Fitting Models to Logarithmic Data | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Transform each observation before plotting

For measured data, a suitable straight-line graph usually passes close to the points, not exactly through all of them. A line of best fit follows their overall trend, with points distributed on both sides; it is not a point-to-point zigzag.

Fit $T=20+Ae^{-bt}$ to a cooling drink, with temperature $T$ in °C and time $t$ in minutes.

Calculate $Y=\ln(T-20)$ for each measured temperature. For example, at $t=2$ , $Y=\ln20.1\approx3.001$ .

Time, measured temperature and transformed temperature
t (min)	T (°C)	ln(T − 20)
0	50.0	3.401
2	40.1	3.001
4	33.5	2.603
6	29.0	2.197
8	26.1	1.808

Use well-separated points on the fitted line; the measurements need not lie exactly on it.

Choose two well-separated points on the fitted line. Using (0,3.40) and (8,1.80):

\begin{gathered}m=\frac{1.80-3.40}{8-0}=-0.20\\-b=-0.20\\b\approx0.20\\\ln A\approx3.40\\A\approx e^{3.40}\approx30\end{gathered}

The model is approximately $T=20+30e^{-0.2t}$ . Check an observation not used to find the gradient: at $t=4$ , it predicts about $33.5^\circ\mathrm C$ , close to the measured value.

Keep extra digits in the transformed data until the final estimates. Points lying on one straight line are called collinear. Measured data may only approximately follow a line.

Your turnPractice

Data follow $y=kx^n$ : (1,12), (2,3), (4,0.75). Which axes make them collinear? Find $k,n$ and check the third point.

Show worked answer

Plot $\ln y$ against $\ln x$ . Collinear means lying on one straight line.

\begin{gathered}n=\frac{\ln3-\ln12}{\ln2-\ln1}=-2\\\ln k=\ln12\Rightarrow k=12\\12(4)^{-2}=0.75\end{gathered}

Return to the original quantities

The fitted cooling model starts at approximately $20+30=50^\circ\mathrm C$ . As time increases, its exponential term tends to zero, so its predicted temperature approaches $20^\circ\mathrm C$ .

Worked example

Estimate when the drink reaches 23°C.

\begin{gathered}23=20+30e^{-0.2t}\\e^{-0.2t}=0.1\\t=\frac{\ln0.1}{-0.2}\\t\approx11.5\text{ minutes}\end{gathered}

This is beyond the observed 0–8 minute interval. It is an extrapolation: a prediction outside the measured range, which relies on the model continuing to apply.

Your turnPractice

A fitted model is $T=18+24e^{-0.3t}$ for $t\ge0$ , with time in minutes. Find the initial temperature, the long-term temperature and when it reaches 24°C.

Show worked answer

\begin{gathered}T(0)=42^\circ\mathrm C\\T\to18^\circ\mathrm C\\24=18+24e^{-0.3t}\\t=\frac{\ln4}{0.3}\\t\approx4.62\text{ minutes}\end{gathered}

The arrow means the temperature approaches 18°C as time increases; the model does not reach it at a finite time.

Transform each observation before plotting

Fit $T=20+Ae^{-bt}$ to a cooling drink, with temperature $T$ in °C and time $t$ in minutes.

Calculate $Y=\ln(T-20)$ for each measured temperature. For example, at $t=2$ , $Y=\ln20.1\approx3.001$ .

Time, measured temperature and transformed temperature
t (min)	T (°C)	ln(T − 20)
0	50.0	3.401
2	40.1	3.001
4	33.5	2.603
6	29.0	2.197
8	26.1	1.808

Use well-separated points on the fitted line; the measurements need not lie exactly on it.

Choose two well-separated points on the fitted line. Using (0,3.40) and (8,1.80):

\begin{gathered}m=\frac{1.80-3.40}{8-0}=-0.20\\-b=-0.20\\b\approx0.20\\\ln A\approx3.40\\A\approx e^{3.40}\approx30\end{gathered}

The model is approximately $T=20+30e^{-0.2t}$ . Check an observation not used to find the gradient: at $t=4$ , it predicts about $33.5^\circ\mathrm C$ , close to the measured value.

Keep extra digits in the transformed data until the final estimates. Points lying on one straight line are called collinear. Measured data may only approximately follow a line.

Your turnPractice

Data follow $y=kx^n$ : (1,12), (2,3), (4,0.75). Which axes make them collinear? Find $k,n$ and check the third point.

Show worked answer

Plot $\ln y$ against $\ln x$ . Collinear means lying on one straight line.

\begin{gathered}n=\frac{\ln3-\ln12}{\ln2-\ln1}=-2\\\ln k=\ln12\Rightarrow k=12\\12(4)^{-2}=0.75\end{gathered}

Return to the original quantities

The fitted cooling model starts at approximately $20+30=50^\circ\mathrm C$ . As time increases, its exponential term tends to zero, so its predicted temperature approaches $20^\circ\mathrm C$ .

Worked example

Estimate when the drink reaches 23°C.

\begin{gathered}23=20+30e^{-0.2t}\\e^{-0.2t}=0.1\\t=\frac{\ln0.1}{-0.2}\\t\approx11.5\text{ minutes}\end{gathered}

This is beyond the observed 0–8 minute interval. It is an extrapolation: a prediction outside the measured range, which relies on the model continuing to apply.

Your turnPractice

A fitted model is $T=18+24e^{-0.3t}$ for $t\ge0$ , with time in minutes. Find the initial temperature, the long-term temperature and when it reaches 24°C.

Show worked answer

\begin{gathered}T(0)=42^\circ\mathrm C\\T\to18^\circ\mathrm C\\24=18+24e^{-0.3t}\\t=\frac{\ln4}{0.3}\\t\approx4.62\text{ minutes}\end{gathered}

The arrow means the temperature approaches 18°C as time increases; the model does not reach it at a finite time.