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Summary

Harder exponential modelling

In a nutshell

When presented with a quantity for which the rate of change of the quantity with respect to an independent variable $t$ , the quantity will be modelled by a exponential function of the form $f(t) = Ae^{kt}$ , where $A$ and $k$ are constants. Such functions have the property that their rate of change with respect to $t$ is linearly proportional to the original function evaluated at $t$ ; in particular, the derivative is $f'(t) = kAe^{kt} = kf(t)$ . You can use the natural logarithms to identify the change in $t$ it takes for the quantity to change in size to a scalar multiple of itself.

Modelling exponential growth

{P = Ae^{kt}}

$P$	A quantity which changes with respect to $t$ at a rate proportional to the quantity itself
$A$	A constant; the value of $P$ when $t = 0$
$k$	A constant; the constant of proportionality between $P$ and $\dfrac{dP}{dt}$
$t$	A variable; in situations like population growth this could be time measured in some unit, e.g. days, but it does not necessarily have to be time.

In many cases, $A$ is positive. Notice that if $A$ is positive and $k$ is positive, $P$ will grow as $t$ increases, and if $A$ is positive but $k$ is negative, $P$ decays as $t$ increases. This is referred to as exponential growth, or exponential decay, respectively.

In situations where $t$ is referring to time (in some specified unit, e.g. days) then $A$ is the initial value of $P$ : when $t = 0$ , you have that $e^{kt} = e^{k \times 0} = 1$ , so $P = Ae^{kt} = A$ .

Example 1

The number of particles in a sample of a radioactive substance is modelled by the equation $N = 10^6e^{-kt}$ , where $t$ is the number of years since initial measurement. The half-life of the substance (the time it takes for the quantity to decay from $N$ to $\frac{N}{2}$ ) is approximately $300000$ years. Give an approximation for the constant $k$ .

You have that the value of $N$ at time $t + 300000$ is half the value of $N$ at time $t$ . Substitute this in an equation, simplify it, apply $\ln()$ to both sides and rearrange:

$\begin{aligned}10^6e^{-k(t + 300000)} &= \frac1210^6e^{-kt}\\e^{-kt - 300000k} &= \frac12e^{-kt}\\e^{-300000k} &= \frac12\\-300000k &= \ln\left(\frac12\right)\\k &= \frac{\ln(\frac12)}{-300000}\\k &= \frac{\ln(2)}{300000}\end{aligned}$

Therefore, $\underline{k = \frac{\ln(2)}{300000}}$ serves as a good approximation.

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