(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A tank with vertical sides has a square cross-section of area 4 ft^{2}. Water is leaving the tank through an orifice of area 5/3 in^{2}. Water also flows into the tank at the rate of 100 in^{3}/s. Show that the water level approaches the value (25/24)^{2}ft above the orifice.

2. Relevant 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

3. 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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# Applied differential equation

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