Take Your Medicine 2
By Brink Harrison
Adapted from Section 9.1 in Hughes-Hawlett, Deborah, et.al; Single Variable Calculus;
John Wiley & Sons, Inc.; New York; 2002
Time:
1 or 2 days
Preparation
none
Time:
Materials:
Number of people infected
How Many People Activity Sheet
Abstract
One of the objectives of the lesson is to write the expanded form of a finite geometric series to
represent the amount of the antibiotic ampicillin in the body as the patient continues to take the
prescribed doses. The students will then use the closed form for the sum of a finite geometric
series to calculate how much of a specific antibiotic remains in the body over time.
Teacher Background
The formula for the sum of the first n terms of a finite geometric series is:
Sn 

a 1 rn
1 r

where a is the first term of the series, r is the common ratio between consecutive terms, and n is
the number of terms in the series. But this time each term of the series,
S n  a  ar  ar 2  ...  ar n 2  ar n 1
represents a dose of the drug and the sum of the series represents the drug level in the body in
the long run. In this situation the value of a is the first dose of the drug, n represents the nth dose
of the drug, and r is equal to the percentage of the original given amount of the antibiotic that is
present at the end of a given time period.
**** Optional: You can make the process more complicated if you wish by having the students
calculate the value of r instead of giving it to them. To calculate what percentage of the original
given amount of the antibiotic is present at the end of a given time period, the students must use
the half-life of the antibiotic along with exponential decay function
Qt   Q0 e kt
For example, the antibiotic doxycycline has a half-life of 16.3 hours. We want to find what percent
of the original given amount of the antibiotic is present at the end of a 24-hour period. First you
need to find the value of k in the above equation. It is easier to assume that you have 100mg in
the beginning. Since 50mg is what is left of the original 100mg after 16.3 hours, the equation then
becomes:
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50  100e16.3k
Solving for k, the equation becomes
50 1
 1  1 
  e16.3k , k  
 ln   0.043
100 2
 16.3  2 
After 24 hours,
Qt   100e 0.04324  35.6
Thus, the percentage of the original amount of doxycycline that remains after 24 hours is
approximately 35.6%. To calculate the sum of a finite number of doses, 35.6% is the number
used as the value of r, the common ratio between the terms, in the closed form of the finite
geometric series. For most of the problems in the homework section, the value of r will be
provided.
By calculating the amount of the drug in the body over time, the students should see that as the
number of doses increases, the amount of the drug in the body right after each dose is taken
approaches a constant value. This “steady state” amount is high enough to kill the bacteria
causing the infection. However, if this level of medication is not maintained for long enough, it can
lead to the development of bacterial resistance.
Antibiotic Resistant Bacteria
Many infectious diseases, including bacterial pneumonia, tuberculosis, and gonorrhea, are
caused by bacteria. Antibiotics work by a variety of mechanisms to kill bacteria and have proven
vital in curing infectious diseases. However, bacteria have the capacity to evolve defense
mechanisms against antibiotics and can become resistant to their effects. When such resistance
develops, bacteria are no longer killed by the antibiotic and, thus, the antibiotic is no longer
capable of treating or curing the disease. The more an antibiotic is used, the more likely that
bacteria will "learn" to evade it.
Natural selection plays a key role in the development of antibiotic resistance. Most bacteria die
when exposed to antibiotics to which they are sensitive. That leaves more space and available
nutrients for surviving bacteria (i.e., for antibiotic-resistant bacteria). As a result, the resistant
bacteria can reproduce and multiply freely and pass on the antibiotic-resistant genes to the next
generation.
Not only can resistant bacteria proliferate after other bacteria are killed off by an antibiotic, but
they also can transfer that resistance to other bacteria that have never been exposed to the
antibiotic. Bacterial cells can join briefly and exchange loops of DNA (called plasmids) that
contain genes that confer antibiotic resistance. For example, if one bacterial species becomes
resistant to a broad-spectrum antibiotic, it could transfer its resistance genes to other bacteria that
have never encountered those antibiotics.
The genes that cause antibiotic resistance function in a number of ways. Some do not permit the
antibiotic to get into the bacteria; others actively pump it out of the bacteria; some produce
enzymes that inactivate the antibiotic; and others modify the antibiotic’s target site in the bacteria.
To make matters worse, many bacteria have become resistant to multiple antibiotics. That
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property can result from the cumulative effect of treating stubborn infections with multiple types of
antibiotics (or acquiring a plasmid with numerous resistance genes).
Antibiotic resistance has increased rapidly in the U.S. and abroad in recent decades.
Streptococcus pneumoniae, a bacterium that can cause ear infections, pneumonia, blood
infections, and meningitis, is becoming increasingly resistant to antibiotic treatment. In 1987,
antibiotic-resistant pneumococci were unknown. By 1997, as many as 40 percent of
pneumococcus isolates were resistant to penicillin and other commonly used antibiotics.
(Source: http://www.cspinet.org/reports/abiotic.htm )
Related and Resource Websites
http://www.cspinet.org/reports/abiotic.htm
Activity
1) Go over the first three problems in yesterday’s homework. Ask the students if they needed to
know the last term of the finite series to be able to calculate the sum. (They did not need it.)
Ask if anybody came up with a real-life application where they need to be able to find the sum of
a finite geometric series. Discuss any answers you might get.
2) Tell the students that one application is in taking antibiotics when they have an infection. Let’s
suppose that they have strep throat and the doctor tells them to take the antibiotic ampicillin
regularly for two weeks. Ampicillin is usually taken in 250 mg doses four times a day, that is
exactly every six hours. It is known that at the end of six hours, due to excretion, about 4% of the
drug is still in the body. How much ampicillin is in the body right after taking the eighth tablet?
What about right after the 48th tablet?
3) Ask the students how they can set up a finite geometric series to represent the situation. Tell
them to let the symbol
Qn
represent the quantity in milligrams of ampicillin in the blood right after having taken the nth
tablet. This means that right after taking the first tablet,
Q1  250
Six hours later, a second tablet is taken, and the amount of the drug in the body becomes
Q2  250  2500.04
because you took a new tablet (250 mg), plus you have 4% of the first tablet as well.
Six hours later, after the third tablet is taken, the amount becomes
Q3  250  2500.04  2500.04
2
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because you took a new tablet (250 mg), plus you have 4% of the second tablet, and (4%)(4%) of
the first tablet as well.
4) The students will hopefully see the pattern. Ask them to give you the next three quantities.
Teacher Cheat sheet:
Q4  250  2500.04  2500.04  2500.04
2
3
Q5  250  2500.04  2500.04  2500.04  2500.04
2
3
4
Q6  250  2500.04  2500.04  2500.04  2500.04  2500.04
2
3
4
5
Point out that the equations are the expanded form of a finite geometric series with a = 250, r =
0.04, and the highest exponent used is (n –1). There are n terms in the series, one for each
tablet.
5) Ask the students to write the expanded form of the finite geometric series that represents the
amount of ampicillin in the blood right after having taken the eighth tablet and the 48 th tablet,
using “+…+”, instead of writing out all of the terms.
Q8  250  250(0.04)  ...  250(0.04) 7
Q48  250  250(0.04)  ...  250(0.04) 47
6) Ask the students if they can think of a short cut for calculating the amount of ampicillin in the
blood right after having taken the 48th tablet instead of writing out all 48 terms and adding them
together. Hopefully somebody will remember how to calculate the sum using the closed form:
Q48 

250 1  0.04
1  0.04
48
  260.42mg
7) Have the students use the closed form to calculate the following amounts of the drug
Q3 , Q4 , Q5 , Q6, and Q8
8) Ask them to describe, as the number of doses increases, what is happening to the amount of
the drug in the body right after a tablet is taken. Have them discuss why this might be happening.
Teacher cheat sheet: Below are answers to the respective amounts.
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









250 1  (0.04) 3
Q3 
 260.4
1  (0.04)
250 1  (0.04) 4
Q4 
 260.416
1  (0.04)
250 1  (0.04) 5
Q5 
 260.41664
1  (0.04)
250 1  (0.04) 6
Q6 
 260.4166656
1  (0.04)
Q8 
250 1  (0.04) 8
 260.4166667
1  (0.04)
Notice that all of these values would round to 260.42mg, the same amount (to two decimal
places) as the amount of the drug in the body right after the 48th tablet. This means that by the
end of the first day (after taking four tablets), the amount of the drug appears to have stabilized at
approximately 260.42mg.
9) Ask the students write the closed form if n tablets are taken. What does this closed form
predict about the long-run level of ampicillin in the body? In other words, what happens to the
level of the drug right after a tablet is taken if a very large number of tablets are taken?
Teacher cheat sheet: The closed form of the series for n tablets is:
Qn 

250 1  (0.04) n
1  (0.04)

If you allow the value of n to get very large,
as n  , 0.04
n

250 1  0.04
 0 and Qn 
1  0.04
n
  250  260.42mg
0.96
Thus, assuming that 250mg continue to be taken every six hours, in the long run the level of the
drug in the body, right after a tablet is taken, appears to be approaching a constant level of
260.42mg.
10) Ask the students how they can set up a finite geometric series to represent the amount of the
drug in the body six hours after having taken a tablet. Remind them that about 4% of the original
amount is left in the body six hours after the tablet is taken. Let the symbol represent the quantity
in milligrams of ampicillin in the blood six hours after having taken the nth
Pn
tablet. This means that six hours after taking the first tablet, or just before taking the 2 nd tablet,
P1  250(0.04)
Six hours after the second tablet is taken, and the amount of the drug in the body becomes
P2  2500.04  2500.04
2
as 4% of the second tablet and (4%)(4%) of the first tablet remain in the body.
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Six hours after the third tablet is taken, the amount becomes
P3  2500.04  2500.04  2500.04
2
3
as there is 4% of the third tablet, (4%)(4%) of the second tablet, and (4%)(4%)(4%) of the first
tablet remain in the body.
11) The students will hopefully see the pattern. Ask them to give you the next three quantities.
Teacher Cheat sheet:
P4  2500.04  2500.04  2500.04   2500.04 
2
3
4
P5  2500.04  2500.04   2500.04  2500.04   2500.04
2
3
4
5
P6  2500.04   2500.04  2500.04  2500.04  2500.04   2500.04
2
3
4
5
6
Point out that the equations are the expanded form of a finite geometric series with a = 250(0.04),
r = 0.04, and the highest exponent used is n. There are n terms in the series, one for what
remains of each tablet six hours after it was taken.
12) Ask the students to write the expanded form of the finite geometric series that represents the
amount of ampicillin in the blood six hours after having taken the eighth tablet or the 48 th tablet,
using “+…+”, instead of writing out all of the terms.
P8  2500.04  250(0.04) 2  ...  250(0.04) 8
P48  2500.04  250(0.04) 2  ...  250(0.04) 48
13) Ask the students if they can think of a short cut for calculating the amount of ampicillin in the
blood six hours after having taken the 48th tablet instead of writing out all 48 terms and adding
them together. Hopefully somebody will remember how to calculate the sum using the closed
form:

2500.04 1  0.04
P48 
1  0.04
48
  10.42mg
14) Have the students use the closed form to calculate the following amounts of the drug
P3 , P4 , P5 , P6, and P8
15) Ask them to describe, as the number of doses increases, what is happening to the amount of
the drug in the body right after a tablet is taken. Have them discuss why this might be happening.
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Teacher cheat sheet: Below are answers to the respective amounts.










P3 
2500.04 1  (0.04) 3
 10.4
1  (0.04)
P4 
2500.04 1  (0.04) 4
 1o.416
1  (0.04)
P5 
2500.04 1  (0.04) 5
 10.41664
1  (0.04)
P6 
2500.04 1  (0.04) 6
 10.4166656
1  (0.04)
P8 
2500.04 1  (0.04) 8
 10.4166667
1  (0.04)
Notice that all of these values would round to 10.42mg, the same amount (to two decimal places)
as the amount of the drug in the body six hours after the 48th tablet.
16) Ask the students write the closed form six hours after the nth tablet is taken. What does this
closed form predict about the long-run level of ampicillin in the body six hours after each tablet is
taken?
Teacher cheat sheet: The closed form of the series for six hours after the nth tablet is:

2500.04 1  (0.04) n
Pn 
1  (0.04)

If you allow the value of n to get very large,
as n  , 0.04
n

2500.04 1  0.04
 0 and Pn 
1  0.04
n
  2500.04  10.42mg
0.96
Thus, assuming that 250mg continue to be taken every six hours, in the long run the level of the
drug in the body, six hours after a tablet is taken, appears to be approaching a constant level of
10.42mg.
17) Have the students make a graph of amount of drug in the body versus time (in six hours
intervals). Assume that the first dose is taken at time t = 0.
Teacher Cheat Sheet: the first seven points would be as follows:
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Q1  250mg
(0, 250)
5.9,10
6, 260
P1  (250)(0.04)  10mg
(11.9,10.4)
P2  (250)(0.04)  (250)(0.04) 2  10.4mg
(12, 260.4)
Q3  250  10.4  260.4mg
Q2  250  10  260
(17.9,10.416) P3  (250)(0.04)  (250)(0.04) 2  (250)(0.04) 3  10.416  10.42mg
(18, 260.42)
Q4  250  10.42  260.42mg
All the rest of the y-values for Q will be 260.42mg and all the rest of the y-values for P will be
10.42mg.
18) Have the students make a graph of amount of drug in the body versus number of doses.
Assume that no drug is in the body at time t = 0.
Teacher Cheat Sheet: the first seven points would be as follows:
no drug is in the body at t  0
(0, 0)
Q1  250mg
(1, 250)
2, 260
Q2  250  10  260
(3, 260.4)
Q3  250  10.4  260.4mg
(4, 260.42)
Q4  250  10.42  260.42mg
(5, 260.42)
Q5  250  10.42  260.42mg
(6, 260.42)
Q6  250  10.42  260.42mg
(7, 260.42)
Q7  250  10.42  260.42mg
19) Looking at the graph of amount of drug versus dose number, ask the students what would
happen if they forgot to take a tablet and went 12 hours in between doses? What would happen if
they stopped taking the antibiotic before completing the full regimen they were given?
Teacher Cheat Sheet: See teacher background.
Homework
The drug erythromycin may also be taken for streptococcal infections. Erythromycin is given in
400mg doses every six hours. At the end of six hours, there is 6.25% of the amount taken left in
the body.
1. Write the expanded finite geometric series for the amount of the drug in the body right after
having taken the nth tablet.
Answer :Qn  400  4000.0625  4000.0625  ...  4000.0625
2
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n 1
2. Write the closed finite geometric series for the amount of the drug in the body right after
having taken the nth tablet.

400 1  0.0625
Answer : Qn 
1  0.0625
n

3. Write the expanded finite geometric series for the amount of the drug in the body six hours
after having taken the nth tablet.
Answer : Pn  4000.0625  4000.0625  ...  4000.0625
2
n
4. Write the closed finite geometric series for the amount of the drug in the body right after
having taken the nth tablet.
Answer : Pn 

4000.0625 1  0.0625
1  0.0625
n

5. Calculate the amount of erythromycin in the body right after taking a tablet over the long-run.
Answer : as n  , Qn 


400 1  (0.0625) n
400

 426.67mg
1  (0.0625)
1  (0.0625)
6. Calculate the amount of erythromycin in the body six hours after taking a tablet over the longrun.


4000.0625 1  0.0625
4000.0625
Answer : as n  , Pn 

 26.67mg
1  0.0625
1  0.0625
n
7. What is the difference between the numbers calculated in problems 5 and 6? Why does this
occur?
Answer: 426.67-26.67=400mg, which is the amount of each individual dose of
erythromycin taken. This occurs because once the level of the drug in the body right after
a dose is taken reaches 426.67mg, the steady state amount, the body excretes 400mg
over six hours, (426.67)(.9375) = 400, because taking one more tablet raises the level
back to the steady state amount.
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