Limit question

1. Oct 22, 2006

laminar

This is biological application question, so the limit can't be negative infinity, which is what I would use as my answer for a regular question.

The rate at which substance is eliminated from the body is proportional to the amount of substance present, and satisfies the differentail equation dx/dt=a-bx, where a and b are positive constants. What is the limiting concentration of substance(limit as t goes to infinity of x(t))?

Can't integrate for this because it's a course on differential calculus only. Any pointers?

2. Oct 22, 2006

arildno

Well, what is the rate of change of the amount once the amount has reached a constant level?

3. Oct 22, 2006

laminar

Rate of change is zero... So, x=a/b when the limiting concentration is reached. What do I do with this?

Thank you for replying so fast

4. Oct 22, 2006

arildno

Well, so what is then the limiting concentration you were asked about?

5. Oct 22, 2006

Office_Shredder

Staff Emeritus
laminar, the rate of change is zero as t goes to infinity (this should be intuitive).

6. Oct 22, 2006

laminar

But then another part of the question asks the time it takes to reach half of the limiting concentration... I'm lost.

7. Oct 22, 2006

arildno

You DO know that the solution of the diff.eq f'=af for some constant f is an expontential function, right?

8. Oct 22, 2006

laminar

Yes, I know that I would find the time by x=x0e^kt.

9. Oct 22, 2006

arildno

No, you don't, because your diff.eq is slightly different!
First, do the following rewriting of your diff.eq:
$$\frac{dx}{dt}=-b(x-\frac{a}{b})$$
with initial condition: $$x(0)=x_{0}$$

Then, introduce the new variable [itex]y(t)=x(t)-\frac{a}{b}\to\frac{dy}{dt}=\frac{dx}{dt}[/tex]

Thus, your diff.eq can be written as:
$$y(t)=-by(t)\to{y}(t)=Ae^{-bt}\to{x}(t)=\frac{a}{b}+Ae^{-bt}$$
where A is a constant to be determined by the initial condition:
$$x_{0}=x(0)=\frac{a}{b}+A\to{A}=x_{0}-\frac{a}{b}$$

$$x(t)=\frac{a}{b}+(x_{0}-\frac{a}{b})e^{-bt}$$

Now, you are to find the time T, so that $$x(T)=\frac{x_{0}}{2}$$

We therefore have to solve the following equation for T in terms of the other parameters:
$$\frac{x_{0}}{2}=\frac{a}{b}+(x_{0}-\frac{a}{b})e^{-bT}$$
Can you do that?

10. Oct 22, 2006

arildno

I'm sorry. I thought the question was to find the time T so that you have half the INITIAL concentration.
In order to find the time t* to half the limiting concentration, adjust the last step accordingly.

Last edited: Oct 22, 2006
11. Oct 22, 2006

laminar

Ah, much thanks. Forgive me, I'm an arts student taking math courses.

12. Oct 22, 2006

arildno

13. Oct 22, 2006

laminar

Yeah I get how to find t, but I still don't get how to find the limiting concentration.