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Functions in modelling questions
By this point the course has covered several families of function, and modelling is a matter of choosing the one whose shape matches the situation, fitting its constants to the data, and stating where the model stops being reasonable. Tides call for a trigonometric model and cooling for an exponential one, and justifying the choice is itself examined.
7 original questions · 24 marks · the functions in modelling notes · Algebra and functions
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Data set one grows by equal amounts in equal time steps; data set two grows by equal factors; data set three repeats every 24 hours. Name the function family suited to each.
Worked answer
Equal steps added: linear. Equal factors: exponential. Repeating on a fixed cycle: trigonometric. B1 B1 B1 for the three families. Diagnosis comes before fitting; the behavioural signature picks the family, and only then do constants get estimated.The depth of water in a harbour is modelled by h = 6 + 1.8 sin (30t)°, in metres and hours. Write down the maximum and minimum depths.
Worked answer
The sine runs between ±1, so h runs from 6 − 1.8 = 4.2 m up to 6 + 1.8 = 7.8 m. B1 B1 for the two depths. Centre line 6, swing 1.8: the two numbers in the model are exactly these two facts.The depth of water in a harbour is modelled by h = 6 + 1.8 sin (30t)°, in metres and hours. Find the period of the tide and the first time high tide occurs.
Worked answer
The sine completes a cycle when 30t reaches 360, so the period is 12 hours. High tide needs sin (30t)° = 1, first at 30t = 90, so t = 3 hours. B1 for the period, M1 for 30t = 90, A1 for t = 3. Both answers come from reading the coefficient, not from any solving.The value of a machine is modelled by V = 400 + 5600e−0.2t pounds after t years. State the initial value and the long-term value, and find the value after 5 years to 3 significant figures.
Worked answer
At t = 0 the value is 400 + 5600 = £6000. As t grows the exponential dies away, leaving the £400 floor: scrap value. At t = 5: V = 400 + 5600e−1 = £2460. B1 for £6000, B1 for £400, M1 for substituting t = 5, A1 for £2460. The two constants carry the two ends of the story: start, and long run.The value of a machine is modelled by V = 400 + 5600e−0.2t pounds after t years. Find when the machine's value reaches £1200, to 3 significant figures.
Worked answer
400 + 5600e−0.2t = 1200 gives e−0.2t = 800/5600 = 1/7. Taking logs: 0.2t = ln 7, so t = 9.73 years. M1 for isolating the exponential, A1 for 1/7, M1 for taking logs, A1 for 9.73 years. The £400 must be moved across before the log is taken; logging a sum term by term is not a thing.A bacterial culture is modelled by P = 20e0.3t after t hours, fitted to data from the first 10 hours. Explain why the model may serve well at t = 12 but should not be trusted at t = 100, referring to both the mathematics and the biology.
Worked answer
Predicting inside the fitted range is interpolation and is anchored on both sides. Everything past t = 10 is extrapolation, and its safety falls away with distance. At t = 12 the model has stepped only just outside its window and the growth conditions plausibly still hold, so the prediction is worth having. At t = 100 it claims around 2 × 1014 bacteria, and unbounded exponential growth needs unlimited food and space, which no dish supplies. The mathematics extends for ever. The conditions that justified it do not, and a model's authority stops where its conditions do. B1 for naming extrapolation, B1 for t = 12 sitting just outside the fitted range, B1 for the size of the prediction at t = 100, B1 for the biological limits. An answer that says only “the number is too big” gets one mark of the four; the marks want the fitted range named and the biology brought in.For the harbour model h = 6 + 1.8 sin (30t)°, find the length of time in each cycle for which the depth exceeds 7 m. Give your answer to 3 significant figures.
Worked answer
Set 6 + 1.8 sin (30t)° > 7, so sin (30t)° > 5/9 = 0.5556. The principal angle is 30t = 33.75°, and sine stays above that value until its partner 180° − 33.75° = 146.25°. So 30t runs from 33.75° to 146.25°, giving t from 1.12 to 4.88 hours, a stretch of 3.75 hours in each 12-hour cycle. M1 for the inequality in sine, A1 for 33.75°, M1 for the second angle, A1 for 3.75 hours. The second angle is where the marks are lost: an inequality in sine needs both ends of the window, and a calculator hands over only the first. Sketching one arch of the curve and marking the line h = 7 across it settles the direction of the inequality in a moment.
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