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 #Q# milligrams (per pill) and that the patient is supposed to take the drug every #h# 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 #T# hours (for simplicity, assume the drug enters the person's bloodstream instantaneously).

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

Let's seek a pattern: #P_{1}=0#, #Q_{1}=0+Q=Q#, #P_{2}=Q\cdot 2^{-h/T}#, #Q_{2}=Q\cdot 2^{-h/T}+Q#, #P_{3}=(Q\cdot 2^{-h/T}+Q)\cdot 2^{-h/T}=Q(2^{-2h/T}+2^{-h/T})#, #Q_{3}=Q(2^{-2h/T}+2^{-h/T})+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})#, #Q_{4}=Q(2^{-3h/T}+2^{-2h/T}+2^{-h/T})+Q#, etc...

The patterns indicate that #P_{n}=Q\sum_{k=1}^{n-1}2^(-\frac{kh}{T})# and #Q_{n}=Q\sum_{k=0}^{n-1}2^(-\frac{kh}{T})#.

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

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