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Functions in modelling questions
By now the course has a shelf of function families, and modelling is a matter of choosing the right one, fitting its constants to the situation, and stating where it stops working. Tides call for trig and cooling calls for exponentials, and the choice itself is what gets 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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