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| Math Central | Quandaries & Queries |
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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. 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, |
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
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