# Differentiation rate of flow from a cylinder

1. Oct 8, 2012

### azzarule

1.
Water flows out of a cylindrical tank under gravity via a tap, the height h(t) of the water column above the tap satisfies the differential equation in the form
dh/dt = -2k√h
where k is some positive constant.
The water column has a height initially of 25m. The tap is turned on and after 50 minutes its height is 24 metres.
Want to compute the height of the column after t minutes.

Then water collected during the first 12 minutes was measured to have volume of 303.9 m3. Calculate how much water in volume was originally in the tank.

2. Relevant equations are above

3. Attempt:
dh/dt = -2k√h
dh/2√h = -kdt
1/2*dh/√h = -kdt
integrate both sides
1/2*2/√h = -kt + C
2/2√h = -kt+c
when h=25 t = 0
2/2√25 = -k0+c
c=0.2
so
2/2√h = -kt+0.2
now I want in the form h=......
I have tried this a few different ways the k constant is confusing me when there is also constant c
although this is worded like a physics problem it is for a mathematical modelling undergraduate subject.
Another solution that I got made sense but could not be solved for the initial condition t=0 it was undefined

2. Oct 8, 2012

### slider142

Your integration of $\frac{\,dh}{2\sqrt{h}}$ is not quite correct. Remember you can write this as $\frac{1}{2}h^{-\frac{1}{2}}\,dh$ and use the power rule. Once you have it integrated, you can correctly use the point h(0) = 25 to solve for C. Use the second point h(50) = 24 to solve for k.

Last edited: Oct 8, 2012
3. Oct 8, 2012

### azzarule

would the correct integration of dh/2√h just be √h then?

4. Oct 10, 2012

Yes. :)