How does calculus relate to medicine?

1 Answer
May 20, 2015

One important area of application is to deciding drug dosages.

Suppose the dose of a drug is QQ milligrams (per pill) and that the patient is supposed to take the drug every hh hours. Futhermore, suppose the half-life of the drug (the amount of time for the amount of the drug to decay to 50% of the starting amount) in a person's bloodstream is TT hours (for simplicity, assume the drug enters the person's bloodstream instantaneously).

Let Q_{n}Qn be the amount of the drug in the body right after the n^{\mbox{th}}nth dose and let P_{n}Pn be the amount of the drug in the body right before the n^{th}nth dose so that Q_{n}=P_{n}+QQn=Pn+Q.

Let's seek a pattern: P_{1}=0P1=0, Q_{1}=0+Q=QQ1=0+Q=Q, P_{2}=Q\cdot 2^{-h/T}P2=Q2hT, Q_{2}=Q\cdot 2^{-h/T}+QQ2=Q2hT+Q, P_{3}=(Q\cdot 2^{-h/T}+Q)\cdot 2^{-h/T}=Q(2^{-2h/T}+2^{-h/T})P3=(Q2hT+Q)2hT=Q(22hT+2hT), Q_{3}=Q(2^{-2h/T}+2^{-h/T})+QQ3=Q(22hT+2hT)+Q, P_{4}=(Q(2^{-2h/T}+2^{-h/T})+Q)\cdot 2^{-h/T}=Q(2^{-3h/T}+2^{-2h/T}+2^{-h/T})P4=(Q(22hT+2hT)+Q)2hT=Q(23hT+22hT+2hT), Q_{4}=Q(2^{-3h/T}+2^{-2h/T}+2^{-h/T})+QQ4=Q(23hT+22hT+2hT)+Q, etc...

The patterns indicate that P_{n}=Q\sum_{k=1}^{n-1}2^(-\frac{kh}{T})Pn=Qn1k=12khT and Q_{n}=Q\sum_{k=0}^{n-1}2^(-\frac{kh}{T})Qn=Qn1k=02khT.

Here's the calculus-related part. As n->\inftyn, it can be shown that P_{n}->\frac{Q}{2^{h/T}-1}PnQ2hT1 and Q_{n}->\frac{Q2^{h/T}}{2^{h/T}-1}QnQ2hT2hT1.

What's the application to medicine? You want to choose hh and QQ so that \frac{Q}{2^{h/T}-1}Q2hT1 is large enough to be effective in the patient's body and so that \frac{Q2^{h/T}}{2^{h/T}-1}Q2hT2hT1 is small enough that it is not dangerous to the patient.