Gravel is being dumped from a conveyor belt at a rate of 25 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 6 ft high? (Round your answer to two decimal places.)
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To solve this problem, we need to apply the related rates concept.
Let's label the variables:
- V = Volume of the pile of gravel (in ft^3)
- r = Radius of the base of the cone (in ft)
- h = Height of the pile of gravel (in ft)
We are given that the gravel is being dumped from the conveyor belt at a rate of 25 ft^3/min. This means that dV/dt = 25 ft^3/min.
The volume of a cone is given by the formula:
V = (1/3) * π * r^2 * h
Since the cone's base diameter and height are always equal, we know that the base radius is equal to half the height (r = h/2).
Let's differentiate the volume equation with respect to time (t) using the chain rule:
dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)
To find how fast the height (dh/dt) is changing when the pile is 6 ft high, we need to solve for dh/dt.
First, let's find the values of r and V when h = 6 ft.
r = h/2 = 6/2 = 3 ft
V = (1/3) * π * r^2 * h = (1/3) * π * 3^2 * 6 = 18π ft^3
Now, substitute these values into the differentiated volume equation and solve for dh/dt:
25 ft^3/min = (1/3) * π * (2 * 3 * dr/dt * 6 + 3^2 * dh/dt)
25 ft^3/min = (2π * 18 + 9π * dh/dt)/3
25 ft^3/min = (36π + 9π * dh/dt)/3
75 = 36π + 9π * dh/dt
9π * dh/dt = 75 - 36π
dh/dt = (75 - 36π) / (9π)
Now, we can evaluate dh/dt by substituting the value of π ≈ 3.14:
dh/dt ≈ (75 - 36 * 3.14)/(9 * 3.14)
dh/dt ≈ 2.22 ft/min
Therefore, the height of the pile is increasing at a rate of approximately 2.22 ft/min when the pile is 6 ft high.
since h = 2r, r = h/2
v = pi/3 r^2h = pi/12 h^3
So,
dv/dt = pi/4 h^2 dh/dt
When h=6, we have
25 = 9pi dh/dt
Now just do the math