Applied Calculus

• We have following variables::

D = dose of drug

V = volume of distributed drug in body

C = Concentration of the drug at time \t"

F = Fraction of dose whitch has been absorbed (Bioavailibility)

A = absorption rate constant

E = Elimination rate constant

t = time

1. Absorption Part :

• This depends on the amount of the drug given, the fraction that has been absorbed and the absorption rate constant. It decreases as time goes on.
• The expression for absorption is given by :

2. Elimination Part :

The elimination is affected by the elimination constant, the distributed volume in the body and the concentration left of the drugs.

The expression of this phenomenon is :

Now, one need to subtract the elimination part from the absorption part. For this we can write following diff erential equation :

• Here I give an example of this process:

EXAMPLE:

• In above equation, we can put following values:

V=15, A=0.5, F=2, D=800, E=0.4 (Here, C is variable)

• By solving this differential equation,it gives the concentration of drug at time "t".
• This value is as below:

• Here is graph of these phenomenons.

• We can see that the portion where the concentration increases( upto approx. t = 2) and then level decreases to almost zero at t =24.
• So,it's all about how we can use calculus in our life.