Question

A pool in the shape of a cylinder with a height of 6 m and a base radius of 7 m is drained for repair. If the rate at which the pool is being drained is 3 m3/min, find the rate at which the water level is falling when the water is 4 m deep to the nearest hundredth.

Answer 1

The rate at which the water level is falling when the water is 4 meters deep to the nearest hundredth is 0.02 meters per minute.

Explanation:

The volume of a cylinder (
`V`), measured in cubic meters, is represented by the following formula:

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

Where:

`r` - Radius, measured in meters.

`h` - Height, measured in meters.

Then, we differentiate (1) in time:

`\dot V = 2\pi \cdot r\cdot h \cdot \dot r +\pi \cdot r^(2)\cdot \dot h` (2)

Where:

`\dot V` - Rate of change of the volume, measured in cubic meters per minute.

`\dot r` - Rate of change of the radius, measured in meters per minute.

`\dot h` - Rate of change of the height, measured in meters per minute.

Then, we clear the rate of change of the height:

`\pi \cdot r^(2)\cdot \dot h = \dot V-2\pi \cdot r\cdot h \cdot \dot r`

`\dot h = (\dot V - 2\pi \cdot r\cdot h\cdot \dot r)/(\pi \cdot r^(2))`

`\dot h = (\dot V)/(\pi \cdot r^(2))-(2\cdot h\cdot \dot r)/(r)`

`\dot h = (1)/(r)\cdot \left((\dot V)/(\pi \cdot r)-2\cdot h\cdot \dot r \right)` (3)

If we know that
`\dot V = -3\,(m^(3))/(min)`,
`r = 7\,m`,
`h = 4\,m` and
`\dot r = 0\,(m)/(min)`, then the rate of change of the height is:

`\dot h = \left((1)/(7\,m) \right)\cdot \left[(-3\,(m^(3))/(min) )/(\pi\cdot (7\,m)-2\cdot (4\,m)\cdot \left(0\,(m)/(min) \right)) \right]`

`h = -0.019\,(m)/(min)`

The rate at which the water level is falling when the water is 4 meters deep to the nearest hundredth is 0.02 meters per minute.