Gravel is being dumped from a conveyor belt at a rate of 10 ft^3/min. 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 23 feet high?
Im working on same problem..... sorry no help here
volume of cone = (1/3)pi(r^2)h
but h=2r or r = h/2
then volume = (1/12)pi(h^3)
dV/dt = (1/4)pi (h^2) dh/dt
10 = (1/4)pi (23^2)dh/dt solve for dh/dt
To find the rate at which the height of the pile is increasing, we need to apply the concept of related rates. Let's denote the height of the cone as h and the radius of its base as r.
We know that the base diameter and height are always the same, which means the radius is half the height. Thus, we have r = h/2.
The volume of a cone is given by the formula V = (1/3)πr^2h. We can substitute r with h/2 to express the volume in terms of height only:
V = (1/3)π(h/2)^2h = (1/3)π(h^3)/4
Differentiating both sides of this equation with respect to time (t) gives us the related rate equation:
dV/dt = (1/3)π(3h^2 dh/dt)/4
Given that the rate of change of volume (dV/dt) is 10 ft^3/min, we can substitute this value in the equation:
10 = (1/3)π(3h^2 dh/dt)/4
Next, we need to find dh/dt, the rate at which the height is increasing when the height is 23 ft.
Substituting h = 23 into the equation:
10 = (1/3)π(3(23)^2 dh/dt)/4
Simplifying the equation gives us:
10 = (1/3)π(3(529) dh/dt)/4
10 = 1117π dh/dt / 4
Now, let's solve for dh/dt:
dh/dt = (10 * 4) / (1117π)
dh/dt ≈ 0.0113 ft/min
Therefore, the height of the pile is increasing at a rate of approximately 0.0113 ft/min when the pile is 23 ft high.
To find how fast the height of the pile is increasing, we need to use related rates. We need to find an equation that relates the height and the volume of the cone-shaped pile.
We are given that the rate at which gravel is being dumped is 10 ft^3/min. This means the volume is increasing at a rate of 10 ft^3/min.
The volume of a cone can be calculated using the formula: V = (1/3) * π * r^2 * h, where V is the volume, r is the radius of the base, and h is the height.
Since the problem states that the base diameter and height are always the same, we know that the radius is equal to half the diameter of the base.
Let's assume that the radius and height of the cone pile at any given time are r and h.
Since the base diameter and height are the same, we can also say that the radius and height are the same. Therefore, r = h.
Now, we can express the volume equation in terms of a single variable, h:
V = (1/3) * π * h^3.
To find how fast the height is changing with respect to time, we differentiate both sides of this equation with respect to time (t):
dV/dt = (1/3) * π * 3h^2 * dh/dt.
We know that dV/dt is given as 10 ft^3/min, so we have:
10 = (1/3) * π * 3h^2 * dh/dt.
Simplifying the equation, we have:
10 = π * h^2 * dh/dt.
Now, we need to find the value of dh/dt when the height of the pile is 23 feet. We can substitute h = 23 into the equation and solve for dh/dt.
10 = π * (23)^2 * dh/dt.
Solving for dh/dt, we have:
dh/dt = 10 / (π * 23^2).
Calculating this value, we find:
dh/dt ≈ 10 / (π * 529) ≈ 0.006 ft/min.
Therefore, the height of the pile is increasing at a rate of approximately 0.006 ft/min when the pile is 23 feet high.