IB Maths AI HLModelling with FunctionsPaper 2Tides, Ferris wheels, oscillation~9 min read
Sinusoidal Models
Any quantity that oscillates between a maximum and a minimum on a fixed cycle — tides, daylight hours, Ferris-wheel heights, temperatures, biological rhythms — can be modelled by y = a sin(b(x − c)) + d or its cosine cousin. The four parameters fall out of the physical setup.
📘 What you need to know
Standard form: y = a sin(b(x − c)) + d or y = a cos(b(x − c)) + d.
Amplitude |a| = (max − min) / 2.
Principal axisy = d = (max + min) / 2.
Period = 360°/b (degrees) or 2π/b (radians); read it as the cycle length from the context.
Phase shiftc: horizontal translation. Use a cosine starting at the max, or a sine starting on the axis going up — whichever matches the time of a known feature.
Predict by substitution; solve for time by setting y equal to a target and using the GDC’s intersect or the inverse trig function.
Reading the four parameters from context
A typical Paper 2 question gives you the maximum and minimum values and the time between two key events. From those alone you can build the whole model. The principal axis sits halfway between max and min — that’s d. The amplitude is the gap from the axis to either extreme — that’s a. The period equals twice the time between a maximum and the next minimum (or directly, the time between two consecutive maxima). And the phase shiftc aligns the curve so that the maximum (for cosine) or the rising axis crossing (for sine) lands at the right time.
Choosing sine or cosine
Both forms describe identical curves — the choice is purely about which feature you’d like to anchor at t = c. Use cosine when the maximum is at the cleanest time (so the wave starts at the top with no shift, or c is small). Use sine when the function is at the principal axis going up at a clean time. If the wave starts at the minimum, use −a cos(bt) + d: the negative coefficient flips cosine upside down so it starts at the trough.
Tide model d(t) = 3 cos(30(t − 6)°) + 6: principal axis 6 m, amplitude 3 m, period 12 h, max at 6 a.m.
Sinusoidal model at a glancey = a sin(b(x − c)) + d or y = a cos(b(x − c)) + da = max − min2 · d = max + min2 · b = 360°period
Solving for time
“When does the tide first reach 7 m?” “At what time is the greenhouse 25°C?” Both are solve-for-t problems. Set the model equal to the target, isolate the trigonometric expression, then apply the inverse function (sin−1 or cos−1) — or simply use the GDC’s intersect feature, which avoids the principal-value pitfalls entirely. Always look at the graph to make sure you’ve picked the correct branch and that the time lies inside the domain of the model.
Sketch first, solve second. A quick sketch shows roughly where the curve crosses the target line, so you know whether you want the first, second or third solution from the GDC.
🧠Recipe — sinusoidal modelling
Identify max, min, and the period from the context.
Compute a and d: a = (max − min)/2, d = (max + min)/2.
Compute b = 360°/period (or 2π/period for radians).
Pick sine or cosine and find c to align the curve with a known feature in time (max for cosine; rising axis crossing for sine).
Apply: substitute a time for prediction, or set equal to a target and use GDC intersect / inverse trig to solve.
Worked examples
WE 1
Ferris wheel height
A Ferris wheel has radius 12 m and its centre 14 m above the ground; one full revolution takes 8 minutes. A passenger boards at the lowest point at t = 0. (a) Write h(t) using cosine. (b) Find h(3) (2 d.p.).
(a) parametersamp = 12, axis d = 14, period 8b = 360/8 = 45°starts at minimum ⇒ use −cosh(t) = −12 cos(45t°) + 14(b) h(3) = −12 cos(135°) + 14cos(135°) = −√2/2h = −12(−√2/2) + 14 = 6√2 + 14h(3) ≈ 22.49 mat t = 4 (half a cycle) the passenger is at the top, h = 26 m.
WE 2
Building a tide model from max/min
The depth d (m) of water in a harbour varies sinusoidally. The maximum depth, 9 m, occurs at 6 a.m. and the minimum, 3 m, at noon. Let t be hours after midnight. (a) Find a, d (axis) and the period. (b) Write d(t) using cosine. (c) Find d at 9 a.m.
(a) amp = (9 − 3)/2 = 3axis = (9 + 3)/2 = 6max-to-min = 6 h ⇒ period = 12 hb = 360/12 = 30°(b) max at t = 6 ⇒ shift cos by 6d(t) = 3 cos(30(t − 6)°) + 6(c) d(9) = 3 cos(30 · 3°) + 6= 3 cos(90°) + 6 = 0 + 6d(9 am) = 6 m9 a.m. is halfway between max (6 a.m.) and min (noon) — on the principal axis falling.
WE 3
Daylight hours through the year
The daylight hours D in a city are modelled by D(t) = 4 sin(360365(t − 80)°) + 12, where t is the day of the year. (a) State maximum and minimum daylight. (b) Find the day on which the maximum occurs. (c) Find D on day 200 (2 d.p.).
(a) amp 4, axis 12max = 12 + 4 = 16; min = 12 − 4 = 8max 16 h · min 8 h(b) sin = 1 when arg = 90°(360/365)(t − 80) = 90t − 80 = 90 · 365/360 = 91.25day t = 171.25 (~ summer solstice)(c) D(200) = 4 sin((360/365)(120)°) + 12arg = 43200/365 ≈ 118.36°sin(118.36°) ≈ 0.8800D ≈ 4(0.880) + 12D(200) ≈ 15.52 h
WE 4
Greenhouse temperature: solve for time
A greenhouse temperature (°C) over 24 hours is modelled by T(t) = 6 cos(15(t − 14)°) + 22 (t in hours after midnight). (a) Maximum, minimum and time of maximum. (b) T at noon (t = 12, 2 d.p.). (c) First time after midnight when T = 25°C.
(a) amp 6, axis 22max = 28 °C at t = 14 (2 p.m.)min = 16 °C at t = 2 (2 a.m.)(b) T(12) = 6 cos(−30°) + 22= 6(√3/2) + 22 = 3√3 + 22T(12) ≈ 27.20 °C(c) solve 25 = 6 cos(15(t − 14)°) + 22cos(15(t − 14)°) = 3/6 = 0.515(t − 14) = ±60°t = 10 or t = 18first occurrence: t = 10 (10 a.m.)T(0) ≈ 16.80 °C falls to min at t = 2, then rises to 25 at t = 10.
WE 5
Wind turbine blade tip
The hub of a wind turbine is 30 m above the ground; the blade is 12 m long; one full rotation takes 4 seconds. At t = 0 the tip is at hub height and rising. (a) Write h(t) using sine. (b) Find h(0.5) (2 d.p.).
(a) parametersamp = 12, axis = 30, period = 4b = 360/4 = 90°starts at axis rising ⇒ pure sine, no shifth(t) = 12 sin(90t°) + 30(b) h(0.5) = 12 sin(45°) + 30= 12(√2/2) + 30 = 6√2 + 30h(0.5) ≈ 38.49 mat t = 1 the tip is at the top (42 m); at t = 3 it’s at the bottom (18 m).
WE 6
Oscillating spring
A mass on a vertical spring oscillates so that its displacement h (cm) from rest at time t (s) is h(t) = 8 sin(90t°). (a) State the amplitude, period, max and min. (b) Find h at t = 0.5 (2 d.p.). (c) Find the first time after t = 0 when h = 4 cm.
(a) read offamp = 8, b = 90 ⇒ period = 360/90 = 4 smax = 8, min = −8(b) h(0.5) = 8 sin(45°)= 8(√2/2) = 4√2h(0.5) ≈ 5.66 cm(c) solve 4 = 8 sin(90t°)sin(90t°) = 0.5first positive: 90t = 30° ⇒ t = 30/90t = 1/3 s ≈ 0.33 sthe second solution within one period is t = (180 − 30)/90 = 5/3 s.
💡 Top tips
Build the four parameters from words: max + min gives a and d; cycle length gives b; the timing of a known feature gives c.
Cosine for max-at-known-time, sine for axis-rising-at-known-time, −cos for min-at-known-time — pick to make c small.
Always match the angle mode (degrees vs radians) of your function before pressing any GDC keys.
For “first time” questions, sketch the model first so you see which solution branch you want.
Round only at the end; carry full precision through the working.
âš Common mistakes
Amplitude as max instead of (max − min)/2 — double-counting the swing.
Period and frequency swapped: b = 360°/period, not period/360°.
Wrong sign for c: in b(t − c) the shift is +c (to the right); b(t + c) shifts left.
Picking the wrong inverse-trig branch: only the principal value is returned; check the other solution within the period.
Wrong angle mode: a model written in degrees evaluated in radians gives nonsense values.
Next up: Logistic Models — S-shaped curves P(t) = L/(1 + Ae−kt) for populations that start exponentially but flatten toward a carrying capacity.
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