{"id":14246,"date":"2018-09-27T16:42:10","date_gmt":"2018-09-27T16:42:10","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/precalculus\/chapter\/modeling-with-trigonometric-equations\/"},"modified":"2020-05-21T05:11:01","modified_gmt":"2020-05-21T05:11:01","slug":"modeling-with-trigonometric-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/chapter\/modeling-with-trigonometric-equations\/","title":{"raw":"Section 5.6: Modeling with Trigonometric Equations","rendered":"Section 5.6: Modeling with Trigonometric Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li style=\"font-weight: 400\">Determine the amplitude and period of sinusoidal functions.<\/li>\r\n \t<li style=\"font-weight: 400\">Model equations and graph sinusoidal functions.<\/li>\r\n \t<li style=\"font-weight: 400\">Model periodic behavior.<\/li>\r\n \t<li style=\"font-weight: 400\">Model harmonic motion functions.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Determining the Amplitude and Period of a Sinusoidal Function<\/h2>\r\nAny motion that repeats itself in a fixed time period is considered <strong>periodic motion<\/strong> and can be modeled by a <strong>sinusoidal function<\/strong>. The <strong>amplitude<\/strong> of a sinusoidal function is the distance from the midline to the maximum value, or from the midline to the minimum value. The <strong>midline<\/strong> is the average value. Sinusoidal functions oscillate above and below the midline, are periodic, and repeat values in set cycles. Recall from <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/precalctwoxmaster\/chapter\/graphs-of-the-sine-and-cosine-functions\/\" target=\"_blank\" rel=\"noopener\">Graphs of the Sine and Cosine Functions<\/a> that the <strong>period<\/strong> of the sine function and the cosine function is [latex]\\text{ }2\\pi .\\text{ }[\/latex] In other words, for any value of [latex]\\text{ }x[\/latex],\r\n<div style=\"text-align: center\">[latex]\\sin \\left(x\\pm 2\\pi k\\right)=\\sin x\\text{ and }\\cos \\left(x\\pm 2\\pi k\\right)=\\cos x\\text{ where }k\\text{ is an integer}[\/latex]<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Standard Form of Sinusoidal Equations<\/h3>\r\nThe general forms of a sinusoidal equation are given as\r\n<div style=\"text-align: center\">[latex]y=A\\sin \\left(Bt-C\\right)+D\\text{ or }y=A\\cos \\left(Bt-C\\right)+D[\/latex]<\/div>\r\nwhere [latex]\\text{amplitude}=|A|,B[\/latex] is related to period such that the [latex]\\text{ period}=\\frac{2\\pi }{B},C\\text{ }[\/latex] is the phase shift such that [latex]\\text{ }\\frac{C}{B}\\text{ }[\/latex] denotes the horizontal shift, and [latex]\\text{ }D\\text{ }[\/latex] represents the vertical shift from the graph\u2019s parent graph.\r\n\r\nNote that the models are sometimes written as [latex]\\text{ }y=a\\sin \\left(\\omega t\\pm C\\right)+D\\text{ }[\/latex] or [latex]\\text{ }y=a\\cos \\left(\\omega t\\pm C\\right)+D[\/latex], and period is given as [latex]\\text{ }\\frac{2\\pi }{\\omega }[\/latex].\r\n\r\nThe difference between the sine and the cosine graphs is that the sine graph begins with the average value of the function and the cosine graph begins with the maximum or minimum value of the function.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1: Showing How the Properties of a Trigonometric Function Can Transform a Graph<\/h3>\r\nShow the transformation of the graph of [latex]\\text{ }y=\\sin x\\text{ }[\/latex] into the graph of [latex]\\text{ }y=2\\sin \\left(4x-\\frac{\\pi }{2}\\right)+2[\/latex].\r\n\r\n[reveal-answer q=\"308183\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"308183\"]\r\n\r\nConsider the series of graphs in\u00a0Figure 2\u00a0and the way each change to the equation changes the image.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164121\/CNX_Precalc_Figure_07_06_0102.jpg\" alt=\"Five graphs, side by side, each showing a manipulation to the former. (A) has y=sin(x). (B) has y=2sin(x), which has double the amplitude. (C) has y=2sin(4x), which quadrupled the frequency (or quartered the period). (D) has y=2sin(4x-pi\/2), which shifted it on the x-axis by pi\/2. (E) has y=2sin(4x-pi\/2) + 2, which shifted it on the y-axis by 2.\" width=\"975\" height=\"545\" \/> <b>Figure 2.<\/b> (a) The basic graph of [latex]\\text{ }y=\\sin x\\text{ }[\/latex] (b) Changing the amplitude from 1 to 2 generates the graph of [latex]\\text{ }y=2\\sin x.\\text{ }[\/latex] (c) The period of the sine function changes with the value of [latex]\\text{ }B[\/latex], such that [latex]\\text{ period}=\\frac{2\\pi }{B}[\/latex]. Here we have [latex]\\text{ }B=4[\/latex], which translates to a period of [latex]\\text{ }\\frac{\\pi }{2}[\/latex]. The graph completes one full cycle in [latex]\\text{ }\\frac{\\pi }{2}\\text{ }[\/latex] units. (d) The graph displays a horizontal shift equal to [latex]\\text{ }\\frac{C}{B}[\/latex], or [latex]\\text{ }\\frac{\\frac{\\pi }{2}}{4}=\\frac{\\pi }{8}[\/latex]. (e) Finally, the graph is shifted vertically by the value of [latex]\\text{ }D[\/latex]. In this case, the graph is shifted up by 2 units.[\/caption][\/hidden-answer]<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2: Finding the Amplitude and Period of a Function<\/h3>\r\nFind the amplitude and period of the following functions and graph one cycle.\r\n<ol>\r\n \t<li>[latex]y=2\\sin \\left(\\frac{1}{4}x\\right)[\/latex]<\/li>\r\n \t<li>[latex]y=-3\\sin \\left(2x+\\frac{\\pi }{2}\\right)[\/latex]<\/li>\r\n \t<li>[latex]y=\\cos x+3[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"269067\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"269067\"]\r\n\r\nWe will solve these problems according to the models.\r\n<ol>\r\n \t<li>[latex]y=2\\sin \\left(\\frac{1}{4}x\\right)\\text{ }[\/latex] involves sine, so we use the form\r\n<div style=\"text-align: center\">[latex]y=A\\sin \\left(Bt+C\\right)+D[\/latex]<\/div>\r\nWe know that [latex]\\text{ }|A|\\text{ }[\/latex] is the amplitude, so the amplitude is 2. Period is [latex]\\text{ }\\frac{2\\pi }{B}[\/latex], so the period is\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\frac{2\\pi }{B}&amp;=\\frac{2\\pi }{\\frac{1}{4}} \\\\ &amp;=8\\pi \\end{align}[\/latex]<\/div>\r\nSee the graph in Figure 3.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164124\/CNX_Precalc_Figure_07_06_0022.jpg\" alt=\"Graph of y=2sin(1\/4 x) from 0 to 8pi, which is one cycle. The amplitude is 2, and the period is 8pi.\" width=\"487\" height=\"217\" \/> <b>Figure 3<\/b>[\/caption]<\/li>\r\n \t<li>[latex]y=-3\\sin \\left(2x+\\frac{\\pi }{2}\\right)\\text{ }[\/latex] involves sine, so we use the form\r\n<div style=\"text-align: center\">[latex]y=A\\sin \\left(Bt-C\\right)+D[\/latex]<\/div>\r\nAmplitude is [latex]\\text{ }|A|[\/latex], so the amplitude is [latex]\\text{ }|-3|=3[\/latex]. Since [latex]\\text{ }A\\text{ }[\/latex] is negative, the graph is reflected over the <em>x<\/em>-axis. Period is [latex]\\text{ }\\frac{2\\pi }{B}[\/latex], so the period is\r\n<div style=\"text-align: center\">[latex]\\frac{2\\pi }{B}=\\frac{2\\pi }{2}=\\pi [\/latex]<\/div>\r\nThe graph is shifted to the left by [latex]\\text{ }\\frac{C}{B}=\\frac{\\frac{\\pi }{2}}{2}=\\frac{\\pi }{4}\\text{ }[\/latex] units.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164126\/CNX_Precalc_Figure_07_06_0032.jpg\" alt=\"Graph of y=-3sin(2x+pi\/2) from -pi\/4 to 3pi\/2, one cycle. The amplitude is 3, and the period is pi.\" width=\"487\" height=\"292\" \/> <b>Figure 4<\/b>[\/caption]<\/li>\r\n \t<li>[latex]y=\\cos x+3\\text{ }[\/latex] involves cosine, so we use the form\r\n<div style=\"text-align: center\">[latex]y=A\\cos \\left(Bt\\pm C\\right)+D[\/latex]<\/div>\r\nAmplitude is [latex]\\text{ }|A|,\\text{ }[\/latex] so the amplitude is 1. The period is [latex]\\text{ }2\\pi .\\text{ }[\/latex]\u00a0This is the standard cosine function shifted up three units.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164128\/CNX_Precalc_Figure_07_06_0042.jpg\" alt=\"Graph of y=cos(x) + 3 from -pi\/2 to 5pi\/2. The amplitude and period are the same as the normal y=cos(x), but the whole graph is shifted up on the y-axis by 3.\" width=\"487\" height=\"273\" \/> <b>Figure 5<\/b>[\/caption]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nWhat are the amplitude and period of the function [latex]\\text{ }y=3\\cos \\left(3\\pi x\\right)?[\/latex]\r\n\r\n[reveal-answer q=\"934998\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"934998\"]\r\n\r\nThe amplitude is [latex]3[\/latex], and the period is [latex]\\frac{2}{3}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Finding Equations and Graphing Sinusoidal Functions<\/h2>\r\nOne method of graphing sinusoidal functions is to find five key points. These points will correspond to intervals of equal length representing [latex]\\text{ }\\frac{1}{4}\\text{ }[\/latex] 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 <em>x<\/em>-intercepts. For example, suppose we want to graph the function [latex]\\text{ }y=\\cos \\theta [\/latex]. We know that the period is [latex]2\\pi [\/latex], so we find the interval between key points as follows.\r\n<div style=\"text-align: center\">[latex]\\frac{2\\pi }{4}=\\frac{\\pi }{2}[\/latex]<\/div>\r\nStarting with [latex]\\text{ }\\theta =0[\/latex], we calculate the first <em>y-<\/em>value, add the length of the interval [latex]\\text{ }\\frac{\\pi }{2}\\text{ }[\/latex] to 0, and calculate the second <em>y<\/em>-value. We then add [latex]\\text{ }\\frac{\\pi }{2}\\text{ }[\/latex] 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 the one below, we can see these key points clearly on the graph shown in Figure 6.\r\n<table id=\"Table_07_07_01\" summary=\"Two rows, six columns. The table has ordered pairs of these column values: (theta, y=cos(theta)), (0,1), (i\/2, 0), (pi,-1), (3pi\/2, 0), (2pi,1).\"><colgroup> <col \/> <col \/> <col \/> <col \/> <col \/> <col \/> <\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]\\theta [\/latex] <\/strong><\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\r\n<td>[latex]\\pi [\/latex]<\/td>\r\n<td>[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\r\n<td>[latex]2\\pi [\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong> [latex]y=\\cos \\theta [\/latex] <\/strong><\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]-1[\/latex]<\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164130\/CNX_Precalc_Figure_07_07_0022.jpg\" alt=\"Graph of y=cos(x) from -pi\/2 to 5pi\/2.\" width=\"487\" height=\"217\" \/> <b>Figure 6<\/b>[\/caption]\r\n\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3: Graphing Sinusoidal Functions Using Key Points<\/h3>\r\nGraph the function [latex]\\text{ }y=-4\\cos \\left(\\pi x\\right)\\text{ }[\/latex] using amplitude, period, and key points.\r\n\r\n[reveal-answer q=\"34331\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"34331\"]\r\n\r\nThe amplitude is [latex] |-4|=4. [\/latex]\u00a0The period is [latex]\\text{ }\\frac{2\\pi }{\\omega }=\\frac{2\\pi }{\\pi }=2.\\text{ }[\/latex] (Recall that we sometimes refer to [latex]\\text{ }B\\text{ }[\/latex] as [latex]\\omega[\/latex].) One cycle of the graph can be drawn over the interval [latex]\\text{ }\\left[0,2\\right].\\text{ }[\/latex] To find the key points, we divide the period by 4. Make a table similar to the one below, starting with [latex]\\text{ }x=0\\text{ }[\/latex] and then adding [latex]\\text{ }\\frac{1}{2}\\text{ }[\/latex] successively to [latex]\\text{ }x\\text{ }[\/latex] and calculate [latex]\\text{ }y.\\text{ }[\/latex] See the graph in Figure 7.\r\n<table id=\"Table_07_07_02\" summary=\"Two rows, six columns. The table has ordered pairs of these column values: (theta, y=-4cos(pi*x)), (0,-4), (1\/2, 0), (1,4), (3\/2, 0), (2, -4).\"><colgroup> <col \/> <col \/> <col \/> <col \/> <col \/> <col \/> <\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]\\frac{3}{2}[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong> [latex]y=-4\\cos \\left(\\pi x\\right)[\/latex] <\/strong><\/td>\r\n<td>[latex]-4[\/latex]<\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]4[\/latex]<\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]-4[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164132\/CNX_Precalc_Figure_07_07_0032.jpg\" alt=\"Graph of y=-4cos(pi*x) using the five key points: intervals of equal length representing 1\/4 of the period. Here, the points are at 0, 1\/2, 1, 3\/2, and 2.\" width=\"487\" height=\"375\" \/> <b>Figure 7<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nGraph the function [latex]\\text{ }y=3\\sin \\left(3x\\right)\\text{ }[\/latex] using the amplitude, period, and five key points.\r\n\r\n[reveal-answer q=\"476139\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"476139\"]\r\n<table id=\"fs-id2353042\" class=\"unnumbered\" summary=\"Graph of y=3sin(3x) using the five key points: intervals of equal length representing 1\/4 of the period. Here, the points are at 0, pi\/6, pi\/3, pi\/2, and 2pi\/3.\">\r\n<thead>\r\n<tr>\r\n<th>x<\/th>\r\n<th>[latex]3\\sin \\left(3x\\right)[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>0<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\frac{\\pi }{6}[\/latex]<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\frac{\\pi }{3}[\/latex]<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\r\n<td>[latex]-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\frac{2\\pi }{3}[\/latex]<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164212\/CNX_Precalc_Figure_07_07_0042.jpg\" alt=\"Graph of y=3sin(3x) using the five key points: intervals of equal length representing 1\/4 of the period. Here, the points are at 0, pi\/6, pi\/3, pi\/2, and 2pi\/3. \" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Modeling Periodic Behavior<\/h2>\r\nWe will now apply these ideas to problems involving periodic behavior.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4: Modeling an Equation and Sketching a Sinusoidal Graph to Fit Criteria<\/h3>\r\nThe average monthly temperatures for a small town in Oregon are given in the table below. Find a sinusoidal function of the form [latex]y=A\\sin \\left(Bt-C\\right)+D[\/latex] that fits the data (round to the nearest tenth) and sketch the graph.\r\n<table id=\"fs-id1336940\" summary=\"Thirteen rows, two columns. The table has ordered pairs of these row values: (Month, Temperature in degrees 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).\"><colgroup> <col \/> <col \/> <\/colgroup>\r\n<thead>\r\n<tr>\r\n<th>Month<\/th>\r\n<th>Temperature, [latex]{}^{\\text{o}}\\text{F}[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>January<\/td>\r\n<td>42.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>February<\/td>\r\n<td>44.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>March<\/td>\r\n<td>48.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>April<\/td>\r\n<td>52.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>May<\/td>\r\n<td>58<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>June<\/td>\r\n<td>63<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>July<\/td>\r\n<td>68.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>August<\/td>\r\n<td>69<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>September<\/td>\r\n<td>64.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>October<\/td>\r\n<td>55.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>November<\/td>\r\n<td>46.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>December<\/td>\r\n<td>43.5<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"270264\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"270264\"]\r\n\r\nRecall that amplitude is found using the formula\r\n<p style=\"text-align: center\">[latex]A=\\frac{\\text{largest value }-\\text{smallest value}}{2}[\/latex]<\/p>\r\nThus, the amplitude is\r\n<p style=\"text-align: center\">[latex]\\begin{align} |A|&amp;=\\frac{69 - 42.5}{2} \\\\ &amp;=13.25 \\end{align}[\/latex]<\/p>\r\nThe data covers a period of 12 months, so [latex]\\frac{2\\pi }{B}=12[\/latex] which gives [latex]B=\\frac{2\\pi }{12}=\\frac{\\pi }{6}[\/latex].\r\n\r\nThe vertical shift is found using the following equation.\r\n<p style=\"text-align: center\">[latex]D=\\frac{\\text{highest value}+\\text{lowest value}}{2}[\/latex]<\/p>\r\nThus, the vertical shift is\r\n<p style=\"text-align: center\">[latex]\\begin{align}D&amp;=\\frac{69+42.5}{2} \\\\ &amp;=55.8\\end{align}[\/latex]<\/p>\r\nSo far, we have the equation [latex]y=13.3\\sin \\left(\\frac{\\pi }{6}x-C\\right)+55.8[\/latex].\r\n\r\nTo find the horizontal shift, we input the [latex]x[\/latex] and [latex]y[\/latex] values for the first month and solve for [latex]C[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}42.5=13.3\\sin \\left(\\frac{\\pi }{6}\\left(1\\right)-C\\right)+55.8 \\\\ -13.3=13.3\\sin \\left(\\frac{\\pi }{6}-C\\right) \\\\ -1=\\sin \\left(\\frac{\\pi }{6}-C\\right) \\\\ \\sin \\theta =-1\\to \\theta =-\\frac{\\pi }{2}\\end{gathered}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align} \\frac{\\pi }{6}-C&amp;=-\\frac{\\pi }{2}\\\\ \\frac{\\pi }{6}+\\frac{\\pi }{2}&amp;=C \\\\ &amp;=\\frac{2\\pi }{3} \\end{align}[\/latex]<\/p>\r\nWe have the equation [latex]y=13.3\\sin \\left(\\frac{\\pi }{6}x-\\frac{2\\pi }{3}\\right)+55.8[\/latex]. See the graph in Figure 8.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164135\/CNX_Precalc_Figure_07_06_0112.jpg\" alt=\"Graph of the equation y=13.3sin(pi\/6 x - 2pi\/3) + 55.8. The average value is a dotted horizontal line y=55.8, and the amplitude is 13.3\" width=\"487\" height=\"331\" \/> <b>Figure 8<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 5: Describing Periodic Motion<\/h3>\r\nThe 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 [latex]y[\/latex] equal the distance from the tip of the hour hand to the ceiling [latex]x[\/latex] hours after noon. Find the equation that models the motion of the clock and sketch the graph.\r\n\r\n[reveal-answer q=\"45024\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"45024\"]\r\n\r\nBegin by making a table of values as shown in the table below.\r\n<table id=\"fs-id1992294\" summary=\"Six rows, three columns. The table has ordered pairs of these row values: (X, y, Points to plot), (Noon, 30 in, (0,30)), (3 PM, 54 in, (3,54)), (6 PM, 78 in, (6,8)), (9 PM, 54 in, (9,54)), (Midnight, 30 in, (12,30)).\"><colgroup> <col \/> <col \/> <col \/> <\/colgroup>\r\n<thead>\r\n<tr>\r\n<th>[latex]x[\/latex]<\/th>\r\n<th>[latex]y[\/latex]<\/th>\r\n<th>Points to plot<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Noon<\/td>\r\n<td>30 in<\/td>\r\n<td>[latex]\\left(0,30\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3 PM<\/td>\r\n<td>54 in<\/td>\r\n<td>[latex]\\left(3,54\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>6 PM<\/td>\r\n<td>78 in<\/td>\r\n<td>[latex]\\left(6,78\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>9 PM<\/td>\r\n<td>54 in<\/td>\r\n<td>[latex]\\left(9,54\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Midnight<\/td>\r\n<td>30 in<\/td>\r\n<td>[latex]\\left(12,30\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nTo model an equation, we first need to find the amplitude.\r\n<p style=\"text-align: center\">[latex]\\begin{align}|A|&amp;=\\left\\rvert\\frac{78 - 30}{2}\\right\\rvert \\\\ &amp;=24 \\end{align}[\/latex]<\/p>\r\nThe clock\u2019s cycle repeats every 12 hours. Thus,\r\n<p style=\"text-align: center\">[latex]\\begin{align}B&amp;=\\frac{2\\pi }{12} \\\\ &amp;=\\frac{\\pi }{6} \\end{align}[\/latex]<\/p>\r\nThe vertical shift is\r\n<p style=\"text-align: center\">[latex]\\begin{align}D&amp;=\\frac{78+30}{2} \\\\ &amp;=54 \\end{align}[\/latex]<\/p>\r\nThere is no horizontal shift, so [latex]C=0[\/latex]. Since the function begins with the minimum value of [latex]y[\/latex] when [latex]x=0[\/latex] (as opposed to the maximum value), we will use the cosine function with the negative value for [latex]A[\/latex]. In the form [latex]y=A\\cos \\left(Bx\\pm C\\right)+D[\/latex], the equation is\r\n<p style=\"text-align: center\">[latex]y=-24\\cos \\left(\\frac{\\pi }{6}x\\right)+54[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164137\/CNX_Precalc_Figure_07_06_0132.jpg\" alt=\"Graph of the function y=-24cos(pi\/6 x)+54 using the five key points: (0,30), (3,54), (6,78), (9,54), (12,30).\" width=\"487\" height=\"392\" \/> <b>Figure 9<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 6: Determining a Model for Tides<\/h3>\r\nThe 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.\r\n\r\n[reveal-answer q=\"705973\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"705973\"]\r\n\r\nAs the water level varies from 7 ft to 15 ft, we can calculate the amplitude as\r\n<p style=\"text-align: center\">[latex]\\begin{align}|A|&amp;=\\left\\rvert\\frac{\\left(15 - 7\\right)}{2}\\right\\rvert \\\\ &amp;=4 \\end{align}[\/latex]<\/p>\r\nThe cycle repeats every 12 hours; therefore, [latex]B[\/latex] is\r\n<p style=\"text-align: center\">[latex]\\frac{2\\pi }{12}=\\frac{\\pi }{6}[\/latex]<\/p>\r\nThere is a vertical translation of [latex]\\frac{\\left(15+8\\right)}{2}=11.5[\/latex]. Since the value of the function is at a maximum at [latex]t=0[\/latex], we will use the cosine function, with the positive value for [latex]A[\/latex].\r\n<p style=\"text-align: center\">[latex]y=4\\cos \\left(\\frac{\\pi }{6}\\right)t+11[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164139\/CNX_Precalc_Figure_07_06_0152.jpg\" alt=\"Graph of the function y=4cos(pi\/6 t) + 11 from 0 to 12. The midline is y=11, three key points are (0,15), (6,7), and (12, 15).\" width=\"487\" height=\"330\" \/> <b>Figure 10<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nThe daily temperature in the month of March in a certain city varies from a low of [latex]24^\\circ\\text{F}[\/latex] to a high of [latex]40^\\circ\\text{F}[\/latex]. Find a sinusoidal function to model daily temperature and sketch the graph. Approximate the time when the temperature reaches the freezing point [latex]32^\\circ\\text{F}[\/latex]. Let [latex]t=0[\/latex] correspond to noon.\r\n\r\n[reveal-answer q=\"130521\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"130521\"]\r\n\r\n[latex]y=8\\sin \\left(\\frac{\\pi }{12}t\\right)+32[\/latex]\r\nThe temperature reaches freezing at noon and at midnight.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164214\/CNX_Precalc_Figure_07_06_0162.jpg\" alt=\"Graph of the function y=8sin(pi\/12 t) + 32 for temperature. The midline is at 32. The times when the temperature is at 32 are midnight and noon.\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]152828[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 7: Interpreting the Periodic Behavior Equation<\/h3>\r\nThe average person\u2019s blood pressure is modeled by the function [latex]f\\left(t\\right)=20\\sin \\left(160\\pi t\\right)+100[\/latex], where [latex]f\\left(t\\right)[\/latex] represents the blood pressure at time [latex]t[\/latex],\u00a0measured in minutes. Interpret the function in terms of period and frequency. Sketch the graph and find the blood pressure reading.\r\n\r\n[reveal-answer q=\"393443\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"393443\"]\r\n\r\nThe period is given by\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{2\\pi }{\\omega }&amp;=\\frac{2\\pi }{160\\pi } \\\\ &amp;=\\frac{1}{80} \\end{align}[\/latex]<\/p>\r\nIn a blood pressure function, frequency represents the number of heart beats per minute. Frequency is the reciprocal of period and is given by\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{\\omega }{2\\pi }&amp;=\\frac{160\\pi }{2\\pi }\\\\ &amp;=80\\end{align}[\/latex]<\/p>\r\nSee the graph in Figure 11.\r\n\r\nThe blood pressure reading on the graph is [latex]\\frac{120}{80}\\text{ }\\left(\\frac{\\text{maximum}}{\\text{minimum}}\\right)[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164141\/CNX_Precalc_Figure_07_06_0172.jpg\" alt=\"Graph of the function f(t) = 20sin(160 * pi * t) + 100 for blood pressure. The midline is at 100.\" width=\"487\" height=\"475\" \/> <b>Figure 11<\/b>[\/caption]\r\n<h4>Analysis of the Solution<\/h4>\r\nBlood pressure of [latex]\\frac{120}{80}[\/latex] is considered to be normal. The top number is the maximum or systolic reading, which measures the pressure in the arteries when the heart contracts. The bottom number is the minimum or diastolic reading, which measures the pressure in the arteries as the heart relaxes between beats, refilling with blood. Thus, normal blood pressure can be modeled by a periodic function with a maximum of 120 and a minimum of 80.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div>\r\n\r\n<span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Modeling Harmonic Motion Functions<\/span>\r\n\r\n<\/div>\r\nHarmonic motion is a form of periodic motion, but there are factors to consider that differentiate the two types. While general <strong>periodic motion<\/strong> applications cycle through their periods with no outside interference, <strong>harmonic motion<\/strong> requires a restoring force. Examples of harmonic motion include springs, gravitational force, and magnetic force.\r\n<h3>Simple Harmonic Motion<\/h3>\r\nA type of motion described as <strong>simple harmonic motion<\/strong> involves a restoring force but assumes that the motion will continue forever. Imagine a weighted object hanging on a spring, When that object is not disturbed, we say that the object is at rest, or in equilibrium. If the object is pulled down and then released, the force of the spring pulls the object back toward equilibrium and harmonic motion begins. The restoring force is directly proportional to the displacement of the object from its equilibrium point. When [latex]t=0,d=0[\/latex].\r\n<div class=\"textbox\">\r\n<h3>A General Note: Simple Harmonic Motion<\/h3>\r\nWe see that <strong>simple harmonic motion<\/strong> equations are given in terms of displacement:\r\n<p style=\"text-align: center\">[latex]d=a\\cos \\left(\\omega t\\right)\\text{ or }d=a\\sin \\left(\\omega t\\right)[\/latex]<\/p>\r\nwhere [latex]|a|[\/latex] is the amplitude, [latex]\\frac{2\\pi }{\\omega }[\/latex] is the period, and [latex]\\frac{\\omega }{2\\pi }[\/latex] is the frequency, or the number of cycles per unit of time.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 8: Finding the Displacement, Period, and Frequency, and Graphing a Function<\/h3>\r\nFor the given functions,\r\n<ol>\r\n \t<li>Find the maximum displacement of an object.<\/li>\r\n \t<li>Find the period or the time required for one vibration.<\/li>\r\n \t<li>Find the frequency.<\/li>\r\n \t<li>Sketch the graph.\r\n<ol>\r\n \t<li>[latex]y=5\\sin \\left(3t\\right)[\/latex]<\/li>\r\n \t<li>[latex]y=6\\cos \\left(\\pi t\\right)[\/latex]<\/li>\r\n \t<li>[latex]y=5\\cos \\left(\\frac{\\pi }{2}t\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"65116\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"65116\"]\r\n<ol>\r\n \t<li>[latex]y=5\\sin \\left(3t\\right)[\/latex]\r\n<ol>\r\n \t<li>The maximum displacement is equal to the amplitude, [latex]|a|[\/latex], which is 5.<\/li>\r\n \t<li>The period is [latex]\\frac{2\\pi }{\\omega }=\\frac{2\\pi }{3}[\/latex].<\/li>\r\n \t<li>The frequency is given as [latex]\\frac{\\omega }{2\\pi }=\\frac{3}{2\\pi }[\/latex].<\/li>\r\n \t<li>See Figure 12. The graph indicates the five key points.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164143\/CNX_Precalc_Figure_07_07_0062.jpg\" alt=\"Graph of the function y=5sin(3t) from 0 to 2pi\/3. The five key points are (0,0), (pi\/6, 5), (pi\/3, 0), (pi\/2, -5), (2pi\/3, 0). \" width=\"487\" height=\"443\" \/> <b>Figure 12<\/b>[\/caption]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>[latex]y=6\\cos \\left(\\pi t\\right)[\/latex]\r\n<ol>\r\n \t<li>The maximum displacement is [latex]6[\/latex].<\/li>\r\n \t<li>The period is [latex]\\frac{2\\pi }{\\omega }=\\frac{2\\pi }{\\pi }=2[\/latex].<\/li>\r\n \t<li>The frequency is [latex]\\frac{\\omega }{2\\pi }=\\frac{\\pi }{2\\pi }=\\frac{1}{2}[\/latex].<\/li>\r\n \t<li>See Figure 13.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164145\/CNX_Precalc_Figure_07_07_0072.jpg\" alt=\"Graph of the function y=6cos(pi t) from 0 to 3.\" width=\"487\" height=\"290\" \/> <b>Figure 13.<\/b>[\/caption]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>[latex]y=5\\cos \\left(\\frac{\\pi }{2}\\right)t[\/latex]\r\n<ol>\r\n \t<li>The maximum displacement is [latex]5[\/latex].<\/li>\r\n \t<li>The period is [latex]\\frac{2\\pi }{\\omega }=\\frac{2\\pi }{\\frac{\\pi }{2}}=4[\/latex].<\/li>\r\n \t<li>The frequency is [latex]\\frac{1}{4}[\/latex].<\/li>\r\n \t<li>See Figure 14.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164148\/CNX_Precalc_Figure_07_07_0082.jpg\" alt=\"Graph of the function y=5cos(pi\/2 t) from 0 to 4.\" width=\"487\" height=\"519\" \/> <b>Figure 14<\/b>[\/caption]<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Damped Harmonic Motion<\/h2>\r\nIn reality, a pendulum does not swing back and forth forever, nor does an object on a spring bounce up and down forever. Eventually, the pendulum stops swinging and the object stops bouncing and both return to equilibrium. Periodic motion in which an energy-dissipating force, or damping factor, acts is known as <strong>damped harmonic motion<\/strong>. Friction is typically the damping factor.\r\n\r\nIn physics, various formulas are used to account for the damping factor on the moving object. Some of these are calculus-based formulas that involve derivatives. For our purposes, we will use formulas for basic damped harmonic motion models.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Damped Harmonic Motion<\/h3>\r\nIn <strong>damped harmonic motion<\/strong>, the displacement of an oscillating object from its rest position at time [latex]t[\/latex] is given as\r\n<p style=\"text-align: center\">[latex]f\\left(t\\right)=a{e}^{-ct}\\sin \\left(\\omega t\\right)\\text{or} f\\left(t\\right)=a{e}^{-ct}\\cos \\left(\\omega t\\right)[\/latex]<\/p>\r\nwhere [latex]c[\/latex] is a damping factor, [latex]|a|[\/latex] is the initial displacement and [latex]\\frac{2\\pi }{\\omega }[\/latex] is the period.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 9: Modeling Damped Harmonic Motion<\/h3>\r\nModel 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 [latex]0.5[\/latex] cycles per second. Both have an initial displacement of 10 cm. The first has a damping factor of [latex]0.5[\/latex] and the second has a damping factor of [latex]0.1[\/latex].\r\n\r\n[reveal-answer q=\"574944\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"574944\"]\r\n\r\nAt time [latex]t=0[\/latex], the displacement is the maximum of 10 cm, which calls for the cosine function. The cosine function will apply to both models.\r\n\r\nWe are given the frequency [latex]f=\\frac{\\omega }{2\\pi }[\/latex] of 0.5 cycles per second. Thus,\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{\\omega }{2\\pi }&amp;=0.5 \\\\ \\omega &amp;=\\left(0.5\\right)2\\pi \\\\ &amp;=\\pi \\end{align}[\/latex]<\/p>\r\nThe first spring system has a damping factor of [latex]c=0.5[\/latex]. Following the general model for damped harmonic motion, we have\r\n<p style=\"text-align: center\">[latex]f\\left(t\\right)=10{e}^{-0.5t}\\cos \\left(\\pi t\\right)[\/latex]<\/p>\r\nFigure 15 models the motion of the first spring system.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164150\/CNX_Precalc_Figure_07_07_0092.jpg\" alt=\"Graph of the first spring system, f(t) = 10(e^(-.5t))cos(pi*t), which begins with a high amplitude and quickly decreases.\" width=\"487\" height=\"632\" \/> <b>Figure 15<\/b>[\/caption]\r\n\r\nThe second spring system has a damping factor of [latex]c=0.1[\/latex] and can be modeled as\r\n<p style=\"text-align: center\">[latex]f\\left(t\\right)=10{e}^{-0.1t}\\cos \\left(\\pi t\\right)[\/latex]<\/p>\r\nFigure 16\u00a0models the motion of the second spring system.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164152\/CNX_Precalc_Figure_07_07_0102.jpg\" alt=\"Graph of f(t) = 10(e^(-.1t))cos(pi*t), which begins with a high amplitude and slowly decreases (but has a high frequency).\" width=\"487\" height=\"522\" \/> <b>Figure 16<\/b>[\/caption]\r\n<h4>Analysis of the Solution<\/h4>\r\nNotice the differing effects of the damping constant. The local maximum and minimum values of the function with the damping factor [latex]c=0.5[\/latex] decreases much more rapidly than that of the function with [latex]c=0.1[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 10: Finding a Cosine Function that Models Damped Harmonic Motion<\/h3>\r\nFind and graph a function of the form [latex]y=a{e}^{-ct}\\cos \\left(\\omega t\\right)[\/latex] that models the information given.\r\n<ol>\r\n \t<li>[latex]a=20,c=0.05,p=4[\/latex]<\/li>\r\n \t<li>[latex]a=2,c=1.5,f=3[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"923862\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"923862\"]\r\n\r\nSubstitute the given values into the model. Recall that period is [latex]\\frac{2\\pi }{\\omega }[\/latex] and frequency is [latex]\\frac{\\omega }{2\\pi }[\/latex].\r\n<ol>\r\n \t<li>[latex]y=20{e}^{-0.05t}\\cos \\left(\\frac{\\pi }{2}t\\right)[\/latex]. See Figure 17.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164155\/CNX_Precalc_Figure_07_07_0112.jpg\" alt=\"Graph of f(t) = 20(e^(-.05t))cos(pi\/2 * t), which begins with a high amplitude and slowly decreases.\" width=\"487\" height=\"361\" \/> <b>Figure 17<\/b>[\/caption]<\/li>\r\n \t<li>[latex]y=2{e}^{-1.5t}\\cos \\left(6\\pi t\\right)[\/latex]. See Figure 18.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164157\/CNX_Precalc_Figure_07_07_0122.jpg\" alt=\"Graph of f(t) = 2(e^(-1.5t))cos(6pi * t), which begins with a small amplitude and quickly decreases to almost a straight line.\" width=\"487\" height=\"513\" \/> <b>Figure 18<\/b>[\/caption]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nThe following equation represents a damped harmonic motion model: [latex]\\text{ }f\\left(t\\right)=5{e}^{-6t}\\cos \\left(4t\\right)[\/latex]\r\nFind the initial displacement, the damping constant, and the frequency.\r\n\r\n[reveal-answer q=\"533970\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"533970\"]\r\n\r\ninitial displacement =6, damping constant = -6, frequency = [latex]\\frac{2}{\\pi }[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 11: Finding a Sine Function that Models Damped Harmonic Motion<\/h3>\r\nFind and graph a function of the form [latex]y=a{e}^{-ct}\\sin \\left(\\omega t\\right)[\/latex] that models the information given.\r\n<ol>\r\n \t<li>[latex]a=7,c=10,p=\\frac{\\pi }{6}[\/latex]<\/li>\r\n \t<li>[latex]a=0.3,c=0.2,f=20[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"293857\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"293857\"]\r\n\r\nCalculate the value of [latex]\\omega [\/latex] and substitute the known values into the model.\r\n<ol>\r\n \t<li>As period is [latex]\\frac{2\\pi }{\\omega }[\/latex], we have\r\n<div style=\"text-align: center\">[latex]\\begin{gathered}\\frac{\\pi }{6}=\\frac{2\\pi }{\\omega } \\\\ \\omega \\pi =6\\left(2\\pi \\right) \\\\ \\omega =12 \\end{gathered}[\/latex]<\/div>\r\nThe damping factor is given as 10 and the amplitude is 7. Thus, the model is [latex]y=7{e}^{-10t}\\sin \\left(12t\\right)[\/latex]. See Figure 19.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164200\/CNX_Precalc_Figure_07_07_0132.jpg\" alt=\"Graph of f(t) = 7(e^(-10t))sin(12t), which spikes close to t=0 and quickly becomes almost a straight line.\" width=\"487\" height=\"291\" \/> <b>Figure 19<\/b>[\/caption]<\/li>\r\n \t<li>As frequency is [latex]\\frac{\\omega }{2\\pi }[\/latex], we have\r\n<div style=\"text-align: center\">[latex]\\begin{gathered}20=\\frac{\\omega }{2\\pi } \\\\ 40\\pi =\\omega \\end{gathered}[\/latex]<\/div>\r\nThe damping factor is given as [latex]0.2[\/latex] and the amplitude is [latex]0.3[\/latex]. The model is [latex]y=0.3{e}^{-0.2t}\\sin \\left(40\\pi t\\right)[\/latex]. See Figure 20.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164202\/CNX_Precalc_Figure_07_07_0142.jpg\" alt=\"Graph of f(t) = .3e^(-.2t)sin(40pi*t), which has a small amplitude but quickly decreases to the appearance of a straight line. The frequency is so high that, in this scaling, the function looks like a solid shape. The zoom in cut out of the graph shows the actual sinusoidal image of the function.\" width=\"487\" height=\"258\" \/> <b>Figure 20<\/b>[\/caption]<\/li>\r\n<\/ol>\r\n<h4>Analysis of the Solution<\/h4>\r\nA comparison of the last two examples illustrates how we choose between the sine or cosine functions to model sinusoidal criteria. We see that the cosine function is at the maximum displacement when [latex]t=0[\/latex], and the sine function is at the equilibrium point when [latex]t=0[\/latex]. For example, consider the equation [latex]y=20{e}^{-0.05t}\\cos \\left(\\frac{\\pi }{2}t\\right)[\/latex] from Example 9. We can see from the graph that when [latex]t=0,y=20[\/latex], which is the initial amplitude. Check this by setting [latex]t=0[\/latex] in the cosine equation:\r\n<p style=\"text-align: center\">[latex]\\begin{align}y&amp;=20{e}^{-0.05\\left(0\\right)}\\cos \\left(\\frac{\\pi }{2}\\right)\\left(0\\right) \\\\ &amp;=20\\left(1\\right)\\left(1\\right) \\\\ &amp;=20\\end{align}[\/latex]<\/p>\r\nUsing the sine function yields\r\n<p style=\"text-align: center\">[latex]\\begin{align}y&amp;=20{e}^{-0.05\\left(0\\right)}\\sin \\left(\\frac{\\pi }{2}\\right)\\left(0\\right) \\\\ &amp;=20\\left(1\\right)\\left(0\\right) \\\\ &amp;=0 \\end{align}[\/latex]<\/p>\r\nThus, cosine is the correct function.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nWrite the equation for damped harmonic motion given [latex]a=10,c=0.5[\/latex], and [latex]p=2[\/latex].\r\n\r\n[reveal-answer q=\"605857\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"605857\"]\r\n\r\n[latex]y=10{e}^{-0.5t}\\cos \\left(\\pi t\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173786[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 12: Modeling the Oscillation of a Spring<\/h3>\r\nA 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 [latex]t[\/latex] seconds after being released.\r\n\r\n[reveal-answer q=\"958106\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"958106\"]\r\n\r\nThe amplitude begins at 5 in. and deceases 30% each second. Because the spring is initially compressed, we will write <em>A<\/em> as a negative value. We can write the amplitude portion of the function as\r\n<p style=\"text-align: center\">[latex]A\\left(t\\right)=5{\\left(1 - 0.30\\right)}^{t}=[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]A\\left(t\\right)=5{\\left(0.70\\right)}^{t}=[\/latex]<\/p>\r\nWe put [latex]{\\left(0.70\\right)}^{t}[\/latex] in the form [latex]{e}^{ct}[\/latex] as follows:\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}0.7={e}^{c} \\\\ c=\\mathrm{ln}(0.7) \\\\ c=-0.357 \\end{gathered}[\/latex]<\/p>\r\nNow let\u2019s 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.\r\n<p style=\"text-align: center\">[latex]\\begin{gathered} 3=\\frac{2\\pi }{\\omega }\\\\ \\omega =\\frac{2\\pi }{3}\\end{gathered}[\/latex]<\/p>\r\nThe 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\r\n<p style=\"text-align: center\">[latex]y=\\cos \\left(\\frac{2\\pi }{3}t\\right)+10[\/latex]<\/p>\r\nFinally, we put both functions together. Our the model for the position of the spring at [latex]t[\/latex] seconds is given as\r\n<p style=\"text-align: center\">[latex]y=-5{e}^{-0.357t}\\cos \\left(\\frac{2\\pi }{3}t\\right)+10[\/latex]<\/p>\r\nSee the graph in Figure 21.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164204\/CNX_Precalc_Figure_07_06_0182.jpg\" alt=\"Graph of the function y = -5e^(-.35t)cos(2pi\/3 t) + 10 from 0 to 24. It starts out as waves with a high amplitude and decreases to almost a straight line very quickly.\" width=\"487\" height=\"282\" \/> <b>Figure 21<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nA mass suspended from a spring is raised a distance of 5 cm above its resting position. The mass is released at time [latex]t=0[\/latex] and allowed to oscillate. After [latex]\\frac{1}{3}[\/latex] 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.\r\n\r\n[reveal-answer q=\"453909\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"453909\"]\r\n\r\n[latex]y=5\\cos \\left(6\\pi t\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 13: Finding the Value of the Damping Constant <em>c<\/em> According to the Given Criteria<\/h3>\r\nA 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.\r\n\r\n[reveal-answer q=\"619520\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"619520\"]\r\n\r\nThe displacement factor represents the amplitude and is determined by the coefficient [latex]a{e}^{-ct}[\/latex] in the model for damped harmonic motion. The damping constant is included in the term [latex]{e}^{-ct}[\/latex]. It is known that after 3 seconds, the local maximum measures one-half of its original value. Therefore, we have the equation\r\n<p style=\"text-align: center\">[latex]a{e}^{-c\\left(t+3\\right)}=\\frac{1}{2}a{e}^{-ct}[\/latex]<\/p>\r\nUse algebra and the laws of exponents to solve for [latex]c[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{align} a{e}^{-c\\left(t+3\\right)}&amp;=\\frac{1}{2}a{e}^{-ct} \\\\ {e}^{-ct}\\cdot {e}^{-3c}&amp;=\\frac{1}{2}{e}^{-ct} &amp;&amp; \\text{Divide out }a.\\\\ {e}^{-3c}&amp;=\\frac{1}{2}&amp;&amp; \\text{Divide out }{e}^{-ct}. \\\\ {e}^{3c}&amp;=2 &amp;&amp; \\text{Take reciprocals}. \\end{align}[\/latex]<\/p>\r\nThen use the laws of logarithms.\r\n<p style=\"text-align: center\">[latex]\\begin{align}{e}^{3c}&amp;=2 \\\\ 3c&amp;=\\mathrm{ln}2 \\\\ c&amp;=\\frac{\\mathrm{ln}2}{3} \\end{align}[\/latex]<\/p>\r\nThe damping constant is [latex]\\frac{\\mathrm{ln}2}{3}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Bounding Curves in Harmonic Motion<\/h2>\r\nHarmonic motion graphs may be enclosed by bounding curves. When a function has a varying <strong>amplitude<\/strong>, such that the amplitude rises and falls multiple times within a period, we can determine the bounding curves from part of the function.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 14: Graphing an Oscillating Cosine Curve<\/h3>\r\nGraph the function [latex]f\\left(x\\right)=\\cos \\left(2\\pi x\\right)\\cos \\left(16\\pi x\\right)[\/latex].\r\n\r\n[reveal-answer q=\"634563\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"634563\"]\r\n\r\nThe graph produced by this function will be shown in two parts. The first graph will be the exact function [latex]f\\left(x\\right)[\/latex], and the second graph is the exact function [latex]f\\left(x\\right)[\/latex] plus a bounding function. The graphs look quite different.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164207\/CNX_Precalc_Figure_07_07_0152.jpg\" alt=\"Graph of f(x) = cos(2pi*x)cos(16pi*x), a sinusoidal function that increases and decreases its amplitude periodically.\" width=\"487\" height=\"145\" \/> <b>Figure 22<\/b>[\/caption]\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164209\/CNX_Precalc_Figure_07_07_0162.jpg\" alt=\"Graph of f(x) = cos(2pi*x)cos(16pi*x), a sinusoidal function that increases and decreases its amplitude periodically. There is also a bonding function drawn over it in red, which makes the whole image look like a DNA (double helix) piece stretched along the x-axis.\" width=\"487\" height=\"218\" \/> <b>Figure 23<\/b>[\/caption]\r\n<h4>Analysis of the Solution<\/h4>\r\nThe curves [latex]y=\\cos \\left(2\\pi x\\right)[\/latex] and [latex]y=-\\cos \\left(2\\pi x\\right)[\/latex] are bounding curves: they bound the function from above and below, tracing out the high and low points. The harmonic motion graph sits inside the bounding curves. This is an example of a function whose amplitude not only decreases with time, but actually increases and decreases multiple times within a period.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div>\r\n\r\n<span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Key Equations<\/span>\r\n\r\n<\/div>\r\n<table id=\"fs-id2608518\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td>Standard form of sinusoidal equation<\/td>\r\n<td>[latex]y=A\\sin \\left(Bt-C\\right)+D\\text{or}y=A\\cos \\left(Bt-C\\right)+D[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simple harmonic motion<\/td>\r\n<td>[latex]d=a\\cos \\left(\\omega t\\right)\\text{ or }d=a\\sin \\left(\\omega t\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Damped harmonic motion<\/td>\r\n<td>[latex]f\\left(t\\right)=a{e}^{-c}{}^{t}\\sin \\left(\\omega t\\right)\\text{or}f\\left(t\\right)=a{e}^{-ct}\\cos \\left(\\omega t\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>Sinusoidal functions are represented by the sine and cosine graphs. In standard form, we can find the amplitude, period, and horizontal and vertical shifts.<\/li>\r\n \t<li>Use key points to graph a sinusoidal function. The five key points include the minimum and maximum values and the midline values.<\/li>\r\n \t<li>Periodic functions can model events that reoccur in set cycles, like the phases of the moon, the hands on a clock, and the seasons in a year.<\/li>\r\n \t<li>Harmonic motion functions are modeled from given data. Similar to periodic motion applications, harmonic motion requires a restoring force. Examples include gravitational force and spring motion activated by weight.<\/li>\r\n \t<li>Damped harmonic motion is a form of periodic behavior affected by a damping factor. Energy dissipating factors, like friction, cause the displacement of the object to shrink.<\/li>\r\n \t<li>Bounding curves delineate the graph of harmonic motion with variable maximum and minimum values.<\/li>\r\n<\/ul>\r\n<div>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id2617942\" class=\"definition\">\r\n \t<dt>damped harmonic motion<\/dt>\r\n \t<dd id=\"fs-id2617946\">oscillating motion that resembles periodic motion and simple harmonic motion, except that the graph is affected by a damping factor, an energy dissipating influence on the motion, such as friction<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id2617951\" class=\"definition\">\r\n \t<dt>simple harmonic motion<\/dt>\r\n \t<dd id=\"fs-id2617956\">a repetitive motion that can be modeled by periodic sinusoidal oscillation<\/dd>\r\n<\/dl>\r\n<\/div>\r\n&nbsp;\r\n<h2 style=\"text-align: center\">Section 5.6 Homework Exercises<\/h2>\r\n1. Explain what types of physical phenomena are best modeled by sinusoidal functions. What are the characteristics necessary?\r\n\r\n2.\u00a0What 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.\r\n\r\n3. 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?\r\n\r\n4.\u00a0Explain the effect of a damping factor on the graphs of harmonic motion functions.\r\n\r\nFor the following exercises, find a possible formula for the trigonometric function represented by the given table of values.\r\n\r\n5.\r\n<table id=\"fs-id1879688\" class=\"unnumbered\" summary=\"Two columns, eight rows. The table has ordered pairs of these row values: (x,y), (0,-4), (3,-1), (6,2), (9,-1), (12,-4), (15,-1), (18,2).\">\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]-4[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3[\/latex]<\/td>\r\n<td>[latex]-1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]6[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]9[\/latex]<\/td>\r\n<td>[latex]-1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]12[\/latex]<\/td>\r\n<td>[latex]-4[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]15[\/latex]<\/td>\r\n<td>[latex]-1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]18[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n6.\r\n<table id=\"fs-id1514649\" class=\"unnumbered\" summary=\"Two columns, eight rows. The table has ordered pairs of these row values: (x,y), (0,5), (2,1), (4,-3), (6,1), (8,5), (10,1), (12,-3).\">\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]4[\/latex]<\/td>\r\n<td>[latex]-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]6[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]8[\/latex]<\/td>\r\n<td>[latex]5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]10[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]12[\/latex]<\/td>\r\n<td>[latex]-3[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n7.\r\n<table id=\"fs-id2202906\" class=\"unnumbered\" summary=\"Two columns, eight rows. The table has ordered pairs of these row values: (x,y), (0,2), (pi\/4, 7), (pi\/2, 2), (3pi\/4, -3), (pi, 2), (5pi\/4, 7), (3pi\/2, 2).\">\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\frac{\\pi }{4}[\/latex]<\/td>\r\n<td>[latex]7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\frac{3\\pi }{4}[\/latex]<\/td>\r\n<td>[latex]-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\pi [\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\frac{5\\pi }{4}[\/latex]<\/td>\r\n<td>[latex]7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n8.\r\n<table id=\"fs-id1379283\" class=\"unnumbered\" summary=\"Two columns, eight rows. The table has ordered pairs of these row values: (x,y), (0,-1), (pi\/4, 2), (pi\/2, 5), (3pi\/4, 2), (pi, -1), (5pi\/4, 2), (3pi\/2, 5).\">\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\frac{\\pi }{4}[\/latex]<\/td>\r\n<td>[latex]7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\frac{3\\pi }{4}[\/latex]<\/td>\r\n<td>[latex]-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\pi [\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\frac{5\\pi }{4}[\/latex]<\/td>\r\n<td>[latex]7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n9.\r\n<table id=\"fs-id2112947\" class=\"unnumbered\" summary=\"Two columns, eight rows. The table has ordered pairs of these column values: (x,y), (0,1), (1,-3), (2,-7), (3,-3), (4,1), (5,-3), (6,-7).\">\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]-7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3[\/latex]<\/td>\r\n<td>[latex]-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]4[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]5[\/latex]<\/td>\r\n<td>[latex]-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]6[\/latex]<\/td>\r\n<td>[latex]-7[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n10.\r\n<table id=\"fs-id3809810\" class=\"unnumbered\" summary=\"Two columns, eight rows. The table has ordered pairs of these row values: (x,y), (0,-2), (1,4), (2,10), (3,4), (4,-2), (5,4), (6,10).\">\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]-2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]4[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]10[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3[\/latex]<\/td>\r\n<td>[latex]4[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]4[\/latex]<\/td>\r\n<td>[latex]-2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]5[\/latex]<\/td>\r\n<td>[latex]4[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]6[\/latex]<\/td>\r\n<td>[latex]10[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n11.\r\n<table id=\"fs-id2795269\" class=\"unnumbered\" summary=\"Two columns, eight rows. The table has ordered pairs of these row values: (x,y), (0,5), (1,-3), (2,5), (3,13), (4,5), (5,-3), (6,5).\">\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3[\/latex]<\/td>\r\n<td>[latex]13[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]4[\/latex]<\/td>\r\n<td>[latex]5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]5[\/latex]<\/td>\r\n<td>[latex]-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]6[\/latex]<\/td>\r\n<td>[latex]5[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n12.\r\n<table id=\"fs-id2760414\" class=\"unnumbered\" summary=\"Two columns, eight rows. The table has ordered pairs of these row values: (x,y), (-3, -1 - rad2), (-2, -1), (-1, 1 - rad2), (0,0), (1, rad2 - 1), (2,1), (3, rad2 + 1).\">\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-3[\/latex]<\/td>\r\n<td>[latex]-1-\\sqrt{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2[\/latex]<\/td>\r\n<td>[latex]-1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1[\/latex]<\/td>\r\n<td>[latex]1-\\sqrt{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]\\sqrt{2}-1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3[\/latex]<\/td>\r\n<td>[latex]\\sqrt{2}+1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n13.\r\n<table id=\"fs-id2796936\" class=\"unnumbered\" summary=\"Two columns, eight rows. The table has ordered pairs of these row values: (x,y), (-1, rad3 - 2), (0,0), (1, 2- rad3), (2, rad3 \/ 3), (3,1), (4, rad3), (5, 2 + rad3).\">\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1[\/latex]<\/td>\r\n<td>[latex]\\sqrt{3}-2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]2-\\sqrt{3}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]4[\/latex]<\/td>\r\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]5[\/latex]<\/td>\r\n<td>[latex]2+\\sqrt{3}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFor the following exercises, graph the given function, and then find a possible physical process that the equation could model.\r\n\r\n14. [latex]f\\left(x\\right)=-30\\cos \\left(\\frac{x\\pi }{6}\\right)-20{\\cos }^{2}\\left(\\frac{x\\pi }{6}\\right)+80\\left[0,12\\right][\/latex]\r\n\r\n15. [latex]f\\left(x\\right)=-18\\cos \\left(\\frac{x\\pi }{12}\\right)-5\\sin \\left(\\frac{x\\pi }{12}\\right)+100[\/latex] on the interval [latex]\\left[0,24\\right][\/latex]\r\n\r\n16. [latex]f\\left(x\\right)=10-\\sin \\left(\\frac{x\\pi }{6}\\right)+24\\tan \\left(\\frac{x\\pi }{240}\\right)[\/latex] on the interval [latex]\\left[0,80\\right][\/latex]\r\n\r\nFor the following exercise, construct a function modeling behavior and use a calculator to find desired results.\r\n\r\n17. A city\u2019s 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.\r\n\r\nFor the following exercises, construct a sinusoidal function with the provided information, and then solve the equation for the requested values.\r\n\r\n18. Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the high temperature of [latex]105\\text{^\\circ F}[\/latex] occurs at 5PM and the average temperature for the day is [latex]85\\text{^\\circ F}\\text{.}[\/latex] Find the temperature, to the nearest degree, at 9AM.\r\n\r\n19. Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the high temperature of [latex]84\\text{^\\circ F}[\/latex] occurs at 6PM and the average temperature for the day is [latex]70\\text{^\\circ F}\\text{.}[\/latex] Find the temperature, to the nearest degree, at 7AM.\r\n\r\n20.\u00a0Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the temperature varies between [latex]47\\text{^\\circ F}[\/latex] and [latex]63\\text{^\\circ F}[\/latex] during the day and the average daily temperature first occurs at 10 AM. How many hours after midnight does the temperature first reach [latex]51\\text{^\\circ F?}[\/latex]\r\n\r\n21. Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the temperature varies between [latex]64^\\circ\\text{F}[\/latex] and [latex]86^\\circ\\text{F}[\/latex] during the day and the average daily temperature first occurs at 12 AM. How many hours after midnight does the temperature first reach [latex]70^\\circ\\text{F?}[\/latex]\r\n\r\n22.\u00a0A Ferris wheel is 20 meters in diameter and boarded from a platform that is 2 meters above the ground. The six o\u2019clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 6 minutes. How much of the ride, in minutes and seconds, is spent higher than 13 meters above the ground?\r\n\r\n23. A Ferris wheel is 45 meters in diameter and boarded from a platform that is 1 meter above the ground. The six o\u2019clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 10 minutes. How many minutes of the ride are spent higher than 27 meters above the ground? Round to the nearest second\r\n\r\n24.\u00a0The sea ice area around the North Pole fluctuates between about 6 million square kilometers on September 1 to 14 million square kilometers on March 1. Assuming a sinusoidal fluctuation, when are there less than 9 million square kilometers of sea ice? Give your answer as a range of dates, to the nearest day.\r\n\r\n25. The sea ice area around the South Pole fluctuates between about 18 million square kilometers in September to 3 million square kilometers in March. Assuming a sinusoidal fluctuation, when are there more than 15 million square kilometers of sea ice? Give your answer as a range of dates, to the nearest day.\r\n\r\n26.\u00a0During a 90-day monsoon season, daily rainfall can be modeled by sinusoidal functions. If the rainfall fluctuates between a low of 2 inches on day 10 and 12 inches on day 55, during what period is daily rainfall more than 10 inches?\r\n\r\n27. During a 90-day monsoon season, daily rainfall can be modeled by sinusoidal functions. A low of 4 inches of rainfall was recorded on day 30, and overall the average daily rainfall was 8 inches. During what period was daily rainfall less than 5 inches?\r\n\r\n28.\u00a0In a certain region, monthly precipitation peaks at 8 inches on June 1 and falls to a low of 1 inch on December 1. Identify the periods when the region is under flood conditions (greater than 7 inches) and drought conditions (less than 2 inches). Give your answer in terms of the nearest day.\r\n\r\n29. In a certain region, monthly precipitation peaks at 24 inches in September and falls to a low of 4 inches in March. Identify the periods when the region is under flood conditions (greater than 22 inches) and drought conditions (less than 5 inches). Give your answer in terms of the nearest day.\r\n\r\nFor the following exercises, find the amplitude, period, and frequency of the given function.\r\n\r\n30. The displacement [latex]h\\left(t\\right)[\/latex] in centimeters of a mass suspended by a spring is modeled by the function [latex]h\\left(t\\right)=8\\sin \\left(6\\pi t\\right)[\/latex], where [latex]t[\/latex] is measured in seconds. Find the amplitude, period, and frequency of this displacement.\r\n\r\n31. The displacement [latex]h\\left(t\\right)[\/latex] in centimeters of a mass suspended by a spring is modeled by the function [latex]h\\left(t\\right)=11\\sin \\left(12\\pi t\\right)[\/latex], where [latex]t[\/latex] is measured in seconds. Find the amplitude, period, and frequency of this displacement.\r\n\r\n32.\u00a0The displacement [latex]h\\left(t\\right)[\/latex] in centimeters of a mass suspended by a spring is modeled by the function [latex]h\\left(t\\right)=4\\cos \\left(\\frac{\\pi }{2}t\\right)[\/latex], where [latex]t[\/latex] is measured in seconds. Find the amplitude, period, and frequency of this displacement.\r\n\r\nFor the following exercises, construct an equation that models the described behavior.\r\n\r\n33. The displacement [latex]h\\left(t\\right)[\/latex], in centimeters, of a mass suspended by a spring is modeled by the function [latex]h\\left(t\\right)=-5\\cos \\left(60\\pi t\\right)[\/latex], where [latex]t[\/latex] is measured in seconds. Find the amplitude, period, and frequency of this displacement.\r\n\r\nFor the following exercises, construct an equation that models the described behavior.\r\n\r\n34. A deer population oscillates 19 above and below average during the year, reaching the lowest value in January. The average population starts at 800 deer and increases by 160 each year. Find a function that models the population, [latex]P[\/latex], in terms of months since January, [latex]t[\/latex].\r\n\r\n35. A rabbit population oscillates 15 above and below average during the year, reaching the lowest value in January. The average population starts at 650 rabbits and increases by 110 each year. Find a function that models the population, [latex]P[\/latex], in terms of months since January, [latex]t[\/latex].\r\n\r\n36.\u00a0A muskrat population oscillates 33 above and below average during the year, reaching the lowest value in January. The average population starts at 900 muskrats and increases by 7% each month. Find a function that models the population, [latex]P[\/latex], in terms of months since January, [latex]t[\/latex].\r\n\r\n37. A fish population oscillates 40 above and below average during the year, reaching the lowest value in January. The average population starts at 800 fish and increases by 4% each month. Find a function that models the population, [latex]P[\/latex], in terms of months since January, [latex]t[\/latex].\r\n\r\n38.\u00a0A spring attached to the ceiling is pulled 10 cm down from equilibrium and released. The amplitude decreases by 15% each second. The spring oscillates 18 times each second. Find a function that models the distance, [latex]D[\/latex], the end of the spring is from equilibrium in terms of seconds, [latex]t[\/latex], since the spring was released.\r\n\r\n39. A spring attached to the ceiling is pulled 7 cm down from equilibrium and released. The amplitude decreases by 11% each second. The spring oscillates 20 times each second. Find a function that models the distance, [latex]D[\/latex], the end of the spring is from equilibrium in terms of seconds, [latex]t[\/latex], since the spring was released.\r\n\r\n40.\u00a0A spring attached to the ceiling is pulled 17 cm down from equilibrium and released. After 3 seconds, the amplitude has decreased to 13 cm. The spring oscillates 14 times each second. Find a function that models the distance, [latex]D[\/latex], the end of the spring is from equilibrium in terms of seconds, [latex]t[\/latex], since the spring was released.\r\n\r\n41. A spring attached to the ceiling is pulled 19 cm down from equilibrium and released. After 4 seconds, the amplitude has decreased to 14 cm. The spring oscillates 13 times each second. Find a function that models the distance, [latex]D[\/latex], the end of the spring is from equilibrium in terms of seconds, [latex]t[\/latex], since the spring was released.\r\n\r\nFor the following exercises, create a function modeling the described behavior. Then, calculate the desired result using a calculator.\r\n\r\n42. A certain lake currently has an average trout population of 20,000. The population naturally oscillates above and below average by 2,000 every year. This year, the lake was opened to fishermen. If fishermen catch 3,000 fish every year, how long will it take for the lake to have no more trout?\r\n\r\n43. Whitefish populations are currently at 500 in a lake. The population naturally oscillates above and below by 25 each year. If humans overfish, taking 4% of the population every year, in how many years will the lake first have fewer than 200 whitefish?\r\n\r\n44.\u00a0A spring attached to a ceiling is pulled down 11 cm from equilibrium and released. After 2 seconds, the amplitude has decreased to 6 cm. The spring oscillates 8 times each second. Find when the spring first comes between [latex]-0.1[\/latex] and [latex]0.1\\text{ cm,}[\/latex] effectively at rest.\r\n\r\n45. A spring attached to a ceiling is pulled down 21 cm from equilibrium and released. After 6 seconds, the amplitude has decreased to 4 cm. The spring oscillates 20 times each second. Find when the spring first comes between [latex]-0.1[\/latex] and [latex]0.1\\text{ cm,}[\/latex] effectively at rest.\r\n\r\n46.\u00a0Two springs are pulled down from the ceiling and released at the same time. The first spring, which oscillates 8 times per second, was initially pulled down 32 cm from equilibrium, and the amplitude decreases by 50% each second. The second spring, oscillating 18 times per second, was initially pulled down 15 cm from equilibrium and after 4 seconds has an amplitude of 2 cm. Which spring comes to rest first, and at what time? Consider \"rest\" as an amplitude less than [latex]0.1\\text{ cm}\\text{.}[\/latex]\r\n\r\n47. Two springs are pulled down from the ceiling and released at the same time. The first spring, which oscillates 14 times per second, was initially pulled down 2 cm from equilibrium, and the amplitude decreases by 8% each second. The second spring, oscillating 22 times per second, was initially pulled down 10 cm from equilibrium and after 3 seconds has an amplitude of 2 cm. Which spring comes to rest first, and at what time? Consider \"rest\" as an amplitude less than [latex]0.1\\text{ cm}\\text{.}[\/latex]\r\n\r\n48.\u00a0A plane flies 1 hour at 150 mph at [latex]{22}^{\\circ }[\/latex] east of north, then continues to fly for 1.5 hours at 120 mph, this time at a bearing of [latex]{112}^{\\circ }[\/latex] east of north. Find the total distance from the starting point and the direct angle flown north of east.\r\n\r\n49. A plane flies 2 hours at 200 mph at a bearing of [latex]\\text{ }{60}^{\\circ }[\/latex], then continues to fly for 1.5 hours at the same speed, this time at a bearing of [latex]{150}^{\\circ }[\/latex]. Find the distance from the starting point and the bearing from the starting point. Hint: bearing is measured counterclockwise from north.\r\n\r\nFor the following exercises, find a function of the form [latex]y=a{b}^{x}+c\\sin \\left(\\frac{\\pi }{2}x\\right)[\/latex] that fits the given data.\r\n\r\n50.\r\n<table id=\"fs-id2196327\" class=\"unnumbered\" summary=\"Two rows, five columns. The table has ordered pairs of these column values: (x,y), (0,6), (1,29), (2, 96), (3, 379).\">\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\r\n<td>6<\/td>\r\n<td>29<\/td>\r\n<td>96<\/td>\r\n<td>379<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n51.\r\n<table id=\"fs-id1229818\" class=\"unnumbered\" summary=\"Two rows, five columns. The table has ordered pairs of these column values: (x,y), (0,6), (1,34), (2, 150), (3, 746).\">\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\r\n<td>6<\/td>\r\n<td>34<\/td>\r\n<td>150<\/td>\r\n<td>746<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n52.\r\n<table id=\"fs-id2427395\" class=\"unnumbered\" summary=\"Two rows, five columns. The table has ordered pairs of these column values: (x,y), (0,4), (1,0), (2,16), (3,-40).\">\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\r\n<td>4<\/td>\r\n<td>0<\/td>\r\n<td>16<\/td>\r\n<td>-40<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFor the following exercises, find a function of the form [latex]y=a{b}^{x}\\cos \\left(\\frac{\\pi }{2}x\\right)+c[\/latex] that fits the given data.\r\n\r\n53.\r\n<table id=\"fs-id1673271\" class=\"unnumbered\" summary=\"Two rows, five columns. The table has ordered pairs of these column values: (x,y), (0,11), (1,3), (2,1), (3,3).\">\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\r\n<td>11<\/td>\r\n<td>3<\/td>\r\n<td>1<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n54.\r\n<table id=\"fs-id2909366\" class=\"unnumbered\" summary=\"Two rows, five columns. The table has ordered pairs of these column values: (x,y), (0,4), (1,1), (2,-11), (3,1).\">\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\r\n<td>4<\/td>\r\n<td>1<\/td>\r\n<td>\u221211<\/td>\r\n<td>1<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li style=\"font-weight: 400\">Determine the amplitude and period of sinusoidal functions.<\/li>\n<li style=\"font-weight: 400\">Model equations and graph sinusoidal functions.<\/li>\n<li style=\"font-weight: 400\">Model periodic behavior.<\/li>\n<li style=\"font-weight: 400\">Model harmonic motion functions.<\/li>\n<\/ul>\n<\/div>\n<h2>Determining the Amplitude and Period of a Sinusoidal Function<\/h2>\n<p>Any motion that repeats itself in a fixed time period is considered <strong>periodic motion<\/strong> and can be modeled by a <strong>sinusoidal function<\/strong>. The <strong>amplitude<\/strong> of a sinusoidal function is the distance from the midline to the maximum value, or from the midline to the minimum value. The <strong>midline<\/strong> is the average value. Sinusoidal functions oscillate above and below the midline, are periodic, and repeat values in set cycles. Recall from <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/precalctwoxmaster\/chapter\/graphs-of-the-sine-and-cosine-functions\/\" target=\"_blank\" rel=\"noopener\">Graphs of the Sine and Cosine Functions<\/a> that the <strong>period<\/strong> of the sine function and the cosine function is [latex]\\text{ }2\\pi .\\text{ }[\/latex] In other words, for any value of [latex]\\text{ }x[\/latex],<\/p>\n<div style=\"text-align: center\">[latex]\\sin \\left(x\\pm 2\\pi k\\right)=\\sin x\\text{ and }\\cos \\left(x\\pm 2\\pi k\\right)=\\cos x\\text{ where }k\\text{ is an integer}[\/latex]<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Standard Form of Sinusoidal Equations<\/h3>\n<p>The general forms of a sinusoidal equation are given as<\/p>\n<div style=\"text-align: center\">[latex]y=A\\sin \\left(Bt-C\\right)+D\\text{ or }y=A\\cos \\left(Bt-C\\right)+D[\/latex]<\/div>\n<p>where [latex]\\text{amplitude}=|A|,B[\/latex] is related to period such that the [latex]\\text{ period}=\\frac{2\\pi }{B},C\\text{ }[\/latex] is the phase shift such that [latex]\\text{ }\\frac{C}{B}\\text{ }[\/latex] denotes the horizontal shift, and [latex]\\text{ }D\\text{ }[\/latex] represents the vertical shift from the graph\u2019s parent graph.<\/p>\n<p>Note that the models are sometimes written as [latex]\\text{ }y=a\\sin \\left(\\omega t\\pm C\\right)+D\\text{ }[\/latex] or [latex]\\text{ }y=a\\cos \\left(\\omega t\\pm C\\right)+D[\/latex], and period is given as [latex]\\text{ }\\frac{2\\pi }{\\omega }[\/latex].<\/p>\n<p>The difference between the sine and the cosine graphs is that the sine graph begins with the average value of the function and the cosine graph begins with the maximum or minimum value of the function.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Showing How the Properties of a Trigonometric Function Can Transform a Graph<\/h3>\n<p>Show the transformation of the graph of [latex]\\text{ }y=\\sin x\\text{ }[\/latex] into the graph of [latex]\\text{ }y=2\\sin \\left(4x-\\frac{\\pi }{2}\\right)+2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q308183\">Show Solution<\/span><\/p>\n<div id=\"q308183\" class=\"hidden-answer\" style=\"display: none\">\n<p>Consider the series of graphs in\u00a0Figure 2\u00a0and the way each change to the equation changes the image.<\/p>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164121\/CNX_Precalc_Figure_07_06_0102.jpg\" alt=\"Five graphs, side by side, each showing a manipulation to the former. (A) has y=sin(x). (B) has y=2sin(x), which has double the amplitude. (C) has y=2sin(4x), which quadrupled the frequency (or quartered the period). (D) has y=2sin(4x-pi\/2), which shifted it on the x-axis by pi\/2. (E) has y=2sin(4x-pi\/2) + 2, which shifted it on the y-axis by 2.\" width=\"975\" height=\"545\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2.<\/b> (a) The basic graph of [latex]\\text{ }y=\\sin x\\text{ }[\/latex] (b) Changing the amplitude from 1 to 2 generates the graph of [latex]\\text{ }y=2\\sin x.\\text{ }[\/latex] (c) The period of the sine function changes with the value of [latex]\\text{ }B[\/latex], such that [latex]\\text{ period}=\\frac{2\\pi }{B}[\/latex]. Here we have [latex]\\text{ }B=4[\/latex], which translates to a period of [latex]\\text{ }\\frac{\\pi }{2}[\/latex]. The graph completes one full cycle in [latex]\\text{ }\\frac{\\pi }{2}\\text{ }[\/latex] units. (d) The graph displays a horizontal shift equal to [latex]\\text{ }\\frac{C}{B}[\/latex], or [latex]\\text{ }\\frac{\\frac{\\pi }{2}}{4}=\\frac{\\pi }{8}[\/latex]. (e) Finally, the graph is shifted vertically by the value of [latex]\\text{ }D[\/latex]. In this case, the graph is shifted up by 2 units.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Finding the Amplitude and Period of a Function<\/h3>\n<p>Find the amplitude and period of the following functions and graph one cycle.<\/p>\n<ol>\n<li>[latex]y=2\\sin \\left(\\frac{1}{4}x\\right)[\/latex]<\/li>\n<li>[latex]y=-3\\sin \\left(2x+\\frac{\\pi }{2}\\right)[\/latex]<\/li>\n<li>[latex]y=\\cos x+3[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q269067\">Show Solution<\/span><\/p>\n<div id=\"q269067\" class=\"hidden-answer\" style=\"display: none\">\n<p>We will solve these problems according to the models.<\/p>\n<ol>\n<li>[latex]y=2\\sin \\left(\\frac{1}{4}x\\right)\\text{ }[\/latex] involves sine, so we use the form\n<div style=\"text-align: center\">[latex]y=A\\sin \\left(Bt+C\\right)+D[\/latex]<\/div>\n<p>We know that [latex]\\text{ }|A|\\text{ }[\/latex] is the amplitude, so the amplitude is 2. Period is [latex]\\text{ }\\frac{2\\pi }{B}[\/latex], so the period is<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}\\frac{2\\pi }{B}&=\\frac{2\\pi }{\\frac{1}{4}} \\\\ &=8\\pi \\end{align}[\/latex]<\/div>\n<p>See the graph in Figure 3.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164124\/CNX_Precalc_Figure_07_06_0022.jpg\" alt=\"Graph of y=2sin(1\/4 x) from 0 to 8pi, which is one cycle. The amplitude is 2, and the period is 8pi.\" width=\"487\" height=\"217\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<\/li>\n<li>[latex]y=-3\\sin \\left(2x+\\frac{\\pi }{2}\\right)\\text{ }[\/latex] involves sine, so we use the form\n<div style=\"text-align: center\">[latex]y=A\\sin \\left(Bt-C\\right)+D[\/latex]<\/div>\n<p>Amplitude is [latex]\\text{ }|A|[\/latex], so the amplitude is [latex]\\text{ }|-3|=3[\/latex]. Since [latex]\\text{ }A\\text{ }[\/latex] is negative, the graph is reflected over the <em>x<\/em>-axis. Period is [latex]\\text{ }\\frac{2\\pi }{B}[\/latex], so the period is<\/p>\n<div style=\"text-align: center\">[latex]\\frac{2\\pi }{B}=\\frac{2\\pi }{2}=\\pi[\/latex]<\/div>\n<p>The graph is shifted to the left by [latex]\\text{ }\\frac{C}{B}=\\frac{\\frac{\\pi }{2}}{2}=\\frac{\\pi }{4}\\text{ }[\/latex] units.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164126\/CNX_Precalc_Figure_07_06_0032.jpg\" alt=\"Graph of y=-3sin(2x+pi\/2) from -pi\/4 to 3pi\/2, one cycle. The amplitude is 3, and the period is pi.\" width=\"487\" height=\"292\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<\/li>\n<li>[latex]y=\\cos x+3\\text{ }[\/latex] involves cosine, so we use the form\n<div style=\"text-align: center\">[latex]y=A\\cos \\left(Bt\\pm C\\right)+D[\/latex]<\/div>\n<p>Amplitude is [latex]\\text{ }|A|,\\text{ }[\/latex] so the amplitude is 1. The period is [latex]\\text{ }2\\pi .\\text{ }[\/latex]\u00a0This is the standard cosine function shifted up three units.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164128\/CNX_Precalc_Figure_07_06_0042.jpg\" alt=\"Graph of y=cos(x) + 3 from -pi\/2 to 5pi\/2. The amplitude and period are the same as the normal y=cos(x), but the whole graph is shifted up on the y-axis by 3.\" width=\"487\" height=\"273\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 5<\/b><\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>What are the amplitude and period of the function [latex]\\text{ }y=3\\cos \\left(3\\pi x\\right)?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q934998\">Show Solution<\/span><\/p>\n<div id=\"q934998\" class=\"hidden-answer\" style=\"display: none\">\n<p>The amplitude is [latex]3[\/latex], and the period is [latex]\\frac{2}{3}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Finding Equations and Graphing Sinusoidal Functions<\/h2>\n<p>One method of graphing sinusoidal functions is to find five key points. These points will correspond to intervals of equal length representing [latex]\\text{ }\\frac{1}{4}\\text{ }[\/latex] 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 <em>x<\/em>-intercepts. For example, suppose we want to graph the function [latex]\\text{ }y=\\cos \\theta[\/latex]. We know that the period is [latex]2\\pi[\/latex], so we find the interval between key points as follows.<\/p>\n<div style=\"text-align: center\">[latex]\\frac{2\\pi }{4}=\\frac{\\pi }{2}[\/latex]<\/div>\n<p>Starting with [latex]\\text{ }\\theta =0[\/latex], we calculate the first <em>y-<\/em>value, add the length of the interval [latex]\\text{ }\\frac{\\pi }{2}\\text{ }[\/latex] to 0, and calculate the second <em>y<\/em>-value. We then add [latex]\\text{ }\\frac{\\pi }{2}\\text{ }[\/latex] 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 the one below, we can see these key points clearly on the graph shown in Figure 6.<\/p>\n<table id=\"Table_07_07_01\" summary=\"Two rows, six columns. The table has ordered pairs of these column values: (theta, y=cos(theta)), (0,1), (i\/2, 0), (pi,-1), (3pi\/2, 0), (2pi,1).\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/> <\/colgroup>\n<tbody>\n<tr>\n<td><strong> [latex]\\theta[\/latex] <\/strong><\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<td>[latex]\\pi[\/latex]<\/td>\n<td>[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\n<td>[latex]2\\pi[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong> [latex]y=\\cos \\theta[\/latex] <\/strong><\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]-1[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164130\/CNX_Precalc_Figure_07_07_0022.jpg\" alt=\"Graph of y=cos(x) from -pi\/2 to 5pi\/2.\" width=\"487\" height=\"217\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/b><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Graphing Sinusoidal Functions Using Key Points<\/h3>\n<p>Graph the function [latex]\\text{ }y=-4\\cos \\left(\\pi x\\right)\\text{ }[\/latex] using amplitude, period, and key points.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q34331\">Show Solution<\/span><\/p>\n<div id=\"q34331\" class=\"hidden-answer\" style=\"display: none\">\n<p>The amplitude is [latex]|-4|=4.[\/latex]\u00a0The period is [latex]\\text{ }\\frac{2\\pi }{\\omega }=\\frac{2\\pi }{\\pi }=2.\\text{ }[\/latex] (Recall that we sometimes refer to [latex]\\text{ }B\\text{ }[\/latex] as [latex]\\omega[\/latex].) One cycle of the graph can be drawn over the interval [latex]\\text{ }\\left[0,2\\right].\\text{ }[\/latex] To find the key points, we divide the period by 4. Make a table similar to the one below, starting with [latex]\\text{ }x=0\\text{ }[\/latex] and then adding [latex]\\text{ }\\frac{1}{2}\\text{ }[\/latex] successively to [latex]\\text{ }x\\text{ }[\/latex] and calculate [latex]\\text{ }y.\\text{ }[\/latex] See the graph in Figure 7.<\/p>\n<table id=\"Table_07_07_02\" summary=\"Two rows, six columns. The table has ordered pairs of these column values: (theta, y=-4cos(pi*x)), (0,-4), (1\/2, 0), (1,4), (3\/2, 0), (2, -4).\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/> <\/colgroup>\n<tbody>\n<tr>\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]\\frac{3}{2}[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong> [latex]y=-4\\cos \\left(\\pi x\\right)[\/latex] <\/strong><\/td>\n<td>[latex]-4[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]4[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]-4[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164132\/CNX_Precalc_Figure_07_07_0032.jpg\" alt=\"Graph of y=-4cos(pi*x) using the five key points: intervals of equal length representing 1\/4 of the period. Here, the points are at 0, 1\/2, 1, 3\/2, and 2.\" width=\"487\" height=\"375\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 7<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Graph the function [latex]\\text{ }y=3\\sin \\left(3x\\right)\\text{ }[\/latex] using the amplitude, period, and five key points.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q476139\">Show Solution<\/span><\/p>\n<div id=\"q476139\" class=\"hidden-answer\" style=\"display: none\">\n<table id=\"fs-id2353042\" class=\"unnumbered\" summary=\"Graph of y=3sin(3x) using the five key points: intervals of equal length representing 1\/4 of the period. Here, the points are at 0, pi\/6, pi\/3, pi\/2, and 2pi\/3.\">\n<thead>\n<tr>\n<th>x<\/th>\n<th>[latex]3\\sin \\left(3x\\right)[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\frac{\\pi }{6}[\/latex]<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\frac{\\pi }{3}[\/latex]<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<td>[latex]-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\frac{2\\pi }{3}[\/latex]<\/td>\n<td>0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164212\/CNX_Precalc_Figure_07_07_0042.jpg\" alt=\"Graph of y=3sin(3x) using the five key points: intervals of equal length representing 1\/4 of the period. Here, the points are at 0, pi\/6, pi\/3, pi\/2, and 2pi\/3.\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Modeling Periodic Behavior<\/h2>\n<p>We will now apply these ideas to problems involving periodic behavior.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 4: Modeling an Equation and Sketching a Sinusoidal Graph to Fit Criteria<\/h3>\n<p>The average monthly temperatures for a small town in Oregon are given in the table below. Find a sinusoidal function of the form [latex]y=A\\sin \\left(Bt-C\\right)+D[\/latex] that fits the data (round to the nearest tenth) and sketch the graph.<\/p>\n<table id=\"fs-id1336940\" summary=\"Thirteen rows, two columns. The table has ordered pairs of these row values: (Month, Temperature in degrees 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).\">\n<colgroup>\n<col \/>\n<col \/> <\/colgroup>\n<thead>\n<tr>\n<th>Month<\/th>\n<th>Temperature, [latex]{}^{\\text{o}}\\text{F}[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>January<\/td>\n<td>42.5<\/td>\n<\/tr>\n<tr>\n<td>February<\/td>\n<td>44.5<\/td>\n<\/tr>\n<tr>\n<td>March<\/td>\n<td>48.5<\/td>\n<\/tr>\n<tr>\n<td>April<\/td>\n<td>52.5<\/td>\n<\/tr>\n<tr>\n<td>May<\/td>\n<td>58<\/td>\n<\/tr>\n<tr>\n<td>June<\/td>\n<td>63<\/td>\n<\/tr>\n<tr>\n<td>July<\/td>\n<td>68.5<\/td>\n<\/tr>\n<tr>\n<td>August<\/td>\n<td>69<\/td>\n<\/tr>\n<tr>\n<td>September<\/td>\n<td>64.5<\/td>\n<\/tr>\n<tr>\n<td>October<\/td>\n<td>55.5<\/td>\n<\/tr>\n<tr>\n<td>November<\/td>\n<td>46.5<\/td>\n<\/tr>\n<tr>\n<td>December<\/td>\n<td>43.5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q270264\">Show Solution<\/span><\/p>\n<div id=\"q270264\" class=\"hidden-answer\" style=\"display: none\">\n<p>Recall that amplitude is found using the formula<\/p>\n<p style=\"text-align: center\">[latex]A=\\frac{\\text{largest value }-\\text{smallest value}}{2}[\/latex]<\/p>\n<p>Thus, the amplitude is<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align} |A|&=\\frac{69 - 42.5}{2} \\\\ &=13.25 \\end{align}[\/latex]<\/p>\n<p>The data covers a period of 12 months, so [latex]\\frac{2\\pi }{B}=12[\/latex] which gives [latex]B=\\frac{2\\pi }{12}=\\frac{\\pi }{6}[\/latex].<\/p>\n<p>The vertical shift is found using the following equation.<\/p>\n<p style=\"text-align: center\">[latex]D=\\frac{\\text{highest value}+\\text{lowest value}}{2}[\/latex]<\/p>\n<p>Thus, the vertical shift is<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}D&=\\frac{69+42.5}{2} \\\\ &=55.8\\end{align}[\/latex]<\/p>\n<p>So far, we have the equation [latex]y=13.3\\sin \\left(\\frac{\\pi }{6}x-C\\right)+55.8[\/latex].<\/p>\n<p>To find the horizontal shift, we input the [latex]x[\/latex] and [latex]y[\/latex] values for the first month and solve for [latex]C[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}42.5=13.3\\sin \\left(\\frac{\\pi }{6}\\left(1\\right)-C\\right)+55.8 \\\\ -13.3=13.3\\sin \\left(\\frac{\\pi }{6}-C\\right) \\\\ -1=\\sin \\left(\\frac{\\pi }{6}-C\\right) \\\\ \\sin \\theta =-1\\to \\theta =-\\frac{\\pi }{2}\\end{gathered}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align} \\frac{\\pi }{6}-C&=-\\frac{\\pi }{2}\\\\ \\frac{\\pi }{6}+\\frac{\\pi }{2}&=C \\\\ &=\\frac{2\\pi }{3} \\end{align}[\/latex]<\/p>\n<p>We have the equation [latex]y=13.3\\sin \\left(\\frac{\\pi }{6}x-\\frac{2\\pi }{3}\\right)+55.8[\/latex]. See the graph in Figure 8.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164135\/CNX_Precalc_Figure_07_06_0112.jpg\" alt=\"Graph of the equation y=13.3sin(pi\/6 x - 2pi\/3) + 55.8. The average value is a dotted horizontal line y=55.8, and the amplitude is 13.3\" width=\"487\" height=\"331\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 8<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Describing Periodic Motion<\/h3>\n<p>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 [latex]y[\/latex] equal the distance from the tip of the hour hand to the ceiling [latex]x[\/latex] hours after noon. Find the equation that models the motion of the clock and sketch the graph.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q45024\">Show Solution<\/span><\/p>\n<div id=\"q45024\" class=\"hidden-answer\" style=\"display: none\">\n<p>Begin by making a table of values as shown in the table below.<\/p>\n<table id=\"fs-id1992294\" summary=\"Six rows, three columns. The table has ordered pairs of these row values: (X, y, Points to plot), (Noon, 30 in, (0,30)), (3 PM, 54 in, (3,54)), (6 PM, 78 in, (6,8)), (9 PM, 54 in, (9,54)), (Midnight, 30 in, (12,30)).\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/> <\/colgroup>\n<thead>\n<tr>\n<th>[latex]x[\/latex]<\/th>\n<th>[latex]y[\/latex]<\/th>\n<th>Points to plot<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Noon<\/td>\n<td>30 in<\/td>\n<td>[latex]\\left(0,30\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>3 PM<\/td>\n<td>54 in<\/td>\n<td>[latex]\\left(3,54\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>6 PM<\/td>\n<td>78 in<\/td>\n<td>[latex]\\left(6,78\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>9 PM<\/td>\n<td>54 in<\/td>\n<td>[latex]\\left(9,54\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Midnight<\/td>\n<td>30 in<\/td>\n<td>[latex]\\left(12,30\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>To model an equation, we first need to find the amplitude.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}|A|&=\\left\\rvert\\frac{78 - 30}{2}\\right\\rvert \\\\ &=24 \\end{align}[\/latex]<\/p>\n<p>The clock\u2019s cycle repeats every 12 hours. Thus,<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}B&=\\frac{2\\pi }{12} \\\\ &=\\frac{\\pi }{6} \\end{align}[\/latex]<\/p>\n<p>The vertical shift is<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}D&=\\frac{78+30}{2} \\\\ &=54 \\end{align}[\/latex]<\/p>\n<p>There is no horizontal shift, so [latex]C=0[\/latex]. Since the function begins with the minimum value of [latex]y[\/latex] when [latex]x=0[\/latex] (as opposed to the maximum value), we will use the cosine function with the negative value for [latex]A[\/latex]. In the form [latex]y=A\\cos \\left(Bx\\pm C\\right)+D[\/latex], the equation is<\/p>\n<p style=\"text-align: center\">[latex]y=-24\\cos \\left(\\frac{\\pi }{6}x\\right)+54[\/latex]<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164137\/CNX_Precalc_Figure_07_06_0132.jpg\" alt=\"Graph of the function y=-24cos(pi\/6 x)+54 using the five key points: (0,30), (3,54), (6,78), (9,54), (12,30).\" width=\"487\" height=\"392\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 9<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 6: Determining a Model for Tides<\/h3>\n<p>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.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q705973\">Show Solution<\/span><\/p>\n<div id=\"q705973\" class=\"hidden-answer\" style=\"display: none\">\n<p>As the water level varies from 7 ft to 15 ft, we can calculate the amplitude as<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}|A|&=\\left\\rvert\\frac{\\left(15 - 7\\right)}{2}\\right\\rvert \\\\ &=4 \\end{align}[\/latex]<\/p>\n<p>The cycle repeats every 12 hours; therefore, [latex]B[\/latex] is<\/p>\n<p style=\"text-align: center\">[latex]\\frac{2\\pi }{12}=\\frac{\\pi }{6}[\/latex]<\/p>\n<p>There is a vertical translation of [latex]\\frac{\\left(15+8\\right)}{2}=11.5[\/latex]. Since the value of the function is at a maximum at [latex]t=0[\/latex], we will use the cosine function, with the positive value for [latex]A[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]y=4\\cos \\left(\\frac{\\pi }{6}\\right)t+11[\/latex]<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164139\/CNX_Precalc_Figure_07_06_0152.jpg\" alt=\"Graph of the function y=4cos(pi\/6 t) + 11 from 0 to 12. The midline is y=11, three key points are (0,15), (6,7), and (12, 15).\" width=\"487\" height=\"330\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 10<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>The daily temperature in the month of March in a certain city varies from a low of [latex]24^\\circ\\text{F}[\/latex] to a high of [latex]40^\\circ\\text{F}[\/latex]. Find a sinusoidal function to model daily temperature and sketch the graph. Approximate the time when the temperature reaches the freezing point [latex]32^\\circ\\text{F}[\/latex]. Let [latex]t=0[\/latex] correspond to noon.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q130521\">Show Solution<\/span><\/p>\n<div id=\"q130521\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y=8\\sin \\left(\\frac{\\pi }{12}t\\right)+32[\/latex]<br \/>\nThe temperature reaches freezing at noon and at midnight.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164214\/CNX_Precalc_Figure_07_06_0162.jpg\" alt=\"Graph of the function y=8sin(pi\/12 t) + 32 for temperature. The midline is at 32. The times when the temperature is at 32 are midnight and noon.\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm152828\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=152828&theme=oea&iframe_resize_id=ohm152828\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 7: Interpreting the Periodic Behavior Equation<\/h3>\n<p>The average person\u2019s blood pressure is modeled by the function [latex]f\\left(t\\right)=20\\sin \\left(160\\pi t\\right)+100[\/latex], where [latex]f\\left(t\\right)[\/latex] represents the blood pressure at time [latex]t[\/latex],\u00a0measured in minutes. Interpret the function in terms of period and frequency. Sketch the graph and find the blood pressure reading.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q393443\">Show Solution<\/span><\/p>\n<div id=\"q393443\" class=\"hidden-answer\" style=\"display: none\">\n<p>The period is given by<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{2\\pi }{\\omega }&=\\frac{2\\pi }{160\\pi } \\\\ &=\\frac{1}{80} \\end{align}[\/latex]<\/p>\n<p>In a blood pressure function, frequency represents the number of heart beats per minute. Frequency is the reciprocal of period and is given by<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{\\omega }{2\\pi }&=\\frac{160\\pi }{2\\pi }\\\\ &=80\\end{align}[\/latex]<\/p>\n<p>See the graph in Figure 11.<\/p>\n<p>The blood pressure reading on the graph is [latex]\\frac{120}{80}\\text{ }\\left(\\frac{\\text{maximum}}{\\text{minimum}}\\right)[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164141\/CNX_Precalc_Figure_07_06_0172.jpg\" alt=\"Graph of the function f(t) = 20sin(160 * pi * t) + 100 for blood pressure. The midline is at 100.\" width=\"487\" height=\"475\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 11<\/b><\/p>\n<\/div>\n<h4>Analysis of the Solution<\/h4>\n<p>Blood pressure of [latex]\\frac{120}{80}[\/latex] is considered to be normal. The top number is the maximum or systolic reading, which measures the pressure in the arteries when the heart contracts. The bottom number is the minimum or diastolic reading, which measures the pressure in the arteries as the heart relaxes between beats, refilling with blood. Thus, normal blood pressure can be modeled by a periodic function with a maximum of 120 and a minimum of 80.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<p><span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Modeling Harmonic Motion Functions<\/span><\/p>\n<\/div>\n<p>Harmonic motion is a form of periodic motion, but there are factors to consider that differentiate the two types. While general <strong>periodic motion<\/strong> applications cycle through their periods with no outside interference, <strong>harmonic motion<\/strong> requires a restoring force. Examples of harmonic motion include springs, gravitational force, and magnetic force.<\/p>\n<h3>Simple Harmonic Motion<\/h3>\n<p>A type of motion described as <strong>simple harmonic motion<\/strong> involves a restoring force but assumes that the motion will continue forever. Imagine a weighted object hanging on a spring, When that object is not disturbed, we say that the object is at rest, or in equilibrium. If the object is pulled down and then released, the force of the spring pulls the object back toward equilibrium and harmonic motion begins. The restoring force is directly proportional to the displacement of the object from its equilibrium point. When [latex]t=0,d=0[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Simple Harmonic Motion<\/h3>\n<p>We see that <strong>simple harmonic motion<\/strong> equations are given in terms of displacement:<\/p>\n<p style=\"text-align: center\">[latex]d=a\\cos \\left(\\omega t\\right)\\text{ or }d=a\\sin \\left(\\omega t\\right)[\/latex]<\/p>\n<p>where [latex]|a|[\/latex] is the amplitude, [latex]\\frac{2\\pi }{\\omega }[\/latex] is the period, and [latex]\\frac{\\omega }{2\\pi }[\/latex] is the frequency, or the number of cycles per unit of time.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 8: Finding the Displacement, Period, and Frequency, and Graphing a Function<\/h3>\n<p>For the given functions,<\/p>\n<ol>\n<li>Find the maximum displacement of an object.<\/li>\n<li>Find the period or the time required for one vibration.<\/li>\n<li>Find the frequency.<\/li>\n<li>Sketch the graph.\n<ol>\n<li>[latex]y=5\\sin \\left(3t\\right)[\/latex]<\/li>\n<li>[latex]y=6\\cos \\left(\\pi t\\right)[\/latex]<\/li>\n<li>[latex]y=5\\cos \\left(\\frac{\\pi }{2}t\\right)[\/latex]<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q65116\">Show Solution<\/span><\/p>\n<div id=\"q65116\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]y=5\\sin \\left(3t\\right)[\/latex]\n<ol>\n<li>The maximum displacement is equal to the amplitude, [latex]|a|[\/latex], which is 5.<\/li>\n<li>The period is [latex]\\frac{2\\pi }{\\omega }=\\frac{2\\pi }{3}[\/latex].<\/li>\n<li>The frequency is given as [latex]\\frac{\\omega }{2\\pi }=\\frac{3}{2\\pi }[\/latex].<\/li>\n<li>See Figure 12. The graph indicates the five key points.\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164143\/CNX_Precalc_Figure_07_07_0062.jpg\" alt=\"Graph of the function y=5sin(3t) from 0 to 2pi\/3. The five key points are (0,0), (pi\/6, 5), (pi\/3, 0), (pi\/2, -5), (2pi\/3, 0).\" width=\"487\" height=\"443\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 12<\/b><\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/li>\n<li>[latex]y=6\\cos \\left(\\pi t\\right)[\/latex]\n<ol>\n<li>The maximum displacement is [latex]6[\/latex].<\/li>\n<li>The period is [latex]\\frac{2\\pi }{\\omega }=\\frac{2\\pi }{\\pi }=2[\/latex].<\/li>\n<li>The frequency is [latex]\\frac{\\omega }{2\\pi }=\\frac{\\pi }{2\\pi }=\\frac{1}{2}[\/latex].<\/li>\n<li>See Figure 13.\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164145\/CNX_Precalc_Figure_07_07_0072.jpg\" alt=\"Graph of the function y=6cos(pi t) from 0 to 3.\" width=\"487\" height=\"290\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 13.<\/b><\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/li>\n<li>[latex]y=5\\cos \\left(\\frac{\\pi }{2}\\right)t[\/latex]\n<ol>\n<li>The maximum displacement is [latex]5[\/latex].<\/li>\n<li>The period is [latex]\\frac{2\\pi }{\\omega }=\\frac{2\\pi }{\\frac{\\pi }{2}}=4[\/latex].<\/li>\n<li>The frequency is [latex]\\frac{1}{4}[\/latex].<\/li>\n<li>See Figure 14.\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164148\/CNX_Precalc_Figure_07_07_0082.jpg\" alt=\"Graph of the function y=5cos(pi\/2 t) from 0 to 4.\" width=\"487\" height=\"519\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 14<\/b><\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h2>Damped Harmonic Motion<\/h2>\n<p>In reality, a pendulum does not swing back and forth forever, nor does an object on a spring bounce up and down forever. Eventually, the pendulum stops swinging and the object stops bouncing and both return to equilibrium. Periodic motion in which an energy-dissipating force, or damping factor, acts is known as <strong>damped harmonic motion<\/strong>. Friction is typically the damping factor.<\/p>\n<p>In physics, various formulas are used to account for the damping factor on the moving object. Some of these are calculus-based formulas that involve derivatives. For our purposes, we will use formulas for basic damped harmonic motion models.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Damped Harmonic Motion<\/h3>\n<p>In <strong>damped harmonic motion<\/strong>, the displacement of an oscillating object from its rest position at time [latex]t[\/latex] is given as<\/p>\n<p style=\"text-align: center\">[latex]f\\left(t\\right)=a{e}^{-ct}\\sin \\left(\\omega t\\right)\\text{or} f\\left(t\\right)=a{e}^{-ct}\\cos \\left(\\omega t\\right)[\/latex]<\/p>\n<p>where [latex]c[\/latex] is a damping factor, [latex]|a|[\/latex] is the initial displacement and [latex]\\frac{2\\pi }{\\omega }[\/latex] is the period.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 9: Modeling Damped Harmonic Motion<\/h3>\n<p>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 [latex]0.5[\/latex] cycles per second. Both have an initial displacement of 10 cm. The first has a damping factor of [latex]0.5[\/latex] and the second has a damping factor of [latex]0.1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q574944\">Show Solution<\/span><\/p>\n<div id=\"q574944\" class=\"hidden-answer\" style=\"display: none\">\n<p>At time [latex]t=0[\/latex], the displacement is the maximum of 10 cm, which calls for the cosine function. The cosine function will apply to both models.<\/p>\n<p>We are given the frequency [latex]f=\\frac{\\omega }{2\\pi }[\/latex] of 0.5 cycles per second. Thus,<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{\\omega }{2\\pi }&=0.5 \\\\ \\omega &=\\left(0.5\\right)2\\pi \\\\ &=\\pi \\end{align}[\/latex]<\/p>\n<p>The first spring system has a damping factor of [latex]c=0.5[\/latex]. Following the general model for damped harmonic motion, we have<\/p>\n<p style=\"text-align: center\">[latex]f\\left(t\\right)=10{e}^{-0.5t}\\cos \\left(\\pi t\\right)[\/latex]<\/p>\n<p>Figure 15 models the motion of the first spring system.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164150\/CNX_Precalc_Figure_07_07_0092.jpg\" alt=\"Graph of the first spring system, f(t) = 10(e^(-.5t))cos(pi*t), which begins with a high amplitude and quickly decreases.\" width=\"487\" height=\"632\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 15<\/b><\/p>\n<\/div>\n<p>The second spring system has a damping factor of [latex]c=0.1[\/latex] and can be modeled as<\/p>\n<p style=\"text-align: center\">[latex]f\\left(t\\right)=10{e}^{-0.1t}\\cos \\left(\\pi t\\right)[\/latex]<\/p>\n<p>Figure 16\u00a0models the motion of the second spring system.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164152\/CNX_Precalc_Figure_07_07_0102.jpg\" alt=\"Graph of f(t) = 10(e^(-.1t))cos(pi*t), which begins with a high amplitude and slowly decreases (but has a high frequency).\" width=\"487\" height=\"522\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 16<\/b><\/p>\n<\/div>\n<h4>Analysis of the Solution<\/h4>\n<p>Notice the differing effects of the damping constant. The local maximum and minimum values of the function with the damping factor [latex]c=0.5[\/latex] decreases much more rapidly than that of the function with [latex]c=0.1[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 10: Finding a Cosine Function that Models Damped Harmonic Motion<\/h3>\n<p>Find and graph a function of the form [latex]y=a{e}^{-ct}\\cos \\left(\\omega t\\right)[\/latex] that models the information given.<\/p>\n<ol>\n<li>[latex]a=20,c=0.05,p=4[\/latex]<\/li>\n<li>[latex]a=2,c=1.5,f=3[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q923862\">Show Solution<\/span><\/p>\n<div id=\"q923862\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute the given values into the model. Recall that period is [latex]\\frac{2\\pi }{\\omega }[\/latex] and frequency is [latex]\\frac{\\omega }{2\\pi }[\/latex].<\/p>\n<ol>\n<li>[latex]y=20{e}^{-0.05t}\\cos \\left(\\frac{\\pi }{2}t\\right)[\/latex]. See Figure 17.\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164155\/CNX_Precalc_Figure_07_07_0112.jpg\" alt=\"Graph of f(t) = 20(e^(-.05t))cos(pi\/2 * t), which begins with a high amplitude and slowly decreases.\" width=\"487\" height=\"361\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 17<\/b><\/p>\n<\/div>\n<\/li>\n<li>[latex]y=2{e}^{-1.5t}\\cos \\left(6\\pi t\\right)[\/latex]. See Figure 18.\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164157\/CNX_Precalc_Figure_07_07_0122.jpg\" alt=\"Graph of f(t) = 2(e^(-1.5t))cos(6pi * t), which begins with a small amplitude and quickly decreases to almost a straight line.\" width=\"487\" height=\"513\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 18<\/b><\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>The following equation represents a damped harmonic motion model: [latex]\\text{ }f\\left(t\\right)=5{e}^{-6t}\\cos \\left(4t\\right)[\/latex]<br \/>\nFind the initial displacement, the damping constant, and the frequency.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q533970\">Show Solution<\/span><\/p>\n<div id=\"q533970\" class=\"hidden-answer\" style=\"display: none\">\n<p>initial displacement =6, damping constant = -6, frequency = [latex]\\frac{2}{\\pi }[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 11: Finding a Sine Function that Models Damped Harmonic Motion<\/h3>\n<p>Find and graph a function of the form [latex]y=a{e}^{-ct}\\sin \\left(\\omega t\\right)[\/latex] that models the information given.<\/p>\n<ol>\n<li>[latex]a=7,c=10,p=\\frac{\\pi }{6}[\/latex]<\/li>\n<li>[latex]a=0.3,c=0.2,f=20[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q293857\">Show Solution<\/span><\/p>\n<div id=\"q293857\" class=\"hidden-answer\" style=\"display: none\">\n<p>Calculate the value of [latex]\\omega[\/latex] and substitute the known values into the model.<\/p>\n<ol>\n<li>As period is [latex]\\frac{2\\pi }{\\omega }[\/latex], we have\n<div style=\"text-align: center\">[latex]\\begin{gathered}\\frac{\\pi }{6}=\\frac{2\\pi }{\\omega } \\\\ \\omega \\pi =6\\left(2\\pi \\right) \\\\ \\omega =12 \\end{gathered}[\/latex]<\/div>\n<p>The damping factor is given as 10 and the amplitude is 7. Thus, the model is [latex]y=7{e}^{-10t}\\sin \\left(12t\\right)[\/latex]. See Figure 19.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164200\/CNX_Precalc_Figure_07_07_0132.jpg\" alt=\"Graph of f(t) = 7(e^(-10t))sin(12t), which spikes close to t=0 and quickly becomes almost a straight line.\" width=\"487\" height=\"291\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 19<\/b><\/p>\n<\/div>\n<\/li>\n<li>As frequency is [latex]\\frac{\\omega }{2\\pi }[\/latex], we have\n<div style=\"text-align: center\">[latex]\\begin{gathered}20=\\frac{\\omega }{2\\pi } \\\\ 40\\pi =\\omega \\end{gathered}[\/latex]<\/div>\n<p>The damping factor is given as [latex]0.2[\/latex] and the amplitude is [latex]0.3[\/latex]. The model is [latex]y=0.3{e}^{-0.2t}\\sin \\left(40\\pi t\\right)[\/latex]. See Figure 20.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164202\/CNX_Precalc_Figure_07_07_0142.jpg\" alt=\"Graph of f(t) = .3e^(-.2t)sin(40pi*t), which has a small amplitude but quickly decreases to the appearance of a straight line. The frequency is so high that, in this scaling, the function looks like a solid shape. The zoom in cut out of the graph shows the actual sinusoidal image of the function.\" width=\"487\" height=\"258\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 20<\/b><\/p>\n<\/div>\n<\/li>\n<\/ol>\n<h4>Analysis of the Solution<\/h4>\n<p>A comparison of the last two examples illustrates how we choose between the sine or cosine functions to model sinusoidal criteria. We see that the cosine function is at the maximum displacement when [latex]t=0[\/latex], and the sine function is at the equilibrium point when [latex]t=0[\/latex]. For example, consider the equation [latex]y=20{e}^{-0.05t}\\cos \\left(\\frac{\\pi }{2}t\\right)[\/latex] from Example 9. We can see from the graph that when [latex]t=0,y=20[\/latex], which is the initial amplitude. Check this by setting [latex]t=0[\/latex] in the cosine equation:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}y&=20{e}^{-0.05\\left(0\\right)}\\cos \\left(\\frac{\\pi }{2}\\right)\\left(0\\right) \\\\ &=20\\left(1\\right)\\left(1\\right) \\\\ &=20\\end{align}[\/latex]<\/p>\n<p>Using the sine function yields<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}y&=20{e}^{-0.05\\left(0\\right)}\\sin \\left(\\frac{\\pi }{2}\\right)\\left(0\\right) \\\\ &=20\\left(1\\right)\\left(0\\right) \\\\ &=0 \\end{align}[\/latex]<\/p>\n<p>Thus, cosine is the correct function.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Write the equation for damped harmonic motion given [latex]a=10,c=0.5[\/latex], and [latex]p=2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q605857\">Show Solution<\/span><\/p>\n<div id=\"q605857\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y=10{e}^{-0.5t}\\cos \\left(\\pi t\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173786\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173786&theme=oea&iframe_resize_id=ohm173786\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 12: Modeling the Oscillation of a Spring<\/h3>\n<p>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 [latex]t[\/latex] seconds after being released.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q958106\">Show Solution<\/span><\/p>\n<div id=\"q958106\" class=\"hidden-answer\" style=\"display: none\">\n<p>The amplitude begins at 5 in. and deceases 30% each second. Because the spring is initially compressed, we will write <em>A<\/em> as a negative value. We can write the amplitude portion of the function as<\/p>\n<p style=\"text-align: center\">[latex]A\\left(t\\right)=5{\\left(1 - 0.30\\right)}^{t}=[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]A\\left(t\\right)=5{\\left(0.70\\right)}^{t}=[\/latex]<\/p>\n<p>We put [latex]{\\left(0.70\\right)}^{t}[\/latex] in the form [latex]{e}^{ct}[\/latex] as follows:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}0.7={e}^{c} \\\\ c=\\mathrm{ln}(0.7) \\\\ c=-0.357 \\end{gathered}[\/latex]<\/p>\n<p>Now let\u2019s 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.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered} 3=\\frac{2\\pi }{\\omega }\\\\ \\omega =\\frac{2\\pi }{3}\\end{gathered}[\/latex]<\/p>\n<p>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<\/p>\n<p style=\"text-align: center\">[latex]y=\\cos \\left(\\frac{2\\pi }{3}t\\right)+10[\/latex]<\/p>\n<p>Finally, we put both functions together. Our the model for the position of the spring at [latex]t[\/latex] seconds is given as<\/p>\n<p style=\"text-align: center\">[latex]y=-5{e}^{-0.357t}\\cos \\left(\\frac{2\\pi }{3}t\\right)+10[\/latex]<\/p>\n<p>See the graph in Figure 21.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164204\/CNX_Precalc_Figure_07_06_0182.jpg\" alt=\"Graph of the function y = -5e^(-.35t)cos(2pi\/3 t) + 10 from 0 to 24. It starts out as waves with a high amplitude and decreases to almost a straight line very quickly.\" width=\"487\" height=\"282\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 21<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>A mass suspended from a spring is raised a distance of 5 cm above its resting position. The mass is released at time [latex]t=0[\/latex] and allowed to oscillate. After [latex]\\frac{1}{3}[\/latex] 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.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q453909\">Show Solution<\/span><\/p>\n<div id=\"q453909\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y=5\\cos \\left(6\\pi t\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 13: Finding the Value of the Damping Constant <em>c<\/em> According to the Given Criteria<\/h3>\n<p>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.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q619520\">Show Solution<\/span><\/p>\n<div id=\"q619520\" class=\"hidden-answer\" style=\"display: none\">\n<p>The displacement factor represents the amplitude and is determined by the coefficient [latex]a{e}^{-ct}[\/latex] in the model for damped harmonic motion. The damping constant is included in the term [latex]{e}^{-ct}[\/latex]. It is known that after 3 seconds, the local maximum measures one-half of its original value. Therefore, we have the equation<\/p>\n<p style=\"text-align: center\">[latex]a{e}^{-c\\left(t+3\\right)}=\\frac{1}{2}a{e}^{-ct}[\/latex]<\/p>\n<p>Use algebra and the laws of exponents to solve for [latex]c[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align} a{e}^{-c\\left(t+3\\right)}&=\\frac{1}{2}a{e}^{-ct} \\\\ {e}^{-ct}\\cdot {e}^{-3c}&=\\frac{1}{2}{e}^{-ct} && \\text{Divide out }a.\\\\ {e}^{-3c}&=\\frac{1}{2}&& \\text{Divide out }{e}^{-ct}. \\\\ {e}^{3c}&=2 && \\text{Take reciprocals}. \\end{align}[\/latex]<\/p>\n<p>Then use the laws of logarithms.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}{e}^{3c}&=2 \\\\ 3c&=\\mathrm{ln}2 \\\\ c&=\\frac{\\mathrm{ln}2}{3} \\end{align}[\/latex]<\/p>\n<p>The damping constant is [latex]\\frac{\\mathrm{ln}2}{3}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Bounding Curves in Harmonic Motion<\/h2>\n<p>Harmonic motion graphs may be enclosed by bounding curves. When a function has a varying <strong>amplitude<\/strong>, such that the amplitude rises and falls multiple times within a period, we can determine the bounding curves from part of the function.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 14: Graphing an Oscillating Cosine Curve<\/h3>\n<p>Graph the function [latex]f\\left(x\\right)=\\cos \\left(2\\pi x\\right)\\cos \\left(16\\pi x\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q634563\">Show Solution<\/span><\/p>\n<div id=\"q634563\" class=\"hidden-answer\" style=\"display: none\">\n<p>The graph produced by this function will be shown in two parts. The first graph will be the exact function [latex]f\\left(x\\right)[\/latex], and the second graph is the exact function [latex]f\\left(x\\right)[\/latex] plus a bounding function. The graphs look quite different.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164207\/CNX_Precalc_Figure_07_07_0152.jpg\" alt=\"Graph of f(x) = cos(2pi*x)cos(16pi*x), a sinusoidal function that increases and decreases its amplitude periodically.\" width=\"487\" height=\"145\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 22<\/b><\/p>\n<\/div>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164209\/CNX_Precalc_Figure_07_07_0162.jpg\" alt=\"Graph of f(x) = cos(2pi*x)cos(16pi*x), a sinusoidal function that increases and decreases its amplitude periodically. There is also a bonding function drawn over it in red, which makes the whole image look like a DNA (double helix) piece stretched along the x-axis.\" width=\"487\" height=\"218\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 23<\/b><\/p>\n<\/div>\n<h4>Analysis of the Solution<\/h4>\n<p>The curves [latex]y=\\cos \\left(2\\pi x\\right)[\/latex] and [latex]y=-\\cos \\left(2\\pi x\\right)[\/latex] are bounding curves: they bound the function from above and below, tracing out the high and low points. The harmonic motion graph sits inside the bounding curves. This is an example of a function whose amplitude not only decreases with time, but actually increases and decreases multiple times within a period.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<p><span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Key Equations<\/span><\/p>\n<\/div>\n<table id=\"fs-id2608518\" summary=\"..\">\n<tbody>\n<tr>\n<td>Standard form of sinusoidal equation<\/td>\n<td>[latex]y=A\\sin \\left(Bt-C\\right)+D\\text{or}y=A\\cos \\left(Bt-C\\right)+D[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simple harmonic motion<\/td>\n<td>[latex]d=a\\cos \\left(\\omega t\\right)\\text{ or }d=a\\sin \\left(\\omega t\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Damped harmonic motion<\/td>\n<td>[latex]f\\left(t\\right)=a{e}^{-c}{}^{t}\\sin \\left(\\omega t\\right)\\text{or}f\\left(t\\right)=a{e}^{-ct}\\cos \\left(\\omega t\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>Sinusoidal functions are represented by the sine and cosine graphs. In standard form, we can find the amplitude, period, and horizontal and vertical shifts.<\/li>\n<li>Use key points to graph a sinusoidal function. The five key points include the minimum and maximum values and the midline values.<\/li>\n<li>Periodic functions can model events that reoccur in set cycles, like the phases of the moon, the hands on a clock, and the seasons in a year.<\/li>\n<li>Harmonic motion functions are modeled from given data. Similar to periodic motion applications, harmonic motion requires a restoring force. Examples include gravitational force and spring motion activated by weight.<\/li>\n<li>Damped harmonic motion is a form of periodic behavior affected by a damping factor. Energy dissipating factors, like friction, cause the displacement of the object to shrink.<\/li>\n<li>Bounding curves delineate the graph of harmonic motion with variable maximum and minimum values.<\/li>\n<\/ul>\n<div>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id2617942\" class=\"definition\">\n<dt>damped harmonic motion<\/dt>\n<dd id=\"fs-id2617946\">oscillating motion that resembles periodic motion and simple harmonic motion, except that the graph is affected by a damping factor, an energy dissipating influence on the motion, such as friction<\/dd>\n<\/dl>\n<dl id=\"fs-id2617951\" class=\"definition\">\n<dt>simple harmonic motion<\/dt>\n<dd id=\"fs-id2617956\">a repetitive motion that can be modeled by periodic sinusoidal oscillation<\/dd>\n<\/dl>\n<\/div>\n<p>&nbsp;<\/p>\n<h2 style=\"text-align: center\">Section 5.6 Homework Exercises<\/h2>\n<p>1. Explain what types of physical phenomena are best modeled by sinusoidal functions. What are the characteristics necessary?<\/p>\n<p>2.\u00a0What 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.<\/p>\n<p>3. 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?<\/p>\n<p>4.\u00a0Explain the effect of a damping factor on the graphs of harmonic motion functions.<\/p>\n<p>For the following exercises, find a possible formula for the trigonometric function represented by the given table of values.<\/p>\n<p>5.<\/p>\n<table id=\"fs-id1879688\" class=\"unnumbered\" summary=\"Two columns, eight rows. The table has ordered pairs of these row values: (x,y), (0,-4), (3,-1), (6,2), (9,-1), (12,-4), (15,-1), (18,2).\">\n<tbody>\n<tr>\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\n<\/tr>\n<tr>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]-4[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3[\/latex]<\/td>\n<td>[latex]-1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]6[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]9[\/latex]<\/td>\n<td>[latex]-1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]12[\/latex]<\/td>\n<td>[latex]-4[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]15[\/latex]<\/td>\n<td>[latex]-1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]18[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>6.<\/p>\n<table id=\"fs-id1514649\" class=\"unnumbered\" summary=\"Two columns, eight rows. The table has ordered pairs of these row values: (x,y), (0,5), (2,1), (4,-3), (6,1), (8,5), (10,1), (12,-3).\">\n<tbody>\n<tr>\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\n<\/tr>\n<tr>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]4[\/latex]<\/td>\n<td>[latex]-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]6[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]8[\/latex]<\/td>\n<td>[latex]5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]10[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]12[\/latex]<\/td>\n<td>[latex]-3[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>7.<\/p>\n<table id=\"fs-id2202906\" class=\"unnumbered\" summary=\"Two columns, eight rows. The table has ordered pairs of these row values: (x,y), (0,2), (pi\/4, 7), (pi\/2, 2), (3pi\/4, -3), (pi, 2), (5pi\/4, 7), (3pi\/2, 2).\">\n<tbody>\n<tr>\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\n<\/tr>\n<tr>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\frac{\\pi }{4}[\/latex]<\/td>\n<td>[latex]7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\frac{3\\pi }{4}[\/latex]<\/td>\n<td>[latex]-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\pi[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\frac{5\\pi }{4}[\/latex]<\/td>\n<td>[latex]7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>8.<\/p>\n<table id=\"fs-id1379283\" class=\"unnumbered\" summary=\"Two columns, eight rows. The table has ordered pairs of these row values: (x,y), (0,-1), (pi\/4, 2), (pi\/2, 5), (3pi\/4, 2), (pi, -1), (5pi\/4, 2), (3pi\/2, 5).\">\n<tbody>\n<tr>\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\n<\/tr>\n<tr>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\frac{\\pi }{4}[\/latex]<\/td>\n<td>[latex]7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\frac{3\\pi }{4}[\/latex]<\/td>\n<td>[latex]-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\pi[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\frac{5\\pi }{4}[\/latex]<\/td>\n<td>[latex]7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>9.<\/p>\n<table id=\"fs-id2112947\" class=\"unnumbered\" summary=\"Two columns, eight rows. The table has ordered pairs of these column values: (x,y), (0,1), (1,-3), (2,-7), (3,-3), (4,1), (5,-3), (6,-7).\">\n<tbody>\n<tr>\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\n<\/tr>\n<tr>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]-7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3[\/latex]<\/td>\n<td>[latex]-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]4[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]5[\/latex]<\/td>\n<td>[latex]-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]6[\/latex]<\/td>\n<td>[latex]-7[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>10.<\/p>\n<table id=\"fs-id3809810\" class=\"unnumbered\" summary=\"Two columns, eight rows. The table has ordered pairs of these row values: (x,y), (0,-2), (1,4), (2,10), (3,4), (4,-2), (5,4), (6,10).\">\n<tbody>\n<tr>\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\n<\/tr>\n<tr>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]-2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]4[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]10[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3[\/latex]<\/td>\n<td>[latex]4[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]4[\/latex]<\/td>\n<td>[latex]-2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]5[\/latex]<\/td>\n<td>[latex]4[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]6[\/latex]<\/td>\n<td>[latex]10[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>11.<\/p>\n<table id=\"fs-id2795269\" class=\"unnumbered\" summary=\"Two columns, eight rows. The table has ordered pairs of these row values: (x,y), (0,5), (1,-3), (2,5), (3,13), (4,5), (5,-3), (6,5).\">\n<tbody>\n<tr>\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\n<\/tr>\n<tr>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3[\/latex]<\/td>\n<td>[latex]13[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]4[\/latex]<\/td>\n<td>[latex]5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]5[\/latex]<\/td>\n<td>[latex]-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]6[\/latex]<\/td>\n<td>[latex]5[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>12.<\/p>\n<table id=\"fs-id2760414\" class=\"unnumbered\" summary=\"Two columns, eight rows. The table has ordered pairs of these row values: (x,y), (-3, -1 - rad2), (-2, -1), (-1, 1 - rad2), (0,0), (1, rad2 - 1), (2,1), (3, rad2 + 1).\">\n<tbody>\n<tr>\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\n<\/tr>\n<tr>\n<td>[latex]-3[\/latex]<\/td>\n<td>[latex]-1-\\sqrt{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-2[\/latex]<\/td>\n<td>[latex]-1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1[\/latex]<\/td>\n<td>[latex]1-\\sqrt{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]\\sqrt{2}-1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3[\/latex]<\/td>\n<td>[latex]\\sqrt{2}+1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>13.<\/p>\n<table id=\"fs-id2796936\" class=\"unnumbered\" summary=\"Two columns, eight rows. The table has ordered pairs of these row values: (x,y), (-1, rad3 - 2), (0,0), (1, 2- rad3), (2, rad3 \/ 3), (3,1), (4, rad3), (5, 2 + rad3).\">\n<tbody>\n<tr>\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\n<\/tr>\n<tr>\n<td>[latex]-1[\/latex]<\/td>\n<td>[latex]\\sqrt{3}-2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]2-\\sqrt{3}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]4[\/latex]<\/td>\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]5[\/latex]<\/td>\n<td>[latex]2+\\sqrt{3}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>For the following exercises, graph the given function, and then find a possible physical process that the equation could model.<\/p>\n<p>14. [latex]f\\left(x\\right)=-30\\cos \\left(\\frac{x\\pi }{6}\\right)-20{\\cos }^{2}\\left(\\frac{x\\pi }{6}\\right)+80\\left[0,12\\right][\/latex]<\/p>\n<p>15. [latex]f\\left(x\\right)=-18\\cos \\left(\\frac{x\\pi }{12}\\right)-5\\sin \\left(\\frac{x\\pi }{12}\\right)+100[\/latex] on the interval [latex]\\left[0,24\\right][\/latex]<\/p>\n<p>16. [latex]f\\left(x\\right)=10-\\sin \\left(\\frac{x\\pi }{6}\\right)+24\\tan \\left(\\frac{x\\pi }{240}\\right)[\/latex] on the interval [latex]\\left[0,80\\right][\/latex]<\/p>\n<p>For the following exercise, construct a function modeling behavior and use a calculator to find desired results.<\/p>\n<p>17. A city\u2019s 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.<\/p>\n<p>For the following exercises, construct a sinusoidal function with the provided information, and then solve the equation for the requested values.<\/p>\n<p>18. Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the high temperature of [latex]105\\text{^\\circ F}[\/latex] occurs at 5PM and the average temperature for the day is [latex]85\\text{^\\circ F}\\text{.}[\/latex] Find the temperature, to the nearest degree, at 9AM.<\/p>\n<p>19. Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the high temperature of [latex]84\\text{^\\circ F}[\/latex] occurs at 6PM and the average temperature for the day is [latex]70\\text{^\\circ F}\\text{.}[\/latex] Find the temperature, to the nearest degree, at 7AM.<\/p>\n<p>20.\u00a0Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the temperature varies between [latex]47\\text{^\\circ F}[\/latex] and [latex]63\\text{^\\circ F}[\/latex] during the day and the average daily temperature first occurs at 10 AM. How many hours after midnight does the temperature first reach [latex]51\\text{^\\circ F?}[\/latex]<\/p>\n<p>21. Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the temperature varies between [latex]64^\\circ\\text{F}[\/latex] and [latex]86^\\circ\\text{F}[\/latex] during the day and the average daily temperature first occurs at 12 AM. How many hours after midnight does the temperature first reach [latex]70^\\circ\\text{F?}[\/latex]<\/p>\n<p>22.\u00a0A Ferris wheel is 20 meters in diameter and boarded from a platform that is 2 meters above the ground. The six o\u2019clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 6 minutes. How much of the ride, in minutes and seconds, is spent higher than 13 meters above the ground?<\/p>\n<p>23. A Ferris wheel is 45 meters in diameter and boarded from a platform that is 1 meter above the ground. The six o\u2019clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 10 minutes. How many minutes of the ride are spent higher than 27 meters above the ground? Round to the nearest second<\/p>\n<p>24.\u00a0The sea ice area around the North Pole fluctuates between about 6 million square kilometers on September 1 to 14 million square kilometers on March 1. Assuming a sinusoidal fluctuation, when are there less than 9 million square kilometers of sea ice? Give your answer as a range of dates, to the nearest day.<\/p>\n<p>25. The sea ice area around the South Pole fluctuates between about 18 million square kilometers in September to 3 million square kilometers in March. Assuming a sinusoidal fluctuation, when are there more than 15 million square kilometers of sea ice? Give your answer as a range of dates, to the nearest day.<\/p>\n<p>26.\u00a0During a 90-day monsoon season, daily rainfall can be modeled by sinusoidal functions. If the rainfall fluctuates between a low of 2 inches on day 10 and 12 inches on day 55, during what period is daily rainfall more than 10 inches?<\/p>\n<p>27. During a 90-day monsoon season, daily rainfall can be modeled by sinusoidal functions. A low of 4 inches of rainfall was recorded on day 30, and overall the average daily rainfall was 8 inches. During what period was daily rainfall less than 5 inches?<\/p>\n<p>28.\u00a0In a certain region, monthly precipitation peaks at 8 inches on June 1 and falls to a low of 1 inch on December 1. Identify the periods when the region is under flood conditions (greater than 7 inches) and drought conditions (less than 2 inches). Give your answer in terms of the nearest day.<\/p>\n<p>29. In a certain region, monthly precipitation peaks at 24 inches in September and falls to a low of 4 inches in March. Identify the periods when the region is under flood conditions (greater than 22 inches) and drought conditions (less than 5 inches). Give your answer in terms of the nearest day.<\/p>\n<p>For the following exercises, find the amplitude, period, and frequency of the given function.<\/p>\n<p>30. The displacement [latex]h\\left(t\\right)[\/latex] in centimeters of a mass suspended by a spring is modeled by the function [latex]h\\left(t\\right)=8\\sin \\left(6\\pi t\\right)[\/latex], where [latex]t[\/latex] is measured in seconds. Find the amplitude, period, and frequency of this displacement.<\/p>\n<p>31. The displacement [latex]h\\left(t\\right)[\/latex] in centimeters of a mass suspended by a spring is modeled by the function [latex]h\\left(t\\right)=11\\sin \\left(12\\pi t\\right)[\/latex], where [latex]t[\/latex] is measured in seconds. Find the amplitude, period, and frequency of this displacement.<\/p>\n<p>32.\u00a0The displacement [latex]h\\left(t\\right)[\/latex] in centimeters of a mass suspended by a spring is modeled by the function [latex]h\\left(t\\right)=4\\cos \\left(\\frac{\\pi }{2}t\\right)[\/latex], where [latex]t[\/latex] is measured in seconds. Find the amplitude, period, and frequency of this displacement.<\/p>\n<p>For the following exercises, construct an equation that models the described behavior.<\/p>\n<p>33. The displacement [latex]h\\left(t\\right)[\/latex], in centimeters, of a mass suspended by a spring is modeled by the function [latex]h\\left(t\\right)=-5\\cos \\left(60\\pi t\\right)[\/latex], where [latex]t[\/latex] is measured in seconds. Find the amplitude, period, and frequency of this displacement.<\/p>\n<p>For the following exercises, construct an equation that models the described behavior.<\/p>\n<p>34. A deer population oscillates 19 above and below average during the year, reaching the lowest value in January. The average population starts at 800 deer and increases by 160 each year. Find a function that models the population, [latex]P[\/latex], in terms of months since January, [latex]t[\/latex].<\/p>\n<p>35. A rabbit population oscillates 15 above and below average during the year, reaching the lowest value in January. The average population starts at 650 rabbits and increases by 110 each year. Find a function that models the population, [latex]P[\/latex], in terms of months since January, [latex]t[\/latex].<\/p>\n<p>36.\u00a0A muskrat population oscillates 33 above and below average during the year, reaching the lowest value in January. The average population starts at 900 muskrats and increases by 7% each month. Find a function that models the population, [latex]P[\/latex], in terms of months since January, [latex]t[\/latex].<\/p>\n<p>37. A fish population oscillates 40 above and below average during the year, reaching the lowest value in January. The average population starts at 800 fish and increases by 4% each month. Find a function that models the population, [latex]P[\/latex], in terms of months since January, [latex]t[\/latex].<\/p>\n<p>38.\u00a0A spring attached to the ceiling is pulled 10 cm down from equilibrium and released. The amplitude decreases by 15% each second. The spring oscillates 18 times each second. Find a function that models the distance, [latex]D[\/latex], the end of the spring is from equilibrium in terms of seconds, [latex]t[\/latex], since the spring was released.<\/p>\n<p>39. A spring attached to the ceiling is pulled 7 cm down from equilibrium and released. The amplitude decreases by 11% each second. The spring oscillates 20 times each second. Find a function that models the distance, [latex]D[\/latex], the end of the spring is from equilibrium in terms of seconds, [latex]t[\/latex], since the spring was released.<\/p>\n<p>40.\u00a0A spring attached to the ceiling is pulled 17 cm down from equilibrium and released. After 3 seconds, the amplitude has decreased to 13 cm. The spring oscillates 14 times each second. Find a function that models the distance, [latex]D[\/latex], the end of the spring is from equilibrium in terms of seconds, [latex]t[\/latex], since the spring was released.<\/p>\n<p>41. A spring attached to the ceiling is pulled 19 cm down from equilibrium and released. After 4 seconds, the amplitude has decreased to 14 cm. The spring oscillates 13 times each second. Find a function that models the distance, [latex]D[\/latex], the end of the spring is from equilibrium in terms of seconds, [latex]t[\/latex], since the spring was released.<\/p>\n<p>For the following exercises, create a function modeling the described behavior. Then, calculate the desired result using a calculator.<\/p>\n<p>42. A certain lake currently has an average trout population of 20,000. The population naturally oscillates above and below average by 2,000 every year. This year, the lake was opened to fishermen. If fishermen catch 3,000 fish every year, how long will it take for the lake to have no more trout?<\/p>\n<p>43. Whitefish populations are currently at 500 in a lake. The population naturally oscillates above and below by 25 each year. If humans overfish, taking 4% of the population every year, in how many years will the lake first have fewer than 200 whitefish?<\/p>\n<p>44.\u00a0A spring attached to a ceiling is pulled down 11 cm from equilibrium and released. After 2 seconds, the amplitude has decreased to 6 cm. The spring oscillates 8 times each second. Find when the spring first comes between [latex]-0.1[\/latex] and [latex]0.1\\text{ cm,}[\/latex] effectively at rest.<\/p>\n<p>45. A spring attached to a ceiling is pulled down 21 cm from equilibrium and released. After 6 seconds, the amplitude has decreased to 4 cm. The spring oscillates 20 times each second. Find when the spring first comes between [latex]-0.1[\/latex] and [latex]0.1\\text{ cm,}[\/latex] effectively at rest.<\/p>\n<p>46.\u00a0Two springs are pulled down from the ceiling and released at the same time. The first spring, which oscillates 8 times per second, was initially pulled down 32 cm from equilibrium, and the amplitude decreases by 50% each second. The second spring, oscillating 18 times per second, was initially pulled down 15 cm from equilibrium and after 4 seconds has an amplitude of 2 cm. Which spring comes to rest first, and at what time? Consider &#8220;rest&#8221; as an amplitude less than [latex]0.1\\text{ cm}\\text{.}[\/latex]<\/p>\n<p>47. Two springs are pulled down from the ceiling and released at the same time. The first spring, which oscillates 14 times per second, was initially pulled down 2 cm from equilibrium, and the amplitude decreases by 8% each second. The second spring, oscillating 22 times per second, was initially pulled down 10 cm from equilibrium and after 3 seconds has an amplitude of 2 cm. Which spring comes to rest first, and at what time? Consider &#8220;rest&#8221; as an amplitude less than [latex]0.1\\text{ cm}\\text{.}[\/latex]<\/p>\n<p>48.\u00a0A plane flies 1 hour at 150 mph at [latex]{22}^{\\circ }[\/latex] east of north, then continues to fly for 1.5 hours at 120 mph, this time at a bearing of [latex]{112}^{\\circ }[\/latex] east of north. Find the total distance from the starting point and the direct angle flown north of east.<\/p>\n<p>49. A plane flies 2 hours at 200 mph at a bearing of [latex]\\text{ }{60}^{\\circ }[\/latex], then continues to fly for 1.5 hours at the same speed, this time at a bearing of [latex]{150}^{\\circ }[\/latex]. Find the distance from the starting point and the bearing from the starting point. Hint: bearing is measured counterclockwise from north.<\/p>\n<p>For the following exercises, find a function of the form [latex]y=a{b}^{x}+c\\sin \\left(\\frac{\\pi }{2}x\\right)[\/latex] that fits the given data.<\/p>\n<p>50.<\/p>\n<table id=\"fs-id2196327\" class=\"unnumbered\" summary=\"Two rows, five columns. The table has ordered pairs of these column values: (x,y), (0,6), (1,29), (2, 96), (3, 379).\">\n<tbody>\n<tr>\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\n<td>0<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\n<td>6<\/td>\n<td>29<\/td>\n<td>96<\/td>\n<td>379<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>51.<\/p>\n<table id=\"fs-id1229818\" class=\"unnumbered\" summary=\"Two rows, five columns. The table has ordered pairs of these column values: (x,y), (0,6), (1,34), (2, 150), (3, 746).\">\n<tbody>\n<tr>\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\n<td>0<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\n<td>6<\/td>\n<td>34<\/td>\n<td>150<\/td>\n<td>746<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>52.<\/p>\n<table id=\"fs-id2427395\" class=\"unnumbered\" summary=\"Two rows, five columns. The table has ordered pairs of these column values: (x,y), (0,4), (1,0), (2,16), (3,-40).\">\n<tbody>\n<tr>\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\n<td>0<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\n<td>4<\/td>\n<td>0<\/td>\n<td>16<\/td>\n<td>-40<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>For the following exercises, find a function of the form [latex]y=a{b}^{x}\\cos \\left(\\frac{\\pi }{2}x\\right)+c[\/latex] that fits the given data.<\/p>\n<p>53.<\/p>\n<table id=\"fs-id1673271\" class=\"unnumbered\" summary=\"Two rows, five columns. The table has ordered pairs of these column values: (x,y), (0,11), (1,3), (2,1), (3,3).\">\n<tbody>\n<tr>\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\n<td>0<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\n<td>11<\/td>\n<td>3<\/td>\n<td>1<\/td>\n<td>3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>54.<\/p>\n<table id=\"fs-id2909366\" class=\"unnumbered\" summary=\"Two rows, five columns. The table has ordered pairs of these column values: (x,y), (0,4), (1,1), (2,-11), (3,1).\">\n<tbody>\n<tr>\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\n<td>0<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td><strong> [latex]y[\/latex] <\/strong><\/td>\n<td>4<\/td>\n<td>1<\/td>\n<td>\u221211<\/td>\n<td>1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-14246\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-14246","chapter","type-chapter","status-publish","hentry"],"part":14191,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters\/14246","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":10,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters\/14246\/revisions"}],"predecessor-version":[{"id":17607,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters\/14246\/revisions\/17607"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/parts\/14191"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters\/14246\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/media?parent=14246"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=14246"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/contributor?post=14246"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/license?post=14246"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}