HELP Diff Eqs, Torricelli's law

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SUMMARY

The discussion centers on applying Torricelli's law to solve a differential equation related to the draining of a cylindrical tank. The equation derived for the volume of water in the tank after time t is V = V0 [1 - (t/T)]^2, where V0 is the initial volume and T is the total draining time. The user expresses confusion regarding the application of Torricelli's law, which states that the velocity of water exiting a hole is proportional to the square root of the water's depth. The solution involves equating the rates of change of volume and height, leading to a solvable differential equation.

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HELP! Diff Eqs, Torricelli's law

Hi, my crazy diff eq professor for some reason decided to assign us a project accompanied by some book problems right in the middle of exam week, due friday (day of my last exam). This is easily the most redic. thing I've ever seen in my college career. Anyhow, I'm trying to get all the book problems done right now, and only have 1 left, that I've been trying to wrap my head around for awhile now. I just really want to get this out of the way so I can go back to studying for my exams!

The question is: Suppose that a cylindrical tank initially containing V0 gallons of water drains (through a bottom hole) in T minutes. Use Torricelli's law to show that the volume of water in the tank after t<=T minutes is V= V0 [1-(t/T)]^2

I'm very lost on this, as we have not touched Torricelli's law at all in class.
 
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Toricelli's law states the velocity of water through the opening is proportional to the square root of the depth of the water. So if h is the depth of the water at time t, v is the velocity of the water through the hole, Ahole is the area of the hole, Ab is the area of the base of the cylinder, and V is the volume of water at time t, you have

\frac {dV}{dt} = A_{hole}\frac{dv}{dt} = -A_{hole} k\sqrt h = -c\sqrt h

V = A_b h\hbox{ so }\frac{dV}{dt}<br /> = A_b\frac{dh}{dt}

Equating the two expressions for dV/dt gives

A_b\frac{dh}{dt}= -c\sqrt h

so lumping the constants together you can write

\frac{dh}{dt}=-C\sqrt h

Solve that equation for h. You have two facts given -- one that the cylinder is empty when t = T and the other that the volume is V0 when t = 0. These will allow you to figure out C and the constant of integration you get when you solve. And, of course, once you know h you know V.
 
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