Calculate change in height of fluid in a cylinder based on flow out

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

A cylinder with radius five has water in it. The water flows out of the cylinder with a rate of 5π units cubed per minute. At what rate does the height of the fluid in the cylinder change?

2. Relevant equations

volume of a cylinder = $πr^{2}h$

3. The attempt at a solution

I know the height of the cylinder is constant. I think using $\int f(t) \mathrm{d} t = \frac{5}{2}πt^{2} + C$ would help me, where $f(t) = 5πt$, or the amount of water released in t minutes.

Thanks for any help.
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 So you already have that the volume of the water is $$V = \pi r^2h$$ And you're given that the rate of loss of water is $\displaystyle 5 units^3min^{-1}$ so you have $$\frac{dV}{dt} = 5$$ Now what is $\displaystyle \frac{dV}{dt}$ ?
 So that tells me that the change in volume with respect to time is 5π, no? For example, after t minutes, $5πt$ water would have been released.

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"I know the height of the cylinder is constant." But the height of the column of water is not. In fact, it is dh/dt that you are asked to find. The radius is constant. Yes, $V=\pi r^2h$. So what is dV/dt as a function of dh/dt?
 Forgive me if I'm completely wrong, been studying for finals all day and night and it's late here. From what I've calculated, $\frac{dV}{dt} = 25π\frac{dh}{dt}$. Thanks for the help.