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 :
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 :
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:
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.