Growth Unit

UNIT 8—Growth and Decay

Teacher Materials


CLICK THE  SYMBOL OF EACH SECTION HEADER TO RETURN HERE.
TEKS Support
Teacher Notes
 Context Overview
 Mathematical Development
 Teaching Suggestions
 Preparation Reading—"dos... age (do¢ sij) n."
 Activity 1—First Steps
 Homework 1—We Need Input, Stephanie!
 Activity 2—Tell Me If It Hurts!
 Homework 2—Let’s Play Doctor
 Activity 3—The ‘K’ in ‘Decay’
 Homework 3—Grrrrrrr-Rate!
 Activity 4—You Grow Some, You Decay Some
 Homework 4—Stay Between the Lines!
 Activity 5—Close to a Final Form
 Assessment—Dosage Deliberations
 Handout 1—Prescription Model
 Transparency 1—Prescription Model
 Transparency 2—Prescription Model
Annotated Student Materials
 Preparation Reading—dos... age (do¢ sij) n.
 Activity 1—First Steps
 Homework 1—We Need Input, Stephanie!
 Activity 2—Tell Me If It Hurts!
 Homework 2—Let’s Play Doctor
 Activity 3—The ‘K’ in ‘Decay’
 Homework 3—Grrrrrrr-Rate!
 Activity 4—You Grow Some, You Decay Some
 Homework 4—Stay Between the Lines!
 Activity 5—Close to a Final Form
 Assessment—Dosage Deliberations
 Unit Summary
 Mathematical Summary
 Key Concepts
Solution to Short Modeling Practice
 Solution for Modeling Bacteria Growth
Solutions to Practice and Review Problems


TEKS Support

This unit contains activities that support the following knowledge and skills elements of the TEKS.

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The mathematical prerequisites for this unit are

The mathematical topics included or taught in this unit are

The equipment list for this unit is

Teacher Notes



"Simple Model"—Context Overview

The context of pharmacology and prescription medicine forms the basis for this unit. The modeling process is applied to identify and isolate various parameters that are relevant to understanding the context. These parameters include the minimum effective level, the maximum level of safe use, the rate of dissipation, the initial dosage, the response time and the deadline for the completion of the treatment. Students are asked to consider these parameters in various combinations, and led to increasingly complicated situations.




Mathematical Development

Linear and exponential functions are reviewed as the delivery of oral medication (pills) is an additive process, while the rate at which the body uses up the medication is a multiplicative process. The distinction between discrete and continuous functions is brought up by assumptions on how the pills dissolve. Sequence notation plays a big part in describing the understanding of how the two disjoint growth processes work in tandem. Working with exponents and roots are brought into consideration, and solving equations using logarithms is introduced. Graphs and tables form the basis for studying the changes in various growth processes over time.




Teaching Suggestions

Note: The development of the materials were made with the assumption that graphing calculators with "home-screen iteration" capability would be available to students for examining long-term projections. It would be preferable to show students how to build spreadsheets, and to explore a lot of those problems with spreadsheet models. As you go through these teacher notes, remember that as an option, if your class has access to computer work stations.




Preparation Reading—"dos... age (do¢ sij) n."

The context is one that is probably familiar to your students, but the chances are that they haven’t thought about where the determination of dosage and time interval come from. The entire first day’s class is devoted to moving the students from the background that they bring on prescription medication and dosage to becoming active modelers of the problem.

Given that goal for the first day, encourage students to participate with their own experiences. Continue to bring them back to the modeling process and the questions that must be addressed, but try to elicit information or ideas on the subject from them, rather than just tell them "This is how we’re going to do it!"




Activity 1—First Steps
   

This is a fairly structured group/class activity, which will guide the students into the first stage of the modeling process. For each question of Activity 1, students should be given the opportunity to read, think and discuss the question in their groups, and then draft a response. That provides a lot of potential input for a class discussion, which will provide students an opportunity to formulate their ideas and suggestions, and then allow them to share those in a forum environment. Finally, assist the class to resolve what the "best" answer would be, and what direction to take in modeling the problem.




Homework 1—We Need Input, Stephanie!
   

The purpose of this assignment is to provide additional clarification on some of the terms brought into the previous discussion, and to build a basis for understanding how the parameters relate to each other. Also, there is an opportunity for students to go find out more about the context area, and to network with professionals that might be involved in pharmacology or medical practice. Students that can bring in information and related facts should be given the opportunity to share what they’ve found ongoing throughout the unit.

Be sure to review the answers to Question 5 of Part 1 thoroughly. The drug amount affects the concentration levels, and the assumptions made in this problem establish a "unit conversion" that will be consistently applied throughout the model.

Some students will see that these dosage problems are proportional and simply multiply to obtain the answer. For example, 600 mg is 6 times the initial does of 100 mg. Therefore, the concentration is 6 times higher. Others will need to do the proportion.

Here is how the proportion will look for problem 5a):



Here is how the proportion will look for problem 5b):



Students may also ask why pills are less effective than shots. There are many answers to this question. The most important factor is that pills travel through the digestive system, most notably the stomach. Stomach acids, and the process of digestion limit the amount of active ingredient that makes it to the bloodstream. Therefore, the doses for pills are often higher than for shots to achieve the same concentraiton. This is also an opportunity to discuss how being a doctor is often a compromise between treating illness, and the comfort and convenience of patients.




Activity 2—Tell Me If It Hurts!
   

This activity focuses solely on the additive nature of the delivery of medicine by repetitive ingestion of pills. The modeling process begins by making unrealistic assumptions on the problem, simply so that a beginning reference can be established for understanding and describing the situation. In this context, we will assume (temporarily) that the chemicals don’t dissipate in the blood, and develop the mathematics that describes additive growth processes.

Various situations and strategies are considered, and students are encouraged to put themselves in the role of a doctor as they write out prescriptions for the various problem solutions. Remind students to associate given information in a particular problem with the problem parameters identified in the handout Prescription Parameters. Some of what the students will be doing in the unit is quite complex and can be confusing. As a strategy for you in helping students understand what they are doing or need to do, you can also draw the connection to those parameters. If students experience difficulty in answering parts of the assignment (or any assignments from the unit), remind them that the tools for exploration are tables and graphs, even if they are not specifically requested in the problems! Point out to students that the traditional names for arithmetic sequences are a0, a1, a2, a3, a4, a5, etc.

Note: 1 µg = 10–6g




Homework 2—Let’s Play Doctor
   

This set of problems still uses the assumption that the chemicals stay in the body indefinitely, and provides students more opportunity to solve problems in manipulating the various parameters. Question 2d) is a chance for your stronger students to apply the "Method of Generalization" and develop a solution in terms of the various parameters, rather than specific value solutions.




Activity 3—The ‘K’ in ‘Decay’
   

Activity 3 is where students focus their attention on the rate of dissipation as a parameter, and the mathematics behind exponential decay processes. Start with a discussion of the modeling process. The assumption that chemicals never dissipate was very unrealistic; students’ experiences will be to the contrary. Rather than add in an additional consideration, this activity tables the consideration of the dosage, and explores what happens to quantities over time once they are ingested. Make sure students understand the reasoning behind the decision that "Simpler is better", until you are able to put everything together.

There may be a need to go over how to convert from a percent of decline to a growth factor. In addition, recursive equations will be extensively used during the rest of the unit, so there should be an emphasis placed on students understanding what the subscripts do for the equation. Point out to students that the traditional names for geometric sequences are b0, b1, b2, b3, b4, b5, etc. A review of exponents, how they represent repeated multiplication operations and how to evaluate them, might be in order.

In debriefing Activity 3, be sure to go over Questions 5-8. Half-life is a useful way to describe exponential decay sequences, and all the various methods introduced for solving equations to find the half-life should be reviewed with the students.




Homework 3—Grrrrrrr-Rate!
   

Practice problems from Activity 3 form the majority of this set of questions. It can be an opportunity for you to find out if your students really understood the material brought out in the Activity. Question 6 asks students to think about how to mix additive growth and exponential decay processes together. In asking the question, it is setting the table for the understanding of how to describe those kind of patterns, and preparing students for the next Activity. Be sure to discuss the students’ findings with them on this.




Activity 4—You Grow Some, You Decay Some
   

In modeling the contextual problem, we’ve considered the effect caused by the oral prescription alone, and we’ve considered the effect caused by the body metabolizing the medication alone. In Activity 4, you now consider what happens when both are going on simultaneously. Be sure students understand that one action is being treated as a discrete process (assumption that the pill dissolves immediately causes "quantum" jumps in concentration), while the other is a continuous process (gradual dissipation of the medicine in the body). This shows up in the graphs as repeated spikes of increase, and classic exponential decay curves.

Mixed growth sequences form the basis for this activity. Students start off by building a sequence that balances the additive growth with the exponential decay. Then, a couple of examples in which the growth processes are not balanced are examined. Change of scale is introduced as a way of focusing on the top or bottom of the exponential curves over time, instead of having to deal with disjoint pieces. Finally, students arrive at the recursive expression defining mixed growth sequences; the logic that describes how to go from one time interval to the next is pretty nice, especially compared to the closed-form equations. Make sure your students understand the logic behind these expressions before moving forward. Also, make sure that the students understand the distinction between the pattern formed by the peak concentrations (pill first, then measure concentration) and the "bottoming-out" concentrations (measure concentration first).




Homework 4—Stay Between the Lines!
   

Students return to the model in this assignment as they juggle parameter values and try and solve problems based on the various conditions and constraints. Consideration for the boundary values of the partial sums is brought out in this set of problems, although not formalized. Again, remind students to refer to the Handout 1—Prescription Model. These problems are beginning to get conceptually more demanding, and there is a greater need for table building, home-screen iteration on the calculator, exploration using a spreadsheet template, and making graphs. Another tool that students can use is to sketch an x- and y- axes, and a rectangular "window" in the first quadrant, to represent the conditions for each of the problems. Encourage them to visualize how concentration would respond under each constraint condition.




Activity 5—Close to a Final Form
   

In this activity, a mathematical approach for finding the theoretical maximum value of a mixed-sequence growth is explored. The modeling approach is to try and establish a more general solution to the problem, than to explore each one as a "new" adventure. Point out to students that they are getting close to the crux of the problem, and that it is always more difficult sledding.

The idea is to analyze the computations for a mixed-sequence growth, in order to determine what’s going on. The first pill’s dosage goes through a decay process repeatedly with each time interval, and the second one as well, although it hasn’t been in the body as long. Each pill’s contribution to the total concentration is expressed using the algebra of exponents, since two time intervals would have a growth factor of k2 (if one time interval has a growth factor of k).

Depending on the students you have, you might want to discuss Questions 2, 3 and 4 in a "whole-class" forum, especially the parts requiring intensive algebra. Make sure the students see the intuitive development of the infinite series as a limit as the number of terms goes to infinity, and how that reduces the partial sums formula to the infinite series formula. Also, make sure they recognize that the infinite series formula has conditions on when it converges to that expression, namely that the growth factor must be less than one.




Assessment—Dosage Deliberations
   

This is the final stage of the work on developing the model. Students take the work done from Activity 5 in developing expressions for calculating the dosage and/or time interval from the theoretical maximum concentration allowed in the problem. They have to adjust their thinking to build in considerations for threshold and tolerance levels, and for response time. Strategies for generalizing the entire problem are hinted at, or at least improper strategies are examined and critiqued. Between the examples that students must work through, and the situations that students need to analyze, there should be enough foundation for students to arrive at a general process for determining dosage and time intervals.

Question 3, 4 and 9 should be discussed with the entire class. If students are able to formulate the process in English, help them to express it in mathematical notation. At any rate, after completion of this assignment, students should review the work done in the unit to try and summarize what was done. A writing assignment requesting that process work could be a nice way of finishing off the unit core material, prior to any assessment of the learning in the unit.




Handout 1—Prescription Model

 




Transparency 1—Prescription Model

 


Transparency 2—Prescription Model



Annotated Student Materials



Preparation Reading—dos... age (do¢ sij) n

Have you ever been given a prescription for some medication? It’s likely that the label on the bottle did not say, "Take two, and call me in the morning!" If it’s an oral medication (pills), probably the number of tablets in the bottle is somewhere on the label. There should also be some way of identifying what substance is in the container, and how much prescribed medicine is in each tablet. And most importantly, there should be instructions on how many pills to take each day, and how often. In other words, what is the dosage?

The American Heritage Dictionary of the American Language has the following definition for dosage: "1. The administration of a therapeutic agent in prescribed amounts. 2. The amount administered." Have you ever wondered how doctors come up with "the amount to be administered"? What kind of thinking and decision-making does it take to determine how to administer something in prescribed amounts? It may be as simple as looking up information in a table, or plugging some numbers into a formula that’s as old as medical knowledge. But it also is a rich problem context in which to explore and to practice mathematical modeling.

In this unit, you are going to place yourself in the role of a medical practitioner and practice the skills you’ve developed in learning to model. What is the question? What are some assumptions that need to be made? What are the variables? How should we begin this exploration? When you get through all that, you still have to figure out what kind of mathematics applies here. You may even need to learn more mathematics in order to solve your problem. That can be a lot to deal with for a single problem and may seem impossible, but there is one thing that is reassuring about the situation. Doctors solve this kind of problem on a daily basis, so it has to be possible. And their work can provide a kind of reality-check for our mathematical prediction, so we can test how effective our model is.

Take a few moments to think about the field of prescriptive medicine. How do they do that, anyway?




Activity 1—First Steps

 = 1, 10, 11

  1. Discuss the reading in your group; share experiences you may have had with the actual context. Then state a problem that would be "worthy" of modeling.


    2. Group:Answers will vary; refer to Teacher Notes for discussion ideas.

    Class: How can you determine the amount of medication to prescribe, and the time interval between dosages (or the number of dosages to give in a predetermined time period)?
  1. What are some of the factors to consider? What limitations must be considered?


    2. Group: Answers will vary; refer to Teacher Notes for discussion ideas.

    Class: Factors to consider include: the size of the patient, the kind of chemical being administered and its rate of absorption and dissipation in the body, whether an initial treatment is applied, whether the drug is administered orally or intravenously, whether the body manufactures the chemical.

    Limitations include: the amount of medicine needed to be effective, the amount that begins to cause harm to the patient, the time interval in which you must begin treatment, the time interval in which you must have completed treatment.
  1. In developing a model for this situation, what assumptions will be initially made?

    2. Group: Answers will vary; refer to Teacher Notes for discussion ideas.

    Class: The critical factor to monitor is the concentration level. Assume that the body weight and blood volume are constant, so that the amount of medicine present directly affects the concentration level. There’s no prior amount of the medicine present before administering the first dosage, and the body doesn’t manufacture the chemical naturally.
  1. What are the variables for the model?


    2. Group: Answers will vary; refer to Teacher Notes for discussion ideas.

    Class: Input (independent) variables will be dosage and time between doses, and the output (dependent) variable will be concentration.
  1. What might be a good "next step" to take in developing the model?


    2. Group: Answers will vary; refer to Teacher Notes for discussion ideas.

    Class: Consider a simpler situation in which only one of the parameters is considered at a time; i.e., where the chemicals don’t dissipate, where there isn’t an initial treatment, or where there is only an initial treatment.



Homework 1—We Need Input, Stephanie!
   

Part 1: Understanding the Problem

In the discussion that took place in Activity 1, input/output variables and parameters for the problem were introduced. For the following questions, Handout 1, Prescription Model, will be useful.

    1. What happens if the concentration level C(t) doesn’t get as high as Cmin?


      2. The amount of the drug won’t be great enough to do any good, and certainly won’t treat the condition.
    2. What happens if the concentration level C(t) gets higher than Cmax?


      3. There could be side effects (i.e., hives), overdose symptoms (i.e., falling asleep), and possibly a new condition brought on from toxic levels of the chemical (i.e., kidney failure).
    3. What does it mean for the response time to be fairly small, like say a couple of hours?

      4. It means that you’d better see the doctor really quickly, or there may be complications. For example, if you have pneumonia, antibiotics can treat the situation fairly easily. But if you go past the response time, you might get fluid in the lungs, which would be very serious.
    4. What does it mean when the deadline for treatment is fairly big, like say a month?


      5. It might mean that there’s no rush to treat the condition, as long as it’s taken care of during that time, but it may mean that the treatment needs to be stretched out over a long time period. What is definite is that the treatment needs to be finished by that moment in time.
    5. What would happen if Cmin were bigger than Cmax?


      6. The condition can’t be treated; or if it has to be treated, there will be some complications, since the smallest amount required to take care of the problem will be more than what the body can handle safely.
  1. Use the axes below to record your answers to parts a) and b). Introduce a scale and use numbers if necessary to draw the lines in the correct places, but indicate what numbers you are using for each of the values.

    1. Sketch in two vertical lines to represent the response time to and the treatment deadline time tf.
    2. Sketch in two horizontal lines to represent the threshold level Cmin and the treatment deadline time Cmax.
  1. If you were using the graphing calculator, what feature would represent the box that you just drew in the previous problem? What would to, tf, Cmin and Cmax represent?


    2. Basically, that’s establishing a WINDOW for viewing the important part of the problem, with xmin = to, xmax = tf, ymin = Cmin and ymax = Cmax. See the shaded part of the graph.
  1. There is a relationship between the response time, the treatment deadline time, the number of doses and the time between doses. For now, we won’t worry about the output concentration, so the actual amount in each pill is not important. Work out a couple of examples first:


    1. to = 6 hrs, tf = 36 hrs, and D = 12 pills. Find Dt.


      2. Note: Review the use and meaning of "D."

      Total treatment time = 36 – 6 = 30 hours. Break that time up into 12 intervals, and you get 2.5 hours in between pills.
    2. to = 2 hrs, tf = 36 hrs, and Dt = 2 hrs. Find D.


      3. 36 – 2 = 34 hours 1/2 2 hour intervals = 17 pills.

    3. Write a general relationship between to, tf, D and Dt.


      4. There will be various ways to express this, but they should all be equivalent to:
    4. What assumptions did you make in answering parts a) and b)? (Those become assumptions on using the relationship you just derived!)


      5. We assumed that the first pill would be given at the beginning of the treatment time interval (at to), and that the last pill to be given would "lose its effect) at the treatment deadline time. In other words, a pill would not be given at the deadline time, since its effects would be after the deadline.

  2. When considering prescriptions, doctors have to be concerned with two factors:

    First, it is not the amount of the dose that treats the problem, but the concentration of the substance that matters. The concentration of the drug is equal to the amount of the substance divided by the blood volume of the patient or. This concentration is usually measured in micrograms over milliliters (µg/ml). Blood volume is strongly related to weight, so most dosages increase with the weight of the patient. The doctor may also consider other factors like gender, metabolism, and general health when they prescribe a drug.

    Second, a part of any drug is used up in dissolving pills, digestion, and other factors. Therefore pills and injections require different dosages to reach the same concentrations in the blood stream. Generally injections are more effective than pills, and require smaller doses. Pills, however, are more convenient for patients to take on a regular basis.

    1. Let's assume that a typical adult male with a blood volume of 2,000 ml is given an injection of 100 mg. After losses getting to the blood stream, this causes a change in concentration of 50 µg/ml. What would the change be for a 600 mg shot?

      (600 mg)/C = (100 mg)/(50 µg/ml)

      C = 300 µg/ml

    2. Let's assume that the same adult male is given a 100 mg pill. After losses getting to the bloodstream, this causes a change in concentration of 10 µg/ml. What would the change in concentration be for a 250 mg pill?

      (250 mg)/C = (100 mg)/(10 µg/ml)

      C = 25 µg/ml

      Note: In developing this model, we will continually work with this patient and drug combination. Therefore, the following assumptions are needed:

      • The drug distributes itself evenly throughout the body.

      • A 100-mg shot will change the concentration of drug by 50 µg/ml.

      • A 100-mg pill will change the concentration of drug by 10 µg/ml.

      You may want to investigate how answers might change if you varied the type of drug administered (which will change the difference in effectiveness between the shot and the pill, the weight of the patient (which will change the blood volume), or other factors you might identify).

      Note:

      1 millogram =

      1 mg = (1/1,000,000)g = 10–6 g

      1 milligram = 1 mg = (1/1,000)g = 10–3 g

      1000 mg = 1 mg



Part 2: Gaining Familiarity With the Problem

It would be nice to know if our parameters, assumptions and variables form a "total package", so we need to check in with some relative experts on the subject. There are several sources of information.

Bring in whatever you can find to share with the class. Happy hunting!




Activity 2—Tell Me If It Hurts!
   

 = 7, 12, 13, 16

In the discussion on modeling the problem of prescription dosage that just took place, a lot of factors were brought in to consideration. A lot of times, it can be overwhelming to consider all of them at the same time, so we’re going to consider a simpler situation. For the duration of this activity, we’ll assume that the chemical never breaks down in the body, but simply accumulates over time as more of it is put into the body.

Remember the assumptions for our patient:

  1. Consider a situation in which a patient has a tolerance level of 100mg for a certain chemical, but a threshold level of 40mg is necessary to treat a condition. As the doctor, you decide to give the patient a single shot, and not bother with a prescription.


    1. How much of the chemical should be in the shot?

      2. Answers will vary, but should range from 80 to 200 mg. Probably the midrange value of 140 mg would be the safest answer.

      A/(70 µg/ml) = 100 mg/(50 µg/ml)

      A = 140 mg
    2. List the various parameter values that relate to this problem and their values. We’ll call this a "control panel" for the problem.

      3. D = 0; Cmin = 40; Cmax = 100; k = 0;

      Co = whatever the answer to part a) was = 70 µg/ml.

    3. Sketch a graph of the situation in the space provided above. Indicate where all the important information is located.

    4. Why is the input variable D not considered? Why is the input variable Dt not considered? Why are the parameters to and tf not considered?

      5. D isn’t considered because we’re assuming that the treatment will consist of a single shot, in this case containing 70 mg. Dt, to and tf are not important since we’ve assumed the chemical doesn’t dissipate over time, and the shot is being given at time t = 0.

    5. Write an equation describing this situation. What would the general equation for these assumptions look like?

      6. In this particular case, C(t) = 70.

      In general, C(t) = n, where n is some number in the interval [Cmin, Cmax].
  1. Let’s try a different approach with that same problem now. Suppose the treatment favors a gradual increase of the concentration level over time, until it just reaches the tolerance level at the treatment deadline time of 48 hours. Furthermore, being anxious to begin treating the symptoms of the condition, you decide on a single shot to bring the chemical concentration up to the threshold level, and then oral medication applied at regular time intervals to finish the job.

    1. How much medicine should be in the shot? How much more medicine needs to be provided from the pills?

      2. 80 mg in the shot; 600 mg from the pills
    2. If the patient were to take a pill every hour for the entire 48 hours of treatment, what would have to be the dosage (amount of each pill)? How did you determine that answer?

      3. Dpill = 12.5 mg. Divide the amount the patient still needs by the number of hours.
    3. It’s more convenient to have the patient take pills twice a day, or four times a day, than to be woken up in the middle of the night every hour, taking a pill. Have the patient take a pill every 6 hours. How many pills, and at what dosage, will the pharmacist provide? What’s the prescription going to read?

      4. Quantity: 8 pills. Dosage: 75 mg. Prescription: Take 1 pill every 6 hours.
    4. Make a table of values for the total amount of medication at 6-hour intervals. Then sketch the graph of concentration of medicine vs. time.


      5. Time
      (hr)

      Concentration
      (µg/ml)

      0

      40

      6

      47.5

      12

      55

      18

      62.5

      24

      70

      30

      77.5

      36

      85

      42

      92.5

      48

      100

    5. In describing this model, what is the closed-form equation for C in terms of t? For C in terms of n (the number of pills taken)?

      6. Closed form: C(t) = 1.25t + 40, where t is measured in hours and C in mg/ml.

      C(n) = 7.5n + 40
    6. Which equation makes more sense to use in this situation? Explain.

      7. It depends on the assumptions. The first one implies that the medication is being released at the same rate during each moment of the 48 hour period. The second one says that the concentration "jumps" every time we take a pill, which isn’t real, but we can "assume" that the pill dissolves instantly.

      The concentration was affected by the number of pills taken, and formed a pattern that was created by repeating an addition operation. That kind of pattern is called an arith-metic sequence; describing such a sequence involves associating the values with its place in the list of numbers.
    7. Fill in the table with the appropriate values. Notice the names used to describe each of the numbers in the sequence.


      8. Term Name

      c0

      c1

      c2

      c3

      c4

      c5

      c6

      c7

      c8

      No. of Pills Taken

      0

      1

      2

      3

      4

      5

      6

      7

      8

      Concentration

      40

      47.5

      55

      62.5

      70

      77.5

      85

      92.5

      100
    1. Now, describe the sequence in recursive form—using one statement to describe what the original value is, and the other statement to describe how to get any value from the previous term of the sequence.

      2. c0 = 40; cn = c n–1 + 7.5 (0 £ n £ 8)

    2. How could you have calculated the value of the 8th term?

      3. Start with 40 and add 7.5 to it a total of 8 times, or

      40 + (7.5)(8).

  1. There may not be a need for immediate treatment, or building up concentration over time with pills may be preferred to giving a shot to the patient. Let’s look at an example of that kind of administration strategy. Suppose that the threshold level of 180 mg must be reached at the response time, which is 10 hours, and that you want to reach the tolerance level of 450 mg by the treatment deadline, which is 36 hours.

    1. How much medicine needs to be provided per hour in order to reach the threshold level on time? Break into 2 time intervals:

      2. 0®10 to reach threshold

      10®36 to reach maximum

      1800 mg / 10 hr = 180 mg/hr
    2. Once you achieve that concentration, how much more medicine needs to be given per hour in order to reach the tolerance level on time?

      3. Amount = 4500 – 1800 = 2700 mg

      Rate = 2700 µg/26 hr = 10.38 mg/hr

    3. How are you going to have to address the fact that the two previous answers aren’t the same?

      4. You need to use two different prescriptions
    4. Let’s say you see the patient at 9:00 a.m., and you decide that taking a pill every two hours is desirable until you reach the threshold level, and after that you want a pill every 6 hours. Write out the prescription.

      5. Pink pills: quantity 5, dosage 360 mg, take one every 2 hours until gone, then switch over to the purple pills

      Purple pills: quantity 4, dosage 675 mg, take one every 6 hours until gone
    5. Explain how you arrived at the prescription you gave.

      6. Because the time interval for the first stage was determined to be 2 hours, I multiplied the hourly amount by two to get the dosage. The quantity was calculated by taking the number of total hours and dividing it by the time interval chosen. For the second stage, the time remaining (26 hrs) didn’t divide by the time interval (6 hours) evenly. The number of complete 6-hr time intervals became the quantity, and the amount that was left to be delivered was divided by the quantity to determine the dosage.
    1. Fill in the table, and then make a graph of C(t) vs. t.


      2. Term

      Time (hrs)

      Concentration
      (mg/ml)

      c0

      0

      0

      c1

      2

      36

      c2

      4

      72

      c3

      6

      108

      c4

      8

      144

      c5

      10

      180

      c6

      16

      247.5

      c7

      22

      315

      c8

      28

      382.5

      c9

      34

      450

    2. Write a recursive equation for how the pills increase the concentration for each part of the prescription strategy of this problem.

      3. c0 = 0, cn = cn–1 + 36 (0 £ n £ 5) and c5 = 180, cn = cn–1 + 67.5 (5 £ n £ 9)

    3. What assumptions have been made about the way in which the pills dissolve?

      4. The pills are completely dissolved by the end of the time period, and we either don’t know or don’t need to be concerned by what’s going on during each time period.
    4. Write closed form equations that describe how concentration changes over time for each part of the prescription strategy of this problem.

      5. C(t) = 18t + 0 (0 £ t £ 10) and C(t) = 11.25(t – 10) + 180 (10 £ t £ 34)

    5. What assumptions have we now made about the way in which the pills dissolve?

      6. The pills are "time-delayed" and dissolve a little bit at a time, and at a constant rate.
    6. Explain how the numbers that make up the closed-form equations are determined by the problem conditions.

      7. In the first equation, the domain is determined by to, the y-intercept by C0, and the slope by Cmin. In the second equation, the domain is determined by the expression tf – to, the y-intercept by C0, and the slope by both tf – to and Cmax – Cmin.



Homework 2—Let’s Play Doctor
   

You wish to compare the strategy of giving a shot to raise the concentration up to the threshold level immediately with one in which you use oral medication to bring the concentration up to the threshold level by the response time. Let’s use a threshold level of 200mg/ml and a tolerance level of 500 mg/ml. The response time will be 12 hrs and the treatment deadline will be 48 hrs.

  1. Consider the strategy of giving an initial shot first. Calculate the amount needed for the shot, and write a prescription for the oral medication.

    The amount needed by the shot is 400 mg, since C0 = Cmin = 200 µg/ml. The additional amount to be delivered by pills is 3000 mg. This is to be done over a 48-hr period, but at 6-hr intervals. Therefore, the prescription will read: Quantity: 8. Dosage: 375 mg. Take 1 every 6 hours.
  2. Now, consider the strategy of treating the condition only with pills. Write the prescription for the oral medication.

    3. The threshold level of 200mg/ml should be reached within 12 hours, in 6-hr doses, so the dosage for the pills has to be 1000 mg each. At that rate, the tolerance level will be reached after 5 pills. So, the prescription will read:

    Quantity: 5. Dosage: 1000 mg. Take 1 every 6 hours.
  3. Let’s change the time between doses to 4 hours. How do the prescriptions change for each of the strategies?

    4. Strategy 1 (with initial shot): Amount of shot: 400 mg. Prescription: Quantity: 12. Dosage: 250 mg. Take 1 every 4 hours.

    Strategy 2 (no initial shot): Prescription: Quantity: 12. Dosage: 416.7 mg. Take 1 every 4 hours.
  4. Now, let’s put everything together. Describe the prescriptions for each of the two strategies in the general case, where the time between doses is Dt, the amount of dosage D, the number of pills n, the amount of the shot C0, the threshold level Cmin, the tolerance level Cmax, the response time to and the deadline time tf.

    5. Strategy 1 (with initial shot): Amount of shot C0 = Cmin

    Quantity: n = integer value of (tf / Dt)

    Dosage: D = (Cmax – Cmin) / n

    Take 1 every Dt hours

    Strategy 2 (with no shot): Initial amount of medicine C0 = 0

    If Cmin / to £ Cmax / tf, then:

    Quantity: n = integer value of (tf / Dt )

    Dosage: D = Cmax / n

    Take 1 every Dt hours

    If Cmin / to > Cmax / tf, then:

    Quantity: n = integer value of ( (Cmax / Cmin) * (to / Dt ) )

    Dosage: D = Cmax / n

    Take 1 every Dt hours




Activity 3—The ‘K’ in ‘Decay’
   

 = 2, 4, 6

Our model of how medicine is introduced into the bloodstream was based on the assumption that the medicine stayed in the body forever. Now we want to focus on how the body uses up, or metabolizes, medicine. We will make two assumptions at this time. The first assumption is that no additional medicine is taken. This assumption will be changed in a later activity.

The second assumption is that a constant fraction, or constant percentage of medicine is metabolized per unit time. This fractional rate of change, which we will designate as r, is constant. The fraction of medicine remaining after a unit time interval is also constant and is designated by K. The two constants are related as:

K = (1 – r)

  1. Let’s start with a situation in which there is 800 mg/ml of a certain medicine present at first, and it decays at a rate of 10% per hour.


    1. What would the concentration be after 1 hour? Explain how you got that answer.

      2. r = 0.10
      K = (1 – r) = 0.90
      C1 hr = 800 µg/ml (0.90) = 720 µg/ml
    2. What would the concentration be after 2 hours? Explain how you got that answer.

      3. At the start of the second hour we have 720 µg/ml. We will lose another 10% the second hour. We still have r = 0.10 and K = 0.90.

      C2 hr = 720 µg/ml (0.90) = 648 µg/ml
    3. Fill in the table with concentrations for the first 10 hours, then make a graph of C(t) vs. t.


      4. Term

      Time
      (hr)

      Concentration
      (mg/ml)

      c0

      0

      800

      c1

      1

      720

      c2

      2

      648

      c3

      3

      583.2

      c4

      4

      524.88

      c5

      5

      472.39

      c6

      6

      425.15

      c7

      7

      382.64

      c8

      8

      344.37

      c9

      9

      309.94

      c10

      10

      278.94

    1. Write a recursive expression for this sequence that describes how it decays each hour.

      2. C0 = 800; Ct = Ct–1 – (0.10) C t–1 or Ct = C t–1 * 0.90
    2. Fill in the table with the values that describe how the chemical decays over 2-hour time periods, and write a recursive expression for this sequence.


      3. No. of 2-hr Time Intervals (n)

      0

      1

      2

      3

      4

      5

      Concentration (C)

      800

      648

      524.88

      425.15

      344.37

      278.94

      C0 = 800; Cn = Cn–1 – (0.19)Cn–1 or Cn = Cn–1 * 0.81
  1. Fill in the table with the values that describe how the chemical decays over 5-hour time periods, and write a recursive expression for this sequence.


    2. No. of 2-Hr. Time Intervals (n)

    0

    1

    2

    Concentration (C)

    800

    472.39

    278.94

    C0 = 800; Cn = Cn–1 – (0.40951)Cn–1 or Cn = Cn–1 * 0.59049
    1. Explain how you determined the numbers describing the last two decay processes.

      2. I took repeated subtractions of 10%, and then determining the percent of the original amount, or I just multiplied 0.90 by itself as many times as the number of hours in the time interval, or refer to table in 1c.

      K5 hr = b5/b0 = 472.39/800

      K5 hr = 0.59

    2. Here’s a closed-form equation for the concentration drop-off on an hourly basis:

      3. C(t) = 800*(0.90)t.

      The order of operations is to raise (0.90) to the power specified by the value of the variable t, and then multiply that answer by 800. Verify that the answer you wrote as the value for the concentration at the end of the 8th hour is correct, by using t = 8 in the closed-form equation.

      C(8) = 800 * (0.90)8 = 800(0.90)(0.90)(0.90)(0.90)(0.90)(0.90)(0.90)(0.90) = 344.37. (It’s the same!)
    3. Write closed-form equations for the other two patterns, which keep track of the concentration in 2-hour and 5-hour intervals.

      4. C(t) = 800*(0.81)t and C(t) = 800*(0.59049)t.
  1. We’re going to go back to the original problem and change the rate of dissipation to see how that affects the concentration level over time. This time, the 800 mg/ml of medicine that is in the body decays at only 3% per hour.


    1. What’s the growth factor, the number by which you multiply the previous hour amount to get the next hour amount?

      2. 0.97
    2. Fill in the table and sketch the graph of concentration C(t) vs. time t.


      3. Time
      (hr)

      Concentration
      (µg/ml)

      0

      800

      1

      776

      2

      752.72

      3

      730.14

      4

      708.23

      5

      686.99

      6

      666.38

      7

      646.39

      8

      627

      9

      608.18

      10

      589.94

    3. Describe this decay pattern in both recursive-form and closed-form expressions.

      4. Recursive: C0 = 800; Cn = Cn–1 * 0.97

      Closed Form: C(t) = 800 (0.97)t
  1. Now, repeat the same calculations as in the previous problem, only allow the medicine to decay at 25% per hour.


    1. What is the growth factor in this case?

      2. 0.75
    2. Fill in the table, and sketch the graph of concentration vs. time.


      3. Time
      (hr)

      Concentration
      (µg/ml)

      0

      800

      1

      600

      2

      450

      3

      337.5

      4

      253.13

      5

      189.84

      6

      142.38

      7

      106.79

      8

      80.09

      9

      60.07

      10

      45.05

    3. What are the recursive and closed-form equations for this decay process?

      4. Recursive: C0 = 800; Cn = Cn–1 * 0.75

      Closed Form: C(t) = 800 (0.75)t
  1. Describe what the various values of the growth factor did for the table and graph of the three situations that were examined.

    2. In each case, the table values for concentration got smaller over time; the graph was a curve that started at 800, dropped toward the x-axis and curved in the direction of the x-axis. As k, the decay rate, got larger, the more quickly the graph curved towards the horizontal axis; the table values got smaller faster as well.
  1. Sometimes, the rate of dissipation (or any other decay process that behaves like this) is described in terms of its half-life, the time it takes for half of the amount to decay. Refer back to the work on the medicine that decays at the rate of 10% per hour.


    1. How many hours it will take for half of the original concentration to be used up? Explain how you got that answer.

      2. Between 6 and 7 hours. Look in the table for how long it takes the 800 mg/ml to decay down to 400 mg/ml.

    2. If the rate of dissipation is larger than 10%, will the half-life go up or down?

      3. It will go down, since the concentrations will get smaller faster.
  1. Let’s examine one of the ways in which you could determine the half-life for a substance, if you knew the rate at which it was being consumed. Let’s assume that the 800 mg/ml of medicine under study is used up in the body at the rate of 15% per hour.


    1. You could get the half-life (at least a "ball-park" figure) by extending the table until you have half the amount that you started with. Complete the table for this decay sequence, and determine the half-life:


      2. Time (t)

      0

      1

      2

      3

      4

      5

      6

      7

      Concentration (C)

      800

      680

      578

      491.3

      417.6

      355

      301.7

      256.5

      The half life would be somewhere between 4 and 5 hours, since that’s the interval in which the substance decreases to only 400 mg/ml.

    2. You could even narrow down the answer, if you figured out a growth factor over a different interval of time. Earlier, we calculated growth factors over two or five hours by (growth factor)2 and (growth factor)5. If we broke down each hour into tenths, what would the growth factor be for each of those intervals?

      3. It’s a solution to the equation x10 = 0.85 (the growth factor in this case).

      So, the new growth factor would be: » 0.9839
    3. Now, fill in the table for this same situation between t = 4 and t = 5 hours, using the growth factor you just calculated. What is your new estimate for the half-life?


      4. Time (t)

      4.0

      4.1

      4.2

      4.3

      4.4

      4.5

      4.6

      4.7

      4.8

      4.9

      5.0

      Concentration (C)

      417.6

      410.9

      404.3

      397.8

      391.3

      385

      378.8

      372.7

      366.7

      360.8

      355

      The half-life for this medicine will be between 4.2 and 4.3 hours.
  1. Another way to determine the half-life would be to use the calculator’s ability to graph functions. Enter the closed-form equation describing this decay sequence as your first function:

    2. y1 = 800(0.85)x,

    and enter the condition for determining when half the substance has decayed as your second function:

    y2 = 400.



    1. What are the WINDOW settings?

      2. Xmin = 0, Xmax = 6, Xscl = 1, Ymin = 0, Ymax = 1000, Yscl = 50
    2. Now, GRAPH and TRACE to the intersection point. What’s the half-life?

      3. Somewhere around 4.28 hours.
    3. Now, use the INTERSECT feature from the CALC menu. Select the first curve and press <ENTER>; select the second function and press <enter>; then move near the intersection point and press <ENTER> a third time. What’s the half-life?

      4. Somewhere around 4.265 hours.



Homework 3—Grrrrrrr-Rate!
   
  1. Let’s look at a table of values that describe a medication that starts out with a concentration level of 600 mg/ml and is used up at a rate of 8.00% per hour.


    2. Term

    Time
    (t) (hr)

    Concentration
    (C) (µg/ml)

    c0

    0

    600

    c1

    1

    552

    c2

    2

    508

    c3

    3

    467

    c4

    4

    430

    c5

    5

    396


    1. What does the ratio c1 / c0 equal?

      2. c1 / c0 = K = 0.92
    2. What does the ratio c3 / c2 equal?

      3. c3 / c2 = K = 0.92
    3. What does the ratio c4 / c3 equal?

      4. c4 / c3 = K = 0.92
    4. What do your answers for parts a)–c) have to do with the original problem?

      5. Reducing the concentration by 8% per hour means that r = –0.08 and K = 0.92. Calculating the ratios of successive terms verifies that the same number was used each time in the multiplying.
    5. What does the ratio c5 / c0 equal? What does that value have to do with the original problem?

      6. 0.6591. Successive multiplying by 0.92 for 5 hours is the same as multiplying by (0.92)5, which equals 0.6591.
    6. Why do you think the rate of dissipation is given the variable ‘k’?

      7. The concentration from one period to the next is found by multiplying the previous concentration by K.
  1. A certain medication decays in the body at the rate of 20.0% per hour. Suppose we start with an initial concentration of 360 mg/ml.


    1. Fill in the table of concentrations for the first 10 hours, then sketch the graph.


      2. Time
      (t) (hr)

      Concentration
      (C) (µg/ml)

      0

      360

      1

      288

      2

      230.4

      3

      184.3

      4

      147.5

      5

      118

      6

      94.4

      7

      75.5

      8

      60.4

      9

      48.3

      10

      38.7

    2. What is the growth factor for this decay sequence?

      3. 0.80
    3. Write the recursive and closed-form equations for this decay sequence.

      4. Recursive: C0 = 360 mg/ml. Ct = Ct–1 * 0.80

      Closed Form: C(t) = 360 (0.80)t
    4. Predict the concentration after 15 hours.

      5. 360 (0.80)15 » 12.67 mg/ml

  1. A certain type of medicine has a half-life of 9 hours. What is the rate of dissipation? (What percent of the amount decays in each hour?)

    2. If you assume that there is 100 µg/ml initially, the equation to solve becomes 50 = 100(x)9.
    50/100 = x9
    = x
    x = 0.926, which corresponds to a drop of around 7.4%.

  1. If a shot with 1200 mg of medicine is administered, and the kind of medicine used decays at a rate of 8.00% per hour, how many hours will it take for the concentration level to drop down to 200 mg/ml?

    2. Build a table, and find that it reaches 200 mg/ml sometime during the fourteenth hour.

    Graph y1 = 600(0.92)x and y2 = 200, and trace to the intersection point, and find that it takes around 13.2 hours to reach 200 mg/ml.

    Time
    (hr)(t)

    Concentration
    (C) (µg/ml)

    0

    600

    1

    552

    2

    508

    3

    467

    4

    430

    5

    395

    6

    364

    7

    335

    8

    308

    9

    283

    10

    261

    11

    240

    12

    221

    13

    203

    14

    187


  1. Find the rate of dissipation for a certain type of medicine if 640 mg are administered at the start, and 100 µg/ml are still remaining after 5 hours.

    2. The equation describing the problem is: 100 = 320 (k)5

    Divide by 320: 0.3125 = (k)5

    Take a 5th root: = k

    k » 0.792.
  1. A patient is given a prescription that reads as follows: Quantity 8, Dosage 500 mg, take 1 pill every 6 hours. The medicine is metabolized in the body at the rate of 3% per hour. Assume that the medicine doesn’t begin to decay until after the last pill is dissolved. Make a table and a graph of this growth and decay sequence.


    2. Time
    (t) (hr)

    Concentration
    (C) (µg/ml)

    0

    0

    6

    50

    12

    100

    18

    150

    24

    200

    30

    250

    36

    300

    42

    350

    48

    400

    54

    333.19

    60

    277.54

    66

    231.18

    72

    192.57

    78

    160.40

    84

    133.61

    90

    111.29




Activity 4—You Grow Some, You Decay Some

 = 3, 8, 10

Our first step at understanding the model that describes concentration over a period of time assumed that there was no rate of dissipation. This allowed us to "concentrate" on the effect of continually adding medicine, and allowed us to describe it mathematically. The second step was to focus our attention on the rate of decay, but it was also assumed that there was no medicine being added. We learned to describe how the body consumes an initial shot of medicine, but still leaves us the task of putting it together.

  1. Let’s begin by working with a situation in which a patient receives a shot of medicine in the amount of 1000 mg. The medicine decays at the rate of 5% per hour, but the patient is given a prescription for 1324.5-mg pills to be taken every 6 hours.


    1. Fill in the table with the amount present each hour. Then sketch the graph.


      2. Time
      (t) (hr)

      Concentration Loss From Decay
      (C) (µg/ml)

      Concentration Gain From Dose
      (µg/ml)

      Concentration Level
      (C) (µg/ml)

      0

      ---

      ---

      500

      1

      25

       

      475

      2

      23.75

       

      451.25

      3

      22.56

       

      428.69

      4

      21.43

       

      407.26

      5

      20.36

       

      386.90

      6

      19.35

      132.45

      500

      7

      25

       

      475

      8

      23.75

       

      451.25

      9

      22.56

       

      428.69

      10

      21.43

       

      407.26

      11

      20.36

       

      386.90

      12

      19.35

      132.45

      500

    2. There are three patterns in the table (or graph). Explain what is going on in each of them.

      3. First, there is the decay of the chemical over a 6-hour period, in which 5% is removed each hour. Second, there is the increase from taking the pill every 6 hours. Finally, there is the long-term pattern that will be a concentration level of 500 µg/ml every 6 hours until the patient runs out of pills.
    3. How could you determine that adding 1324.5 mg would just cancel out the effects of the dissipation of the chemical?

      4. Calculate 500*(0.95)6, which gives 367.55 µg/ml. That’s the concentration that’s left after six hours. Then subtract that from the original amount of 500 µg, to figure out how much to add (in order to restore it to the original value).
  1. The pattern discovered in the previous problem can be called an "equilibrium" state, since it maintains a balance between the dissipation of the chemical in the body and the increases caused by taking pills. Use the same situation as before, except change the prescription to take the pills every 4 hours instead.


    1. Fill in the table with the amount present each hour. Then, sketch the graph.


      2. Time
      (t) (hr)

      Concentration Loss From Decay
      (C) (µg/ml)

      Concentration Gain From Dose
      (µg/ml)

      Concentration Level
      (C) (µg/ml)

      0

      ---

      ---

      500

      1

      25

       

      475

      2

      23.75

       

      451.25

      3

      22.56

       

      428.69

      4

      21.43

      132.45

      539.71

      5

      26.99

       

      512.72

      6

      25.64

       

      487.08

      7

      24.35

       

      462.73

      8

      23.14

      132.45

      572.04

      9

      28.60

       

      543.44

      10

      27.17

       

      516.27

      11

      25.81

       

      490.46

      12

      24.52

      132.45

      598.39

       

       

       

       

    2. Does the result you found in part a) agree with your "intuition" about what should happen if you shorten Dt?

      3. Yes, because the amount contained in the pill just balanced out the effects of chemical dissipation over a 6-hour period. Not as much would dissipate in a 4-hour period, so you’re slowly going to increase the concentration over time.
    3. What would the dosage have to be, in order to achieve that "equilibrium" pattern again?

      4. 500 – 500(0.95)4 = 92.75 µg/ml, so the dosage would be 927.5 mg.
  1. Look over the work done in Question 2 again. Refer back to the table to answer the following questions:


    1. How much did the concentration go up during the first four-hour interval (from t = 0 to t = 4)? During the second four-hour interval? During the third four-hour interval?

      2. 39.71 µg/ml; 32.33 µg/ml; 26.35 µg/ml.

    2. If the pills continue to be taken at 4-hour intervals, does it seem likely that the concentration will continue to go up? Explain.

      3. It will continue to increase, but the amount of increase will get smaller over time. Eventually, it will reach a "top" value or upper limit on the sequence.
    3. Do you think the dissipation rate affects whether or not the concentration will continue to go up? Explain.

      4. If the dissipation rate is a fairly large percentage, so that the amount being added from taking pills is smaller than the amount being used up by the body, then the sequence won’t go up at all. And if the dissipation rate is really small, then the growth will seem like our original additive model.
  1. Work with the same situation as before, except change the prescription to take the pills every 12 hours instead.


    1. Fill in the table with the amount present every 2 hours. Then, sketch the graph.


      2. Time
      (t) (hr)

      Concentration Loss From Decay
      (C) (µg/ml)

      Concentration Gain From Dose
      (µg/ml)

      Concentration Level
      (C) (µg/ml)

      0

      ---

      ---

      500

      2

      48.75

       

      451.25

      4

      44.0

       

      407.25

      6

      39.71

       

      367.55

      8

      35.84

       

      331.71

      10

      32.34

       

      299.37

      12

      29.19

      132.45

      402.63

      14

      39.26

       

      363.37

      16

      35.43

       

      327.94

      18

      31.97

       

      295.97

      20

      28.86

       

      267.11

      22

      26.04

       

      241.07

      24

      23.50

      132.45

      350.02

      26

      34.13

       

      315.89

      28

      30.80

       

      285.09

      30

      27.79

       

      257.30

      32

      25.09

       

      232.21

      34

      22.64

       

      209.57

      36

      20.43

      132.45

      321.59

      38

      31.36

       

      290.23

      40

      28.29

       

      261.94

      42

      25.54

       

      236.40

      44

      23.05

       

      213.35

      46

      20.80

       

      192.55

      48

      18.77

      132.45

      306.23

    2. What would the dosage have to be, in order to achieve that "equilibrium" pattern again?

      3. 500 – 500(0.95)12 = 229.82 µg/ml, so the dosage would be 2289.2 mg.

    3. What would be the rate of dissipation over a 12-hour period?

      4. (0.95)12 » 0.54
    4. It’s nice to keep track of time in 12-hour intervals, since that determines the peak amounts. (Don’t forget that the concentration drops off during the 12 hours!) Fill in the table with the amount present after each time interval, and sketch the graph.


      5. Time
      (t) (hr)

      Concentration Loss From Decay
      (C) (µg/ml)

      Concentration Gain From Dose
      (µg/ml)

      Concentration Level
      (C) (µg/ml)

      0

      ---

      ---

      500

      12

      230

      132.45

      402.50

      24

      185.15

      132.45

      349.77

      36

      160.89

      132.45

      321.33

      48

      147.81

      132.45

      305.97

  1. Look over the work done in Question 4 again. Refer back to the table to answer the following questions:


    1. How much did the concentration go down during the first twelve-hour interval? During the second twelve-hour interval? During the third interval? During the fourth interval?

      2. 97.5 µg/ml; 52.73 µg/ml; 28.44 µg/ml; 15.36 µg/ml.
    2. If the pills continue to be taken at 12-hour intervals, does it seem likely that the concentration will continue to keep going down? Explain.

      3. It will continue to decrease, but the amount of decrease will get smaller over time. Eventually, it reaches a "bottom" value or lower limit on the sequence.
    3. Write a recursive formula to describe the same growth pattern found in the 12-hr. interval chart from Question 4.

      4. C0 = 500; Cn = Cn–1 – Cn–1*(0.95)12 + 132.45 or Cn = Cn–1*[1 – (0.95)12] + 132.45
    4. Now, use home-screen iteration to determine whether it "bottoms" out or goes to zero. If it does have a lower limit, find it!

      5. The value does have a lower limit. The calculator answer is: 245.11 µg/ml.



Homework 4—Stay Between the Lines!
   

We’re going to try and put together aspects of our model by specifying values for various parameters. Remember that the concentration must be above Cmin after the response time t0, and must stay below Cmax at all times. Each problem will involve a rate of dissipation (k), the option of an initial shot to elevate concentration levels (C0), and ask you to determine dosage (D) or time between doses (Dt).

  1. A doctor needs to treat a patient’s condition with a chemical having the following properties: the threshold level is 400 µg/ml, the tolerance level 750 µg/ml, and the rate of dissipation 4% per hour.


    1. Start by administering a shot that will bring the concentration of the chemical up to the tolerance level. How much chemical will be in the shot?

      2. 1500 mg
    2. Choose a convenient time period for the pills (6, 8 or 12 hours, for example). Then determine a dosage that will keep the concentration level at an equilibrium between the amount being consumed and the amount added from the pills taken.

      3. Answers will vary.

      If Dt = 6 hrs, then DC = 750 – 750 * (0.96)6 » 162.93 µg/ml, so D = 1629.3 mg.

      If Dt = 8 hrs, then DC = 750 – 750 * (0.96)8 » 208.96 µg/ml, so D = 2089.6 mg.

      If Dt = 12 hrs, then DC = 750 – 750 * (0.96)12 » 290.47 µg/ml, so D = 2904.7 mg.
  1. A similar situation occurs with a different patient chemistry in which the threshold level is 680 µg/ml, the tolerance level 800 µg/ml, and the dissipation rate given as having a half-life of 9 hours.


    1. Find the hourly rate at which the chemical is consumed.

      2. 50 = 100(x)9, so x = » 0.926
    2. What would be the dosage and time between doses in order to maintain an equilibrium level between the amount being used up and the amount contained in the pills being taken?

      3. DC = 800 – 800 * (0.96)2 » 114.02 µg/ml, so D = 1140.2 mg.
  1. A patient has a prescription for a chemical that has a rate of dissipation r = 6% per hour. A shot with 250 mg of the substance is delivered initially, and a prescription of pills is filled, with dosage 500 mg to be taken every 6 hours. Will the concentration get above a threshold level of 150 µg/ml? Will the concentration go over the tolerance level of 180 µg/ml?

    2. The recursive equation for this is: C0 = 250; Cn = Cn–1 * (0.946) + 500, and the sequence reaches almost 160 µg/ml, but doesn’t reach all the way to 180 µg/ml.
  1. Let’s examine a pattern that ends up with an upper limit to the values in the sequence. Suppose you start with C0 = 250 µg/ml, and the chemical has a dissipation rate of 8% per hour, and every 6 hours, you take a pill containing 3000 mg.


    1. Calculate the concentration values for the first 6 hours. Record in the table.


      2. Time (hr)

      0

      1

      2

      3

      4

      5

      6

      Concentration (µg/ml)

      250

      230

      211.6

      194.67

      179.1

      164.77

      451.59


    2. Calculate the concentration values in 6-hour intervals, after each pill has been taken. Record in the table.


      3. No. of 6-hr blocks

      0

      1

      2

      3

      4

      5

      6

      Concentration (µg/ml)

      250

      451.59

      573.82

      647.94

      692.88

      720.13

      736.66


    3. Write a recursive equation describing the values in this growth pattern right after taking the pills.

      4. C0 = 250. Cn = Cn–1 * (0.92)6 + 300
    4. Calculate the concentration values in 6-hour intervals, just before each pill has been taken. Record in the table.


      5. No. of 6-hr blocks

      0

      1

      2

      3

      4

      5

      6

      Concentration (µg/ml)

      ---

      151.59

      273.82

      347.94

      392.88

      420.13

      436.66


    5. Write a recursive equation describing the values in the growth pattern just before taking the pills.

      6. C0 = –50. Cn = (Cn–1 + 300)* (0.92)6
    6. What is the largest concentration possible in this medical treatment?

      7. 762.11 µg/ml.
  1. For a certain chemical, the threshold level is 480 µg/ml, while the tolerance level is 630 µg/ml. If the rate of dissipation is 3% per hour, can you find a prescription that will eventually keep its values "in between the lines" determined by the threshold and tolerance levels?

    2. Answers will vary. Possible answers include:

    C0 = 200 µg/ml; Dt = 4 hrs; D = 600 mg — maximum concentration 524 µg/ml

    C0 = 200 µg/ml; Dt = 6 hrs; D = 1000 mg — maximum concentration 599 µg/ml

    C0 = 400 µg/ml; Dt = 12 hrs; D = 1500 mg — maximum concentration 490 µg/ml




Activity 5—Close to a Final Form
   

 = 5, 11, 14, 15

In combining the breakdown of the medicine in the body over time with the pills taken from a prescription, we found that the growth tends to level off after a long time. If the dosage of the prescription doesn’t at least match the amount of medicine that is consumed during the time between doses, the total concentration drops off, but decays by smaller amounts over time and eventually reaches a lowest point. If the dosage exceeds the amount consumed during that time, the concentration goes up, but in smaller and smaller amounts until it finally reaches a highest concentration.

  1. Suppose the medicine decays at a rate of 6% per hour, and that the prescription dosage is 125 mg. The patient takes the first pill right in the pharmacy, and an additional pill every 8 hours after that.


    1. The concentration before taking the pill will be different from just after taking the pill. In the table, record the concentrations just before taking the pill, during the first two days:


      2. Pill Number (term)

      a0

      a1

      a2

      a3

      a4

      a5

      a6

      Time Elapsed (hrs)

      0

      8

      16

      24

      32

      40

      48

      Concentration (125 µg/ml)

      0

      76.20

      122.64

      150.96

      168.21

      178.73

      185.15


    2. Now, record the concentrations just after taking the pill, during the first two days:


      3. Pill Number (term)

      A0

      A1

      A2

      A3

      A4

      A5

      A6

      Time Elapsed (hrs)

      0

      8

      16

      24

      32

      40

      48

      Concentration (125 µg/ml)

      125

      201.20

      247.64

      275.96

      293.21

      303.73

      310.15


    3. What do you notice about the two sets of sequences above: {a0, a1, a2, a3, …., an} and {A0, A1, A2, A3, …., An}?

      4. For each pill taken, the concentration just before taking the pill is 125 µg/ml less than just after taking the pill.
    4. Continue calculating the two sequence values until they reach a maximum value. What are those values?

      5. Before taking pill: 195.16 µg/ml

      After taking pill: 320.16 µg/ml
    5. What factors in the problem contribute to those values being the maximum?

      6. The decay rate, the time interval between doses and the amount of dosage
  1. It would be nice to be able to predict the maximum value that a particular set of conditions would have. To derive the relationship, we need to examine calculations that have been performed.


    1. Write an expression that determines the concentration just before taking the second pill in terms of the dosage and the dissipation rate.

      2. 125(0.94)8 or 125(0.6096)
    2. Write the expression (not the value) that determines the concentration just after taking the second pill.

      3. 125(0.6096) + 125
    3. Here’s the key step for understanding: write the expression that determines the concentration just before taking the third pill. Simplify, but don’t evaluate it.

      4. [125(0.6096) + 125]*0.6096 = 125(0.6096)2 + 125(0.6096)
    4. Now, write the expression that determines the concentration just after taking the third pill. Simplify, but don’t evaluate it.

      5. 125(0.6096)2 + 125(0.6096) + 125
    5. There are three terms in the simplified expression. Why? What do the three terms have in common? What’s different about them? In relation to the problem, what does this mean?

      6. There are three terms because three pills have been taken at this point in time. They all have a factor of 125, the dosage of the pills. Two of the terms are being multiplied by a power of 0.6096, the dissipation rate over an 8-hour time interval. The exponents are the number of 8-hour time intervals that each pill has been decaying.
    6. Now, evaluate the expression to see if it matches the value you recorded in the table as A3.

      7. 125(0.6096)2 + 125(0.6096) + 125 » 247.65 µg/ml. Yes, it matches.
    7. Finally, simplify the expression you wrote as the previous answer by factoring out whatever is common to all the terms.

      8. 125 [(0.6096)2 + (0.6096) + 1]
  1. Expressions like the kind just discovered can be expressed in a more convenient form through algebraic manipulation.


    1. Multiply out the following expressions:

      2. i) (1 – x) (1 + x)

      1 – x2

      ii) (1 – x) (1 + x + x2)

      1 – x3
    2. Now, use the previous results to simplify the following expressions:

      3. i) (1 – x2)/(1 – x)

      1 + x

      ii) (1 – x3)/(1 – x)

      1 + x + x2

      It turns out that expressions which add up all the powers of a particular number can be evaluated according to the following closed-form expression:

      1 + x + x2 + x3 + x4 + x5 + ….. + xn = (1 – xn+1) / (1 – x)

      where x is the base, or the number that is being raised to the various powers, and n is the number of terms that are being considered for the pattern. In mathematics, this expression is referred to as a partial sum of a geometric sequence.

  1. Go back to the last answer of Question 2.


    1. What value for x would you want to use in the formula above?

      2. (0.94)8 » 0.6096

    2. What value for n would you want to use?

      3. 2
    3. What value would the expression (1 – xn + 1) / (1 – x) give?

      4. 1.9812
    4. What else needs to be done to calculate the concentration just after taking the third pill? What answer do you get? How does it compare to the original value for a3?

      5. Multiply by 125. You get 247.65 µg/ml, which is the same value as previously obtained in the table.

    5. Predict the concentration just after taking the fourth pill, according to the formula. Then compare it to the table value for A4.

      6. A4 = 125 (1 – 0.60963) / (1 – .6096) » 247.65 µg/ml. Within round-off error, it is the same value.

    6. Predict the concentration just after the twentieth pill.

      7. A20 = 125 (1 – 0.609619) / (1 – 0.6096) » 320.16 µg/ml.

  1. Suppose that a medicine is being administered that has a dissipation rate of 6% per hour, and a physician is considering a prescription of 2000 mg every 6 hours. Calculate the concentration that will build up in the patient after one week using the closed-form expression, then check the answer by home screen iteration.

    2. At an hourly rate of 6%, the amount remaining after 6 hours is around 69%.

    A28 = 200 (1 – 0.6928) / (1 – 0.69) » 645.1 µg/ml. Home-screen iteration yields the same answer.

  1. It gets even better! In the closed-form expression, one of the terms is affected by the number of pills taken. The others have a constant value for a particular problem.


    1. Using the information from the previous problem, calculate the expression (0.69)n as n (the number of pills taken) increases.


      2. No. of pills (n)

      0

      1

      2

      3

      4

      5

      6

      7

      8

      9

      10

      Value of (0.69)n

      1

      .690

      .476

      .329

      .227

      .156

      .108

      .074

      .051

      .035

      .024


    2. Describe the pattern in the values for the expression (0.69)n as n increases.

      3. It gets smaller and smaller, but it appears to be getting closer and closer to zero.
    3. How does that affect the value of the closed-form expression used to calculate the total concentration:

      4. (1 – (0.69)n+1) / (1 – 0.69) as n increases?

      As the term (0.69)n+1 gets smaller and smaller, the value in the numerator goes toward a value of 1 (1 – 0). The value in the denominator stays the same, since 1 – 0.69 is 0.31 (always has, and always will!). So the value of the closed-form expression becomes 1 / 0.31, or 3.226.
    4. How can we use the fact that the closed-form expression tends toward that value to determine the maximum concentration possible for the prescription given in Question 5?

      5. Multiply the concentration by the value from the closed-form expression. In this case, 200 * 3.226 » 645.2 µg/ml.

      Rather than forming a partial sum of a geometric sequence, like we did in Question 5, we are forming an infinite series of a geometric sequence. In this context, it would be like taking pills forever, and in mathematics, it is adding an infinite number of terms together which keep getting smaller and smaller, but never getting to zero (or below). Under these circumstances, the closed-form expression can be rewritten in this way:

      1 / (1 – r),

      where r is the percent of dissipation over the time interval between doses.

      As long as r < 1, as it is in these kinds of problems, you can use the formula and get an answer to something that should take forever to get. If r ³ 1, then the series gets infinitely big and you can’t apply the formula.

  1. A patient needs to take some medicine that has a dissipation rate of 8% per hour.


    1. Calculate the maximum concentration for a prescription of 1800 µg/ml taken every 8 hours.

      2. 180 / (1 – 0.928) » 180 / (1 – 0.513) = 180 / 0.487 = 369.6 µg/ml.

    2. Calculate the dosage that would level out the concentration at 600 µg/ml.

      3. Solving the equation: 600 = C0 / (1 – 0.513), you get C0 = 292.2 µg/ml. Therefore, the dosage has to be 2922 mg.

    3. How long would it take the prescription in part b) to reach a threshold concentration of 450 µg/ml?

      4. This is a partial sums problem now, so either use the closed-form expression or build a table. When you take the third pill, the concentration will rise as high as 519 µg/ml.

    4. How long will it take that prescription to stay above the threshold concentration of 450 µg/ml?

      5. Each time a pill is taken, the concentration peaks at a value that eventually reaches 600 µg/ml. But in between doses, the medicine is dissipating to roughly half of the concentration at the beginning of the time interval. So, this prescription will never stay above a threshold concentration of 450 µg/ml.




Assessment—Dosage Deliberations
   

In the last activity, a mathematical approach was developed for figuring out the maximum concentration from taking pills in a prescription. While the problem of determining prescriptions for medicine is complex, all of the components have now been analyzed. It’s time to put all the pieces together to this modeling problem.

  1. Consider a medicine that has a dissipation rate of 5% per hour, a medical condition that has a response time of 24 hours, and a patient with a threshold level of 200 µg/ml and a tolerance level of 450 µg/ml.


    1. If the time between doses is chosen to be 8 hours, what dosage amount would be needed to have the highest concentration possible approach the tolerance level?

      2. Solve the equation: 450 = D * 1 / (1 – 0.958); the dosage needed would have to be 1514.6 mg.

    2. Does this treatment keep the concentration above the threshold level by the end of the response time?

      3. Yes, since the concentration just before taking the fourth pill (after 24 hours) is 211.37 µg/ml.

    3. How would your prescription change if the response time were shorter?

      4. Possibly adjust the time interval (make it shorter) or provide an initial shot to elevate concentration levels more quickly.
  1. Let’s change our strategy and give an initial shot of 400 mg, so that the response time is not as much of a concern. Assume that since a shot is given, the patient won’t take any pills until the end of the first time interval.


    1. Keeping Dt = 8 hours and the same dosage as in the previous problem, how long does it take the treatment to keep the concentration above the threshold level?

      2. Approximately 18.5 hours
    2. In general, it would be nice to not have to worry about response time at all. For this particular situation, how much medicine should be in the initial shot? How did you calculate that concentration?

      3. 351.5 µg/ml. Since the dosage amount is 151.5 m g, just determine the shot amount by taking the threshold concentration and adding one dose (this is equivalent to having the patient take the first pill at time t = 0).

  1. Generalize the situation from Activity 5; make the prescription amount D, the time between doses Dt, and the dissipation rate k. What would be the formula (the closed-form expression) for the highest concentration possible with the prescription alone?

    2. Since it is an infinite geometric series, the highest concentration is given by the formula: Cmax = D / (1 – rDt)

  1. Explain how to use the results from Question 3 to determine the dosage for a patient having tolerance level Cmax for a medicine with dissipation rate r.

    2. Multiply Cmax * (1 – rDt), for a given time between doses to determine dosage.
  1. What might be wrong with the following thinking: start by specifying a time interval, and calculate the dosage that will establish the highest concentration level for the prescription to be at the threshold level?

    2. The dosage amount might not be available in stock quantities (not a serious problem since a pharmacist could fill an individual prescription order), but there is also constraints of being above the threshold level by the response time, and having to remain above that level during the treatment time.
  1. During one interval between doses, what is the relationship between the peak concentration (right after the pill is taken) and the lowest concentration (just before the next pill is taken)?

    2. The amount of the dosage, since it jumps up by that amount immediately after a pill is taken.
  1. After having taken a lot of pills, what is the relationship between the highest concentration possible for that prescription and the lowest concentration reached during a time interval?

    2. The amount of dissipation over the time interval is the difference between the highest and lowest concentrations, but after a large number of pills are taken, that difference approaches the dosage amount in value.
  1. Examine the following strategy: give an initial shot to raise the concentration up to the threshold level immediately, and establish the dosage amount as being the difference between the threshold and tolerance levels. Is this a good plan for treatment of the patient? Explain.

    2. It solves the problem of being at the threshold level by the response time, and the concentration level remains above the threshold level. But in order for the concentration to approach the tolerance level over time, the peak concentrations will go above the tolerance level.
  1. In the general case, we are using Cmin, Cmax, r, D and Dt to represent the variables and constraints to the problem of determining prescription amounts. How much medicine should be in an initial shot, so that the concentration always remains above the threshold level, and over time will gradually approach the tolerance level?

    2. D = Cmax * (1 – rDt). Make the initial shot have concentration Cmin + D.




Unit Summary

In this unit, the contextual problem of trying to determine a prescription to provide treatment for an illness was explored. Input variables for the problem included the amount of dosage and the time between doses. Parameters for modeling the problem included: the threshold level (the minimum concentration level to be effective), the tolerance level (the concentration at the maximum safety level), the rate of dissipation, whether to provide an initial dose prior to the prescription medication, the response time (the time at which treatment needed to be at some effective level) and the treatment deadline. The output for the model was a mathematical description of the concentration of a substance in the bloodstream over time. Assumptions for the model included that all patients were the same size, weight, age and sex, and that any pills would dissolve immediately.

In modeling the problem, several distinct phases of investigation transpired. At first, the assumption was that the substance either didn’t decay over time, or we just weren’t interested in looking at that portion of the problem. That opened up several possible strategies for delivering the medicine: a "super pill" or initial shot as interventions that only needed one treatment, and an additive model for administering the drug by oral prescriptions. The second phase corresponded to the assumption that there would be no oral prescriptions, but we would consider the decay of the concentration of medicine over time from body metabolism. It was also during this phase that a consideration for the threshold and tolerance levels was temporarily put aside. This led to a multiplicative model, in which the dissipation rate of the substance and the time between doses determined the number used in the repetitive multiplication.

Combining those two parts together led to a mixed growth pattern, where the amount of increase got smaller over time, and eventually the concentration reached a maximum level. The maximum level was affected by the dosage, the time between doses and the dissipation rate of the substance, and being able to predict just how those factors affected the maximum level was the key to a general strategy for determining prescriptions. Finally, the threshold and tolerance levels were brought back into the picture, so that the model could be more realistic.




Mathematical Summary

The mathematics of sequences played a big role in the development of the unit, and in solving the contextual problem. Arithmetic sequences were formed by the repetitive addition of concentration amounts from pills. The notation and graphical representation was explored, and recursive and closed-form descriptions developed. Disjoint patterns were introduced to handle the additional constraint caused by considering a response time.

Geometric sequences were formed by patterns caused from the gradual dissipation of medicine in the body over time. The rate of dissipation was found to be a percent of the amount of chemical present in the body, leading to repetitive multiplication by growth factors less than 1. It was more convenient to consider the decay rate over the time interval between doses, rather than on a "per-hour" basis. Problems involving the algebra of exponents and the handling of logarithms came up naturally, as did the properties of the graphs of these kind of exponential functions.

Mixed sequences, ones in which repetitive multiplication and addition are combined, came into consideration. Patterns of exponential decay were exhibited during each time interval between doses, but concentrations at both the beginning and the end of those time intervals showed gradual increases over time. But the amount of increase became smaller with each pill taken, and finally the concentrations leveled off.

Partial sums and infinite series were also explored, in trying to describe the total concentration in terms of the contributions of individual pills and dissipation over time of the various quantities. A closed-form expression was developed for predicting the actual concentration at the beginning or end of each time interval, as well as the theoretical maximum concentration possible by taking repetitive doses over time.




Key Concepts

Arithmetic Sequence: Pattern formed by repetitive addition.

Closed-Form Equation: Equations that allow you to find the value of one variable given the value of another variable. Equations of the form c = 2p + 4, y = 3x + 2, and x =2 + 5t are examples of closed-form equations.

Concentration: Amount of medicine relative to the amount of blood in the bloodstream.

C = (Amount of medicine) / (Volume of blood)

Decay: Pattern in which the amount of decrease is a percent of the amount present.

Dosage Amount: Amount of medicine contained in an individual pill or shot.

Geometric Sequence: Pattern formed by repetitive multiplication.

Half-Life: Time it takes for half of a substance undergoing a decay to be used up.

Infinite Series: Adding up all of the terms of a geometric sequence.

Initial Dose: First dose of a medicine to be administered as a sequence, or series of doses.

Partial Sum: Adding up some consecutive terms of a sequence.

Rate of Dissipation: Measures how much of the medicine is metabolized by the body during a time interval; it can be measured as a percent of the total amount per hour.

Recursive Equation: Equations that indicate the relationship between the current value of a variable based on a previous value of the same variable. These equations require a statement of the initial value.

xcurrent = 2.5 xprevious or xcurrent = xprevious + C

Response Time: Interval of time in which the treatment must reach the threshold level.

Threshold Level: Minimum concentration level for the medicine to be effective.

Time Interval Between Doses: Time interval between individual pills, usually expressed in terms of the number of pills to be taken each day.

Tolerance Level: Maximum concentration level that can be considered "safe" for use.

Treatment Deadline: Interval of time in which the treatment must be completed.



Solution to Short Modeling Practice



Solution for Modeling Bacteria Growth

A study of certain bacterial reproduction gives the cell counts shown here.



A

B

C

1

Time (hr)

Cell Count

2

0

597

3

2

893

4

4

1339

5

6

1995

6

8

2976

7

10

4433

8

12

6612

9

14

9865

10

16

14719

11

18

21956

12

20

32763

Graph Cell Count vs Time using a graphing calculator or spread sheet. What mathematical functions have a shape similar to this graph?

From the graph, carefully estimate the time required for the cell count to double from 5000 to 10,000. Carefully estimate the time required for the cell count to double from 10,000 to 20,000.

A good estimate of the doubling time from this graph is 3.46, but answers may vary. The number of periods, n, for any time, t, is found by dividing time by the time required for the cell count to double. n = t/3.46. The recursive exponential model is then

The cell count predicted using the model and values above are shown in the table. Note that the prediction for 20 hours is very close to the original data. If the prediction was not a close match, the estimate of the doubling time should be adjusted until there is a close match.


A

B

C

1

Time (hr)

Predicted

2

0

597

3

2

891.2101

4

4

1330.411

5

6

1986.057

6

8

2964.814

7

10

4425.917

8

12

6607.071

9

14

9863.13

10

16

14723.82

11

18

21979.93

12

20

32811.96

13

22

48982.15

14

24

73121.26

15

26

109156.5

This graph of the predicted cell count indicates that the count should reach 100,000 cells in slightly more than 25 hours.

Solutions to Practice and Review Problems

Exercise 1

    I = 3.5 (T + 1)(–0.64)



    a.

    T (min)

    I (in/hr)

    0

    3.5

    10

    0.75

    20

    0.50

    30

    0.39

    40

    0.33

    50

    0.28

    60

    0.25

    b) through d) The students’ graphs should appear generally as shown below.

    From the graph or the table, the infiltration rate when irrigation begins is 3.5 in/hr.

    From the graph, the infiltration rate will decrease to 0.3 in/hr between 40 and 50 minutes after irrigation begins.

Exercise 2

    Balancen = Deposit × (1 + i)n

    Balance10 = $100 × (1 + 0.06)10

    Balance10 = $179.08

    Balance20 = $100 × (1 + 0.06)20

    Balance20 = $320.71

Exercise 3

    L1 = 0.03 + 1.96

    L1 = 0.0367

    L1 – L2 = 0.0367 – 0.0233, or about 1.34%, which agrees.

Exercise 4

  1. After the second year, Number of stores = 32

    After the third year, Number of stores = 33

    After the nth year, Number of stores = 3n

  1. After the third year (n = 3), Number of stores = 33

    Number of stores = 27 stores

    After the sixth year (n = 6), Number of stores = 36

    Number of stores = 729 stores

  1. You could generate a graph of the formula and look for the year when the number of stores reaches the desired level. Or a simple "trial and error" approach could be used to find the year in which the total exceeds 50,000.

Exercise 5

  1. Number of bacteria = 2n

  2. Two hours are equal to 6 twenty-minute periods

    (120 min ÷ 20 min per period),

    Number of bacteria = 26

    Number of bacteria = 64

    Five hours are equal to 15 twenty-minute periods

    (300 min ÷ 20 min per period),

    Number of bacteria = 215

    Number of bacteria = 32,768

  1. A graph of the formula could be drawn, and the number of 20-minute periods elapsed where the population reaches one million noted. Another way is the "trial and error" method of simply evaluating the formula until a value exceeding 1 million is obtained. (As a matter of fact, after 20 periods, or almost 7 hours, the population exceeds one million.)

Exercise 6

Activity = 5000 decays per second ×

  1. Activity = 5000 decays per second ×

    Activity = 2500 decays per second

  1. Activity = 5000 decays per second ×

    Activity = 1502 decays per second (rounded)

  1. You could construct a graph of the declining activity versus days since the injection was given. Then read from the graph the time where the activity drops to 5% of 5000, or 250 decays per second. Or, a "trial and error" approach may be used to obtain an approximate time, trying successive values for t.

Exercise 7

    T = –0.48868252 + 19,873.14503 V – 218,614.5353 V2

    + 11,569,199.78 V3 – 264,917,531.4 V4 + 2,018,441.314 V5

    For V = 0.025 volt,

    T = –0.48868252 + 19,873.14503 (0.025) – 218,614.5353 (0.025)2 + 11,569,199.78 (0.025)3 – 264,917,531.4 (0.025)4 + 2,018,441.314 (0.025)5

    T = 437°C

    This problem requires more precision than is typically available in handheld calculators. In particular, the fourth-order term requires entering a number on the order of 200 million (nine whole-number digits), which is impossible for 8-digit calculators (without factoring).

    For students who like a challenge, have them rewrite the fourth-order coefficient as two factors (e.g., 264,917.5314 × 1000). This will permit them to work the problem. Since their calculators still allow them to enter only 8 digits, their answer won’t agree exactly with more precise devices. However, rounded to three significant digits, T = 437°C, the answer will be quite acceptable.

Exercise 8

  1. F = (0.49) (87.3) + (0.45) (6.2) – (6.36) (6.5) + 8.71 or about 12.9% body fat.

  2. Convert each of the units to cm by multiplying by 2.54.

    F = (0.49) (86.36) + (0.45) (12.7) – (6.36) (6.35) + 8.71 or about 16.4% body fat.

Exercise 9

  1. The y-intercept is 187.

  2. The slope is (187 – 153) / (0 – 40) = –0.85.

  3. H = –0.85A + 187

  4. H = (–0.85) (44) + 187 = 149.6 beats per minute.

Exercise 10

  1. After two weeks, or on day 14, it looks like there won’t be very much sap left in the tree! Thus, you would probably get essentially zero containers of sap on day 14, based on the fact that the numbers are getting quite small on day 9. Of course, this is based on the common-sense notion that the number of containers won’t "become negative"!

  2. The students’ graphs should appear generally as the graph shown here. They may have the points joined with a curve, but that is not really correct since there is no value for this relationship at Day Number 2.5, for example.


  3. The student’s table should appear generally as given here.


    4. Day Number

    Number of Containers

    Total Number of Containers

    1

    2.5

    2.5

    2

    2

    4.5

    3

    2

    6.5

    4

    1.5

    8.0

    5

    1

    9.0

    6

    1

    10.0

    7

    0.8

    10.8

    8

    0.7

    11.5

    9

    0.5

    12.0


  1. The student’s graph should appear generally as the graph shown above. As in Part a, after two weeks it is apparent that not much sap will be flowing in the tree. Thus the total yield from this tree is likely to be just barely more than 12 containers, perhaps 13 or 14—hardly worth the extra effort for those last 5 or 6 days, not to mention the possible damage to the tree.

    The graph shown above is a cumulative graph, indicating the accumulated output from the tree. The students should notice that, whereas the graph for Part b approached a constant value of zero (near the x-axis), the graph above approaches a constant value equal to the total output from the tree. The graph for Part b has a descending value (negative slope), while this graph has an increasing value (positive slope).

Exercise 11

  1. The student’s table should appear generally as shown here.


  2. You have put in $1200.00, while the bank has put in a total of $39.73 in interest


  3. Substitute D = $100 and i = 0.005, as suggested in the problem, and simplify.

    4. B = $100

    B = $20,000 (1.005n – 1)



    Month

    End-of-month Balance

    Interest Paid by Bank

    1

    100.00

    0.50

    2

    200.50

    1.00

    3

    301.50

    1.51

    5

    505.03

    2.53

    6

    607.56

    3.04

    7

    710.60

    3.55

    8

    814.15

    4.07

    9

    918.22

    4.59

    10

    1022.81

    5.11

    11

    1127.92

    5.64

    12

    1233.56

    6.17

    Some of your students may get "excited" because of the "$20,000" in the formula. However, the second factor (in parentheses) will always be a fraction much less than 1.

  1. To use the formula for a period of four years, n = 4 yr × 12 mo/yr = 48 mo.

    2. B = $20,000 (1.00548 – 1)

    B = $5409.78

Exercise 12

  1. The student’s ordered pairs should appear as shown below.

    2. { (1, 437), (2, 494), (3, 1200), (4, 2422), (5, 2790), (6, 3202), (7, 3266), (8, 3687) }

  1. The student’s graph should appear generally as shown here. We’ve drawn a straight line for simplicity, and added the extrapolation called for in Part c.

  1. Based on the extrapolation of our graph, it appears that the plant will be able to meet the schedule by the 15th week. The students may interpret their graphs differently, so you’ll have to examine their graphs and their conclusions.

Exercise 13

The equation c = 20t + 40 represents the relationship between the time worked (t) and the charge for the work (c).

Exercise 14

  1. (1,3)
    (2,7)
    (3,11)
    (4,15)
    (5,19)

  2. See chart.
  3. See chart.

  4. y = 4x – 1

  5. (6,23)
    (7,27)
    (8,31)

Exercise 15

One method is to substitute the given dosage available (1800 mg) and solve for w.

D = 0.1 w2 + 5w

1800 = 0.1 w2 + 5w

0.1 w2 + 5w – 1800 = 0

The values for w that satisfy the above equation can be found using the quadratic formula, where a = 0.1, b = 5, and c = –1800.

Of course, a negative weight is meaningless, so the correct answer is 111.5 kg, or about 245 lb.

The students could also graph the equation for D, as shown above, and read the graph to determine a weight of about 110 kg. While not yielding as accurate a result, this method is not incorrect and should be complimented as being "creative."

Exercise 16

Solution:

The current water usage is:

w0 = (71 gal/per/day)(97,300 per)

w0 = 6.9 × 106 gal/day

The water use will increase 0.095 per year so the ratio of one year's usage to the next is:

wi = wi–1(1.095)

The sequence for the next four years will be:

w1 = 7.6 × 106, w2 = 8.3 × 106, w3 = 9.1 × 106, w4 = 9.9 × 106

The water demand will exceed the present capacity in 5 years. w5 = 11 × 106 gallons per day.