Exponential Growth and Decay from a Data-Pairs Approach

header_expgrwthdcyMy last two posts discussed the mathematics of linear growth and decay. If you have not read those posts, you might find it helpful to read them before continuing. This post focuses on finding an exponential equation that expresses a relationship between two variables by first constructing a table of data-pairs to better understand the relationship and see the pattern in the relationship.

Most exponential growth/decay relationships involve a time variable t and the amount A of some quantity at time t. Amount could be the current value of an investment account, population of a city, remaining kilograms of radioactive material, assessed value of a truck, etc. The text box and observations below explain how and why the basic fundamental exponential growth/decay formula A = A0*bt/k works, and the role that the parameters A0, b, and k play in the equation. Periodic growth factor is another way to think of the base multiplier b.

txtbx1_expgrwthdcy

Some observations about A = A0*bt/k where b > 0:
• The point (0, A0) is the intercept on the vertical axis of the graph.
• Base multiplier b is a periodic growth or decay factor.
• If 0 < b < 1, the equation models exponential decay.
• If b > 1, the equation models exponential growth.
• Exponential growth/decay is about repeated multiplication by growth/decay factor b.
• A0 and any other point on the graph determines a unique exponential equation.
• If A0 is positive, the graph is above and asymptotic to the horizontal axis.
• If A0 is negative, the graph is below and asymptotic to the horizontal axis.

Most discussions about finding the equation of an exponential relationship don’t start by looking at data-pairs in a table. After only a couple of demonstrations of how to apply the data-pairs approach, students quickly develop the ability to find the three key parameters of an exponential growth/decay relationship. Exponential equations of the form A = A0*bt/k where base b is a rational number are much easier to comprehend than equations of the form A = A0*ekt where e is the irrational math constant = 2.718281828459045 . . .  . I will use four familiar math problems that involve an exponential relationship to illustrate the table data-pairs approach. In the comments section of this post, you will find an example that further clarifies my reason for expressing most exponential growth/decay equations as A = A0*bt/k where b is a rational number, and the reason that the solutions of population and radioactive growth/decay problems tend to be expressed in terms of base e only.

Problem 1: Consider a population of bacteria that is growing exponentially 50% every 4 hours and the current population is 60 bacteria. Let t = number of hours in the future and N = the number of bacteria after t hours.
(a) Find an equation that expresses N as a function of t.
(b) Find the population after 10 hours and 45 minutes ago.
(c) Express N as a function of t if the population is increasing 5% every 15 minutes.

The solution is given in the text box below. Problem solvers should carefully read the problem, create a table of data-pairs, determine the equation parameters, and then write the equation that models the problem situation. A companion exponential growth graph with a series of slope/rate triangles is provided to show the role that the equation parameters play in the relationship. Of course, the problem solver should always check the solution by using a computer graphing program to graph the equation.

txtbx2_expgrwthdcy

fig1_expgrwthdcy

Problem 2: Suppose a person invests $10,000 in a CD that will earn interest at 6%/year and interest is compounded monthly. Let t = the number of years in the future and V = the value of the investment after t years.
(a) Express V as a function of t.
(b) Find the value of the investment after 10 years and 20 years.
(c) Express V as a function of t if interest is compounded 360 times per year.

txtbx3_expgrwthdcy

Problem 3: The half-life of a radioactive substance equals the time it takes (20 days, 149 years, 5,700 years, etc.) for the substance to lose half its mass. Consider a radioactive substance with a half-life of 60 days that currently has a 100 kg mass. Let time t = the number of days in the future and A = the mass of the remaining substance in kg at time t. Refer to table and companion graph below.


(a) Find a formula that expresses A as a function of time t in days.
(b) Find the mass of the substance after 135 days.
(c) Find a formula for A(t) if the half-life = 6 hours instead of 60 days.

txtbx5_expgrwthdcyfig2_expgrwthdcy

Problem 4: The two exponential growth/decay graphs along with key points on the graphs are shown below.
(a) For graph A: Write an equation that expresses y as a function of x.
(b) For graph B: Write an equation that expresses y as a function of x.

fig3_expgrwthdcy

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Here are four exercises that you can give to your students. The graphs are a mixture of linear and exponential growth/decay graphs. Using the points on the graph, find the equation of the graph. If you wish, remind them that they should first create a table of data-pairs. Let them do the exercises with a partner and then check their answers by using a computer to graph the equations. We want to create a save environment in which kids feel free to experiment and check their answers for understanding. It’s OK to make a mistake, just fix it. If the first attempt to fix a mistake fails, so what? Try again. This is how real people learn to do anything that is worthwhile. The solutions are given at the end of this post.

fig4_expgrwthdcy

Comments:
• Consider the two mathematically equivalent equations below that model the population growth of a small town where t equals the number of years after 2010.

P = 5,200(1.08)t/4  and  P = 5,200e0.019240260t

The first equation immediately tells us the population of the town was 5,200 in 2010, and the population is increasing 8% every 4 years. The second equation tells us the population of the town was 5,200 in 2010, but by just inspecting the second equation, only God can figure out that the population is increasing 8% every four years. (Increasing 8% every 4 years is slightly less than increasing 2% every year.)

• It’s a snap to find the derivative of functions of the form y = Aekt. To find the derivative of functions of the form y = Abx/k where base b is a rational number requires a little more work. I suspect this is the reason that the solutions of population and radioactive grow/decay problems tend to be expressed in terms of base e only. From my point of view, this is not a sufficient reason to do so because converting an exponential function from one base to another base is a simple procedure. My free handout Logarithmic Base Conversion shows how to do this.

• All modern physicists know that the equations they discovered can only give us an approximation of how nature’s laws work. The brilliant physicist Richard Feynman, over and over again, stated this fundamental fact in his lectures and talks. In reference to problem (3) above, if we conducted an experiment with a radioactive material by measuring the remaining mass of the material at various points in time, we would find a discrepancy between the experimental results and the predicted results. No matter how accurately we measure mass and time, the errors can’t be taken out of the experimental observations. We can only say that the remaining mass of radioactive material at time t lies in an area of uncertainly which is the area under a probability distribution curve. This is why least-squares regression equations are used to describe the relationship between two variables.

• The formula for calculating the future value of an account after t years when interest is compounded continuously is FV = Pert where P = the principal and r = the annual interest rate expressed as a decimal. It’s impossible to express this relationship with a base that is a rational number. In a future post, I will give a derivation of this formula in a manner that does not require an understanding of concepts in calculus.

• I have used the handout, Introduction to Exponential Growth and Decay, with college algebra students, pre-calculus students, and as a review for more advanced students. To download the free student and teacher versions of the handout, go to mathteachersresource.com/instructional-content.html. There are other free handouts on properties of exponents, properties of logarithms, solving exponential/logarithmic equations, and logarithmic base conversion.

• All graphs in this post were created with my software, Basic Trig Functions. I designed the software to help teachers quickly make custom content for their classrooms. This software allows you to easily copy any graphic and then import it directly into a document (e.g. lesson plan, class handout, test) or further manipulate it in various graphic processing programs.

My next post will show how to solve Newton’s Law of Cooling problems without understanding differential calculus.

Solutions to exercises:
Graph A: y = -2x + 40
Graph B: y = 40*0.5x/2.5
Graph C: y = 20*1.5x/5
Graph D: y = x + 10

Teaching Slope and the Equation of a Line – Part 2

header_lingrwthpt2My last post discussed the mathematics of linear growth/decay from an analytic geometry point of view. If you have not read that post, you might find it helpful to read it before continuing. This post will focus on finding a linear equation that expresses a relationship between two variables by first creating a table of data-pairs. After your students inspect the table, encourage them to write the slope-intercept or the point-slope equation of a line that contains the data. Once the equation is found, it can be manipulated and used to answer a wide variety of questions. The general procedure for finding the linear relationship between independent variable I and dependent variable D is described below:

  • Set up a table of data values so that the independent variable I is in ascending order.
  • From the table, determine the steady rate of change between I and D which equals slope m.
  • From the table, determine equation constant b by finding D when I = 0.
  • The equation D = mI + b expresses the relationship between D and I.
  • The equation I = (D – b)/m expresses the inverse relationship between I and D.

If it is not possible to determine equation constant b from the table, pick any point (p, q) in the table. The equations D – q = m(I – p)  or D = mI – mp + q express the linear relationship. Of course, if the data pairs in the table don’t show a steady rate of change, then the relationship between the variables is not linear.

Most discussions about finding the equation of a linear relationship don’t start by looking at data-pairs in a table. From my past experience, the table-data-pair approach works because it enables students to immediately see and understand the relationship. I routinely use this approach when writing computer programs. It helps me better understand the problem and helps to clarify my thinking. I have selected four familiar linear relation type math problems to demonstrate the table data-pairs approach.

Problem 1: Brian’s electricity provider charges him $0.14 per kWh (kilowatt-hour) of electricity, plus a basic connection charge of $17.50 per month. Let n = the number of kilowatt-hours of electricity Brian used in a month and c = the amount charged for n kilowatt-hours of electricity.
(a) Find an equation that expresses the relationship between the variables n and c.
(b) If Brian used 861 kWh of electricity in July, how much was he billed for July?
(c) If Brian was billed $155.82 for August, how many kWh of electricity did he use in August?

Problem 1 Solution: Use the information stated in the problem to construct a data-pair-table similar to the table shown in the text box below. The table immediately tells us that the steady rate of change constant m = $0.14/kWh, and c = 17.50 when n = 0. Using the slope-intercept form of the equation of a line, the equations c = 0.14n + 17.50 and n = (c – 17.50)/0.14 follow. If Brian used 861 kWh of electricity in July, he was billed $138.04 for July. If Brian was billed $155.82 for August, he used 988 kWh of electricity in August. (Refer to the text box below.)

txtbx1_lingrwthpt2

Problem 2: Bubba’s doctor put Bubba on a strict diet that was designed to lose weight at a steady average rate of 10 pounds every 4 weeks. When Bubba started the diet, he weighed 405 pounds. Let n = the number of weeks that Bubba is on the diet and w = Bubba’s expected weight after n weeks on the diet. (Expected weight approximately equals the actual weight on the diet.)
(a) Find an equation that expresses the relationship between the variables n and w.
(b) Find Bubba’s expected weight after 1 year (52 weeks) on the diet.
(c) At one point in his diet, Bubba weighed 335 pounds. Approximately how many weeks was he on the diet?

Problem 2 Solution: Use the information stated in the problem to construct a table of data-pairs similar to the table shown in the text box below. The table tells us that the steady weight loss rate m = -10/4 = -2.5 pounds per week and w = 405 pounds when n = 0. Using the slope-intercept form, the equations w = -2.5n + 405 and n = -(w – 405)/2.5 = -w/2.5 + 162 follow. Bubba’s expected weight after one year of the diet = 275 pounds. If Bubba weighed 355 pounds at one point in his diet, he had been on the diet about 20 weeks. (Refer to the text box below.)

txtbx2a_lingrwthpt2

Problem 3: A partial handicap chart for a bowling league is shown in the text box below. Bowlers who average more than 210 pins per game receive no handicap, and bowlers who average 210 or less pins per game receive a handicap as indicated in the chart. The variable A represents the bowler’s current average, and H represents the bowler’s handicap.

txtbx3_lingrwthpt2
(a) Based on the data in the chart above, find a formula that expresses H as a function of A.
(b) Find a bowler’s handicap if his average score is 147. Round down to nearest integer.
(c) Find a bowler’s average score if he has a 100 pin handicap. Round down to nearest integer.

Problem 3 Solution: The table tells us that the steady rate of change constant m = 9/(-10) = -0.9. This tells us that the handicap drops 0.9 pins for every point increase in the bowler’s average score. Using the point-slope form of the equation of a line with the point (210, 9), the relationship between A and H can be expressed as H – 9 = -0.9(A – 210) or H = -0.9A + 198. Solving the equation for A, we have the equation A = (198 – H)/0.9. The handicap for a 147 bowler = -0.9*147 + 198 = 65.7 = 65 pins. If a bowler has a 100 pin handicap, his average bowling score = (198 – 100)/0.9 = 108.88889 = 108.

Problem 4: At 320F and 00C water freezes. At 2120F and 1000C water boils. Use the fact that the relationship between F and C is linear to find the following:
(a) Find an equation that expresses F as a function of C.
(b) Find an equation that expresses C as a function of F.
(c) Convert 680F (average room temperature) to degrees Celsius.
(d) Convert 98.60F (average body temperature) to degrees Celsius.

Problem 4 Solution: Use the information stated in the problem to construct a data-pair-table similar to the table shown in the text box below. The table immediately tells us that the steady rate of change m = 180/100 = 9/5 and F = 32 when C = 0. Using the slope-intercept form of the equation of a line, the equations F = 9C/5 + 32 and C = 5(F – 32)/9 follow. When F = 680F, C = 200C. When F = 98.60F, C = 370C.

txtbx4a_lingrwthpt2

Comments and Suggestions:

  • The sample problems in this post are stated so that the questions are very direct and specific. All variables in the problems are well defined. The problems that students will typically encounter are stated in a less direct manner. Understanding the problem is the first rule of problem solving. Students need to carefully read the problem for understanding, and then write a clear description of the variables in the problem. If they learn to do this step well, they will understand the problem they are trying to solve and have much better chance of becoming a successful problem solver. When I ask struggling students what the problem is about or what the problem is asking them to find, invariably they can’t give me a simple answer. Of course, every experienced math teacher or tutor understands what I’m talking about.
  • Initially present problems to students in very direct and specific manner. As they become more competent problem solvers, present problems in a less direct or general manner that is more representative of real-life problem solving scenarios.
  • Before presenting the lesson, review the basics of solving for a variable in an equation.
  • The manner in which the sample problem solutions are stated models how I would like my students to answer a question. Getting students to do this is not easy. Example: Refer to part (b) of problem 2. The response that I want from a student requires him/her to write a complete sentence. Requiring only a short response like “275 pounds” does not help students become better writers. At what age can we realistically expect students to respond with a complete sentence? I don’t know.
  • Checking answers is another principle of problem solving. I tell my students that I always check my answer in order to catch a mistake. When writing this post, I caught a simple arithmetic error (and was able to fix it before you caught it).
  • Remind more mathematically mature students that, in many cases, the equation we write to model a given problem situation is only an approximate model. In reference to problem (2) above, people on a diet won’t lose the exact same amount of weight in equal time periods. This is why least-squares regression equations are used to describe the relationship between two variables. Statisticians are forced to use least-squares regression equations to describe the relationship between two or more variables, at a specific level of confidence, because the statistical data was obtained from a random sample of some population.
  • I use the handout Linear Growth and Decay Models with my beginning algebra students, and as a review for more advanced students. To download the free student and teacher versions of the handout, go to mathteachersresource.com/instructional-content.html.

Teaching Slope and the Equation of a Line – Part 1

top_lingrwthpt1The mathematics of linear growth/decay, exponential growth/decay, inverse variation, and joint variation relationships are some of the most important core concepts that high school students can learn. With these concepts mastered, students will have the necessary foundation to better comprehend more advanced work in mathematics and science. Newton’s famous inverse square law of physics, F = km1m2/d2, is an example of a joint and inverse variation relationship. I once told a class of computer programming students that knowing how to find the equation that expresses the relationship between two loop control variables is one of the more important skills they can learn. When writing computer programs, I create functions that map math coordinates to computer screen coordinates and vice versa. All of these functions are just applications of linear growth/decay.

This post discusses the mathematics of linear growth/decay from an analytic geometry point of view and how I teach these concepts to my students. The statements that I make about the slope of a line, tangent of an angle, slopes of parallel lines, slopes of perpendicular lines, and equations of lines are familiar to many. If there shall be any novelty in this post, it will be in the manner in which the content is presented. Many text books treat some of these relationships as a simple mathematical fact of life. At the teachable moment, it’s nice to be able to give a curious student a straight forward explanation of why a not so simple fact is true.

All graphs in this post were created with my program Basic Trig Functions. Equations can be entered in any format. It’s not necessary to express equations as explicit functions of x. Go to mathteachersresource.com to view multiple screen shots of the program’s modules. Click the ‘learn more’ button in the TRIGONOMETRIC FUNCTIONS section. Teachers will find useful comments at the bottom of each screen shot.

I will start by discussing the concept of slope of a line. I tell my students that the slope of a line describes the steepness of a line. It would be easy to fall off a house roof that has a slope > 1. The slopes or grades of train tracks are gentle, and are usually well under 10% (1/10). A train on a track with a 1% (1/100) slope can only pull half or less of the load that it can pull on a zero slope track. After driving down a road with a 6% (3/50) grade, you fully appreciate the message on the warning sign. A vertical line is infinitely steep, and therefore we say that its slope is undefined or it has no slope. No slope lines are NOT the same as zero slope lines. As I discussed in a previous post, the brilliant math educator, Zalman Usiskin, had a profound effect on my understanding of slope and how I teach these concepts to my students. Diagram (1) below illustrates the concept of slope and the text box gives the definition of a slope of a line. Most students immediately understand that positive slope lines describe linear growth and negative slope lines describe linear decay.

diagram1_lingrwthpt1txtbx1_lingrwthpt1

Most students can guess that parallel lines have equal slopes. Based on theorems from Euclidean geometry, diagram (2) below illustrates why parallel lines have equal slopes. Line AF is a transversal for parallel lines AB and DE, and angles BAC and EDF are corresponding angles. Of course, each of two parallel horizontal lines have slopes equal to zero, and each of two parallel vertical lines have undefined slopes.

diagram2a_lingrwthpt1Almost all students can’t guess that the product of the slopes of two oblique perpendicular lines equals -1 (Oblique lines are not horizontal or vertical). Many text books treat this relationship as a simple mathematical fact of life. The reason that this relationship is true can be explained in terms of theorems in Euclidean geometry, or in terms of a trigonometric identity. A theorem in Euclidean geometry states the following: “The altitude to the hypotenuse of a right triangle is the mean proportional between the segments into which it divides the hypotenuse.” Diagram (3) below shows the setup for showing why the product of slopes rule works for perpendicular lines. The text box below gives the formal explanation why this relationship is true. It is sufficient to demonstrate that this relationship is true for perpendicular lines that intersect at (0, 0) because sliding the intersection point to any point in the x-y coordinate plane preserves this relationship. Although the second demonstration is like using a sledge hammer to drive a tack, it’s an interesting way to demonstrate the relationship.

diagram3_lingrwthpt1txtbx2_lingrwthpt1

Diagram (4) below shows the setup for deriving the slope-intercept form and the point-slope form of the equation of a line, and the text box below gives the derivation of the equations.

diagram4_lingrwthpt1txtbx3_lingrwthpt1

The equation of a line comes in six flavors which are described in the text box below. When you know slope m and y-intercept (0, b) of a line, the slope-intercept equation is the flavor of choice. If you know the slope m and one point (x1, y1) on the line, find the point-slope equation of the line, and then you can convert it to the slope-intercept form of the equation. The standard equation Ax + By = C where the constants A, B, and C are relatively prime integers, no common factors, is useful when setting up to solve a system of linear equations. Some definitions of the standard form of an equation require the constant A to be a nonnegative integer.

txtbx4_lingrwthpt1

Now for a short quiz to give to your students as a fun exercise with slope and equations of lines. Diagram (5) below shows the graphs of six linear equations. Have your students find the equation of each graph. If the line is oblique (not horizontal or vertical), write the equation in slope-intercept form. The answers are given at the end of this post. If you decide to take the quiz now, promise me that you will not take a peek.

diagram5_lingrwthpt1

Comments and Suggestions:

  • Similar to the graphs above, give students the graphs of linear equations and require them to find the equations of the graphs in different formats. This exercise is a good way to improve algebraic manipulation skills. Let them test their equations by entering the equations into a computer graphing program that allows them to enter equations in any format. If their equation is incorrect, they immediately see why and fix it. They now have a reason to do some algebra, and the computer becomes an experimental tool.
  • Learning to find the equation of a line in different formats is not so easy for many students. Remind them that most students initially make many errors and mistakes, and they are no different. There is no shame in that. They should think of the exercises as good mental gymnastics. It just takes time to sort things out.
  • I use the handout Introduction to Slope and Equation of a Line (student version) with my beginning algebra students and as a review for more advanced students. You can also download the teacher version of the handout. To download other free math handouts and exercises in a variety of topics for your classroom, go to mathteachersresource.com/instructional-content.html.

My next post will focus on setting up a table of data values, recognizing a linear relationship pattern in the table, finding an equation that expresses the relationship between two variables, and using the equation to make predictions. A major goal, of course, should be to get students to the point where they really understand that the slope of a line is a unit rate and be able to give a simple interpretation of what the slope of a line means in a given problem situation.

Quiz Solutions:
Graph A:  y = 6
Graph B:  y = -3/4x + 3
Graph C:  x = -7
Graph D:  y = 3x + 18
Graph E:  y = -2x + 16
Graph F:  y = 4/3x – 8