IB Analysis and Approaches HL · Functions · Section 15 of 15

Applications of circular functions

3 worked examples · By Álvaro López

Key idea

Many phenomena repeat regularly: the tides, the height of a Ferris wheel car, the temperature over the year. They are modelled with $y = a sin (b (x - c)) + d$ or with the cosine, and each letter has a visual meaning on the graph. In the IB exam you are asked to find those letters from a graph or from two data points, and to use the model to answer context questions.

What each parameter does

Start from $y = sin x$ and apply the transformations you already know. Each letter in $y = a sin (b (x - c)) + d$ is one of them, and the same holds for $y = a cos (b (x - c)) + d$ .

Consequences that are used constantly:

the maximum is $d + ∣ a ∣$ and the minimum is $d - ∣ a ∣$ ;
translating the graph, horizontally or vertically, leaves the period and the amplitude unchanged;
a vertical stretch does not change the period, and a horizontal one does not change the amplitude.

Graph of y = 2 sin(2(x - pi/4)) + 1: midline y = 1, amplitude 2 between the midline and a maximum, and period pi measured between two consecutive maxima

In the drawing, $a = 2$ , $b = 2$ , $c = \frac{π}{4}$ , $d = 1$ : the midline is $y = 1$ , the maximum is $3$ and the minimum is $- 1$ , and the period is $\frac{2 π}{2} = π$ (the distance between two consecutive maxima).

To describe the graph as a transformation of $y = sin x$ , the natural order is: vertical stretch by a factor of $∣ a ∣$ (and reflection if $a < 0$ ), horizontal stretch by a factor of $\frac{1}{∣ b ∣}$ , horizontal translation by $c$ units and vertical translation by $d$ units.

Finding the parameters from a graph

There are several valid formulas for the same curve (for example $c$ and $c + P$ give the same one), and sine and cosine turn into each other with a shift. If the question fixes the form, keep it.

A quick example: if a maximum is at $(2, 8)$ and the next minimum at $(5, 2)$ , then $a = \frac{8 - 2}{2} = 3$ and $d = \frac{8 + 2}{2} = 5$ ; from a maximum to a minimum there is half a period ( $5 - 2 = 3$ ), so the period is $6$ and $b = \frac{2 π}{6} = \frac{π}{3}$ ; and since the maximum is at $x = 2$ , with the cosine $c = 2$ : $f (x) = 3 cos (\frac{π}{3} (x - 2)) + 5$ .

Periodic models

A context problem usually has three parts: identify what $a$ , $d$ and the period mean (for example, the mean height of the water, the tidal range and the hours of a full cycle), evaluate the model at an instant, and solve an inequality, of the type "for how long is the height above a certain value?".

Worked examples

Example 1 Basic

Reading the parameters from the formula

Let $f (x) = 5 sin (4 (x - \frac{π}{6})) - 2$ .

a) Find its amplitude.

b) Find its period.

c) Find its maximum value.

d) Find its minimum value.

Solution

The formula has the form $a sin (b (x - c)) + d$ with $a = 5$ , $b = 4$ , $c = \frac{π}{6}$ and $d = - 2$ .

a) The amplitude is $∣ a ∣ = 5$ .

b) The period is $\frac{2 π}{∣ b ∣} = \frac{2 π}{4} = \frac{π}{2}$ .

c) The maximum is $d + ∣ a ∣ = - 2 + 5 = 3$ .

d) The minimum is $d - ∣ a ∣ = - 2 - 5 = - 7$ .

The horizontal translation $\frac{π}{6}$ does not affect any of the four values.

Example 2 Medium

Formula from a maximum and a minimum

Part of the graph of a function $f (x) = a cos (b (x - c)) + d$ , with $a > 0$ , $b > 0$ and $c = 1$ , is shown in the figure. The point $A (1, 10)$ is a maximum and the point $B (4, 2)$ is the next minimum.

Part of the graph of a sinusoidal function with a maximum at the point A(1, 10) and the next minimum at the point B(4, 2)

a) Find the value of $a$ .

b) Find the value of $d$ .

c) Find the value of $b$ (in terms of $π$ ).

d) Calculate $f (0)$ .

Solution

a) The amplitude comes from halving the difference between the maximum and the minimum: $a = \frac{10 - 2}{2} = 4$ .

b) The midline is halfway: $d = \frac{10 + 2}{2} = 6$ .

c) Between the maximum $A$ and the minimum $B$ there is half a period: $4 - 1 = 3$ , so the period is $P = 6$ . Then $b = \frac{2 π}{P} = \frac{2 π}{6} = \frac{π}{3}$ .

d) The function is $f (x) = 4 cos (\frac{π}{3} (x - 1)) + 6$ . Check with $A$ : $f (1) = 4 cos 0 + 6 = 10$ . Then

$f (0) = 4 cos (- \frac{π}{3}) + 6 = 4 \cdot \frac{1}{2} + 6 = 8$

Example 3 Hard

The tide in a harbour

The depth of the water, in metres, at the entrance to a harbour, $t$ hours after midnight, is modelled by

$D (t) = 7 + 3 cos (\frac{π}{6} (t - 2)), 0 \leq t \leq 24$

a) Find the maximum and minimum depth. Give the pair (maximum, minimum).

b) Calculate the depth at 5 am.

c) A boat needs at least $8$ metres of water to enter. At what time (in hours after midnight, exact value) can it no longer enter for the first time?

d) For how many consecutive hours, in each cycle, is the depth at least $8$ metres? Give the exact value.

Solution

a) The amplitude is $3$ and the midline is $d = 7$ . The maximum is $7 + 3 = 10$ m and the minimum is $7 - 3 = 4$ m.

b) $D (5) = 7 + 3 cos (\frac{π}{6} \cdot 3) = 7 + 3 cos \frac{π}{2} = 7$ metres.

c) First the period is found: $\frac{2 π}{π /6} = 12$ hours. Solve $D (t) = 8$ :

$3 cos (\frac{π}{6} (t - 2)) = 1 ⟹ cos u = \frac{1}{3}, u = \frac{π}{6} (t - 2)$

The principal value is $u = arccos \frac{1}{3} \approx 1.231$ rad. The solutions of $cos u = \frac{1}{3}$ are $u = \pm arccos \frac{1}{3}$ (plus periods). In terms of $t$ : $t = 2 \pm \frac{6}{π} arccos \frac{1}{3}$ , that is, $t \approx - 0.35$ and $t \approx 4.35$ .

Since $t \geq 0$ , the depth at $t = 0$ is $D (0) = 7 + 3 cos (- \frac{π}{3}) = 8.5 > 8$ , so at first the boat can enter, and the depth falls to $8$ metres at

$t = 2 + \frac{6}{π} arccos \frac{1}{3} \approx 4.35 h (at around 4 : 21)$

d) The water is above $8$ metres between the two solutions $t = 2 \pm \frac{6}{π} arccos \frac{1}{3}$ of the same cycle. The duration is the difference:

$2 \cdot \frac{6}{π} arccos \frac{1}{3} = \frac{12}{π} arccos \frac{1}{3} \approx 4.70 h$

that is, about $4$ h $42$ min in each $12$ -hour cycle.

In the exam: in a problem of this type, the set-up of the equation, the solution with the calculator in the correct mode (here, radians) and the interpretation in context (the time, not the value of $u$ ) are marked separately. A sketch of the period with the line $D = 8$ helps avoid mixing up the intervals.

Common mistakes

Summary

In $y = a sin (b (x - c)) + d$ : amplitude $∣ a ∣$ , period $\frac{2 π}{∣ b ∣}$ (or $\frac{36 0 ^{\circ}}{∣ b ∣}$ ), horizontal shift $c$ and midline $y = d$ ; maximum $d + ∣ a ∣$ and minimum $d - ∣ a ∣$ .
From a graph: $a = \frac{M - m}{2}$ , $d = \frac{M + m}{2}$ , the period between two maxima and $b = \frac{2 π}{P}$ ; $c$ is the $x$ -coordinate of a maximum (cosine).
A periodic model is identified, evaluated, and $f (t) = k$ is solved to find for how long the function exceeds a value.
The GDC plots, solves by intersection and fits a sinusoidal curve to a table; watch the radian/degree mode.

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