Tides, hours of daylight, the height of a Ferris wheel seat: each rises and falls on a regular schedule. A sinusoid with a shift and a midline can model all of them. The work is translating the story into four numbers.
Key idea
From the story, find: the midline $k = \frac{\max + \min}{2}$; the amplitude $|A| = \frac{\max - \min}{2}$; the period, and from it $B = \frac{2\pi}{\text{period}}$; and a starting point. Use $\cos$ if a maximum happens at the shift, $-\cos$ if a minimum does, and $\sin$ if the quantity crosses the midline going up.
Why it works
The midline is the average of the highest and lowest values, halfway between them. The amplitude is the distance from the midline to either extreme, which is half the full swing. Cosine is convenient because $\cos 0 = 1$ puts a peak right at the shift.
Worked examples
Worked example: A Ferris wheel
A wheel $40$ m across has its lowest seat $3$ m above the ground and turns once every $12$ minutes. You board at the bottom. Model your height, then find it after $4$ minutes.
- Center: $3 + 20 = 23$ m (midline). Radius $20$ m (amplitude). Period $12$, so $B = \frac{2\pi}{12} = \frac{\pi}{6}$.
- You start at the minimum, so use $-\cos$: $h(t) = 23 - 20\cos\left(\frac{\pi}{6}t\right)$.
- At $t = 4$: $\cos\frac{2\pi}{3} = -\frac{1}{2}$, so $h = 23 + 10 = 33$ m.
Answer: $h(t) = 23 - 20\cos\left(\frac{\pi}{6}t\right)$; $33$ m after $4$ minutes
Worked example: Tides
High tide is $9$ ft at $2$ a.m. and the next low tide is $3$ ft at $8$ a.m. Write a model for the depth $t$ hours after midnight.
- Midline $\frac{9 + 3}{2} = 6$; amplitude $\frac{9 - 3}{2} = 3$.
- High to low is half a cycle, $6$ hours, so the period is $12$ and $B = \frac{\pi}{6}$.
- A maximum at $t = 2$ suggests cosine shifted right $2$.
Answer: $d(t) = 6 + 3\cos\left(\frac{\pi}{6}(t - 2)\right)$
Worked example: Daylight
A city has $15$ hours of daylight on its longest day (day $172$ of the year) and $9$ hours on its shortest. Write a model with a period of $365$ days.
- Midline $12$, amplitude $3$, $B = \frac{2\pi}{365}$.
- The maximum is on day $172$, so shift a cosine right $172$.
Answer: $$D(t) = 12 + 3\cos\left(\frac{2\pi}{365}(t - 172)\right)$$
Common mistake
Starting with the wrong function. If you use $\sin$ for a rider who boards at the bottom, the model says they start halfway up. Always check $t = 0$ (or the given time) in your model against the story.
Different correct models exist. A sine with a different shift can describe the same wave; what matters is that the graphs agree.