If a cylindrical tank standing upright (with one circular base on the ground) has a radius of 12 cm for the base and is being drained at 23 cm³/sec, the water level in the tank is dropping at [tex]0.038cm/s[/tex]
The formula for calculating the volume of the cylinder is expressed as:
[tex]V = \pi r^2 h[/tex] where:
r is the radius of the tank
h is the height of the tank
Get the rate of change of the volume
[tex]\frac{dV}{dt} =\frac{dV}{dr}\times \frac{dr}{dt}\\23 = 2 \pi rh\times \frac{dr}{dt}[/tex]
Given that radius is 12cm, let the height be 8cm
Substitute the given values into the formula
[tex]\frac{dV}{dt} =\frac{dV}{dr}\times \frac{dr}{dt}\\23 = 2(3.142) (8)(12)\times \frac{dr}{dt}\\\frac{dr}{dt}=\frac{23}{603.264}\\ \frac{dr}{dt}=0.038cm/s[/tex]
Hence the water level in the tank is dropping at [tex]0.038cm/s[/tex]
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