cp ~bfarr/Drug_start.mws ~You can copy the worksheet now, but you should read through the lab before you load it into Maple. Once you have read through the exercises, start up Maple, load the worksheet

where is the decay constant, and is a property of the particular drug being used. It is usually obtained experimentally. The worksheet

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 initial 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.

during the treatment.

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

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

to get

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 as

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
is

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

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

and

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 6.2 mg/ml. Four hours later, the concentration had
dropped to 3.4 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?
- Consider drug A again. Suppose that the time between doses is
reduced to two hours. Can you find a dose such that the
concentration stays below and above for at least 36
hours? If you have trouble, go to the next exercise.
- Consider drug A again with . First, compute the minimum
initial dose that will keep the concentration above the
minimum effective level for for the first two hours. Then, using
this value for , compute the maximum concentration
. Does this explain why you had trouble in the
previous exercise?

2003-01-21