# 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