Question

At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 16 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate (in ft/min) is the height of the pile changing when the pile is 2 feet high

Answer 1

The rate at which the height of the pile is changing when the height is 2 ft. is 0.5659 ft./min. Computed using differentiation.

The sand is falling onto a conical pile, thus we get the pile in the shape of a cone, whose volume is given as:

V = (1/3)πr²h,

where V represents the volume of the cone, r represents its radius, and h represents its height.

In the question, we are given that the rate of piling is 16 ft.³/min, which implies that:

dV/dt = 16 ... (i)

We are also given that the diameter is three times the altitude (height), which can be shown as:

2r = 3h,

or, r  = (3/2)h.

Thus, the volume can now be shown as:

V = (1/3)π{(3/2)h}²h,

or, V = (3πh³)/4.

Differentiating this with respect to time, we get:

dV/dt = (3π/4)(3h²)dh/dt.

Now, we need to calculate the rate of change of height, when the height is  2 feet, thus we take h = 2

Using dV/dt = 16 from (i), we can write:

16 = (3π/4)(3(2)²)dh/dt.,

or, 16 = 9π(dh/dt),

or, dh/dt = 16/9π = 0.5659 ft./min.

Thus, the rate at which the height of the pile is changing when the height is 2 ft. is 0.5659 ft./min.

Learn more about differentiation at

brainly.com/question/13461339

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