Harbour tide models: harmonics, general solutions and timetable risk
The harbour master at a small port models the water depth over the bar as $d_1(t) = 3.2 + 2.4\sin!\left(\dfrac{\pi t}{6}\right)$ metres, where $t$ is hours after midnight and the tide repeats every 12 hours. The vehicle ferry needs at least 4.5 m of water to cross safely. The printed timetable has the first crossing departing at 01:00 and the last return leaving at 04:30. A new survey shows a problem with $d_1$: the depth logged at midnight was 4.8 m, but $d_1(0) = 3.2$ m. The harbour master therefore proposes $d_2(t) = 3.2 + 1.8\sin!\left(\dfrac{\pi t}{6}\right) + 1.6\cos!\left(\dfrac{\pi t}{6}\right)$, which reproduces the midnight reading. Before reprinting the timetable she wants to know how much the two models disagree about the safe windows, and whether the 04:30 return can still be advertised.
1. Using $d_1$, find the times (to the nearest minute) at which the depth first reaches 4.5 m and next falls back to 4.5 m, and state the length of the safe window.
Read the worked answer
Reduce to a sine equation. $3.2 + 2.4\sin!\left(\dfrac{\pi t}{6}\right) = 4.5$ gives $\sin!\left(\dfrac{\pi t}{6}\right) = \dfrac{1.3}{2.4} = 0.54167$.
General solution. With $\alpha = \arcsin(0.54167) = 0.5723$ rad,
$$\frac{\pi t}{6} = 0.5723 + 2k\pi \quad\text{or}\quad \frac{\pi t}{6} = \pi - 0.5723 + 2k\pi = 2.5693 + 2k\pi .$$
First two solutions in $[0, 24)$. Solving each for $t = \dfrac{6}{\pi}(\text{angle})$:
$$t = \frac{6 \times 0.5723}{\pi} = 1.093\ \text{h} \approx 01{:}06, \qquad t = \frac{6 \times 2.5693}{\pi} = 4.907\ \text{h} \approx 04{:}54 .$$
Adding the 12-hour period gives the repeats at 13:06 and 16:54.
Window. $d_1$ is above 4.5 m between 01:06 and 04:54, a window of $4.907 - 1.093 = 3.814\ \text{h} \approx 3\ \text{h}\ 49\ \text{min}$. Displacement does not affect this length, so every window on this model is about 3 h 49 min long.
2. Show that $d_2(t) = 3.2 + R\sin\!\left(\dfrac{\pi t}{6} + \alpha\right)$ for suitable constants $R$ and $\alpha$, giving $R$ (3 s.f.) and $\alpha$ (3 s.f., radians). Hence state the maximum depth predicted by the new model.
Read the worked answer
Expand the target form. $R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha$, so with $\theta = \dfrac{\pi t}{6}$ we need
$$R\cos\alpha = 1.8, \qquad R\sin\alpha = 1.6 .$$
Amplitude. $R = \sqrt{1.8^2 + 1.6^2} = \sqrt{3.24 + 2.56} = \sqrt{5.80} = 2.4083 \approx 2.41$.
Phase. $\tan\alpha = \dfrac{1.6}{1.8} = 0.8889$, and both components are positive so $\alpha$ is in the first quadrant: $\alpha = 0.7266 \approx 0.727$ rad.
So $d_2(t) = 3.2 + 2.41\sin!\left(\dfrac{\pi t}{6} + 0.727\right)$.
Check. At $t = 0$: $3.2 + 2.4083\sin(0.7266) = 3.2 + 1.600 = 4.80$ m, matching the logged midnight depth - this is exactly what the phase shift was for.
Maximum depth. The sine term peaks at 1, so the maximum depth is $3.2 + 2.408 = 5.61$ m, occurring when $\dfrac{\pi t}{6} + 0.7266 = \dfrac{\pi}{2}$, i.e. $t = 2.61$ h, about 02:37. A single harmonic term is more informative than two separate sine and cosine terms because the amplitude, phase and maximum all read off directly.
3. Using $d_2$, find the first time after midnight at which the depth falls below 4.5 m and the next time it rises back to 4.5 m. Use this to advise whether the 04:30 return departure can be advertised.
Read the worked answer
Set up. $3.2 + 2.4083\sin(X) = 4.5$ with $X = \dfrac{\pi t}{6} + 0.7266$ gives $\sin X = 0.53980$, so $\beta = \arcsin(0.53980) = 0.5701$ rad and
$$X = 0.5701 + 2k\pi \quad\text{or}\quad X = \pi - 0.5701 + 2k\pi = 2.5715 + 2k\pi .$$
Locate midnight. At $t = 0$, $X = 0.7266$, which lies between $0.5701$ and $2.5715$; the depth is therefore already above 4.5 m at midnight ($d_2(0) = 4.8$). So the first crossing after midnight is the downward one, at $X = 2.5715$:
$$t = \frac{6(2.5715 - 0.7266)}{\pi} = \frac{6 \times 1.8449}{\pi} = 3.523\ \text{h} \approx 03{:}31 .$$
Next crossing. $X = 0.5701 + 2\pi = 6.8533$, giving $t = \dfrac{6(6.8533 - 0.7266)}{\pi} = 11.70\ \text{h} \approx 11{:}42$.
Advice. Under $d_2$ the bar is passable until about 03:31, so a 04:30 departure would cross in about 3.3 m of water (check: $d_2(4.5) = 3.2 + 2.4083\sin(3.083) = 3.34$ m) and must be withdrawn. $d_1$ promised safety until 04:54 - a 1 h 23 min overrun that could ground the ferry. The midnight reading rules out $d_1$ (it is out by 1.6 m at $t=0$), so the timetable should be rebuilt on $d_2$ with a safety margin, not on the old model.