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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)

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Answer:

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.

Explanation:

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


V = (\pi)/(3)\cdot r^(2)\cdot h (1)

Where:


r - Radius, in feet.


h - Height, in feet.

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


(r)/(h) = k (2)

By applying (2) in (1):


h = (r)/(k)


V = (\pi)/(3\cdot k)\cdot r^(3) (3)

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


\dot V = (\pi)/(k)\cdot r^(2)\cdot \dot r (4)

Where:


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


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


\dot r = (k\cdot \dot V)/(\pi\cdot r^(2))

If we know that
k = (1)/(3),
\dot V = 24\,(ft^(3))/(s) and
r = 2\,ft, then the rate of change of the radius of the surface of the water is:


\dot r = (\left((1)/(3) \right)\cdot \left(24\,(ft^(3))/(s) \right))/(\pi\cdot (2\,ft)^(2))


\dot r = 0.637\,(ft)/(s)

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.

User Adriaan
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