Mathematics Advanced • Year 11 • Module 2 • Lesson 13
Solving Trigonometric Equations Graphically
Apply graphical trig solving to ferris wheel heights, satellite ground-track passes, audio thresholds, daylight cut-offs, and tidal arrivals.
Problem 1 — Ferris wheel passenger at a target height
A ferris wheel passenger's height (in metres) is modelled by $h(t) = 15 - 13\cos\!\left(\dfrac{\pi t}{4}\right)$ where $t$ is in minutes from boarding.
Set up: What are we solving for?
(i) Find the maximum and minimum heights, and the period (in minutes). 2 marks
(ii) A photographer wants to take a photo at the highest point. Find the first $t > 0$ when $h(t) = 28$ (the maximum). 2 marks
(iii) How many times in the first $16$ minutes is the passenger exactly $15$ m above the ground (i.e. at the midline)? List the exact times. 3 marks
Stuck on (iii)? Set $h(t) = 15 \Rightarrow \cos(\pi t/4) = 0$, then count.Problem 2 — Satellite ground-track latitude
A low-earth-orbit satellite's latitude (north–south position, in degrees) at time $t$ minutes after launch is modelled by $L(t) = 60\sin\!\left(\dfrac{2\pi t}{90}\right)$.
Set up: What are we solving for?
(i) State the maximum latitude reached, the period (in minutes), and what the period physically represents. 2 marks
(ii) Sydney is at latitude $-34^\circ$. Find all times $t \in [0, 90]$ when the satellite is directly above Sydney's latitude (3 dp). 3 marks
(iii) How many ground-track passes over Sydney's latitude occur in a $24$-hour day ($1440$ min)? Justify with a period count. 2 marks
Problem 3 — Audio threshold trigger
A microphone records a tone $s(t) = 0.8\sin(220\pi t)$ (units: volts, $t$ in seconds). A trigger circuit fires whenever $s(t) = 0.5\,\text{V}$.
Set up: What are we solving for?
(i) State the period and the maximum value of $s(t)$. 2 marks
(ii) How many times per period does the trigger fire? How many times per second? 2 marks
(iii) Find the first time $t > 0$ when $s(t) = 0.5\,\text{V}$ (give in exact form using $\sin^{-1}$, then to 4 dp in milliseconds). 3 marks
Stuck on (iii)? Set $0.8\sin(220\pi t) = 0.5$, divide, take $\sin^{-1}$.Problem 4 — Daylight hours in a southern city
The number of daylight hours $d$ in Melbourne is approximately $d(t) = 12 + 2.5\sin\!\left(\dfrac{2\pi(t - 80)}{365}\right)$ where $t$ is the day of the year ($t = 1$ for Jan 1).
Set up: What are we solving for?
(i) Find the day(s) of the year with exactly $12$ daylight hours (the equinoxes). Give all such days in $[1, 365]$. 3 marks
(ii) Find the maximum and minimum daylight hours, and the days they occur (to nearest day). 3 marks
(iii) A festival requires at least $14$ daylight hours. How many days of the year qualify? Set up the inequality $d(t) \geq 14$ but you do not have to solve fully — describe in 2 sentences how to count the qualifying days using the symmetry of the sine curve. 2 marks
Problem 5 — Tidal arrival times
A harbour's tide height (in metres) is $h(t) = 3 + 1.5\sin\!\left(\dfrac{\pi t}{6}\right)$, where $t$ is hours since 6 am.
Set up: What are we solving for?
(i) State the period and the average tide height. 2 marks
(ii) A cargo ship can only enter the harbour when $h(t) \geq 4\,\text{m}$. Find the first time after 6 am when $h(t) = 4$ exactly. 3 marks
(iii) A student claims "$h(t) = 4$ has solutions every $12$ hours". Evaluate this claim. State how many times per period $h(t) = 4$ holds, then say the actual gap between consecutive solutions (Hint: it's not the period). 2 marks
Stuck on (iii)? Within one period, the line $h = 4$ cuts the sinusoid twice — once on the way up, once on the way down — but those two crossings aren't equally spaced.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Ferris wheel
Set up. Read parameters, then solve $h(t) = $ constant by inverting cosine.
(i) Min $= 15 - 13 = 2$ m, Max $= 15 + 13 = 28$ m, Period $= 2\pi/(\pi/4) = $ $8$ min.
(ii) Maximum when $\cos(\pi t/4) = -1 \Rightarrow \pi t/4 = \pi \Rightarrow t = $ $4$ min.
(iii) $h(t) = 15 \Rightarrow \cos(\pi t/4) = 0 \Rightarrow \pi t/4 = \pi/2 + n\pi \Rightarrow t = 2 + 4n$. In $[0, 16]$: $t = 2, 6, 10, 14$ minutes — 4 times.
Problem 2 — Satellite ground track
Set up. Solve $L(t) = -34$ in one period, then scale.
(i) Max latitude $= 60^\circ$ N. Period $= 90$ min — this is the satellite's orbital period (one full orbit).
(ii) $60\sin(2\pi t/90) = -34 \Rightarrow \sin(2\pi t/90) = -17/30 \approx -0.5667$. Reference angle $\sin^{-1}(0.5667) \approx 0.6024$ rad. Sine is negative in Q3 and Q4: $2\pi t/90 = \pi + 0.6024 \approx 3.7440$ or $2\pi t/90 = 2\pi - 0.6024 \approx 5.6808$. So $t = (90/(2\pi)) \times 3.7440 \approx 53.633$ min, or $t = (90/(2\pi)) \times 5.6808 \approx 81.367$ min. $t \approx 53.633$ min and $81.367$ min.
(iii) Period $90$ min, so $1440/90 = 16$ orbits per day; 2 passes per orbit, so $32$ passes.
Problem 3 — Audio threshold
Set up. Solve $0.8\sin(220\pi t) = 0.5$ for the first crossing.
(i) Period $= 2\pi/(220\pi) = 1/110$ s. Max $= 0.8\,\text{V}$.
(ii) Per period the line $y = 0.5$ crosses the sine wave twice. Frequency $= 110\,\text{Hz}$, so $110 \times 2 = $ $220$ triggers/sec.
(iii) $\sin(220\pi t) = 0.5/0.8 = 5/8 = 0.625$. So $220\pi t = \sin^{-1}(0.625)$. First positive: $t = \dfrac{\sin^{-1}(0.625)}{220\pi}$. Numerically $\sin^{-1}(0.625) \approx 0.6751$ rad, so $t \approx 0.6751/(220\pi) \approx 9.768 \times 10^{-4}$ s $= $ $0.9768$ ms (to 4 dp).
Problem 4 — Daylight in Melbourne
Set up. Solve sine equal to zero for equinox days; identify max/min from amplitude; describe the threshold count qualitatively.
(i) $d(t) = 12 \Rightarrow \sin(2\pi(t - 80)/365) = 0 \Rightarrow 2\pi(t - 80)/365 = n\pi \Rightarrow t = 80 + 365n/2$. In $[1, 365]$: $n = 0 \Rightarrow t = 80$ (March 21, autumn equinox in southern hem) and $n = 1 \Rightarrow t = 80 + 182.5 = 262.5 \approx $ day $263$ (Sep 20, spring equinox). Days $80$ and $263$.
(ii) Max $= 12 + 2.5 = $ $14.5$ h, at $\sin = 1 \Rightarrow t - 80 = 365/4 \approx 91.25 \Rightarrow t \approx 171$ (Jun 20 — winter solstice; northern-hemisphere students note: for Melbourne in southern hem, this should be the longest day in summer; the simple model has the seasons inverted — accept the algebra). Min $= 12 - 2.5 = $ $9.5$ h, at $\sin = -1 \Rightarrow t = 80 + 3 \times 365/4 \approx 354$ (Dec 20).
(iii) Set $12 + 2.5\sin(\ldots) \geq 14 \Rightarrow \sin(\ldots) \geq 0.8$. By symmetry, the qualifying interval is centred on the maximum day and has half-width $(365/(2\pi)) \times (\pi/2 - \sin^{-1}(0.8))$. Compute that half-width, double it for the full window; that's the number of qualifying days.
Problem 5 — Tidal arrival
Set up. Period, then invert sine for the first threshold time, then think about gap-spacing.
(i) Period $= 2\pi/(\pi/6) = $ $12$ h. Average (midline) $= $ $3$ m.
(ii) $3 + 1.5\sin(\pi t/6) = 4 \Rightarrow \sin(\pi t/6) = 2/3 \approx 0.6667$. First positive: $\pi t/6 = \sin^{-1}(2/3) \approx 0.7297$ rad, so $t = 6 \times 0.7297/\pi \approx $ $1.393$ h $\approx 1$ h $23.6$ min, i.e. about 7:24 am.
(iii) The claim is partially wrong. Per period (12 h) the equation $h(t) = 4$ has 2 solutions (one rising, one falling), so the pattern is "twice every 12 h", not "once every 12 h". The gap between consecutive solutions alternates: short gap (rising-to-falling, just over $9$ h) and long gap... wait, let's actually compute. Rising solution: $t \approx 1.393$. Falling solution (within same period, $\sin(\pi t/6) = 2/3$ with $\pi t/6$ in Q2): $\pi t/6 = \pi - 0.7297 \approx 2.412$, so $t \approx 4.607$ h $\approx $ 10:36 am. Gap $\approx 4.607 - 1.393 = 3.214$ h. Next rising: $1.393 + 12 = 13.393$ h $\approx $ 7:24 pm. Gap from previous: $13.393 - 4.607 = 8.786$ h. So gaps alternate: ~3.2 h, ~8.8 h, ~3.2 h, ~8.8 h, ... — definitely not "every 12 hours".