Consider a tank in the shape of an inverted right circular cone that is leaking water. The dimensions of the conical tank are a height of 12 ft and a radius of 10 ft. How fast does the depth of the water change when the water is 10 ft high if the cone leaks at a rate of 8 cubic feet per minute?

Using implicit differentiation, it is found that the depth of water changes at a rate of -0.0764 feet per minute.

The volume of a cone of radius r and height h is given by:

Applying implicit differentiation, we can find it's rate of change, thus:

In this problem:

Then:

The depth of water changes at a rate of -0.0764 feet per minute.

A similar problem is given at link

Step-by-step explanation:

Given

Represent radius with r and height with h

when

Express radius in terms of height

First, we need to determine the volume of the cone in terms of height.

Substitute

Differentiate both sides w.r.t time (t)

Solve for

At this point, and

So, we have:

Hence:

The rate is

7 ft 6 in means 7 1/2, because 6 in is 1/2 of a foot. 7 ft 8 in means 7 2/3 because 8 in is 2/3 of a foot. 23/3 = 7 2/3. 15/2 = 7 1/2.

23/3   *   15/2   = 57 1/2

step-by-step explanation: