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This lab concerns a model for a drug being given to a patient at
regular intervals. As the drug is broken down by the body, its
concentration in the bloodstream decreases. However, it doesn't
disappear completely before the next dose is given. This means that
there is a tendency for the average drug concentration to increase
over time. It turns out that a geometric series is the key to
understanding how high the concentration can go.
To assist you, there is a worksheet associated with this lab that you can copy into your home directory by going to your computer's start menu and choose run. In the run field type:
when you hit enter, you can then choose MA1023 and then choose the worksheet
Remember to immediately save it in your own home directory. Once you've copied and saved the worksheet, read through the background on the internet and the background of the worksheet before starting the exercises.
In this section, we describe an exponential decay model for the
concentration of a drug in a patient's bloodstream. We assume that the drug
is administered intravenously, so that the concentration of the drug
in the bloodstream jumps almost immediately to its highest level. The
concentration of the drug then decays exponentially. If we use
to represent the concentration at time t, and to represent the
concentration just after the first dose is administered then our exponential
decay model would be given by
where is the decay constant, and is a property of the particular
drug being used. It is usually obtained experimentally. The worksheet
Drug_start.mws has examples of how to compute from
Now suppose that an additional dose of the drug is given to the
patient. Since we are assuming that when the drug is
administered it is diffused so rapidly throughout the bloodstream
that, for all
practical purposes, it reaches its highest concentration
instantaneously, we would see a jump in the concentration of the drug
when the new dose is given, as shown in the graph below.
After the additional dose is given, the concentration again decays over time.
A problem facing physicians is the fact that for most drugs, there is
a concentration, , below which the drug is ineffective and a
concentration, , above which the drug is dangerous. Thus the
physician would like the have the concentration satisfy
This requirement helps determine the initial dose of a drug and when
the next dose should be administered. For example, the first dose
should never raise the concentration above . That is, we must have
. To get a handle on the time between doses, we can
calculate the maximum possible time between doses. That is, suppose an
dose is given such that the
concentration immediately after the dose is , the maximum
safe dose. If we calculate the time at which the concentration has
decayed to , then this gives the maximum time interval between
doses. This gives us an upper bound on the time between doses. The
worksheet contains examples of this kind of calculation. Note
that many factors could be important in
determining the time between doses that is actually used, including
practical considerations like
hospital schedules and shift changes.
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, , and a minimum effective concentration, . We say
that a treatment program of equal, regularly-spaced doses is safe and
effective if the concentration of the drug satisfies
during the treatment.
In the first part of this lab, we presented
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 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 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
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.
We can calculate the
concentration just before the second dose is administered by setting
in our equation
where by we mean the
Now, when the second dose is administered the concentration jumps by
an increment so that the concentration just after the second
dose is given is
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 , instead of the more familiar
case of . One way to handle this is to write the exponential term
so that at , the exponent is . If we do this, then we can
write the concentration as a function of time as
This function is only valid after the second dose is administered and
before the third dose is given. That is, for .
Now, suppose that a third dose of the drug is given at . The
concentration just before the third dose is given is , which
which we can also write as
When the third dose is given, the concentration would jump again by
and the concentration just after the third dose would be
This process can be continued and leads to the following two formulas.
The first is the concentration just before the
dose of the drug. This is
The second result we need is the concentration just after the
dose, which is
At this point, you are probably wondering how geometric series fit
into this lab. The answer should be a lot clearer if we define a
Note that , since and are both positive constants.
The properties of the exponential function can be used to show that
where is a non-negative integer.
We can write our two fromulas for the concentration just
before and after the dose in terms of as
where the formula for the partial sum of a geometric series has been
used to obtain the last equality in each of the equations above.
Now, suppose a treatment program is to be continued indefinitely. The
formulas above show that and both increase
with . This means that the minimum concentration is the
concentration just before the second dose or
and that the maximum concentration occurs just after the last
dose. Thus we have that
- Suppose that for a certain drug, which we'll refer to as drug A,
the following results were
obtained. Immediately after the drug was administered, the
concentration was 3.3 mg/ml. Six hours later, the concentration had
dropped to 1.55 mg/ml. Determine the value of for this drug.
- Suppose that for drug A, the maximum
safe level is
and the minimum effective level is
. What is the maximum possible time between doses
for this drug?
- Consider drug A, assuming that doses are
given every six hours, or . Compute the minimum initial dose
that will keep the concentration above the minimum effective
level for the first six hours, i.e before the second dose is given.
- Consider drug A again, with doses to be given
every six hours. Can you find a dose such that the concentration
stays below and above for at least 72 hours?
- Trials of another drug produced the following data. The concentration just after the drug was administered was 6 mg/ml and 5 hours later the concentration was 1.2 mg/ml. If the maximum safe concentration is 10 mg/ml and the minimum effective concentration is 2 mg/ml, find one set of values of and that provide a safe and effective treatment program. (Hint: Copy commands from exercises 1, 3, and 4 and make the appropriate changes. You should also change the variable name for the constant from to .)
- Suppose that new research with drug A shows that the maximum safe level is and the minimum effective dose is Can you find values of and that produce a safe and effective treatment program? Try at least two sets of values of and . If you have trouble, go to the next exercise.
- If you take the ratio of formulas for
, then cancels out and you get the following.
Using a plot, or otherwise, find the smallest value for the expression on the right-hand side of the equation above (i.e ) for . It turns out that if is smaller than this value, there is no way to come up with a safe and effective treatment program.
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Dina J. Solitro-Rassias