Edexcel A level Maths (9MA0) · Pure mathematics › Exponentials and logarithms
Practise Exponential growth and decay models. 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy7 marks
The value, \(V\) pounds, of a car is modelled by \[V = 18{,}000\mathrm{e}^{-0.25t},\] where \(t\) is the time in years after the car was bought.
(a) State the value of the car when it was bought.[1]
(b) Find the value of the car after 5 years, to the nearest pound.[2]
(c) Find the time taken for the value of the car to fall to £5400. Give your answer to 3 significant figures.[3]
(d) Explain the significance of the value 0.25 in the model.[1]
Show the answer and mark scheme
(a)Answer: £18,000
B1 for £18,000
Worked solution: When \(t = 0\), \(\mathrm{e}^{0} = 1\), so \(V = 18{,}000\).
(b)Answer: £5157
M1 for substituting \(t = 5\)
A1 for £5157
Worked solution: \(V = 18{,}000\mathrm{e}^{-1.25} = 5157.086\), i.e. £5157.
(c)Answer: \(t = 4.82\) years
M1 for \(18{,}000\mathrm{e}^{-0.25t} = 5400\)
M1 for taking logarithms correctly: \(-0.25t = \ln \frac{3}{10}\)
A1 for awrt \(4.82\)
Worked solution: \(\mathrm{e}^{-0.25t} = \frac{3}{10} \Rightarrow t = -\frac{1}{0.25}\ln \left(\frac{3}{10}\right) = 4.82\) years.
(d)Answer: It is the rate constant: the rate of decrease is 0.25 times the current value (per year).
B1 for a correct interpretation, e.g. \(\frac{\mathrm{d}V}{\mathrm{d}t} = -0.25V\), so the rate of decrease is proportional to the current value with constant 0.25
Worked solution: \(\frac{\mathrm{d}V}{\mathrm{d}t} = -0.25V\): the value decreases at a rate proportional to its current size, and 0.25 is the constant of proportionality.
Question 2Medium6 marks
The value of a van was recorded when it was new and at the end of each of the first four years.
[object Object]
(a) A model \(V = 32\,000\mathrm{e}^{-kt}\) is proposed for the value, \(V\) pounds, after \(t\) years. Use the value when \(t = 2\) to find \(k\), giving your answer to 3 significant figures.[2]
(b) Use the model to predict the value of the van after 4 years.[2]
(c) Comment on the accuracy of the model, using the data in the table.[1]
(d) Give a reason why the model may not be suitable for predicting the value of the van after 20 years.[1]
Show the answer and mark scheme
(a)Answer: \(k = 0.178\)
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.).
(b)Answer: £15,680
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.
(c)Answer: The recorded value after 4 years is £15,800, so the prediction is only about £120 (under 1%) too low: the model fits the data well.
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.
(d)Answer: e.g. the model predicts a value of about £900, but the van would probably be scrapped or have only scrap value; the rate of depreciation may change with age.
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.
Question 3Hard10 marks
A cup of tea is made with water at 100 °C and is then left to cool in a room. A model for the temperature, \(T\) °C, of the tea \(t\) minutes later is \[T = \theta + (100 - \theta)\mathrm{e}^{-kt}\]where \(\theta\) (the room temperature in °C) and \(k\) are positive constants. The temperature of the tea is 60 °C after 10 minutes and 40 °C after 20 minutes.
(a) Show that \(\theta = 20\).[4]
(b) Find the exact value of \(k\).[2]
(c) Find the time taken for the tea to cool to 25 °C.[2]
(d) The temperature of the tea is actually measured as 29 °C after 40 minutes. Suggest a reason for the difference, and suggest how the model could be refined.[2]
(d)Answer: e.g. the rate of cooling slows more than the model assumes (a lid, a thick mug, or the room warming up); refine by fitting \(\theta\) and \(k\) to more data (e.g. including the 40-minute reading) or allowing the room temperature to vary.
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.