Gravel is being dumped from a conveyor belt at a rate of 20 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 11 ft high

Step-by-step explanation:

Since the gravel being dumped is in the shape of a cone, we will use the formula for calculating the volume of a cone.

Volume of a cone V = πr²h/3

If the diameter and the height are equal, then r = h

V = πh²h/3

V = πh³/3

If the gravel is being dumped from a conveyor belt at a rate of 20 ft³/min, then dV/dt = 20ft³/min

Using chain rule to get the expression for dV/dt;

dV/dt = dV/dh * dh/dt

From the formula above, dV/dh = 3πh²/3

dV/dh =  πh²

dV/dt = πh²dh/dt

20 = πh²dh/dt

To calculate how fast the height of the pile is increasing when the pile is 11 ft high, we will substitute h = 11 into the resulting expression and solve for dh/dt.

20 = π(11)²dh/dt

20 = 121πdh/dt

dh/dt = 20/121π

dh/dt = 20/380.133

dh/dt = 0.0526ft/min

This means that the height of the pile is increasing at  0.0526ft/min

what is the question here

step-by-step explanation:

answer: -8.2562234e+13