# differential modelling

• Oct 11th 2008, 01:19 PM
lakuzy100
differential modelling
Can someone pls help me out with this question.
As water floes out from a straight cylindrical vessel, the rate of change of the height of the free surface level is directly proportional to the square root of the height.It os observed that the height at t=o is h02.25m and at t=90s h=1.75m.When will be height be h=1.25m
• Oct 11th 2008, 04:22 PM
Chris L T521
Quote:

Originally Posted by lakuzy100
Can someone pls help me out with this question.
As water floes out from a straight cylindrical vessel, the rate of change of the height of the free surface level is directly proportional to the square root of the height.It os observed that the height at t=o is h02.25m and at t=90s h=1.75m.When will be height be h=1.25m

The statement

Quote:

the rate of change of the height of the free surface level is directly proportional to the square root of the height
tells us how to set up the differential equation.

It tells us that the rate of change of the height $\displaystyle \left(\frac{\,dh}{\,dt}\right)$ is directly proportional to the square root of the height $\displaystyle \left(\sqrt{h}\right)$

Thus, our DE is $\displaystyle \frac{\,dh}{\,dt}=k\sqrt{h}$, where $\displaystyle k$ is some constant of proportionality.

To solve, apply the separation of variables technique:

$\displaystyle \frac{\,dh}{\,dt}=k\sqrt{h}\implies \frac{\,dh}{\sqrt{h}}=k\,dt$

Integrating both sides, we see that

$\displaystyle 2\sqrt{h}=kt+C\implies h=\left(\tfrac{1}{2}kt+C\right)^2$

Use the first condition to find C.

Then use the second condition to find k.

Finally, find the time it takes to reach a height of 1.25.

Can you take it from here?

--Chris