Modelling with Sinusoidal Functions

1. Find the maximum and minimum values described (or read off a table/graph).
2. Compute the axis, $c=\frac{\max +\min }{2}$, and the amplitude, $a=\frac{\max -\min }{2}$.
3. Find the period (how long for one full repeat), then compute $k=\frac{360°}{\text{period}}$.
4. Decide whether the situation starts at a maximum, a minimum, or the axis — that choice decides whether to build from cosine (starts at max/min naturally) or sine (starts at the axis naturally), and fixes the sign and value of $d$.

Why does each step work? The axis is the AVERAGE of the maximum and minimum because a sinusoid spends exactly as long above its midline as below it, and rises just as far above it as it falls below it; the midline sits halfway between the extremes. The amplitude is HALF the range, $\frac{\max -\min }{2}$, because the range covers the climb from the axis to the top plus the drop from the axis to the bottom, two equal amplitudes. For $k$, remember that the angle inside the cosine must sweep through $360°$ during one cycle of the real situation. If one cycle takes $40$ seconds, the angle has to grow by $\frac{360°}{40}=9°$ every second, and that rate is $k$. Finally $d$ is just a TIME: the moment the chosen parent function starts its cycle (a maximum for $\cos$, a minimum for $-\cos$, a rising axis crossing for $\sin$).

A Ferris wheel is the cleanest model because every parameter is a piece of the wheel. The centre height is the axis, since the rider is above the centre for half the turn and below it for the other half. The radius is the amplitude, since the rider is never more than one radius above or below the centre. The time for one rotation is the period, and the wheel turns through $\frac{360°}{\text{period}}$ each second, which is $k$. Where the rider boards decides the parent: boarding at the bottom means the height starts at its minimum, so $-\cos$ fits with $d=0$. For the wheel in the figure, $h(t)=-10\cos (9t)+12$. Read the graph against the wheel: a quarter-turn ($10$ s) puts the rider level with the centre at $12$ m, a half-turn ($20$ s) puts them at the top, $22$ m, and three quarters ($30$ s) brings them back to $12$ m on the way down. If the wheel's platform is raised, only $c$ changes; a bigger wheel changes $a$ (and usually $c$ too, because the centre must rise to keep the bottom off the ground); a faster wheel changes only $k$.

Ferris Wheel Trig Problem

Real data rarely arrives as a tidy description. Given a table, scan for the largest and smallest readings (these give $c$ and $|a|$), find how long it takes for the pattern to repeat (the period, then $k$), and note WHEN the largest reading occurs (that time is $d$ for a $+\cos$ model). The monthly average temperatures below peak at $22$ °C in month $7$ (July) and bottom out at $-6$ °C in month $1$ (January), repeating every $12$ months. So $c=\frac{22+(-6)}{2}=8$, $|a|=\frac{22-(-6)}{2}=14$, $k=\frac{360°}{12}=30$ and, using the July peak, $T(t)=14\cos (30(t-7))+8$. Using the January low instead gives $T(t)=-14\cos (30(t-1))+8$, the same curve. Real measurements wobble a little, so a model fitted from data is judged by how closely it tracks the table, not by hitting every value exactly.

Many exam questions run the model backwards: given a height or depth, find the TIME. On a graph, draw the horizontal line at that value and read the $t$-values where it cuts the curve; between the maximum and minimum, every value is reached TWICE per cycle, once rising and once falling. With the equation, isolate the trig ratio and use special angles and the CAST rule. For the Ferris wheel, when is the rider at $17$ m? Set $-10\cos (9t)+12=17$, so $-10\cos (9t)=5$ and $\cos (9t)=-0.5$. The related angle is $60°$, and cosine is negative in quadrants II and III, so $9t=120°$ or $9t=240°$, giving $t\approx 13.3$ s (rising) and $t\approx 26.7$ s (falling). Notice the two times sit symmetrically around $t=20$ s, the moment the rider is at the top; symmetry about a maximum or minimum is a quick check on any pair of answers.