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- anonymous

An environmental hydrologist needs to know the total input, C, measured in cubic feetper minute, of water into a shallow circular lake with known parabolic cross-section y = 0.00001x^2, , where the lake is formed by rotatingthe parabola around the vertical axis. This quantity C is dicult to measure directly,as the lake is fed by both surface run-o and underground streams, and depleted byevaporation (also measured in cubic feet per minute) equal to 0.01 times the area ofthe lake surface at any time. The hydrologist is seeking your mathematical help tond C. Find h(t).

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- anonymous

- schrodinger

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- dumbcow

here is what i was able to come up with
For Volume of the lake at any point:
\[V = 100,000\pi \int\limits_{0}^{h}y dy = 50,000\pi h^{2}\]
Area of surface of lake at any point:
\[A = 100,000\pi h\]
where h is height of the water at some time t
C is rate volume is increasing or dV/dt
\[\frac{dV}{dt} = \frac{dh}{dt}*\frac{dV}{dh}\]
\[\frac{dV}{dh} = 100,000 \pi h\]
dV/dt or C is equal to rate of intake - rate of evaporation
Let K be some constant representing rate of intake from streams and run-off
\[\frac{dV}{dt} = K - 1,000\pi h\]
put it together:
\[K-1,000\pi h = \frac{dh}{dt}(100,000\pi h)\]
\[\rightarrow \frac{dh}{dt} = \frac{K-1,000\pi h}{100,000\pi h}\]
this turns out to be a crazy nonlinear differential equation
so i used wolfram to get solution for h(t)
http://www.wolframalpha.com/input/?i=dh%2Fdt+%3D+%28k-1000*pi*h%29%2F%28100%2C000*pi*h%29
hope this helps

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