Exponential Growth and Decay Models - Algebraic View
This lesson on exponential growth and decay models provides an unified
algebraic and view of discrete and continuous growth and decay models -
discrete compound, continous compound, half-life and doubling time
models. All are or can be expressed in terms of the natural logarithm
and its inverse, the exponential. But half-life and doubling-time
models may be expressed or analysed with the aid logarithms to base 2
and the 2x exponential function. Which logarithms and
exponentials are employed in the forward and backward use of these
models is a matter of taste, guided by ideas of what is simplest.
The general form is \[A(t)= A_0 e^{kt} \] for some constant k where t
usually denotes time - a continuous variable. All the growth and decay
formulas can be expressed in this form, and vice versa.
If k is positive, the formula represents exponential growth in time. If
k is negative, the formula represents exponential decay in time.
In the event time is given in discrete units, say years or months, a
letter N favoured for the representation of discrete variable may be used.
In this case \[A(n)= A_0 e^{kN} \] where N denotes the whole number of
elapsed units of time.
Solving for k
The natural logarithm of $A(t)$ \[ y = \ln(A(t)) = \ln(A_0) + k \cdot
t \] is a linear function of time t. So its graph is a straight with slope
or rate of change \begin{eqnarray*} k&=&\frac{y_2-y_1}{t_2-t_1} \\
&=& \frac{\ln(A(t_2)) -\ln(A(t_1)) }{t_2-t_1} \\ &=& \frac
1{t_2-t_1} \ln \left(\frac{A(t_2)}{A(t_1)} \right) \end{eqnarray*} The
special case $t_2=t$ and $t_1=0$ gives \[ k = \frac 1t \ln \left( \frac{A(t
)}{A_0} \right) \]
Model Recognition
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If the graph of $y = \ln(A)$ versus t lies on a straight line y = kt
+ b then \[ A =\exp(kt+b ) = e^b e^{kt}= A_0e^{kt} \] where $A_0 =
e^b$.
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If the graph of $y = \log_c(A)$ versus t lies on a straight line y =
kt + b then \[ A =c^{kt+b } = c^b c^{kt}= A_0c^{kt} \] where $A_0 =
c^b.$ The cases where c is 2, e or 10 are of the most interest.
Exponential model recognition provides one reason of the use of
logarithmic graph paper. In 1976, in a digression from mathematics
studies, I developed an exponential decay model for the action of an ore
grinding machine, and then asked for data to help get the model
parameters - a grinding constant k. The engineer I was helping then
provided a log paper of data points falling on a straight line. The
development was my independent rediscovery of the work of others.
Examples subject to limitations
-
Initial Population growth where food is plentiful. The model fails
when the limit of food or resourse is met. The net result may be
population stability - no growth - if mortality rates equal birth
rates; or a population collapse due to a shortage of food or
resources. For further thought, investigate bacteria growth in a
petri dish, human population growth of your region or country;
population growth of bacteria and animal life in the oceans.
-
The continuous growth an investment over time in which growth or
interested in compounded continously.
-
Decay and half-life of radio active elements.
-
The amount of particles with diameter large than a given size d in a
grinding machine.
-
Economic growth models of government planners whose optimism knows no
bounds. In Quebec, the Mirabel white elephant Airport was justified
by predictions of continued economic growth, exponential style,
without bounds.
Compound Growth and Decay Models
The discrete compound growth or decay model has the form
\[A = P(1+r)^N \]
with $A_0=$ where N is a number of periods, and r is the change period.
Here r > -1 may be given by a decimal, a fraction or a percentage. The
value of N may be determined from the values of A, P and r by taking the
natural logarithm of both sides
Here
\[A = A_0(1+r)^N \] can be written in the form $A = A_0e^{kN}$ with $k
=\ln(1+r)$ because \[A = A_0(1+r)^N = A_0 \exp( n \ln(1+r))\]
Here $1+r = \exp(k)$ allows r to be computed from the calculation of k
above, or in a similar, more direct manner. In the case of compound growth
and decay, the graph of \[ y = \ln(A(N)) = \ln(P) + N \cdot \ln(1+r) \]
versus the discrete variable n consists of discrete points on a straight
line. with slope $\ln(1+r)$. The same equation may be used to approximate N
when A, P and r are given.
Practical Question: What do you do when the approximation does not
give a whole number, but is close to a whole number?
In practice the formula $A = P(1+r)^N $ may represent
-
a compound interest bank account with interest rate r given as a
number, a fraction or most often as a percentage. In years gone by,
Swiss bank accounts use to charge a negative rate of interest for the
storage of large sums of money in private. Whence $A = P(1+r)^N $
represents a compound decay or shrinkage model. I have with regrets
never had such an account, albeit with bank charges, the effective
interest rate on small bank accounts may be zero or negative.
-
Bird or whale or other plant or animal populations with growth rate r
usually expressed as a percentage. In the case of Beluga whales, the
percentage r appears to be a negative 1 tor 2 percent per year.
Whence $A = P(1+r)^N $ represents a compound decay or shrinkage
model.
All good things come to an end. Typically, the formula $A = P(1+r)^N $
represents growth or decay over a short period of time because the compound
growth rate may vary.
Exercise 1 : $A = P(1+r)^N $ in the form $A = P 2^{\frac NT}$ Hint:
Observe $ y = 2^{log_c(y)}$ and let $y=\frac AP$
Exercise 2 : Given r > 0, how may years T is needed for the amount A
to double $A = P(1+r)^N $ ? Give an exact formula for T.
Continous Doubling Time Models
\[A(t) = A_0 2^{\frac tT} \] The property $A(t+T) = 2 A(t)$ (why?)
implies T is the doubling time. Use a whole number approximation for T.
Here \[A(t) = A_0 2^{\frac tT} \] can be written in the form $A =
A_0e^kt$ because \[A(t) = A_0 2^{\frac tT} = A_0 \exp( \frac tT \ln(2))\]
The graph of $y=\log_2(A)$ versus time t has slope $m = \frac1T$
Continous Half-Life Models
\[A(t) = A_0 2^{-\frac tT} \] The property $A(t+T) = \frac12 A(t)$
(why?) implies T is a half-life - in fact the half-life.
Here \[A(t) = A_0 2^{-\frac tT} \] can be written in the form $A =
A_0e^kt$ because \[A(t) = A_0 2^{-\frac tT} = A_0 \exp(- \frac tT
\ln(2))\] The graph of $y=\log_2(A)$ versus time t has slope $m =
-\frac1T$ with a negative value.
Carbon Dating Before Atomic Bomb Testing 1950s onward
In the upper atmostsphere, beyond the reach of human activities, the
sun radiation in the form of high energy particles, gamma rays [?], hits
air molecules, including those of Carbon-Dioxide [$CO_2$] The radiation
converts some of the Carbon-12 in the atmosphere into the radio-active form
Carbon-14. The latter has a half-life of 5,730 years.
The density of Carbon-14 in the atmosphere stabilises when the decay of
Carbon-14 present balances the generation of Carbon 14. In the last
several thousand years before the industrialization and associated
burning of fossil fuels released Carbon Dioxide into the atmosphere, we
assume that the density of Carbon 14 in the atmosphere, its proportion to
normal Carbon-12, was constant.
A living plant on land absorbs atmospheric carbon while growing, with the
absorption stopping at the time of death. At the time of death, the
fraction $f(t) = A(t)$ of Carbon-14 to Carbon-12 in the plant tissue
equals that of the atmosphere. But over time, the Carbon 14 and hence the
ratio decays. Here
\[f(t) \approx A_0 2^{-\frac tT} \]
where T = 5,730 years, t = time since death, and $A_0$ = the [average]
ratio of Carbon-14 to Carbon 12 during the life of the plant. We assume
that ratio was constant before, with regrets, the advent of human-kind's
atomic age.
In this model, the time since death of the plant is unknown, but T, $A_0$
and the current value $f(t)$ are known or measureable. Whence \[
\log_2(\frac{f(t)}{A_0}) = -\frac t{T}=-\frac t{5,730 \hbox{ years}} \]
provides an linear equation in t to solve for the latter's value. The
Carbon-14 dating is reliable for upto 60000 years according to the
reference. That being said, in my youth, I heard this methods was
reliable for less than 15000 years. The technology may have advanced
while I was not looking.
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