Exponential growth and decayAQA A-Level Maths: Revision notes

A quantity shows exponential growth or decay when it changes by the same factor in equal time intervals. The standard models are $y=A\mathrm{e}^{kt}\quad\text{or}\quad y=Ab^t,$ where $A$ is the initial value (at $t=0$). For $\mathrm{e}^{kt}$: $k>0$ gives growth and $k<0$ gives decay. The two forms are linked by $b=\mathrm{e}^k$, so $y=80\mathrm{e}^{-0.05t}$ is the same as $y=80\times0.951^t$. The reason $\mathrm{e}$ is natural is that the rate of change of $y$ is proportional to $y$ itself: the larger the quantity, the faster it grows or decays.

To find a value, substitute $t$: $M=80\mathrm{e}^{-0.05\times10}=48.5$. To find a time, set the model equal to the target and rearrange: $80\mathrm{e}^{-0.05t}=10\Rightarrow\mathrm{e}^{-0.05t}=\tfrac18\Rightarrow t=\frac{\ln8}{0.05}=41.6.$ Divide by the coefficient first, then take natural logarithms. The question may ask for the answer in particular units (days, hours, years), so state them.

For decay, the half-life is the time for the quantity to halve. Solve $\mathrm{e}^{kt}=\frac12$ with $k<0$: $t=\frac{\ln2}{|k|}$. For growth, the doubling time is $\frac{\ln2}{k}$. These times are constant: they do not depend on the starting amount or the moment you start measuring. Proof: the ratio $\frac{y(t+T)}{y(t)}=\mathrm{e}^{kT}$ contains no $t$. Example: $k=-0.3$ gives half-life $\frac{\ln2}{0.3}=2.31$ hours for a drug concentration.

With interest compounded continuously at annual rate $r$ (as a decimal), the value of an investment $P$ after $t$ years is $V=P\mathrm{e}^{rt}$. It is the limiting case of compounding more and more often. For £6000 at 4%, $V=6000\mathrm{e}^{0.04t}$: after 5 years $V=7328$. This is slightly more than annual compounding ($6000\times1.04^5=7300$). The equivalent annual rate is $\mathrm{e}^r-1$, here $\mathrm{e}^{0.04}-1=4.08\%$. Doubling time $=\frac{\ln2}{r}=17.3$ years.

Radioactive decay: $N=N_0\mathrm{e}^{-\lambda t}$, with half-life $\frac{\ln2}{\lambda}$. Drug concentration: $C=C_0\mathrm{e}^{-kt}$ falls as the body clears the drug; a drug is only effective while $C$ stays above a threshold, found by solving $C_0\mathrm{e}^{-kt}=\text{threshold}$. Population growth: $P=P_0\mathrm{e}^{kt}$ with $k>0$. In each case interpret the constants: $A$ is the starting amount and $k$ is the continuous rate of change per unit time. A growth constant $0.12$ means the population grows at a rate $0.12P$, i.e. 12% of its size per year.

An exponential growth model predicts unlimited growth, which is unrealistic for populations and other real quantities: food, space, disease and predators limit growth. For example, deer on an island cannot exceed what the food supply supports. Refinements include a model in which growth slows as the population approaches a maximum value (a carrying capacity), such as a logistic model, or a model with different values of $k$ in different periods. Exponential decay never reaches zero in the model, but a real quantity is made of whole atoms or molecules, so it eventually reaches zero. Evaluate a model by comparing its predictions with data and by asking whether the assumptions (constant rate, no limits) hold in the context.