Mathematics: 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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A cone has a radius of 5 ft and a height of 15

Ответ:

samanthabutryn

15.08.2021

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:

$V = \frac{\pi}{3}\cdot r^{2}\cdot h$ (1)

Where:

- Radius, in feet.

- Height, in feet.

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

$\frac{r}{h} = k$ (2)

By applying (2) in (1):

$h = \frac{r}{k}$

$V = \frac{\pi}{3\cdot k}\cdot r^{3}$ (3)

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

$\dot V = \frac{\pi}{k}\cdot r^{2}\cdot \dot r$ (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.

$\dot r = \frac{k\cdot \dot V}{\pi\cdot r^{2}}$

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

$\dot r = \frac{\left(\frac{1}{3} \right)\cdot \left(24\,\frac{ft^{3}}{s} \right)}{\pi\cdot (2\,ft)^{2}}$

$\dot r = 0.637\,\frac{ft}{s}$

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.

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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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