Mathematics Advanced • Year 11 • Module 2 • Lesson 14
Modelling with Trigonometric Functions
Apply sinusoidal modelling end-to-end to ferris wheels, ocean tides, breathing cycles, blood pressure, and pendulum motion — extracting parameters then making predictions.
Problem 1 — Ferris wheel ride (kinematic)
A ferris wheel of diameter $40\,\text{m}$ has its centre $25\,\text{m}$ above the ground. It completes one revolution every $6$ minutes. A rider boards at the lowest point at $t = 0$ minutes.
Set up: What are we solving for?
(i) State the max height, min height, period, and find $a$, $b$, $d$. 3 marks
(ii) Write a cosine model $h(t)$ (Hint: starts at min $\Rightarrow$ use $-\cos$). Verify at $t = 0$. 2 marks
(iii) Find the height at $t = 4\,\text{min}$ (exact form, then to 2 dp). At this moment, is the rider rising or falling? Justify. 3 marks
Stuck? Differentiating mentally: $-\cos$ has derivative $\sin$, which tells you whether the rider is rising.Problem 2 — Tide window for a boat
The tide height at a port is $h(t) = 2 + 1.5\cos\!\left(\dfrac{2\pi t}{12.4}\right)$ metres, where $t$ is hours after high tide. A boat needs at least $h \geq 2.8\,\text{m}$ to enter.
Set up: What are we solving for?
(i) State the high tide and low tide values, and the period. 2 marks
(ii) Find the first time $t > 0$ when $h(t) = 2.8$. Give to 3 dp. 3 marks
(iii) Using symmetry of the cosine curve about $t = 0$, find the total time per cycle during which the boat can enter. 3 marks
Problem 3 — Lung volume during breathing
An adult's lung volume during quiet breathing varies between $2.2\,\text{L}$ (end of exhale) and $2.7\,\text{L}$ (peak of inhale). One full breath takes $5$ seconds. At $t = 0$ the person is at end-exhale.
Set up: What are we solving for?
(i) Build a sinusoidal model $V(t)$. (Starts at minimum $\Rightarrow$ use $-\cos$.) 3 marks
(ii) Find $V$ at $t = 1.25\,\text{s}$ (quarter period — should be the midline). 2 marks
(iii) Tidal volume (the amount of air moved per breath) is max − min. State the tidal volume from your model and check it against the typical adult value ($\approx 0.5\,\text{L}$). 2 marks
Problem 4 — Blood pressure model
An arterial blood pressure trace varies between a systolic peak of $120\,\text{mmHg}$ and a diastolic trough of $80\,\text{mmHg}$. The heart rate is $72\,\text{bpm}$ (so period $= 60/72 = 5/6$ s). At $t = 0$ the trace is at the diastolic minimum.
Set up: What are we solving for?
(i) Write $P(t)$ in the form $-a\cos(bt) + d$ (matching the "start at min" condition). 3 marks
(ii) Find the first time $t > 0$ when $P(t) = 100$ mmHg (the textbook "mean arterial pressure" approximation). 3 marks
(iii) State one limitation of using a pure sinusoid to model blood pressure (real traces have a sharper rise than fall — the "dicrotic notch"). 1 mark
Problem 5 — Pendulum motion (SHM)
A pendulum of length $1\,\text{m}$ swings with small-angle simple harmonic motion. Its horizontal displacement (from rest) is $x(t) = 0.15\sin(\pi t)$ metres, where $t$ is in seconds.
Set up: What are we solving for?
(i) State the amplitude, period (in seconds), and the position at $t = 0$. 2 marks
(ii) Find the first time $t > 0$ at which the bob passes through $x = 0.075\,\text{m}$ (half the amplitude). Give exact form. 3 marks
(iii) A student claims the bob passes through $x = 0.075$ exactly twice per second. Evaluate this claim — count the crossings per period correctly, then compute the actual rate per second. 2 marks
Stuck on (iii)? Per period the line $x = 0.075$ crosses the sine wave twice; period is $2$ s, so $2/(2) = 1$ crossing per second on average. The student double-counted.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Ferris wheel
Set up. Extract parameters from a ferris wheel description (centre and diameter give max/min), build a starts-at-min model, then evaluate.
(i) Max $= 25 + 20 = 45$ m, min $= 25 - 20 = 5$ m. Period $= 6$ min. $a = (45 - 5)/2 = 20$, $d = (45 + 5)/2 = 25$, $b = 2\pi/6 = \pi/3$.
(ii) Starts at min, so $-\cos$: $h(t) = 25 - 20\cos(\pi t/3)$. Verify: $h(0) = 25 - 20 = 5$ ✓.
(iii) $h(4) = 25 - 20\cos(4\pi/3) = 25 - 20(-1/2) = 25 + 10 = $ $35$ m. At $t = 4$, the argument $4\pi/3$ is in Q3 (between $\pi$ and $3\pi/2$). $\dfrac{d}{dt}[-\cos(\pi t/3)] = (\pi/3)\sin(\pi t/3)$. At $t = 4$, $\sin(4\pi/3) = -\sqrt{3}/2 < 0$, so derivative is negative — rider is falling. (Makes sense: max was at $t = 3$, so by $t = 4$ the rider is past the peak and coming down.)
Problem 2 — Tide window
Set up. Tide model is symmetric about $t = 0$; find where $h$ first drops to threshold $2.8$, then double by symmetry to get the entry window per cycle.
(i) High $= 2 + 1.5 = 3.5$ m, low $= 2 - 1.5 = 0.5$ m, period $= 12.4$ h.
(ii) $2 + 1.5\cos(2\pi t/12.4) = 2.8 \Rightarrow \cos(2\pi t/12.4) = 0.8/1.5 = 8/15 \approx 0.5333$. First positive $t$: $2\pi t/12.4 = \cos^{-1}(8/15) \approx 1.0064$ rad. So $t \approx 12.4 \times 1.0064/(2\pi) \approx $ $1.986$ h $\approx 1$ h $59.1$ min.
(iii) By symmetry of cosine about $t = 0$, $h(t) \geq 2.8$ for $t \in [-1.986, +1.986]$, total length $\approx $ $3.972$ h $\approx 3$ h $58$ min per cycle.
Problem 3 — Breathing
Set up. Build start-at-min cosine model, evaluate at midline time, compute tidal volume.
(i) $a = (2.7 - 2.2)/2 = 0.25$. $d = (2.7 + 2.2)/2 = 2.45$. Period $5$ s $\Rightarrow b = 2\pi/5$. Starts at min, so $-\cos$: $V(t) = 2.45 - 0.25\cos(2\pi t/5)$.
(ii) $V(1.25) = 2.45 - 0.25\cos(2\pi(1.25)/5) = 2.45 - 0.25\cos(\pi/2) = 2.45 - 0 = $ $2.45$ L (midline — quarter-period into the breath, going up).
(iii) Tidal volume $= \text{max} - \text{min} = 2.7 - 2.2 = $ $0.5$ L ✓ (matches the typical $\approx 500$ mL).
Problem 4 — Blood pressure
Set up. Parameter extraction, start-at-min, then solve for the mean.
(i) $a = (120 - 80)/2 = 20$, $d = (120 + 80)/2 = 100$, $b = 2\pi/(5/6) = 12\pi/5$. $P(t) = 100 - 20\cos(12\pi t/5)$.
(ii) $100 - 20\cos(12\pi t/5) = 100 \Rightarrow \cos(12\pi t/5) = 0 \Rightarrow 12\pi t/5 = \pi/2 \Rightarrow t = 5/(24) = $ $5/24$ s $\approx 0.208$ s.
(iii) Limitations (any one): real arterial waveform rises faster than it falls (sinusoidal model is symmetric, so it misses the rapid systolic upstroke and the dicrotic notch); a pure sine wave doesn't model variability in heart rate (heart-rate variability); reflected waves and vessel compliance create extra peaks that a single sinusoid cannot represent.
Problem 5 — Pendulum
Set up. Read parameters, solve $0.15\sin(\pi t) = 0.075$, then count rates carefully.
(i) Amplitude $0.15$ m, period $T = 2\pi/\pi = 2$ s, $x(0) = 0$ m (centre, going right).
(ii) $0.15\sin(\pi t) = 0.075 \Rightarrow \sin(\pi t) = 0.5 \Rightarrow \pi t = \pi/6 \Rightarrow t = $ $1/6$ s.
(iii) Claim is incorrect. Per period of $2$ s, the line $x = 0.075$ crosses the sine wave 2 times (once rising at $t = 1/6$ s, once falling at $t = 5/6$ s — by symmetry). So the rate is $2$ crossings per $2$ s $= $ $1$ crossing per second on average, not $2$. The student likely confused "crossings per period" with "crossings per second".