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Homework Statement
A tank with vertical sides has a square cross-section of area 4 ft2. Water is leaving the tank through an orifice of area 5/3 in2. Water also flows into the tank at the rate of 100 in3/s. Show that the water level approaches the value (25/24)2 ft above the orifice.
Homework Equations
Rate of discharge of volume through the orifice is [itex]4.8A_0 \sqrt{y}[/itex] cubic feet per second, where [itex]A_0[/itex] = size of orifice in square feet
The Attempt at a Solution
[tex]\dfrac{dV}{dt}=100/12^3-4.8A_0 \sqrt{y}[/tex]
Also, [itex]V(y)=\int_0^y 4 du[/itex] so
[tex]\dfrac{dV}{dy}=4[/tex]
By the chain rule, [itex]\dfrac{dV}{dt}=\dfrac{dV}{dy} \dfrac{dy}{dt}[/itex] so
[tex]100/12^3-4.8A_0 \sqrt{y} = 4 \dfrac{dy}{dt}[/tex]
The problem right before this was the same except water was not being added at all, and that was an easily solvable differential equation. I am stuck on this, and when I got the answer from WolframAlpha it did not look encouraging that I was on the right track.
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