let [tex]A[/tex] be the amount of a drug present [tex]t[/tex] units of time after its administration. this amount could be in terms of drug concentration.(adsbygoogle = window.adsbygoogle || []).push({});

the basic assumption i made was that the rate of change of [tex]A[/tex] is the difference between the rate of absorption and the rate of elimination where the rate of absorption is constant until the initial amount administered is gone and the rate of elimination is proportional to the amount present:

[tex]\frac{dA}{dt}=k_{1}u\left( k_{2}-t\right) -k_{3}A\left( t\right) [/tex].

here, [tex]u[/tex] is the unit step function defined so that [tex]u\left( t\right) =1[/tex] if [tex]t\geq 0[/tex] and is 0 otherwise. [tex]k_{1}[/tex] is related to the rate of absorption and [tex]k_{3}[/tex] is related to the rate of elimination. the absorption aspect

remains activated until [tex]t=k_{2}[/tex] which corresponds to when the intitial amount administered runs out.

the solution to this with [tex]A\left( 0\right) =0[/tex] is [tex]A\left( t\right) =\frac{e^{-k_{3}t}k_{1}}{k_{3}}\left( -1+e^{k_{3}t}+\left( e^{k_{2}k_{3}}-e^{k_{3}t}\right) u\left( t-k_{2}\right) \right) [/tex].

it became clear that [tex]A[/tex] must achieve a maximum value when [tex]t=k_{2};[/tex] here, [tex]A^{\prime }[/tex] is undefined whereas otherwise it is nonzero. let [tex]t_{peak}=k_{2}[/tex]. [tex]\ A\left( t_{peak}\right) =\frac{k_{1}}{k_{3}}\left(

1-e^{-t_{peak}k_{3}}\right) [/tex]; this is the peak amount present. let [tex]A_{peak}=A\left( t_{peak}\right) [/tex].

if we let [tex]t_{half}[/tex] be the element in the set [tex]\left\{ t>t_{peak}:A\left( t\right) =\frac{1}{2}A\left( t_{peak}\right) \right\} [/tex] and solve, we obtain that [tex]t_{half}=\frac{\log 2+t_{half}k_{3}}{k_{3}}[/tex], where the log is the natural log.

now that we know [tex]t_{peak}[/tex], [tex]A_{peak}[/tex], and [tex]t_{half}[/tex] in terms of [tex]k_{1}[/tex], [tex]k_{2}[/tex], and [tex]k_{3}[/tex], we can solve for [tex]k_{1}[/tex], [tex]k_{2}[/tex], and [tex]k_{3}[/tex] to get

[tex]k_{1}=\frac{A_{peak}\log 2}{\left( 2^{\frac{t_{peak}}{t_{peak}-t_{half}}}-1\right) \left( t_{peak}-t_{half}\right) }[/tex]

[tex]k_{2}=t_{peak}[/tex]

[tex]k_{3}=\frac{\log 2}{t_{half}-t_{peak}}[/tex].

putting these together, we get [tex]A\left( t\right) =\frac{A_{peak}\left( 1-2^{\frac{t}{t_{peak}-t_{half}}}+\left( 2^{\frac{t-t_{peak}}{t_{peak}-t_{half}}}-1\right) u\left( t-t_{peak}\right) \right) }{1-2^{\frac{t_{peak}}{t_{peak}-t_{half}}}}[/tex]. the graph of this looks like something of the form [tex]ate^{bt}[/tex] though that formula doesn't allow one to control the time to reach peak absorption, the peak amount, and the half-life which makes sense because it only has two parameters while the defining differential equation for the other formula has three parameters.

one can now construct a sum where the terms added are of the form [tex]A\left(t-kt_{int}\right) u\left( t-kt_{int}\right) [/tex] where [tex]k[/tex] are natural numbers and [tex]t_{int}[/tex] is the fixed time interval in between administrations of the drug though one could, of couse, add things of the form [tex]A\left(t-t_{k}\right) u\left( t-t_{k}\right) [/tex] where [tex]\left[ t_{k},t_{k+1}\right] [/tex] are not intervals of constant length. if the intervals are fixed, one can study the long term behavior of the drug.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Drug absorption/elimination formulas

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**