Gravel is being dumped from a conveyor belt at a rate of 30 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 15 ft high? (Round your answer to two decimal places.)
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Thanks :)
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Volume of a cone is (1/12)*pi*h*d^2
Given that the diameter and the height are equivalent, give them a single name (x)
V = (1/12)*pi*x^3
The derivative of V with respect to x - this is dV/dx
You already know dV/dt = 30 ft3/min
You want to know dx/dt at a specific point
Write a little equation: dV/dt = (dV/dx)*(dx/dt)
So, dx/dt = (dV/dt)/(dV/dx)
dV/dx can be evaluated from the above equation: dV/dx = (3*pi/12)*x^2
Evaluate all this at x = 15 ft:
dV/dx = 176.71 ft2/min
dx/dt = (30 ft3/min)/(176.71 ft2/min) = 0.17 ft/min
Given that the diameter and the height are equivalent, give them a single name (x)
V = (1/12)*pi*x^3
The derivative of V with respect to x - this is dV/dx
You already know dV/dt = 30 ft3/min
You want to know dx/dt at a specific point
Write a little equation: dV/dt = (dV/dx)*(dx/dt)
So, dx/dt = (dV/dt)/(dV/dx)
dV/dx can be evaluated from the above equation: dV/dx = (3*pi/12)*x^2
Evaluate all this at x = 15 ft:
dV/dx = 176.71 ft2/min
dx/dt = (30 ft3/min)/(176.71 ft2/min) = 0.17 ft/min