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 14 ft high?

The rate at which the height of the pile is increasing when the pile is 14 ft high is;

dh/dt = 0.1624

Let h be the height of the cone.

We are told that the diameter and height are always equal.

Thus; d = h

Radius;r = d/2 = h/2

Now, formula for volume of a cone is;

V = ⅓πr²h

Putting h/2 for r to get;

V = ⅓π(h/2)²h

V = (1/12)πh³

Now, let's differentiate with respect to t since we are given rate at which volume is increasing and we want to find rate at which height of 14 ft is increasing.

Thus;

dV/dt = (3/12)πh²(dh/dt)

We are given; dV/dt = 25 ft³/min and h = 14 ft

Thus;

25 = ¼π(14²)(dh/dt)

dh/dt = 100/196π

dh/dt = 0.1624

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feet per minute.

Step-by-step explanation:

Let x represent height of the cone.

We have been given that gravel is being dumped from a conveyor belt at a rate of 25 ft^3/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.

Since radius is half the diameter, so radius of cone would be .

Now we will use volume of cone formula.

Now, we will take derivative with respect to time.

Upon substituting our given values, we will get:

Therefore, the height of the pile is increasing at a rate of feet per minute.