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

Answer:

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.

Step-by-step 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:

$\pi \cdot r^{2}\cdot \dot h = \dot V-2\pi \cdot r\cdot h \cdot \dot r$

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

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

$\dot h = \frac{1}{r}\cdot \left(\frac{\dot V}{\pi \cdot r}-2\cdot h\cdot \dot r \right)$ (3)

If we know that $\dot V = -3\,\frac{m^{3}}{min}$, $r = 7\,m$, $h = 4\,m$ and $\dot r = 0\,\frac{m}{min}$, then the rate of change of the height is:

$\dot h = \left(\frac{1}{7\,m} \right)\cdot \left[\frac{-3\,\frac{m^{3}}{min} }{\pi\cdot (7\,m)-2\cdot (4\,m)\cdot \left(0\,\frac{m}{min} \right)} \right]$

$h = -0.019\,\frac{m}{min}$

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.