Exponential growth and decay modelsEdexcel International A Level Maths: Revision notes

A quantity that changes at a rate proportional to its size follows an exponential model: $N=Ae^{kt},\qquad \frac{dN}{dt}=kN.$

• $k>0$: exponential growth (for example bacteria, populations).
• $k<0$: exponential decay (for example radioactive mass, with $m=50e^{-0.02t}$). The constant $A$ is the initial value: the value when $t=0$, since $e^0=1$.

To find a value, substitute $t$. To find a time, solve for $t$ using natural logarithms. Example: $P=400e^{0.05t}$. When $t=10$, $P=400e^{0.5}=659$. Doubling time: $e^{0.05t}=2$ so $t=\frac{\ln2}{0.05}=13.9$. For decay, $m=50e^{-0.02t}$ falls to 20 when $e^{-0.02t}=0.4$, so $t=\frac{\ln0.4}{-0.02}=45.8$. The rate of change is found by differentiating: $\frac{dm}{dt}=-0.02\times50e^{-0.02t}$, which is $-1$ when $t=0$. The negative sign shows the mass is decreasing.

Use the initial value to find $A$, then a second data point to find $k$. Example: $P=Ae^{kt}$ with $P=25$ at $t=0$ and $P=31$ at $t=5$. $A=25$. Then $31=25e^{5k}$, so $e^{5k}=1.24$ and $k=\frac{\ln1.24}{5}=0.0430$. Prediction at $t=15$: $P=25(1.24)^3=47.7$. Keep $k$ unrounded (or use $1.24^3$) in the next step to avoid rounding errors.

Consider what the model predicts as $t\to\infty$ and whether the values make sense.

• Growth ($k>0$): $N\to\infty$. No real population or colony can grow without limit, so the model is only valid for a limited range.
• Decay ($k<0$): $N\to0$ from above, but never reaches 0, since $e^{kt}>0$.
• Negative $t$ or very large $t$ may give values that are not physically sensible.

A second model can fix a weakness in the first. A model of the form $y=B+Ce^{-kt}\quad(k>0)$ tends to the limit $B$ as $t\to\infty$. Example: tea cooling in a room at $20^\circ$C, $\theta=20+70e^{-0.04t}$. Initially $\theta=90$; as $t\to\infty$, $\theta\to20$, the room temperature. This is realistic, whereas $\theta=90e^{-0.04t}$ would cool towards $0^\circ$C. Population: $P=60-35e^{-0.04t}$ levels off at 60 thousand, while $P=25e^{kt}$ grows without limit. To compare two models, evaluate both at the same time and comment on the difference and on long-term behaviour.