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.

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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

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.