Alternative Format — Lesson 6: Applications of Sinusoidal Functions

Let's Start Thinking 

Sinusoidal Functions in the Real World

In the last lesson of this unit, we will examine real-world situations that can be modelled by sinusoidal functions. There will be a little bit of astronomy and weather, some mechanics and ecology, as well as some fun experiences.

Sources: Sunrise Over Earth - shulz/E+/Getty Images;
Canadian Lynx - jimkruger/iStock/Getty Images; Ferris Wheel - Nikada/iStock/Getty Images

These scenarios are merely a taste of some applications were sinusoidal functions can be used to predict behavior and gain insight into the properties of these situations.

Try This

The average daily temperatures for each month in Winnipeg, Manitoba, Canada are in the table below.

1. Write sinusoidal functions to model the average daily high temperatures and the average daily low temperatures.
2. Compare the properties of the functions. Describe why some properties are similar and others are different.

Applications

In this section, we examine natural phenomena and engineered products that can be modelled by sinusoidal functions.

Ecology: Predator-Prey Interaction

In the wild, the population of two species may be related.

Note: The degree symbol is not used in many of the sinusoidal functions in this lesson as the context of the application may require different units.

Example 1

Solution — Part A

For a sinusoidal function of the form \(y=a\sin\left(b\left(x-h\right)\right)+k\), the maximum value of the function is \(k+\lvert a\rvert\).

From the model, \(H\left(t\right)=20\sin\left(45t\right)+60\), the maximum population was \(20~000+60~000=80~000\) hare.

To determine in which years the hare population was at a maximum, we need to solve for \(t\) in the equation:

The base sine function reaches its maximum value of \(1\) when the input is \(90\) and repeats regularly since for all \(n\in\mathbb{Z}\), \(\sin\left(\left(90+360n\right)\right)=1\). Equating the inputs of the two equations, we get

Therefore, the hare population was at a maximum in the years 2002, 2010, 2018, and every \(8\) years afterwards.

Solution — Part B

Method 1: Compare Period and Phase Shift

Comparing the hare model \(H\left(t\right)=20\sin\left(45t\right)+60\) and the lynx model \(L\left(t\right)=4.5\sin\left(45\left(t-2\right)\right)+8.5\), the two sine functions have the same \(b\) value but different \(h\) values. The period of both models is \(\dfrac{360}{45}=8\) years but the lynx model has a phase shift of \(2\) years to the right. This means that the lynx will reach their maximum population \(2\) years after the hare reach their maximum population, or in the years 2004, 2012, 2020, and every \(8\) years afterwards.

Method 2: Solve Equation

The maximum lynx population of \(4500+8500=13~000\) occurs when 

Therefore, the maximum lynx population occurs in the years 2004, 2012, 2020, and every \(8\) years afterwards.

Solution — Part C

Predator-prey interaction says that the more prey, the healthier the predators may become to catch prey and to reproduce. This implies that there is a delay between when the hare reach their maximum population and food is abundant for the lynx and when the lynx population reproduces to reach its maximum. From the models, this delay is \(2\) years.

Ocean Cycles

While our Earth is close to spherical, the world's oceans bulge in two areas, one around the sublunar point and the other around the antipodal point. These points are constantly moving as the Earth makes its daily rotation on its axis, and as the Moon rotates around the Earth every 28 days. It turns out that a particular point on Earth will be a sublunar point every 24 hours and 50 minutes.

Image not to scale.

Sources: Moon - Kativ/E+/Getty Images; Earth - loops7/iStock/Getty Images

Example 2

Solution — Part A

From the equation \(D\left(t\right)=4.3\cos\left(29t\right)+7\), we know \(a=4.3\), \(b=29\), and \(k=7\).

• The value \(k=7\) means that the average water depth in the harbour is \(7\) m.
• The amplitude is \(4.3\) so the height of the water could be up to \(4.3\) m away from the average or between \(7-4.3=2.7\) m and \(7+4.3=11.3\) m.

• The period is \(\dfrac{360}{29}\approx12.41\) hours. This means that high tide occurs approximately every \(12\) hours and \(25\) minutes.

Solution — Part B

1. Since high tide occurred at 5:00 AM, 9:00 AM is \(4\) hours later.\[\begin{align*} D\left(4\right)&=4.3\cos\left(29\left(4\right)\right)+7\\ &=5.115\ldots \end{align*}\]

   At 9:00 AM, the water was approximately \(5.1\) m deep.

2. From 5:00 AM to 2:00 PM is \(9\) hours, so 2:45 PM is \(9 + 0.75 = 9.75\) hours after high tide.\[\begin{align*} D\left(9.75\right)&=4.3\cos\left(29\left(9.75\right)\right)+7\\ &=7.948\ldots \end{align*}\]

   At 2:45 PM, the water was approximately \(7.9\) m deep.

Check Your Understanding 1

Question

The position of a rider on a Ferris wheel is modelled by the sinusoidal function. 

where \(h\left(t\right)\) is the rider's height in metres above the ground and \(t\) is the time in minutes from when the ride begins.

1. If the ride is \(12\) minutes long, how many revolutions does the rider make on the wheel? 
2. What was the rider's height after \(9\) minutes? Round your answer to the nearest tenth of a metre.

Answer

1. Number of revolutions \(=3\)
2. \(16\)

Feedback

1. The period of the function is \(\dfrac{360}{90}=4\) minutes for one revolution, so the number of revolutions of the ride is \(12 \div 4=3\).
2. To find the rider's height after \(9\) minutes, substitute and evaluate

   \(\begin{align*}h(9)=-14 \cos (90(9))+16=16 \end{align*}\)

   The rider's height was \(16.0\) m.

Electric Facts

Example 3

Solution — Part A

Voltage

The maximum voltage of the model is the amplitude, \(170\) V.

According to the Did You Know, 

The described voltage for the model function is approximately \(120\) V.

Frequency

The model function has \(b=21~600\), so the period of the model is \(\dfrac{360}{\lvert b\rvert}=\dfrac{360}{21\,600}=\dfrac1{60}\) seconds. In one second, the model would have \(60\) cycles or a frequency of \(60\) Hz.

Solution — Part B

Voltage

For a described voltage of \(220\) V,

The maximum voltage is approximately \(311\) V, which will be the amplitude of the model function.

Frequency

The frequency of \(50\) Hz is \(50\) cycles in one second so the period of the model function will be the reciprocal of \(50\) or \(\dfrac{1}{50}\) of a second.

Assuming \(b\) is positive,  

An equation to model the voltage of a \(220\) V, \(50\) Hz alternating current is \(W\left(t\right)=311\sin\left(18\,000t\right)\).

Combustion Engines

Internal combustion engines are used for automobiles and other machinery. Power to drive an engine is created by compressing fuel inside a cylinder then igniting it. The compression is the result of a piston that moves up and down within the cylinder. The piston in turn is attached to the engine's rotating crankshaft.

Example 4

Solution — Part A

From the given function, \(a=50\) and \(k=200\).

The maximum distance of the piston head from the crankshaft is \(k+\lvert a\rvert=200+50=250\) mm. The minimum distance is \(k-\lvert a\rvert=200-50=150\) mm.

Solution — Part B

The period of the function is \(\dfrac{360}{\lvert b\rvert}=\dfrac{360}{14\,400}=\dfrac1{40}\) seconds. The function has \(40\) cycles in one second. Since the fuel is ignited once every \(2\) cycles, it is ignited \(20\) times per second. Therefore, in one minute, the fuel will be ignited \(20 \times 60 = 1200\) times.

Solution — Part C

If the piston represented by \(P\left(t\right)\) is in the intake stroke of the combustion cycle, then the piston in the power stroke of the cycle would also be in the same position at the same time so the function \(P\left(t\right)\) could represent it as well.

The other two pistons, in the compression and exhaust strokes, have the same positions at the same time so their equations would be the same. Their positions have the same amplitude, axis, and period as \(P\left(t\right)\) but they would be moving in the opposite direction. Their positions could be represented by a vertical reflection of \(P\left(t\right)\) so their equation would be \(Q\left(t\right)=-50\cos\left(14\,400t\right)+200\).

Modelling Data

Explore This 1

Question — Part A

The number of hours of daylight in 2018 for five worldwide locations are shown. Which graphs can be modelled by a sinusoidal function? Match each graph with the function that best models each curve on the graph.

Answer — Part A

1. C. \(y=-2.18\sin\left(\dfrac{72}{73}\left(x - 80\right)\right) + 12.15\)
2. B. \(y=0.19\sin\left(\dfrac{72}{73} \left(x - 80\right)\right) + 12.12\)
3. E. \(y=3.18\sin\left(\dfrac{72}{73}\left(x - 80\right)\right) + 12.18\)
4. D. \(y=4.40\sin\left(\dfrac{72}{73} \left(x - 80\right)\right) + 12.23\)
5. A. Not a sinusoidal function

Feedback — Part A

To match, compare the amplitude of the graphs \(\left(\dfrac{\text{Max} - \text{Min}}{2}\right)\) and the amplitude of the model functions \(\left(\lvert a\rvert\right)\).

Question — Part B

The locations where the graphical data in part a) was collected are shown on the world map. Can you match each place name with its location?

Answer — Part B

1. 3. London, Ontario, Canada, \(43^\circ\) N, \(81^\circ\) W
2. 5. London, United Kingdom, \(52^\circ\) N, \(0^\circ\) W
3. 4. Tromso, Norway, \(70^\circ\) N, \(19^\circ\) E
4. 1. East London, South Africa, \(33^\circ\) S, \(28^\circ\) E
5. 2. Kuala Lumpur, Malaysia, \(3^\circ\) N, \(102^\circ\) E

Feedback — Part B

Latitude is the angular distance of a place north or south of the Earth's equator.

Longitude is the angular distance of a place east or west of the prime meridian.

The locations marked on the map are as follows:

1. London, Ontario, Canada, \(43^\circ\) N, \(81^\circ\) W
2. London, United Kingdom, \(52^\circ\) N, \(0^\circ\) W
3. Tromso, Norway, \(70^\circ\) N, \(19^\circ\) E
4. East London, South Africa, \(33^\circ\) S, \(28^\circ\) E

5. Kuala Lumpur, Malaysia, \(3^\circ\) N, \(102^\circ\) E

Question — Part C

What is the relationship between the function modelling the number of hours of daylight and the location where it is measured?

Observe the location of each city, the associated graph, and equation for each location.

A. London, Ontario, Canada \(43^\circ\) N

\(y=3.18\sin\left(\dfrac{72}{73}\left(x - 80\right)\right) + 12.18\)

B. London, United Kingdom, \(52^\circ\) N

\(y=4.40\sin\left(\dfrac{72}{73} \left(x - 80\right)\right) + 12.23\)

C. Tromso, Norway, \(70^\circ\) N

Not a sinusoidal function

D. East London, South Africa, \(33^\circ\) S

\(y=-2.18\sin\left(\dfrac{72}{73}\left(x - 80\right)\right) + 12.15\)

E. Kuala Lumpur, Malaysia, \(3^\circ\) N

\(y=0.19\sin\left(\dfrac{72}{73} \left(x - 80\right)\right) + 12.12\)

Source: Map - prospective56/iStock/Getty Images

Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0.

Interactive Version — Part C

Modelling Daylight Data

References

• United States Naval Observatory (USNO). (2018). [Duration of Daylight/Darkness Data for 2018]. Retrieved from https://aa.usno.navy.mil/data/docs/Dur_OneYear.php

Explore This Summary

In the Explore This activity, you might have noticed that one of the five hours of daylight graphs could not be modelled using a sinusoidal function. 

The one anomaly was Graph 5, which had horizontal parts to its function. This graph matched to Tromso, Norway, which is north of the Arctic Circle, and has \(24\) hours of daylight during part of the summer, and \(24\) hours of darkness for part of the winter.

You might have noticed that the phase shifts and the vertical translations for the four sinusoidal models are almost identical. That is, the phase shift was \(80\) for all models and the vertical translations varied from \(12.12\) to \(12.23\).

The \(80\)^th day of 2018 was March 21, the first day of spring, and the average number of daylight hours anywhere in the world was a bit more than \(12\) hours a day.

Of the four sinusoidal graphs, you might have noticed that Graph 1 was the only one whose model had a reflection in the \(x\)-axis.

East London, South Africa is the only location that is south of the equator. 

You might have noticed that there appears to be a relation between the amplitude of the function and the latitude of the location. The greater the amplitude, the farther the location is from the equator.

Model/Function 2 has an amplitude of \(0.19\), and Kuala Lumpur, Malaysia is closest to the equator at \(3^\circ\) N.

Model/Function 1 has an amplitude of \(2.18\), and East London South Africa is \(33^\circ\) S of the equator. 

Model/Functions 3 and 4 have amplitudes of \(3.18\) and \(4.4\), respectively. And their location latitudes are \(43^\circ\) and \(52^\circ\) N.

Collecting data and modelling the data with functions can lead to comparisons and insight, such as the connection between the locations' latitude and the amplitude of the sinusoidal model. If you were wondering why the average number of daylight hours is more than \(12\), do a little research for a scientific explanation about the sun, the horizon, and the Earth's atmosphere.

References

• United States Naval Observatory (USNO). (2018). [Duration of Daylight/Darkness Data for 2018]. Retrieved from https://aa.usno.navy.mil/data/docs/Dur_OneYear.php

Bungee Jumping

Bungee jumpers usually jump from a fixed platform, like a crane or a bridge, while connected to a long elastic cord. The first part of a jump is free fall until the bungee cord stretches and sends the jumper rebounding upwards again. Several rebounds can occur before the jumper is lifted back to the platform.

 Did You Know?

Commercial bungee jumping began in the 1980's in New Zealand.

Source: Bungee Jumper - matejmm/E+/Getty Images

Example 5

Solution — Part A

Here is a graph of Hollam's data:

The first part of a bungee jump is free fall and is not influenced by the bungee cord.

From the data, Hollam's height decreases more slowly after \(5\) seconds which suggests that he was no longer free falling and the bungee cord was beginning to stretch.

We will use the domain \(\left\{ t\in\mathbb{R}\mid t\ge 5\right\}\) for our sinusoidal model. 

Solution — Part B

Hollam's minimum height was \(9\) m at \(8\) seconds and his maximum rebound height was \(69\) m at \(14\) seconds. This could be half of a sinusoidal cycle with an amplitude of \(\lvert a\rvert=\dfrac{69-9}{2}=30\) m and an average height of  \(k=\dfrac{69+9}{2}=39\) m. The period is \(2\left(14-8\right)=12\) seconds, so \(b=\dfrac{360}{12}=30\).

There are infinitely many functions that could model Hollam's height. Here are some sample functions.

The base sine function has a \(y\)-intercept on its axis and the function is increasing. If Hollam's function is 

• a sine function, using the increasing axis point \(\left(11,39\right)\) then \(h=11\) and \(a\gt0\), \(h\left(t\right)=30\sin\left(30\left(t-11\right)\right)+39,~t\ge 5\);
• a sine function, using the decreasing axis point \(\left(5,39\right)\) then \(h=5\) and \(a\lt0\), \(h\left(t\right)=-30\sin\left(30\left(t-5\right)\right)+39,~t\ge 5\).

The base cosine function has a \(y\)-intercept at its maximum, so if Hollam's function is

• a cosine function, using the maximum point \(\left(14,69\right)\) then \(h=14\) and \(a\gt0\), \(h\left(t\right)=30\cos\left(30\left(t-14\right)\right)+39,~t\ge 5\);
• a cosine function, using the minimum point \(\left(8,9\right)\) then \(h=8\) and \(a\lt0\), \(h\left(t\right)=-30\cos\left(30\left(t-8\right)\right)+39,~t\ge 5\).

Try This Revisited

Solution — Part A

The outdoor temperature depends on the seasons so a sinusoidal model with a period of \(12\) months and \(b=\dfrac{360}{12}=30\) is reasonable.

Average Daily High Temperature

Let \(H\left(m\right)\) represent the average daily high temperature, in \(^\circ\)C, of month \(m\), where \(m=1\) is January and \(m=12\) is December.

From the data, the amplitude of the model sinusoidal function could be \(\dfrac{26-\left(-10\right)}{2}=18\).

The average daily high temperature for the year could be \(\dfrac{26+\left(-10\right)}{2}=8\).

The maximum of the data occurs in July, so a cosine function model could be \(H\left(m\right)=18\cos\left(30\left(m-7\right)\right)+8\).

Average Daily Low Temperature

Let \(L\left(m\right)\) represent the average daily low temperature, in \(^\circ\)C, of month \(m\).

The amplitude of the model sinusoidal function could be \(\dfrac{16-\left(-18\right)}{2}=17\).

The average daily low temperature for the year could be \(\dfrac{16+\left(-18\right)}{2}=-1\).

The maximum of the data occurs in July, so a cosine function model could be \(L\left(m\right)=17\cos\left(30\left(m-7\right)\right)-1\).

Solution — Part B

The amplitude, period, and phase shift of the two models are almost identical. This implies that when the daily high temperature is warmer, the daily low temperature is also warmer. This is reasonable as warmer days typically correspond to warmer nights as well.

The vertical translations for the models are different, \(+8\) compared to \(-1\). Since the rest of the model parameters are similar, the \(k\) values suggest that throughout the year, the daily highs are about \(8-\left(-1\right)=9^\circ\)C warmer than the daily lows. Looking at the data, we can see that differences between the highs and lows for each month is between \(7^\circ\) and \(10^\circ\). This is useful when you leave your home in the early afternoon and want to know whether or not you will need a jacket in the evening.

Angle of Elevation

Example 6

Solution — Part A

• We will assume we are modelling the angle of elevation for a non-leap year. If the function has a period of \(365\) days, then \(b=\dfrac{360}{365}=\dfrac{72}{73}\).
• From the data, the maximum angle is \(91.3^\circ\) and the minimum angle is \(52.7^\circ\).
• The amplitude for the model is \(\dfrac{91.3-52.7}{2}=19.3\) and \(k=\dfrac{91.3+52.7}{2}=72\).

Using a cosine function and the maximum data point \(\left(181,91.3\right)\), the model function is \(A\left(d\right)=19.3\cos\left(\dfrac{72}{73}\left(d-181\right)\right)+72\), where \(A\left(d\right)\) is the angle of elevation of the sun in degrees and \(d\) is the day of the year, day \(1\) is January 1^st.

Alternatively, for a sine function model, we could use the increasing axis point one-quarter of the period or \(\dfrac{365}4=91.25\) days before the maximum at \(\left(89.75,72\right)\) so \(h=89.75\). The parameters \(a\), \(b\), and \(k\) would be the same as in previous cosine model, so the model function would be \(A\left(d\right)=19.3\sin\left(\dfrac{72}{73}\left(d-89.75\right)\right)+72\).

Solution — Part B

To determine the angle of elevation on March 21^st, we need to determine the day number, and then evaluate the model function. Using the cosine model,

• January has \(31\) days and February has \(28\) days, so March 21^st is the \(80\)^th day of the year.\[A\left(80\right)=19.3\cos\left(\dfrac{72}{73}\left(80-181\right)\right)+72\approx68.7759\]
• On March 21^st, the angle of elevation that the panel should be facing is approximately \(68.8^\circ\).

Check Your Understanding 2

Question — Version 1

A daily interest index for the term "math" on a search engine is the percent of the number of searches for "math" on a particular day compared to the highest number of searches for "math" in a day. The interest index for \(15\) consecutive days is shown.

Identify the period to be used in a sinusoidal model of the data. Think about what the period reveals about the trend pattern of searches for "math" and why that pattern might exist.

Answer — Version 1

The period is \(7\) days.

Feedback — Version 1

There appears to be a weekly trend in the search interest. Days \(1\), \(8\), and \(15\) could be the beginning of the weekends when there might be less interest in school.

Therefore, a sinusoidal model of this data would have a period of \(7\) days.

Question — Version 2

The horizontal distance a swinging pendulum is from a wall is measured at various points in time. The resulting data is summarized in the following table: 

Identify the period to be used in a sinusoidal model of the data. Think about what the period tells you about the swinging pendulum.

Answer — Version 2

The period is \(8\) seconds.

Feedback — Version 2

The pendulum is closest to the wall at ‌\(0\) and ‌\(8\) seconds and furthest from the wall at ‌\(4\) and ‌\(12\) seconds.

Therefore, it takes \(8\) seconds for a complete swing of the pendulum, so a sinusoidal model of this data would have a period of \(8\) seconds.

Wrap-Up

Take It With You

You board the ride at the bottom of a wheel when the arm is vertical.

1. Sketch a graph of your height above the ground during the ride.
2. Write an equation to model your height above the ground during the ride.