Answer:
Water is being pumped at a rate of [tex]x=9972.07cm^{3}/min[/tex]
Explanation:
We have volume of right circular cone is given by
[tex]V=\frac{1}{3}\pi r^{2}h[/tex]
Differentiating with respect to time we get
[tex]V=\frac{1}{3}\pi r^{2}h\\\\\frac{dV}{dt}=\frac{\pi }{3}(2rh\times \frac{dr}{dt}+\pi r^{2}\frac{dh}{dt})[/tex]
Now let the rate at which water is entering the conical vessel be [tex]xcm^{3}/min[/tex]
The cumulative change in volume is given by [tex](10000-x)[/tex]
By similar angle criterion of the triangle we have
[tex]\frac{2}{6}=\frac{r(h)}{h}\\\\\therefore r(h)=\frac{h}{3}[/tex]
[tex]\frac{dr(h)}{dt}=\frac{dh(t)}{3dt}[/tex]
Using all the above results we in the above equation we have
[tex]10000+x=\frac{\pi }{3}(\frac{2^{3}\times 20}{9}+\frac{4\times 20}{9})[/tex]
Solving for x we get
[tex]x=9972.07cm^{3}/min[/tex]
