Mathematics Advanced • Year 11 • Module 2 • Lesson 14
Modelling with Trigonometric Functions
Practise HSC-style writing on building models, using them to predict, and evaluating their limitations against real data.
1. Short-answer questions
1.1 A daylight model gives $D(t) = 13 + 2\sin\!\left(\dfrac{2\pi t}{12}\right)$ where $D$ is hours of daylight and $t$ is months from March 21. State: amplitude, period (in months), midline, maximum daylight, and the month when maximum occurs. 3 marks Band 3
1.2 The temperature in a city varies between $5^\circ$C and $25^\circ$C with annual period; the minimum occurs in July ($t = 6$ months after January). Using $T(t) = a\cos(b(t - c)) + d$:
(a) Find $a$, $b$, $c$, $d$.
(b) Write $T(t)$.
(c) Find the temperature in October ($t = 9$ months after January), to 1 dp. 4 marks Band 3-4
1.3 A ferris wheel of radius $10\,\text{m}$ has its centre $12\,\text{m}$ above the ground; one revolution takes $4\,\text{min}$. A rider boards at the bottom at $t = 0$. (a) Write a model for the rider's height $h(t)$ in metres. (b) Find the first time $t > 0$ when $h(t) = 17\,\text{m}$ (the average viewing height, which is half-way between centre and max). 4 marks Band 4
Stuck on 1.3? Min height = 2, max = 22, midline = 12. Use $-\cos$ for "starts at min".2. Extended response
2.1 A wind turbine's blade tip traces a vertical circle. At time $t$ seconds the height (in metres) of one labelled blade tip is $h(t) = 50 + 25\sin\!\left(\dfrac{\pi t}{3}\right)$.
(a) State the amplitude, period, midline, maximum and minimum heights, and explain in one sentence what each of these means physically for the turbine.
(b) Find the first time $t > 0$ when the blade tip is at height $60\,\text{m}$. Show your working — set up the equation, rearrange, and apply $\sin^{-1}$.
(c) A wind technician claims that at $t = 3\,\text{s}$ the blade tip is moving fastest. Evaluate this claim by considering the geometry of a sinusoid (where does it have maximum slope?), and state at which $t$-value(s) in the first period the blade tip is actually moving slowest. State one limitation of the model (real turbines speed up and slow down due to wind variability). 8 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — amplitude $25$, period $6$ s, midline $50$ m, max $75$ m, min $25$ m.
• 1 mark — physical meanings (amplitude = blade radius; period = time per revolution; midline = hub height; etc.).
Part (b) — 3 marks
• 1 mark — sets up $50 + 25\sin(\pi t/3) = 60 \Rightarrow \sin(\pi t/3) = 2/5$.
• 1 mark — first positive: $\pi t/3 = \sin^{-1}(2/5) \approx 0.4115$ rad.
• 1 mark — solves: $t = 3 \times 0.4115/\pi \approx 0.393$ s.
Part (c) — 3 marks
• 1 mark — Sinusoids have maximum slope at the midline crossings, not at the maximum. At $t = 3$, $\sin(\pi) = 0$ — the blade is at the midline going down (slope $-25\pi/3 \approx -26.18\,\text{m/s}$, the maximum absolute value of $h'$). So the student is actually correct in magnitude but only by coincidence.
• 1 mark — slowest at the maximum and minimum, i.e. $t = 1.5$ s and $t = 4.5$ s (where slope is $0$). Note that the student's claim is right but for the wrong reason.
• 1 mark — limitation: real wind turbines have variable speed due to wind gusts, blade pitch control, and inertia.
Your response:
Stuck on (c)? The derivative of $\sin(\pi t/3)$ is $(\pi/3)\cos(\pi t/3)$. Maximum magnitude when $|\cos| = 1$, minimum when $\cos = 0$.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Daylight model parameters (3 marks)
Sample response. Amplitude $= 2$ h, period $= 2\pi/(2\pi/12) = 12$ months, midline $= 13$ h, max daylight $= 13 + 2 = 15$ h. Max occurs when $\sin(2\pi t/12) = 1 \Rightarrow t = 3$ months after March 21, i.e. around June 21.
Marking notes. 1 mark for amplitude + midline. 1 mark for period. 1 mark for max value + month identification.
1.2 — Temperature with phase shift (4 marks)
Sample response.
(a) $a = (25 - 5)/2 = 10$, $d = (25 + 5)/2 = 15$, $b = 2\pi/12 = \pi/6$. Min at $t = 6$: for $-a\cos(b(t - c))$, the min is at $t = c$ when using $+\cos$ with $a$ negative... easier: use $a\cos(b(t - c))$ with $a$ negative and $c = 0$, so min is at $t = c$. With $c = 6$ and $a = -10$: $T(t) = -10\cos(\pi(t-6)/6) + 15$. Equivalently $T(t) = 10\cos(\pi t /6) + 15$ shifted... cleaner: $T(t) = 15 - 10\cos(\pi(t - 6)/6)$ [2].
(b) Same expression above [1].
(c) $T(9) = 15 - 10\cos(\pi(9-6)/6) = 15 - 10\cos(\pi/2) = 15 - 0 = $ $15.0^\circ$C (midline — October is half-way between min-July and max-January) [1].
Marking notes. 2 marks for parameters, 1 for the equation, 1 for the substitution and answer. Common error: students use $+\cos$ without realizing the $-c$ in the bracket needs $c$ set to where the max (not min) is.
1.3 — Ferris wheel (4 marks)
Sample response. (a) Min $= 12 - 10 = 2$, max $= 12 + 10 = 22$, $d = 12$, $a = 10$, $b = 2\pi/4 = \pi/2$. Starts at min, so $-\cos$: $h(t) = 12 - 10\cos(\pi t/2)$ [2].
(b) $12 - 10\cos(\pi t/2) = 17 \Rightarrow -10\cos(\pi t/2) = 5 \Rightarrow \cos(\pi t/2) = -1/2$. First positive: $\pi t/2 = 2\pi/3 \Rightarrow t = $ $4/3$ min ($\approx 1$ min $20$ s) [2].
Marking notes. (a) 1 for parameters, 1 for the correct $-\cos$ form. (b) 1 for setting up the cosine equation, 1 for the correct exact $t$.
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a) — Parameters. Amplitude $= 25$ m (the blade-tip radius), period $= 2\pi/(\pi/3) = $ $6$ s (one revolution every $6$ s), midline $= 50$ m (the height of the hub), max $= 75$ m (tip at top of circle), min $= 25$ m (tip at bottom). [1 mark — values] Physical meaning: amplitude is the blade radius, period is the rotational period, midline is the hub height, max/min are top/bottom of the swept circle. [1 mark — interpretation]
Part (b) — First time at $60$ m. Set $50 + 25\sin(\pi t/3) = 60 \Rightarrow \sin(\pi t/3) = 10/25 = 2/5$. [1 mark]
First positive solution: $\pi t/3 = \sin^{-1}(2/5) \approx 0.4115$ rad. [1 mark]
Therefore $t = (3/\pi) \times 0.4115 \approx $ $0.393$ s. [1 mark]
Part (c) — Speed analysis. The rate of change of height is $h'(t) = 25 \times (\pi/3)\cos(\pi t/3) = (25\pi/3)\cos(\pi t/3)$. Maximum speed occurs when $|\cos(\pi t/3)| = 1$, i.e. when $\pi t/3 = 0, \pi, 2\pi, \ldots$ — at $t = 0, 3, 6, \ldots$ s. So the student's claim that $t = 3$ s is a fastest-moving moment is correct, but the physical reason is not "because that's when the tip is at the maximum height" — at $t = 3$ s, $h(3) = 50 + 25\sin(\pi) = 50$ m, i.e. the tip is on the midline, going down. Maximum vertical speed occurs at midline crossings, not at the top/bottom. [1 mark — correctly diagnosing the student's reasoning vs the actual answer]
Slowest vertical speed: $\cos(\pi t/3) = 0 \Rightarrow \pi t/3 = \pi/2$ or $3\pi/2$ $\Rightarrow$ $t = 1.5$ s and $t = 4.5$ s — exactly at max ($t = 1.5$) and min ($t = 4.5$). [1 mark — minima]
Limitation: real wind turbines have variable rotational speed because wind gusts, blade-pitch control, and rotor inertia all modulate the rotation. A constant-period sinusoid cannot capture this — a more accurate model would use a time-varying $b(t)$. [1 mark — limitation]
Total: 8/8.
Band descriptors for marker.
Band 3: Parameter values in (a) without physical interpretation; sets up (b) but uses an incorrect quadrant for $\sin^{-1}$; agrees with the student in (c) without analysis. ≈ 3-4 marks.
Band 4: All of (a) and (b) correct; student's claim in (c) is partly evaluated but reasoning is muddled. ≈ 5-6 marks.
Band 5: Full analysis in (c) including correct slowest-times; explicit derivative argument. ≈ 7 marks.
Band 6: All of the above plus the nuanced point that the student's claim is "right but for the wrong reason" and an explicit limitation referencing variable speed / pitch control. 8/8.