Questions on functions in modelling

Question 1

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.

Accepted Answer

Equal steps added: linear. Equal factors: exponential. Repeating on a fixed cycle: trigonometric. B1 B1 B1 for the three families.

Question 2

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.

Accepted 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.

Question 3

Name the four stages of the problem solving cycle in the order they run, and state what sends the work round the cycle a second time.

Accepted Answer

Specify the problem, collect the information, process and represent it, interpret the result. B1 for the four stages in that order. The work goes round again when the interpretation shows the model to be wrong, or the question to have been too loose, and the cycle returns to the specification rather than patching the result. B1 for that.

Question 4

The time T hours that a pump takes to empty a tank is modelled as inversely proportional to the pumping rate r litres per minute. When r = 40, T = 6. Find T when r = 60.

Accepted Answer

Inverse proportion is T = k/r. Substituting the known pair, 6 = k/40, so k = 240 and the model is T = 240/r. At r = 60, T = 240/60 = 4 hours. M1 for T = k/r with the pair substituted, A1 for k = 240, A1 for T = 4. The behaviour that identifies the family is the constant product: rate × time is 240 whichever pump is used, and since T is in hours and r in litres per minute that is 240 litre-hours per minute, which is 240 × 60 = 14 400 litres once the hours are turned into minutes: the volume of the tank in disguise. Writing T = kr gives a pump that takes longer the faster it runs.

Question 5

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.

Accepted 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.

Question 6

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.

Accepted 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.

Question 7

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.

Accepted 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.

Question 8

Three pieces of work from a harbour study are listed: (i) reading the depth gauge at the harbour mouth every hour for a fortnight and recording the values; (ii) fitting h = 5 + 2.4 sin (30t)° to the readings; (iii) reporting that the depth is at least 4 m for about fifteen hours in every day. Name the stage of the problem solving cycle each belongs to.

Accepted Answer

(i) is collecting the information: taking the readings and writing them down is the gathering of the data, carried out to a plan that was fixed earlier. (ii) is processing and representing: a family has been chosen and its constants fitted. (iii) is interpreting the result: the model's output has been turned back into hours and metres in the harbour. B1 B1 B1 for the three stages. None of the three is specifying, which happened earlier, when a vague question about safety became a question about a depth, a threshold and a window, and the decision to measure the depth hourly for a fortnight was made. Naming stage (i) as specifying, because the readings follow a plan, mixes up carrying out the plan with making it; naming it as processing, on the grounds that hourly readings are data, mixes up gathering the data with doing something to it afterwards.

Question 9

A council asks whether a tidal swimming beach is safe to use. Rewrite this as a modelling question precise enough for a model to answer, naming a quantity, a threshold and a time window, and state what should be measured and how often, giving a reason for the frequency.

Accepted Answer

A usable version: for how many hours in each day is the depth of water at the swimming area at least 1.5 m, across a fortnight? B1 for naming a measurable quantity, depth of water at a stated place. B1 for a threshold and a window, at least 1.5 m, hours per day over a fortnight, so that an answer is a number the council can act on. Measure the depth at the swimming area itself, hourly, for two weeks. B1 for the measurement and its frequency. B1 for the reason: the tide turns roughly twice a day, so hourly readings resolve a twelve-hour cycle, whereas one reading a day cannot show a tide at all; and a fortnight covers both spring and neap tides.

Question 10

In a town the longest day has 16.5 hours of daylight and the shortest has 7.5 hours, and the pattern repeats every 12 months. The number of hours of daylight D is modelled by D = a + b sin (km)°, with m in months after the spring equinox. Find a, b and k, and the value of D two months after the spring equinox, to 3 significant figures.

Accepted Answer

The midline is the average of the extremes, a = (16.5 + 7.5)/2 = 12, and the amplitude is half their difference, b = (16.5 − 7.5)/2 = 4.5. A twelve-month period needs 12k = 360, so k = 30. B1 for a = 12 and b = 4.5, B1 for k = 30. At m = 2, D = 12 + 4.5 sin 60° = 12 + 3.897 = 15.9 hours. M1 for substituting m = 2 into the fitted model, A1 for 15.9. The model is D = 12 + 4.5 sin (30m)°. Fitting b as 9, the full gap between longest and shortest, puts the longest day at 21 hours.

Question 11

The mass M grams of a radioactive sample after t days is modelled by M = 80e^−kt. After 5 days the mass is 60 g. Find k to 3 significant figures, and the mass after 20 days to 3 significant figures.

Accepted Answer

At t = 5: 60 = 80e^−5k, so e^−5k = 3/4 and 5k = ln (4/3), giving k = 0.0575. M1 for substituting the data pair and isolating the exponential, A1 for k = 0.0575. Then at t = 20, M = 80e^−20k. Since 20 days is four lots of 5, e^−20k = (3/4)⁴, so M = 80 × 81/256 = 25.3 g. M1 for substituting t = 20, A1 for 25.3. Using the rounded k = 0.0575 in the calculator gives 25.33, so either route lands on 25.3, but carrying the exact k, or the ratio 3/4, is the safer habit: a k rounded to one figure, 0.06, gives 24.1 g.

Question 12

A bacterial culture is modelled by P = 20e^0.3t 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.

Accepted 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. 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. Saying only that the number is too big is not enough: name the fitted range and bring in the biology.

Question 13

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.

Accepted 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. An inequality in sine needs both ends of the window, and a calculator returns only the principal angle, so find the second angle as well. Sketching one arch of the curve and marking the line h = 7 across it settles the direction of the inequality in a moment.

Question 14

A drink is left in a room at 20 °C. A student models its temperature by T = 80e^−0.05t, with T in °C and t in minutes. State what this model predicts for large t and explain why that is wrong for the drink. Write down a refined model of the form T = c + Ae^−kt that starts at 80 °C and has the correct long-term behaviour, and given that the drink is at 50 °C after 20 minutes find k to 3 significant figures. State which stage of the problem solving cycle the refinement returns the work to.

Accepted Answer

As t grows, e^−0.05t tends to zero, so the model predicts T tends to 0 °C. B1. A drink cools towards the temperature of its surroundings and not below it, so the long-term value should be 20 °C, not 0 °C. B1 for the reason. The refined model is T = 20 + 60e^−kt: the constant 20 is the room, and 20 + 60 = 80 keeps the starting value. B1 for the form with both constants. At t = 20, 20 + 60e^−20k = 50, so e^−20k = 1/2 and 20k = ln 2, giving k = ln 2/20 = 0.0347. M1 for isolating the exponential, A1 for k. The refinement returns the work to the specifying stage: the assumption that failed, cooling towards nothing, is written into a sharper specification that includes the surroundings, and only then is a constant fitted again. B1 for naming the stage. Writing the refined model as 20 + 80e^−kt keeps the room but starts the drink at 100 °C, so both constants need checking against the two facts, the start and the limit.

Question 15

The population of a town was recorded every ten years: 5000, 6000, 7200 and 8640 at t = 0, 10, 20 and 30 years. Decide, with a reason from the data, whether a linear model or an exponential model P = 5000e^kt describes the growth. Fit the model you choose, giving any constant to 3 significant figures, and use it to predict the population at t = 50 to 3 significant figures.

Accepted Answer

Test the two signatures. The differences are 1000, 1200 and 1440, which are not equal, so the growth is not linear. The ratios are 6000/5000, 7200/6000 and 8640/7200, each equal to 1.2, so each decade multiplies the population by the same factor. M1 for testing differences or ratios, B1 for choosing the exponential model because the ratios are constant. Fit it: e^10k = 1.2, so k = ln 1.2/10 = 0.0182. M1 for e^10k = 1.2 or an equivalent statement, A1 for k. At t = 50, five decades on, P = 5000 × 1.2⁵ = 12 441.6, so about 12 400. A1. The choice comes before the fit: a linear model fitted to the first two points gives 10 000 at t = 50, and its error grows with every decade, because equal steps cannot reproduce equal factors. A fitted model with a constant ratio does not prove the growth will continue at 20% a decade; that is an assumption to state when the prediction is reported.