



 
Hi Jason, If you look at Penny's response to a similar problem you will see an expression for the Volume of the tank. I am going to change her notation slightly and let $L$ be the length of the tank and $h$ the height of the water in the tank. With this change the expression is \[V = L \left(r^2 cos^{1}\frac{r  h}{r}  (r  h) \sqrt{2 h r  h^2} \right). \] In this expression $L$ and $r$ are constants and $V$ and $h$ vary with time. The rates at which $V$ and $h$ vary with time are measured by their derivatives with respect to time $t$.Differentiating both sides with respect to $t$ and some simplification gives me \[\frac{dV}{dt} = L \frac{dh}{dt} \left( \frac{r^2}{\sqrt{2 h r  h^2}} + \sqrt{2 h r  h^2}  \frac{(r  h)^2}{\sqrt{2 h r  r^2}} \right). \] I think this is the expression you need, $\large \frac{dV}{dt}$ is the rate of change in the volume (the flow rate, which is negative) and $\large \frac{dh}{dt}$ is the rate of change of the height of the liquid. You need to take care with the units however when you use it.Let me suppose that you are measuring $L, r \mbox{ and } h$ in inches then you need to measure the flow ($\frac{dV}{dt},$ rate of change in the volume) in cubic inches per minute. It looks like you are measuring the flow in gallons per minute and there are 231 cubic inches in a US gallon so you need to multiply the rate in gallons per minute by 231. I hope this helps,  


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