Maths › Trigonometry › Trigonometric modelling

Trigonometric modelling

Anything that turns, swings or breathes in cycles is trigonometry with its sleeves rolled up: wheels, tides, daylight, alternating current. This lesson reads real periodic models, centre line first, amplitude second, period third, and lets the harmonic form tame the messy ones with two trig terms.

Year 12-13EDEXCEL 9MA0 5.9

Builds on Compound angles and the harmonic form and Functions in modelling.

IN THIS TOPIC

  • Build and read wheel-style models of the form k − R cos ωt, in radians.
  • Extract centre, amplitude, period and phase from any periodic model.
  • Use the harmonic form to analyse models with two trig terms.

WHAT YOU PROBABLY THINK

A bigger amplitude means a faster oscillation.

A seat on a wheel

A point on a turning wheel is the cleanest periodic model there is. With the axle at height k, radius R and angular speed ω radians per second, a seat starting at the bottom has height h = k − R cos ωt. The cosine starts at 1, so the model starts at its minimum, exactly where a boarding platform wants it.

A big wheel of radius 10 metres with its axle 12 metres up, beside the height-time graph of one seat: h equals 12 minus 10 cos of pi t over 15, starting at 2 metres, peaking at 22, and repeating every 30 secondsradius 10 m, axle at 12 mseat starts at 2 mh = 12 − 10 cos (πt/15)top: 22 mone turn every 30 s
FIG. 1Axle 12 m up, radius 10 m: h = 12 − 10 cos (πt/15). Boarding at 2 m, top of the turn at 22 m, one revolution every 30 s.

WORKED EXAMPLE

Reading the wheel

A wheel's seat height is modelled by h = 12 − 10 cos (πt/15), in metres and seconds. Find the period, the greatest and least heights, and the first time the seat is 17 m up.

The period is 2π ÷ (π/15) = 30 s, and h runs from 12 − 10 = 2 m to 12 + 10 = 22 m.

Setting h = 17: cos (πt/15) = −½, so πt/15 = 2π/3 and t = 10 s.

Centre line from the axle, amplitude from the radius, period from the coefficient: every number in the formula is a physical fact about the wheel, and questions test each one.

The opening lie muddles the two dials a periodic model owns. Amplitude sets how far the oscillation swings, the period how often it repeats, and the wheel shows them living in different parts of the formula: R in front of the cosine, ω inside it.

Two terms, one tide

Real data often arrives as a sine term plus a cosine term, and last lesson's harmonic form is the tool that makes such models readable, collapsing them to a single wave with visible amplitude and phase.

A water depth model 6 plus 2 sine of 0.5 t plus 1.5 cos of 0.5 t: harmonic form collapses it to amplitude 2.5 about depth 6, so the depth runs between 3.5 and 8.5 metres, first peaking near t equal to 1.85high water 8.5 m at t = 1.85mean 6 mlow water 3.5 m
FIG. 2d = 6 + 2 sin (0.5t) + 1.5 cos (0.5t): harmonic form gives amplitude 2.5 about the mean depth 6, so high water is 8.5 m, first at t = 1.85.

YOUR TURN

Collapse, then read

Water depth is modelled by d = 6 + 2 sin (0.5t) + 1.5 cos (0.5t), in metres and hours. Find the greatest and least depths, and the first time of high water, before opening the working.

Show the working

The trig part folds: R = √(4 + 2.25) = 2.5 and tan α = 1.5/2 gives α = 0.644, so d = 6 + 2.5 sin (0.5t + 0.644).

Depth runs from 6 − 2.5 = 3.5 m to 6 + 2.5 = 8.5 m.

High water needs 0.5t + 0.644 = π/2, so t = 1.85 hours, and every 4π hours after.

The constant 6 never joined the fold; harmonic form applies to the oscillating part alone, about whatever centre line the model carries.

TRY IT UNSEEN

Criticise the daylight model

Hours of daylight are modelled by D = 12 + 4.5 sin (2πt/365), with t in days after the spring equinox. State the period and the longest day, and give one reason the model needs refining for a real town.

Show the working

The period is 2π ÷ (2π/365) = 365 days, and the longest day is 12 + 4.5 = 16.5 hours, a quarter of a period in.

One honest criticism: the amplitude depends on latitude, so a single fixed 4.5 cannot serve both Penzance and Aberdeen; the refinement is fitting the amplitude to local daylight data.

Model-criticism marks here mirror the modelling lesson's: name the assumption, say where it fails, propose the bounded fix.

THE EXAM BIT

  • Read models centre first, amplitude second, period third; each is one mark and they never move house within the formula.
  • Period is 2π over the coefficient of t, in radians; check the mode before anything else.
  • Solve height and depth equations with the widen-substitute-translate routine, keeping exact multiples of π while you can.
  • Two trig terms mean harmonic form; fold the oscillating part and leave the centre line alone.
  • “First time” means the smallest positive solution; state the later ones with the period if asked.

CHECK YOURSELF

A buoy's height above mean sea level is modelled by y = 1.2 sin (0.8t), in metres and seconds. Find the amplitude, the period, and the first time the buoy is 0.6 m above mean level.

Show a hint

sin = ½ has a familiar exact angle.

Show the answer

Amplitude 1.2 m; period 2π/0.8 = 7.85 s.

0.6 m needs sin (0.8t) = ½, so 0.8t = π/6 and t = 0.654 s.

The next crossing follows at 0.8t = 5π/6; periodic models always carry their whole solution family behind the first answer.

Centre line, amplitude, period: three dials, three separate homes in the formula.

Fold two-term models with the harmonic form, then read the extremes straight off R.

CHECK YOUR PROGRESS

Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device only.

  • Build and read wheel-style models of the form k − R cos ωt, in radians.
  • Extract centre, amplitude, period and phase from any periodic model.
  • Use the harmonic form to analyse models with two trig terms.

No animated video for this topic yet; these notes stand alone.