Mathematics and Logic - Skill and Concept Development

Exponential Growth and Decay Models - Algebraic View

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.

Solving for k

Model Recognition

• 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 $A_0 = e^b$.

• 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

1. 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.

2. The continuous growth an investment over time in which growth or interested in compounded continously.

3. Decay and half-life of radio active elements.

4. The amount of particles with diameter large than a given size d in a grinding machine.

5. 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 \[A = A_0(1+r)^N = A_0 \exp( n \ln(1+r))\]

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

1. 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.

2. 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.

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

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

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

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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