Sinusoidal Modelling

This is where everything in the unit comes together. Periodic situations in the real world, like a Ferris wheel, the tides, the seasons, or the length of the day, can be described by sinusoidal equations. With a model, you can predict values and answer questions like “when will the water be deep enough?” All angles are in degrees.

Key ideas

Building a model

Use the same four steps as for equations from graphs, with the real quantities:

1. Amplitude $a = \dfrac{\text{max} - \text{min}}{2}$
2. Axis $c = \dfrac{\text{max} + \text{min}}{2}$
3. $k = \dfrac{360}{\text{period}}$, with the period measured in the units of the input (seconds, hours, days, months). Then $k$ is “degrees per unit of time”.
4. Phase shift $d$: the time of a maximum (cosine) or of an upward crossing of the axis (sine). Use a negative $a$ with cosine when the situation starts at a minimum, like a rider boarding a Ferris wheel at the bottom.

Using a model

• Predict a value: substitute the time.
• Find when a value occurs: set the equation equal to the value, isolate the cosine (or sine), and find the angles. You can also read it from a graph or use a graphing calculator.
• Interpret the parameters in context (for example, the axis is the height of the wheel’s centre).

Real data

Real measurements won’t fit a curve perfectly. Plot the data, estimate the maximum, minimum, and period from it, and build a curve that fits as closely as you can. Models only make sense for realistic inputs: time usually starts at $0$.

Worked examples

Example 1: A Ferris wheel

A Ferris wheel has a diameter of $20$ m, its centre is $12$ m above the ground, and it turns once every $40$ s. A rider boards at the lowest point at $t = 0$.

(a) Write a model for the rider’s height $h$ (in metres) after $t$ seconds.

(b) Find the height after $15$ s.

(c) When, during the first turn, is the rider $17$ m above the ground?

Solution.

(a) $a = 10$ (the radius), $c = 12$ (the centre), and $k = \tfrac{360}{40} = 9$. The rider starts at the minimum, so use a negative cosine:

$$h(t) = -10\cos(9t) + 12$$

(b)

$$h(15) = -10\cos 135^\circ + 12 \approx -10(-0.7071) + 12 \approx 19.07$$

About $19.07$ m.

(c) Solve $-10\cos(9t) + 12 = 17$:

$$-10\cos(9t) = 5 \quad\Rightarrow\quad \cos(9t) = -0.5$$

In the first turn, $9t$ goes from $0^\circ$ to $360^\circ$. $\cos\theta = -0.5$ at $\theta = 120^\circ$ and $240^\circ$, so $9t = 120$ or $9t = 240$:

$$t \approx 13.3 \text{ s} \qquad \text{or} \qquad t \approx 26.7 \text{ s}$$

The rider is at $17$ m on the way up at about $13.3$ s, and on the way down at about $26.7$ s.

Example 2: Tides

At a harbour, high tide is $8.4$ m at 3:00 a.m., and the next low tide, $1.6$ m, is $6.2$ hours later. Write a model for the water depth $t$ hours after midnight, and estimate the depth at noon.

Solution. $a = \tfrac{8.4 - 1.6}{2} = 3.4$ and $c = \tfrac{8.4 + 1.6}{2} = 5$. High to low is half a period, so the period is $12.4$ h and $k = \tfrac{360}{12.4} \approx 29.03$. The maximum is at $t = 3$:

$$D(t) = 3.4\cos\big(29.03(t - 3)\big) + 5$$

At noon, $t = 12$. Using $k = \tfrac{360}{12.4}$ exactly on the calculator:

$$D(12) = 3.4\cos\left(\frac{360}{12.4} \times 9\right) + 5 \approx 4.49$$

The depth at noon is about $4.5$ m.

Example 3: Monthly temperatures

The average monthly temperatures in a Canadian city are shown below (month $1$ is January).

Month	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
Temp. (°C)	$-8$	$-6$	$-1$	$6$	$13$	$18$	$20$	$18$	$13$	$6$	$-1$	$-6$

Write a sinusoidal model and use it to describe the temperature in April.

Solution. Maximum $20$ (July), minimum $-8$ (January). So $a = 14$, $c = 6$, and the period is $12$ months, giving $k = \tfrac{360}{12} = 30$. The maximum is at month $7$:

$$T(m) = 14\cos\big(30(m - 7)\big) + 6$$

April is $m = 4$: $T(4) = 14\cos(-90^\circ) + 6 = 6$°C, right on the axis, as the warming spring temperatures pass the yearly average.

(A sine version also works: $T(m) = 14\sin\big(30(m - 4)\big) + 6$.)

Example 4: Changing the conditions

How would the Ferris wheel model in Example 1 change if the wheel turned once every $30$ s instead?

Solution. Only the period changes, so only $k$ changes: $k = \tfrac{360}{30} = 12$.

$$h(t) = -10\cos(12t) + 12$$

The graph has the same maximum, minimum, and axis, but it’s horizontally compressed: each cycle takes $30$ s instead of $40$ s.

Common mistakes

Using $360$ as the period. The period is the real time for one cycle (like $40$ s). $k$ is $360$ divided by that.

Starting at the wrong point. A rider who boards at the bottom starts at a minimum: use $-a\cos$, or a sine with a shift.

Rounding $k$ too early. In Example 2, using $k = 29.03$ instead of $\tfrac{360}{12.4}$ changes the answer in the second decimal place. Keep it exact on the calculator.

Finding only one time. In a full cycle, the height usually reaches a given value twice: once going up and once going down.

Calculator in radian mode. Every model here uses degrees.

Practice

1. (Warm-up) Find $k$ for a period of $60$ seconds.

Solution

$k = \tfrac{360}{60} = 6$.

2. (Warm-up) A Ferris wheel seat goes between $1$ m and $25$ m above the ground. Find $a$ (positive) and $c$.

Solution

$a = 12$, $c = 13$.

3. (Warm-up) A model is $h(t) = 5\sin(18t) + 7$, with $h$ in metres and $t$ in seconds. State the amplitude, period, and axis, with units.

Solution

Amplitude $5$ m, period $\tfrac{360}{18} = 20$ s, axis $h = 7$ m.

4. (Core) A Ferris wheel has a radius of $15$ m, its centre is $17$ m above the ground, and it turns once every $60$ s. A rider boards at the bottom at $t = 0$. Write a model and find the rider’s height after $10$ s.

Solution

$h(t) = -15\cos(6t) + 17$.

$h(10) = -15\cos 60^\circ + 17 = -7.5 + 17 = 9.5$ m.

5. (Core) A weight on a spring bobs between $10$ cm and $30$ cm above the floor, completing one bounce every $2$ s. It starts at its highest point. Write a model and find its height after $0.5$ s.

Solution

$a = 10$, $c = 20$, $k = \tfrac{360}{2} = 180$, starting at a maximum:

$$h(t) = 10\cos(180t) + 20$$

$h(0.5) = 10\cos 90^\circ + 20 = 20$ cm (passing through the middle).

6. (Core) In one city, the longest day of the year (day $172$) has $15.6$ hours of daylight, and the shortest has $8.6$ hours. Using a period of $365$ days, write a model for the hours of daylight on day $t$, and estimate the daylight on day $80$.

Solution

$a = 3.5$, $c = 12.1$, $k = \tfrac{360}{365} \approx 0.986$, with a maximum at day $172$:

$$D(t) = 3.5\cos\big(0.986(t - 172)\big) + 12.1$$

$D(80) \approx 12.1$ hours, close to the axis, which makes sense: day $80$ (late March) is near the spring equinox, when day and night are about equal.

7. (Core) For the Ferris wheel in Question 4, during the first turn, between which times is the rider more than $20$ m above the ground? For how long is that?

Solution

Solve $-15\cos(6t) + 17 = 20$: $\cos(6t) = -0.2$, so $6t \approx 101.5^\circ$ or $258.5^\circ$, giving $t \approx 16.9$ s or $43.1$ s.

The rider is above $20$ m between about $16.9$ s and $43.1$ s, for about $26.2$ s.

8. (Challenge) The wheel in Question 4 is sped up so it turns once every $45$ s. Write the new model. How do its graph and the time spent above $20$ m change?

Solution

$k = \tfrac{360}{45} = 8$, so $h(t) = -15\cos(8t) + 17$.

The graph is compressed horizontally by a factor of $\tfrac{45}{60} = \tfrac{3}{4}$, with the same maximum, minimum, and axis. Every time scales by $\tfrac{3}{4}$. Solving again, $\cos(8t) = -0.2$ gives $8t \approx 101.5^\circ$ or $258.5^\circ$, so $t \approx 12.7$ s or $32.3$ s: the rider is above $20$ m for about $19.6$ s.

9. (Challenge) A bike pedal is $17$ cm from the crank’s centre, which is $30$ cm above the ground. It turns once every $0.75$ s and starts at its lowest point. Write a model for the pedal’s height, and check it at $t = 0.375$ s.

Solution

$a = 17$, $c = 30$, $k = \tfrac{360}{0.75} = 480$, starting at a minimum:

$$h(t) = -17\cos(480t) + 30$$

At $t = 0.375$ s (half a turn): $-17\cos 180^\circ + 30 = 47$ cm, the highest point. ✓