Gravel is being dumped from a conveyor belt at a rate of 35 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 9 ft high? (Round your answer to two decimal places.)

  dh/dt ≈ 0.55 ft/min

Step-by-step explanation:

The volume is given by the formula ...

  V = (1/3)πr²h

We have r = h/2, so the volume as a function of height is ...

  V = (1/3)π(h/2)²h = (π/12)h³

Then the rates of change are related by ...

  dV/dt = (π/4)h²·dh/dt

  dh/dt = (4·dV/dt)/(πh²) = 4(35 ft³/min)/(π(9 ft)²)

  dh/dt ≈ 0.55 ft/min

answer: you need to organize this better it does not make sense.

step-by-step explanation: ?