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The periodic nature of trigonometric functions can be applied to model real-life events and phenomena that are repetitive and periodic. This lesson will consider a fire department, where such models are not only helpful but absolutely crucial to understand. Additionally, how a sinusoidal regression can model periodic data will be understood.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Explore

Phenomena Modeled by Trigonometric Functions

During a math lesson, Ali realized many real-life situations that repeat at regular intervals could be modeled by periodic functions. His friends do not believe him. To prove to his friends that trigonometric functions are used in real life, he scoured the internet and found several interesting examples of these functions.

Function rule	What does the function model?
$R (t) = \frac{9 0 0 sin 2 t}{3 2}$	The horizontal distance from a player to a basketball thrown with an initial velocity of $30$ feet per second at an angle with measure $t .$ Negative distance means that the direction of the throw changes.
$s (t) = 3.5 sin \frac{π}{1 8 0} (t + 4) + 15.7$	The average wind speed in a certain city, measured in miles per hour.
$P (t) = 100 - 20 cos \frac{8 π}{3} t$	The blood pressure of a person at rest, measured in millimeters of mercury.
$N (t) = 3.7 sin (\frac{4 t}{5} - 0.7) + 20$	The number of thousands of employees at a sports company.

Ali wanted to share the examples with his class, so he printed out the graphs of the functions.

Unfortunately, he forgot to label the graphs. Graph the given functions using a graphing calculator and help Ali match the graphs with their corresponding function rules.

Challenge

How to Model Average Monthly Water Temperature?

Ali's brother Davontay is a volunteer firefighter. At the end of each year, he likes to analyze certain data related to fire-fighting and rescue services for his department. This year, he received a special assignment from the Fire Chief.

External credits: Tomas Eidsvold

Davontay was asked to collect data about the average monthly water temperature $D,$ in degrees Fahrenheit, as a function of the month of the year $t .$ The table's data then needs to be analyzed. Later, this analysis will test for a relationship between the temperatures and number of rescue boat operations. By analyzing this table, the Fire Chief will let Davontay drive the rescue boat!

	Average Monthly Water Temperatures
$t$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
$D (^{\circ} F)$	$70$	$73$	$75$	$78$	$82$	$85$	$86$	$85$	$84$	$81$	$75$	$73$

The Fire Chief outlines the mission in more detail, here.

Make a scatter plot of the given data.
Find a periodic function that models the average water temperature.
Lastly, graph the function on the scatter plot and find its period.

Example

Using an Electrocardiogram to Calculate a Pulse Rate

Davontay explains that he and his squad frequently cooperate with medical rescue teams during road accidents. When providing first aid to victims, paramedics measure their vital signs — the electrical activity of a heart in millivolts — using an electrocardiogram (ECG).

A pulse rate of a person is the number of times their heart beats in one minute. Each cycle in the given graph represents one heartbeat.

a Davontay states that each horizontal unit corresponds to

0.15

seconds. What is the pulse rate of this accident victim?

b Find the amplitude of the given ECG knowing that one vertical unit corresponds to

1

millivolt.

Hint

a Start by calculating the period of the periodic function.

b The amplitude is half the difference of the maximum and minimum values of a periodic function.

Solution

a To find the pulse rate, calculate the number of heart beats in

60

seconds. Begin by finding the period of the given periodic function. Davontay highlighted one cycle — the shortest repeating portion of the graph. He states that the peaks can be used as a reference.

This is only one example portion of the graph that could be chosen as one cycle. The horizontal length of one cycle is

4

units. Because

1

unit corresponds to

0.15

seconds, the period can be determined by multiplying these numbers.

4 \cdot 0.15 = 0.6 seconds

The period is

0.6

seconds. This means that one heartbeat lasts

0.6

seconds. Let

p

be the pulse rate of the person. Therefore,

p

is the number of times that the heart beats in one minute. If

0.6

is multiplied by

p,

the result is

60

seconds.

0.6 p = 60

▼

Solve for

p

WriteAsFrac

Write as a fraction

\frac{3}{5} p = 60

MultEqn

$LHS \cdot \frac{5}{3} = RHS \cdot \frac{5}{3}$

p = 60 (\frac{5}{3})

MoveLeftFacToNum

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

p = \frac{3 0 0}{3}

CalcQuot

Calculate quotient

p = 100

The pulse rate of the accident victim is

100

beats per minute.

b The amplitude is half the difference of the maximum and minimum values of the periodic function.

A = \frac{y _{max} - y _{min}}{2}

Davontay points out that each repeating portion of the graph is identical, therefore it is sufficient to analyze only one cycle. Determine the difference between the maximum and minimum electrical activity in the graph.

Because one vertical unit corresponds to

1

millivolt, the difference

y_{max} - y_{min}

3

millivolts. Divide this value by

2

to find the amplitude of the function.

A = \frac{3}{2} mV \Leftrightarrow A = 1.5 mV

Example

Periodic Change in the Distance of a Sound Beam

Despite analogue radio receivers, SMS, and phone calls, electronic sirens are still used in many fire stations. They create a loud modulated sound that can be heard within a few miles.

External credits: @freepik

Davontay's fire station has a siren that makes $3$ rotations per minute and whose sound beam can be heard within a $4$ mile radius. It so happens that Davontay's house is exactly $4$ miles away from the fire station. Let $D (t)$ be the periodic function that represents the distance from the end of the sound beam to Davontay's house in terms of the time $t,$ measured in seconds.

a What is the period of the function in seconds?

b Make an example graph of the function for

0 \leq t \leq 60 .

Assume that the end of the sound beam is given by the farthest point the beam reaches. Additionally, the siren begins its rotation from the direction of Davontay's house.

Answer

20

seconds

b Example Solution:

Hint

a Find the time in which the siren completes one full cycle.

b What are the maximum and minimum distances of the sound beam's end in relation to Davontay's house? Use the period found in Part A to graph the function.

Solution

a It is given that the siren makes

3

rotations per minute. Because of the rotation, the distance of the sound beam varies periodically in time. To find the period of the function, it is sufficient to calculate the time that it takes to complete one rotation.

Number of Rotations	Time
$3$	$1 minute = 60 seconds$
$1$	$\frac{6 0}{3} = 20$ seconds

Therefore, the siren makes one cycle in $20$ seconds, which is the period of the function.

b It is a given that Davontay's house is

4

miles away from the siren. Davontay looks through some files and provides the following graph, which shows the rotation of the sound beam. He points out that the beam begins in the direction of his house.

Because the radius of the sound beam is

4

miles and the distance from the siren to Davontay's house is also

4

miles, take the difference of those two numbers to find that the minimum distance between the end of the sound beam and the house is

0

miles. This happens when the siren is directed towards the house.

Minimum Value of D (t) : 0 miles

Conversely, when the siren is facing towards the opposite direction, the maximum distance of the sound beam's end is

8

miles. See the applet for a visual demonstration of the maximum distance.

Maximum Value of D (t) : 8 miles

Each cycle of the function fluctuates between these two values in time. Because the period of the function is

20

seconds, each cycle is repeated every

20

seconds. It is also given that the siren starts rotating from the direction of Davontay's house. Therefore, the value of the function at

t = 0

0 .

Extra

Determining the Function Rule

Although it was sufficient to make an example graph of the distance function, the function rule can also be found. Begin by recalling the coordinates of the points on a unit circle.

These coordinates can be generalized to the coordinates of the points on a circle of different radius. They are known as polar coordinates. Let $r$ be the radius of a circle. The coordinates of a generalized circle are multiplied by $r .$

In the given situation

r

is equal to

4 .

The angle of rotation depends on the time

t .

Recall that the period of the function is

20

seconds. This means that the siren makes a full turn — a rotation of

2 π

radians — in

20

seconds. Next, the angle

θ_{1}

by which the siren rotates in

1

second can be found.

θ_{1} = \frac{2 π}{2 0} \Leftrightarrow θ_{1} = \frac{π}{1 0}

Therefore, the angle of rotation in

t

seconds is the product of

\frac{π}{1 0}

and

t .

θ = \frac{π}{1 0} t

Now, the coordinates of a point that lies on the circle after

t

seconds can be written. Assume that the fire station is in the origin and the Davontay's house is at

(4, 0) .

The distance between the house and the sound beam can be calculated by using the Distance Formula.

D (t) = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstitutePoints

Substitute $(4, 0)$ & $(4 c o s \frac{π}{1 0} t, 4 s i n \frac{π}{1 0} t)$

D (t) = (4 c o s \frac{π}{1 0} t - 4)^{2} + (4 s i n \frac{π}{1 0} t - 0)^{2}

▼

Simplify right-hand side

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

D (t) = (4 cos \frac{π}{1 0} t)^{2} - 2 (4 cos \frac{π}{1 0} t) (4) + 4^{2} + (4 sin \frac{π}{1 0} t - 0)^{2}

SubTerm

Subtract term

D (t) = (4 cos \frac{π}{1 0} t)^{2} - 2 (4 cos \frac{π}{1 0} t) (4) + 4^{2} + (4 sin \frac{π}{1 0} t)^{2}

PowProdII

$(a b)^{m} = a^{m} b^{m}$

D (t) = 4^{2} cos^{2} \frac{π}{1 0} t - 2 (4 cos \frac{π}{1 0} t) (4) + 4^{2} + 4^{2} sin^{2} \frac{π}{1 0} t

CalcPowProd

Calculate power and product

D (t) = 16 cos^{2} \frac{π}{1 0} t - 32 cos \frac{π}{1 0} t + 16 + 16 sin^{2} \frac{π}{1 0} t

CommutativePropAdd

Commutative Property of Addition

D (t) = 16 cos^{2} \frac{π}{1 0} t + 16 sin^{2} \frac{π}{1 0} t - 32 cos \frac{π}{1 0} t + 16

FactorOut

Factor out $16$

D (t) = 16 (cos^{2} \frac{π}{1 0} t + sin^{2} \frac{π}{1 0} t) - 32 cos \frac{π}{1 0} t + 16

In future lessons, Pythagorean identities will be presented. One of these identities states that the sum of the square of the sine and the square of the cosine of the same angle is always equal to

1 .

cos^{2} \frac{π}{1 0} t + sin^{2} \frac{π}{1 0} t = 1

This means that the function rule can be further simplified.

D (t) = 16 (cos^{2} \frac{π}{1 0} t + sin^{2} \frac{π}{1 0} t) - 32 cos \frac{π}{1 0} t + 16

▼

Simplify right-hand side

$sin^{2} (θ) + cos^{2} (θ) = 1$

D (t) = 16 (1) - 32 cos \frac{π}{1 0} t + 16

IdPropMult

Identity Property of Multiplication

D (t) = 16 - 32 cos \frac{π}{1 0} t + 16

AddTerms

Add terms

D (t) = 32 - 32 cos \frac{π}{1 0} t

SplitIntoFactors

Split into factors

D (t) = 16 (2) - 16 (2) cos \frac{π}{1 0} t

FactorOut

Factor out $16$

D (t) = 16 (2 - 2 cos \frac{π}{1 0} t)

SqrtProd

$a \cdot b = a \cdot b$

D (t) = 4 2 - 2 cos \frac{π}{1 0} t

Finally, the exact distance function has been obtained.

Discussion

Sinusoids

The graphs of the sine function and the cosine function are called sinusoids. Sinusoids describe smooth periodic cycles.

The function rules for the sine and cosine functions can be written as follows.

Sine y = a sin b (x - h) + k Cosine y = a cos b (x - h) + k

In these trigonometric functions,

∣ a ∣

is the amplitude,

\frac{2 π}{∣ b ∣}

is the period,

h

is the horizontal shift, and

k

is the vertical shift.

Example

Graphing an Earthquake Wave

Davontay originally volunteered as a firefighter because they live near a volcano along the coast of a seismic zone — an area where earthquakes usually occur. A few years ago, on a dark cloudy day there was danger of a looming earthquake. Davontay's fire department had no choice but to organize the evacuation of all the town's homes.

Davontay was told that the predicted earthquake wave had a frequency of $\frac{1}{6}$ hertz and an amplitude — displacement of the ground — of $5$ millimeters.

a Write an equation of a sinusoid that represents the displacement of the ground

d (t),

measured in millimeters, in terms of the time

t,

measured in seconds. Davontay says that the point

(0, 0)

lies on the sinusoid and it is halfway between the lowest and the highest point of the wave.

b Davontay lays on the table a few more interesting sinusoids to consider.

Together, Davonty figures that by graphing the function obtained in Part A, the graph that corresponds to this function can be determined. Give it a try!

c Davontay recorded that at precisely

28

seconds after the quake hit, it was as if he could hear the ground tremble! What is the absolute displacement of the ground after

28

seconds? Write the answer in exact form.

Hint

a Start by determining which function — sine or cosine — describes this sinusoid. Then, find the period of the function to write its equation.

b Start by drawing the midline of the function. Then, plot some key points on the graph and connect them with a smooth curve.

c Evaluate the function at

t = 28 .

Solution

a Sinsusoids are graphs of sine or cosine functions. Start by determining which function is more suitable for the given situation.

It is given that the point

(0, 0)

lies on the sinusoid and is halfway between the lowest and the highest point. Therefore, the sine function without any translation should be chosen for the equation of the sinusoid. Recall the format of a sine function.

y = a sin b x

Here,

∣ a ∣

and

\frac{2 π}{∣ b ∣}

are the amplitude and the period of the graph, respectively. Since it is given that the amplitude is

5

millimeters,

∣ a ∣

is equal to

5 .

Therefore, there are two possible values for

a

— positive and negative.

∣ a ∣ = 5 ⇓ a = 5 or a = - 5

For simplicity, the positive value

a = 5

will be chosen. Next, the period of the function will be determined. Recall that the period of a periodic function is the reciprocal of its frequency.

Period = \frac{1}{Frequency}

By knowing that the frequency is

\frac{1}{6},

the period can be found.

Period = \frac{1}{Frequency}

Substitute

$Frequency = \frac{1}{6}$

Period = \frac{1}{\frac{1}{6}}

▼

Evaluate right-hand side

DivByFracD

$\frac{a}{b / c} = \frac{a \cdot c}{b}$

Period = \frac{1 ( 6 )}{1}

IdPropMult

Identity Property of Multiplication

Period = \frac{6}{1}

DivByOne

$\frac{a}{1} = a$

Period = 6

The period of the sinusoid is

6 .

To calculate the value of

b

in the function rule, use the fact that the period is

\frac{2 π}{∣ b ∣} .

Period = \frac{2 π}{∣ b ∣}

Substitute

$Period = 6$

6 = \frac{2 π}{∣ b ∣}

▼

Solve for

b

MultEqn

$LHS \cdot ∣ b ∣ = RHS \cdot ∣ b ∣$

6 ∣ b ∣ = 2 π

DivEqn

$LHS / 6 = RHS / 6$

∣ b ∣ = \frac{2 π}{6}

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

∣ b ∣ = \frac{π}{3}

$b \geq 0 : b = \frac{π}{3} b < 0 : b = - \frac{π}{3} (I) (II)$

b = \pm \frac{π}{3}

Therefore,

b

can be either positive or negative. For simplicity, keep the positive value of

b .

Finally, the equation of the sinusoid can be written. Keep in mind that the input of the function is

t .

y = a sin b x ⇓ d (t) = 5 sin \frac{π}{3} t

b To graph a sine function, there are four steps to follow.

Find the amplitude, period, and translation of the function
Draw the midline
Plot some key points on the graph
Draw the graph

It is given that the amplitude is

5

millimeters. In Part A, it was also obtained that the period is

6

seconds.

Equation : Amplitude : Period : d (t) = 5 sin \frac{π}{3} t 5 millimeters 6 seconds

Looking at the equation, it is seen that its graph is not translated in any direction. Next, since the sine function is not translated, the midline of the graph is the horizontal line

y = 0 .

Graph this line on a coordinate plane.

Now, key points ranging over at least one cycle will be plotted. These points are the maximums, minimums, and intersections with the midline. The maximums of

y = sin x

occur once every cycle at the following

x -

coordinates.

x = - \frac{3 π}{2}, \frac{π}{2}, \frac{5 π}{2}, \dots

The period of the considered sinusoid is

6,

which is

\frac{3}{π}

times the period

2 π

of the parent sine function. Because the graph of

d (t)

is not translated horizontally, the maximums are not shifted to the right or to the left. This means that the

t -

coordinates of the maximums are as follows.

t = \frac{3}{π} (- \frac{3 π}{2}), \frac{3}{π} (\frac{π}{2}), \frac{3}{π} (\frac{5 π}{2}), \dots ⇕ t = - \frac{9}{2}, \frac{3}{2}, \frac{1 5}{2}, \dots

Since the midline is

y = 0

and the amplitude is

5,

the maximum value of the function is

0 + 5 = 5 .

Plot the maximum points on the graph with the midline.

Similarly, the minimums of

d (t)

are located halfway between the maximums at the following

t -

coordinates.

t = - \frac{1 5}{2}, - \frac{3}{2}, \frac{9}{2}, \dots

The minimum value of the function is

0 - 5 = - 5 .

Plot the minimums in the same coordinate plane.

Next, between every neighboring maximum and minimum are the intersections with the midline.

t = - 6, - 3, 0, 3, 6, \dots

Because these points lie on the midline, their

y -

coordinate is

0 .

Finally, by connecting the points with a smooth curve and continuing it periodically in both directions, the sinusoid can be drawn.

c To find the absolute displacement after

28

seconds, start by evaluating the function at

t = 28 .

d (t) = 5 sin \frac{π}{3} t

Substitute

$t = 28$

d (28) = 5 sin \frac{π}{3} (28)

▼

Evaluate right-hand side

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

d (28) = 5 sin \frac{2 8 π}{3}

WriteSum

Write as a sum

d (28) = 5 sin \frac{4 π + 2 4 π}{3}

WriteSumFrac

Write as a sum of fractions

d (28) = 5 sin (\frac{4 π}{3} + \frac{2 4 π}{3})

CalcQuot

Calculate quotient

d (28) = 5 sin (\frac{4 π}{3} + 8 π)

SplitIntoFactors

Split into factors

d (28) = 5 sin (\frac{4 π}{3} + 4 \cdot 2 π)

$sin (θ) = sin (θ + n \cdot 2 π)$

d (28) = 5 sin \frac{4 π}{3}

d (28) = 5 (- \frac{3}{2})

MultPosNeg

$a (- b) = - a \cdot b$

d (28) = - 5 \cdot \frac{3}{2}

MoveLeftFacToNum

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

d (28) = - \frac{5 3}{2}

The absolute displacement of the ground after

28

seconds is the absolute value of

- \frac{5 3}{2} .

∣ d (28) ∣ = ∣ ∣ ∣ ∣ ∣ ∣ - \frac{5 3}{2} ∣ ∣ ∣ ∣ ∣ ∣ ⇓ ∣ d (28) ∣ = \frac{5 3}{2}

Therefore, the absolute displacement is

\frac{5 3}{2}

millimeters.

Example

Cosine Model of a Sound Wave

On the weekends, Ali likes to help Davontay do some cleaning at the fire station. This particular time, Davontay asks his brother to download a mobile app to tune a guitar left at the station by a firefighter who plays it during some downtime. Not everything at a station is related to an emergency!

External credits: Roberta Sorge @robertina

The standard tuning of a guitar is A– $440,$ which means that the same note A on a keyboard will vibrate $440$ times per second. After downloading the app, the guitar player showed up and excitedly came over to talk about sound waves with Ali.

a The amplitude of a sound wave, measured in decibels, determines its relative loudness. The Fire Chief often yells at the guitar player to keep the amplitude of the guitar's sound wave at

60

decibels, which is the same volume as a normal conversation. Together with Davontay, determine the function rule of a non-translated cosine funtion that models this sound wave.

b Consider one cycle of the following sinusoids.

The guitar player asks which sinusoid represents the model obtained in Part A?

c Ali discovered that the frequency of a sound wave influences the pitch of the sound. Suppose the frequency of some other note is two times smaller. The guitar player wonders if the amplitude and period of the sound wave increase, decrease, or remain the same? Tell her the correct answer.

Hint

a Find the period of the function to write its function rule.

b Start by drawing the midline of the function. Then, plot some key points on the graph and connect them with a smooth curve.

c What is the relationship between the period and the frequency of a periodic function? Does the frequency influence the amplitude of the function?

Solution

a Consider the general function rule of a cosine function.

y = a cos b (x - h) + k

Because it is given that the function is not translated, both

h

and

k

are equal to

0 .

In this function rule

∣ a ∣

is the amplitude and

\frac{2 π}{∣ b ∣}

is the period. The amplitude of a sound wave is

60

decibels. This means that

∣ a ∣

is equal to

60 .

∣ a ∣ = 60 ⇓ a = 60 or a = - 60

There are two possible values of

a .

For example, the positive value

a = 60

will be chosen. Next, the period of the function will be determined. Recall that the period of a periodic function is the reciprocal of its frequency.

Period = \frac{1}{Frequency}

Because the given frequency is

440

hertz, the period of the function is

\frac{1}{4 4 0}

seconds. To calculate the value of

b

in the function rule, use the fact that the period is

\frac{2 π}{∣ b ∣} .

\frac{1}{4 4 0} = \frac{2 π}{∣ b ∣}

▼

Solve for

b

MultEqn

$LHS \cdot ∣ b ∣ = RHS \cdot ∣ b ∣$

\frac{1}{4 4 0} \cdot ∣ b ∣ = 2 π

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

\frac{∣ b ∣}{4 4 0} = 2 π

MultEqn

$LHS \cdot 440 = RHS \cdot 440$

∣ b ∣ = 880 π

$b \geq 0 : b = 880 π b < 0 : b = - 880 π (I) (II)$

b = \pm 880 π

This answer indicates that

b

can be either positive or negative. Because a cosine function that is not translated is an even function, both values give the same function. Not to complicate the formula, keep the positive value of

b .

Finally, the function rule for the cosine function can be written.

y = a cos b x ⇓ y = 60 cos (880 π x)

b To graph a cosine function, there are four steps to follow.

Find the amplitude, period, and translation of the function.
Draw the midline.
Plot some key points on the graph.
Draw the graph.

It is given that the amplitude is $60$ decibels. In Part A, it was also obtained that the period is $\frac{1}{4 4 0}$ seconds. Next, since the cosine function is not translated, the midline of this function is $y = 0 .$ Graph this line on a coordinate plane.

Now, key points ranging over one cycle will be plotted. These points are the maximums, minimums, and intersections with the midline. The maximums of

y = cos x

occur once every cycle at the following

x -

coordinates.

x = 0, 2 π, \dots

The period of the considered sinusoid is

\frac{1}{4 4 0},

which is

\frac{1}{8 8 0 π}

of the period

2 π

of the parent function. Because the function is not translated horizontally, the maximums are not shifted to the right or to the left. This means that the

x -

coordinates of the maximums are as follows.

x = \frac{1}{8 8 0 π} \cdot 0, \frac{1}{8 8 0 π} \cdot 2 π, \dots ⇓ x = 0, \frac{1}{4 4 0}, \dots

Since the midline is

y = 0

and the amplitude is

60,

the maximum value of the function is

0 + 60 = 60 .

Plot the maximum points on the graph with the midline.

The minimum of the function is located halfway between the maximums at $x = \frac{1}{8 8 0} .$ The minimum value is $0 - 60 = - 60 .$ Plot the minimum in the same coordinate plane.

Next, between every neighboring maximum and minimum the sinusoid intersects with the midline.

x = \frac{1}{1 7 6 0}, \frac{3}{1 7 6 0}, \dots

Because these points lie on the midline, their

y -

coordinate is

0 .

Finally, by connecting the points with a smooth curve, the sinusoid can be drawn. Although the graph may not represent the whole sound wave, keep in mind that only one cycle of each graph is given in the options.

This graph corresponds to option B.

c The guitar player asks how the change in frequency influences the period and the amplitude of a sound wave. Analyze each measure one at a time.

Period

As mentioned in Part A, the period of a periodic function is the reciprocal of its frequency.

Period = \frac{1}{Frequency}

Therefore, the period and frequency represent an inverse variation where the constant of variation is

1 .

This means that when the frequency decreases two times, the period doubles. Then, the pitch of the sound is noticeably lower.

Amplitude

The amplitude of a periodic function is half the difference between the maximum and minimum values, while the period is the length of one cycle.

The amplitude is related to neither period nor frequency. Therefore, when the frequency decreases $2$ times, the amplitude does not change. This means that the pitch of a sound does not influence its volume, which makes perfect sense. The guitar player is thankful for this understanding.

Example

How Do Trigonometric Functions Model Chest Compressions?

One of the most important and valuable fire brigade operations is qualified first aid. Ali admires his brother for conducting first aid training for people that do not have professional knowledge or experience in this field. The basic emergency procedure taught during training is cardiopulmonary resuscitation (CPR) — chest compressions often combined with artificial breaths.

Trainees practice CPR on a medical-training mannequin. CPR for an adult person consists of cycles of $30$ chest compressions and two breaths. Compressions should be $5$ to $6$ centimeters into the mannequin's or person's chest at a rate of $100$ to $120$ compressions per minute.

a Write a trigonometric function that represents the height of the compression at time

t

seconds. Assume that compressions are started at

t = 0

and that the initial height is

0

centimeters. Use the maximum values of the depth and the rate of compressions for the function.

b Consider the following sinusoids that Davontay shows the trainees.

By graphing the function obtained in Part A, choose its corresponding graph.

Hint

a Start by selecting the sinusoid that matches the situation. Then, find its midline, amplitude, and period.

b Identify the transformations done on the parent function and then graph the given function.

Solution

a The maximum depth of compressions is

6

centimeters and the maximum rate is

120

compressions per minute. To write the function that represents the height of compressions, Davontay suggests to start by choosing the sinusoid that matches the situation.

Because compressions are started at

t = 0,

the maximum height is also at

t = 0 .

This means that a cosine function will be used.

Cosine h (t) = a cos b (t - h) + k

In this function,

∣ a ∣

is the amplitude,

\frac{2 π}{∣ b ∣}

is the period,

h

is the horizontal shift, and

k

is the vertical shift. Now, write the equation for the midline. Recall that the midline lies between the maximum and minimum values. It is given that the maximum height is

0

centimeters and the depth is

6

centimeters.

Maximum Height : Minimum Height : 0 cm 0 - 6 = - 6 cm

The equation for the midline is given by the average of these values. This equation gives the vertical translation of the function.

Midline y = \frac{0 + ( - 6 )}{2} \Leftrightarrow y = - 3

Since the midline is

y = - 3,

the vertical shift is

k = - 3 .

Next, find the amplitude of the function by calculating the difference between the midline value and the maximum value.

∣ a ∣ = ∣ 0 - (- 3) ∣

SubNeg

$a - (- b) = a + b$

∣ a ∣ = ∣ 0 + 3 ∣

IdPropAdd

Identity Property of Addition

∣ a ∣ = ∣ 3 ∣

AbsPos

$∣ 3 ∣ = 3$

∣ a ∣ = 3

$a \geq 0 : a = 3 a < 0 : a = - 3 (I) (II)$

a = \pm 3

Since compressions are started at

t = 0

and then the height is maximum, choose a positive value of

a .

The rate of

120

compressions per minute means that there are

2

compressions every second, which gives a frequency of

2

hertz. The period of the function is the reciprocal of its frequency. Therefore, the period is

\frac{1}{2}

seconds. Write the equation for the period and solve it for

b .

\frac{2 π}{∣ b ∣} = \frac{1}{2}

▼

Solve for

b

MultEqn

$LHS \cdot$

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