3.2 Exponential Models and Applications

Introduction

Imagine that you have a delicious mug of creamy, homemade hot cocoa that is, unfortunately, much too hot to drink. How long should you leave it on your kitchen counter so that it cools but doesn't lose all of its comforting warmth?

We're not ready to answer that question yet, but it should not surprise you that it has to do with exponential functions! In this section, we will develop variations of the basic exponential function that are used in applications as varied as radiation therapy, population growth and, of course, the cooling of hot cocoa.

But before we discuss these specific examples, we must first learn how to interpret the base of an exponential function and incorporate a few standard transformations.

Growth Factor and Growth Rate

You learned in the previous section that the base $b$ of an exponential function $f(x)=b^{x}$ is a growth factor. For instance, if $f(x)=1.08^{x}$, then the growth factor is $b=1.08$, meaning the function is multiplied by $1.08$ every time $x$ increases by $1$.

We can also say that each new value of $f(x)=1.08^{x}$ is $8 \text{%}$ larger than the current one. This second interpretation focuses on the growth rate $r=b-1$ rather than the growth factor $b$.

In real world applications the growth rate is given much more frequently than the growth factor. For instance, business commonly report that sales are up by $3 \text{%}$ or down by $5 \text{%}$, instead of announcing growth factors of $1.03$ or $0.95$.

For this reason, it is often useful to make the substitution $b = 1+r$ and write exponential functions in the form

[ f(x)=(1+r)^{x} ]

Note that growth occurs when $r>0$ and decay occurs if $ -1<r<0$.

Initial Value

Another way to change the basic exponential function $f(x)=b^{x}$ is to apply a vertical stretch. You might recall that a vertical stretch is caused by multiplying the function by some constant $a$, which has the effect of multiplying each y-value by $a$. Use the figure below to explore the how exponential functions respond to vertical stretches. Make a note of your observations before continuing.

You should have noticed that multiplying an exponential function $f(x)=b^{x}$ by a constant $a$ causes the y-intercept to change from $(0, 1)$ to $(0, a)$. For instance, if $a=4$, then the y-intercept will be $(0, 4)$.

This is actually a consequence of the rule of exponents that says $b^{0}=1$, so $f(0)=a \cdot b^{0} = a$ for any valid base.

The Standard Exponential Model

By including a vertical stretch with the exponential form $f(x)=(1+r)^{x}$, we obtain a more general function

[ f(x)=a(1+r)^{x} ]

that starts with an initial value of $a$ and grows (or decays) at a constant rate $r$. We will call this equation the standard exponential model. All other exponential functions are variations of this equation.

Let's revisit an earlier example. In the final stage of the 2010 FIFA World Cup, 16 teams competed for the world championship. After each round, one-half of the teams were eliminated and eventually Spain emerged as the champion.

We now have the tools needed to model this scenario with the standard exponential model $f(x)=a(1+r)^{x}$.

Since the final stage of the World Cup starts with 16 teams, we know that $a=16$. And if $50\text{%}$ the teams are eliminated after each round, then $r=-0.5$ and the function

[ f(x)=16 (1-0.5)^{x}=16(0.5)^{x} ]

models the number of teams left after $x$ rounds have been played.

Half-Life

Our first variation of the standard model will enable us to work with applications that involve half lives and doubling times.

The half-life of a function is the amount of time needed for the value to be reduced by half. Every exponential decay function has a half-life.

For instance, the exponential decay function in this graph models the value of a sports car. The initial value of $60,000 is cut in half every 3 years, so we would say it has a half-life of 3 years.

To make the standard model fit this situation we must stretch the $50\text{%}$ decay out over a three-year period. In other words, we must stretch the function horizontally by a factor of $3$. So we can model the value of this car with the function

[ f(x)=60000(1-0.5)^{x/3} ]

where $a=60000$ is the initial value and $r=-0.5$ is the growth rate.

Doubling Time

On the other hand, the doubling time of a function is the amount of time needed for the value to double. Every exponential growth function has a doubling time.

Consider, for example, the value of a collectible toy car like the one shown in this graph. We say that the value of this toy car has a doubling time of 5 years, since its value doubles every 5 years.

To make the standard model fit this situation we must stretch the $100\text{%}$ growth out over a five-year period. In other words, we need to stretch the function horizontally by a factor of $5$. This means we can model the value of this toy car with the function

[ f(x)=2(1+1.00)^{x/5} ]

where $a=2$ is the initial value and $r=1.00$ is the growth rate.

Recognize Doubling Time and half-life Models

Let's generalize these last two examples. If an initial quantity $a$ grows exponentially at a rate of $r$ every time $x$ increases by $c$, then that growth can be modeled by an exponential function of the form

[ f(x)=a (1+r)^{x/c} ]

For instance, the function $f(x)=18(1+0.64)^{x/7}$ could model an initial quantity of $18$ that increases by $64\text{%}$ every time $x$ increases by $7$.

This new form $f(x)=a (1+r)^{x/c}$ gives us greater flexibility when modeling realistic scenarios. Of particular interest is the case when $r=1$, which gives us the doubling time model, and the case where $r=-1/2$, which is called the half-life model.

Doubling Time Model	Half-Life Model
$f(x)=a(2)^{x/c}$	$f(x)=a\left(\frac{1}{2}\right)^{x/c}$

As an example, when technology innovator, Intel, produced its first microprocessor in 1971, it contained 2300 transistors. Intel's founder, Gordon Moore, predicted that the number of transistors on a computer chip would double every two years. His prediction is known as Moore's law and can be modeled by

[ f(x)=2300(2)^{x/2} ]

where $x$ is the number of years since 1971.

In 2011, Intel's Sandy Bridge processor had more than 2.2 billion transistors. How does this compare with Moore's prediction for 2011? Since 2011 is 40 years after 1971, we evaluate the above function with $x=40$

Calculator Evaluation of $f(40) = 2300(2)^{40/2}$
2300*2^(40/2) = 2411724800

which is very close to the actual value. When evaluating expressions like this on a calculator, make sure to include parenthesis around the exponent.

Periodic Compound Interest

Another variation of the standard model $f(x)=a(1+r)^{x}$ arises from applications in banking and finance.

An essential concept of investing is taught in the classic movie Mary Poppins. In one scene, Mr. Dawes tries to convince young Michael to invest his tuppence (two pennies) in the bank with the following words of wisdom:

If you invest your tuppence, wisely in the bank
Safe and sound
Soon that tuppence, safely invested in the bank,
Will compound.

Interest is compounded when it is added to the initial account balance, or principal. Once that interest has been added to the principal it starts earning interest as well, compounding the effect.

In the financial world, it is common for interest to be compounded periodically throughout the year, sometimes on a monthly, weekly or even a daily basis.

If interest is compounded $n$ times a year, only a fraction $\frac{r}{n}$ of the annual interest is earned during each of those periods. If there are $n$ interest periods per year, the number of periods in $t$ years would be $n \cdot t$. With this in mind, we modify the standard exponential model to create a function for working with periodic compound interest.

[ A(t) = P\left(1 + \frac{r}{n}\right)^{n \cdot t} ]

where

• $A(t)$ is the account balance after t years
• $P$ is the principal (initial balance)
• $r$ is the annual interest rate (written as a decimal)
• $n$ is then number of compounding periods per year (ie. 12 for monthly, 52 for weekly, 365 for daily, etc.)
• $t$ is time in years

For example, if $3000 is invested at 6% compounded monthly for 8 years, the account balance would be

[ A(8) = 3000\left(1+\frac{0.06}{12}\right)^{12 \cdot 8}\approx 4842.43 ]

When using a calculator, remember to include parenthesis around the exponent.

Continuous Compound Interest

It is logical to wonder what might happen if we were to increase the number of compounding periods $n$ over the course of a year. Let's investigate.

Suppose you invest $\$25,000$ at $5\text{%}$ interest for one year. Using $P=25000$, $r=0.05$ and $t=1$, the accompanying table shows the account balance $A$ for ever increasing values of $n$.

$n$	$A(1)=25000 \left(1+\frac{0.05}{n}\right)^{n \cdot1}$
1	$26,250.00
12	$26,279.05
52	$26,281.15
365	$26,281.69
1,000	$26,281.74
10,000	$26,281.77
100,000	$26,281.78
1,000,000	$26,281.78

The table suggests that more frequent compounding does not produce unlimited growth over the course of a year. For instance, compounding interest 365 times a year (every day) gives a balance of $26,281.69, which is only very slightly better than compounding just 12 times a year (every month).

It also appears that no matter how many times we compound interest, the balance never gets above $26,281.78.

To explain why more frequent compounding does not produce unlimited wealth, consider what happens to the equation $A(t)=P\left(1+\frac{r}{n}\right)^{n t}$ as $n \rightarrow \infty$. If $n$ gets larger, then $\frac{r}{n} \rightarrow 0$.

In the past we've used the letter $h$ for quantities that get close to zero. We will follow that convention and make the substitution $h= \frac{r}{n}$. Note that this implies $n=\frac{r}{h}$. With these changes we rewrite $A(t)$ as follows:

$A(t) =P\left(1+\frac{r}{n}\right)^{n t}$	Start with the periodic compound interest function.
$A(t) = P(1+h)^{\left(\frac{r}{h} \right)t}$	Substitute $h= \frac{r}{n}$ and $n = \frac{r}{h}$.
$A(t) = P(1+h)^{\left(\frac{1}{h}\right) r t}$	Rewrite $\left(\frac{r}{h}\right) t$ as $\left(\frac{1}{h}\right) r t$.
$ = P\big[(1+h)^{\frac{1}{h}}\big]^{ r t}$	Rewrite using the rule of exponents $x^{m n} = (x^m)^{n}$.

The expression in brackets $[(1+h)^{1/h}]$ might remind you of something from the previous section. If $h \rightarrow 0$ then $(1+h)^{1/h} \rightarrow e \approx 2.71828$. This is the reason for the limit. Compounding interest more and more frequently makes the base closer and closer to $e$. The resulting function

[ A(t)=P e^{r \thinspace t} ]

is known as the continuous compound interest formula and is widely used in finance, business and economics.

Work with Continuous Growth

Aside from compound interest, there are many other quantities that change continuously. For instance, the number of bacteria in a petri dish does not increase suddenly at the end of a day, week or month. Rather the population increases continuously at a rate that is proportional to the size of population.

Similarly, the amount of carbon-14 in an object decays continuously.

When used in different settings such as these, the $A(t)=P e^{ r t}$ formula is often referred to as "continuous growth".

In general, we will say a quantity that experiences continuous exponential growth (or decay) can be modeled by a function of the form

[ f(t)=a e^{ k t} ]

where $a$ is the initial quantity when $x=0$ and $k$ is the continuous rate of growth.

The continuous exponential growth function is often written with different letters to match specific application. You will need to be able to recognize continuous growth regardless of how the function is written. For example,

[ \begin{align} Q(x) &= Q_0 e^{ k x} \newline n(t) &= n_0 e^{ r t} \newline y &= C e^{ r x} \end{align} ]

are all continuous growth models.

Unlimited Growth

In general, exponential growth cannot be sustained unless there are unlimited resources. For this reason it is sometimes called unlimited growth. A historic example of this is "tulipmania".

During the 1630's, Holland was crazy for tulips and their value skyrocketed exponentially. At one point, tulip prices were as much as 20-times higher than they had been just months before. Some economists estimate that half of all the money in the Dutch economy was wrapped up in the tulip market. Such a huge exponential rate of growth was not sustainable and soon the tulip market crashed, bankrupting many Dutch citizens.

In the next few examples we look at situations where, instead of exploding, the growth is limited and eventually levels off. Surprisingly, exponential functions are at the center of those applications as well.

Newton's Law of Cooling

Suppose you put a pizza in a 450°F oven. The temperature of the pizza will increase, but in a limited way since it will never exceed the temperature of the oven. Isaac Newton studied situations like this and discovered a law that modeled the heating or cooling of an object.

Newton's law of cooling says that rate of temperature change in an object is proportional to the temperature difference between the object and its surroundings. Based on this, Newton found that the temperature of an object at time $t$ is given by

[ T(t) = S + (T_{0} - S)e^{-k \thinspace t} ]

where $S$ is the temperature of the surroundings (or the temperature of the oven, in this case) $T_{0}$ is the initial temperature of the object (or the original temperature of the pizza), and $ k $ is a positive constant that depends on the nature of the object.

This function models cooling when $T_0 > S $ and heating if $T_0 < S$.

In either case, $y = S$ is a horizontal asymptote. Also notice that the resulting curves are nothing more than exponential functions that have been reflected and/or shifted vertically.

Let's consider a cooling example. Suppose you put a 160°F hot cup of cocoa on the counter in a 70°F kitchen. The cocoa will cool according to Newton's law of cooling as it approaches room temperature.

Assuming $k=0.08$, what will be the temperature of the cocoa in 20 minutes?

[ \begin{align} T(t) &= S+(T_0-S)e^{-k \thinspace t} \newline &=70 + (160-70)e^{-0.08 \cdot 20} \newline &=70 + 90e^{-1.6} \newline &\approx 88.17 \end{align} ]

We conclude that after 20 minutes the temperature of the cocoa is roughly 88°F.

Logistic Growth

Our final variation of the standard exponential model has its roots in population growth.

Imagine that a small number of trout are introduced into a large lake. When the population is small, the amount of food and resources available appears unlimited, and the population grows exponentially. But as the population increases, resources become more scarce and the growth levels off.

In 1838, Belgian mathematician Pierre-Francois Verhulst created a function that has this very behavior. His function, now called the logistic function, assumes growth is proportional to both the current population size and the available resources. The equation and graph of a logistic function are given below. The values $a$, $b$ and $c$ are constants.

The number $c$ is a horizontal asymptote of the function and is often called the carrying capacity. In the trout example, it would represent the maximum number of trout that the lake can support.

Looking Ahead

In this section we have used transformations to create a wide variety of exponential models.

Standard model	$f(x)=a(1+r)^{x}$
half-life	$f(x)=a\left(\frac{1}{2}\right)^{x/c}$
Doubling Time	$f(x)=a(2)^{x/c}$
Compound Interest	$A(t)=P\left(1+\frac{r}{n}\right)^{n \thinspace t}$
Continuous Growth	$f(x)=a e^{k\thinspace x}$
Newton's Law of Cooling	$T(t)=S+(T_0-S)e^{-k \thinspace t}$
Logistic	$f(x)=\frac{c}{1+ae^{-b \thinspace x}}$

We will return to these models later when we solve exponential equations. The main tool for solving exponential equations is called the logarithm, and it just happens to be the focus of the next section.