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

1. Dec 13, 2011

### mrwall-e

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.

2. Dec 14, 2011

### JHamm

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 $\displaystyle \frac{dV}{dt}$ ?

3. Dec 15, 2011

### mrwall-e

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.

4. Dec 15, 2011

### HallsofIvy

Staff Emeritus
"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?

5. Dec 15, 2011

### mrwall-e

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.