Modelling with sine and cosine

Covers AS91575

The sinusoidal model

Many real quantities rise and fall periodically — tides, daylight hours, temperature, the height of a seat on a Ferris wheel.
These are modelled with a sine or cosine function:

$y = a\sin\!\big(b(t - c)\big) + d$

$a$ — the amplitude: half the distance from the lowest to the highest value.
$b$ — sets the period $T = \dfrac{2\pi}{b}$ (the time for one full cycle).
$c$ — the horizontal (phase) shift: how far the curve is shifted along the $t$ -axis.
$d$ — the vertical shift, the midline the curve oscillates about.

Building a model from a description

Midline: $d = \dfrac{\text{max} + \text{min}}{2}$ .
Amplitude: $a = \dfrac{\text{max} - \text{min}}{2}$ .
$b$ from the period: $b = \dfrac{2\pi}{T}$ .
$c$ from where a known feature (a maximum, or a rising midline crossing) happens. Choosing cosine lets you put a maximum at $t = 0$ , which often makes $c = 0$ .

Using the model

Predict a value by substituting a time $t$ .
Find when the quantity reaches a level by solving $y = \text{target}$ — a trig equation, so give all the times in the interval asked for.

Worked ExampleTide depth at a wharf

At a wharf the water is $2$ m deep at low tide and $6$ m deep at high tide. High tides are $12$ hours apart, and there is a high tide at $t = 0$ hours. Model the depth and find the depth $3$ hours after high tide.

Step 1 — Midline and amplitude

$d = \frac{6 + 2}{2} = 4, \qquad a = \frac{6 - 2}{2} = 2$

Step 2 — Find $b$ from the period

$T = 12 \;\Rightarrow\; b = \frac{2\pi}{12} = \frac{\pi}{6}$

Step 3 — Write the model

A maximum sits at $t = 0$ , so use cosine (no shift needed):

$y = 2\cos\!\left(\frac{\pi}{6}t\right) + 4$

Step 4 — Evaluate at $t = 3$

$y = 2\cos\!\left(\frac{\pi}{6}\cdot 3\right) + 4 = 2\cos\frac{\pi}{2} + 4 = 2(0) + 4$

$\boxed{\text{Depth} = 4 \text{ m}}$

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A model for temperature is $y = 5\sin\!\big(\tfrac{\pi}{12}t\big) + 18$ . State the amplitude, the midline, and the period.

Merit

In a NZ town the number of daylight hours varies between $9$ (mid-winter) and $15$ (mid-summer) over a $12$ -month cycle. Taking $t$ in months with a maximum at $t = 0$ , write a cosine model for the daylight hours.

Excellence

Using the tide model $y = 2\cos\!\big(\tfrac{\pi}{6}t\big) + 4$ (depth in metres, $t$ in hours from a high tide), find the first two times the depth is $5$ m.

Modelling with sine and cosine

Covers AS91575

The sinusoidal model

Many real quantities rise and fall periodically — tides, daylight hours, temperature, the height of a seat on a Ferris wheel.
These are modelled with a sine or cosine function:

$y = a\sin\!\big(b(t - c)\big) + d$

$a$ — the amplitude: half the distance from the lowest to the highest value.
$b$ — sets the period $T = \dfrac{2\pi}{b}$ (the time for one full cycle).
$c$ — the horizontal (phase) shift: how far the curve is shifted along the $t$ -axis.
$d$ — the vertical shift, the midline the curve oscillates about.

Building a model from a description

Midline: $d = \dfrac{\text{max} + \text{min}}{2}$ .
Amplitude: $a = \dfrac{\text{max} - \text{min}}{2}$ .
$b$ from the period: $b = \dfrac{2\pi}{T}$ .
$c$ from where a known feature (a maximum, or a rising midline crossing) happens. Choosing cosine lets you put a maximum at $t = 0$ , which often makes $c = 0$ .

Using the model

Predict a value by substituting a time $t$ .
Find when the quantity reaches a level by solving $y = \text{target}$ — a trig equation, so give all the times in the interval asked for.

Worked ExampleTide depth at a wharf

Step 1 — Midline and amplitude

$d = \frac{6 + 2}{2} = 4, \qquad a = \frac{6 - 2}{2} = 2$

Step 2 — Find $b$ from the period

$T = 12 \;\Rightarrow\; b = \frac{2\pi}{12} = \frac{\pi}{6}$

Step 3 — Write the model

A maximum sits at $t = 0$ , so use cosine (no shift needed):

$y = 2\cos\!\left(\frac{\pi}{6}t\right) + 4$

Step 4 — Evaluate at $t = 3$

$y = 2\cos\!\left(\frac{\pi}{6}\cdot 3\right) + 4 = 2\cos\frac{\pi}{2} + 4 = 2(0) + 4$

$\boxed{\text{Depth} = 4 \text{ m}}$

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A model for temperature is $y = 5\sin\!\big(\tfrac{\pi}{12}t\big) + 18$ . State the amplitude, the midline, and the period.

Merit

Excellence

Using the tide model $y = 2\cos\!\big(\tfrac{\pi}{6}t\big) + 4$ (depth in metres, $t$ in hours from a high tide), find the first two times the depth is $5$ m.

Modelling with sine and cosine

The sinusoidal model

Building a model from a description

Using the model

Step 1 — Midline and amplitude

Step 2 — Find bbb from the period

Step 3 — Write the model

Step 4 — Evaluate at t=3t = 3t=3

Test yourself

Practice by grade

Modelling with sine and cosine

The sinusoidal model

Building a model from a description

Using the model

Step 1 — Midline and amplitude

Step 2 — Find bbb from the period

Step 3 — Write the model

Step 4 — Evaluate at t=3t = 3t=3

Test yourself

Practice by grade

Step 2 — Find $b$ from the period

Step 4 — Evaluate at $t = 3$

Step 2 — Find $b$ from the period

Step 4 — Evaluate at $t = 3$