Exponential growth and decay models

• B1 for 200 g

Worked solution: When \(t = 0\), \(\mathrm{e}^{0} = 1\), so \(m = 200\).

• M1 for substituting \(t = 20\)
• A1 for 110 g

Worked solution: \(m = 200\mathrm{e}^{-0.6} = 109.762\), i.e. 110 g.

• M1 for \(200\mathrm{e}^{-0.03t} = 100\)
• M1 for taking logarithms correctly: \(-0.03t = \ln \frac{1}{2}\)
• A1 for awrt \(23.1\)

Worked solution: \(\mathrm{e}^{-0.03t} = \frac{1}{2} \Rightarrow t = -\frac{1}{0.03}\ln \left(\frac{1}{2}\right) = 23.1\) days.

• B1 for a correct interpretation, e.g. \(\frac{\mathrm{d}m}{\mathrm{d}t} = -0.03m\), so the rate of decrease is proportional to the current mass with constant 0.03

Worked solution: \(\frac{\mathrm{d}m}{\mathrm{d}t} = -0.03m\): the mass decreases at a rate proportional to its current size, and 0.03 is the constant of proportionality.

• M1 for \(22\,400 = 32\,000\mathrm{e}^{-2k}\) leading to \(k = \frac{1}{2}\ln\frac{32\,000}{22\,400}\)
• A1 for awrt 0.178

Worked solution: \(\mathrm{e}^{-2k} = \frac{22\,400}{32\,000} = 0.7 \Rightarrow k = -\frac{1}{2}\ln 0.7 = 0.178\) (3 s.f.).

• M1 for \(32\,000\mathrm{e}^{-4k}\) (or \(32\,000 \times 0.7^{2}\))
• A1 for £15,680 (allow £15,700 from using \(k = 0.178\))

Worked solution: \(V = 32\,000\mathrm{e}^{-4k} = 32\,000 \times \left(\mathrm{e}^{-2k}\right)^{2} = 32\,000 \times 0.7^{2} = 15\,680\), so £15,680.

• B1 for a comparison with the recorded value £15,800 (difference about £120, under 1%), e.g. the model fits the data well

Worked solution: The recorded value is £15,800, so the model underestimates by £120 (about 0.8%): a good fit for the first four years.

• B1 for a sensible reason, e.g. extrapolating far beyond the data; the value may level off at a scrap value; the van may no longer be working

Worked solution: The model predicts about \(32\,000 \times 0.7^{10} \approx 900\) pounds after 20 years, a long extrapolation from four years of data; in practice the van’s value depends on its condition and may level off at a scrap value.

• M1 for forming two equations: \(60 = \theta + (100 - \theta)\mathrm{e}^{-10k}\) and \(40 = \theta + (100 - \theta)\mathrm{e}^{-20k}\)
• M1 for using \(\mathrm{e}^{-20k} = (\mathrm{e}^{-10k})^2\) to eliminate \(k\)
• M1 for \((40 - \theta)(100 - \theta) = (60 - \theta)^2\) and expanding
• A1* for \(4000 - 140\theta = 3600 - 120\theta\), so \(\theta = 20\)

Worked solution: \(\mathrm{e}^{-10k} = \dfrac{60 - \theta}{100 - \theta}\) and \(\mathrm{e}^{-20k} = \dfrac{40 - \theta}{100 - \theta}\). Since \(\mathrm{e}^{-20k} = (\mathrm{e}^{-10k})^2\):
\(\dfrac{40 - \theta}{100 - \theta} = \dfrac{(60 - \theta)^2}{(100 - \theta)^2} \Rightarrow (40 - \theta)(100 - \theta) = (60 - \theta)^2\)
\(4000 - 140\theta + \theta^2 = 3600 - 120\theta + \theta^2 \Rightarrow 400 = 20\theta \Rightarrow \theta = 20\).

• M1 for \(\mathrm{e}^{-10k} = \frac{40}{80} = \frac{1}{2}\)
• A1 for \(k = \frac{\ln 2}{10}\)

Worked solution: \(60 = 20 + 80\mathrm{e}^{-10k} \Rightarrow \mathrm{e}^{-10k} = \frac{1}{2} \Rightarrow k = \dfrac{\ln 2}{10}\).

• M1 for \(80\mathrm{e}^{-kt} = 5\), i.e. \(\mathrm{e}^{-kt} = \frac{1}{16}\)
• A1 for 40 (minutes)

Worked solution: \(25 = 20 + 80\mathrm{e}^{-kt} \Rightarrow \mathrm{e}^{-kt} = \frac{1}{16} = \left(\frac{1}{2}\right)^4\). Since \(\mathrm{e}^{-10k} = \frac{1}{2}\), \(t = 40\) minutes.

• B1 for a sensible reason in context, e.g. the room temperature is not constant (the room warms up), heat loss is not exactly proportional to the temperature difference, or conditions changed (the cup was covered, milk added)
• B1 for a sensible refinement, e.g. refit the constants using more measurements (including later times), or use a model in which \(\theta\) varies with time

Worked solution: The tea cooled more slowly than predicted: perhaps the room warmed up, or heat is not lost at a rate exactly proportional to the temperature difference (e.g. the mug itself stores heat). The model could be refined by fitting \(\theta\) and \(k\) using more readings over a longer time, or by letting the room temperature vary.