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Find the amplitude and period of the following functions and graph one cycle.
We will solve these problems according to the models.
We know that $\text{}\left|A\right|\text{}$ is the amplitude, so the amplitude is 2. Period is $\text{}\frac{2\pi}{B},$ so the period is
See the graph in [link] .
Amplitude is $\text{}\left|A\right|,$ so the amplitude is $\text{}\left|-3\right|=3.$ Since $\text{}A\text{}$ is negative, the graph is reflected over the x -axis. Period is $\text{}\frac{2\pi}{B},$ so the period is
The graph is shifted to the left by $\text{}\frac{C}{B}=\frac{\frac{\pi}{2}}{2}=\frac{\pi}{4}\text{}$ units. See [link] .
Amplitude is $\text{}\left|A\right|,\text{}$ so the amplitude is 1. The period is $\text{}2\pi .\text{}$ See [link] . This is the standard cosine function shifted up three units.
What are the amplitude and period of the function $\text{}y=3\text{\hspace{0.17em}}\mathrm{cos}(3\pi x)?$
The amplitude is $\text{}3,$ and the period is $\text{}\frac{2}{3}.$
One method of graphing sinusoidal functions is to find five key points. These points will correspond to intervals of equal length representing $\text{}\frac{1}{4}\text{}$ of the period. The key points will indicate the location of maximum and minimum values. If there is no vertical shift, they will also indicate x -intercepts. For example, suppose we want to graph the function $\text{}y=\mathrm{cos}\text{\hspace{0.17em}}\theta .$ We know that the period is $\text{\hspace{0.17em}}2\pi ,$ so we find the interval between key points as follows.
Starting with $\text{}\theta =0,$ we calculate the first y- value, add the length of the interval $\text{}\frac{\pi}{2}\text{}$ to 0, and calculate the second y -value. We then add $\text{}\frac{\pi}{2}\text{}$ repeatedly until the five key points are determined. The last value should equal the first value, as the calculations cover one full period. Making a table similar to [link] , we can see these key points clearly on the graph shown in [link] .
$\theta $ | $0$ | $\frac{\pi}{2}$ | $\pi $ | $\frac{3\pi}{2}$ | $2\pi $ |
$y=\mathrm{cos}\text{\hspace{0.17em}}\theta $ | $1$ | $0$ | $\mathrm{-1}$ | $0$ | $1$ |
Graph the function $\text{}y=\mathrm{-4}\text{\hspace{0.17em}}\mathrm{cos}\left(\pi x\right)\text{}$ using amplitude, period, and key points.
The amplitude is $|-4|=4.$ The period is $\text{}\frac{2\pi}{\omega}=\frac{2\pi}{\pi}=2.\text{}$ (Recall that we sometimes refer to $\text{}B\text{}$ as $\text{}\omega .)\text{}$ One cycle of the graph can be drawn over the interval $\text{}\left[0,2\right].\text{}$ To find the key points, we divide the period by 4. Make a table similar to [link] , starting with $\text{}x=0\text{}$ and then adding $\text{}\frac{1}{2}\text{}$ successively to $\text{}x\text{}$ and calculate $\text{}y.\text{}$ See the graph in [link] .
$x$ | $0$ | $\frac{1}{2}$ | $1$ | $\frac{3}{2}$ | $2$ |
$y=\mathrm{-4}\text{\hspace{0.17em}}\mathrm{cos}\left(\pi x\right)$ | $\mathrm{-4}$ | $0$ | $4$ | $0$ | $\mathrm{-4}$ |
Graph the function $\text{}y=3\text{\hspace{0.17em}}\mathrm{sin}(3x)\text{}$ using the amplitude, period, and five key points.
x | $3\mathrm{sin}\left(3x\right)$ |
---|---|
0 | 0 |
$\frac{\pi}{6}$ | 3 |
$\frac{\pi}{3}$ | 0 |
$\frac{\pi}{2}$ | $\mathrm{-3}$ |
$\frac{2\pi}{3}$ | 0 |
We will now apply these ideas to problems involving periodic behavior.
The average monthly temperatures for a small town in Oregon are given in [link] . Find a sinusoidal function of the form $\text{\hspace{0.17em}}y=A\text{\hspace{0.17em}}\mathrm{sin}\left(Bt-C\right)+D\text{\hspace{0.17em}}$ 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 |
Recall that amplitude is found using the formula
Thus, the amplitude is
The data covers a period of 12 months, so $\text{\hspace{0.17em}}\frac{2\pi}{B}=12\text{\hspace{0.17em}}$ which gives $\text{\hspace{0.17em}}B=\frac{2\pi}{12}=\frac{\pi}{6}.$
The vertical shift is found using the following equation.
Thus, the vertical shift is
So far, we have the equation $\text{\hspace{0.17em}}y=13.3\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\pi}{6}x-C\right)+\mathrm{55.8.}$
To find the horizontal shift, we input the $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ values for the first month and solve for $\text{\hspace{0.17em}}C.$
We have the equation $\text{\hspace{0.17em}}y=13.3\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\pi}{6}x-\frac{2\pi}{3}\right)+\mathrm{55.8.}\text{\hspace{0.17em}}$ See the graph in [link] .
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