



 
Hi Carolyn, Since water is flowing out of the tank $\large \frac{dV}{dt}$ is negative so there is a positive constant $k$ so that \[\frac{dV}{dt} = k \sqrt{h}.\] Since the tank has a constant crosssectional area $A$ the volume is given by $V = A h$ and $\large \frac{dV}{dt} = \normalsize A \large \frac{dh}{dt}.$ Thus \[k \sqrt{h} = A \frac{dh}{dt}.\] or \[ A \frac{dh}{h^{1/2}} = k dt.\] Integrate both sides and solve for $h.$ I got the expression in your question but with an $A$ in it. Can you see how to complete the problem now? Write back and tell us what you did if you need more assistance. Penny  


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 