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.

- 2 + 10x³ - 21x² + 19x - 4

step-by-step explanation:

each term in the second factor is multiplied by each term in the first factor

x²(- 2x² + 4x - 1) - 3x(- 2x² + 4x - 1) + 4(- 2x² + 4x - 1) ← distribute

= - 2 + 4x³ - x² + 6x³ - 12x² + 3x -8x² + 16x - 4

collect like terms

= - 2 + 10x³ - 21x² + 19x - 4 ← in standard form

c is the correct answer

step-by-step explanation: