Modelling with functionsEdexcel A-Level Maths: Revision notes

A mathematical model uses a function to describe a real situation. Match the function to the behaviour:

• Exponential growth $N=Ae^{kt}$ ($k>0$): the rate of increase is proportional to the size, as for bacteria.
• Exponential decay $T=a+be^{-kt}$: a quantity falls towards a limit $a$, as for cooling, with $T\to a$.
• Trigonometric $d=a+b\sin(ct)$: repeating cycles, as for tides and hours of sunlight. The mean is $a$, the amplitude is $b$ and the period is $\frac{360^\circ}{c}$ (or $\frac{2\pi}{c}$).
• Reciprocal $P=\frac kV$: inverse proportion, as for pressure and volume at constant temperature.

For cooling $T=20+70e^{-0.05t}$: initial value $T(0)=90$; limit $T\to20$ as $t\to\infty$. Solving for time: $50=20+70e^{-0.05t}\Rightarrow e^{-0.05t}=\frac37\Rightarrow t=\frac{\ln(7/3)}{0.05}=16.9$ minutes. Isolate the exponential first, then take natural logs. Rate of change: for $N=500e^{0.4t}$, $\frac{dN}{dt}=200e^{0.4t}$, which is $0.4N$. At $t=5$ the rate is $200e^2\approx1480$ bacteria per hour. Doubling or target size: $500e^{0.4t}=10000$ gives $t=\frac{\ln20}{0.4}=7.49$ hours.

For tides $d=6+2.5\sin(30t)^\circ$: maximum $8.5$ m, minimum $3.5$ m, period $\frac{360}{30}=12$ hours. To find when $d=4.75$: $\sin(30t)^\circ=-0.5$, so $30t=210$ and $t=7$ (07:00). Use the smallest positive angle for the first time, then add $360^\circ$ or use symmetry for later times. For gas pressure, use the given data to find the constant: $P=\frac kV$ with $P=120$ at $V=250$ gives $k=30000$, so $P=\frac{30000}{V}$. Then $V=80$ gives $P=375$ kPa. Halving the volume doubles the pressure.

Every model rests on assumptions, and an examiner expects you to evaluate them.

• Exponential growth predicts unlimited growth: $500e^{0.4t}$ gives about $8\times10^7$ bacteria at $30$ hours, which is impossible with finite food and space. A refined logistic model such as $N=\frac{20000}{1+39e^{-0.4t}}$ matches the start ($N(0)=500$) but levels off at $20000$.
• The cooling model assumes the surroundings stay at $20\,^\circ\text{C}$ and the cooling follows a simple exponential curve.
• A tide model assumes a regular cycle, but weather and the lunar cycle change the heights, so add another term or fit new data.
• $P=\frac{k}{V}$ assumes constant temperature and an ideal gas; at very high pressure the model over-predicts, which suggests a refinement such as $P(V-b)=k$. To compare a model with data, calculate the prediction and a percentage difference: a prediction of $1500$ against a measured $1200$ is $25\%$ too high.