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 Question from Carolyn, a parent: The flow of water out of a hole in a tank is known to be proportional to the square root of the height of water above the hole. That is, dV/dt (proportional to) sq root (h) The tank has a constant cross-sectional area A, show that the height of water in the tank is given by h = ((-kt+C)/2)^2 If the tank is 9 metres high, and it takes 5 hours for it to drain from full to half full, how much longer will we have to wait until it is completely empty?

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 cross-sectional 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.