Drug Dosage Modeling

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April 30th 2011, 01:14 PM
Relmiw

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?
April 30th 2011, 01:47 PM
pickslides

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?
April 30th 2011, 02:00 PM
Relmiw

Quote:

Originally Posted by pickslides

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.
April 30th 2011, 02:13 PM
pickslides

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 $\displaystyle e^{K}=C_0$

$\displaystyle C= C_0 e^{-kt}$
May 1st 2011, 05:12 PM
Relmiw

Hey, thank you. I bet that is what they did.
August 2nd 2012, 10:40 AM
davidsmith

Re: Drug Dosage Modeling

this should need time to figure it out...

www.drugaddiction-recovery.com