Determine drug concentration over time, given its halflife and dosage

Question

I want to calculate which of two doses is going to have the most active ingredient over the total time of an experiment.

So as an example let's say I have a drug which has a halflife of 5 hours, and I am considering two doses, a single 10mg every 48 hours, or two 5mg doses taken 24 hours apart. Which one of these is going to have the most active ingredient taking into account the half life/decay of the drug over the 48 hour timeframe? I guess this would be the average concentration over 48 hours.

If someone give me a step by step example of the math I need that would be great!

Answer 1

There are different solutions to different problems.

You can be willing to know which solution will provide you with the maximum average drug in the system during 48 hours (problem 1).

Or you can be willing to know which solution will provide with the highest drug quantity remaining in the system when the 48 hours are expired (problem 2).

In any case you need to know the amount of drug vs time.

It is of the form $$D(t)=D_0 \times e^{-\frac tT}$$ with $T$ a factor such that $e^{-\frac 5T}=\dfrac 12$ (the amount of drug is divided by 2 after 5 hours).

Then you solve (problem 2) very easily, since you have to compare both $D(48)$ (just pay attention to the fact that in one case you add 5mg only after 24 hours, so you may have to add two terms)

The (problem 1) requires to compute the average drug quantity in the system during the period. This requires to do integral calculus.

Let's call $M$ the average drug quantity in the system over the period $\tau$

$$M=\dfrac{1}{\tau} \int_{0}^{\tau}D(t)dt$$

Of course you have two different conditions in one case, so in that case you need to do the math for the first 24 hours with the initial conditions, and the last 24 hours with the new conditions (no such things when you put from the beginning the 10mg of drug).

Can you take it from here?

Answer 2

The figur represents the quantity $N$ of drug in the two cases.

In blue the first case, that is represented by the equation $$ N=10 \times 2^{-t/5} $$ for $0\le t \le 48$ the red represent the secon case in wich the second summininstration adds to the residual of the first, so that for the first $24 h$ we have: $$ N=5\times 2^{-t/5} $$ for $0\le t \le 24$ and than: $$ N=5\times \left (2^{-(t)/5}+2^{-(t-24)/5} \right) $$ for $24<t\le 48$.

Answer 3

I'm going to model this after radioactive decay, which follows the differential equation that the activity is proportional to the number of particles $N(t)$, so $$\frac{dN}{dt}=-kN$$ which has the solution (try to solve it with separation of variables yourself!) $$N(t)=N_0e^{-kt}$$ where $N_0$ is the number of particles present at $t=0$ and $k$ is a constant that can be determined from the halflife $$T_{1/2}=\frac{\ln(2)}{k}$$ (Btw, let me know if you want me to show any of these results.)

So now to your concrete question: I'm going to assume that we can ignore any dose-response relationship here, as you haven't specified any, which makes the task a lot easier. To find the total activity of the drug we simply need to integrate over time:

For the single dosage of 10 mg, you simply calculate $$(1):\int_0^{T}N_0e^{-kt}dt=\frac{N_0}{k} \left(1-e^{-kT} \right)$$ where $N_0$ is the 10 mg dose and $k$ can be found using the knowledge of the halflife (as shown above) and where $T$ is the 48 hours. In the case of the two doses, you simply halve the dose $N_0$ and integrate the first dose from $0$ to $T$ and add that the integration over the second dose from $T/2$ to $T$:

$$(2): \int_0^{T}\frac{N_0}{2}e^{-kt}dt+\int_{\frac{T}{2}}^T\frac{N_0}{2}e^{-kt}dt=\frac{N_0}{2k} \left[ 1-2e^{-kT} + e^{-kT/2} \right]$$

Whichever og $(1)$ or $(2)$ is larger is the one you should choose (I can't make that calculation here, since I don't know the halflife). Feel free to ask any additional questions if it is still not completely clear.

Answer 4

$10$ mg every other day or $5$ mg daily are equivalent on average. Both $5$ mg/day.

(There is a neglectible difference if you stop averaging after two days, as in the case of split doses, the second dose decays less time.)