Consider a water tank with vertical sides and with cross-section a square of area 4 feet. Water exits the tank through a hole which as area equal to \frac{5}{3} square inches, and water is added to the tank at a rate of 100 cubic inches per second. Show that the water level approaches the value of feet above the hole no matter what the initial water level was.

**Incomplete.**

*Related*

Can’t say I didn’t try on this one, but it’s got me stumped…

We know from the given information that the rate of change of the volume of the water in the tank can be written as:

Dividing by the cross-sectional area A(y) = 4 ft^2 gives us the change in y over t

Which is a separable equation. Rearranging terms and integrating gives us

And this is where I got stuck. I guess if you send t to +infinity, the only y that would satisfy the equation would be a number that sends 25 – 24 sqrt(y) to 0, which is

Not sure if that holds though…

I see the problem with this conclusion is that as the time goes to infinity, the level grows infinitely, and the integral grows infinitely too. So, in a way infinity = infinity

As far as I see, there is no need to go past dV/dt. The derivative is 0 if h=(25/24)^2. If it is negative, it means that the hight is above the equilibrium, so the volume decreases until it reaches the equilibrium. If it is positive, it means that the height is below the equilibrium, so the volume increases until it reaches the equilibrium.