the rate of dumping gravel is given as 50 cubic feet per minute, so we have:
dV/dt = 50 ft^3/min
Now we can solve for dh/dt:
50 ft^3/min = 576π * dh/dt
Divide both sides of the equation by 576π:
(50 ft^3/min) / (576π) = dh/dt
Approximately, the rate at which the height of the pile is increasing when the pile is 24 feet high is:
dh/dt ≈ 0.028 ft/min
To solve this problem, we need to find the rate at which the height of the pile is changing when the pile is 24 feet high.
Let's denote the height of the pile as h (in feet) and the radius of the base of the pile as r (in feet). Since the base diameter and height are always the same, the radius of the base is equal to half of the height of the pile.
We are given that the gravel is being dumped at a rate of 50 cubic feet per minute. This means that the rate of change of volume, dV/dt, is 50 ft^3/min. We want to find the rate of change of height, dh/dt, when h = 24 ft.
We can start by finding an equation that relates the height, the radius, and the volume of the cone.
The volume of a right circular cone is given by the formula:
V = (1/3)(pi)(r^2)(h)
Since we know that the base diameter and height are always the same, we can write r = h/2.
Substituting this into the volume equation, we get:
V = (1/3)(pi)((h/2)^2)(h)
V = (1/12)(pi)(h^3)
Now, we can differentiate both sides of this equation with respect to time t to find the derivative of the volume with respect to time, dV/dt, in terms of dh/dt:
dV/dt = (3/12)(pi)(h^2)(dh/dt)
Since dV/dt is given as 50 ft^3/min, we have:
50 = (3/12)(pi)(h^2)(dh/dt)
We need to find the value of dh/dt when h = 24 ft. Plugging in these values, we can solve for dh/dt:
50 = (3/12)(pi)(24^2)(dh/dt)
dh/dt = (50*12)/(3*pi*(24^2))
dh/dt ≈ 0.0714 ft/min
Therefore, the height of the pile is increasing at a rate of approximately 0.0714 feet per minute when the pile is 24 feet high.