Exponential modelling

In a nutshell

There is a wide range of situations wherein you will want to study quantities for which the rate of growth or decay is proportional to the quantity itself. Functions of the form $$f(t) = Ae^{kt}$$, where $$A$$ and $$k$$ are constants, can be used to model such quantities, as they have the property that their rate of change is proportional to the original function; in particular, the derivative is $$f'(t) = kAe^{kt} = kf(t)$$.

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Modelling exponential growth

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

A petri dish contains a tiny colony of bacteria. Two scientists approximate the population of the colony, and then decide to model the population in the dish going forward by the equation $$P = 1000000e^{0.02t}$$, where $$t$$ is the time in minutes since the scientists initially approximated the population of the bacteria. What is the number $$1000000$$​ in this model? Approximately what will be the population in the dish $$6$$ hours after the scientists make their approximation? What is $$\dfrac{dP}{dt}$$​? Sketch the graph of $$P$$ against $$t$$ (choosing the scales of your axes appropriately).

The number $$1000000$$ is the constant $$A$$ in this model - it refers to the initial value, i.e. the value of $$P$$​ when $$t = 0$$​, which here is the approximation made by the scientists.

There are $$60 \times 6 = 360$$ minutes in $$6$$ hours. Using your calculator, find that the model predicts the population at time $$t = 360$$ to be $$1000000e^{0.02 \times 360}\approx1,339,430,764$$.

​$$\dfrac{dP}{dt}$$​ is equal to:

​$$\begin{aligned}\dfrac{dP}{dt} &= \dfrac{d}{dt}(1000000e^{0.02t})\\&= 1000000\dfrac{d}{dt}(e^{0.02t})\\&= 1000000 \times 0.02e^{0.02t}\\&= 20000e^{0.02t}\end{aligned}$$​​

The graph of this function looks something like this:

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