Modeling with Trigonometric Functions

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Show the transformation of the graph of \(y=\sin \ x\) into the graph of \(y=2\ \sin (4x-\frac{\pi }{2})+2.\)

   Kuratidza mhinduro

   Consider the series of graphs in and the way each change to the equation changes the image.

2. Find the amplitude and period of the following functions and graph one cycle.

   1. ⓐⓐ \(y=2\ \sin (\frac{1}{4}x)\)
   2. ⓑⓑ \(y=-3\ \sin (2x+\frac{\pi }{2})\)
   3. ⓒⓒ \(y=\cos \ x+3\)
   Kuratidza mhinduro

   We will solve these problems according to the models.

   1. ⓐ \(y=2\ \sin (\frac{1}{4}x)\) involves sine, so we use the form \[y=A\ \sin (Bt+C)+D\]

      We know that \(|A|\) is the amplitude, so the amplitude is 2. Period is \(\frac{2\pi }{B},\) so the period is

      \[\begin{array}{l}\frac{2\pi }{B}=\frac{2\pi }{\frac{1}{4}} \\ =8\pi \end{array}\]

      See the graph in .

   2. ⓑⓑ \(y=-3\ \sin (2x+\frac{\pi }{2})\) involves sine, so we use the form \[y=A\ \sin (Bt-C)+D\]

      Amplitude is \(|A|,\) so the amplitude is \(|-3|=3.\) Since \(A\) is negative, the graph is reflected over the x-axis. Period is \(\frac{2\pi }{B},\) so the period is

      \[\frac{2\pi }{B}=\frac{2\pi }{2}=\pi\]

      The graph is shifted to the left by \(\frac{C}{B}=\frac{\frac{\pi }{2}}{2}=\frac{\pi }{4}\) units. See .

   3. ⓒ \(y=\cos \ x+3\) involves cosine, so we use the form \[y=A\ \cos (Bt\pm C)+D\]

      Amplitude is \(|A|,\) so the amplitude is 1. The period is \(2\pi .\) See . This is the standard cosine function shifted up three units.

3. What are the amplitude and period of the function \(y=3\ \cos (3\pi x)?\)

   Kuratidza mhinduro

   The amplitude is \(3,\) and the period is \(\frac{2}{3}.\)

4. Graph the function \(y=-4\ \cos (\pi x)\) using amplitude, period, and key points.

   Kuratidza mhinduro

   The amplitude is \(|-4|=4.\) The period is \(\frac{2\pi }{\omega }=\frac{2\pi }{\pi }=2.\) (Recall that we sometimes refer to \(B\) as \(\omega .)\) One cycle of the graph can be drawn over the interval \([0,2].\) To find the key points, we divide the period by 4. Make a table similar to , starting with \(x=0\) and then adding \(\frac{1}{2}\) successively to \(x\) and calculate \(y.\) See the graph in .

   \(x\)	\(0\)	\(\frac{1}{2}\)	\(1\)	\(\frac{3}{2}\)	\(2\)
   \(y=-4\ \cos (\pi x)\)	\(-4\)	\(0\)	\(4\)	\(0\)	\(-4\)

5. Graph the function \(y=3\ \sin (3x)\) using the amplitude, period, and five key points.

   Kuratidza mhinduro

   x	\(3\sin (3x)\)
   0	0
   \(\frac{\pi }{6}\)	3
   \(\frac{\pi }{3}\)	0
   \(\frac{\pi }{2}\)	\(-3\)
   \(\frac{2\pi }{3}\)	0

6. The average monthly temperatures for a small town in Oregon are given in . Find a sinusoidal function of the form \(y=A\ \sin (Bt-C)+D\) that fits the data (round to the nearest tenth) and sketch the graph.

   Month	Temperature, \({}^{\text{o}}\text{F}\)
   January	42.5
   February	44.5
   March	48.5
   April	52.5
   May	58
   June	63
   July	68.5
   August	69
   September	64.5
   October	55.5
   November	46.5
   December	43.5

   Kuratidza mhinduro

   Recall that amplitude is found using the formula

   \[A=\frac{\text{largest value }-\text{smallest value}}{2}\]

   Thus, the amplitude is

   \[\begin{array}{l}|A|=\frac{69-42.5}{2} \\ =13.25\end{array}\]

   The data covers a period of 12 months, so \(\frac{2\pi }{B}=12\) which gives \(B=\frac{2\pi }{12}=\frac{\pi }{6}.\)

   The vertical shift is found using the following equation.

   \[D=\frac{\text{highest value}+\text{lowest value}}{2}\]

   
   Thus, the vertical shift is

   \[\begin{array}{l}\begin{array}{l}D=\frac{69+42.5}{2} \\ =55.8\end{array}\end{array}\]

   So far, we have the equation \(y=13.3\ \sin (\frac{\pi }{6}x-C)+55.8.\)

   To find the horizontal shift, we input the \(x\) and \(y\) values for the first month and solve for \(C.\)

   \[\begin{array}{ll}\ 42.5=13.3\ \sin (\frac{\pi }{6}(1)-C)+55.8 & \\ -13.3=13.3\ \sin (\frac{\pi }{6}-C) & \\ -1=\sin (\frac{\pi }{6}-C) & \sin \theta =-1\to \theta =-\frac{\pi }{2} \\ \frac{\pi }{6}-C=-\frac{\pi }{2} & \\ \frac{\pi }{6}+\frac{\pi }{2}=C & \\ =\frac{2\pi }{3} & \end{array}\]

   We have the equation \(y=13.3\ \sin (\frac{\pi }{6}x-\frac{2\pi }{3})+55.8.\) See the graph in .

7. The hour hand of the large clock on the wall in Union Station measures 24 inches in length. At noon, the tip of the hour hand is 30 inches from the ceiling. At 3 PM, the tip is 54 inches from the ceiling, and at 6 PM, 78 inches. At 9 PM, it is again 54 inches from the ceiling, and at midnight, the tip of the hour hand returns to its original position 30 inches from the ceiling. Let \(y\) equal the distance from the tip of the hour hand to the ceiling \(x\) hours after noon. Find the equation that models the motion of the clock and sketch the graph.

   Kuratidza mhinduro

   Begin by making a table of values as shown in .

   \(x\)	\(y\)	Points to plot
   Noon	30 in	\((0,30)\)
   3 PM	54 in	\((3,54)\)
   6 PM	78 in	\((6,78)\)
   9 PM	54 in	\((9,54)\)
   Midnight	30 in	\((12,30)\)

   To model an equation, we first need to find the amplitude.

   \[\begin{array}{l} \\ \begin{array}{l}|A|=|\frac{78-30}{2}| \\ =24\end{array}\end{array}\]

   The clock’s cycle repeats every 12 hours. Thus,

   \[\begin{array}{l} \\ \begin{array}{l}B=\frac{2\pi }{12} \\ =\frac{\pi }{6}\end{array}\end{array}\]

   The vertical shift is

   \[\begin{array}{l} \\ \begin{array}{l}D=\frac{78+30}{2} \\ =54\end{array}\end{array}\]

   There is no horizontal shift, so \(C=0.\) Since the function begins with the minimum value of \(y\) when \(x=0\) (as opposed to the maximum value), we will use the cosine function with the negative value for \(A.\) In the form \(y=A\ \cos (Bx\pm C)+D,\) the equation is

   \[y=-24\ \cos (\frac{\pi }{6}x)+54\]

   See .

8. The height of the tide in a small beach town is measured along a seawall. Water levels oscillate between 7 feet at low tide and 15 feet at high tide. On a particular day, low tide occurred at 6 AM and high tide occurred at noon. Approximately every 12 hours, the cycle repeats. Find an equation to model the water levels.

   Kuratidza mhinduro

   As the water level varies from 7 ft to 15 ft, we can calculate the amplitude as

   \[\begin{array}{l} \\ \begin{array}{l}|A|=|\frac{(15-7)}{2}| \\ =4\end{array}\end{array}\]

   The cycle repeats every 12 hours; therefore, \(B\) is

   \[\frac{2\pi }{12}=\frac{\pi }{6}\]

   There is a vertical translation of \(\frac{(15+7)}{2}=11.\) Since the value of the function is at a maximum at \(t=0,\) we will use the cosine function, with the positive value for \(A.\)

   \[y=4\ \cos (\frac{\pi }{6}\ t)+11\]

   See .

9. The daily temperature in the month of March in a certain city varies from a low of \(24\text{ ^{\circ}F}\) to a high of \(40\text{ ^{\circ}F}\text{.}\) Find a sinusoidal function to model daily temperature and sketch the graph. Approximate the time when the temperature reaches the freezing point \(32\text{ ^{\circ}F}\text{.}\) Let \(t=0\) correspond to noon.

   Kuratidza mhinduro

   \(y=8\sin (\frac{\pi }{12}t)+32\)
   The temperature reaches freezing at noon and at midnight.

10. The average person’s blood pressure is modeled by the function \(f(t)=20\ \sin (160\pi t)+100,\) where \(f(t)\) represents the blood pressure at time \(t,\) measured in minutes. Interpret the function in terms of period and frequency. Sketch the graph and find the blood pressure reading.

    Kuratidza mhinduro

    The period is given by

    \[\begin{array}{l}\begin{array}{l}\frac{2\pi }{\omega }=\frac{2\pi }{160\pi } \\ =\frac{1}{80}\end{array}\end{array}\]

    In a blood pressure function, frequency represents the number of heart beats per minute. Frequency is the reciprocal of period and is given by

    \[\begin{array}{l}\frac{\omega }{2\pi }=\frac{160\pi }{2\pi } \\ =80\end{array}\]

    See the graph in .

11. For the given functions,

    1. Find the maximum displacement of an object.
    2. Find the period or the time required for one vibration.
    3. Find the frequency.
    4. Sketch the graph.

    1. ⓐ \(y=5\ \sin (3t)\)
    2. ⓑ \(y=6\ \cos (\pi t)\)
    3. ⓒ \(y=5\ \cos (\frac{\pi }{2}t)\)
    Kuratidza mhinduro

    1. ⓐ \(y=5\ \sin (3t)\)
       1. The maximum displacement is equal to the amplitude, \(|a|,\) which is 5.
       2. The period is \(\frac{2\pi }{\omega }=\frac{2\pi }{3}.\)
       3. The frequency is given as \(\frac{\omega }{2\pi }=\frac{3}{2\pi }.\)
       4. See . The graph indicates the five key points.
    2. ⓑ \(y=6\ \cos (\pi t)\)
       1. The maximum displacement is \(6.\)
       2. The period is \(\frac{2\pi }{\omega }=\frac{2\pi }{\pi }=2.\)
       3. The frequency is \(\frac{\omega }{2\pi }=\frac{\pi }{2\pi }=\frac{1}{2}.\)
       4. See .
    3. ⓒ \(y=5\ \cos (\frac{\pi }{2})\ t\)
       1. The maximum displacement is \(5.\)
       2. The period is \(\frac{2\pi }{\omega }=\frac{2\pi }{\frac{\pi }{2}}=4.\)
       3. The frequency is \(\frac{1}{4}.\)
       4. See .

12. Model the equations that fit the two scenarios and use a graphing utility to graph the functions: Two mass-spring systems exhibit damped harmonic motion at a frequency of \(0.5\) cycles per second. Both have an initial displacement of 10 cm. The first has a damping factor of \(0.5\) and the second has a damping factor of \(0.1.\)

    Kuratidza mhinduro

    At time \(t=0,\) the displacement is the maximum of 10 cm, which calls for the cosine function. The cosine function will apply to both models.

    We are given the frequency \(f=\frac{\omega }{2\pi }\) of 0.5 cycles per second. Thus,

    \[\begin{array}{l}\ \frac{\omega }{2\pi }=0.5 \\ \omega =(0.5)2\pi \\ =\pi \end{array}\]

    The first spring system has a damping factor of \(c=0.5.\) Following the general model for damped harmonic motion, we have

    \[f(t)=10{e}^{-0.5t}\cos (\pi t)\]

    models the motion of the first spring system.

    The second spring system has a damping factor of \(c=0.1\) and can be modeled as

    \[f(t)=10{e}^{-0.1t}\cos (\pi t)\]

    models the motion of the second spring system.

13. Find and graph a function of the form \(y=a{e}^{-ct}\cos (\omega t)\) that models the information given.

    1. ⓐ \(a=20,c=0.05,p=4\)
    2. ⓑ \(a=2,c=1.5,f=3\)
    Kuratidza mhinduro

    Substitute the given values into the model. Recall that period is \(\frac{2\pi }{\omega }\) and frequency is \(\frac{\omega }{2\pi }.\)

    1. ⓐ \(y=20{e}^{-0.05t}\cos (\frac{\pi }{2}t).\) See .
    2. ⓑ \(y=2{e}^{-1.5t}\cos (6\pi t).\) See .

14. The following equation represents a damped harmonic motion model: \(f(t)=5{e}^{-6t}\cos (4t)\) Find the initial displacement, the damping constant, and the frequency.

    Kuratidza mhinduro

    initial displacement =6, damping constant = -6, frequency = \(\frac{2}{\pi }\)

15. Find and graph a function of the form \(y=a{e}^{-ct}\sin (\omega t)\) that models the information given.

    1. ⓐ \(a=7,c=10,p=\frac{\pi }{6}\)
    2. ⓑ \(a=0.3,c=0.2,f=20\)
    Kuratidza mhinduro

    Calculate the value of \(\omega\) and substitute the known values into the model.

    1. ⓐ As period is \(\frac{2\pi }{\omega },\) we have \[\begin{array}{l}\ \frac{\pi }{6}=\frac{2\pi }{\omega } \\ \omega \pi =6(2\pi ) \\ \omega =12\end{array}\]

       The damping factor is given as 10 and the amplitude is 7. Thus, the model is \(y=7{e}^{-10t}\sin (12t).\) See .

    2. ⓑ As frequency is \(\frac{\omega }{2\pi },\) we have \[\begin{array}{l}\ 20=\frac{\omega }{2\pi } \\ 40\pi =\omega \end{array}\]

       The damping factor is given as \(0.2\) and the amplitude is \(0.3.\) The model is \(y=0.3{e}^{-0.2t}\sin (40\pi t).\) See .

16. Write the equation for damped harmonic motion given \(a=10,c=0.5,\) and \(p=2.\)

    Kuratidza mhinduro

    \(y=10{e}^{-0.5t}\cos (\pi t)\)

17. A spring measuring 10 inches in natural length is compressed by 5 inches and released. It oscillates once every 3 seconds, and its amplitude decreases by 30% every second. Find an equation that models the position of the spring \(t\) seconds after being released.

    Kuratidza mhinduro

    The amplitude begins at 5 in. and deceases 30% each second. Because the spring is initially compressed, we will write A as a negative value. We can write the amplitude portion of the function as

    \[A(t)=5{(1-0.30)}^{t}\]

    We put \({(1-0.30)}^{t}\) in the form \({e}^{ct}\) as follows:

    \[\begin{array}{l}0.7={e}^{c} \\ c=\ln .7 \\ c=-0.357\end{array}\]

    Now let’s address the period. The spring cycles through its positions every 3 seconds, this is the period, and we can use the formula to find omega.

    \[\begin{array}{l} \\ 3=\frac{2\pi }{\omega } \\ \omega =\frac{2\pi }{3}\end{array}\]

    The natural length of 10 inches is the midline. We will use the cosine function, since the spring starts out at its maximum displacement. This portion of the equation is represented as

    \[y=\cos (\frac{2\pi }{3}t)+10\]

    Finally, we put both functions together. Our the model for the position of the spring at \(t\) seconds is given as

    \[y=-5{e}^{-0.357t}\cos (\frac{2\pi }{3}t)+10\]

    See the graph in .

18. A mass suspended from a spring is raised a distance of 5 cm above its resting position. The mass is released at time \(t=0\) and allowed to oscillate. After \(\frac{1}{3}\) second, it is observed that the mass returns to its highest position. Find a function to model this motion relative to its initial resting position.

    Kuratidza mhinduro

    \(y=5\cos (6\pi t)\)

19. A guitar string is plucked and vibrates in damped harmonic motion. The string is pulled and displaced 2 cm from its resting position. After 3 seconds, the displacement of the string measures 1 cm. Find the damping constant.

    Kuratidza mhinduro

    The displacement factor represents the amplitude and is determined by the coefficient \(a{e}^{-ct}\) in the model for damped harmonic motion. The damping constant is included in the term \({e}^{-ct}.\) It is known that after 3 seconds, the local maximum measures one-half of its original value. Therefore, we have the equation

    \[a{e}^{-c(t+3)}=\frac{1}{2}\ a{e}^{-ct}\]

    Use algebra and the laws of exponents to solve for \(c.\)

    \[\begin{array}{lllll}a{e}^{-c(t+3)}=\frac{1}{2}a{e}^{-ct} & \\ {e}^{-ct}⋅{e}^{-3c}=\frac{1}{2}{e}^{-ct}\begin{array}{llll} & & & \end{array} & \text{Divide out }a. \\ {e}^{-3c}=\frac{1}{2} & \text{Divide out }{e}^{-ct}. \\ {e}^{3c}=2 & \text{Take reciprocals}.\end{array}\]

    Then use the laws of logarithms.

    \[\begin{array}{l}{e}^{3c}=2 \\ 3c=\ln \ (2) \\ c=\frac{\ln \ (2)}{3}\end{array}\]

    The damping constant is \(\frac{\ln \ (2)}{3}.\)

20. Graph the function \(f(x)=\cos (2\pi x)\cos (16\pi x).\)

    Kuratidza mhinduro

    The graph produced by this function will be shown in two parts. The first graph will be the exact function \(f(x)\) (see ), and the second graph is the exact function \(f(x)\) plus a bounding function (see . The graphs look quite different.

21. Explain what types of physical phenomena are best modeled by sinusoidal functions. What are the characteristics necessary?

    Kuratidza mhinduro

    Physical behavior should be periodic, or cyclical.

22. What information is necessary to construct a trigonometric model of daily temperature? Give examples of two different sets of information that would enable modeling with an equation.

23. If we want to model cumulative rainfall over the course of a year, would a sinusoidal function be a good model? Why or why not?

    Kuratidza mhinduro

    Since cumulative rainfall is always increasing, a sinusoidal function would not be ideal here.

24. Explain the effect of a damping factor on the graphs of harmonic motion functions.

25. \(x\)	\(y\)
    \(0\)	\(-4\)
    \(3\)	\(-1\)
    \(6\)	\(2\)
    \(9\)	\(-1\)
    \(12\)	\(-4\)
    \(15\)	\(-1\)
    \(18\)	\(2\)

    Kuratidza mhinduro

    \(y=-3\cos (\frac{\pi }{6}x)-1\)

26. \(x\)	\(y\)
    \(0\)	\(5\)
    \(2\)	\(1\)
    \(4\)	\(-3\)
    \(6\)	\(1\)
    \(8\)	\(5\)
    \(10\)	\(1\)
    \(12\)	\(-3\)

27. \(x\)	\(y\)
    \(0\)	\(2\)
    \(\frac{\pi }{4}\)	\(7\)
    \(\frac{\pi }{2}\)	\(2\)
    \(\frac{3\pi }{4}\)	\(-3\)
    \(\pi\)	\(2\)
    \(\frac{5\pi }{4}\)	\(7\)
    \(\frac{3\pi }{2}\)	\(2\)

    Kuratidza mhinduro

    \(5\sin (2x)+2\)

28. \(x\)	\(y\)
    \(0\)	\(1\)
    \(1\)	\(-3\)
    \(2\)	\(-7\)
    \(3\)	\(-3\)
    \(4\)	\(1\)
    \(5\)	\(-3\)
    \(6\)	\(-7\)

    Kuratidza mhinduro

    \(y=4-6\cos (\frac{x\pi }{2})\)

29. \(x\)	\(y\)
    \(0\)	\(-2\)
    \(1\)	\(4\)
    \(2\)	\(10\)
    \(3\)	\(4\)
    \(4\)	\(-2\)
    \(5\)	\(4\)
    \(6\)	\(10\)

30. \(x\)	\(y\)
    \(0\)	\(5\)
    \(1\)	\(-3\)
    \(2\)	\(5\)
    \(3\)	\(13\)
    \(4\)	\(5\)
    \(5\)	\(-3\)
    \(6\)	\(5\)

    Kuratidza mhinduro

    \(y=\tan (\frac{x\pi }{8})\)

31. \(x\)	\(y\)
    \(-3\)	\(-1-\sqrt{2}\)
    \(-2\)	\(-1\)
    \(-1\)	\(1-\sqrt{2}\)
    \(0\)	\(0\)
    \(1\)	\(\sqrt{2}-1\)
    \(2\)	\(1\)
    \(3\)	\(\sqrt{2}+1\)

32. \(x\)	\(y\)
    \(-1\)	\(\sqrt{3}-2\)
    \(0\)	\(0\)
    \(1\)	\(2-\sqrt{3}\)
    \(2\)	\(\frac{\sqrt{3}}{3}\)
    \(3\)	\(1\)
    \(4\)	\(\sqrt{3}\)
    \(5\)	\(2+\sqrt{3}\)

    Kuratidza mhinduro

    \(\tan (\frac{x\pi }{12})\)

33. \(f(x)=-30\ \cos (\frac{x\pi }{6})-20\ {\cos }^{2}(\frac{x\pi }{6})+80\ [0,12]\)

34. \(f(x)=-18\ \cos (\frac{x\pi }{12})-5\ \sin (\frac{x\pi }{12})+100\) on the interval \([0,24]\)

35. \(f(x)=10-\sin (\frac{x\pi }{6})+24\ \tan (\frac{x\pi }{240})\) on the interval \([0,80]\)

36. A city’s average yearly rainfall is currently 20 inches and varies seasonally by 5 inches. Due to unforeseen circumstances, rainfall appears to be decreasing by 15% each year. How many years from now would we expect rainfall to initially reach 0 inches? Note, the model is invalid once it predicts negative rainfall, so choose the first point at which it goes below 0.

37. Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the high temperature of \(105\text{^{\circ}F}\) occurs at 5PM and the average temperature for the day is \(85\text{^{\circ}F}\text{.}\) Find the temperature, to the nearest degree, at 9AM.

    Kuratidza mhinduro

    75 °F

38. Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the high temperature of \(84\text{^{\circ}F}\) occurs at 6PM and the average temperature for the day is \(70\text{^{\circ}F}\text{.}\) Find the temperature, to the nearest degree, at 7AM.

39. Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the temperature varies between \(47\text{^{\circ}F}\) and \(63\text{^{\circ}F}\) during the day and the average daily temperature first occurs at 10 AM. How many hours after midnight does the temperature first reach \(51\text{^{\circ}F?}\)

    Kuratidza mhinduro

    8 a.m.

40. Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the temperature varies between \(64\text{^{\circ}F}\) and \(86\text{^{\circ}F}\) during the day and the average daily temperature first occurs at 12 AM. How many hours after midnight does the temperature first reach \(70\text{^{\circ}F?}\)

Tarisa yako

Parts of this page are adapted from OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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