Answer:
the rate of change is 3.8m/s
Step-by-step explanation:
The volume of the right circular cone is
\(V= \frac{\pi*r^2h}{3}.\)
The rate of change of volume is
\(\frac{dV}{dt} =\frac{d}{dt}( \frac{\pi*r^2h}{3})=\frac{\pi}{3} \frac{d}{dt}(r^2h)=\frac{\pi}{3}(2rh\frac{dr}{dt}+r^2\frac{dh}{dt})\)
In order to proceed further we have to define r in terms of h so tht we can compute the derivative above.
The ratio between h and r is
\(\frac{h}{r}=\frac{6}{3} =2\)
Therefore \(r=\frac{h}{2}\).
We plug that into the derivative above and get:
\(2rh\frac{dr}{dt}+r^2\frac{dh}{dt}=\frac{h^2}{2}\frac{dh}{dt}+\frac{h^2}{4}\frac{dh}{dt}=\frac{3h^2}{4}\frac{dh}{dt}.\)
Thus
\(\frac{dV}{dt}=\frac{\pi}{3}\frac{3h^2}{4}\frac{dh}{dt}\)
Now for the numerical part.
The rate of change of volume \(\frac{dV}{dt}\) is \(12\frac{m^3}{s}\), so when the water is 2 meters deep \(h=2\), therefore:
\(12=\frac{\pi*3*2^2}{4*3} \frac{dh}{dt} \\\\\therefore \boxed{ \frac{dh}{dt}=3.8m/s.}\)
