# Drug Dosage Modeling

• Apr 30th 2011, 02:14 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?
• Apr 30th 2011, 02:47 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?
• Apr 30th 2011, 03:00 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.
• Apr 30th 2011, 03:13 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, 06:12 PM
Relmiw
Hey, thank you. I bet that is what they did.
• Aug 2nd 2012, 11:40 AM
davidsmith
Re: Drug Dosage Modeling
this should need time to figure it out...