\chapter{Exponential Growth and Decay}

\section{Continuously Compounded Interest}

Let's return to the \index{bank interest problem} bank interest problem and suppose a more realistic
annual interest rate of $k$ (this is a percent written in decimal form,
for example 8\%$=0.08$ --- in this problem and many that follow, the
units of $k$ are reciprocal time, i.e., $0.08\frac{1}{\text{yr}}$ annual
interest rate). We again want to find out how our principal $P$ grows in
$t$ years. We again assume continuous compounding. But this time we
shall use a different method to derive the function $y(t)$ that tells
the amount of money in the bank after $t$ years.


The method is to set up and solve a certain ``\index{differential equation} differential equation.''
Recall that a {\bf differential equation} is an equation for finding an
unknown function in which the derivative of the unknown function (and
perhaps its second derivative or higher derivatives) appears. For
example, (1) the differential equation $y'=f(x)$, where $f(x)$ is a
given function, is satisfied by an antiderivative of $f(x)$; (2) the
differential equation $\ddot{y}=-g$ is satisfied by the falling body
formula $y(t)=-\frac{1}{ 2}gt^2+v_0t+y_0$.

\smallskip

Differential equations are fundamental in applications, because often we
have to get at a function $y(t)$ indirectly, using some information about
the rate of change of $y(t)$ that comes from translating some basic
principle (of physics or finances, for instance) into mathematics.  For
example, the differential equation $\ddot{y}=-g$ is simply the statement
that gravity near the surface of the earth imparts a constant acceleration.

\smallskip

In the case of bank interest continuously compounded at rate $k$, the basic
principle is that {\it at any instant interest is accumulating at a rate in
dollars per year equal to $k$ times the current amount in the bank}.  For
example, if $k=8\%=0.08$ and you put \$1000 in the bank, then at first your
money increases at the rate of \$80.00 per year; but a little while later,
when you have (say) \$1010 in the bank, at that instant it is increasing at
the rate of \$80.80 per year.  That is, you are ``earning interest on the
interest'' as well as on the original principal.  The basic principle in
italics can be written mathematically as follows: $$\frac{dy}{ dt}=ky,$$
i.e., the function $y(t)$ telling the number of dollars in the bank at time
$t$ years increases at a rate equal to $k$ times the present value of this
function.

\smallskip

Our task now is to solve this equation to find an expression for $y$ in
terms of $t$.  A useful technique in this and many other differential
equations is called ``\index{separation of variables} separation of variables.''  This means that we get
everything involving $y$ on one side of the equation and everything
involving $t$ on the other.  Here the $dy$ and $dt$ are handled just like
algebraic quantities.  (When discussing the chain rule and also the method
of substitution in integrals, we saw that it often makes sense to split up
expressions like $dy/dx$ or $dy/dt$, treating $dy$, $dx$, $dt$ separately.)
In the above differential equation we separate variables by multiplying
both sides by $dt$ and dividing both sides by $y$.  After that the second
step is to put an integral sign on both sides: $$\int\frac{dy}{ y}=\int
k\,dt.$$ On the left we have an indefinite integral with respect to the
variable of integration $y$, and on the right we have an indefinite
integral with respect to $t$.  On the left we get $\ln|y|+C_1$.  Since
$y(t)$ is a positive function, we drop the absolute value sign.  Also, we
move the constant of integration to the right of the equality, subtracting
it from the constant of integration we get on the right from integrating
$\int k\,dt=kt+C_2$.  That is, we set $C=C_2-C_1$.  Thus, we have: $$\ln
y=kt+C.$$ Finally, to solve for $y$ we exponentiate both sides (this means:
raise $e$ to the expression on each side of the equality).  Since $e^{\ln
y}=y$, this gives us a formula for $y$: $$y(t)=e^{kt+C}=e^Ce^{kt}.$$ We are
not done, however, because we must determine the constant $e^C$ in front of
$e^{kt}$.  As in acceleration--velocity--distance problems, we do this by
using initial information.  In the bank interest problem, the initial
information is that we had $P$ dollars in the bank at time $t=0$: $y(0)=P$.
Substituting $t=0$ in the equation $y(t)=e^Ce^{kt}$ gives the result:
$P=y(0)=e^Ce^0=e^C$.  Thus, we conclude that $$y(t)=Pe^{kt}.$$ (The special
case of this formula when $k=100\%=1$ was obtained in the last section by a
different method.)

\smallskip 

A function of the form $y=Pe^{kt}$ is called {\bf exponential growth}.  Its
graph is shown below (in the case $k=8\%=0.08$).

\begin{center}
%-figure LA -----------------------------
\begin{psfigure}{4in}{1.5in}{figLA.ps}
%\putcrosshair{0,0}
\puttext{0, 20.}{\small $\displaystyle y = P\, e^{0.08 t}$}
\end{psfigure}\\
%------------------------------
\end{center}

\subsection{Effective annual interest rate} 
By the {\bf effective} interest rate we mean the percent by which the money
actually increases if left in the bank 1 year.  The increase is from $P$ to
$Pe^{k\cdot 1}$, i.e., $P(e^k-1)$.  Thus, the percent increase is $e^k-1$
(which we usually convert to a percent by multiplying by 100):
$$\text{effective~annual~interest~rate}=e^k-1.$$ For example, if $k=0.08$,
then the \index{effective annual interest rate} effective rate is $e^{0.08}-1=0.0833=8.33\%$.  If $k=100\%$, then
the effective rate is $e-1=171.83\%$.  In general, if $k$ is small, then
the effective rate is only slightly more than $k$; while if $k$ is large,
then the effective rate is much greater, because it takes into account the
interest earned on the interest.  Sometimes people say ``annual interest
rate'' when they really mean ``effective annual interest rate.''  Sometimes
$k$ is called the {\bf \index{instantaneous interest rate} instantaneous interest rate}.  

\smallskip

\noindent
{\bf Note.} {\em Whenever we say simply ``interest rate'' we shall mean the
instantaneous rate $k$, not the effective rate $e^k-1$}.

\smallskip

\subsection{Doubling time} 
At time $t$ we have $Pe^{kt}$ in the bank.  Let us ask how long it will
take for the money to double.  This time period, which we call $D$, must
satisfy the equation $Pe^{k(t+D)}=2Pe^{kt}$.  Dividing both sides by
$Pe^{kt}$ and then taking the $\ln$ of both sides, we obtain $kD=\ln 2$,
i.e., $D=(\ln 2)/k$.  Notice that the \index{doubling time} doubling time $D$ does not depend on
the starting time $t$ or on the principal $P$: it depends only on the
interest rate.  For example, at 8\% interest rate the doubling time is
$(\ln 2)/0.08=8$ 2/3 years.


\section[Other examples]{Other Examples of Exponential Growth.} 

Any time we have something whose rate of increase is proportional to its
present value (i.e., $\frac{dy}{ dt}=ky$), we obtain the exponential function
$y(t)=y_0e^{kt}$, where $k$ is the constant of proportionality between the
rate of increase and the present value, and $y_0$ is the initial value of
our function.  Among the examples of this situation are:

\smallskip

\subsection{Inflation}
Rate of increase in cost of living is measured as an annual percent of the
current cost of living.  If the \index{rate of inflation} rate of inflation remains constant, then
this cost of living follows the law of exponential growth.

\begin{example}
%{\bf Example 1.} 
In the six-month period from June to December 1990 in
Nicaragua, the c\'ordoba deflated from an exchange rate of about 200000 per
dollar to 2000000 per dollar.  Find both the effective annual rate and the
instantaneous rate $k$ of inflation of the price in \index{c\'ordobas} c\'ordobas of an item
whose dollar cost remains constant.

\smallskip
Here the inflation was by a factor of 10 in 1/2 year, and hence by a factor
of 100 in 1 year.  So the cost increased by $99=9900\%$.  Thus, the
effective rate is $99=e^k-1$.  To find $k$, we add 1 to both sides and take
$\ln$, obtaining $k=\ln 100=4.605= 460.5\%$.
\end{example}



\subsection{Population}  
If the birth and death rates remain constant (and there is no immigration
or emigration), then the population as a function of time follows the law
of exponential growth, in which $k$ is the difference: $k=$ birth rate $-$
death rate.

\begin{example}
%{\bf Example 2.} 
If a bacteria culture has a 4\% per hour \index{instantaneous
birth rate} instantaneous
birth rate and a 2.5\% instantaneous death rate, what is its doubling
time?

\smallskip
Here $k=4\%-2.5\%=1.5\%$, and so $D=(\ln 2)/0.015=46.2$ hours.
\end{example}


\section{Exponential Decay} 

Here we have a function $y(t)$ which {\bf decreases} at a rate proportional
to its current value, with proportionality constant $k$: $\frac{dy}{
dt}=-ky$.  Mathematically speaking, nothing much changes: to solve this
differential equation we simply replace $k$ by $-k$ in our earlier
derivation, thereby obtaining the function $y(t)=y_0e^{-kt}$.  However, the
graph of $y_0e^{-kt}$ as $t$ increases looks very different from the graph
of $y_0e^{kt}$, namely, it ends up decreasing rapidly toward 0:

\begin{center}
%-figure LB -----------------------------
\begin{psfigure}{4in}{1.5in}{figLB.ps}
%\putcrosshair{0,0}
\puttext{0, 13.3}{\small $\displaystyle y = y_{0} e^{-0.08\, t}$}
\puttext{-133.3, 33.3}{\Small$y_{0}$}
\end{psfigure}\\
%------------------------------
\end{center}

\subsection{Half-life} An exponential decay function $y(t)=y_0e^{-kt}$ has a
``\index{half-life} half-life'' rather than a doubling time.  This is the length of time $H$
that it takes to decrease to half its current value.  To find $H$, we set
$y_0e^{-k(t+H)}$ equal to half of $y_0e^{-kt}$.  Dividing this equation
through by $y_0e^{-kt}$ and then taking ~ln~ gives: $-kH=\ln(1/2)=-\ln 2$,
and so $H=(\ln 2)/k$.  Note that this is the same as the doubling time
formula in the case of exponential growth.


\section{Examples of Exponential Decay.} 

\subsection{Radioactive decay}


\begin{example}
%{\bf Example 3.} 
Some future visitors to Earth find a planet that was
rendered uninhabitable by a catastrophe that occurred many thousands of
years before.  By reconstructing the accident, they determine how much
radioactive Plutonium 239 was released at that time.  They also measure the
amount of \index{Plutonium 239} Plutonium 239 remaining, and find that only 1/500 of the
radioactive Plutonium is still present.  Using the value 24400 years for
the half-life of Plutonium 239, determine how much time elapsed between the
accident and the arrival of the visitors.

\smallskip

We first find $k$, using the relation $kH=\ln 2$: $k=\ln 2/24400=
0.0000284$.  Then we find $t$ by setting $y(t)=\frac{1}{ 500}y_0$, i.e.,
$\frac{1}{ 500}y_0=y_0e^{-kt}$.  Dividing by $y_0$ and taking $\ln$ gives
$-kt=\ln(1/500)=-\ln 500$, so that $t=(\ln 500)/0.0000284=218800$ years.

\end{example}


\subsection{Population when the death rate exceeds the birth rate}


\subsection{``Real'' salary (measured in inflation-adjusted dollars, i.e., buying
power of what you earn) if the rate of salary increase is less than the
inflation rate}


\subsection*{}
Finally, we should emphasize that real-world phenomena follow the law of
exponential growth or decay only if the thing being measured satisfies the
differential equation $y'=ky$ or $y'=-ky$.  In practice, when we suppose
this to be the case we are making a simplifying assumption that may be
quite inaccurate.  For example, most human or animal populations are
influenced by many factors and cannot be described at all accurately by the
exponential growth model.
% 
