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 (in ft/min) is the height of the pile increasing when the pile is 15 ft high? (Round your answer to two decimal places.)

The height of the pile is increasing at a rate of 1.78 ft/min

Step-by-step explanation:

As we can see, we have the rate at which the volume is increasing

Mathematically, the volume of a cone can be calculated by the formula;

1/3 * π * r^2 * h

but here, we are dealing with base diameter which is d

and r is d/2

So we have the formula as;

1/3 * π * (d/2)^2 * h

since h is same as d for this particular cone, we can replace d with h so that we now have;

1/3 * π * h^3/4

= πh^3/12

what we have in the question is dv/dt

what we want to calculate is dh/dt

Mathematically;

dh/dt = dh/dv * dv/dt

dh/dv is 1/dv/dh

dv/dh = πh^2/36 where h = 15

dv/dh = 225 π/36

dh/dv is now;

36/225π

So, finally;

dh/dt = 36/225 π * 35

dh/dt = 1.78 ft/min

idk

step-by-step explanation: