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".