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MATH 220 Biocalculus
Project 1: Drug Concentration: Modeling with Functions
Goals
• To explore functions that describe mathematical models
• To explore models involving exponential functions
• To provide an introduction to geometric series
PROBLEM 1: Drug Dosage
We will now look at a simple model for drug administration. We will consider a drug that is administered
intravenously. In this situation, we are able to assume that the drug reaches its maximum concentration
in the bloodstream as soon as it is injected. We will also assume that the drug a metabolized at a rate
proportional to the concentration of the drug in the bloodstream. This assumption is equivalent to stating
the concentration of a single dose of the drug decays exponentially in the bloodstream. In this exercise, we
will investigate the a fictitious drug that behaves in the manner described above called Calctor.
1. Let C(t) be the concentration of Calctor in the bloodstream measured in µg/ml t hours after the
drug is administered. If concentration of the dosage is C0 = C(0) = 200 µg/ml, and 80% of the
drug is metabolized in six hours, determine the value of the elimination constant r in the formula
for C(t) = C0 e−rt and print out a plot of the graph of C(t) using Derive for the twelve-hour period
beginning the instant the drug is administered. (Make sure your graph is labeled, and the axes of your
plot are labeled with appropriate units. Make sure that you choose an appropriate window for your
graph.) In Derive, set the constant r equal to the quantity determined above.
2. If C0 = 200, when does the concentration reach a level of 50 µg/ml? What if C0 = 250?
3. There are two very important concentration levels of drug that must be considered to determine an
effective dosing schedule for the drug. These are the minimum effective concentration (MEC) and the
minimum toxic concentration (MTC). The MEC is the minimum concentration level of the drug that
will have therapeutic benefits for the patient. The MTC is the concentration level above which the
drug becomes toxic to the patient. (Even though, the MEC and MTC can vary from patient to patient,
reasonable levels can be estimated for a wider population based on factors including weight and age.)
For a drug to be effective and safe, it is important to keep the concentration of the drug between the
MEC and MTC. We will assume that it has been determined that the MEC for Calctor is 50 µg/ml
and that the MTC is 300 µg/ml.
Our goal is will be to determine an effective dosing schedule assuming that the patient will remain
on Calctor indefinitely. Two quantities will have to be determined in the following steps: a dosage
amount C0 (which we will assume to be constant) and the constant time period between each dose, T
hours. That is, the dosage concentration C0 = C(0) and the frequency T now will now be treated as
parameters. THERE ARE NO CALCULATIONS STEPS IN THIS ITEM.
4. The Model. Assume that a dosage with a concentration of C0 is administered every T hours. What
is the concentration C(T ) after T hours? (Write this expression down in your report.)
Notice that because C(t) is an exponentially decreasing function, C(t) always will be greater than zero.
That is, there will always be a residual amount of the drug in the bloodstream. (Of course, after a long
enough period of time, the drug will no longer be detectable.) If T hours is the period of time between
doses, the residual concentration of this first dose is C(T ) µg/ml when t = T hours. Similarly, after
kT hours since the initial dose, the residual concentration of the initial dose is C(kT ) µg/ml. At the
instant the kth dose is administered ((k − 1)T hours after the initial dose, there is
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C((k − 1)T ) = C0 e−rT (k−1)) µg/ml remaining from the initial dose,
C((k − 2)T ) = C0 e−rT (k−2) µg/ml remaining from the second dose,
..
.
C(2T ) = C0 e−2rT µg/ml remaining from the (k − 2)nd dose,
C(T ) = C0 e−rT µg/ml remaining from the (k − 1)st dose,
in addition to the new dose of C0 .
Hence the residual concentration of the drug in the bloodstream just before the the kth dose is administered is
rk =
k−1
C0 e−rT i = C0 e−rT + · · · + C0 e−rT (k−2) + C0 e−rT (k−1) =
i=1
C0 e−rT (1 − e−rT (k−1) )
µg/ml.
1 − e−rT
Hence the total concentration of the drug in the bloodstream at the instant the kth dose is administered
is
Ck =
k
C0 e−rT (i−1) = C0 + C0 e−rT + · · · + C0 e−rT (k−2) + C0 e−rT (k−1) µg/ml.
i=1
The expression above is the kth partial sum of the geometric series with initial term C0 and common
ratio e−rT . As rT > 0, we see that 0 < e−rT < 1. We now will show that if we were to administer
the drug indefinitely, the total concentration would never exceed a particular maximal level M , which
will be expressed in terms of C0 and T . Once we have this maximum possible concentration, M we
can control both C0 and T so that the drug concentration remains between the MEC and the MTC
throughout the treatment.
We now introduce the mathematics we need to calculate M and develop a convenient formula for a
general model. In general, the geometric series with initial term of a and common ration of R is defined
to be the limit
k
aRi−1 = a + aR + aR2 + · · · .
S = lim
k→∞
i=1
The kth partial sum of this series is
Sk =
k
aRi−1 = a + aR + aR2 + · · · + aRk−2 + aRk−1 .
i=1
When |R| < 1, we will now show that
S = lim Sk = lim
k→∞
k→∞
k
i=1
aRi−1 =
a
.
1−R
If we multiply both sides of the equation
Sk =
k
aRi−1 = a + aR + aR2 + · · · + aRk−2 + aRk−1
i=1
2
by R, we obtain
RSk =
k
aRi = aR + aR2 + aR3 + · · · + aRk−1 + aRk .
i=1
Now, subtract this last equation from the previous one to obtain
(1 − R)Sk
=
k
i=1
=
k
aRi−1 −
k
aRi
i=1
(aRi−1 − aRi )
i=1
= (a − aR) + (aR − aR2 ) + (aR2 − aR3 ) + · · · + (aRk−2 − aRk−1 ) + (aRk−1 − aRk )
= a + (−aR + aR) + (−aR2 + aR2 ) + · · · (−aRk−2 + aRk−2 ) + (−aRk−1 + aRk−1 ) − aRk
= a(1 − Rk ).
Hence
Sk =
a(1 − Rk )
.
1−R
Now since 0 < |R| < 1, we know that limk→∞ Rk = 0, and therefore
a
a(1 − Rk )
=
.
k→∞
1−R
1−R
S = lim Sk = lim
k→∞
We now go back to the case of our drug Calctor. In this case, a = C0 and R = e−rT . Over each dosing
interval, the concentration decreases and then jumps at the instant the next dose is administered.
This implies that the maximum concentration level could only be achieved at the instant a dose is
administered. The discussion above describes how the residual amount of the drug measured at the
instant new doses are administered increases with successive doses. The calculation, however, implies
that this amount never exceeds a total concentration
M=
C0
µg/ml.
1 − e−rT
We therefore must ensure that the frequency of dosing T hours and the dosage concentration C0 µg/ml
are determined so that drug concentration never exceeds the MTC of 300 µg/ml. We can accomplish
this with the following equation:
MT C = M =
C0
µg/ml.
1 − e−rT
Another conclusion we can draw from the facts that the concentration decreases over each dosing interval and the minimum (residual) concentration at the end of the dosing intervals increases with successive
doses is that the minimum concentration occurs at the instant (before) the second dose administered.
Hence the MEC should never drop below this level. We can accomplish this by setting up the equation:
M EC = C(T ) = C0 e−rT µg/ml.
We are now in a position to determine C0 and T but we will look at the graph of C(t) first.
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5. If we administer the drug once every T hours, the concentration function will not be a continuous
function because each time we administer the drug, the concentration level jumps C0 µg/ml. We can
write down this function as a piecewise-defined function as follows:
C(t) =

C0 e−rt




(C0 e−rT + C0 )e−r(t−T )



−rT
−2rT

)
 C0 e (1−e
+ C0 e−r(t−2T )
−rT
0≤t<T
T ≤ t < 2T
·· ·




C0 e−rT (1−e−krT )

+ C0 e−r(t−kT )

−rT

1−e


···
kT ≤ t < (k + 1)T
1−e
2T ≤ t < 3T
We can input this expression into Derive as follows:
f(r, t) := e^(−rt)
W(r, n, k) := SUM(e^(−rnt), t, 1, k)
C(r, n, a, t, i) := IF((n(i − 1) ≤ t) ∧ (t < in), a(1 + W(r, n, i − 1))f(r, t − n(i − 1)))
To plot 40 doses, enter the expression VECTOR([C(r, n, a, t, i)], i, 1, 40)
and then click on = .
Note that we have made some notational changes to make inputting into to Derive easier: n = T ,
a = C0 .
We are quite ready to plot yet. We need to enter slider bars to give values to n, the period of time
between doses in hours, and a, the concentration of the dose in µg/ml. In the 2D Plot Window, insert
a slider bar for n, with a minimum of 1, a maximum of 12, and 22 intervals, and a slider bar for a, with
a minimum of 50, maximum of 250, and 20 intervals. (Be sure the select “update plot while sliding.”)
Now in the Algebra Window enter the values of MEC of MTC, which will be plotted as horizontal
lines. Finally, highlight these two values and the evaluated VECTOR command and plot them in the
2D Plot Window. Play with the slider bars and experiment with different values of a and n.
6. Adjust the sliders to illustrate the following conditions. In each case, print out the plot window showing
the graph of C, y = 300, y = 50 on a grid of dimensions [−1, 81] × [−10, 310].
(a) Dosage concentration: 200µg/ml, frequency: once every 6 hours.
(b) Dosage concentration: 200µg/ml, frequency: once every 5 hours.
(c) Dosage concentration: 150µg/ml, frequency: once every 4 hours.
(d) Dosage concentration: 200µg/ml, frequency: once every 5 hours.
(e) Dosage concentration: 240µg/ml, frequency: once every 6 hours.
(f) Dosage concentration: 60µg/ml, frequency: once every hour.
7. Each case above, either explain why the dosing strategy does not work, or show why the dosing strategy
0
µg/ml and
will always work. In each case, compute the upper bound of the concentration, 1−eC−rT
−rT
the lower bound of the concentration C(T ) = C0 e
µg/ml. In cases that do not work, qualitatively
explain how close the strategy is to being within the desired bounds. That is, comment if a “close”
strategy would be viable.
8. In the previous step, you analyzed some possible dosing strategies. We will now compute the ideal
strategy from solving our equations
MT C =
C0
µg/ml,
1 − e−rT
M EC = C(T ) = C0 e−rT µg/ml.
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Using MTC = 300 and MEC = 50, solve these equations in Derive or solve them by hand. (In Derive,
Click on Solve, System, and proceed from there.) You will obtain two solutions. Which of these
two ideal strategies would you chose. *(Think about the cost of the drug.) Neither of these solutions
provide “nice” numbers. What would you recommend for a reasonable dosing strategy? That is, let’s
assume that the dosage will be in multiples of 5 and the frequency of the doses will be on a whole
number of hours. If there are multiple solutions, which solution would be more effective? Explain your
answer carefully.
9. The dosing method discussed throughout this lab can lead to oscillatory growth, which may not be
desirable. For instance, it may take several doses before the drug remains within the range between
the MET and MTC. We can avoid the oscillatory growth behavior by considering an initial large dose
of the drug followed by a constant dose C0 that administered every T hours. We now investigate what
this initial dose should be.
Recall that the residual concentration of the drug in the bloodstream just before the the kth dose is
administered is
rk =
C0 e−rT (1 − e−rT (k−1) )
µg/ml.
1 − e−rT
Just before a dose, the residuals approach
ρ=
Notice that
C0 e−rT
µg/ml.
1 − e−rT
ρ = M e−rT
and that
ρ + C0 = M.
If give an initial dose of M set
M EC = ρ = M e−rT µg/ml
and
M T C = M = ρ + C0 µg/ml,
we can solve for values of C0 and T so that the drug concentration remains between the MEC and the
MTC. Describe why this is the case. Find these values of C0 and T . Create a plot in such an instance.
How does this result compare to the result using constant dosing for all doses?
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