# Inverses of Functions and Relations

My previous post discussed the mathematical concepts of function and relation. Because the content of this post heavily depends on an understanding of the ideas presented in that post, you may find it helpful to read it before continuing.

The concept of the inverse of a relation is a natural extension of the important concept of a relation. The central idea is that an inverse relation is about reversing a relationship by exchanging variables, reversing/undoing an operation, or reversing/undoing a series of operations in a specific order. The following five questions and situations illustrate how a person uses the concept of the inverse of a relation to solve a problem.

(1) If we know a formula to convert Fahrenheit temperatures to Celsius, what formula converts Celsius temperatures to Fahrenheit?

(2) If we know a formula that tells us how to calculate the area of a circle from its radius, what formula will tell us how to calculate the radius of the circle from its area?

(3) A diner in a restaurant uses the restaurant’s menu function in inverse mode to determine what food items on the menu he/she can afford.

(4) A criminal investigator uses the one-to-one function that matches people with DNA molecules in inverse mode to match a sample of DNA molecules with a criminal.

(5) When solving for the sides and angles of a triangle, a trig student uses the inverse trig functions on his/her calculator to find the measure of an angle that has a specific trig function value.

The purpose of this post is to discuss inverse functions and relations when the matching rule is given by an x-y variable equation where both the domain and range is a subset of real numbers. These concepts will be discussed from algebraic and geometric points of view.

I will begin by looking at inverses of functions and relations from a geometric point of view. The two text boxes below summarize the geometric relationships between a relation and the inverse of a relation. The companion graphs illustrate the geometric relationships described in the text boxes. Notice that exchanging the variables in an equation gives us the equation of the inverse relation. These observations, of course, follow from the definition of the inverse of a relation, midpoint formula, definition of slope, and the fact that the product of the slopes of two perpendicular lines equals -1.

The text box below shows examples of elementary functions and the corresponding inverse relation which may or may not be a function. Notice that the inverse of the functions y = x2 and y = |x| are relations, but not functions since y = x2 and y = |x| are not one-to-one functions. As a reminder, the symbol √(x) means take the positive square root of x, and positive real numbers have a positive square root and a negative square root. Also note that the function y = Sin(x) is not one-to-one, and therefore the inverse relation is not a function. Calculators get around this problem by restricting the range of the function Sin-1(x) to values that range from –π/2 to π/2.

The next part explains how I teach the inverse of trig functions y = Sin(x) and y = Cos(x). Initially, students struggle with the definitions of the inverse trig functions. Consider the equations listed in the edit box and graphs below. Because the trig functions are periodic, there are infinitely many solutions for each equation. Because the calculator keys Cos-1(x) and Sin-1(x) are function keys, the calculator should display only one of the infinitely possible output values. When x ranges from 0 to π, Cos(x) is one-to-one in adjacent quadrants I and II, and all possible output values of Cos(x) from -1 to 1 can be generated in quadrants I and II. Therefore Cos-1(x) is a function if the output is restricted to range values from 0 to π radians. When x ranges from – π/2 to π/2, Sin(x) is one-to-one in adjacent quadrants I and IV, and all possible output values of Sin(x) from -1 to 1 can be generated in quadrants I and IV. Therefore Sin-1(x) is a function if the output is restricted to range values from – π/2 to π/2 radians.  I have my trig students find six solutions of simple trig equations.  Example: Find six angles β in degrees in quadrant III, 3 positive and 3 negative, such that Cos(β) = -0.951056516. Round solutions to the nearest tenth of a degree.

I will conclude this post my showing you how I teach my students to find the inverse of a function when the function is composed of basic functions. The steps in the algorithm involve applying inverse operations in the reverse order of the order of operation rules. Exercises of this type reinforce concepts and are a good way to practice algebra skills. If you want to add some rigor to your course, have students check their solution by showing f(f-1(x)) = f -1(f(x)) = x. I remind students that an initial equation like x = y/(3y – 4) is an equation of the inverse relation, but it’s not expressed as a function of x. When a relationship is expressed as a function of x, we can graph the relation with a graphing utility. This is one of the reasons that we teach kids to solve an equation for a given variable. Sometimes I tell students to rearrange the equation for some variable because it makes more sense to them.

Useful tools from Math Teacher’s Resource:

•   The graphs in my posts are created with my software, Basic Trig Functions. I think that you will find it very useful for teaching mathematical concepts in your classroom and developing custom instructional content. Relations can be entered as an explicitly defined function of x, an explicitly defined function of y, or as an implicitly defined x-y variable relation. Check it out at mathteachersresource.com/trigonometry.

•   There are a wide variety of free handouts that teachers can use to create lessons or give to students as a handy reference handout. Among these handouts are Inverse Relations and Functions, Even and Odd Functions, and Relations and Functions Introduction handouts. Go to mathteachersresource.com/instructional-content to download MTR handouts. All content is available for immediate download. No sign-up required; no strings attached!

•   Some readers wanted to know the equation of the lead graph in my previous post. The equation of the graph is Cos(x) + Cos(y) >= 0.4 where both x and y range from -15 to 15. In view of the fact that Cos(x) is an even function, it should be no surprise that the graph has symmetry with respect to the x-axis, the y-axis, and the origin.

•   The equation of the strange graph at the end of my previous post is 2xSin(3x) + 2y <= 3yCos(x + 2y) + 1. If you are skeptical, here are six solutions that you can plug into the equation to verify that the equation really does have solutions that satisfy the equality relationship. Just make sure that your calculator is in radian angle mode.

(-5.4, 5.195 577 636)

(5.5, 5.976 946 313)

(8.680 865 276, -5.2)

(-6.8, -6.786 215 284)

(0.578 827 17, -3)

(0.051 781 64, 5.8)

# Functions, Relations, and One-To-One Functions

Other than the concept of a number, I can’t think of a more important, pervasive, useful, and unifying mathematical concept than that of function and relation.

In a previous post, I explained how I introduce the concepts of functions and relations to my students in a concrete manner by using examples that illustrate how people use these concepts every day without realizing it. I have found this approach to be effective with beginning and advanced students. One of the important ideas that I want the reader to get from that post is that functions and relations are much more than some equation or formula such as y = f(x) = 2Sin(3x) – x,  y = g(x) = (x2-9)/x2 or |x| + |y| = 10. An equation is just one of many different types of matching rules of a relation. When a mathematician or scientist writes an equation, the equation is just the matching rule of some relation that tells us how to match domain elements with range elements. When a child is learning how to add two whole numbers, the child is learning how to use the addition function to match a pair of whole numbers with a single whole number. A diner in a restaurant uses the restaurant’s menu function to match food items with prices, determine what food items are in the restaurant’s domain, and the price range of the restaurant’s food items.

The purpose of this post is to discuss functions and relations when the matching rule is given by an x-y variable equation where both the domain and range is a subset of real numbers. Equations give us an algebraic description of the relation, and the graph of a relation gives us a geometric description of the relation.

I will begin by explaining the difference between a function and relation where the x-variable represents all possible numerical values in the domain of the relation, and the y-variable represents all possible numerical values in the range of the relation. All functions are relations, but not every relation is a function. For a relation to be a function of x, every value of x in the domain of the relation is matched with only one y value. The text box and the companion graph below give examples of relations where y is not a function of x. An ordered pair of real numbers (p, q) on the graph of a relation tells us that p matches with q. In each relation below, there is at least one x-value that matches with two or more different y-values which in turn tells us that there is a vertical line that intersects the graph of the relation in two or more points. All of these relations fail the vertical line test and therefore they are not functions of x. When the graph of a relation is given, it’s easy to tell whether or not the relation is also a function; just apply the vertical line test.

I will now discuss the symbols that are used with functions. For beginning students, these symbols are very abstract. If a student never learns the meaning of these symbols, the student will never learn calculus. When we write an equation of the form y = f(x), we are saying that the variable y is an explicitly defined function of x such that every input value of the function matches with exactly one output value. The y variable represents an output value of the function, and is said to be a dependent variable because its value depends on the value of the x variable. The x variable represents an input value of the function, and is said to be an independent variable because its value can be any freely selected value in the domain of the function. The mathematical expression for the symbol f(x) tells us how to calculate the output value of the function for every x in the domain of the function. I routinely tell my students that the symbols y and f(x) represent the same thing, and the point (x, y) = (x, f(x)) tells us how high above or below the x-axis the point is.

Now let’s take a look at one-to-one functions. A function y = f(x) is one-to-one if and only if f(a) ≠ f(b) whenever a ≠ b. In other words, the output values of one-to-one functions are always different if the input values are different. A restaurant menu function is not one-to-one because different food items map to the same price. The Social Security function that matches people with valid social security numbers is one-to-one because two different people are always matched to different valid social security numbers. If a function is not one-to-one, there are two different numbers p and q in the domain such that f(p) = f(q) which in turn tells us there is a horizontal line that intersects the graph of the function in two or more points. When the graph of a relation is given, it’s easy to tell whether or not the relation is a one-to-one function. First apply the vertical line test to see if the relation is a function. If the relation is a function, apply the horizontal line test to see if the function is one-to-one.

The text box and companion graphs below describe a variety of explicitly defined functions of x. Notice that all of the one-to-one functions are either strictly increasing or strictly decreasing functions. If a function is even, it follows from the definition of an even function that the function is not a one-to-one function.

The text box below shows how the concept of a one-to-one function can be used to solve a logarithmic equation. Step 3 in the solution follows from the fact that all Log functions are one-to-one. This type of reasoning is definitely a step up in mathematical maturity and sophistication for most students. Initially, some students think that both sides of the equation in step 2 were divided by Log. Of course, experienced math teachers know what I’m talking about. This reminds me of an old math joke. Question: What is sin x / n? Answer: sin x / n = six = 6.

Useful tools from Math Teacher’s Resource

I will conclude this post by showing you the graph of a relation that only myself and some of my students have seen. The equation for this relation has nothing to do with anything. It’s just a somewhat random inequality that popped in my head. I show this relation to students to remind them that the graph of a mathematical relation can be any set of points in the x-y coordinate plane.

# Comparing Circular Trig Functions with Hyperbolic Trig Functions

This post does a geometric compare and contrast of the circular functions Cos(x) and Sin(x) with the hyperbolic trig functions Cosh(x) and Sinh(x). I will do this by showing points of the form (Cos(t), Sin(t)) are points on the unit circle, and by showing points of the form (Cosh(t), Sinh(t)) are points on the unit hyperbola.

I will start the discussion by doing a quick review of radian angle measure and the definitions of the circular trigonometric functions Cos(θ), Sin(θ), and Tan(θ) where the input variable θ is the radian measure of a central angle of a circle with center at (0, 0) and radius equal to r. The text box below gives the definitions of the circular functions Cos(θ), Sin(θ), and Tan(θ) along with the radian measure of an angle and the formula for the area of a sector of a circle for angle θ. The formula for the area of a sector of a circle follows from the definition of radian angle measure and the fact that the area of a circular sector is proportional to the length of the arc that subtends the central angle.

The graph below illustrates the circular trig relationships in the text box above when r = 25 units, θ = α = 1.2870 radians and θ = β = -2.8578 radians. α, β, arc lengths and areas of circular sectors are rounded to 4 decimal places.

When the circular trig functions are defined in terms of the unit circle, r = 1, the input variable of the circular trig functions can be treated as an angle with angle measure = t radians, arc length = t length units, or time = t time units. This makes it possible to model periodic motion and periodic processes with circular trig functions. The text box below gives the definitions of Cos(t), Sin(t), and Tan(t) in terms of the unit circle. The unit circle graph illustrates the unit circle relationships when t = 2.58 units and t = -1.6 units. Notice that the area of the circular sector of arc length t = |t|/2 units2.

It’s now time to take a look at the unit hyperbola and the hyperbolic trig functions x = Cosh(t) and y = Sinh(t). My previous post discussed the derivation, the graphs, and some identities for hyperbolic trig functions. If you have not read that post, I strongly suggest that you read it before continuing. The text box below summarizes the key points that are pertinent to this discussion. The graph illustrates the unit hyperbola relationships when t = 2.2 and t = -1.6. An understanding of integral calculus is required to understand the derivation of some hyperbolic relationships.

If you are interested in Einstein’s theory of special relatively, you will find The Geometry of Special Relatively by Tevian Dray a must read. The author shows an elegant way to express the Lorentz transformations in terms of Cosh(β) and Sinh(β) where β is implicitly defined by the equation Tanh(β) = v/c where v is some constant velocity and c is the speed of light. To my amazement, I learned that the Lorentz transformations are just hyperbolic rotations!

Dray’s book inspired me to think more deeply about the circular trig functions, hyperbolic trig functions, and the difference between the two geometries. Dray warns the reader that hyperbola geometry should not be confused with hyperbolic geometry which is the curved geometry of the two-dimensional unit hyperboloid. I will conclude this post with some of my personal observations about circular trig functions and hyperbolic trig functions, and finish with a wonderful example from Dray’s book that really clarified for me the difference between Euclidean geometry and hyperbola geometry.

The diagram below shows a right triangle in hyperbola geometry. The results are astonishing when you consider the definitions of Cos(θ), Sin(θ) and Tan(θ) for Euclidean right triangles. Of course, the triangle below would be impossible in Euclidean geometry, but possible in hyperbola geometry. Just think of the diagram as a clever way to picture time and space quantities in spacetime physics. In hyperbola geometry, a triangle is a right triangle if and only if the lengths of the sides satisfy the Pythagorean Theorem of hyperbola geometry, a2 – b2 = c2 where a = c*Cosh(β), b = c*Sinh(β), and β is an angle of the triangle. From the formulas for the inverse hyperbolic trig functions, it follows that β = arcCosh(5/4) = arcSinh(3/4) = arcTanh(3/5) = 0.693147181. Notice that the hypotenuse of the hyperbolic right triangle is not the longest side.

# Hyperbolic Functions Cosh(x), Sinh(x) and Tanh(x)

This post discusses how the function f(x) = ex is used to create the hyperbolic trig functions Cosh(x), Sinh(x), and Tanh(x). Trig students immediately recognize the remarkable similarity between identities for the functions Cos(x), Sin(x), and Tan(x), and identities for the functions Cosh(x), Sinh(x), and Tanh(x). The hyperbolic trig functions have many important applications in many branches of mathematics and science. A couple of great examples are provided later in this post. I find these functions fun and interesting to play with, and I continue to find new ways of looking at and understanding these functions.

I will start the discussion by defining hyperbolic trig functions Cosh(x), Sinh(x), and Tanh(x) in terms of the functions y = f(x) = ex / 2 and y = f(-x) = e-x / 2 which are neither even nor odd. My last post discussed some of the properties and characteristics of even and odd functions. I concluded the post by showing how to create an even function and an odd function from any function which is not necessarily even or odd. If you have not read that post, you may want to read it before continuing. In view of the results obtained in my previous post, it follows that Cos(x) and Cosh(x) are even functions, and Sin(x), Sinh(x), Tan(x), and Tanh(x) are odd functions. The text box below gives the definitions of the three main hyperbolic trig functions. Graphs A, B, C, and D show the graphs of f(x), f(-x), Cos(x), Cosh(x), Sin(x), Sinh(x), and Tanh(x). As a reminder, the functions Cos(x), Sin(x), and Tan(x) are periodic, but the functions Cosh(x), Sinh(x), and Tanh(x) are not.

The text box below gives a comparison of some standard trigonometric identities and hyperbolic trig identities. Go online and check out the other hyperbolic trig identities. You will be amazed. The other day I found out that the derivative of an even function is an odd function, and the derivative of an odd function is an even function. Being able to understand and recognize even functions, odd functions, one-to-one functions, and the inverse of a function gives a student a whole new level of mathematical maturity and sophistication.

The textbox below shows the infinite Taylor series expansion of the functions Cos(x), Cosh(x), Sin(x), and Sinh(x). It’s interesting to see how close and yet very different the infinite series expansions of the functions are. Notice that the Taylor series expansion of Cos(x) and Cosh(x) are sums and differences of even functions! Also notice that the Taylor series expansion of Sin(x) and Sinh(x) are sums and differences of odd functions! The function ex is the sum of even and odd functions, and therefore it’s neither even nor odd.

I find the infinite series expansion of the inverse functions for the circular trig functions and the hyperbolic trig functions very interesting. The similarities are striking. One can deduce whether or not the inverse of a function is an even or odd function by just doing a simple inspection the infinite series expansion of the function.

In doing research for this post, I discovered an interesting relationship between a catenary curve and a parabolic curve. Imagine a piece of chain, rope, or cable that is hanging from its endpoints to form a U-shaped curve. Galileo (1564 – 1642) thought that this U-shaped curve was parabolic. In 1691, the mathematicians Leibniz, Huygens and Johann Bernoulli showed that the U-shaped curve is described by the hyperbolic cosine function. Freely-hanging electric power cables, silk threads on a spider’s web, or suspension bridge cables have the U-shaped catenary curve. The Gateway Arch in St. Louis, Missouri is said to be an inverted catenary. All catenary curves are the result of sliding, stretching, rotating, or reflecting the graph of y = Cosh(x). Shown below are the equations and graphs of three curves. The results speak for themselves.

My next post will do a geometric compare and contrast of the circular functions Cos(x) and Sin(x) with the hyperbolic trig functions Cosh(x) and Sinh(x). I will do this by defining Cos(x) and Sin(x) in terms of the unit circle, and by defining Cosh(x) and Sinh(x) in terms of the unit hyperbola.

I will end this post my showing you the graph of the first eight terms of a Fourier approximation of a square sine wave. You will notice that the graph is the graph of an odd function and the Fourier approximation is the sum of eight odd functions. Those even and odd functions are everywhere!

# Even and Odd Functions

The concepts of even and odd functions are usually introduced in advanced high school algebra, college algebra, trigonometry, or precalculus courses. Trig students learn how to apply the concepts of even and odd functions to simplify trigonometric expressions. Calculus students learn how to apply the concepts of even and odd functions to simplify the calculation of a definite integral. When students understand and can recognize even and odd functions, they are usually amazed how often these functions appear in application problems.

This post will discuss even and odd functions from both an algebraic and geometric point of view. Readers are encouraged to download my free handout Even and Odd Functions which is a handy reference that teachers can give to students.

I will start the discussion by describing even functions. The text box below gives a description of an even function from several points of view. Graph A illustrates what we mean when we say that a graph has symmetry with respect to the y-axis.

The text box below gives a description of an odd function from several points of view. Graph B illustrates what we mean when we say that a graph has symmetry with respect to the origin.

The text box below gives some basic observations about even and odd functions. My free handout Even and Odd Functions gives a more in depth list of the important properties of these functions.

The text box below shows the infinite Taylor series expansion of the function y = Cos(x). Graph C shows the graph of y = Cos(x) and the graph of the first five terms of the Taylor series expansion of Cos(x). Notice that the Taylor series expansion of Cos(x) is the sum and difference of even functions!

The text box below shows the infinite Taylor series expansion of the function y = Sin(x). Graph D shows the graph of y = Sin(x) and the graph of the first five terms of the Taylor series expansion of Sin(x). Notice that the Taylor series expansion of Sin(x) is the sum and difference of odd functions!

I will end this post by showing you how to create an even or odd function from any function y = f(x) that’s not necessarily even or odd. The text box below shows how and why this can be done. Graph E shows the results of creating an even function and an odd function from the function y = f(x) = 0.25(x – 4)2 – 3Sin(x – 4) – 5. This result explains why the hyperbolic functions cosh(x) and sinh(x) are even and odd functions respectively.

More useful tools from Math Teacher’s Resource

•   The graphs in this post were created with my software, Basic Trig Functions. I think that you will find it very useful for teaching mathematical concepts in your classroom and developing custom instructional content. Check it out at mathteachersresource.com/trigonometry.

•   In addition to the Even and Odd Functions Handout linked in this post, Math Teacher’s Resource offers a wide variety of free math handouts, lessons, and exercises available at mathteachersresource.com/instructional-content. All content is available for immediate download. No sign-up required; no strings attached!

# Modeling Limited Population Growth with the Logistic Function

Because of limits on food, living space, disease, current technology, war, and other factors, most populations have limited growth as opposed to unlimited exponential growth which is modeled by the classic exponential growth equation P = P0bt/k. A limited growth population starts growing almost exponentially, but reaches a critical point in time where its growth rate slows, and the population starts to asymptotically approach an upper limit as time increases. There are several models that are used to describe limited growth of a population.

In this post, I will discuss the logistic function which was used by the Belgian mathematician Pierre Francois Verhulst (1804-1849) to study limited population growth. The logistic function also has applications in artificial neural networks, biology, chemistry, demography, ecology, economics, biomathematics, geoscience, mathematical psychology, sociology, political science, probability, and statistics.

The two text boxes below describes the key parameters and relationships between the parameters of a logistic function. Graph (A) shows a typical logistic function curve and how equation parameters can be calculated from known characteristics of the population. If pmax, p0, and tc are known, then a, b, and k can be calculated. Likewise, if a, b, and k are known, then pmax, p0, and tc can be calculated. Keep in mind that all limited growth models can only give us a good approximation of a population value at some point in time.

In most cases, the key parameters of a logistic equation are unknown, but an observed set of data-pairs is known. The least-squares logistic equation of a data set is the best of all possible logistic equations that describes the relationship between the data-pair variables. Best possible equation means that the sum of the squared errors (difference between observed value and predicted value) is minimized. Modern graphing calculators have the capability of findings a least-squares equation for a variety of models such as linear, quadratic, cubic, quartic, sinusoidal, log, exponential, and logistic. When given a logistic type data set, I will use a graphing calculator to find the least-squares logistic equation of the data set, and then calculate various characteristics and properties of the resulting logistic model. I will now take a look at two problems that illustrate how the logistic function can be used to describe limited population growth.

Problem 1 solution: Use math software to do a scatter plot of the data, find the least-squares logistic equation p = 12.0121 / (1 + 10.6694e-0.023856x) of the data set, and then do the appropriate calculations. Refer to graph (B) below. From graph (B) we see that the world population growth rate started to slow in 1999, and the upper limit of the world population is about 12 billion. Keep in mind that this least-squares equation is our current best description of world population growth. Future unknowable events will alter this model.

Problem 2: The logistic function N(t) = 3,600 / (1 + 29.4e-0.2t ) models the spread of a disease in a town. N(t) = the total number of people infected at time t, and t = the number of days after the first reported infections.

(a) How many people were initially infected?

(b) How many people were infected after 10 days and after 30 days?

(c) When did the rate of infection start to slow?

(d) What is the upper limit of the number of infected people?

Problem 2 solution: Use math software to graph the equation, and then do the appropriate calculations. Refer to graph (C) below.

(a) About 118 people were initially infected.

(b) After 10 and 30 days, 723 and 3,355 were infected.

(c) About day 17, the infection rate started to slow.

(d) The upper limit of the number of people infected = 3,600.

• It’s fun and interesting to experiment with different logistic function parameters. Experimentation always gives a better learning experience.
• With my graphing calculator, it took about 8 seconds to compute the parameters of a logistic equation. This is an indication of the complexity of the algorithms for computing the parameters of a least-squares equation. I tell my students that they should be ever thankful that they have access to such wonderful computation tools.
• Computer math software allows students to focus on math concepts, and not get lost in gory computational details. This is why graphing calculators have revolutionized the way we teach statistics. Just getting the ‘answer’ is no longer sufficient. Students must be able to interpret and explain the meaning of the answer in the context of the problem.

# Derivation of Continuous Compound Interest Formula without Calculus

Jacob Bernoulli, 1654-1705

My students, like most people, like money and find the topic of compound interest interesting. After completing a unit on simple, compound and continuous compound interest, one of my students told me that math is useful and interesting after all.

This post will discuss the derivation of the formula for the future value of an investment when interest is compounded continuously, FV = Pert. No prior understanding of the limit concept in calculus is required. I will be using the limit concept, but I will give an informal intuitive explanation of the limit concept as it comes up in the discussion. A recent post discussed an approach for deriving an equation that models exponential growth/decay. Problem (2) in that post showed the derivation of the compound interest formula FV = P(1 + r/k)kt where FV = the future value of the investment account, P = principle or one time lump-sum investment, r = annual percent rate of return expressed as a decimal, k = the number of times per year interest is compounded, and time t = the number of years the principal is invested.

Before I can get to the derivation of the equation FV = Pert, I need to explain what continuous compound interest means. Let’s consider an investment where P = \$10,000, average annual rate of return = 7% = 0.07, and the investment collects interest over a period of 20 years. I adopted the standard banking convention rule that 1 year = 360 days. (Whether we use 365 or 360 days in a year makes no significant difference. Apparently banks like 30-day months.) The text box below shows how increasing the number of times per year interest is compounded affects the future value of an investment.

Students immediately notice that there is a point where it makes no difference how often interest is compounded, and they completely understand the difference between simple interest and compound interest. I tell them that the future value of the \$10,000 investment, \$40,552.00 in this example, represents the upper limit of one’s greed. When interest is compounded more times per year (k approaches infinity), and interest is compounded over smaller and smaller time intervals; say every second, every microsecond, or continuously. No matter what the principal is or the annual interest rate, there is always an upper limit of the future value of an investment, and the upper limit is reached when interest is compounded continuously.

In 1683 in the course of his study of continuous compound interest, Jacob Bernoulli (1654-1705) wanted to find the number that was the limiting value of the expression (1+1/n)^n as n approaches infinity. This is the first time that a number is defined as the limiting value of an expression. Bernoulli determined that this special number is bounded and lies between 2 and 3. In 1748 Leonard Euler (pronounced Oil-er) (1707-1783) published a document in which he named this special number e. He showed that e is the limiting value of the expression (1 + 1/n)n as n approaches infinity, and is approximately equal to 2.718281828459045235. He also gave another definition of e as the limiting value of the infinite sum 1 + 1/1! + 1/2! + 1/3! + . . . . Euler is generally given credit as the first to prove e is an irrational number.

To help you better understand the definition of the irrational number e, I will start by comparing the graphs of functions of the form y = (1 + 1/k)x where k is a fixed constant and the graph of the function y = (1 + 1/x)x. Refer to graphs (A) and (B) and the companion text box below. A quantity that approaches infinity means the quantity gets bigger and bigger without any upper boundary. A quantity that approaches a fixed constant means the quantity gets infinitely close to the fixed constant.

The purpose of the above graphs and the comments in the text box is to demonstrate that a subtle difference in the expressions (1 + 1/k)x and (1 + 1/x)x results in far different limiting values as x approaches ∞. The key result needed in the derivation of the continuous compound interest formula is the fact that e = limiting value of (1 + 1/x)x as x approaches ∞ when x is any positive real number. Considering that the expression (1 + 1/n)n is a rational number for every positive integer n, it is astonishing that the expression (1 + 1/n)n approaches an irrational number as n approaches ∞. I can now show you the derivation of the continuous compound interest formula FV = Pert.

• When I did the calculations for compounding every minute and compounding every second with my graphing calculator, I got results that were slightly different than the expected results. When I used double floating point precision real numbers in a computer program, program output agreed with the expected results. We need to constantly remind ourselves that calculator or computer calculations of expressions that involve very large numbers, or require a large number of iterations to arrive at a solution, results may be slightly different than the expected or theoretical value.

• Using problems similar to the examples in this post, I show my students how compound interest works and what continuous compounding of interest means. I have them enter the expressions into their graphing calculator as the lesson progresses. This gives them practice using their calculator and they gain a better understanding and appreciation of what compound interest is all about. They are astonished when I show them \$10,000*e.07*20 = \$40,552.00.

• For a class of curious or advanced students, it’s not wasted class time to show the derivation of the continuous compound interest formula. Less advanced students are usually content with learning how to use the formula. My handout, Basic Financial Formulas, provides an overview of useful financial formulas that you can use in your classroom.

• The derivation of the continuous compound interest formula is a great opportunity to expose advanced high school algebra, college algebra and pre-calculus students to the limit concept in calculus.

• As mentioned earlier, very term of the sequence an = (1 + 1/n)n is a rational number, but the sequence itself converges to the irrational number e. Most calculus students find this very counterintuitive. What a great opportunity to launch a discussion of any number of related math concepts!

• The constants 0, 1, π, e, and i where i2 = -1 are the five most important constants in mathematics because they are widely used in equations that describe relationships in all branches of mathematics and science. The equation eπi + 1 = 0, which is due to Leonhard Euler, is one of the most interesting and intriguing equations in mathematics. Euler used the symbol e for the irrational constant, and in his honor, e is named Euler’s number.

• Both Bernoulli and Euler were prolific mathematical giants. Much of what is routinely used in mathematics and science can be traced back to the work of these two great men. L’Hospital’s Rule in calculus is due to Bernoulli, not L’Hospital. L’Hospital published the rule, but Bernoulli discovered the rule and gave it, for a fee, to L’Hospital.

Because of limits on food, living space, disease, existing technology, war, and other factors, most populations have limited growth as opposed to unlimited exponential growth which is modeled by the classic exponential growth equation P = P0bt/k. A limited growth population starts growing almost exponentially, but it reaches a critical point in time where its growth rate slows, and the population starts to exponentially and asymptotically approach an upper limit. There are several models that are used to describe limited growth of a population. In my next post, I will discuss the logistic function which was used by the Belgium mathematician Pierre Francois Verhulst (1804-1849) to study limited population growth. The logistic function also has applications in artificial neural networks, biology, chemistry, demography, ecology, economics, biomathematics, geoscience, mathematical psychology, sociology, political science, probability, and statistics.

# Solving Newton’s Law of Cooling/Heating Problems without Differential Calculus

Sir Isaac Newton (portrait by Godfrey Kneller, 1689)

My last post discussed how to find an exponential growth/decay equation that expresses a relationship between two variables by first constructing a table of data-pairs to better understand and derive the fundamental grow/decay equation A = A0*bt/k. Because the content of this post depends on the concepts developed in my last post, I strongly suggest that you read that post before continuing.

This post shows how to solve Newton’s law of cooling and heating problems without any understanding of differential calculus, which makes this post different from descriptions found in differential calculus text books. Newton’s Law of Cooling describes the relationship between the temperature of an object and time t when the object is placed in an environment where the ambient (or surrounding) temperature is maintained at a constant temperature. Newton’s law of cooling and heating is described as follows:

(a) If the initial temperature of the object equals the ambient temperature, the temperature of the object remains constant as time t increases.

(b) If the initial temperature of the object is greater than the ambient temperature, the object cools and its temperature exponentially and asymptotically approaches the ambient temperature as time t increases.

(c) If the initial temperature of the object is less than the ambient temperature, the object heats up and its temperature exponentially and asymptotically approaches the ambient temperature as time t increases.

I will use two familiar cooling/heating problems to illustrate how the table data-pair approach can be applied to solve a Newton’s law of cooling or heating problem. The key step in solving a cooling/heating problem is to carefully read the problem and then apply what Newton tells us about cooling and heating to create a rough sketch of the growth/decay graph of the model with key points labeled. Even if you don’t know the equation of the graph, the rough sketch will enable you to determine the parameters of the growth/decay equation. From this rough sketch, recognize that the graph is just the result of a vertical translation of an exponential decay graph in the form A = A0*bt/k. (In view of what Newton tells us about cooling and heating, the rough graph makes perfect sense to students.)

Problem 1: A pot of boiling soup is put into a sink filled with cold water. The temperature of the soup was 1000 C when it was first put into the sink. By adding ice and stirring the water, the temperature of the water was maintained at a constant temperature of 50 C. If the temperature of the soup was 600 C after 10 minutes, how many minutes will it take for the temperature of the soup to reach a room temperature of 200 C?

Solution: Refer to graphs A and B below where x = time t in minutes and y = the temperature of the soup in degrees Celsius. Similar to graph A, first draw a rough sketch of the model with key points labeled. Recognizing that graph A is just the result of a 5 unit vertical translation of an exponential decay graph, use the information from the first rough sketch to draw a rough sketch of the exponential decay graph with key points labeled, similar to graph B. Now use the key points on the sketch of graph B to find the equation of graph B, and then apply the equation transformation rules to find the equation of graph A. To find out how many minutes it will take for the temperature of the soup to reach 200 C, use a computer graphing program to find the intersection point of the graphs y = 20 and y = 95(55/95)x/10 + 5. Graph A tells us the temperature of the soup equals 200 C, when time t = 33.77 minutes or about 34 minutes.

Problem 2: A 400 F roast is put into an oven that is set to bake at 3500 F. After 2 hours, the temperature of the roast is 1250 F. The roast is considered done when its internal temperature reaches 1650 F. How many hours well it take to cook the roast?

Solution: Refer to graphs C and D below where x = time t in minutes and y = the temperature of the roast in degrees Fahrenheit. The strategy is to first draw a rough sketch of the model with key points labeled; similar to graph C below. Recognizing that graph C is the result of a 350 unit vertical translation of an exponential decay graph that was reflected over the x-axis, use the information from the sketch of graph C to draw a rough sketch of the flipped exponential decay graph with key points labeled, similar to graph D. Now use the key points on the sketch of graph D to find the equation of Graph D, and then apply the equation transformation rules to find the equation of graph C. To find out how many hours it will take to cook the roast, use a computer graphing program to find the intersection point of the graphs y = 165 and y = -310(225/310)x/2 + 350. Graph C tells us that it will take 3.22 hours or about 3 hours and 13 minutes to cook the roast.

Here are four exercises that you can give to your students. The solutions are provided. (See my third comment below.) You or your students shouldn’t be too disappointed if you fail to correctly solve all four exercises on your first attempt.

Exercise 1: When first removed from an oven and placed in a 700 F room to cool, the temperature of a cake was 1800 F. Three minutes later the temperature of the cake dropped to 1600 F.

(a) What is the temperature of the cake after 20 minutes? (A: 98.870 F or about 990 F)
(b) How many minutes will take for the cake to cool to 900 F? (A: 25.49 minutes or about 26 minutes)

Exercise 2: The temperature of a very small metal bar was 300 C when it was dropped into a large barrel of hot water having a 750 C temperature. After 1 second, the temperature of the bar was 310 C.

(a) How long will it take for the temperature of the bar to reach 700 C? (A: 97.77 seconds or about 98 seconds)
(b) How long will it take for the temperature of the bar to reach 740 C? (A: 169.39 seconds or about 170 seconds)

Exercise 3: Find the equation of graph A below.  (A: y = 40(1/5)x/10 + 30)

Exercise 4: Find the equation of graph B below. (A: y = -90(2/3)x/5 + 160)

• In the two sample problems above, the final step in the solution involved finding the intersection point of two graphs. This gives us the solution from a geometric point of view. The solution from an algebraic point of view involves log functions which would enable you to find the solution faster. As I mentioned in previous posts, whenever possible, solutions to problems should be understood from both an algebraic and geometric point of view.

• Solving exponential growth and decay problems naturally leads to a need to understand logarithms and log functions.

• All modern physicists know that the equations they discovered can only give us an approximation of how nature’s laws work. In reference to problem (2) above, if we conducted an experiment with a roast by measuring its internal temperature at various points in time, we would find a discrepancy between the experimental results and the predicted results. No matter how accurately we measure the internal temperature of the roast and time, the errors can’t be taken out of the experimental observations. We can only say that the interval temperate of the roast at some specific point in time 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.

• I have used the handout Newton’s Law of Cooling with college algebra and pre-calculus students, and with more advanced students that I tutor. To download the free student and teacher versions of the handout, go to mathteachersresource.com/instructional-content. There are other free handouts on properties of exponents, properties of logarithms, solving exponential/logarithmic equations, and logarithmic base conversion.

• Using the approach presented in my last post and this post, I believe it’s possible to teach how to solve exponential growth/decay problems to younger mathematically capable students. From my own experience, students find these types of problems interesting and practical.

• All graphs in this post were created with my program Basic Trig Functions. I designed the program to make it easy for teachers to create content for their own courses.

My next post will discuss the derivation of the formula for the future value of an investment when interest is compound continuously, FV = Pert. The post will assume that the reader has no understanding of the limit concept in calculus.

# Exponential Growth and Decay from a Data-Pairs Approach

My 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.

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.

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.

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.

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.

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.

• 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

My 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.)

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.)

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.

(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.