Mathematical modelling

bubokribuck · 2011-11-19 17:32:33

I've been given this question to solve:

x(t) is something dissolved in water of w liters. The water with the substance is drained off at rate r.

It asks me to construct a formulation of a continuous time model (which will be a differential equation) and to come up with a general solution to this continuous time model, which I don't even know where to start. Any suggestions please?

Last edited by bubokribuck (2011-11-20 08:19:07)

Bob · 2011-11-19 22:12:55

hi bubokribuck

Just checking this over. I'll edit if I find an error. 10.12 GMT. SEEMS OK. 12.35 GMT

water in at r

substance + water out at r.

=> total volume in tank stays constant at w

assume mixture is 'well stirred'

concentration of substance at time t = x(t)/w

=> substance leaves the tank at r.x(t)/w

so dx/dt = -rx/w

use separation of variables and the initial condition to complete the problem.

Hope that helps.

Bob

Last edited by Bob (2011-11-20 00:34:51)

bubokribuck · 2011-11-20 00:51:37

Thanks, that helps a lot

So dx/dt is the time for the substance to be drained off completely, right?

Bob · 2011-11-20 01:37:01

hi bubokribuck

Not quite. x(t) is the amount of substance in the water. That 't' means it changes over time.

dx/dt is the rate of change of 'x'.

When you solve the differential equation you will get a negative exponential function. That means the amount of 'x' in the water will go down quickly to start with, but then at an ever slowing rate. Because it is an exponential it will never actually reach zero.

That makes sense, because as the concentration gets lower, less and less substance is drained off in each second. It will, according to the model, never be drained off completely. Of course, it's just a model. It doesn't take account of individual molecules. Or the difficulty of measuring how much is left when it gets vey low.

If you would like, post back your working and I'll check it.

Bob

bubokribuck · 2011-11-20 01:55:32

Thank you very much for your help Bob!

But I'm still quite confused. If x(t) is the amount of the substance, then what x and t each stands for? And if the substance will never be drained off completely, what does t=0 mean then? (I need to include x[sub]0[/sub] somewhere in the formula but don't know where to put it).

bubokribuck · 2011-11-20 02:06:07

Ah, I think I'm getting it now. So here's what I thought:

x(t) is the amount of substance at time t. When t=0, x(t) is the initial amount of the substance. When t=1, it is as the substance has been drained off for 1 minute at rate r. (suppose t is in minutes)

Is the above statement correct?

bubokribuck · 2011-11-20 02:30:41

so dx/dt = -rx/w
use separation of variables and the initial condition to complete the problem.

If I'm not wrong, dx/dt = -rx/w can be written as x'(t) = -rx/w, right? If this is true, then what x(t) equals to? Can I write something like x(t) = ??? to start with?

Bob · 2011-11-20 05:04:04

hi bubokribuck

I've been making a wildlife pond so I've missed all your posts until now.

But I'm still quite confused. If x(t) is the amount of the substance, then what x and t each stands for? And if the substance will never be drained off completely, what does t=0 mean then? (I need to include x0 somewhere in the formula but don't know where to put it).

x is the amount of the substance at any time. It will vary as time passes.

t measures the time since t = o, the beginning

x(t) is just another way of saying this.

t = 0 is the start. At that time you are told to make

When t=1, it is as the substance has been drained off for 1 minute at rate r. (suppose t is in minutes)

It is the amount that is left in the tank after some has drained off, because x(t) is the amount of substance not how much is drained off.

You are right to have t in minutes because the rates are in litres per minute.

If I'm not wrong, dx/dt = -rx/w can be written as x'(t) = -rx/w, right? If this is true, then what x(t) equals to? Can I write something like x(t) = ??? to start with?

You can write it like that but that's not how this is solved.

A differential equation is simply any equation that has a dy/dx in it or dp/dt or dw/dq etc.

In this case we have ended up with

That's the differential equation bit done. Now, how do we get x(t) in a formula without any differential bits.

Just think about all those algebraic symbols. Some are constants (r and w) so should not bother us much. The question could have had numbers for those; eg w = 100 litres r = 20 L/m.

The tricky bits are the variables, x and t, and the differential which has dx and dt in it.

'Separation of variables' is a method that works for some differential equations. It works here.

We try to get the 'x bits' on one side of the equation and the 't bits' on the other.

That's why it's called 'separation of variables'; we separate the x from the t.

Here's how it is done:

Some people accept that step quite happily; other might complain that dx and dt shouldn't be separated like this. Strictly, they are correct but the next step makes it OK again so I'll put it down quickly before anyone notices!

Remember the constants? Let's take them outside the integrals.

Now, I'm hoping you can do those integrations; so I'll stop there for you to have a try. Post an answer and then I'll tell you what to do next.

Bob

Last edited by Bob (2011-11-20 05:08:49)

bubokribuck · 2011-11-20 06:47:28

Thank you very much Bob! At first I didn't remember what "Separation of variables" was, but your working out has reminded me. If I haven't done it incorrectly, the final equation should be something like

, for some constant C.

I hope this is right.

Last edited by bubokribuck (2011-11-20 06:50:54)

Bob · 2011-11-20 07:32:19

hi

Yes that's it exactly. If you replace -C by ln(A) for some different constant A, then you can do this:

Why? Because we can raise e to the power of each side :

And the e^ and ln are inverse functions so they 'cancel out' leaving

This neat form for the function has the properties we want. It starts at t = 0 , x = A and gets smaller as t gets bigger.

so

and you have the function that completes the problem.

Below is a graph showing how x changes over time. I've used A = 100, w = 500 and r = 20 in order to 'fix' the constants.

Last edited by Bob (2011-11-20 07:39:14)

bubokribuck · 2011-11-20 08:15:27

Thank you so much Bob! You've saved my life! I've managed to solve the problem eventually. Thank you again! (just can't thank you enough )

Bob · 2011-11-20 08:50:18

You are welcome!

Bob

bubokribuck · 2011-11-21 04:51:13

Hi Bob, sorry that I still have one small problem.

Can you show me how you got dx/dt=-rx/w from r.x(t)/w please.

Thank you!

Bob · 2011-11-21 06:51:11

hi bubokribuck

x and x(t) are the same thing.

x(t) just means the variable x is a function of t.

writing it as x (without the t) is just to make it easier to write.

It's a way of showing that a function is equal to an expression in terms of another variable.

I'll make up a different example to show what I mean.

new example:

another example:

x(t) is the distance travelled by an object.

The velocity is given by

so

Hope that helps.

Bob

bubokribuck · 2011-11-21 09:21:02

I've always thought that f(x) = y = 3x-5
so x(t) needs to be represented by another letter, but looks like it's not necessary.

Thank you very much, now I get it!

Math Is Fun Forum

#1 2011-11-19 17:32:33

Mathematical modelling

#2 2011-11-19 22:12:55

Re: Mathematical modelling

#3 2011-11-20 00:51:37

Re: Mathematical modelling

#4 2011-11-20 01:37:01

Re: Mathematical modelling

#5 2011-11-20 01:55:32

Re: Mathematical modelling

#6 2011-11-20 02:06:07

Re: Mathematical modelling

#7 2011-11-20 02:30:41

Re: Mathematical modelling

#8 2011-11-20 05:04:04

Re: Mathematical modelling

#9 2011-11-20 06:47:28

Re: Mathematical modelling

#10 2011-11-20 07:32:19

Re: Mathematical modelling

#11 2011-11-20 08:15:27

Re: Mathematical modelling

#12 2011-11-20 08:50:18

Re: Mathematical modelling

#13 2011-11-21 04:51:13

Re: Mathematical modelling

#14 2011-11-21 06:51:11

Re: Mathematical modelling

#15 2011-11-21 09:21:02

Re: Mathematical modelling

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