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Math Help - Drug Dosage Modeling

  1. #1
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    Drug Dosage Modeling

    Hopefully this is the right forum for this. I'm going over a Math Modeling book which is dealing with drug dosage modeling and i'm having trouble figuring out how they reached the following conclusion.

    If we assume that C(0) is the concentration at t = 0, then we have the model:
    dC / dt = -k * C
    implies
    C(t) = C(0) * e^(-k * t)

    I've taken Calc. 3, so I feel like I should have some idea of what's going on here, but I can't remember. My idea:

    dC / dt = -k * C
    dC = -k * C * dt
    integral(1)dC = integral(-k * C)dt
    c = -k * c * t + constant
    1 = -k * t + constant
    -k * t = constant.
    Which, is a far cry from what the book has come up with. Could someone give me an idea of what's going on here?
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  2. #2
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    Try this

    \displaystyle \frac{dC}{dt}= -kC

    \displaystyle \frac{dC}{C}= -k~dt

    \displaystyle \int \frac{dC}{C}=\int -k~dt

    Can you finish him from here?
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  3. #3
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    Quote Originally Posted by pickslides View Post
    Try this

    \displaystyle \frac{dC}{dt}= -kC

    \displaystyle \frac{dC}{C}= -k~dt

    \displaystyle \int \frac{dC}{C}=\int -k~dt

    Can you finish him from here?
    ln(C) = -k t
    e^ln(C) = e^(-k t)
    C = e^(-k t)

    That makes sense to me now, thanks. I'm still not sure where the C(t) and C(0) came from though. I will have to investigate this some more.
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  4. #4
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    Don't forget the constant of integration, lets call it K in this case

    \displaystyle \ln C= -kt+K

    \displaystyle C= e^{-kt+K}

    \displaystyle C= e^{-kt}\times e^{K}

    Making \displaystyle e^{K}=C_0

    \displaystyle C= C_0 e^{-kt}
    Last edited by pickslides; April 30th 2011 at 07:06 PM.
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  5. #5
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    Hey, thank you. I bet that is what they did.
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  6. #6
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    Re: Drug Dosage Modeling

    this should need time to figure it out...

    www.drugaddiction-recovery.com
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