Math 30-1 · Sinusoidal models
Waves in the real world
Anything that repeats smoothly — a Ferris wheel, the tide, hours of daylight — can be modelled with sine or cosine. The whole skill is turning a description into four numbers: a, b, c and d.
Reading the parameters
The idea: in y = a sin[b(x − c)] + d or y = a cos[b(x − c)] + d, you can read everything from the maximum, the minimum and the length of one cycle.
| Parameter | From a graph or description |
|---|---|
| Amplitude |a| | (max − min) ÷ 2 |
| Midline d | (max + min) ÷ 2 |
| Period | the time or distance for one full cycle; then b = 2π ÷ period |
| Horizontal shift c | where the cycle starts — factor b out first |
Max 9 and min 1 give amplitude 4 and midline y = 5. A period of 12 gives b = 2π/12 = π/6.
Watch out: in y = 2 sin(3x − π) + 1 the shift is not π. Factor it: 2 sin[3(x − π/3)] + 1, so the shift is π/3 right.
Writing an equation
The idea: pick the function that starts where your situation starts. Cosine starts at a maximum; negative cosine starts at a minimum; sine starts at the midline going up.
Worked example
A graph has a maximum at (2, 7) and the next minimum at (8, 1). Write an equation.
- Amplitude: (7 − 1) ÷ 2 = 3. Midline: (7 + 1) ÷ 2 = 4.
- Max to min is half a cycle: 8 − 2 = 6, so the period is 12 and b = 2π/12 = π/6.
- There is a maximum at x = 2, and cosine starts at a maximum, so use cosine shifted 2 right.
y = 3 cos[π/6 (x − 2)] + 4
Sine and cosine are the same wave
Sine is just cosine shifted right by a quarter period: sin x = cos(x − π/2). So there are always several correct equations for the same graph — any one that matches is right.
Ferris wheels, tides and daylight
The idea: each real situation maps onto the parameters the same way every time.
| Situation | Amplitude | Midline | Period |
|---|---|---|---|
| Ferris wheel | radius | height of the centre | time for one rotation |
| Tides | half the gap between high and low tide | average water level | time from high tide to high tide |
| Daylight hours | half the gap between longest and shortest day | average day length | 365 days |
A wheel with radius 15 m, centre 17 m, 60-second rotation and boarding at the bottom: a = 15, d = 17, b = 2π/60 = π/30, and a negative cosine to start at the minimum. h(t) = −15 cos(πt/30) + 17.
Watch out: using the diameter as the amplitude. The rider only goes one radius above and below the centre.
Using a model
The idea: once you have the equation, questions ask you to plug in a time, or to find the times that give a certain value.
Plugging in
For h(t) = −10 cos(πt/20) + 12 at t = 0: cos 0 = 1, so h = −10 + 12 = 2 m.
Finding when
When is the rider at the top? The top means −10 cos(πt/20) is as big as possible, so cos(πt/20) = −1. That first happens when πt/20 = π, so t = 20 s.
When is the tide h(t) = 2 sin(πt/6) + 5 exactly at 5 m, for 0 ≤ t < 12? That is the midline, so sin(πt/6) = 0: t = 0 and t = 6 hours.
Using a graphing calculator
For values that are not at a maximum, minimum or midline, graph the model and the horizontal line y = value, then find the intersections. Always set the window to match the context — time cannot be negative.
Watch out: giving only the first answer. A repeating model usually hits the same height more than once per cycle — check the whole interval the question asks about.