% Interest/contents.tex
%
\chapter{\mychaptername}
\label{\here}
\chaptermark{Exponential Growth}
\tocnotetoo{
In this chapter we explore how investments and populations grow and
radioactivity decays --- exponentially.
}

\begin{goals}

\begin{goal}{exponentialgrowth}
Understand that exponential growth (or decay) is constant relative change.
\end{goal}

\begin{goal}{interest}  
Understand how compound interest is calculated.
\end{goal}

\begin{goal}{depreciation}
Work with exponential decay.
\end{goal}

\begin{goal}{doublingtimes}
Reason using doubling times, half-lives, rule of 70.
\end{goal}

\begin{goal}{exponentialregression}
Fit exponential models to data.
\end{goal}
\end{goals}


\qrsection[compounding]{Money earns money}

Imagine that you have \$1,000 to invest.  Would you rather earn \$100
per year in interest, or 8\% per year in interest?  In each case the
interest is added into your principal (the balance in your account)
each year, and you never make any withdrawals.

The first scenario is called \myindex{simple interest}. You find the
new balance by adding \$100 each year to the previous balance. After
one year the balance would be \$1,100, after two years \$1,200, and so on. 

The second scenario is a little more complicated. The interest each
year is a fixed percentage of the balance. The one-plus trick finds
the new balance in one step: after one year you would have
$1.08 \times 
\$1,000 = \$1,080$.  After two your balance would  be $1.08 \times \$1,080 =
\$1,166.40$.   This pattern is called 
\myindex{compound interest}.

Table~\ref{fourYears} shows your balance in each case
for the first four years.

\begin{table}
\centering
\begin{tabular}{c c c}
\toprule
 & \multicolumn{2}{c}{Balance \ \ } \\
Year  & Simple interest & Compound interest\\
\midrule
now & \$1,000 & \$1,000.00\\
1 & \$1,100 & \$1,080.00\\
2 & \$1,200 & \$1,166.40\\
3 & \$1,300 & \$1,259.71\\
4 & \$1,400 & \$1,360.49\\
\bottomrule
\end{tabular}
\caption{Simple and compound interest}
\tablesource{Hand built data.}
\label{fourYears}
\end{table}

So far simple interest offers a better return on your investment. What
would the numbers be in 10 years?  We could continue building the
table a year at a time by hand (which is tedious), we could have a
spreadsheet calculate for us (we'll do that in a minute) or we could
look for a pattern and find a formula for each scenario, so that we
can compute for any year we like without having to do the work for all
the years in between.  We'll do that first.

Simple interest leads to a \myindex{linear equation}.  Each year the
balance increases by a fixed amount, \$100, so the slope is
\$100/year. The intercept is the starting value, \$1,000. The linear
function is
%
\begin{equation*}
B=1000 + 100 \times T
\end{equation*}
%
where $B$ represents the balance, in dollars, and $T$ the number of
years. If you leave your money growing until $T=10$ years, your
balance will be $1000 + 100 \times 10 = 2000$ dollars.  

The function describing compound interest isn't linear.
The percentage increase is constant but the amount of interest changes
from year to year. In the first year you earn \$80, while in the
second you earn \$86.40.  To see what kind of function to use, we
unwind the arithmetic in the compound interest column of
Table~\ref{fourYears}. 

\begin{alignat*}{4}
&\text{Year 1:} \quad && 1080.00 && = 1000.00 \times 1.08 && \\
&\text{Year 2:} \quad && 1166.40 && = 1080.00 \times 1.08 && =(1000 \times 1.08)
\times 1.08 =1000 \times 1.08^2\\ 
&\text{Year 3:} \quad && 1259.71 && = 1166.40 \times 1.08 && = (1080 \times
1.08^2) \times 1.08 = 1000 \times 1.08^3 
\end{alignat*}

It's clear that the function describing this growth is
 \begin{equation}\label{eq:exponentialEquation}
B=1000 \times (1.08)^T
\end{equation}
where, as before, $B$ represents the balance, in dollars, and $T$ the
time, in years.
It's an exponential function, because the 
independent variable $T$
is the exponent of 1.08.
The 0.08 in 
$1.08 = 1 + 0.08$ is the constant relative change for each
additional year. The 1000 is where we start: the value of 
$B$ when $T = 0$.
(You may remember but not have enjoyed the fact that $1.08^0 =
1$. If so, perhaps it makes a little more sense in this context. After
0 years you've received no interest, so your balance should be
multiplied just by 1.)


Suppose you want to compare the balances at simple and compound
interest after year 10. With simple interest you will have $1000 + 10
\times 100 = 2000$ dollars. But how can you compute $1000 \times
1.08^{10}$ without boringly multiplying by $1.08$ ten times? The
calculator in Figure~\ref{fig:onlinecalculator}
isn't powerful enough. For that job you need a
\emindex{scientific calculator}, one with a key labeled
\framebox{$y^x$} or \framebox{$x^y$}.

There are many online. Here are two:
\url{www.math.com/students/calculators/source/scientific.html},
\url{web2.0calc.com/}. Each will tell you that at 
the end of year 10 the balance will be about \$2159,
so the exponential growth has caught up with the linear.

You can do the computation with the Google calculator's buttons
but to use the search bar 
or a spreadsheet you need to know how to enter the exponent from the keyboard
without a \framebox{$y^x$} key. Both use the caret character
``$^{\wedge}$'' to raise a number to a power. That's meant to suggest
literally ``raising'' the exponent. You just type
\displayexcel{
1000 * 1.08 $^{\wedge}$ 10
}
\noindent into the Google search bar, or as a formula (preceded by an
equal sign) in a cell in a spreadsheet to check the arithmetic in the previous
paragraphs. 

In \commonsense{} we rarely put things to remember in boxes, but the
moral of this discussion deserves that treatment:

\begin{center}
\framebox{
\pbox{\textwidth}{
In linear growth, the \myindex{absolute change} is constant.  \\
In exponential growth, the \myindex{relative change} is constant.
}
}
\end{center}


Interest isn't the only place exponential growth happens. In
\exref{findexponentials} we ask you to think about others.

\qrsection[excelexponential]
{Exploring exponential growth with a spreadsheet}

You can answer ``what if'' questions about exponential growth by
changing the initial investment and interest rate in the
the spreadsheet \link{exponentialGrowth.xlsx}.
Figure~\ref{fig:twoexponentials}
shows two examples, for the equations
\begin{equation*}
B=1000 \times (1.08)^T
\end{equation*}
and
\begin{equation*}
B=1000 \times (1.16)^T.
\end{equation*}
Each swoops upward at an increasing rate. That shape is the signature
for exponential growth. The graphs look similar, but the scales on the
vertical axes tell a different story.
The spreadsheet has a second tab
(labeled ``Compare two growth trajectories'') that plots two
exponential curves on the same set of
axes. Figure~\ref{fig:twoexponentialstogether} clearly shows
how much faster growth is at 16\% than at 8\%.

\figfile{exp8and16twocropped.pdf}
\begin{figure}
\centering
\includegraphics[width=5in]{\thefigurefilename}
\caption{Two exponential graphs}
\label{fig:twoexponentials}
\end{figure}
\figfile{}
\figfile{exp8and16cropped.pdf}
\begin{figure}
\centering
\includegraphics[height=60mm]{\thefigurefilename}
\caption{Two exponential graphs on the same set of axes}
\figsource{Charts from Excel spreadsheet we wrote.}
\label{fig:twoexponentialstogether}
\end{figure}
\figfile{}

Let's take some time to see how Excel updates the calculations when we
change the constants in the equation. Click on one of the cells
in the exponential column, say \cell{B17}.  The formula bar reads
%
\begin{equation}\label{eq:exponentialEquationExcel}
=\text{{\excel{START*RELCHANGE}}}^{\wedge}\text{{\excel{A17}}},
\end{equation}
%
which is Excel's version of
(\ref{eq:exponentialEquation}).
We labeled cells \cell{A10} and \cell{A11} as
\excel{START} and \excel{RELCHANGE} so that we could use 
(\ref{eq:exponentialEquationExcel}) instead of 
%
\begin{equation*}
=\text{\excel{ \$A\$10*\$A\$11}}^{\wedge}\text{\excel{A19}}.
\end{equation*}
%
The version using cell labels is much
easier to understand than the one with cell references, 
and it doesn't need the dollar signs to tell Excel not to change those
references when we copy from one row to another.

To label a cell, click on it. The
Name Box at the left of the Formatting Toolbar will contain the address
of the cell, so if you click on cell \cell{H4} you will see \cell{H4}
there. You can highlight the
contents of the box and type in your own name. 
\index{Excel, naming cells}
\begin{teacher}
Librecalc also supports cell labels. 
\end{teacher}
Figure~\ref{fig:exponentialScreenshot} shows two screen shots of our
spreadsheet, the first with cell values, the second with cell formulas.

\begin{figure}
\centering
\figfile{excellabels-crop.pdf}
\includegraphics[height=1.8in]{\thefigurefilename}
\caption{Screenshots showing Excel formulas using named cells}
\label{fig:exponentialScreenshot}
\end{figure}
\figfile{}

%\begin{figure}
%\centering
%{
%\figfile{screenshot1.png}
%\includegraphics[height=1.8in]{\thefigurefilename}
%\figfile{screenshot2.png}
%\includegraphics[height=1.8in]{\thefigurefilename}
%}
%\caption{Screenshots showing Excel formulas using named cells}
%\label{fig:exponentialScreenshot}
%\end{figure}
%\figfile{}
%
The numbers in columns \cell{B} and \cell{C} are the same.
Excel computes them in different ways. We've seen how \cell{B17} uses
the algebra in (\ref{eq:exponentialEquationExcel}).
The value in cell \cell{C17} comes from the previous value in
\cell{C18} instead:
\displayexcel{
= C16*RELCHANGE.
}

In the hypothetical investment comparison at the start of this chapter
linear growth starts out better but by year 10 exponential growth
leads to a higher balance. To explore what happens in more detail, use
the spreadsheet {\link{linearExponential.xlsx}}. It extends
Table~\ref{fourYears} to cover 15 years.
The graph in Figure~\ref{fig:linearExponentialPlot}
shows that starting at year 7, the value of the exponential function
is larger than the linear. 
 
\figfile{linearExponentialPlotcropped.pdf}
\begin{figure}
\centering
\includegraphics[height=50mm]{\thefigurefilename}
\caption{Linear vs exponential growth}
\label{fig:linearExponentialPlot}
\end{figure}
\figfile{}

Now we can answer ``what-if'' questions.\index{what if}
Suppose, for example, our money earned 7\% interest instead of 8\%
interest. To redo the calculations we need to change just one number:
replace the 1.08 in cell \cell{A9} with 1.07. 
Excel recomputes the values of the exponential function in column
\cell{C} and redraws the graph. Then you can see that with this lower
interest rate, we have to wait 11 years before the exponential growth
of compound interest gives us a better return. 

\qrsection[exponentialdepreciation]{Depreciation}

It's always easier to think about increases (adding and multiplying)
than decreases (subtracting and dividing) but sometimes things do
decrease.

Suppose you buy a new car for \$20,000. As soon as you drive it out of
the dealer's lot it's worth less. In fact it's worth less each year:
it \emph{depreciates}\index{depreciation}. Its value depends on its
age. 

If the car is a business expense you might choose linear depreciation
for tax purposes --- suppose the value decreases by 
\$1,800 each year. The equation that determines the value $V$ as a
function of the age $A$ is 
%
\begin{equation*}
	V = 20,000 - 1,800A.
\end{equation*}
%
But a more realistic way to model the value of the car is to assume
that the percentage decrease is the same each year. Suppose it's
13\%. Then each year its value is 87\% of what it was the year before.
The corresponding equation is
%
\begin{equation*}
	V = 20,000 \times 0.87^A.
\end{equation*}
%
We can use our old friend \link{linearExponential.xlsx} to draw
Figure~\ref{linearExponentialDepreciation} showing 
what the car is worth over time in each case. Set \excel{START} to 20,000,
\excel{ABSCHANGE} to -1,800, and \excel{RELCHANGE} to $1-0.13=0.87$.
(The relative change is still positive. It's a decrease rather than an
increase because it's less than 1.)

\figfile{linearExponentialDepreciationcropped.pdf}
\begin{figure}
\centering
\includegraphics[height=50mm]{\thefigurefilename}
\caption{Linear vs exponential depreciation}
\figsource{Chart from Excel spreadsheet we wrote.}
\label{linearExponentialDepreciation}
\end{figure}
\figfile{}
When the depreciation is linear the car is worthless (at least on
paper) after about 11 years. Excel doesn't know that, so it continues the
graph on into negative values. If we wanted to use this graph in a
more formal presentation we'd have to prevent that, and change the
labels. Leaving it this way exhibits the power of thinking abstractly
in Excel --- the original spreadsheet can manage shrinking just as
easily as growth. 


\qrsection[doublingtimes]{Doubling times and half-lives}

How long will it take to double your money? (We'll assume you're clever
enough to insist on compound interest.) The answer depends
on the interest rate and the initial balance. The
\link{exponentialGrowth.xlsx} spreadsheet 
shows that at 8\% interest with an initial investment of \$1,000 the
balance is \$2,000 after 9 years (the table shows \$1999.004627, which
is quite close enough to double).

If you change the initial investment to
\$100 then Excel shows a balance of
\$200 after the same 9 years. Experimenting with many different
initial investments always shows the same doubling time. So
the time it takes to double your money does not depend on the
amount you start with.
\begin{teacher}
Of course you don't need to rely on experiments to know that the
doubling time is independent of the initial value. It's very easy to prove
with a little bit of algebra. But this is a book about
quantitative reasoning, not about algebra. For its intended audience
the experiments are more convincing than the more formal
mathematics many people find mysterious.
\end{teacher}

What about a different interest rate? If you use 5\% interest in the
spreadsheet the doubling time seems to be between 14 and 15 years. 
We can do a calculation:  $1.05^{14.5} = 2.028826162$, so 14.5 is a good guess.

At 2\% interest it takes more than 30 years to double your money, so
the spreadsheet doesn't give us the answer. We could find it by adding
some rows, but we'll use another method instead.
We'll try to guess the value of $T$ 
in the equation $1.02^T = 2$ and adjust our guess until we're close enough.
Perhaps the answer is $T= 40$ years:
\begin{equation*}
	1.02^{40} = 2.208039664 .
\end{equation*}
Too big, so we need less time. Try 35:
\begin{equation*}
	1.02^{35} = 1.999889553.
\end{equation*}
Bingo!

We've collected these results and a few more in the second column
of Table~\ref{doublingTime}. 

\begin{table}
\centering
\begin{tabular}{S[table-format=3.0]
                S[table-format=2.0]
                S[table-format=2.0]}
\toprule
{interest rate (\%)} & {approximate doubling time} & {70/rate} \\
\midrule
2 & 35 & 35.0 \\
3 & 24 & 23.3 \\
5 & 14.5 & 14.0 \\
7 & 10.5 & 10.0 \\
8 & 9 & 8.8 \\
10 & 7.5 & 7.0 \\
15 & 5 & 4.7 \\
20 & 4 & 3.5 \\
50 & 1.7 & 1.4 \\
100 & 1 & 0.7 \\
\bottomrule
\end{tabular}
\caption{Double your money}
\tablesource{Hand built data.}
\label{doublingTime}
\end{table}

The third column in that table shows the results from the ``Rule of
70'',\index{rule of 70} that says that you can 
estimate the compound interest doubling time by dividing the magic
number 70 by the annual interest rate as a percent. The approximation
is better when the interest rate isn't too large; those are just the
cases that matter most in everyday investing. The most commonly quoted
consequence of the Rule of 70 is that money invested at 7\% will
double in 10 years. 

Figure~\ref{ruleOf70} shows
how good the Rule of 70 is for interest rates up to 20\%.

\figfile{ruleOf70cropped.pdf}
\begin{figure}
\centering
\includegraphics[height=50mm]{\thefigurefilename}
\caption{Doubling times}
\figsource{Charts from Excel spreadsheet we wrote.}
\label{ruleOf70}
\end{figure}
\figfile{}

When a relative increase occurs repeatedly the
doubling time is independent of the initial value.
So if inflation is 5\% per year, all prices will
double in 14 years. 

Knowing the doubling time helps you make quick calculations. Since 5\%
inflation doubles prices in 14 years it will quadruple them in 28
years. In 42 years they will be eight times as large. The Bureau of
Labor Statistics inflation calculator says that inflation in the 42
years from 1968 to 2010 increased the cost of a \$100 item to
\$626. That's not quite eight times as much, so the average inflation
rate for those years was not quite 5\% per year.

The Rule of 70 applies to depreciation as well --- it tells you the
\emindex{half-life}. That's the equivalent for depreciation 
of the doubling time --- the time until half the original value has
disappeared. Like doubling time, the half-life depends on the
depreciation rate, but not on the original value. For 13\% annual
depreciation it's approximately 
$70/13 = 5.38461538 \approx 5.4$ years.

The term half-life comes from atomic physics, where it describes the
way the quantity of a radioactive element decreases over time. The
following quotation from  
\url{http://archives.nirs.us/factsheets/hlwfcst.htm} provides food for
quantitative thought.

\begin{quotation}
After ten half-lives, one-thousandth of the original
concentration [of a radioactive substance] is
left; after 20 half-lives, one millionth.  Generally 10-20 half-lives
is called the hazardous life of the waste.  Example:   plutonium-239,
which is in irradiated fuel [from a nuclear power plant], has a
half-life of 24,400 years.  It is 
dangerous for a quarter million years, or 12,000 human generations.
\index{nuclear waste}\index{hazardous life}\index{plutonium}%
\begin{csmr}
High-Level Radioactive Waste,
Nuclear Information and Resource Service,
\url{http://archives.nirs.us/factsheets/hlwfcst.htm},\access{March 2, 2020}.
Quoted with permission.
\end{csmr}
\end{quotation}

This is the kind of quotation that begs to have its numbers checked.

First let's look at ``ten half-lives''. After one half-life the
concentration is half what it was at the start. After two half-lives
it's half of a half, or 1/4. After three half-lives it's 1/8. So after
10 half-lives it's 
\begin{equation*} 
\frac12 \times \frac12 \times \cdots \times \frac12 = \frac {1}{2^{10}}
\end{equation*}
of what it was.
\begin{teacher}
If we were teaching algebra and not quantitative
reasoning we might use a negative exponent to write $(1/2)^{10}$ as
$2^{-10}$. But we're not, so we don't, so we avoid the time it would
take to remind students about working with negative exponents.
\end{teacher}

We saw when we studied the metric prefixes that $2^{10} = 1,024 \approx
1,000$. That's why ``kilo'' means ``1,000'' most of the time but
``1,024'' when computers are involved. Now we use the same fact to see
why $1/2^{10}$, which is exactly 1/1,024, is approximately 1/1,000 ---
one one thousandth.

What about twenty half-lives? In that time the original concentration
will be reduced to 1/1,000 of 1/1,000 of what it was at the
start. Since a thousand thousand is a million, that's one
one-millionth.

According the quotation, plutonium-239 will be dangerous for at
least ten 24,000 year half-lives. That's 240,000 years, which is
indeed about a quarter  of a million years. Is it 12,000 generations?
Yes, if you calculate with $240,000/12,000 = 20$ years per
generation. That's perhaps a little low for the developed world, but
good enough to highlight the danger of nuclear waste.

\qrsection[exponentialmodels]{Exponential models}

For compound interest and radioactive decay the equation
for exponential change gives exact answers, just as the linear
equation gives exact answers for simple interest and electricity bills.

When change is approximately linear a regression line may be
useful. When it's approximately exponential, we can construct an {\em
exponential trendline}.  Most elementary
texts discuss the reproduction of \myindex{bacteria} as a toy example
of exponential growth.
Bacteria reproduce by dividing, so each individual gives
rise to 2, then 4, then 8 descendants, and so on. The number of
bacteria grows exponentially. But that can't go on forever. Since
exponential growth curves quickly become steep, 
it can't go on for very long.
Eventually, crowding or diminishing resources cause growth to slow,
perhaps even to reverse as organisms die faster than new ones are
born. 
\begin{teacher}
The material in this section on bacterial growth is at the edge of
what we think students in a quantitative reasoning course need. It
deals with real data, not the artificial doubling time problems in
most books. Carrying through the discussion in sufficient detail so that
students could solve similar problems would take time better spent on
other topics. But if there's time in the syllabus there are ideas here
worth exploring. They tie together all the themes of the chapter.
\end{teacher}

Table~\ref{bacteriaGrowth} records the population of three different
strains of the \emph{Burkholderia cenocepacia}
bacterium in a one day experiment conducted by Professor  
Vaughn Cooper at the University of New Hampshire.
\index{Cooper, Vaughn}

\begin{table}
\centering
\begin{tabular}{S[table-format=2.0]
  S[table-format=3.1]
  S[table-format=3.1]
  S[table-format=3.1]}
  \toprule
     & \multicolumn{3}{c}{{population} } \\
\cmidrule{2-4}
{time} & {W} & {R} & {S} \\
\midrule
4 & 7.9 & 8.5 & 17.5 \\
8 & 17.3 & 13.5 & 42.1 \\
12 & 44.7 & 48.9 & 225.2 \\
17 & 119.3 & 268.7 & 407.3 \\
20 & 41.3 & 98.0 & 149.3 \\
24 & 41.3 & 64.0 & 160.0 \\
\bottomrule
\end{tabular}
\begin{csmr}[Bacteria growth\label{bacteriaGrowth}]
\emph{Growth rates of different Burkholderia cenocepacia mutants}
(reported in \url{doi.org/10.1073/pnas.1207025110}) in minimal
galactose medium provided by the laboratory of Vaughn Cooper.
\end{csmr}
Time: hours\\
Population: millions of organisms per milliliter
\end{table}
The raw data points in the graph on the left in
Figure~\ref{bacteriaGrowthGraph} (built in the spreadsheet
\link{bacteriaGrowth.xlsx}) suggest that the growth of each strain was
exponential until about hour 17. To construct the graph on the right
we plotted the data for each strain for that period. In the resulting
chart we right-clicked on a data point for each strain, selected
\excel{Add Trendline \ldots}, and chose \excel{Exponential} on the
\excel{Type} tab.  As usual, we asked for the equations and the
$R$-squared values.  Those are all pretty close to 1; best for strain W,
worst for strain S.

\figfile{bacteriaGrowthChartscropped.pdf}
\begin{figure}
\centering
\includegraphics[width=5in]{\thefigurefilename}
\caption{Bacteria growth}
\figsource{Charts from Excel spreadsheet we wrote.}
\label{bacteriaGrowthGraph}
\end{figure}
\figfile{}

Let's look at the equation for the exponential trendline for strain W,
which you can see in the figure:

\begin{equation}\label{exp0.2109}
y = 3.3663\ e^{0.2109x}.
\end{equation}

What is the ``$e$'' in this equation?\index{e@\textit{e}} 
The complete answer to that
question calls for much more mathematics than you need to know to
apply common sense to 
quantitative arguments. But since Excel uses it you may encounter it
somewhere so we'll discuss it briefly.

The simplest explanation is that $e$ is just a particular number ---
approximately $2.7183$. Like $\pi \approx 3.1416$
it's one of those numbers whose decimal expansion ``goes on forever'',
so the first few decimal places give just an approximation.
The number $e$ appears naturally in discussions of exponential growth just as
$\pi$ appears in discussions of circles. It was named by the prolific
mathematician Leonhard Euler \index{Euler, Leonhard} (1707-1783) who
was the first to recognize its importance. 

Back to our exponential growth function for the strain W.  We see that
\begin{equation*}
e^{0.2109} \approx (2.7183)^{0.2109} \approx 1.2348.
\end{equation*}
You don't have to remember the value of $e$ to compute with it, since
Excel and LibreCalc and the Google calculator provide the built-in function EXP
to do the job. Entering \excel{=EXP(0.2109)} in a spreadsheet cell or the
same thing without the equal sign in the Google search bar will give
you the same answer. 

Now we can substitute $1.2348$ for $e^{0.2109}$ in 
(\ref{exp0.2109}) and rewrite it as
\begin{equation*}
y = 3.3663\times 1.2348^x.
\end{equation*}
In that form we recognize this behavior as exponential growth at a
constant rate of about 23\% per hour. The Rule of 70 tells us to 
expect a doubling time of about $70/23 \approx 3$ hours. That
matches the data, which does indeed show the strain W population
doubling about every three hours for most of the day.

If you experiment with \excel{EXP} you will find that $e^{0.7} =
2.0137527\ldots \approx 2$. It's the 0.7 in the exponent that leads to
the Rule of 70. To learn just how, go on to take a course in calculus.

Using the exponential trendlines for constant growth rate to predict
the future pop\-u\-la\-tions would fail in this experiment. About two
thirds of the way through the 24 hour day a different reality appears.
The gray bullets in Figure~\ref{bacteriaGrowthGraph} show that all
three populations drop dramatically.  (A newspaper reporter wanting to
emphasize the drama might say, incorrectly, that the populations
dropped ``exponentially.'')

It's the biologist's job to understand why. The analysis of the
exponential growth at the start has told him only how the bacteria
grow when there's lots of food, lots  of room and no competition.

\qrsection[exponentially]{``Exponentially''}

A Google search for a definition of \emph{exponentially} finds this
first meaning:
\begin{teacher}
We occasionally see students write ``exponential'' as
``expodental''. Although we try not to discourage them by correcting
their spelling, this is one we fix.
\end{teacher}

\begin{quotation}
1. (with reference to an increase) more and more rapidly. \\
``our business has been growing exponentially''
\end{quotation}

This is from \theTimes{} in April, 2019.

\begin{quotation}
``The way ads are targeted today is radically different from the way it
was done 10 or 15 years ago,'' said Frederike Kaltheuner, who heads the
corporate exploitation program at Privacy International. ``It's become
exponentially more invasive, and most people are completely unaware of
what kinds of data feeds into the targeting.''%
\begin{csmr}
S. Thompson,
These Ads Think They Know You,
\theTimes{},
April 30, 2019,
\url{www.nytimes.com/interactive/2019/04/30/opinion/privacy-targeted-advertising.html?action=click&module=Opinion&pgtype=Homepage}
\access{April 30, 2019}.
\end{csmr}
\end{quotation}

The \emph{Daily Beast}
\index{Daily Beast} seems particularly fond of the word.

\begin{quotation}
 ``[I]n 2009, the [Sketchbook Museum] project
 moved to New  York and exponentially grew across the globe.''
 \begin{csmr}
   D. Genis,
   Blurred Lines at NY Sketchbook Museum,
   \emph{Daily Beast}, November 1, 2014,
   \url{www.thedailybeast.com/blurred-lines-at-ny-sketchbook-museum}
   \access{April 22, 2019}.
 \end{csmr}
\end{quotation}

\begin{quotation}
``Exponentially powerful technologies are transforming our sphere of
 possibilities.''
 \begin{csmr}
J, Silva,    
On Transhumanism and Why Technology Is Our Silicon Nervous System,
\emph{Daily Beast}, April 26, 2014,
   \url{www.thedailybeast.com/on-transhumanism-and-why-technology-is-our-silicon-nervous-system}
   \access{April 22, 2019}.
\end{csmr}
\end{quotation}

 It's sad to see ``exponentially'' reduced to a bland adjective when
 we understand its  precise mathematical meaning --- the kind of 
 growth captured in the formula for compound interest:
 \begin{equation*}
\text{balance} =\text{start} \times (1 + \text{rate})^{\text{time}},
\end{equation*}

 Even writers wno know that ``exponentially'' must somehow involve an
 exponent can get the mathematics wrong. Victoria Markovitz wrote in a
 \emph{National Geographic}  article  
that

  \begin{quotation}
  [t]he amount of power you can produce [with a wind turbine] is
  determined by the square of   the blade radius. That means
  increasing the size of the turbine has 
  an exponential effect on power.
  \begin{csmr}
V. Markovitz, 
Sizing Up Wind Energy: Bigger Means Greener, Study Says,
National Geographic,
July 20, 2012,
\url{news.nationalgeographic.com/news/energy/2012/07/120720-bigger-wind-turbines-greener-study-says/}
\access{May 1, 2019}.  
\end{csmr}
  \end{quotation}

  To think about this claim we put the power output equation from the
  linked  post at   \url{www.nationalgeographic.com/environment/great-energy-challenge/2012/worlds-largest-wind-turbines-is-bigger-always-better/}
 in spreadsheet \link{turbinepower.xlsx}. Then we drew
Figure~\ref{fig:turbinepower} to see how power production $P$  depends on
blade length $R$ for lengths up to 35 feet. The dotted curve predicts
values using the square of the blade radius. The solid curve that
grows much faster
predicts values using an exponential trendline. The two equations are
essentially 
\begin{equation*}
\text{quadratic}: \quad \text{power} =  0.0016 \: \text{radius}^2 
\end{equation*}
and
\begin{equation*}
\text{exponential}: \quad \text{power} =  0.19 e^{0.07 \: \text{radius}}
\approx 0.19 \: (1.07)^{\text{radius}} .
\end{equation*}
When Markovitz writes that blade length has an ``exponential effect on
power'' she is misusing the word,  

\figfile{turbinepowercropped.pdf}
\begin{figure}
\centering
\includegraphics[width=3in]{\thefigurefilename}
\caption{Turbine power}
\label{fig:turbinepower}
\end{figure}
\figfile{}

\exstart

\begin{exx}{\hassolution\routine\sref{compounding}
\gref{exponentialgrowth}\gref{interest}}
Compound interest computations.

If you invest \$1500 at 7\% interest compounded every year, how much
will you have at the end of 10 years? 15 years? 20 years? Use the
formula for exponential growth; then check your answers with the
\link{exponentialGrowth.xlsx} spreadsheet. 

\begin{sol}
The Google calculator tells me that
\gc[,]{
1 500 * (1.07\caret 10) = 2 950.72704
}
which is about \$2950.

Similarly,
\begin{equation*}
\$1500 \times (1.07)^{15} = \$4138.54731 \approx \$4140,
\end{equation*}
\begin{equation*}
\$1500 \times (1.07)^{20} = \$5804.52669 \approx \$5800.
\end{equation*}
I've rounded the results to the nearest \$10. The spreadsheet gives
the same answers.

\end{sol}

\end{exx}

\begin{exx}[findexponentials]{\hassolution\gref{absolutevsrelativechange}
\sref{compounding}\gref{exponentialgrowth}}
When do you expect exponential growth?

In each of the following situations, explain why you would expect
linear or exponential growth.

Think about whether the change is best described as an absolute rate
(like dollars per hour or gallons per mile) or a percentage (like 
percent per year or percent per washing).

Write the units for the kind of rate you decide on.

\begin{abcd}

\item Price increases from year to year due to inflation.

\item How the amount of gas you use depends on how far you drive.

\item The amount of money left on your public transportation debit
card as the days go by and you commute to school or work.

\item The amount of sales tax you pay, depending on how much you buy.

\item The amount of dirt left in your kid's filthy jeans when you wash
them over and over again.

\item The population of the world as the years go by.

\item Your credit card balance if you stop making payments.
(We will study credit cards in the next chapter.)

\item The height of the snow as it accumulates in a big storm.

\item The number of people sick in the first weeks of the
flu season.

\item The number of subscribers to a hot new social network in its
first days.

\end{abcd}

Think about your answers before you look at the hints.

\begin{hint}
\begin{abcd}
\item Inflation is usually reported as a percent increase.

\item No hint needed.

\item No hint needed.

\item Read this one carefully to think about what depends on
  what. Don't just jump at the word ``percent''.

\item The washing happens over and over again on the same day ---
  that's how dirty they were.

\item No hint needed.

\item Interest on unpaid balances accumulates.

\item What units would the weatherman use to report the rate at which
  snow was accumulating?

\item Think about how the number of people exposed to germs depends on
  the number of people sick.

\item (Electronic) word of mouth generates new subscribers from old ones.

\end{abcd}

\end{hint}

\begin{sol}
\begin{abcd}

\item Price increases from year to year due to inflation.

This should be (approximately) exponential since annual inflation is
measured as a percentage increase, not an absolute amount.

\item How the amount of gas you use depends on how far you drive.

This is a linear relationship. The equation that computes gas used
depending on miles driven has slope gallons/mile.

\item The amount of money left on your public transportation debit
card as the days go by and you commute to school or work.

This is linear, with a negative slope: the rate of decrease is the
number of dollars per day you spend.

\item The amount of sales tax you pay, depending on how much you buy.

This is linear, not exponential. The sales tax $T$ depends on the
amount $S$ of stuff you buy using the formula
\begin{equation*}
	T = rS
\end{equation*}
where $r$ is the sales tax rate. In Massachusetts in 2014, $r = 6.25\%
= 0.0625.$

\item The amount of dirt left in your kid's filthy jeans when you wash
them over and over again.

If each washing gets rid of the same percentage of the remaining dirt
then the amount of dirt left decreases exponentially.

\item The population of the world as the years go by.

Population growth is (approximately) exponential. The difference
between the birth rate and death rate determines the percentage change
each year.

\item Your credit card balance if you stop making payments.

This is exponential growth. The monthly growth rate is the annual
rate divided by 12. In reality, the credit card company won't let you
completely stop making payments. You must always pay some minimum
amount. Your balance will slowly decrease over time.

\item The height of the snow as it accumulates in a big storm.

This is linear. The height depends on the time; the slope is measured
in (inches of snow) per hour.

\item The number of people sick in the first weeks of the
flu season.

This is exponential because each person with the 'flu spreads it to some
number of healthy people he or she comes in contact with. That means
the rate of spread depends on the number of people already sick, so
the rate is increasing.

\item The number of subscribers to a hot new social network in its
first days.

This is like an infection. The number of new subscribers grows faster
and faster since each new subscriber invites his or her friends. This
can only be true when the social network is new. After a while the
people being invited will already have joined.
\end{abcd}

\end{sol}
\end{exx}

\begin{exx}{\hassolution\sref{exponentially}\gref{exponentialgrowth}}
Is it really exponential?

In everyday usage the phrase ``growing exponentially'' is just a
vibrant synonym for ``growing rapidly''. It's rare that it really
means a constant relative change.

Find instances of ``exponential'' growth in the media where what's
meant is just very rapid growth.

\begin{sol}

  From \theTimes{} on May 31, 2019:
  
  \begin{quotation}
  Infields have been altering their positioning based on their
  opponents for decades, but the frequency and precision have risen
  exponentially in recent years.%
  \begin{csmr}
     J. Wagner,
    All the Yankees Need to Know Is Inside Their Caps (or Their
    Pockets),
    \theTimes{}, May 30, 2019,
    \url{https://www.nytimes.com/2019/05/30/sports/yankees-cheat-sheets.html}
    \access{June 1, 2019}.
\end{csmr}
\end{quotation}
  
Searching for ``exponential'' on 
\emph{The Denver Post} website on November 19, 2014 found fifteen
articles for the year to date --- more than one a month. Here are a few
snippets. Only the first is actually about growth.

\begin{itemize}
\item[]
\begin{quotation}
Reverb: It seems like with you [Greensky Bluegrass guitarist Dave
  Bruzza] playing two nights at the Ogden, you have just been
exponentially growing over the last few years here in Denver.
\end{quotation}

\item[]
\begin{quotation}
The two ballot measures [on gaming and on genetically modified
  organisms] show a vivid contrast: one sports competitive fundraising
efforts, and the other displays an exponential gap in contributions
from the two sides.  
\end{quotation}

\item[]
\begin{quotation}
`Any amount of additional education is valuable. It's essential we get
all students to the finish line and get their degrees, but along the
way, students' eyes are opened up,'' Chavez said. ``Every step, every
day, every year further opens up their eyes, and they are also then
sharing these experiences with younger siblings, friends and peers, so
it's an exponential positive benefit to the public.
\end{quotation}

\item[]
\begin{quotation}
The bond is designed to alleviate exponential crowding in 27J schools,
and it will be on the Nov. 4 ballot.  
\end{quotation}

\end{itemize}


Here are the first four entries when I searched \theTimes{} web site
for ``exponential'' on June 9, 2011:

\begin{itemize}
\item[] What's Next? The Cognocene --- Room for Debate\\
But the human influence on Earth systems, while present at low levels
for perhaps millennia, has really only gone exponential with the \ldots\\
May 19, 2011 --- Room For Debate 

\item[] Toyota Tries to Loosen Grip on U.S. Product Development \\
\url{wheels.blogs.nytimes.com/2011/06/01/chastised-toyota-tries-to-loosen-grip-on-product-development-for-u-s/ }\\
I'm in my 19th year at Toyota, and I've definitely seen an
exponential growth in responsibility for North American operations. \ldots\\
June 1, 2011 --- Wheels 

\item[] Ricardo Sanchez, Retired Army General, Files to Run for Senate
in\ldots\\
\url{thecaucus.blogs.nytimes.com/2011/05/11/general-in-abu-ghraib-scandal-files-to-run-for-senate-in-texas/ }\\
The Democratic Senatorial Campaign Committee has identified Texas,
which has experienced exponential growth in Hispanic residents over
the \ldots\\
May 11, 2011 --- The Caucus 

\ldots plan for making money from advertising, finally answering the
question of how the company expects to turn its exponential growth
into revenue. 

\end{itemize}


The first of these really is describing the exponential growth or the
world's population since the Industrial Revolution.

The last three may be accurate assertions about the growth they claim,
but there's no evidence that the authors looked for data.

Here are more, from a student.

\begin{itemize}

\item[]
September 21, 2011 --- \emph{New York Times Online} \\
Title: At Facebook, Exponential Sharing \\
Author: Somini Sengupta	 \\
Quote: ``Facebook's latest features will allow for an exponential
expansion of shared information.''

\item[]
August 4, 2011 --- \emph{New York Times Online} \\
Title: Classical Music Moves From Concert Halls to Cafes \\
Author: Chloe Veltman \\
Quote: ``The exponential growth of Classical Revolution and other
innovative chamber music entities, like the San Francisco Friends of
Chamber Music and Opera on Tap, point to a broader trend in classical
music --- with its traditionally lofty image and high ticket prices --- of
making it more approachable.''

\item[]
October 23, 2011 --- Letters to the Editor {NYT Online} \\
Title: The Road to Ruin \\
Author: Frances Hawxwell \\
Quote: ``The editorial 'self-inflicted misery' (Oct. 17) ridicules the
British coalition government's attempt at tackling the national debt
through controlling our (exponential) rate of increase in government
spending and borrowing.''

\item[]
September 12, 2011 --- \emph{Boston Globe Online} \\
Title: Creativity in the Age of Psychiatric Medication \\
Author: Claudia Gold \\
Quote: ``I discuss the problems associated with the exponential rise
in prescribing of psychiatric medication for children.''

\item[]
October 10th, 2011 --- \emph{Washington Post Online} \\
Title: Why University Presidents Refuse Reform \\
Author: Mark Taylor \\
Quote: ``As the exponential increase in the cost of education
translates into ever more limited resources, it is impossible to avoid
conflicts of interest among these constituencies.''

\item[]
November 13th, 2011 --- \emph{The Miami Herald Online} \\
Title: Climate change, beetle may doom rugged pine \\
Author: Craig Welch \\
Quote: ``We know the incidence of blister rust infection and mountain
pine beetle outbreaks is increasing exponentially [.]''
\end{itemize}

\end{sol}
\end{exx}

\begin{exx}{\hassolution\sref{compounding}\gref{exponentialgrowth}}
Health care spending.

In Chapter~\ref{Percentages}, \exref{healthcarespending}, we used data
from the 2010 National Health Expenditures report to compute the
absolute and relative changes in 
\myindex{health care spending} per person from
2007 to 2008.

\begin{abcd}
\item Use the results of those calculations to build linear and
exponential models for the growth of health care spending per person.

\item Use each model to predict when health care spending will reach
\$10,000 per person per year.

\end{abcd}

\begin{sol}

\begin{abcd}
\item Use the results of those calculations to build linear and
exponential models for the growth of health care spending per person.

Let $H$ represent health care spending, in dollars per person, and $y$
the number of years since 2006. Then the equations are
\begin{equation*}
H = 7421 + 260y
\end{equation*}
and
\begin{equation*}
H = 7421 \times 1.035^y.
\end{equation*}
(The slope for the linear equation is $7681-7421 = 260$. The relative change
for the exponential equation is $7681/7421 = 1.035$.)

\item Use each model to predict when health care spending will reach
\$10,000 per person per year.

I put these numbers into the spreadsheet that compares linear and
exponential growth and found that the linear model predicts that $H$
will reach \$10,000 in ten years from 2006 --- that is, in 2016. The
exponential model predicts the same thing will happen about one year
sooner, in 2015.

I could have found the same answers without even writing down the
equations by asking Excel to make a scatterplot of the points
(0,~7421) and (1,~7681), then constructing both linear and exponential
trendlines and asking for a 10 year forecast. 

The exponential trendline has equation
\begin{equation*}
	H = 7421 \ e^{0.0344y}.
\end{equation*}
I checked with \excel{=EXP(0.0344)} and found that
\begin{equation*}
	e^{0.0344} = 1.034998523 \approx 1.035,
\end{equation*}
%
which matches the 3.5\% relative change I computed in part (a).

\end{abcd}

\end{sol}
\end{exx}

\begin{exx}{\hassolution\sref{compounding}\gref{exponentialgrowth}}
Malthus\index{Malthus, Thomas}.

In 1798 Thomas Malthus, an English economist and clergyman, wrote
``An Essay on the Principle of Population''. He said there:

\begin{quotation}
I think I may fairly make two postulata.

First, That food is necessary to the existence of man.

Secondly, That the passion between the sexes is necessary and will
remain nearly in its present state.

These two laws, ever since we have had any knowledge of mankind, appear
to have been fixed laws of our nature, and, as we have not hitherto
seen any alteration in them, we have no right to conclude that they
will ever cease to be what they now are, without an immediate act of
power in that Being who first arranged the system of the universe, and
for the advantage of his creatures, still executes, according to fixed
laws, all its various operations.

\ldots

Assuming then my postulata as granted, I say, that the power of
population is indefinitely greater than the power in the earth to
produce subsistence for man.

Population, when unchecked, increases in a geometrical ratio.
Subsistence increases only in an arithmetical ratio. A slight
acquaintance with numbers will shew the immensity of the first power in
comparison of the second.

\ldots

The power of population is so superior to the power in the earth to produce
subsistence for man, that premature death must in some shape or other
visit the human race. The vices of mankind are active and able
ministers of depopulation. They are the precursors in the great army of
destruction; and often finish the dreadful work themselves. But should
they fail in this war of extermination, sickly seasons, epidemics,
pestilence, and plague, advance in terrific array, and sweep off their
thousands and ten thousands. Should success be still incomplete,
gigantic inevitable famine stalks in the rear, and with one mighty blow
levels the population with the food of the world.%
\begin{csmr}
T. R. Malthus, An Essay on the Principle of Population.
\url{www.gutenberg.org/etext/4239}
\access{July 17, 2015}.
\csmrcomment{Malthus is in the public domain}
\end{csmr}
\end{quotation}

Malthus claimed that the food supply grows in a linear fashion. As a
unit of food supply he used the amount of food
needed for one person for one year. He estimated food production in
Britain in 1798 as 7,000,000 food units and that food production might
increase by a constant 280,000 units each year. 

Malthus also believed that the population of Britain was growing at a rate
of 2.8\% each year. In 1798, the population was about 7,000,000.

\begin{abcd}
\item Write a linear function that models food production.
\item Write an exponential function that models population growth.
\item
Was there enough food for each individual in Britain in 1798?
\item Using Malthus's models, determine whether there would be enough food
for each individual in Britain in 1800.

\item
Malthus claimed that the population in Britain would eventually
outstrip the food supply --- a prediction we now
call ``the Malthusian dilemma.'' He didn't have Excel to do the arithmetic for
him, but we do. Use it to estimate when Malthus's predicted
disaster would occur. Was Malthus right?

\end{abcd}

\begin{sol}

\begin{abcd}

\item Write a linear function that models food production.

Let $F$ stand for food production, measured in thousands of food
units (so that we don't use too many zeroes) and $T$ for the number of
years since 1798. Then Malthus's model for food production is the equation
\begin{equation*}
    F = 280T + 7000.
\end{equation*}
It's a straight line with slope 280K units/year and intercept 7000K units.

\item Write an exponential function that models population growth.

Let $P$ stand for the population of Britain, in thousands of people,
and $T$ for the number of years since 1798. Then Malthus's model for
population growth is the exponential growth equation
\begin{equation*}
    P = 7000 \times (1.028)^T.
\end{equation*}

\item
Was there enough food for each individual in Britain in 1798?


There was sufficient food in Britain in 1798 since
Malthus arranged his models so that they started out in perfect
agreement --- the population of 700K people needed exactly the 700K food units
produced that year.

\item Using Malthus's models, determine whether there would be enough food
for each individual in Britain in 1800.

I could do this problem with a calculator, plugging in the value $T = 2$
in each of the equations above. But reading ahead to the next problem
I decided to use the LinearExponential.xlsx spreadsheet instead.
I set 


\begin{verbatim}
     7000   START: Starting value (Kunits)
      280   ABSCHANGE: Absolute increase (Kunits/year)
    1.028   RELCHANGE: Relative increase ( /year)
\end{verbatim}

Then the spreadsheet shows

\begin{verbatim}
    year 2     7560 Kunits food    7397 Kpeople
\end{verbatim}

so there's more than enough food for the population.

\item
Malthus claimed that the population in Britain would eventually
outstrip the food supply --- a prediction we now
call ``the Malthusian dilemma.'' He didn't have Excel to do the arithmetic for
him, but we do. Use it to estimate when Malthus's predicted
disaster would occur. Was Malthus right?

By adding a few lines to the table in the spreadsheet from 1 to 2 I
could see that the exponential function caught up with the linear
function at about 26 years. So Malthus predicted that in 1824 people
in Britain would start to go hungry. 

History says that didn't happen --- famine hasn't hit Britain ever
(there were short rations during and right after World War II, but
that's not what Malthus had in mind).
\end{abcd}

\end{sol}

\end{exx}

\begin{exx}{\hassolution\sref{compounding}\gref{exponentialgrowth}
\gref{interest}}
The \myindex{pawn shop} \myindex{business model}.

On April 9, 2011 \theTimes{} reported on a pawn shop that opened in an
ex-Blockbuster store:

\begin{quotation}
The borrowers are given 60 days to pay back the
loan, and La Familia charges a 20 percent interest rate per month. (So
for a \$100 loan, the borrower would need to pay back \$140 after 60
days.)%
\begin{csmr}
S. Clifford,
Other Retailers Find Ex-Blockbuster Stores Just Right,
\theTimes{} (April 8, 2011),
\url{www.nytimes.com/2011/04/09/business/09blockbuster.html}
\access{July 23, 2015}.
\csmrcomment{38 words fair use}
\end{csmr}
\end{quotation}

\begin{abcd}
\item
Explain why 20\% interest per month on a \$100 loan for two months 
would actually require repayment of a little more than \$140.

\item What is the annual interest rate when this business lends money?
\end{abcd}

\begin{sol}

\begin{abcd}
\item
Explain why 20\% interest per month on a \$100 loan for two months 
would actually require repayment of a little more than \$140.

After the first month the debt is $1.2 \times \$100 = \$120$. At the
end of two months it's $1.2 \times \$120 = \$144$. You have to pay
interest on the interest.

\item What is the annual interest rate when this business lends money?

The compound interest formula (coming from the 1+ trick) says that to
find the total owed after a year I should multiply the original loan
amount by $(1.2)^{12}= 8.92$. That corresponds to an interest rate of
792\% --- almost 800\%!

\end{abcd}

\end{sol}

\end{exx}



\begin{exx}{\hassolution\worthy\sref{excelexponential}\gref{exponentialgrowth}
\gref{interest}}
Playing with exponential growth.

Open the spreadsheet \link{exponentialGrowth.xlsx} and
describe what happens to the graph when you make each of the following
experiments. If you can see easily what happens to the numbers,
describe that too.

\begin{abcd}

\item Change the value of \excel{START} from 1,000 to 10, then 100,
then 10,000. Change it to some other random positive values that
aren't as nice.

\item Change the value of \excel{START} from 1,000 to $-1,000$.

\item Change the value of \excel{RELCHANGE} to 1.

\item Change the value of \excel{RELCHANGE} to 1.01 (1\% growth). Fit
a linear trendline to the data. What is the $R$-squared value? What does
it tell you?

\item Change the value of \excel{RELCHANGE} to 2. Why does the graph
look flat at 0 as far as $T = 20$? Is it really flat?

\item Change the value of \excel{RELCHANGE} to 10.

\item Change the value of \excel{RELCHANGE} to 0.9 (a 10\%
decrease ).

\item (Optional) Can you figure out how we got the label on the graph
to incorporate the values of \excel{START} and \excel{RELCHANGE}?

\end{abcd}

\begin{sol}

Here's one student's solution. 

\begin{abcd}

\item  When I change the \excel{START} value to 10, then to
100 and then to 10,000 the graph keeps the exact same shape. The
values also contain the exact same numbers. The decimal place just moves
a place or two depending on the value of \excel{START}. 

\item  When I change the \excel{START} value to $-1,000$ the
graph flips upside down to represent negative numbers moving away from
0 exponentially. 

\item  When I change \excel{RELCHANGE} to 1 the graph becomes a
flat line and all of the values remain 1,000 because there is no
change. 

\item When I change \excel{RELCHANGE} to 1.01 the linear trend line
shows an $R^2$ value of 0.99842. This shows that a relative change of
1.01, while technically exponential growth, is very close to linear
growth because a 1\% growth will take many years to show a large growth
year to year. 

\item  When I change \excel{RELCHANGE} to 2 the graph flat
until $T=20$ but it really isn't. The reason it looks
flat is because exponential growth doubles and the doubling gets so
large at that time that the numbers before it are too small on the
graph to show a visible change. 

\item When I change \excel{RELCHANGE} to 10 the graph is like the
  previous one but  more dramatic, In other words, showing even larger
  increases over time.  

\item  Setting \excel{RELCHANGE} to 0.9 shows exponential decay. The
interesting part is that unlike growth where the numbers start with
relatively small increases and they get larger, with decay the
decrease starts at a fixed large number decreases at a small rate. 

\end{abcd}

\end{sol}
\end{exx}

\begin{exx}{\hassolution\sref{excelexponential}\gref{exponentialgrowth}} 
Five percent.

If you try to use \link{linearExponential.xlsx} to see when
exponential growth at 5\% catches linear you see that it's still
behind at 20 years, which is as far as the table goes.

Modify the spreadsheet to determine when it catches up.

\begin{sol}
When I changed \excel{RELCHANGE} from \excel{1.08} to \excel{1.05}
the curves on the chart no longer crossed.

I selected cells \excel{A34:C34}, copied, then pasted into the next 10
rows. Looking at the numbers, I could see that columns \excel{B} and
\excel{C} were about equal at 27 years.

Unfortunately, the chart didn't update itself! There are several
tricks to make that happen. One is to right click on one of the curves
in the chart, choose \excel{Select Data ...} from the menu, and edit
the \excel{Chart data range:}. Another is to \excel{Insert} a whole
bunch of rows between rows \excel{33} and \excel{34} and
then copy row \excel{33} to those empty rows. 

\end{sol}
\end{exx}

\begin{exx}{\untested\sref{excelexponential}\gref{exponentialgrowth}
\gref{interest}}
Playing with linear vs exponential growth.

Use the spreadsheet \link{linearExponential.xlsx} to answer these questions.

\begin{abcd}

\item How does changing the initial investment change the time it
takes for the exponential function to catch up with the linear
function?

\item Does doubling or tripling both the initial
investment and the absolute change affect the time it takes for the
exponential function to catch up to the linear function?

\end{abcd}

\end{exx}

\begin{exx}{\hassolution\sref{excelexponential}\gref{exponentialgrowth}
\gref{interest}}
Deals you can't believe.

The data in this problem aren't real. But the problem is interesting
and instructive, so it's worth spending time on. 

Suppose you are shopping for a car and find three deals advertised:

\begin{itemize}

\item
Make a \$10,000 down payment and pay only
\$100 per month for two years.  

\item
Just \$5000 down, monthly payments start at a low \$50 and
increase by \$50 each month for two years.

\item
Give me \$1.00 today and take the car home! Pay 1 penny for the
first month. Then double your payment each month. After two
years, the car is yours.  
\end{itemize}

\begin{abcd}

\item Before you do any calculating, which deal do you think is best? Why?

\item  What would your monthly payments be in the second and tenth
months if you take the second dealer's offer? 

\item  What would your monthly payments be in the second and tenth
months if you take the third dealer's offer? 

\item For each deal, write an algebraic expression that gives the
monthly payment. 


\item Use Excel to calculate your total payments for the 24 months. 
Set up four
columns as in Table~\ref{carDeals}.
Then tell Excel how to fill in the columns to 24
months. Finally, use the \excel{SUM} function to add up the payments. 


\begin{table}
\centering
\begin{tabular}{|r|r|r|r|}
\hline
Month & \multicolumn{3}{c|}{Payment} \\
\hline
  &  Deal 1  &  Deal 2  & Deal 3 \\
\hline
 (down)     0 	&   10,000 &     5,000   &    1.00 \\
          1 &         100   &      50  &      0.01 \\
          2 & & & \\
          \vdots & & & \\
         24 & & & \\
\hline
Total & & & \\
\hline
\end{tabular}
\caption{Three car deals}
\tablesource{Hand built data.}
\label{carDeals}
\end{table}

\item
Now use what your calculations tell you to compare the three
deals. Which is best? Which is worst? 

\end{abcd}


\begin{sol}

\begin{abcd}

\item Before you do any calculating, which deal do you think is best? Why?

Most people think that the last deal might be the best since the
monthly payments start out so small.

\item  What would your monthly payments be in the second and tenth
months if you take the second dealer's offer? 

The payments would be \$100 in the second month and \$500 in the tenth.

\item  What would your monthly payments be in the second and tenth
months if you take the third dealer's offer? 

The payments would be \$0.02 in the second month and \$5.12 in the tenth.

\item For each deal, write an algebraic expression that gives the
monthly payment. 

Writing $M$ for the month and $P$ for the payment, the three
expressions are
\begin{equation*}
	P = 100
\end{equation*}
for the first deal,
\begin{equation*}
	P = 50M
\end{equation*}
for the second and
\begin{equation*}
	P = 0.01 \times 2^M 
\end{equation*}
for the third.

\item Use Excel to calculate your total payments for the 24 months. 

The spreadsheet at \slink{UnbelievableDealSolution.xlsx} tells me that
the three totals are \$12,400,	\$20,000 and \$167,773.15 for the
three deals.

\item
Now use what your calculations tell you to compare the three
deals. Which is best? Which worst? 

Clearly the first deal is the best and the third is the worst --- by a
lot!
\end{abcd}

\end{sol}
\end{exx}

\begin{exx}{\hassolution\complex\sref{excelexponential}\gref{exponentialgrowth}}
Green energy in China.
\index{green energy}

In the December 21 \& 28 2009 issue of \theNewYorker{} Evan Osnos
wrote in his essay 
``Green Giant: Beijing's crash program for \myindex{clean energy}'' that
China's spending on R \& D, now seventy billion dollars a year, has
been growing at an annual rate of about twenty percent for two
decades.%
\begin{csmr}
E. Osnos,
Green Giant,
\emph{The New Yorker} (December 21, 2009),
\url{www.newyorker.com/magazine/2009/12/21}
\access{July 29, 2015}.
\csmrcomment{paraphrase, so no permission needed}
\end{csmr}

\begin{abcd}

\item What does ``R\&D'' stand for?

\item Use Excel to build a chart of annual Chinese R\&D expenditures
for the years 1989-2008. 

\item Add a data column showing the annual expenditures adjusted for
inflation (use the United States cost of living index) and display
that data on your chart.

\end{abcd}

\begin{sol}

``R\&D'' stands for ``Research and Development''.

See \slink{ChinaRandDSolution.xlsx}. 

I did a little bit of extra work in
that spreadsheet: I found an exponential trendline for the inflation
adjusted expenditures. That trendline shows that the increase adjusted
for inflation was about 17\% per year, not 20\%. That makes sense,
since the average inflation rate was about 3\% per year.

\end{sol}

\end{exx}

\begin{exx}{\hassolution\sref{exponentialdepreciation}
\gref{exponentialgrowth}}
Car excise tax.
\index{depreciation}

In Massachusetts you pay excise tax each year on the current value of 
your automobile. Assume for the sake of this problem that the
rate is 3\%, so you would pay \$600 in excise tax in the first year
you owned a new \$20,000 car.

Use Excel to answer the following questions.

\begin{abcd}
\item Suppose the car depreciates linearly at a rate of \$1,800 per
year. What formula calculates the amount of excise tax you pay as a
function of the age of the car? 

\item If you own the car for ten years, 
what will the car be worth then and how
much total excise tax will you have paid?

\item Answer the same questions 
if it depreciates at the rate of 13\% per year.

\item Find real data on the way a new car depreciates in value. Is an
exponential model a good approximation?

\end{abcd}

\begin{sol}

\begin{abcd}
\item Suppose the car depreciates linearly at a rate of \$1,800 per
year. What formula calculates the amount of excise tax you pay as a
function of the age of the car?

Writing $E$ for excise tax and $A$ for the age of the car in years,
the formula is
\begin{equation*}
	E = 0.03 \times (20,000 - 1,800 A).
\end{equation*}

\item If you own the car for ten years, 
what will the car be worth then and how
much total excise tax will you have paid?

After 10 years the car will be worth \$3,800 and I will have paid a
total of \$3,570 in excise taxes.

You can see my work in the spreadsheet at
\slink{ExciseTaxSolution.xlsx}.

\item Answer the same questions 
if it depreciates at the rate of 13\% per year.

The formula is
%
\begin{equation*}
	E = 0.03 \times (20,000(0.87)^A).
\end{equation*}

After 10 years the car will be worth \$5,711 and I will have paid a
total of \$3,469 in excise taxes.

That's interesting. The car is worth more and my total tax bill is a
little bit less. That's because the exponential depreciation cuts the
value of the car rapidly at the start, then more slowly.

\item Find real data on the way a new car depreciates in value. Is an
exponential model a good approximation?

The car depreciation calculator at
\url{www.money-zine.com/Calculators/Auto-Loan-Calculators/Car-Depreciation-Calculator/}
says that with a low depreciation rate the first year depreciation a
\$20,000 car would be \$3,600 and the total ten year depreciation
would be \$14,254. That's pretty close to the exponential depreciation
in this problem. If I took the trouble to look at the depreciation a
year at a time I would probably see something like a big hit in the
first year and then perhaps something between 10\% and 12\% a year
after that.
  	  	  	 
\end{abcd}

\end{sol}
\end{exx}


\begin{exx}{\hassolution\sref{doublingtimes}\gref{doublingtimes}}
Iodine 131.\index{iodine 131}

An article in \theTimes{} on April 6, 2011 soon after the
\myindex{Fukushima} disaster discussed levels of
radioactive iodine (iodine 131) in fish caught near Japan.  The
article noted that Japan recently revised the safety limit for iodine 
131 in fish to 2,000 becquerels per kilogram.  (A becquerel is a
measure of radiation.) 

Radioactive iodine has a \myindex{half-life} of about 8 days.

If a fish contained 10,000 becquerels of iodine 131 per kilogram, how
long would it take for the iodine to decay to a ``safe'' level? 

\begin{sol}

If a fish
contained 10,000 becquerels of iodine 131 per kilogram, how long would
it take for the iodine to decay to a ``safe'' level? 

In 8 days the level would be 5,000 becquerels per kg. In 16 days it
would be 2,500, so almost safe. In 24 days it would be only 1,250. So
the answer is somewhere between 16 and 24 days --- I estimate about 19
days. I could find the answer exactly with the formula and trial and
error guesses but this estimate is good enough.

\end{sol}

\end{exx}

\begin{exx}{\untested\sref{doublingtimes}\gref{exponentialgrowth}\gref{doublingtimes}}
Quadrupling time.

\begin{abcd}

\item Explain why the quadrupling time in exponential growth is just
twice the doubling time. 

\item Show that quadrupling time is given by a ``Rule of
140'' analogous to the rule of 70.
\end{abcd}
\end{exx}


\begin{exx}{\hassolution\sref{doublingtimes}\gref{exponentialgrowth}\gref{doublingtimes}\gref{interest}}
Tripling time.

Suppose you invest \$1000 at 10\% interest compounded every
year. (That's a pretty good rate of return if you can get it --- don't
trust a Madoff\index{Madoff, Bernard} promise!)

\begin{abcd}

\item How long will it be until your balance is \$3000? 

\suspend{abcd}

\begin{hint} 
Use the \texttt{ExponentialGrowth.xlsx} spreadsheet, or find the answer
by trying different 
values for $T$ (\# of years)  in the formula until you find one that
gets you close to \$3000.)  
\end{hint}

\resume{abcd}

\item Find the tripling time for some other interest rates.

\item Check that the \myindex{tripling time} in exponential growth is given
(approximately) by a ``Rule of 110.'' 

\item Check that $e^{1.1} \approx 3$.

\end{abcd}

\begin{sol}
\begin{equation*}
\$1000 \times (1.07)^{15} = \$4177.24817
\end{equation*}
so 15 years is too long. Try 10:
\begin{equation*}
\$1000 \times (1.07)^10 = \$2593.74246.
\end{equation*}
That's too short a time. After 11 years:
\begin{equation*}
\$1000 \times (1.07)^11 = \$2853.11671.
\end{equation*}
Close. How about 12 years?
\begin{equation*}
\$1000 \times (1.07)^12 = \$3138.42838
\end{equation*}
so the investment will triple in between 11 and 12 years.

That's consistent with a Rule of 110, which predicts a tripling time
of $110/10 = 11$ years.

\end{sol}

\end{exx}

\begin{exx}{\untested\sref{doublingtimes}\gref{doublingtimes}\gref{interest}}
Compounding very frequently.

\begin{abcd}

\item
Calculate the \myindex{effective rate} for 8\% annual interest when it's
compounded weekly, daily, hourly and once every second.

\item
Estimate the effective rate if the interest is compounded every
instant.

\suspend{abcd}

\begin{hint}
Your computations should suggest that it won't be infinite,
which might have been your first guess.
\end{hint}

\resume{abcd}

\item Redo the calculations starting with a 25\% annual
increase. (Not realistic for interest on a bank account!) Show that
the Rule of 70 for doubling times is more accurate the more frequently
you compound the interest.

\end{abcd}

\end{exx}

\begin{exx}{\hassolution\complex\worthy\sref{doublingtimes}
\gref{exponentialgrowth}\gref{doublingtimes}}
How fast does information double?

In the Preface to the Carnegie Corporation report \emph{Writing to
Read} Vartan Gregorian wrote that he's been told that the amount of
available information doubles every two to three years.%
\begin{csmr}
S, Graham and M. Hebert,
Writing to Read,
Carnegie Corporation (2010),
\url{all4ed.org/wp-content/uploads/2010/04/WritingToRead.pdf}
\access{July 23, 2015}.
\csmrcomment{paraphrase}
\end{csmr}

\begin{abcd*}
\item What growth rate in percent per year would lead to a doubling
time of two to three years?

\item Who is Vartan Gregorian?

\item Can you verify his assertion?

\end{abcd*}

\begin{hint}
You can answer the first question using guess-and-check until you're
close enough. You'll need a web search for the second and a very
good web search for the third, which is open-ended.
\end{hint}
\begin{sol}
\begin{abcd*}
\item What growth rate in percent per year would lead to a doubling
time of two to three years?

I experimented using the exponential growth spreadsheet from the text
and found that a growth rate of about 41\% per year doubles things in
two years. (That's because $1.41 \times 1.41 \approx 2$: the square
root of 2 is the exact answer, but not needed in this problem or this
book.)

For three year doubling the spreadsheet experiment gives a growth rate
of about 26\% per year. (1.26 is close to the cube root of 2.)

So doubling every two to three years corresponds to an annual increase
of between 26 and 41 percent.

\item Who is Vartan Gregorian?

Wikipedia tells me that Vartan Gregorian is the president of Carnegie
Corporation of New York (in 2011).

\item Can you verify his assertion?

From
\url{solace.com/blog/handling-data-growth-as-digital-information-doubles-every-18-24-months/}

\begin{quotation}
Various studies have claimed that the amount of digital information in
the world doubles every 18 to 24 months.  
\end{quotation}

\ldots but where are the actual studies?

Then there's this:

\begin{quotation}
It is projected that just four years from [2010], the world's information
base will be doubling in size every 11 hours.%
\begin{csmr}
D. Farber,
2010: Data doubling every 11 hours,
Between the Lines (February 13, 2007),
\url{www.zdnet.com/blog/btl/2010-data-doubling-every-11-hours/4497}
\access{July 23, 2015}.
\csmrcomment{21 words fair use}
\end{csmr}
\end{quotation}

So the actual doubling time is hard to pin down even approximately.
\end{abcd*}

\end{sol}
\end{exx}


\begin{exx}{\hassolution\sref{exponentialmodels}
\gref{exponentialregression}\gref{doublingtimes}}
Bacteria doubling time.

Find the approximate doubling times for strains
R and S in the bacteria  growth example in
\sref*{exponentialmodels}.


\begin{sol}
I tried to make an estimate just by looking at the tables. The best I
could do was to say that for both the doubling time seemed to be about
3 hours.

To get better estimates I looked at the exponential trendlines. For
strain R the exponential factor is
\begin{equation*}
	e^{0.2732} = 1.314163051,
\end{equation*}
which corresponds to a 31\% increase per hour.
Then I used the Rule of 70 to get the doubling time:
\begin{equation*}
	\frac{70}{31.4} \approx 2.23
\end{equation*}
so the doubling time is about two and a quarter hours, so two hours
and 15 minutes.

For strain S
\begin{equation*}
	e^{0.2558} = 1.291494403
\end{equation*}
which corresponds to a 29\% increase per hour.
Then the Rule of 70 says
\begin{equation*}
	\frac{70}{29.1} \approx 2.40
\end{equation*}
so the doubling time is about two hours and 25 minutes --- just ten
minutes more than for strain R.

\end{sol}
\end{exx}

\begin{exx}{\hassolution\sref{exponentialmodels}\gref{exponentialgrowth}
\gref{exponentialregression}}
When will R catch S?

The population of strain S outnumbers that of strain R for the 
entire first 17 hours of the experiment discussed in 
\sref*{exponentialmodels}. But the exponential trendline equation
shows that strain R is growing faster. If the exponential growth were
to continue (which it didn't) when would strain W catch up?

\begin{hint}
Guess a number of hours, try your guess in the two exponential
equations, then adjust your guess up or down until the answers match.
\end{hint}

\begin{sol}
I built a small spreadsheet with these formulas in columns \cell{A:D}

\begin{center}
\begin{tabular}{|r|r|r|r|}
\hline
Time & S & R & R/S \\
\hline
64 & =6.5913*EXP(0.2558*A4) & =2.13*EXP(0.2732*A4) & =C4/B4 \\
=A4+1 & =6.5913*EXP(0.2558*A5) & =2.13*EXP(0.2732*A5) & =C5/B5 \\
=A5+1 & =6.5913*EXP(0.2558*A6) & =2.13*EXP(0.2732*A6) & =C6/B6 \\
\hline
\end{tabular}
\end{center}

I experimented with the value in cell \cell{A4} to find out where the ratio
R/S switched from less than 1 to greater.

Here are the numbers.

\begin{center}
\begin{tabular}{|r|r|r|r|}
\hline
Time & S & R & R/S \\
\hline
64 & 84897108.9 & 83547173.28 & 0.984099157 \\
65 & 109644141 & 109794608.1 & 1.001372323 \\
66 & 141604794.4 & 144288017.2 & 1.018948672 \\
\hline
\end{tabular}
\end{center}

They show that strain R would catch up at about 65 hours.

\end{sol}
\end{exx}

\begin{exx}{\untested\sref{exponentialmodels}\gref{exponentialregression}}
The magic number $e$.\index{e@\textit{e}} 

\begin{abcd*}

\item
Find the value of $e$ in Excel using the formula \excel{=EXP(1)}.

\item Find the value from Google with the same formula (without the
equal sign). Check that 
the answers agree as far as they go together.

\item Which provides more digits? 

\item Can you get more precision from Excel by formatting the cell in
which the number appears?

\item
Find even more digits with an internet search.
\end{abcd*}

\end{exx}

\begin{exx}{\hassolution\sref{exponentialmodels}\gref{exponentialgrowth}}
Educating mothers saves lives, study says.

On September 17, 2010 \theGlobe{} carried an Associated Press report
on a study that found that the death rate for children under 5 dropped
by nearly 10 percent for every extra year of their mother's
education. That education saved 4.2 million children in the developing
countries in 2009.

The story continued with these baseline numbes:

\begin{quotation}
In 1970, women aged 18 to 44 in developing countries went to school
for about two years. That rose to seven years in 2009.%
\begin{csmr}
Educating mothers saves lives, study says,
Associated Press reported in \theGlobe (September 17, 2010),
\url{www.boston.com/news/world/europe/articles/2010/09/17/educating_mothers_saves_lives_study_says/}
\access{March 30, 2020}.
\end{csmr}
\end{quotation}

\begin{abcd}
\item 
How much did the death rate for children under 5 decline from 1970 to 2009?

\item
Build as much as you can of the exponential  model implicit in this
quotation. What are the independent and dependent variables? What is
the annual relative change?

\end{abcd}

\begin{sol}

\begin{abcd}
\item 
How much did the death rate for children under 5 decline from 1970 to 2009?

Since the number of years of schooling increased by $7-2 = 5$ years
the death rate decreased by 10\% five times. Since $0.9^5= 0.59049$
the death rate in 2009 was only 60\% of what it was in 1970.

\item
Build as much as you can of the exponential  model implicit in this
quotation. What are the independent and dependent variables? What is
the annual relative change?

The independent variable is $y$, the number of years of education for
women. The dependent variable is $D$, the death rate for children
under five.
The annual change is a decrease of 10\%, which means the equation
looks like 
\begin{equation*}
	D = S \times 0.9^{y}.
\end{equation*}
$S$ is the death rate when women don't have any schooling at all.
There isn't enough information to find its value.

\end{abcd}


\end{sol}
\end{exx}

\begin{exx}{\hassolution\sref{exponentialmodels}\gref{exponentialgrowth}
\gref{exponentialregression}}
Email.

On May 30, 20ll Virginia Heffernan blogged in \theTimes{} that the
number of email accounts grew from about 15 million in the early 1990s
to 569 million by December 1999, and that today [when she was writing]
there are [were] more than 3 billion.%
\begin{csmr}
V. Heffernan,
The Trouble With E-Mail,
\theTimes{} (May 29, 2011),
\url{opinionator.blogs.nytimes.com/2011/05/29/the-trouble-with-e-mail/}
\access{July 23, 2015}.
\csmrcomment{paraphrase, no permission needed}
\end{csmr}

\begin{abcd}
\item Is this exponential growth?

\item Can you use these numbers to make predictions?
\end{abcd}

\begin{hint}
Compare the relative change in the number of email accounts for the
two time periods (early 1900s to end of 1999, end of 1999 to the time
of the blog post).
\end{hint}
\begin{sol}
\begin{abcd}
\item Is this exponential growth?

I'll try 1990 for ``the early 1990s''. Then the data tell me that in
those 10 years the number of email accounts grew by a factor of
\begin{equation*}
\frac{569 \text{ million}}{15 \text{ million}} \approx 38.
\end{equation*}
(The \myindex{relative change} is what matters here.)

In the eleven and a half years from the end of 1999 to May of 2011 the
number grew by a factor of only
\begin{equation*}
\frac{3 \text{ billion}}{569 \text{ million}} \approx 5.3.
\end{equation*}

If the growth were truly exponential throughout the time then the
figure for 2011 would have to be more than 38 times as large as
the figure for the end of 1999.

\item Can you use these numbers to make predictions?

No. The number of email accounts grew very rapidly at the beginning
and more slowly after that. I don't have enough data to make
predictions.

\end{abcd}

\end{sol}
\end{exx}



\begin{exx}{\hassolution\sref{exponentialmodels}\gref{exponentialregression}}
When will India pass China?

In an article dated April 1, 2011 on the website About.com you could
read that India, the world's second largest country, had a population
of 1.21 billion. India was expected to pass China by 2030, when it
would have more than 1.53 billion people. China's population then
would be just 1.46 billion.

The article noted that India's growth rate of 1.6\% per year doubles
its population in less than 44 years.%
\begin{csmr}
M. Rosenberg,
India's Population,
About.com (April 1, 2011),
\url{geography.about.com/od/obtainpopulationdata/a/indiapopulation.htm}
\access{July 23, 2015}.
\csmrcomment{paraphrase}
\end{csmr}

\begin{abcd}

\item 	Is the article correct in stating that an annual growth rate
of 1.6\% means India's population will double in 44 years? 

\item 	Assuming that India's growth rate remains 1.6\% annually, what
will its population be in 2030 when it surpasses China's?   

\item Assuming that India's growth rate remains 1.6\% annually from
2011 on, what
will its population be in the year 2100? Compare that figure to the
current population of the world. Do you think India's growth rate can
in fact continue at 1.6\% for the 89 years from 2011 to 2100?

\end{abcd}

\begin{sol}
\begin{abcd}

\item 	Is the article correct in stating that an annual growth rate
of 1.6\% means India's population will double in 44 years? 

The rule of 70 says that a growth rate of 1.6\% per year has a
doubling time of about $70/1.6 = 43.75$ or about 44 years.

\item 	Assuming that India's growth rate remains 1.6\% annually, what
will its population be in 2030 when it surpasses China's?   

2030 is 19 years from 2011. Then India's population will be
$1.21 \times 1.016^{19} = 1.64$ billion.

That's a lot larger than the predicted peak for China in that
year, so I suspect the prediction is based on an assumption that
India's growth rate will slow down, although not enough to keep it
from passing China.

\item Assuming that India's growth rate remains 1.6\% annually from
2011 on, what
will its population be in the year 2100? Compare that figure to the
current population of the world. Do you think India's growth rate can
in fact continue at 1.6\% for the 89 years from 2011 to 2100?

At the 2011 growth rate, India's population in 89 years would be
$1.21 \times 1.016^{89} = 4.97$ billion, or just about 5 billion. The
2011 world population is 7 billion. I don't believe that India
could hold nearly as many people then as the whole world held in 2011, so
I don't think the growth rate can be 1.6\% per year for the rest of
the century.

Here's another interesting way to do this problem. The doubling time is about
44 years. Twice that is 88 years, so almost to the end of the
century. At the 2011 rate of increase, by then the population will
have quadrupled, to about 4.8 billion.

\end{abcd}

\end{sol}

\end{exx}

\begin{exx}{\hassolution\sref{exponentialmodels}
\gref{exponentialregression}}
Health care expenditures grow.

The National Health Expenditures report, released in January 2009,
stated that overall health care spending in the United States rose
from \$7062 per person in 2006 to \$7421 per person in 2007. 

\begin{abcd}
\item Calculate both the absolute change and percentage change in
health care spending per person from 2006 to 2007.

\item Using 2006 as your starting year (2006 = year 0), determine an
exponential equation that calculates the amount of health care
spending over time assuming the annual percentage change stays the
same. Clearly identify the variable names and symbols in your
equation. 

\item Using 2006 as your starting year (2006 = year 0), determine a
linear equation that calculates the amount of health care spending
over time assuming the annual absolute change stays the same. Clearly
identify the variable names and symbols in your equation. 

\item  Create an Excel spreadsheet to compare the two growth models'
predictions for health care spending through the year 2021. Include a
chart showing both models.  

\item Which model first predicts that U.S. health care spending will
reach a level of \$10,000 per person? In what year will that occur? 
\end{abcd}

\begin{sol}
\begin{abcd}
\item Calculate both the absolute change and percentage change in
health care spending per person from 2006 to 2007.

The absolute change in health care spending was
%
\begin{equation*}
\$7421 -  \$7062 = \$359.
\end{equation*}
%
To find the percentage change I computed
\begin{equation*}
\frac{\$7421}{\$7062} = 1.05083546,
\end{equation*}
%
which corresponds to a percentage change of about 5.1\%.

\item Using 2006 as your starting year (2006 = year 0), determine an
exponential equation that calculates the amount of health care
spending over time assuming the annual percentage change stays the
same. Clearly identify the variable names and symbols in your
equation. 

Let $H$ represent health care spending, in dollars per person, and $y$
the number of years since 2006. Then the equation is
%
\begin{equation*}
H = 7062\times 1.0508^y.
\end{equation*}

\item Using 2006 as your starting year (2006 = year 0), determine a
linear equation that calculates the amount of health care spending
over time assuming the annual absolute change stays the same. Clearly
identify the variable names and symbols in your equation. 

Using $H$ and $y$ as above, the equation is
%
\begin{equation*}
H = 7062 + 359y.
\end{equation*}
The slope is the constant absolute change, in dollars per year.

\item  Create an Excel spreadsheet to compare the two growth models'
predictions for health care spending through the year 2021. Include a
chart showing both models.  

I modified the spreadsheet from the book that compares linear and
exponential growth to create \slink{HealthCareSolution.xlsx} where I
drew this chart:

\figfile{HealthCareSolutionChartcropped.pdf}
\begin{center}
\includegraphics[height=60mm]{\thefigurefilename}
\end{center}
\figfile{}

\item Which model first predicts that U.S. health care spending will
reach a level of \$10,000 per person? In what year will that occur? 

The exponential growth model predicts higher values for all years
after 2007, so it predicts \$10,000 sooner --- in 2014. The linear
model predicts \$10,000 a year later.
\end{abcd}

\end{sol}
\end{exx}

\begin{exx}{\hassolution\sref{exponentialmodels}\gref{exponentialgrowth}\gref{exponentialregression}}
Joe Seeley, \emph{in memoriam}.\index{Seeley, Joe}

Joe Seeley died at age 50 in the fall of 2012.

He was a brave and witty blogger at 
\url{joes-blasts.blogspot.com/} throughout his
hospitalization,
creating virtual lemonade from the sourest of lemons. 
I think his words helped him; I know they helped
those who cared for him to cheer him on. They will help the hospital
staff care better for patients who come after him. And they will help
you learn a little mathematics.

Figure~\ref{fig:joeseeleybloodcount} appeared in the blog at a hopeful
moment in his odyssey. It shows Joe's white blood cell
counts on days following a stem cell transplant. 
He chose white for the bars, to symbolize white blood cells,
and red for the background, for blood in general. I wrote him about it:

\figfile{JoeSeeleyBloodCountCropped.pdf}
\begin{figure}
\centering
\includegraphics[height=60mm]{\thefigurefilename}
\caption{Proliferating white blood cells}
\figsource{\url{joes-blasts.blogspot.com/} }
\figcomment{We have permission to use this chart.}
\label{fig:joeseeleybloodcount}
\end{figure}
\figfile{}

\begin{quotation}
March 18, 2011 6:58 AM 

Ethan Bolker said \ldots

Exponential growth is good! \ldots
Will you still have a daily double after the predicted short
dip? May I use your data for my quantitative reasoning class at UMass
Boston? 

March 18, 2011 9:48 AM  

Joseph Seeley said \ldots

I will not see doubling again, unless something is wrong. Over the
next few months, the counts will rise and fall, sometimes for no
reason that the doctors can determine. 

I hereby authorize the use of my blood count data for any and all
educational purposes. 
\end{quotation}

\begin{abcd}

\item Enter the data in Excel. Reproduce Joe's chart. Match the
formatting (labels, colors, sizes, fonts) as well as you can.

\item Create an exponential trendline for the data.

\item Use common sense or your trendline to predict when Joe's white
blood cell count will be in the normal range. 

\item His white blood count on March 17 was 5100. Does that
match your prediction?

\item Modify your chart to include this new information. 
Mark it with a suitable exclamation!

\end{abcd}

\begin{sol}
See \slink{JoeSeeleySolution.xlsx}. I answered all the questions in
the spreadsheet.

\end{sol}
\end{exx}


\begin{exx}{\hassolution\sref{exponentialdepreciation}\gref{depreciation}}
How impatient are you?

\theGlobe{} reported on September 24, 2012 that:

\begin{quotation}
[MIT grad Andy Berkheimer] found that [YouTube] viewers
  start closing out if there's even a two-second delay. Every
one-second delay after that results in a 5.8 percent increase in the
number of people who give up. A 40-second wait costs a video nearly a
third of its audience.\index{YouTube}% 
\begin{csmr}
M. B. Farrell,
MIT grad led team that built faster YouTube player,
\theGlobe{} (September 24, 2012),
\url{www.bostonglobe.com/business/2012/09/23/building-faster-youtube/JqbVsEFUJfa5tpQmgbujkL/story.html}
\access{March 30, 2020}.
\csmrcomment{Globe, OK}
\end{csmr}
\end{quotation}


Show that at this rate more than ninety percent of the viewers would
give up after 40 seconds --- not the ``nearly a third'' in the quote.

\begin{sol}
Since
%
\begin{equation*}
0.942^{40} = 0.0916296677 <  10\%,
\end{equation*}
%
fewer than 10\% of the viewers are still around after 40 seconds. More
than 90\% have given up.

\end{sol}

\end{exx}

\begin{exx}{\untested\sref{compounding}\gref{exponentialgrowth}}
Does compounding always matter?
\index{Silver, Nate}

On page 32 of his excellent and highly recommended
\emph{The Signal and the Noise} Nate Silver wrote
\begin{quotation}
[Over the] 100-year-period from 1896 through 1996
\ldots sale prices of houses had increased by just 6 percent
\emph{total} after inflation, or about 0.06 percent annually.%
\begin{csmr}
N. Silver,
\emph{The Signal and the Noise},
page 32,
Penguin Press (September 27, 2012).
\csmrcomment{27 word quote must count as fair use. I could
  paraphrase it if necessary}
\end{csmr}
\end{quotation}

%CHANGE  precent -> percent
Silver clearly divided six percent by 100 to reach his 0.06 percent
annually conclusion. But that's not how percent increases work. The
0.06 percent annual increase must be compounded. 

Would taking compounding into account change Silver's fundamental
point?

\begin{sol}

Would taking compounding into account change Silver's fundamental
point?

To calculate the result of compounding 0.06 percent per year for 100 years
I found
%
\begin{equation*}
1.0006^100 = 1.0618174413.
\end{equation*}

That rounds to 6 percent. The extra two tenths of a percent from
compounding doesn't really matter in the argument Silver is making.

\end{sol}

\end{exx}


\begin{exx}{\hassolution\sref{compounding}\gref{interest}}
\myindex{Cuba}, you owe us \$7 billion.

On April 18, 2014 Leon Neyfakh wrote in \theglobe{} that property
confiscated by the Cuban government in the 1959 revolution was
\begin{quotation}
\ldots originally valued at \$1.8 billion, which at 6
 percent simple interest translates to nearly \$7 billion today.%
\begin{csmr}
L. Neyfakh,
Cuba, you owe us \$7 billion,
\theGlobe{} (April 18, 2014),
\url{www.bostonglobe.com/ideas/2014/04/18/cuba-you-owe-billion/jHAufRfQJ9Bx24TuzQyBNO/story.html}
\access{March 30, 2020}.
\end{csmr}
\end{quotation}

\begin{abcd}
\item Is the simple interest calculation in the quotation correct?

\item What would the value be today at 6 percent compound interest?

\item What would the value be today simply taking inflation into
  account?

\item Discuss which of the three valuations makes the most sense.
\end{abcd}

\begin{sol}
\begin{abcd}
\item Is the simple interest calculation in the quotation correct?

At 6 percent simple interest the value grows linearly at a rate
(slope) of $0.06 \times 1.8 = 0.108$ billion dollars per year. In the
55 years from 1959 to 2014 it would grow to
%
\begin{equation*}
\$1.8\text{b} 
+ 0.108 \frac{\text{\$b}}{\text{year}} 
\times 55 \text{ years}
= \$7.74\text{b}
\end{equation*}
so the quotation seems to underestimate the 2014 value. It's almost
8 billion dollars.

\item What would the value be today at 6 percent compound interest?

Compounding at a rate of 6 percent per year the value after 55 years
would be
%
\begin{equation*}
\$1.8\text{b} \times 1.06^{55} = \$44.4\text{b}.
\end{equation*}

\item What would the value be today simply taking inflation into
  account?

The inflation calculator tells me that \$1.8 billion in 1959 would be
worth \$14.2 billion in 2014.

\item Discuss which of the three valuations makes the most sense.

Simple interest isn't a realistic way to think about how the value of
an investment changes over time. The ``six percent'' in the compound
interest calculation is a made up growth factor. I think the most
sensible way to think about what that 1.8 billion dollars would be
worth in 2014 is simply to adjust it for inflation.

I could make a case for something larger than the inflation adjusted
figure.
\end{abcd}

\end{sol}


\end{exx}

\begin{exx}{\hassolution\sref{compounding}\gref{interest}}
``As Time Goes By''.

On December 13, 2012 you could read in \theTimes{} that 
the piano from Rick's place in the 1942 movie \emindex{Casablanca} is
up for auction. 

\begin{quotation}
Sotheby's expects [it] to sell from \$800,000 to
\$1.2 million in the auction on Friday. That is between 34 to 48 times
what [Ingrid] Bergman was paid for sharing top billing with Humphrey
Bogart. \index{Bogart, Humphrey}\index{Bergman, Ingrid}%
\begin{csmr}
J. Barron,
As Time Goes By, What's This Piano Worth?,
\theTimes{} (December 13, 2012),
\url{cityroom.blogs.nytimes.com/2012/12/13/as-time-goes-by-whats-this-piano-worth/}
\access{July 23, 2015}.
\csmrcomment{36 words fair use}
\end{csmr}
\end{quotation}

\begin{abcd}
\item How much was Ingrid Bergman paid for her role in the film?
  Calculate this two ways using the data in the quotation and comment
  on what you discover.

\item Would adjusting her pay to take inflation into account allow her
  to bid on the piano in 2012?
\item What compound interest rate would she have to have earned on her
  pay to bid on the piano in 2012?
\item Find out what happened at the auction.

\end{abcd}

\begin{sol}
\begin{abcd}
\item How much was Ingrid Bergman paid for her role in the film?
  Calculate this two ways using the data in the quotation and comment
  on what you discover.

Using the high end estimate, Ingrid Bergman was paid $\$1,200,000/48 =
\$25,000$ for her role as Ilsa in \emph{Casablanca}. 

Just to check, I get the same answer for the low end of the range ---
and found $\$800,000/34 = \$23,529$. That's a weird number, and
doesn't match the nice looking \$25K from the high estimate. I think
that one is right. So I calculated $\$800,000/\$25,000 = 32$! 

I think the \emph{Times} made a typographical error. The ``34'' should
have been ``32''.

\item Would adjusting her pay to take inflation into account allow her
  to bid on the piano in 2012?

The inflation calculator tells me that \$25K in 1942 would be \$352K
in 2012 --- not nearly enough to buy the piano.

\item What compound interest rate would she have to have earned on her
  pay to bid on the piano in 2012?

There were $2012 - 1942 = 70$ years between the film and the auction. 
For the low end, I need to find the value of $r$ that makes
\begin{equation*}
25,00 \times (1+r)^{70} = 800,000, 
\end{equation*}
%
or, equivalently, 
\begin{equation*}
(1+r)^{70} = \frac{800,000}{25,000} = 32.
\end{equation*}

I experimented in Excel. With $r=0.05$ I get 30.43. With $r=0.06$ I
get 59 --- way too big. $r=0.051$ gives me 32.5. Close enough.

\item Find out what happened at the auction.

On December 14, 2012 \theTimes{} reported that

\begin{quote}
``Casablanca'' Piano Is Sold for \$602,500 at Auction
\end{quote}

On November 24, 2014 \theTimes{} reported on 
another Casablanca piano auction, that referred to this one
too:

\begin{quotation}
The other piano, the one in the flashback scene, sold for \$602,500 at
Sotheby's in December 2012. (Dr. Milan once owned that piano, too, but
sold it in 1988 for \$154,000.%
\begin{csmr}
J. Barron,
'Casablanca' Piano Sells for \$3.4 Million at Bonhams,
\theTimes{} (November 24, 2014),
\url{www.nytimes.com/2014/11/25/nyregion/casablanca-piano-to-be-auctioned-at-bonhams.html}
\access{July 23, 2015}.
\csmrcomment{36 words fair use}
\end{csmr}
\end{quotation}

So now we can compute Dr. Milan's gain \ldots and take inflation into account.
\end{abcd}


\end{sol}
\end{exx}

\begin{ReviewExercises}

\begin{rexx}
You invest \$500 in an account that earns \$10 in interest each year.  
\begin{abcd} 
\item  At the end of 24 months, what is the balance?
\item  At the end of 30 months, what is the balance?
\item At the end of 5 years, what is the balance?
\item Find the linear equation that gives the balance after $t$ years.
\end{abcd}
\end{rexx}


\begin{rexx}
You buy a car for  \$15,000 and for tax purposes you depreciate it
at a rate of 11\% per year. 
\begin{abcd} 
\item  At the end of 24 months, what is the value of the car?
\item At the end of 5 years, what is the value of the car?
\item Find the exponential equation that gives the value of the car
after $t$ years. 
\item  Does the value of the car ever reach \$0?  
\end{abcd}
\end{rexx}


\begin{rexx}
Calculate the percentage.
\begin{abcd}
\item  What is  8\% of \$2000?
\item  What is 108\% of \$2000?
\item  What is 3.25\% of \$800?
\item  What is 103.25\% of \$800?
\end{abcd}
\end{rexx}


\begin{rexx}
Use a calculator to evaluate these expressions using  exponents.
(You may find typing into the Google or Bing calculator much faster
than using one that requires you to press keys, either with your
fingers or with a mouse. )
\begin{abcd}
\item  $1.03^4$
\item  $0.89^5$
\item $140\times 1.03^4$
\item  $80\times 0.89^5$
\item $\frac{1}{3^8}$
\item $(\frac13)^8$
\item $1.25^0$
\item $1.25^1$
\item $e^2$
\item $e^{15}$
\end{abcd}
\end{rexx}

\begin{rexx}
In the exponential functions below, identify the relative change and
the initial amount. 
%CHANGE added periods
\begin{abcd}
\item  $P=100\times(1.05)^T$.
\item $y = 400\times(0.88)^x$.
\item  $S = 550\times (1.22)^Q$.
\item $P=96\times(0.50)^T$.
\end{abcd}
\end{rexx}

\begin{rexx}
Excel gives the following best-fit exponential function for a set of
data: 
$y = 2.099\times e^{1.344x}$.  Find the constant growth rate 
and rewrite the function without using $e$.
\end{rexx} 

\end{ReviewExercises}

\begin{NewExercises}

  \begin{exx}{\untested\hassolution\sref{compounding}\gref{exponentialgrowth}\gref{interest}}
    There's more next year.

    Lewis Carroll's Professor talks with Sylvie and Bruno:
    \index{Lewis Carroll}\index{Sylvie and Bruno}
    
  \begin{quotation}
``Come in!'' [said the Professor]

``Only the tailor, Sir, with your little bill,'' said a meek voice
    outside the door. 

``Ah, well, I can soon settle his business,'' the Professor said to the
    children, ``if you'll just wait a minute. How much is it, this
    year, my man?'' The tailor had come in while he was speaking. 

``Well, it's been a doubling so many years, you see,'' the tailor
    replied, a little gruffly, ``and I think I'd like the money
    now. It's two thousand pound, it is!''

``Oh, that's nothing!'' the Professor carelessly remarked, feeling in 
    his pocket, as if he always carried at least that amount about
    with him. ``But wouldn't you like to wait just another year, and
    make it four thousand? Just think how rich you'd be! Why, you
    might be a King, if you liked!''

``I don't know as I'd care about being a King,'' the man said
    thoughtfully. ``But it dew sound a powerful sight o' money! Well,
    I think I'll wait ---''

``Of course you will!'' said the Professor. ``There's good sense in you,
    I see. Good-day to you, my man!''

``Will you ever have to pay him that four thousand pounds?'' Sylvie
    asked as the door closed on the departing creditor. 

``Never, my child!'' the Professor replied emphatically. ``He'll go on
doubling it, till he dies. You see it's always worth while waiting
another year, to get twice as much money!''
\begin{csmr}
Lewis Carroll,
Sylvie and Bruno,
Project Gutenberg,
\url{www.gutenberg.org/files/620/620-h/620-h.htm}
\access{July 27, 2019}
\end{csmr}
\end{quotation}

  \begin{abcd*}
\item Use the fact that $2^{10} \approx 1000$ to
estimate the original tailor's bill if it's been doubling for
five years.

\item Check your estimate with a precise calculation.

\end{abcd*}


  \begin{sol}
\begin{abcd*}
\item Use the fact that $2^{10} \approx 1000$ to
estimate the original tailor's bill if it's been doubling for
five years.

\begin{equation*}
2000 = 2 \ times 1000 \approx 2^{11} = 2^52^6
\end{equation*}
so in five years an original bill of $2^6 = 64$ pounds will grow to
about 2000 pounds.

\item Check your estimate with a precise calculation.

  If $B$ is the original bill then
  $$
  2^5B = 2000
  $$
  so
  $$
  B = \frac{2000}{2^5} = 62.5 \text{ pounds }.
  $$
\end{abcd*}

\end{sol}
  
  \end{exx}

  \begin{exx}{\untested\hassolution}
    Light pollution.
    \index{light pollution}

    On August 18, 2019 Kelsey Johnson wrote in \theTimes{} that

    \begin{quotation}
[T]he global amount of artificial light at night has been growing by at
least 2 percent per year. At this rate the amount of light pollution
originating from Earth-based sources alone will double in less than 50
years.%
\begin{csmr}
  K. Johnson,
  Is the Evening Sky Doomed?,
  \theTimes{} August 18, 2019,
  \url{www.nytimes.com/2019/08/17/opinion/sunday/light-pollution.html}
  \access{August 18, 2019}.
\end{csmr}
    \end{quotation}

    In a more technical discussion linked from that article you can
    read Jeff Hecht's observation that
    
    \begin{quotation}
A new analysis of satellite data from the past four years shows that
the total acreage lit by artificial light at night increased by an
average of 2.2 percent a year. The brightness of the areas lit at the
start of the study also increased by the same rate --- 2.2 percent
annually --- around the globe.
\begin{csmr}
  J. Hecht,
  Awash in Artificial Light, the World Gets 2 Percent Brighter Each
  Year,
  \emph{IEEE Spectrum}, November 22, 2017,
  \url{https://spectrum.ieee.org/energywise/energy/environment/awash-in-artificial-light-the-world-gets-2-percent-brighter-each-year}
  \access{August 18, 2019}.
\end{csmr}
    \end{quotation}

    \begin{abcd}
\item Use the Rule of 70 to find a better estimate of the light
  pollution doubling time.

\item How do you think Johnson arrived at the 50 year doubling time?

  \item Use the data in the quote from Hecht to explain why the annual
    light pollution increase might be better reported as almost four
    and a half percent.

    \end{abcd}
    
    \begin{sol}
\begin{abcd}
\item Use the Rule of 70 to find a better estimate of the light
  pollution doubling time.

  Thirty or 32 years would be a better estimate since $70/2.2 \approx
  31.8$.

\item How do you think Johnson arrived at the 50 year doubling time?

  By thinking that 2 percent per year would lead to 100 percent in 50
  years.

  \item Use the data in the quote from Hecht to explain why the annual
    light pollution increase might be better reported as almost four
    and a half percent.

Since both lit area and the light pollution from lit area are
increasing by 2.2 percent annually
the total annual light pollution increases by a factor of
\begin{equation*}
1.022^2  = 1.044484 ,
\end{equation*}
which
corresponds to almost 4.5 percent. 

    
\end{abcd}
\end{sol}


  \end{exx}
  
  \begin{exx}{\hassolution\untested}
    Thirty four years at seven percent.
    \index{Higgins, George V.}
    
From \emph{Swan Boats at Four}, a novel by George V. Higgins:

\begin{quotation}
Rutledge said ``In other words, if we'd painted over that damned
picture in the summer of nineteen seventy-eight, we would've made the
club, and ourselves individually, liable for a hundred thousand bucks,
plus interest at, say, an average of seven percent per annum,
compounded for thirty-four years. \ldots''

``Offhand,'' [David] said, ``I can't even imagine how much that
would've been.'' 

``At the time, I couldn't either,'' Rutledge said, `` \ldots so we
looked it up --- I don't mean we figured it out. \ldots I don't
recall the exact figure, but it came out to around a million and a
half dollars.''%
\begin{csmr}
George V. Higgins, 
\emph{Swan Boats at Four},
Henry Holt and Company, 1995,
pp. 198-199.
\end{csmr}
\end{quotation}

David is a banker. He would know the Rule of 70 and figure it out
offhand, without pencil and paper. Higgins should have known that. 

\begin{abcd}

\item  Use the Rule of 70 to decide whether Rutledge was right
  when he said the figure was ``around a million   and a half dollars''.

\item Calculate the liability accurately.

\end{abcd}

\begin{sol}
\begin{abcd}

\item  Use the Rule of 70 to decide whether Rutledge was right
  when he said the figure was ``around a million   and a half dollars''.

The rule tells me that seven percent interest doubles
a debt in ten years. In twenty years it will quadruple, in
thirty it will be eight times as large. So the \$100,000 would be
\$800,000 after 30 years. It would be \$1,600,000 after 40
years. That's barely over a million and a half, so I don't think
it would be a million and a half after just 34 years.


\item Calculate the liability accurately.

\begin{equation*}
100,000 \times 1.07^{34} = 997811.353702
\end{equation*}
%
so the \$100,000 debt would grow to just about a million dollars in 34
years.
\end{abcd}

\end{sol}

\end{exx}

  \begin{exx}{\untested\hassolution}
      Disrupting the cow.\index{artificial meat}

    A November 1919 article in \theGlobe{} reported that the 
    cost of precision fermentation (PF), a process for growing meat in
    a laboratory, was falling ``exponentially''

    \begin{quotation}
 --- from \$1
million per kilogram in 2000 to about \$100 today. With the
technologies we have today, we project these costs will fall even
lower --- to \$10 per kilogram by 2023-25.
\begin{csmr}
  Disrupting the cow,
  T. Soba and C. Tubb,
  \theGlobe,
  November 29, 2019,
  \url{www.bostonglobe.com/2019/11/29/opinion/disrupting-cow/}
  \access{December 3, 2019}.
\end{csmr}
    \end{quotation}

\begin{abcd}
\item Use the data from the Globe to verify the statement
  \begin{quotation}
The cost of PF falls by a factor of about $1/100$ per decade.
\end{quotation}

  \item Rewrite the statement in (a) using an assertion about
    percentage change.


  \item
    Explain why the cost of PF is falling by a factor of about $1/10$
    every five years, to 10\$ of what it starts at.

  \item Is the Globe article right when it projects
\$10 per kilogram by 2023-25?

\end{abcd}

  
    \begin{sol}
\begin{abcd}
\item Use the data from the Globe to verify the statement
  \begin{quotation}
The cost of PF falls by a factor of about $1/100$ per decade.
\end{quotation}

  From 2000 to 2019 is about two decades. If the cost falls by a
  factor of $1/100$ in one decade it will fall by
  \begin{equation*}
\frac{1}{100} \times \frac{1}{100} = \frac{1}{10000}
  \end{equation*}
  in two decades. Since a million is six zeroes,
  \begin{equation*}
    \frac{1}{10000} \times \frac{\$1 \text{ million}}{\text{kg}} =
      \frac{\$100}{\text{kg}}.
\end{equation*}
  \item Rewrite the statement in (a) using an assertion about
    percentage change.

    In a decade the cost of PF falls by about 99\% to 1\% of what it
    started at.

  \item
    Explain why the cost of PF is falling by a factor of about $1/10$
    every five years, to 10\$ of what it starts at.

    That's because two five year periods is a decade and
  \begin{equation*}.
\frac{1}{10} \times \frac{1}{10} = \frac{1}{100} .
  \end{equation*}
    
  \item Is the Globe article right when it projects
\$10 per kilogram by 2023-25?

Yes. 2023-2025 is about five years from 2019, so the cost should drop
to 10\% of \$100/kg.
\end{abcd}
\end{sol}
\end{exx}
  \end{NewExercises}

\setexercisecounter{}

\begin{ExtraExercises}

\begin{exx}{\hassolution}
So many words!

On December 1, 2012 R. Alexander Bentley and Michael J. O'Brien
wrote in \theTimes{} that

\begin{quotation}
[F]or the last 300 years, the number of words published
  annually grew exponentially by about 3 percent per year. From about
  20 million words for 1700, the annual word count grew to several
  trillion for 2000.% 
\begin{csmr}
R. A. Bentley and M. J. O'Brien,
The Buzzwords of the Crowd,
\theTimes{},
December 1, 2012,
\url{www.nytimes.com/2012/12/02/opinion/sunday/science-and-buzzwords.html}
\access{July 28, 2019}.
\end{csmr}
\end{quotation}

\begin{abcd}
\item Check the authors' arithmetic.

\item If the growth continues at the same rate how many words will be
published in the year 3000?

\item How much confidence do you have in your prediction?
\end{abcd}


\begin{sol}

\begin{abcd}
\item Check the authors' arithmetic.

The coolest way to do this is with the rule of 70. A 3\% annual
increase has a doubling time of $70/3 \approx 25$ years. The question
asks about $300$ years of growth, which would be $12$ doublings.

But to get from 20 million to 20 trillion you must multiply by $10^6$.
Since $2^{10} \approx 1,000$, that's 20 doublings. It would take 19
doublings to get to 10 trillion, and 16 to get to 1 trillion. So 12 is
not enough.

You can, of course, solve this problem the boring way with some arithmetic:
\begin{equation*}
(20 \text{ million})\times (1.03)^{300} \approx 140,000 \text{ million}
= 140 \text{ billion}.
\end{equation*}
That's at least an order of magnitude short of ``several trillion''.

To get to 2 trillion you'd need 19 doublings in 300 years. That's a
doubling time of about 16 years. Then the rule of 70 says you'd have
to have had an annual  growth rate of about 4.3\%.

\item If the growth continues at the same rate how many words will be
published in the year 3000?

Another 100 years at 4.3\% would be about 6 more doublings, so each
trillion words would grow to $2^6 = 64$ trillion.

\item How much confidence do you have in your prediction?

Not much!

\end{abcd}

\end{sol}

\end{exx}
\begin{exx}{\hassolution} 
World population.
\index{world population}

According to a Harvard School of Public Health press release 
the world's population has grown slowly for most of human history. It
took until 1800 for the population to hit 1 billion. However, in the
past half-century, population jumped from 3 to 7 billion. In 2011,
approximately 135 million people will be born and 57 million will die,
a net increase of 78 million people.  

\begin{abcd}
\item By what percent did world population increase in 2011?

\item Write the equation for a mathematical model for world population
growth for years since 2011 if the annual net increase seen in 2011
remains constant for the remainder of the century. (Use 7 billion as
the 2011 population.) 

\item Write the equation for a mathematical model for world population
growth for years since 2011 if the annual percentage increase seen in
2011 remains constant for the remainder of the century. (Use 7 billion
as the 2011 population.) 

\item Construct an Excel spreadsheet predicting world population for
the years through 2100 using both models. Graph both predictions on
the same chart. 

\item The table below gives the United Nations high estimate for
world population growth during the remainder of this century.

\begin{center}
\begin{tabular}{cS[table-format=2.1]}
\toprule
Year &	{Population} \\
     & {(billions)} \\
\midrule
2011 &	7 \\
2025 & 8.5 \\
2050 &	10.6 \\
2100 &	15.8 \\
\bottomrule
\end{tabular}
\end{center}

Which of your models most closely matches the UN high estimate? 
\end{abcd}

\begin{sol}
\begin{abcd}
\item By what percent did world population increase in 2011?

The percentage increase was
\begin{equation*}
\frac{78 \times 10^6}{7 \times 10^{9}}
= 
0.01114285714
\approx 1.1\%.
\end{equation*}

\item Write the equation for a mathematical model for world population
growth for years since 2011 if the annual net increase seen in 2011
remains constant for the remainder of the century. (Use 7 billion as
the 2011 population.) 

Let $Y$ be the number of years since 2011 and $P$ the world
population, in billions. Then the equation is
\begin{equation*}
P = 7 + 0.078Y.
\end{equation*}

\item Write the equation for a mathematical model for world population
growth for years since 2011 if the annual percentage increase seen in
2011 remains constant for the remainder of the century. (Use 7 billion
as the 2011 population.) 

With the same variables as above, the exponential equation is
\begin{equation*}
P = 7 \times (1.011)^Y.
\end{equation*}

\item Construct an Excel spreadsheet predicting world population for
the years through 2100 using both models. Graph both predictions on
the same chart. 

\item The table below gives the United Nations' high estimate for
world population growth during the remainder of this century.

Which of your models most closely matches the UN high estimate? 

I've added the predictions to the table. Numbers are in billions. The
last two columns show the relative errors in the predictions.

\begin{center}
\begin{tabular}{c
S[table-format=2.1]
S[table-format=2.2]
S[table-format=2.2]
S[table-format=1.2]
S[table-format=1.2]
}
\toprule
Year &	{UN} & {linear} & {exp} & {lin/UN} & {exp/UN} \\
\midrule
2011 & 7.0  & 7.00  & 7.00  & 1.00 & 1.00 \\
2025 & 8.5 & 8.19 & 8.17 & 0.96 & 0.99 \\
2050 & 10.6 & 10.32 & 10.78 & 0.97 & 1.02 \\
2100 & 15.8 & 14.57 & 18.77 & 0.92 & 1.19 \\
\bottomrule
\end{tabular}
\end{center}

The linear and exponential predictions are both pretty close for the
first half of the 21st century. By 2100 the linear prediction is
8\% lower than the UN's, the exponential prediction 19\% higher.
\end{abcd}

\end{sol}

\end{exx}
\begin{exx}{\needsquestions}
Making it into the \myindex{Hall of Fame}.

On January 12, 2013, Nate Silver \index{Silver, Nate} wrote in his
blog at \theTimes{} that individual voting totals for the baseball
Hall of Fame seemed to increase by about ten percent each year.

\begin{quotation}
Thus, a player who received 10 percent of the vote in his first year
would be expected to receive about 11 percent on his second try, while
a player who got 50 percent of the vote would go up to 55 percent.%
\begin{csmr}
N. Silver,
In Cooperstown, a Crowded Waiting Room,
\theTimes{}
(January 12, 2013),
\url{fivethirtyeight.blogs.nytimes.com/2013/01/12/in-cooperstown-a-crowded-waiting-room/}
\access{July 23, 2015}.
\csmrcomment{42 words fair use}
\end{csmr}
\end{quotation}

Explain why this is exponential growth. Look up the original. Make
some projections.

\end{exx}


\begin{exx}{\untested\needsquestions}
Save another one percent.

The interactive web based calculator provided by \theTimes{} at
\url{%
www.nytimes.com/interactive/2010/03/24/your-money/one-pct-more-calculator.html
}
suggests many
questions about savings, compound interest and inflation.


\end{exx}

\begin{exx}{\untested\needsquestions}
The \myindex{London olympics}.

On July 22, 2012 \theGlobe{} wrote about the third Olympic games to be
hosted by London:

\begin{quotation}
The Games have grown geometrically during the past 104
years --- from 2,023 athletes representing 22 countries competing in
109 events in 24 sports in 1908 to 4,064 athletes, 59 countries, 136
events, and 19 sports in 1948 to 10,500 athletes, 204 countries, 302
events and 37 sports in 2012.%
\begin{csmr}
J. Powers,
London ready to complete Olympic triple play,
\theGlobe{} (July 22, 2012),
\url{www.bostonglobe.com/sports/2012/07/22/london-ready-complete-olympic-triple-play/iaAdNViKdg53AyrkHMORwN/story.html}
\access{July 23, 2015}.
\csmrcomment{Globe, OK}
\end{csmr}
\end{quotation}

Note: ``geometric'' is a synonym for ``exponential''.

Possible questions: find exponential regression curves for the numbers
of athletes, countries and events.

\end{exx}

\begin{exx}{\untested\needsquestions\sref{exponentialdepreciation}
\gref{exponentialgrowth}}
Radioactive waste.
\index{radioactive waste}

The web site \url{www.nirs.org/factsheets/hlwfcst.htm} we
quoted earlier offers much more information about radioactive waste.
\index{radioactive waste}\index{plutonium}

Ask and answer some interesting Fermi problems suggested by the data there.
You could consider what it says about a nuclear power
plant near you.%

We could construct the Fermi problems based on this radioactive waste
data ourselves, and ask the students to solve them. But by this time
in the course we hope they can start from the numbers and create their
own.

\end{exx}

\begin{exx}{\untested\needsquestions}
Payday loans 2016.
\index{payday loan}

\begin{quotation}
People who borrow money against their paychecks are generally supposed
to pay it back within two weeks, with substantial fees piled on: A
customer who borrows \$500 would typically owe around \$575, at an
annual percentage rate of 391 percent. 
\begin{csmr}
S. Cowley,
Payday Loans' Debt Spiral to Be Curtailed,
\emph{The New York Times}, June 2, 2016,
\url{www.nytimes.com/2016/06/02/business/dealbook/payday-borrowings-debt-spiral-to-be-curtailed.html}
\access{June 2, 2016}.
\end{csmr}
\end{quotation}

Many more juicy numbers in the article.

\end{exx}

\begin{exx}{\untested\needsquestions}
Was this a good deal on a mortgage?

On July 6, 2016 Max Jacob posted this question at
\url{money.stackexchange.com/questions/66978/was-this-a-good-deal-on-a-mortgage}.

\begin{quotation}
I just graduated college last year, and was looking to buy a
house. Clearly I wouldn't be able to buy it outright, so I was
thinking of getting a mortgage. I have a steady job (~\$55,000 annual
before taxes) and an okay credit score (~740). Here's the deal that a
mortgage consultant gave me: 

\$2,000 per month for 30 years on a \$300,000 home (he didn't mention
any other fees) 

It seems terrible to me because it comes out to a total of \$720,000
over that 30-year period. I'm completely new to the concept, so maybe
I'm just being naive.%
\begin{csmr}
Was this a good deal on a mortgage?
\url{money.stackexchange.com/questions/66978/was-this-a-good-deal-on-a-mortgage}
\access{July 6, 2016},
\ccbysa{}.
\end{csmr}
\end{quotation}

Possible uses: classroom discussion, group homework assignment, think
about the answers informative answers at the site, appended here.

From user Aganju 
(\url{money.stackexchange.com/users/35405/aganju}:
\begin{quotation}
That seems a very bad offer, it borders on fraud.

In the current US economy, you should be able to get between 3 and 4
\% APR (and that number is what you should look at). That means that
for 300000 over 30 years, you'd pay \$1265 to \$1432 per month.   

If you are able to pay more than that monthly rate, you should go for
less than 30 years - 20, 15, 10, whatever you can afford - but don't
overextend yourself. 

Google ``mortgage calculator'' to do your own calculations.
\end{quotation}

From user quid
(\url{money.stackexchange.com/users/22881/quid}:
\begin{quotation}
I'm calculating that to about a 7\% apr, which given loan rates
available today seems a bit high. 

I wouldn't get too caught up on what that equates to over the life of
the loan. There are a lot of forces in play over a 30 year period,
namely the time value of money. 30 years from now a dollar will be
less valuable in real terms due to the forces of inflation. At 2\% per 
year in inflation today's \$1 will be worth about \$0.55 in 30 years. 
\end{quotation}

From user keshlam
(\url{money.stackexchange.com/users/12439/keshlam}):
\begin{quotation}
Some part of the payment is probably also going for tax escrow,
insurance payments, probably PMI if you aren't putting at least 20\%
down. Get a complete breakdown of the costs.  

Remember to budget for upkeep.

And please see past discussion of why buying a home at this point in
your career/life may be very, very premature. 
\end{quotation}

\end{exx}

\begin{exx}{\untested\routine\hassolution}
Get rich quick.

Tad Friend wrote in the October 10, 2016 issue of  \thenewyorker{}
that
\begin{quotation}
YC's gold standard for revenue growth  is ten percent a week, which
compounds to 142x a year.%
\begin{csmr}
T. Friend, 
Adding a Zero,
\emph{The New Yorker}, 
October 10, 2016, p. 77.
\end{csmr}
\end{quotation}

Check his calculation.

\begin{sol}
He's right. The Google calculator tells me 
\begin{equation*}
1.1^{52} = 142.042931984
\end{equation*}
which will multiply any starting amount by 142.
\end{sol}
\end{exx}

\begin{exx}{\untested\needsquestions}
Double, double, $\ldots$

In April, 2017 Tad Friend wrote in \emph{The New Yorker} that

\begin{quotation}
Progress in computers, or anyway in
semiconductors, has been subject to Moore's Law, the exponential
flywheel that has doubled capacity every two years. In linear
progress, after thirty iterations you've advanced thirty steps; in
exponential progress, you've advanced 1.07 billion steps. Our progress
in mapping the human genome looked like it was linear --- and then was
revealed, once the doublings grew significant, as exponential. 
\begin{csmr}
T. Friend,
Silicon Valley's Quest to Live Forever,
\emph{The New Yorker}, April 3, 2017,
\url{www.newyorker.com/magazine/2017/04/03/silicon-valleys-quest-to-live-forever}
\access{April 20, 2017}.
\end{csmr}
\end{quotation}

\begin{abcd}
\item Where does that 1.07 billion come from?
%Answer: It's $2^{30}$.

\item Question about Moore's Law?
\end{abcd}
\end{exx}


\begin{exx}{\untested}
Free meat!

On July 2, 2017 \theGlobe{} published an article about bioengineered
meat grown from animal stem cells.

\begin{quotation}
Steve Myrick, vice president of business development at Memphis
[Meets], said the company is now producing beef, chicken, and duck in
stainless steel tanks for roughly \$2,400 per pound. That'a lot cheaper than it
was even a year ago, and the price is falling about 15 to 20 percent
per month, as the company improves yields and finds less expensive
nutrients.%
\begin{csmr}
D. Scharfenberg,
I'm a vegetarian. Bring on the lab-grown meat,
\theGlobe, July 2, 2017,
\url{bostonglobe.com/ideas/2017/06/29/vegetarian-bring-lab-grown-meat/M8s2vNYxiYZsusZPM8fF8K/story.html}
\access{July 28, 2019}.
\end{csmr}
\end{quotation}

\begin{abcd}
\item Estimate the cost ``a year ago''.

\item When will the cost be competitive with what meat sells for in
  your local market?
\end{abcd}
\end{exx}


\begin{exx}{\untested}
Payday loans 2017.
\index{payday loan}
\index{McDonalds}

On October 5. 2017 \theTimes{} reported that the
 Consumer Financial Protection Bureau announced new rules applicable
 to payday loans. The article read (in part)

\begin{quotation}
Currently, a cash-strapped customer might borrow \$400 from a payday
lender. The loan would be due two weeks later --- plus \$60 in interest
and fees. That is the equivalent of an annual interest rate of more
than 300 percent, far higher than what banks and credit cards charge
for loans. 

\ldots

The payday-lending industry is vast. There are now more payday loan
stores in the United States than there are McDonald's restaurants. The
operators of those stores make around \$46 billion a year in loans,
collecting \$7 billion in fees. Some 12 million people, many of whom
lack other access to credit, take out the short-term loans each year,
researchers estimate.
\begin{csmr}
S. Cowley,
Payday lending faces tough new restrictions by consumer agency,
\emph{The New York Times},  October 5, 2017,
\url{www.nytimes.com/2017/10/05/business/payday-loans-cfpb.html}
\access{October 6, 2017}.
\end{csmr}
\end{quotation}

\begin{abcd}
\item Verify the interest calculation in the first paragraph of the
  quotation.

\item Are the assertions in the second paragraph reasonable? 
The data in Figure~\ref{fig:paydayloanstores} should help answer this
question.

\figfile{paydayloanstores.pdf}
\begin{figure}
\centering
\includegraphics[width=4in]{\thefigurefilename}
\begin{csmr}[Payday Loan Stores 2004-2013\label{fig:paydayloanstores}]
J. Hecht,
Alternative Financial Services, February 27, 2014, slide 31, 
\texttt{cfsaa.com/Portals/0/cfsa2014\_conference/Presentations/CFSA2014\_THURSDAY\_GeneralSession\_JohnHecht\_Stephens.pdf}
(broken link) 
\access{October 5, 2017}.
\end{csmr}
\end{figure}
\figfile{}

\end{abcd}
\end{exx}

%HarryPotterLoan.jpg
%At 239.45% APR, this is evidently equivalent to a payday lender. 

\begin{exx}{\untested\needsquestions}
R\&D in China.
\index{China}
\index{R\&D}

\begin{quotation}
China is the clearest example. Since 2000, China's spending on
research and development has grown by an average of 18 percent each
year, while ours grew by only 4 percent. This has placed China a
decisive second in R \& D expenditures behind the United States, where
the government and private sector together invest far more than any
other country. Even so, the share of R \& D funded by the federal
government declined to about 25 percent from just over 30 percent from
2010 to 2015.%
\begin{csmr}
M. Zuberjan,
Falling Short on Science,
\emph{The New York Times}, January 26, 2018,
\url{www.nytimes.com/2018/01/26/opinion/falling-short-on-science.html}
\access(January 26, 2018)
\end{csmr}

%
\end{quotation}

\end{exx}

\begin{exx}{\untested\hassolution}
Endangered apes.

On April 26, 2018 \emph{The Washington Post} summarized a study from
the journal \emph{Science Advances} that reported

\begin{quotation}
\ldots 
estimates that there were 361,900 gorillas and 128,700 chimpanzees as
of 2013. That's about one-third more gorillas and one-tenth more
chimpanzees than previous surveys estimated, though those calculations
were performed differently and were not designed to count the animals
across their entire range. 

That is the good news. Now the bad: Researchers found that gorilla
populations are dropping faster than they believed, at a rate of
nearly 3 percent per year, said Fiona Maisels, a conservation
scientist with the Wildlife Conservation Society and the University of
Stirling in Scotland. At this rate, half of the world's gorillas could
be gone by about 2040, she said.%
\begin{csmr}
D. Main, 
Vast survey finds far more gorillas in Africa than previously believed
- and some bad news, too,
\emph{The Washington Post}, 
April 26, 2018,
\url{www.washingtonpost.com/news/animalia/wp/2018/04/26/vast-survey-finds-far-more-gorillas-in-african-forests-than-previously-believed-and-some-bad-news-too}
\access{July 28, 2019}
\end{csmr}
\end{quotation}

\begin{abcd}
\item How many gorillas and chimpanzees were estimated in the previous
  surveys?

\item This article updates numbers from 2013. Estimate the gorilla
  population in 2018, when this study was published.

\item Is Dr. Maisel's prediction about the 2040 gorilla population
  reasonable?
\end{abcd}

\begin{sol}
\begin{abcd}
\item How many gorillas and chimpanzees were estimated in the previous
  surveys?

``One third more gorillas'' means that 
\begin{equation*}
\frac{4}{3} \times \text{previous estimate} = 361,900
\end{equation*}
so
\begin{equation*}
\text{previous estimate} = \frac{3}{4} \times 361,900
=271,425 \approx 270,000 \text{ gorillas}.
\end{equation*}
A similar calculation says the previous estimate was about 117,00
chimpanzees.

\item This article updates numbers from 2013. Estimate the gorilla
  population in 2018, when this study was published.

Given three percent per year decrease the 2018 gorilla population is
about
\begin{equation*}
270,000 \times 0.97^5 \approx 232,000 .
\end{equation*}

\item Is Dr. Maisel's prediction about the 2040 gorilla population
  reasonable?

The rule of 70 says that the half life of a population declining at a
rate of $3$ percent per year is about $70/3 \approx 23$ years. Since
2040 is 22 years from now Dr. Maisel's prediction is reasonable.
\end{abcd}

\end{sol}
\end{exx}

\begin{exx}{\untested}
An extra percentage point.

On May 22, 2018 \theGlobe{} reported that

\begin{quotation}
Over the past 18 months, the average rate on 30-year fixed mortgages
has climbed a full percentage point, which translates into an
additional \$100,000 in interest over the life of a \$500,000 loan.%
\begin{csmr}
E. Horowitz,
Gas and mortgages are getting expensive again. Welcome to a normal
economy,
\theGlobe, May 22, 2018.
\url{www.bostonglobe.com/business/2018/05/22/gas-and-mortgages-are-getting-expensive-again-welcome-normal-economy/BAkMj9pm0UvVlJaW1Avm0I/story.html}
\access{July 28, 2019}.
\end{csmr}
\end{quotation}

\begin{abcd}
\item Verify this assertion.

\item Does the difference in interest paid depend on the actual
  mortgage rate, or only on the increase?
\end{abcd}
\end{exx}

\begin{exx}{\untested}
The Internet of Things.
\index{internet of things}

Figure~\ref{fig:connectedHomes} shows the growth of U. S. households
with devices like appliances connected to the internet.


\figfile{connectedHomes.pdf}
\begin{figure}
\centering
\includegraphics[width=4in]{\thefigurefilename}
\begin{csmr}[Connected Homes, 2015-2017\label{fig:connectedhomes}]
There's No Place Like [ A CONNECTED ] Home,
McKinsey\&Company,
\url{www.mckinsey.com/spContent/connected_homes/index.html}
\access{June 23, 2018}.
\end{csmr}
\end{figure}
\figfile{}

\begin{abcd}
\item Is the $31\%$ compound annual growth rate correct?

\item Estimate when every home in the United States will be connected.
\end{abcd}
\end{exx}

\begin{exx}{\untested\worthy\needsquestions}
5 miilion Bostonians?

On August 19, 2019 Martin Finucane wrote in \theGlobe{} that
in 1900 Globe reporter Thomas F. Anderson reported a prediction that
Boston's population would be 5,251,330 in 2000.%
\begin{csmr}
M. Finucane,
5.2 million people crammed into Boston? In 1900, some of the best
minds thought it was a possibility,
theGlobe, August 19, 2018,
\url{www.bostonglobe.com/metro/2018/08/19/million-people-crammed-into-boston-some-best-minds-thought-was-possibility/m2gQBXmqN46Evrv5pb9rwK/story.html}
\access{August 19, 2018}.
\end{csmr}

The original article said
\begin{quotation}
This population is estimated on the rate of increase of the last 20
years, and on the same basis (the population according to the census
of 1900 being 560,892), the figures for consecutive 10-year periods
will probably be as follows. . .
\end{quotation}

The current story continued

\begin{quotation}
Anderson then cited a series of population figures that appeared to
reflect an approximately 25 percent increase per decade for 100
years [\ldots because he] had just seen the population of the city
explode by 54 percent from 1880 to 1900.
\end{quotation}

$1.25^2 = 1.5625$

$1.25^{10} = 9.31$

$560,000 * 1.25 ^10 \approx 5,200,000$
\end{exx}

\end{ExtraExercises}


