Periodic modelling | Year 11

01 / LEARN

Understand the idea.

Turn the context into graph features

For an ideal periodic height, extract the midline, amplitude, period and starting phase from the context. The midline is the average of maximum and minimum, while amplitude is half their difference. State the time and height units before fitting the rule.

Use angular frequency $ω = 2 π / T$ , with the same time unit as the input. Positive cosine suits a starting maximum and negative cosine a starting minimum. Verify the rule against another given observation and restrict it to a sensible modelling interval.

ω = \frac{2 π}{T}

Solve, interpret and evaluate

To solve a threshold question, isolate the trig ratio and account for every branch in a cycle. Equality gives boundary times; an inequality asks for the whole allowed time interval. Dividing an inequality by a negative coefficient reverses its direction.

A single sinusoid assumes regular repetition and stable extremes. Real tides can be affected by weather and additional cycles, while machinery may vary speed. Treat the result as a prediction under stated assumptions and compare it with further observations when available.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Example 1

An idealised tide is 0.5 m at $t = 0$ and 3.5 m at $t = 6$ hours. The period is 12 hours. Model its height.

Worked solution

1
Choose a cosine model with the observed heights as extremes.
$D = \frac{3.5 + 0.5}{2} = 2, A = \frac{3.5 - 0.5}{2} = 1.5$
This choice fits the observations half a cycle apart. The heights alone do not establish that it is the only possible sinusoid.
2
Calculate angular frequency.
$ω = \frac{2 π}{12} = \frac{π}{6} rad h^{- 1}$
The argument must advance by $2 π$ in twelve hours.
3
Choose a starting minimum.
$h (t) = 2 - 1.5 cos (\frac{π t}{6}) m$
Negative cosine starts one amplitude below the midline; t is measured in hours.
4
Verify the given heights.
$h (0) = 0.5 m, h (6) = 3.5 m$
At these times the cosine values are $1$ and $- 1$ , matching both observations.

Answer

$h (t) = 2 - 1.5 cos (π t / 6)$ m.

EXAMPLE 02

Example 2

For this tide, when is $h (t) \geq 2.75$ m in the first 12 hours?

Worked solution

1
Isolate the trigonometric inequality.
$\begin{aligned} 2 - 1.5 cos (\frac{π t}{6}) & \geq 2.75 \\ - 1.5 cos (\frac{π t}{6}) & \geq 0.75 \\ cos (\frac{π t}{6}) & \leq - \frac{1}{2} \end{aligned}$
Dividing by negative $1.5$ reverses the inequality.
2
Find the allowed argument interval in one cycle.
$\frac{2 π}{3} \leq \frac{π t}{6} \leq \frac{4 π}{3}$
Across $0 \leq t \leq 12$ , cosine is at most $- 1 / 2$ between these two unit-circle angles.
3
Convert the argument bounds to time.
$4 \leq t \leq 8 h$
Multiplying by positive $6 / π$ preserves both inequalities; equality is allowed, so both boundary times are included.

Answer

From 4 to 8 hours, inclusive.

03 / YOUR TURN

Try it yourself.

Check your understanding

A wheel turns once every 40 seconds. Find its angular frequency for a model using seconds.

Show answer & reasoning

$\frac{π}{20}$ radians per second.

One revolution covers $2 π$ radians in $40$ seconds. Therefore $ω = \frac{2 π}{40} = \frac{π}{20}$ radians per second.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

A model has maximum $28^{\circ}$ C and minimum $16^{\circ}$ C. Find amplitude and midline.
2 marks
A wheel has radius 8 m, axle 10 m above ground and period 40 s. It starts at the bottom. Give a height rule.
2 marks
For that wheel, find the height after 10 s.
2 marks
For that wheel, find every time at height 14 m on $[0, 40)$ .
2 marks
Why is $cos (\frac{2 π t}{12})$ inappropriate for a 365-day daylight cycle when $t$ is in days?
2 marks
Give two limitations of modelling a tide with a fixed sine wave.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

A model has maximum $28^{\circ}$ C and minimum $16^{\circ}$ C. Find amplitude and midline.
2 marks
Half-difference is 6; average is 22.
Amplitude $6^{\circ}$ C; midline $22^{\circ}$ C.
A wheel has radius 8 m, axle 10 m above ground and period 40 s. It starts at the bottom. Give a height rule.
2 marks
The midline is 10, amplitude 8 and angular frequency $\frac{π}{20}$ ; negative cosine begins at height 2.
$h (t) = 10 - 8 cos (\frac{π t}{20})$ m.
For that wheel, find the height after 10 s.
2 marks
The argument is $\frac{π}{2}$ , whose cosine is 0.
10 m.
For that wheel, find every time at height 14 m on $[0, 40)$ .
2 marks
Solve cosine $- \frac{1}{2}$ , giving arguments $\frac{2 π}{3}, \frac{4 π}{3}$ .
Multiply by $\frac{20}{π}$ .
$t = \frac{40}{3}, \frac{80}{3}$ s.
Why is $cos (\frac{2 π t}{12})$ inappropriate for a 365-day daylight cycle when $t$ is in days?
2 marks
A 365-day cycle needs angular frequency $\frac{2 π}{365}$ per day.
It repeats after 12 days.
Give two limitations of modelling a tide with a fixed sine wave.
2 marks
The fixed amplitude/period assumptions ignore those effects, so predictions must be checked.
Weather can change heights; real successive tidal cycles can have unequal highs or timing.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics, Further Trigonometry: periodic functions and models, trigonometric identities and reciprocal functions; the Library sequence follows the approved school-topic structure.

Official curriculum source ↗

In this lesson

Understand the idea.

Turn the context into graph features

Solve, interpret and evaluate

See the method in action.

Example 1

Choose a cosine model with the observed heights as extremes.

Calculate angular frequency.

Choose a starting minimum.

Verify the given heights.

Example 2

Isolate the trigonometric inequality.

Find the allowed argument interval in one cycle.

Convert the argument bounds to time.

Try it yourself.

Check your understanding

Keep practising