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Limit question

  1. Oct 22, 2006 #1
    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. jcsd
  3. Oct 22, 2006 #2

    arildno

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    Well, what is the rate of change of the amount once the amount has reached a constant level?
     
  4. Oct 22, 2006 #3
    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
     
  5. Oct 22, 2006 #4

    arildno

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    Well, so what is then the limiting concentration you were asked about?
     
  6. Oct 22, 2006 #5

    Office_Shredder

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    laminar, the rate of change is zero as t goes to infinity (this should be intuitive).
     
  7. Oct 22, 2006 #6
    But then another part of the question asks the time it takes to reach half of the limiting concentration... I'm lost.
     
  8. Oct 22, 2006 #7

    arildno

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    You DO know that the solution of the diff.eq f'=af for some constant f is an expontential function, right?
     
  9. Oct 22, 2006 #8
    Yes, I know that I would find the time by x=x0e^kt.
     
  10. Oct 22, 2006 #9

    arildno

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

    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:
    [tex]y(t)=-by(t)\to{y}(t)=Ae^{-bt}\to{x}(t)=\frac{a}{b}+Ae^{-bt}[/tex]
    where A is a constant to be determined by the initial condition:
    [tex]x_{0}=x(0)=\frac{a}{b}+A\to{A}=x_{0}-\frac{a}{b}[/tex]

    Therefore, your solution is:
    [tex]x(t)=\frac{a}{b}+(x_{0}-\frac{a}{b})e^{-bt}[/tex]

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

    We therefore have to solve the following equation for T in terms of the other parameters:
    [tex]\frac{x_{0}}{2}=\frac{a}{b}+(x_{0}-\frac{a}{b})e^{-bT}[/tex]
    Can you do that?
     
  11. Oct 22, 2006 #10

    arildno

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    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
  12. Oct 22, 2006 #11
    Ah, much thanks. Forgive me, I'm an arts student taking math courses.
     
  13. Oct 22, 2006 #12

    arildno

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    Arts or no arts, have you found your answer now?
     
  14. Oct 22, 2006 #13
    Yeah I get how to find t, but I still don't get how to find the limiting concentration.
     
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