At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 4 cubic feet per minute. the diameter of the base of the cone is approximately three times the altitude. at what rate is the height of the pile changing when the pile is 2 feet high? (hint: the formula for the volume of a cone is v = 1 3 πr2h.)

When the pile is 2 feet high, the height is changing at a rate of 0.14 feet per minute.

Step-by-step explanation:

The formula for the volume of a cone is:

Where:

r: radius of the base

h: height of the cone

We've been told that the diameter of the base is three times the height of the cone, which can be written like this:

Then, replacing this in the volume formula:

Here, we've found a formula for the volume in terms of the height.

Given that the height changes in time, then the derivative of volume with respect of the time (the rate of change of the volume in time) will be:

Then the rate of change of the height will be:

So, the rate of change of the height when the pile is 2 ft high can be found using this last formula (dV/dt is the rate of change of the volume, which is 4 cubic feet per minute)

So, when the pile is 2 feet high, the height is changing at a rate of 0.14 feet per minute.

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