A cone has a radius of 5 ft and a height of 15 ft. It is empty and is being filled with water at a constant rate of 24 ft 3 / sec . Find the rate of change of the radius of the surface of the water when the radius of the surface of the water is 2 ft. (You must also include the units)

The rate of change of the radius of the surface of the water when the radius of the surface of the water is 2 feet is approximately 0.637 feet per second.

Step-by-step explanation:

The volume of the cone (), in cubic feet, is defined by the following equation:

(1)

Where:

- Radius, in feet.

- Height, in feet.

And there is the following ratio of the radius to the height is:

(2)

By applying (2) in (1):

(3)

And the rate of change of the radius is found by differentiating on (3):

(4)

Where:

- Rate of change of the volume, in cubic feet per second.

- Rate of change of the surface of the water, in feet per second.

If we know that , and , then the rate of change of the radius of the surface of the water is:

The rate of change of the radius of the surface of the water when the radius of the surface of the water is 2 feet is approximately 0.637 feet per second.

you didn't post any answer choices, so its not 100% clear what your teacher is looking for

however, log(15/7) is equivalent to log(15) - log(7)

i'm using the rule log(x/y) = log(x) - log(y)