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Next: Exercises Up: Exponentials - Part 2 Previous: Purpose



Maximum time between doses

In the first part of this lab, we described an exponential decay model for the concentration of a drug in the blood of a patient. Recall that if the first dose of the drug produces a maximum concentration of C0, then the concentration of the drug is given by

C1(t) = C0 e-kt

We also said that a drug would have a minimum effective concentration, m and a maximum safe concentration M. The problem of devising a safe and effective treatment program is to come up with a dose of the drug and an interval between doses, so that the drug concentration is always above the minimum effective level and below the maximum safe level. For simplicity, we will assume that the doses all supply the same amount of the drug and that the time between doses is fixed.

The first problem is to figure out a reasonable time between doses. One way to get a handle on this is to calculate the maximum time between doses. That is, suppose a dose is given such that the concentration immediately after the dose is given is M, the maximum safe dose. If we calculate the time at which the concentration has decayed to m, then this gives the maximum time interval between doses. For example, consider the drug from the exercises in part 1 of this lab. To help us later on, we'll refer to this drug as drug B. The experimental data given said that the initial concentration was 4.8 mg/ml and the concentration 3 hours later was 2.23 mg/ml. From this data we can determine the value of k with the following command.

  > k2 := solve(2.23=4.8*exp(-3*k),k);

{\it k2} := .2555381108\end{displaymath}\end{maplelatex}
Then we can calculate the maximum time between doses with the command shown below to be about 5.8 hours.
  > solve( 8*exp(-k2*t)=1.8,t);

This calculation tells us that we can try to use any time interval less than 5.8 hours between doses. Many factors could be important in determining the value to use, including practical concerns like hospital schedules and shift changes.

Equal, regularly-spaced doses

We next consider what happens if equal doses of the drug are given at regular time intervals. Recall that a drug has a maximum safe concentration, M, and a minimum effective concentration, m. We say that a treatment program of equal, regularly-spaced doses is safe and effective if the concentration C(t) of the drug satisfies

m \leq C(t) \leq M \end{displaymath}

during the treatment.

In the first part of this lab, we presented the expression

C1(t) = C0 e-kt

for the concentration of the drug after the first dose. This expression is valid as long as only a single dose is given. However, suppose that at t=L a second dose is given and that the amount of the drug administered is the same as the first dose. According to our model, the concentration will jump immediately by an amount equal to C0 when the second dose is given. However, when the second dose is given, there is still some of the drug in the bloodstream remaining from the first dose. This means that to compute the concentration just after the second dose, we have to add the value C0 to the concentration remaining from the first dose. During the time between the second and third doses, the concentration decays exponentially from this value. To find the concentration after the third dose, we would have to repeat this process, but now we have contributions from the first and second doses to include.

Below, we describe how one can find the concentration at any later time, assuming that the doses are equal and that the time between doses is fixed, but it turns out to be rather complicated. To help you visualize what is going on, a Maple procedure called drug has been written. If you give this procedure values for C0, the concentration jump caused by each dose, L, the time between doses, and k, the decay rate for the drug, it returns the concentration as a function of t. Before you can use this command, you must first load it with the first command shown below. If you get an error message, get help from your IA. Once you've successfully loaded the drug command, the second command below will bring up the help page. (The drug command was written at WPI, so you'll have to download it from here if you want to use it with your own copy of the Student version of Maple. Do this only if you don't use Maple over the network using an X terminal or PC-Xware.)

  > readlib(drug):
  > ?drug
To see how to use the drug command, consider the drug from the exercises in part one of this lab. Use the values C0 = 7 and L=3. That is, the initial dose produces a concentration of 7 mg/ml and the time between doses is 3 hours. Recall that the maximum safe dose is M=8 and the minimum effective dose is m=1.8 Then the following commands will define the concentration C(t), evaluate it just after the first dose and just before and just after the second dose, and, finally, plot the concentration over a period of 12 hours.
  > C := drug(C0=7,L=3,k=k2);

\lefteqn{{C} := {t} \rightarrow 13.07392996...
 ...{t} + .7666143324\,{\rm floor}(\,1/
  > C(0);

  > C(2.999);

  > C(3);

  > plot(C(t),t=0..12);

Notice that the concentration just after a dose increases with each successive dose. Notice also that the concentration soon exceeds the maximum safe concentration M=8. This means that the dose is too high, and this is not a safe treatment program.

You can go back and plot the concentration for a longer time interval. You should notice that the values of the peak concentrations just after a dose rise rapidly for the first few doses and then appear to level off. It can be shown that this always happens, so the maximum concentration eventually levels off as more doses are given. The limiting value for an infinite number of doses can be calculated explicitly, using information on geometric series, but that is a topic for Calc 3. To help you understand what is involved in the drug program, we describe how to obtain formulas for the concentration after the first few doses below.

We can calculate the concentration just before the second dose is administered by setting t=L in our equation

C1(t) = C0 e-kt

to get

C1(L) = C0 e-kL

Now, when the second dose is administered the concentration jumps by an increment C0 so that the concentration just after the second dose is given is

C0 + C1(L) = C0+ C0 e-kL = C0 (1 + e-kL)

The concentration then decays from this value according to our exponential decay rule, but with a slight twist. The twist is that the ``initial'' concentration is at t=L, instead of the more familiar case of t=0. One way to handle this is to write the exponential term as


so that at t=L, the exponent is . If we do this, then we can write the concentration as a function of time as

C2(t) = C0 (1 + e-kL) e-k(t-L)

This function is only valid after the second dose is administered and before the third dose is given.

Now, suppose that a third dose of the drug is given at t=2L. The concentration just before the third dose is given is C2(2L), which is

C2(2L) = C0 (1 + e-kL) e-kL

which we can also write as

C2(2L) = C0 (e-kL+ e-2kL)

When the third dose is given, the concentration would jump again by C0 and the concentration just after the third dose would be

C0 (1 +e-kL+ e-2kL)

Finally, an expression for the concentration that would be valid after the third dose is administered would be

C3(t) = C0 (1 +e-kL+ e-2kL) e-k(t-2L)

This process can be continued to get expressions for the concentration for any number of equal, regularly-spaced doses.

next up previous
Next: Exercises Up: Exponentials - Part 2 Previous: Purpose

William W. Farr