In Activity 3.1, we learned to calculate the “per unit” growth factor using an overall percent change between two values. In this case the growth factor is called *discrete*. It means the percent change between values occurs in “chunks” like \(2 \%\) every \(5\) units or \(27 \%\) every \(1\) unit.

For the dicrete exponential model, a \(2 \%\) increase every \(5\) units will actually be \(2 \%\) greater than the previous value at the end of \(5\) units.

In this activity we first explore the meaning of exponential notation in the context of some real situations and learn to directly communicate our observations into an equation.

By the end of this activity we develop a new form of the exponential function called the *continuous* model.

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Subsection3.2.1Interpreting the Exponential Form

Consider a car that travels from Portland, OR to Seattle, WA (about \(200\) miles) in the span of \(4\) hours. The trip is completed at an average speed of \(200/4 = 50\) miles per hour. We can use this information to write equations for the distance, \(D\) miles, from Portland as functions of time, \(t\) hours, in two different ways:

\(D = \frac{200}{4} \cdot t\) or \(D = 50\cdot t\)

Some may argue these are the same formula. But are they?

The first equation states \(200\) miles were covered in the span of \(4\) hours. This is something that actually happened. It’s in the data.

The second equation states that \(50\) miles were covered in \(1\) hour. We don’t know if this actually happened — it’s just an average rate of change. There may not have been any hour during the trip in which the car traveled exactly \(50\) miles.

Both equations will give us the same results, but each one tells a different story.

There is a difference between an obeserved result and a calculated result.

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Example3.2.1

*The value of a particular stock has increased by \(20 \%\) over the last \(5\) years.*

In this example the increase in stock value was actually measured as \(20 \%\) and it actually took \(5\) years for the increase to occur.

When we calculate the *per year* growth factor, “b”

\begin{align*}
b^{5} \amp= 1.2\\
b \amp= 1.2^{1/5}\\
b \amp= 1.0371
\end{align*}
It represents an average increase of approximately a \(3.71 \%\) per year. But that increase may or may not have actually happened in any one year — again, it is only an *average* annual percent change.

Earlier we learned that exponents “count” how many times the growth factor is repeated.

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Example3.2.2

The expression
\begin{equation*}
296(1.013)^{5}
\end{equation*}
states that an initial amount of \(296\) is increased by \(1.3 \%\) compounded \(5\) times.

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Exercise3.2.3“Counting” the Growth Factor

Translating between English and Math languages is actually very useful in practice. We can translate our observations made in English directly into expressions or equations in Math.

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Exercise3.2.4Interpret the meaning of the fractional exponent

Think of fractional exponents as keeping track of the “pieces” of the growth factor

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Exercise3.2.5A decrease of \(20 \%\) every \(5\) years

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Example3.2.6

A chemical biologist measures the breakdown of a pollutant in a lake. She calculates a \(4 \%\) decrease in the amount of pollutant from the \(4890\) ppm (parts per million) initially measured. The decrease occurs over a time span of \(72\) hours.

Without missing a beat, or the need for any algebra, the biologist can immediately write a simple exponential formula as a function of time to model what she just measured.

Using the initial amount of pollutant and letting \(t\) represent the time in hours since her first measurement, she writes
\begin{equation*}
P(t)=4890(0.96)^{\frac{t}{72}}
\end{equation*}

The formula literally says, “An initial amount \(4890\) ppm of pollutant decreases by \(4 \%\) every \(72\) hours.”

Of course, she doesn't know if the breakdown is actually exponential, it's just an assumption. She could have assumed a linear breakdown of the pollutant in which the \(4 \%\) decrease of the initial amount is the same amount each hour. A linear formula would therefore be,
\begin{equation*}
P(t)=4890-4890(0.4 \cdot t)
\end{equation*}

Nobody knows what the correct formula should be. There are many possible formulas that could model the breakdown of the pollutant. But one model may predict past or future data better than others and knowing what model to choose and why is the reason the biologist gets paid the big bucks.

By developing the connection between our literal understanding of a situation and the math notation used to describe it we afford ourselves great flexibility in how we interpret and how we model “real-life” data.

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Example3.2.7

A population of \(300\) bacteria doubles every \(5\) hours. Write an exponential formula for the population size as a function of time, \(t\) hours.

We can write a formula by choosing an algebaic technique.

Our formula will be something like \(P(t)=300\cdot b^{t}\) where \(300\) is the known initial population and \(b\) is the unkown, hourly growth factor.

We know that in \(5\) hours the population will grow from \(300\) to \(600\text{.}\) Therefore,
\begin{gather*}
300*b^{5}=600\\
b^{5}=\frac{600}{300}\\
b^{5}=2\\
b=2^{\frac{1}{5}}\\
b \approx 1.1487
\end{gather*}
so our formula is \(P(t) \approx 300(1.1487)^{t}\)

On the other hand, if you can “speak math”, you can write a formula in about two seconds
\begin{equation*}
P(t)=300(2)^{\frac{t}{5}}
\end{equation*}

The formula literally says, “Start with \(300\) and multiply by \(2\) every \(5\) hours.”

With some knowlege of the properties of exponents we see how the two formulas are actually the same

\begin{gather*}
300(2)^{\frac{t}{5}}\\
=300(2^\frac{1}{5} )^{t}\\
\approx 300(1.1487)^{t}
\end{gather*}

Either way we arrive at an hourly percent increase of about \(14.87 \%\) per hour.

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Subsection3.2.2Finance — Compounded Interest

We now turn our attention towards finance where there are some variations in how exponential patterns are formulated.

Keep in mind that financial formulas quickly get extremely complicated, so our examples are simple in order to make our point.

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Exercise3.2.8

An bank loan of \(\$ 2700\) is charged an annual interest of \(12 \%\)

Is there a way for the bank to make even more money? The bank already made more money by dividing the interest into \(12\) pieces. What if the interest is divided into smaller pieces and compounded more often during the year?

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Exercise3.2.9

The last exercise leaves the bank with some good news and some bad news. The good news is more money is made the more often they compound the \(12 \%\) interest during the year. The bad news is the increases in money made is shrinking.

In changing the compounding from once a year to \(12\) times a year the bank increase the loan value by about \(3042.43-3024= \,\$ 18.43 \)

But in changing the compounding from daily to hourly there was only an increase of \(3044.24-3044.18 = \,\$ 0.06\) which is only \(6\) cents!

We can extend what we concluded in the last exercise with a general form of an expression for compounded interest.

\begin{equation*}
A_{0}(1+\frac{r}{n})^{nt}
\end{equation*}

Here \(A_{0}\) is the initial amount or value, \(r\) is the decimal equivalent of the interest rate, \(n\) is the number of times the interest is compounded and \(t\) is the amount of time the initial amount is left to grow or decay (think stock values over time).

You may already know where this is going. We have to ask, "What is the most amount of money that can be made?" The answer depends on how often the interest is compounded. So, the real question is, "What is the most compounding possible?"

Let's think about that. Compounding the interest is the process of “chopping” up the interest percentage into pieces and calculating the increase or decrease for each piece until all the pieces are counted. Focus on the growth factor for \(t=1\) (notice the exponents in each example below have a multiple of *something*\(\cdot 1\)) year assuming some intitial value \(A_{0}\text{.}\)

*Annual interest compounded **yearly*: the interest is divided into \(1\) piece and \(1\) increase is counted each year.

\begin{equation*}
A_{0}(1+\frac{r}{1})^{1 \cdot 1}
\end{equation*}

*Annual interest compounded **monthly*: the interest is divided into \(12\) pieces and \(12\) increases are counted each year.

\begin{equation*}
A_{0}(1+\frac{r}{12})^{12 \cdot 1}
\end{equation*}

*Annual interest compounded \(daily\)*: the interest is divided into \(365\) pieces and \(365\) increases are counted each year.

\begin{equation*}
A_{0}(1+\frac{r}{365})^{365 \cdot 1}
\end{equation*}

What is the maximum number of times any number can be divided? Infinity (and beyond!)

In math, infinite compounding is called *continuous compounding*. It means the interest is being calculated at every moment in time. In fact, there is no time at which the interest is not being calculated.

When continuous compounding occurs, the general formula we came up with earlier
\begin{equation*}
A_{0}(1+\frac{r}{n})^{nt}
\end{equation*}
changes form. It becomes
\begin{equation*}
A_{0}e^{rt}
\end{equation*}

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Subsection3.2.3From English to Math (and back again)

Compounding interest \(12\) times each year is the same as compounding the interest every \(\frac{1}{12}\) of a year. It is referred to as “compounded monthly” because \(\frac{1}{12}\) of a year is \(1\) month.

Therefore another way to write the same equation would be

\(V(t) = 4280(1.0401)^{\frac{t}{1/12}}\)

Which means the interest is calculated every \(\frac{1}{12}\) of a year, but no one would do this because it is silly looking.

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Example3.2.10

From basic algebra, you may recall that dividing by a fraction is the same as multiplying by the reciprocal of the fraction.

\begin{align*}
\frac{t}{\frac{1}{12}} \amp=t \cdot \frac{12}{1}\\
\amp =12t
\end{align*}

Instead of writing \(\frac{t}{\frac{1}{12}}\) we keep the exponent as \(12t\) which is the same as monthly and the equation is prettier.

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Exercise3.2.11

Comparing interest compounded over many years versus multiple times during a year.