Expert: David Montiel - 8/17/2009

Question
QUESTION: An underground storage tank is being filled with liquid as shown in the diagram.Initially the tank is empty.At time t hours after filling begins,the volume of liquid is v m^3 & the depth of liquid is h m.It is given that v=4/3(h^3).The liquid is poured inat a rate of 20m^3 per hour,but owing to leakage,liquid is lost at a rate proportional to h^2.When h=1,dh/dt=4.95.i)show that hsatisfies the differential equation dh/dt=5/h^2-1/(20). ii)verify that (20h^2)/(100-h^2)= -20 + (2000)/((10-h)(10+h)). iii) Hence solve the differential equation in part (i), obtaining an expression for t in terms of h. How to do the iii) part? I know how to do parts i) & ii). Thanks a lot

ANSWER: Before I answer I have to tell you that my policy is not to give full answers to problems, or even part of them. Put I'll be happy to give you some hints to help you get to the solution.

The first thing you should notice is that we can use separation of variables because the right-hand side of the equation in part (i) doesn't even depend on "t", so all you need to do is write it out as an easier expression to integrate.

The first step is to carry out the fraction sum on this side. We have:

dh/dt = 5/h^2 - 1/(20) = (100-h^2)/(20h^2)  

Notice that we get the inverse of the left hand side of part (ii). So, if we take the inverse of the whole differential equation, we get

dt/dh = (20h^2)/(100-h^2)

which, from part (ii), we can write as

dt/dh = -20 + 2000/((10-h)(10+h))

The next step is to write the second part in terms of sums of fractions (Remember integration by partial fractions?).

You should get

dt/dh = -20 + 100/(10+h) + 100/(10-h)   (you can easily verify this is true)

This equation is now very easy to integrate. You should of course get "t" on the left side and a function of
"h" on the right side. Remember that t=0 when h=0 (the tank is empty in the beginning).

I hope this helps, let me know if there is anything that's not clear.

David





---------- FOLLOW-UP ----------

QUESTION: I can't get the correct terms or solve the differential equation. Please help me. I only got the expression which is t= -20h+In((10-h)/(10+h)). Thanks

Answer
Did you get that expression from this equation?

dt/dh = -20 + 100/(10+h) + 100/(10-h)

If you did that then you have already solved the differential equation. The expression

t= -20h + In((10-h)/(10+h)

IS THE ANSWER to part iii). You don't have to do anything else. Except I think you have the numerator and the denominator in the logarithm reversed, and also a factor of "100" missing.

I got:

t= -20h + 100*In((10+h)/(10-h))

...so check your algebra.

Cheers,
David