# Homework Help: Applied differential equation

1. Sep 27, 2011

### process91

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Rate of discharge of volume through the orifice is $4.8A_0 \sqrt{y}$ cubic feet per second, where $A_0$ = size of orifice in square feet

3. The attempt at a solution

$$\dfrac{dV}{dt}=100/12^3-4.8A_0 \sqrt{y}$$

Also, $V(y)=\int_0^y 4 du$ so
$$\dfrac{dV}{dy}=4$$

By the chain rule, $\dfrac{dV}{dt}=\dfrac{dV}{dy} \dfrac{dy}{dt}$ so
$$100/12^3-4.8A_0 \sqrt{y} = 4 \dfrac{dy}{dt}$$

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.

Last edited: Sep 27, 2011
2. Sep 27, 2011

### process91

I think, perhaps, the solution is not to actually solve the differential equation, but rather to do something like this:

Instead of trying to solve my differential equation, I just examine it at y=(25/24)^2, and see that it is zero. Examining the second derivative, we find that it is always negative so this is a maximum of y. For y less than this, the derivative is positive and above this value the derivative is negative. Hence no matter what value of y the problem starts with, as t increases it will approach (25/24)^2.

Does that seem valid?

3. Sep 27, 2011

### LawrenceC

You can demonstrate the discharge rate at y = 25/24 equals the mass influx so the level won't rise anymore.

Also you could write a DE for the rate of volume of water increase in the tank.

Atank * (dV/dt) = 100/12**3 - 4.8 * A0 * (y**2)** .5

Set dV/dt to zero and solve for y. You should get 25/24.