{"id":14414,"date":"2018-09-27T16:55:56","date_gmt":"2018-09-27T16:55:56","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/precalculus\/chapter\/polar-coordinates-graphs\/"},"modified":"2025-02-05T05:24:12","modified_gmt":"2025-02-05T05:24:12","slug":"polar-coordinates-graphs","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/precalculus\/chapter\/polar-coordinates-graphs\/","title":{"raw":"Polar Coordinates: Graphs","rendered":"Polar Coordinates: Graphs"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Test polar equations for symmetry.<\/li>\r\n \t<li>Graph polar equations by plotting points.<\/li>\r\n<\/ul>\r\n<\/div>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165505\/CNX_Precalc_Figure_08_04_0012.jpg\" alt=\"Illustration of the solar system with the sun at the center and orbits of the planets Mercury, Venus, Earth, and Mars shown.\" width=\"731\" height=\"411\" \/> <b>Figure 1.<\/b> Planets follow elliptical paths as they orbit around the Sun. (credit: modification of work by NASA\/JPL-Caltech)[\/caption]\r\n\r\nThe planets move through space in elliptical, periodic orbits about the sun, as shown in\u00a0Figure 1. They are in constant motion, so fixing an exact position of any planet is valid only for a moment. In other words, we can fix only a planet\u2019s <em>instantaneous <\/em>position. This is one application of <strong>polar coordinates<\/strong>, represented as [latex]\\left(r,\\theta \\right)[\/latex].\u00a0We interpret [latex]r[\/latex] as the distance from the sun and [latex]\\theta [\/latex] as the planet\u2019s angular bearing, or its direction from a fixed point on the sun. In this section, we will focus on the polar system and the graphs that are generated directly from polar coordinates.\r\n<h2>Testing Polar Equations for Symmetry<\/h2>\r\nJust as a rectangular equation such as [latex]y={x}^{2}[\/latex] describes the relationship between [latex]x[\/latex] and [latex]y[\/latex] on a Cartesian grid, a <strong>polar equation <\/strong>describes a relationship between [latex]r[\/latex] and [latex]\\theta [\/latex] on a polar grid. Recall that the coordinate pair [latex]\\left(r,\\theta \\right)[\/latex] indicates that we move counterclockwise from the polar axis (positive <em>x<\/em>-axis) by an angle of [latex]\\theta [\/latex], and extend a ray from the pole (origin) [latex]r[\/latex] units in the direction of [latex]\\theta [\/latex]. All points that satisfy the polar equation are on the graph.\r\n\r\nSymmetry is a property that helps us recognize and plot the graph of any equation. If an equation has a graph that is symmetric with respect to an axis, it means that if we folded the graph in half over that axis, the portion of the graph on one side would coincide with the portion on the other side. By performing three tests, we will see how to apply the properties of symmetry to polar equations. Further, we will use symmetry (in addition to plotting key points, zeros, and maximums of [latex]r[\/latex])\r\nto determine the graph of a polar equation.\r\n\r\nIn the first test, we consider symmetry with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex] (<em>y<\/em>-axis). We replace [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(-r,-\\theta \\right)[\/latex] to determine if the new equation is equivalent to the original equation. For example, suppose we are given the equation [latex]r=2\\sin \\theta [\/latex];\r\n<div style=\"text-align: center;\">[latex]\\begin{align}r&amp;=2\\sin \\theta \\\\ -r&amp;=2\\sin \\left(-\\theta \\right)&amp;&amp; \\text{Replace}\\left(r,\\theta \\right)\\text{with }\\left(-r,-\\theta \\right). \\\\ -r&amp;=-2\\sin \\theta&amp;&amp; \\text{Identity: }\\sin \\left(-\\theta \\right)=-\\sin \\theta. \\\\ r&amp;=2\\sin \\theta&amp;&amp; \\text{Multiply both sides by}-1. \\end{align}[\/latex]<\/div>\r\nThis equation exhibits symmetry with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex].\r\n\r\nIn the second test, we consider symmetry with respect to the polar axis ( [latex]x[\/latex] -axis). We replace [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(r,-\\theta \\right)[\/latex] or [latex]\\left(-r,\\pi -\\theta \\right)[\/latex] to determine equivalency between the tested equation and the original. For example, suppose we are given the equation [latex]r=1 - 2\\cos \\theta [\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{align}r&amp;=1 - 2\\cos \\theta \\\\ r&amp;=1 - 2\\cos \\left(-\\theta \\right)&amp;&amp; \\text{Replace }\\left(r,\\theta \\right)\\text{with}\\left(r,-\\theta \\right).\\\\ r&amp;=1 - 2\\cos \\theta&amp;&amp; \\text{Even\/Odd identity} \\end{align}[\/latex]<\/div>\r\nThe graph of this equation exhibits symmetry with respect to the polar axis.\r\n\r\nIn the third test, we consider symmetry with respect to the pole (origin). We replace [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(-r,\\theta \\right)[\/latex] to determine if the tested equation is equivalent to the original equation. For example, suppose we are given the equation [latex]r=2\\sin \\left(3\\theta \\right)[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{gathered}r=2\\sin \\left(3\\theta \\right)\\\\ -r=2\\sin \\left(3\\theta \\right)\\end{gathered}[\/latex]<\/div>\r\nThe equation has failed the <strong>symmetry test<\/strong>, but that does not mean that it is not symmetric with respect to the pole. Passing one or more of the symmetry tests verifies that symmetry will be exhibited in a graph. However, failing the symmetry tests does not necessarily indicate that a graph will not be symmetric about the line [latex]\\theta =\\frac{\\pi }{2}[\/latex], the polar axis, or the pole. In these instances, we can confirm that symmetry exists by plotting reflecting points across the apparent axis of symmetry or the pole. Testing for symmetry is a technique that simplifies the graphing of polar equations, but its application is not perfect.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Symmetry Tests<\/h3>\r\nA <strong>polar equation<\/strong> describes a curve on the polar grid. The graph of a polar equation can be evaluated for three types of symmetry, as shown in Figure 2.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"923\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165507\/CNX_Precalc_Figure_08_04_002new2.jpg\" alt=\"3 graphs side by side. (A) shows a ray extending into Q 1 and its symmetric version in Q 2. (B) shows a ray extending into Q 1 and its symmetric version in Q 4. (C) shows a ray extending into Q 1 and its symmetric version in Q 3. See caption for more information.\" width=\"923\" height=\"336\" \/> <b>Figure 2.<\/b> (a) A graph is symmetric with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex] (y-axis) if replacing [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(-r,-\\theta \\right)[\/latex] yields an equivalent equation. (b) A graph is symmetric with respect to the polar axis (x-axis) if replacing [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(r,-\\theta \\right)[\/latex] or [latex]\\left(-r,\\mathrm{\\pi -}\\theta \\right)[\/latex] yields an equivalent equation. (c) A graph is symmetric with respect to the pole (origin) if replacing [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(-r,\\theta \\right)[\/latex] yields an equivalent equation.[\/caption]<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a polar equation, test for symmetry.<\/h3>\r\n<ol>\r\n \t<li>Substitute the appropriate combination of components for [latex]\\left(r,\\theta \\right):[\/latex] [latex]\\left(-r,-\\theta \\right)[\/latex] for [latex]\\theta =\\frac{\\pi }{2}[\/latex] symmetry; [latex]\\left(r,-\\theta \\right)[\/latex] for polar axis symmetry; and [latex]\\left(-r,\\theta \\right)[\/latex] for symmetry with respect to the pole.<\/li>\r\n \t<li>If the resulting equations are equivalent in one or more of the tests, the graph produces the expected symmetry.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1: Testing a Polar Equation for Symmetry<\/h3>\r\nTest the equation [latex]r=2\\sin \\theta [\/latex] for symmetry.\r\n\r\n[reveal-answer q=\"995200\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"995200\"]\r\n\r\nTest for each of the three types of symmetry.\r\n<table id=\"eip-id3747579\" summary=\"Three rows and two columns. The first column contains the steps to test for a type of symmetry, and the second column gives an example. The first column, first row tests symmetry with respect to theta= pi\/2. Test: Replacing (r, theta) with (-r, -theta) yields the same result. Thus, the graph is symmetric with respect to the line pi\/2. The example is -r = 2sin(-theta). By the even-odd identity, -r = -2sin(theta). After multiplying by -1, r=2sin(theta), so it passes the test. The next test is symmetry with respect to the polar axis. Test: Replacing theta with -theta does not yield the same equation. Therefore, the graph fails the test and may or may not be symmetric with respect to the polar axis. Example: r=2sin(-theta). By the even-odd identity, r=-2sin(theta). We have then r=-2sin(theta) which does not equal 2 sin(theta), so it fails. Finally, there is symmetry with respect to the pole. Test: Replacing r with -r changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole. Example: -r = 2sin(theta). r=-2sin(theta) which does not equal 2sin(theta), so it fails the test.\">\r\n<tbody>\r\n<tr>\r\n<td>1) Replacing [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(-r,-\\theta \\right)[\/latex] yields the same result. Thus, the graph is symmetric with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex].<\/td>\r\n<td>[latex]\\begin{align}&amp;-r=2\\sin \\left(-\\theta \\right) \\\\ &amp;-r=-2\\sin \\theta&amp;&amp; \\text{Even-odd identity} \\\\ &amp;r=2\\sin \\theta&amp;&amp; \\text{Multiply}\\text{by}-1 \\\\ &amp;\\text{Passed} \\end{align}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2) Replacing [latex]\\theta [\/latex] with [latex]-\\theta [\/latex] does not yield the same equation. Therefore, the graph fails the test and may or may not be symmetric with respect to the polar axis.<\/td>\r\n<td>[latex]\\begin{align}&amp;r=2\\sin \\left(-\\theta \\right) \\\\ r&amp;=-2\\sin \\theta &amp;&amp; \\text{Even-odd identity} \\\\ &amp;r=-2\\sin \\theta \\ne 2\\sin \\theta \\\\ &amp;\\text{Failed} \\end{align}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3) Replacing [latex]r[\/latex] with [latex]-r[\/latex] changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole.<\/td>\r\n<td>[latex]\\begin{align}&amp;-r=2\\sin \\theta \\\\ &amp;r=-2\\sin \\theta \\ne 2\\sin \\theta \\\\ &amp;\\text{Failed} \\end{align}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h4>Analysis of the Solution<\/h4>\r\nUsing a graphing calculator, we can see that the equation [latex]r=2\\sin \\theta [\/latex] is a circle centered at [latex]\\left(0,1\\right)[\/latex] with radius [latex]r=1[\/latex] and is indeed symmetric to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex]. We can also see that the graph is not symmetric with the polar axis or the pole. See 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\/27165509\/CNX_Precalc_Figure_08_04_0032.jpg\" alt=\"Graph of the given circle on the polar coordinate grid. Center is at (0,1), and it has radius 1.\" width=\"487\" height=\"369\" \/> <b>Figure 3<\/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\nTest the equation for symmetry: [latex]r=-2\\cos \\theta [\/latex].\r\n\r\n[reveal-answer q=\"47168\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"47168\"]\r\n\r\nThe equation fails the symmetry test with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex] and with respect to the pole. It passes the polar axis symmetry test.\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]169519[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>\u00a0Graphing Polar Equations by Plotting Points<\/h2>\r\nTo graph in the rectangular coordinate system we construct a table of [latex]x[\/latex] and [latex]y[\/latex] values. To graph in the polar coordinate system we construct a table of [latex]\\theta [\/latex] and [latex]r[\/latex] values. We enter values of [latex]\\theta [\/latex]\u00a0into a <strong>polar equation<\/strong> and calculate [latex]r[\/latex]. However, using the properties of symmetry and finding key values of [latex]\\theta [\/latex] and [latex]r[\/latex] means fewer calculations will be needed.\r\n<h2>Finding Zeros and Maxima<\/h2>\r\nTo find the zeros of a polar equation, we solve for the values of [latex]\\theta [\/latex] that result in [latex]r=0[\/latex]. Recall that, to find the zeros of polynomial functions, we set the equation equal to zero and then solve for [latex]x[\/latex]. We use the same process for polar equations. Set [latex]r=0[\/latex], and solve for [latex]\\theta [\/latex].\r\n\r\nFor many of the forms we will encounter, the maximum value of a polar equation is found by substituting those values of [latex]\\theta [\/latex] into the equation that result in the maximum value of the trigonometric functions. Consider [latex]r=5\\cos \\theta [\/latex]; the maximum distance between the curve and the pole is 5 units. The maximum value of the cosine function is 1 when [latex]\\theta =0[\/latex], so our polar equation is [latex]5\\cos \\theta [\/latex], and the value [latex]\\theta =0[\/latex] will yield the maximum [latex]|r|[\/latex].\r\n\r\nSimilarly, the maximum value of the sine function is 1 when [latex]\\theta =\\frac{\\pi }{2}[\/latex], and if our polar equation is [latex]r=5\\sin \\theta [\/latex], the value [latex]\\theta =\\frac{\\pi }{2}[\/latex] will yield the maximum [latex]|r|[\/latex]. We may find additional information by calculating values of [latex]r[\/latex] when [latex]\\theta =0[\/latex]. These points would be polar axis intercepts, which may be helpful in drawing the graph and identifying the curve of a polar equation.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2: Finding Zeros and Maximum Values for a Polar Equation<\/h3>\r\nUsing the equation in Example 1, find the zeros and maximum [latex]|r|[\/latex] and, if necessary, the polar axis intercepts of [latex]r=2\\sin \\theta [\/latex].\r\n\r\n[reveal-answer q=\"927984\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"927984\"]\r\n\r\nTo find the zeros, set [latex]r[\/latex] equal to zero and solve for [latex]\\theta [\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;2\\sin \\theta =0 \\\\ &amp;\\sin \\theta =0 \\\\ &amp;\\theta ={\\sin }^{-1}0 \\\\ &amp;\\theta =n\\pi&amp;&amp; \\text{where }n\\text{ is an integer} \\end{align}[\/latex]<\/p>\r\nSubstitute any one of the [latex]\\theta [\/latex] values into the equation. We will use [latex]0[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align} &amp;r=2\\sin \\left(0\\right) \\\\ &amp;r=0 \\end{align}[\/latex]<\/p>\r\nThe points [latex]\\left(0,0\\right)[\/latex] and [latex]\\left(0,\\pm n\\pi \\right)[\/latex] are the zeros of the equation. They all coincide, so only one point is visible on the graph. This point is also the only polar axis intercept.\r\n\r\nTo find the maximum value of the equation, look at the maximum value of the trigonometric function [latex]\\sin \\theta [\/latex], which occurs when [latex]\\theta =\\frac{\\pi }{2}\\pm 2k\\pi [\/latex] resulting in [latex]\\sin \\left(\\frac{\\pi }{2}\\right)=1[\/latex]. Substitute [latex]\\frac{\\pi }{2}[\/latex] for [latex]\\mathrm{\\theta .}[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;r=2\\sin \\left(\\frac{\\pi }{2}\\right) \\\\ &amp;r=2\\left(1\\right) \\\\ &amp;r=2 \\end{align}[\/latex]<\/p>\r\n\r\n<h3>Analysis of the Solution<\/h3>\r\nThe point [latex]\\left(2,\\frac{\\pi }{2}\\right)[\/latex] will be the maximum value on the graph. Let\u2019s plot a few more points to verify the graph of a circle.\r\n<table id=\"Table_08_04_01\" summary=\"Eight rows and 3 columns. First column is labeled theta, second column is labeled r=2sin(theta), and third column is labeled r. The table has ordered triples of these column values: (0, r=2sin(0)=0, 0), (pi\/6, r=2sin(pi\/6)=1, 1), (pi\/3, r=2sin(pi\/3) = approx. 1.73, 1.73), (pi\/2, r=2sin(pi\/2) = 2, 2), (2pi\/3, r=2sin(2pi\/3)=approx. 1.73, 1.73), (5pi\/6, r=2sin(5pi\/6)=1, 1), and (pi, r=2sin(pi)=0).\">\r\n<thead>\r\n<tr>\r\n<th>[latex]\\theta [\/latex]<\/th>\r\n<th>[latex]r=2\\sin \\theta [\/latex]<\/th>\r\n<th>[latex]r[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>0<\/td>\r\n<td>[latex]r=2\\sin \\left(0\\right)=0[\/latex]<\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\frac{\\pi }{6}[\/latex]<\/td>\r\n<td>[latex]r=2\\sin \\left(\\frac{\\pi }{6}\\right)=1[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\frac{\\pi }{3}[\/latex]<\/td>\r\n<td>[latex]r=2\\sin \\left(\\frac{\\pi }{3}\\right)\\approx 1.73[\/latex]<\/td>\r\n<td>[latex]1.73[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\r\n<td>[latex]r=2\\sin \\left(\\frac{\\pi }{2}\\right)=2[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\frac{2\\pi }{3}[\/latex]<\/td>\r\n<td>[latex]r=2\\sin \\left(\\frac{2\\pi }{3}\\right)\\approx 1.73[\/latex]<\/td>\r\n<td>[latex]1.73[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\frac{5\\pi }{6}[\/latex]<\/td>\r\n<td>[latex]r=2\\sin \\left(\\frac{5\\pi }{6}\\right)=1[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\pi [\/latex]<\/td>\r\n<td>[latex]r=2\\sin \\left(\\pi \\right)=0[\/latex]<\/td>\r\n<td>[latex]0[\/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\/27165512\/CNX_Precalc_Figure_08_04_0042.jpg\" alt=\"Graph of circle on the polar coordinate grid. The center is at (0,1), and it has radius 1. Six points along the circumference are marked: (0,0), (1, pi\/6), (1.3, pi\/3), (2, pi\/2), (1.73, 2pi\/3), and (1, 5pi\/6).\" width=\"487\" height=\"369\" \/> <b>Figure 4<\/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\nWithout converting to Cartesian coordinates, test the given equation for symmetry and find the zeros and maximum values of [latex]|r|:[\/latex] [latex]r=3\\cos \\theta [\/latex].\r\n\r\n[reveal-answer q=\"16075\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"16075\"]\r\n\r\nTests will reveal symmetry about the polar axis. The zero is [latex]\\left(0,\\frac{\\pi }{2}\\right)[\/latex], and the maximum value is [latex]\\left(3,0\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Graphing Circles and the 5 Classic Polar Curves<\/h2>\r\n<h2>Investigating Circles<\/h2>\r\nNow we have seen the equation of a circle in the polar coordinate system. In the last two examples, the same equation was used to illustrate the properties of symmetry and demonstrate how to find the zeros, maximum values, and plotted points that produced the graphs. However, the circle is only one of many shapes in the set of polar curves.\r\n\r\nThere are five classic polar curves<strong>: cardioids<\/strong>, <strong>lima\u04abons, lemniscates, rose curves<\/strong>, and <strong>Archimedes\u2019 spirals<\/strong>. We will briefly touch on the polar formulas for the circle before moving on to the classic curves and their variations.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Formulas for the Equation of a Circle<\/h3>\r\nSome of the formulas that produce the graph of a circle in polar coordinates are given by [latex]r=a\\cos \\theta [\/latex] and [latex]r=a\\sin \\theta [\/latex], where [latex]a[\/latex] is the diameter of the circle or the distance from the pole to the farthest point on the circumference. The radius is [latex]\\frac{|a|}{2}[\/latex], or one-half the diameter. For [latex]r=a\\cos \\theta , [\/latex] the center is [latex]\\left(\\frac{a}{2},0\\right)[\/latex]. For [latex]r=a\\sin \\theta [\/latex], the center is [latex]\\left(\\frac{a}{2},\\pi \\right)[\/latex]. Figure 5\u00a0shows the graphs of these four circles.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165514\/CNX_Precalc_Figure_08_04_005new2.jpg\" alt=\"&quot;Four\" \/>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3: Sketching the Graph of a Polar Equation for a Circle<\/h3>\r\nSketch the graph of [latex]r=4\\cos \\theta [\/latex].\r\n\r\n[reveal-answer q=\"198958\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"198958\"]\r\n\r\nFirst, testing the equation for symmetry, we find that the graph is symmetric about the polar axis. Next, we find the <strong>zeros<\/strong> and maximum [latex]|r|[\/latex] for [latex]r=4\\cos \\theta [\/latex]. First, set [latex]r=0[\/latex], and solve for [latex]\\theta [\/latex] . Thus, a zero occurs at [latex]\\theta =\\frac{\\pi }{2}\\pm k\\pi [\/latex]. A key point to plot is [latex]\\left(0,\\text{ }\\text{ }\\frac{\\pi }{2}\\right)[\/latex].\r\n\r\nTo find the maximum value of [latex]r[\/latex], note that the maximum value of the cosine function is 1 when [latex]\\theta =0\\pm 2k\\pi [\/latex]. Substitute [latex]\\theta =0[\/latex] into the equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}r=4\\cos \\theta \\\\ r=4\\cos \\left(0\\right)\\\\ r=4\\left(1\\right)=4\\end{gathered}[\/latex]<\/p>\r\nThe maximum value of the equation is 4. A key point to plot is [latex]\\left(4,0\\right)[\/latex].\r\n\r\nAs [latex]r=4\\cos \\theta [\/latex] is symmetric with respect to the polar axis, we only need to calculate <em>r<\/em>-values for [latex]\\theta [\/latex] over the interval [latex]\\left[0,\\pi \\right][\/latex]. Points in the upper quadrant can then be reflected to the lower quadrant. Make a table of values similar to the table below. The graph is shown in Figure 6.\r\n<table id=\"Table_08_04_02\" summary=\"Two rows and ten columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,4), (pi\/6, 3.46), (pi\/4, 2.83), (pi\/3,2), (pi\/2,0), (2pi\/3,-2), (3pi\/4,-2.83), (5pi\/6, -3.46), and (pi,4).\">\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]\\theta [\/latex] <\/strong><\/td>\r\n<td>0<\/td>\r\n<td>[latex]\\frac{\\pi }{6}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\pi }{4}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\pi }{3}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{2\\pi }{3}[\/latex]<\/td>\r\n<td>[latex]\\frac{3\\pi }{4}[\/latex]<\/td>\r\n<td>[latex]\\frac{5\\pi }{6}[\/latex]<\/td>\r\n<td>[latex]\\pi [\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong> [latex]r[\/latex] <\/strong><\/td>\r\n<td>4<\/td>\r\n<td>3.46<\/td>\r\n<td>2.83<\/td>\r\n<td>2<\/td>\r\n<td>0<\/td>\r\n<td>\u22122<\/td>\r\n<td>\u22122.83<\/td>\r\n<td>\u22123.46<\/td>\r\n<td>4<\/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\/27165517\/CNX_Precalc_Figure_08_04_0062.jpg\" alt=\"Graph of 4=4cos(theta) in polar coordinates. Points (0, pi\/2), (-2, 2pi\/3), (4,0), and (2, pi\/3) are marked on the circumference.\" width=\"487\" height=\"364\" \/> <b>Figure 6<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Investigating Cardioids<\/h2>\r\nWhile translating from polar coordinates to Cartesian coordinates may seem simpler in some instances, graphing the classic curves is actually less complicated in the polar system. The next curve is called a cardioid, as it resembles a heart. This shape is often included with the family of curves called lima\u00e7ons, but here we will discuss the cardioid on its own.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Formulas for a Cardioid<\/h3>\r\nThe formulas that produce the graphs of a <strong>cardioid<\/strong> are given by [latex]r=a\\pm b\\cos \\theta [\/latex] and [latex]r=a\\pm b\\sin \\theta [\/latex] where [latex]a&gt;0,b&gt;0[\/latex], and [latex]\\frac{a}{b}=1[\/latex]. The cardioid graph passes through the pole, as we can see in Figure 7.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"923\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165519\/CNX_Precalc_Figure_08_04_007new2.jpg\" alt=\"Graph of four cardioids. (A) is r = a + bcos(theta). Cardioid extending to the right. (B) is r=a-bcos(theta). Cardioid extending to the left. (C) is r=a+bsin(theta). Cardioid extending up. (D) is r=a-bsin(theta). Cardioid extending down.\" width=\"923\" height=\"290\" \/> <b>Figure 7<\/b>[\/caption]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given the polar equation of a cardioid, sketch its graph.<\/h3>\r\n<ol>\r\n \t<li>Check equation for the three types of symmetry.<\/li>\r\n \t<li>Find the zeros. Set [latex]r=0[\/latex].<\/li>\r\n \t<li>Find the maximum value of the equation according to the maximum value of the trigonometric expression.<\/li>\r\n \t<li>Make a table of values for [latex]r[\/latex] and [latex]\\theta [\/latex].<\/li>\r\n \t<li>Plot the points and sketch the graph.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4: Sketching the Graph of a Cardioid<\/h3>\r\nSketch the graph of [latex]r=2+2\\cos \\theta [\/latex].\r\n\r\n[reveal-answer q=\"869763\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"869763\"]\r\n\r\nFirst, testing the equation for symmetry, we find that the graph of this equation will be symmetric about the polar axis. Next, we find the zeros and maximums. Setting [latex]r=0[\/latex], we have [latex]\\theta =\\pi +2k\\pi [\/latex]. The zero of the equation is located at [latex]\\left(0,\\pi \\right)[\/latex]. The graph passes through this point.\r\n\r\nThe maximum value of [latex]r=2+2\\cos \\theta [\/latex] occurs when [latex]\\cos \\theta [\/latex] is a maximum, which is when [latex]\\cos \\theta =1[\/latex] or when [latex]\\theta =0[\/latex]. Substitute [latex]\\theta =0[\/latex] into the equation, and solve for [latex]r[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered} r=2+2\\cos \\left(0\\right) \\\\ r=2+2\\left(1\\right)=4 \\end{gathered}[\/latex]<\/p>\r\nThe point [latex]\\left(4,0\\right)[\/latex] is the maximum value on the graph.\r\n\r\nWe found that the polar equation is symmetric with respect to the polar axis, but as it extends to all four quadrants, we need to plot values over the interval [latex]\\left[0,\\pi \\right][\/latex]. The upper portion of the graph is then reflected over the polar axis. Next, we make a table of values, as in the table below, and then we plot the points and draw the graph. See Figure 8.\r\n<table id=\"Table_08_04_03\" summary=\"Two rows and six columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,4), (pi\/4, 3.41), (pi\/2, 2), (2pi\/3, 1), and (pi, 0).\">\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 }{4}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{2\\pi }{3}[\/latex]<\/td>\r\n<td>[latex]\\pi [\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong> [latex]r[\/latex] <\/strong><\/td>\r\n<td>4<\/td>\r\n<td>3.41<\/td>\r\n<td>2<\/td>\r\n<td>1<\/td>\r\n<td>0<\/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\/27165522\/CNX_Precalc_Figure_08_04_0082.jpg\" alt=\"Graph of r=2+2cos(theta). Cardioid extending to the right. Points on the edge (0,pi), (4,0),(3.4, pi\/4), (2,pi\/2), and (1, 2pi\/3) are shown.\" width=\"487\" height=\"369\" \/> <b>Figure 8<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Investigating Lima\u00e7ons<\/h2>\r\nThe word <em>lima\u00e7on<\/em> is Old French for \"snail,\" a name that describes the shape of the graph. As mentioned earlier, the cardioid is a member of the lima\u00e7on family, and we can see the similarities in the graphs. The other images in this category include the one-loop lima\u00e7on and the two-loop (or inner-loop) lima\u00e7on. <strong>One-loop lima\u00e7ons<\/strong> are sometimes referred to as <strong>dimpled lima\u00e7ons<\/strong> when [latex]1&lt;\\frac{a}{b}&lt;2[\/latex] and <strong>convex lima\u00e7ons<\/strong> when [latex]\\frac{a}{b}\\ge 2[\/latex].\r\n<div class=\"textbox\">\r\n<h3>A General Note: Formulas for One-Loop Lima\u00e7ons<\/h3>\r\nThe formulas that produce the graph of a dimpled <strong>one-loop lima\u00e7on<\/strong> are given by [latex]r=a\\pm b\\cos \\theta [\/latex] and [latex]r=a\\pm b\\sin \\theta [\/latex] where [latex]a&gt;0,b&gt;0,\\text{and 1&lt;}\\frac{a}{b}&lt;2[\/latex]. All four graphs are shown in Figure 9.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"923\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165524\/CNX_Precalc_Figure_08_04_009new2.jpg\" alt=\"Four dimpled lima\u00e7ons side by side. (A) is r=a+bcos(theta). Extending to the right. (B) is r=a-bcos(theta). Extending to the left. (C) is r=a+bsin(theta). Extending up. (D) is r=a-bsin(theta). Extending down. \" width=\"923\" height=\"290\" \/> <b>Figure 9.<\/b> Dimpled lima\u00e7ons[\/caption]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a polar equation for a one-loop lima\u00e7on, sketch the graph.<\/h3>\r\n<ol>\r\n \t<li>Test the equation for symmetry. Remember that failing a symmetry test does not mean that the shape will not exhibit symmetry. Often the symmetry may reveal itself when the points are plotted.<\/li>\r\n \t<li>Find the zeros.<\/li>\r\n \t<li>Find the maximum values according to the trigonometric expression.<\/li>\r\n \t<li>Make a table.<\/li>\r\n \t<li>Plot the points and sketch the graph.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 5: Sketching the Graph of a One-Loop Lima\u00e7on<\/h3>\r\nGraph the equation [latex]r=4 - 3\\sin \\theta [\/latex].\r\n\r\n[reveal-answer q=\"739204\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"739204\"]\r\n\r\nFirst, testing the equation for symmetry, we find that it fails all three symmetry tests, meaning that the graph may or may not exhibit symmetry, so we cannot use the symmetry to help us graph it. However, this equation has a graph that clearly displays symmetry with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex], yet it fails all the three symmetry tests. A graphing calculator will immediately illustrate the graph\u2019s reflective quality.\r\n\r\nNext, we find the zeros and maximum, and plot the reflecting points to verify any symmetry. Setting [latex]r=0[\/latex] results in [latex]\\theta [\/latex] being undefined. What does this mean? How could [latex]\\theta [\/latex] be undefined? The angle [latex]\\theta [\/latex] is undefined for any value of [latex]\\sin \\theta &gt;1[\/latex]. Therefore, [latex]\\theta [\/latex] is undefined because there is no value of [latex]\\theta [\/latex] for which [latex]\\sin \\theta &gt;1[\/latex]. Consequently, the graph does not pass through the pole. Perhaps the graph does cross the polar axis, but not at the pole. We can investigate other intercepts by calculating [latex]r[\/latex] when [latex]\\theta =0[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}r\\left(0\\right)=4 - 3\\sin \\left(0\\right) \\\\ r=4 - 3\\cdot 0=4 \\end{gathered}[\/latex]<\/p>\r\nSo, there is at least one polar axis intercept at [latex]\\left(4,0\\right)[\/latex].\r\n\r\nNext, as the maximum value of the sine function is 1 when [latex]\\theta =\\frac{\\pi }{2}[\/latex], we will substitute [latex]\\theta =\\frac{\\pi }{2}[\/latex]\u00a0into the equation and solve for [latex]r[\/latex]. Thus, [latex]r=1[\/latex].\r\n\r\nMake a table of the coordinates similar to the table below.\r\n<table id=\"Table_08_04_04\" summary=\"Two rows and fourteen columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,4), (pi\/6, 2.5), (pi\/3,1.4), (pi\/2, 1), (2pi\/3, 1.4), (5pi\/6, 2.5), (pi,4), (7pi\/6, 5.5), (4pi\/3,6.6), (3pi\/2, 7), (5pi\/3, 6.6), (11pi\/6, 5.5), and (2pi, 4).\">\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 }{6}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\pi }{3}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{2\\pi }{3}[\/latex]<\/td>\r\n<td>[latex]\\frac{5\\pi }{6}[\/latex]<\/td>\r\n<td>[latex]\\pi [\/latex]<\/td>\r\n<td>[latex]\\frac{7\\pi }{6}[\/latex]<\/td>\r\n<td>[latex]\\frac{4\\pi }{3}[\/latex]<\/td>\r\n<td>[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{5\\pi }{3}[\/latex]<\/td>\r\n<td>[latex]\\frac{11\\pi }{6}[\/latex]<\/td>\r\n<td>[latex]2\\pi [\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong> [latex]r[\/latex] <\/strong><\/td>\r\n<td>4<\/td>\r\n<td>2.5<\/td>\r\n<td>1.4<\/td>\r\n<td>1<\/td>\r\n<td>1.4<\/td>\r\n<td>2.5<\/td>\r\n<td>4<\/td>\r\n<td>5.5<\/td>\r\n<td>6.6<\/td>\r\n<td>7<\/td>\r\n<td>6.6<\/td>\r\n<td>5.5<\/td>\r\n<td>4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe graph is shown in Figure 10.\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\/27165527\/CNX_Precalc_Figure_08_04_0102.jpg\" alt=\"Graph of the lima\u00e7on r=4-3sin(theta). Extending down. Points on the edge are shown: (1,pi\/2), (4,0), (4,pi), and (7, 3pi\/2). \" width=\"487\" height=\"423\" \/> <b>Figure 10.<\/b> One-loop lima\u00e7on[\/caption]\r\n<h4>Analysis of the Solution<\/h4>\r\nThis is an example of a curve for which making a table of values is critical to producing an accurate graph. The symmetry tests fail; the zero is undefined. While it may be apparent that an equation involving [latex]\\sin \\theta [\/latex] is likely symmetric with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex], evaluating more points helps to verify that the graph is correct.\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\nSketch the graph of [latex]r=3 - 2\\cos \\theta [\/latex].\r\n\r\n[reveal-answer q=\"875395\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"875395\"]\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165558\/CNX_Precalc_Figure_08_04_0112.jpg\" alt=\"Graph of the lima\u00e7on r=3-2cos(theta). Extending to the left.\" \/>\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]149412[\/ohm_question]\r\n\r\n<\/div>\r\nAnother type of lima\u00e7on, the <strong>inner-loop lima\u00e7on<\/strong>, is named for the loop formed inside the general lima\u00e7on shape. It was discovered by the German artist Albrecht <strong>D\u00fcrer<\/strong>(1471-1528), who revealed a method for drawing the inner-loop lima\u00e7on in his 1525 book <em>Underweysung der Messing<\/em>. A century later, the father of mathematician Blaise <strong>Pascal<\/strong>, \u00c9tienne Pascal(1588-1651), rediscovered it.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Formulas for Inner-Loop Lima\u00e7ons<\/h3>\r\nThe formulas that generate the <strong>inner-loop lima\u00e7ons<\/strong> are given by [latex]r=a\\pm b\\cos \\theta [\/latex] and [latex]r=a\\pm b\\sin \\theta [\/latex] where [latex]a&gt;0,b&gt;0[\/latex], and [latex]a&lt;b[\/latex]. The graph of the inner-loop lima\u00e7on passes through the pole twice: once for the outer loop, and once for the inner loop. See Figure 11\u00a0for the graphs.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"923\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165529\/CNX_Precalc_Figure_08_04_012new2.jpg\" alt=\"Graph of four inner loop lima\u00e7ons side by side. (A) is r=a+bcos(theta),a&lt;b. Extended to the right. (B) is a-bcos(theta), a&lt;b. Extends to the left. (C) is r=a+bsin(theta), a&lt;b. Extends up. (D) is r=a-bsin(theta), a&lt;b. Extends down.\" width=\"923\" height=\"290\" \/> <b>Figure 11<\/b>[\/caption]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 6: Sketching the Graph of an Inner-Loop Lima\u00e7on<\/h3>\r\nSketch the graph of [latex]r=2+5\\text{cos}\\theta [\/latex].\r\n\r\n[reveal-answer q=\"360339\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"360339\"]\r\n\r\nTesting for symmetry, we find that the graph of the equation is symmetric about the polar axis. Next, finding the zeros reveals that when [latex]r=0,\\theta =1.98[\/latex].\u00a0The maximum [latex]|r|[\/latex] is found when [latex]\\cos \\theta =1[\/latex] or when [latex]\\theta =0[\/latex]. Thus, the maximum is found at the point (7, 0).\r\n\r\nEven though we have found symmetry, the zero, and the maximum, plotting more points will help to define the shape, and then a pattern will emerge.\r\n<table id=\"Table_08_04_05\" summary=\"Two rows and fourteen columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,7), (pi\/6, 6.3.), (pi\/3,4.5), (pi\/2, 2), (2pi\/3, -.5), (5pi\/6, -2.3), (pi,-3), (7pi\/6, -2.3), (4pi\/3,-.5), (3pi\/2, 2), (5pi\/3, 4.5), (11pi\/6, 6.3), and (2pi, 7).\">\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 }{6}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\pi }{3}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{2\\pi }{3}[\/latex]<\/td>\r\n<td>[latex]\\frac{5\\pi }{6}[\/latex]<\/td>\r\n<td>[latex]\\pi [\/latex]<\/td>\r\n<td>[latex]\\frac{7\\pi }{6}[\/latex]<\/td>\r\n<td>[latex]\\frac{4\\pi }{3}[\/latex]<\/td>\r\n<td>[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{5\\pi }{3}[\/latex]<\/td>\r\n<td>[latex]\\frac{11\\pi }{6}[\/latex]<\/td>\r\n<td>[latex]2\\pi [\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong> [latex]r[\/latex] <\/strong><\/td>\r\n<td>7<\/td>\r\n<td>6.3<\/td>\r\n<td>4.5<\/td>\r\n<td>2<\/td>\r\n<td>\u22120.5<\/td>\r\n<td>\u22122.3<\/td>\r\n<td>\u22123<\/td>\r\n<td>\u22122.3<\/td>\r\n<td>\u22120.5<\/td>\r\n<td>2<\/td>\r\n<td>4.5<\/td>\r\n<td>6.3<\/td>\r\n<td>7<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAs expected, the values begin to repeat after [latex]\\theta =\\pi [\/latex]. The graph is shown in Figure 12.\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\/27165532\/CNX_Precalc_Figure_08_04_0132.jpg\" alt=\"Graph of inner loop lima\u00e7on r=2+5cos(theta). Extends to the right. Points on edge plotted are (7,0), (4.5, pi\/3), (2, pi\/2), and (-3, pi).\" width=\"487\" height=\"423\" \/> <b>Figure 12.<\/b> Inner-loop lima\u00e7on[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Investigating Lemniscates<\/h2>\r\nThe lemniscate is a polar curve resembling the infinity symbol [latex]\\infty [\/latex] or a figure 8. Centered at the pole, a lemniscate is symmetrical by definition.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Formulas for Lemniscates<\/h3>\r\nThe formulas that generate the graph of a <strong>lemniscate<\/strong> are given by [latex]{r}^{2}={a}^{2}\\cos 2\\theta [\/latex] and [latex]{r}^{2}={a}^{2}\\sin 2\\theta [\/latex] where [latex]a\\ne 0[\/latex]. The formula [latex]{r}^{2}={a}^{2}\\sin 2\\theta [\/latex] is symmetric with respect to the pole. The formula [latex]{r}^{2}={a}^{2}\\cos 2\\theta [\/latex] is symmetric with respect to the pole, the line [latex]\\theta =\\frac{\\pi }{2}[\/latex], and the polar axis. See Figure 13\u00a0for the graphs.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"923\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165535\/CNX_Precalc_Figure_08_04_014new2.jpg\" alt=\"Four graphs of lemniscates side by side. (A) is r^2 = a^2 * cos(2theta). Horizonatal figure eight, on x-axis. (B) is r^2 = - a^2 * cos(2theta). Vertical figure eight, on y axis. (C) is r^2 = a^2 * sin(2theta). Diagonal figure eight on line y=x. (D) is r^2 = -a^2 *sin(2theta). Diagonal figure eight on line y=-x.\" width=\"923\" height=\"282\" \/> <b>Figure 13<\/b>[\/caption]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 7: Sketching the Graph of a Lemniscate<\/h3>\r\nSketch the graph of [latex]{r}^{2}=4\\cos 2\\theta [\/latex].\r\n\r\n[reveal-answer q=\"721985\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"721985\"]\r\n\r\nThe equation exhibits symmetry with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex], the polar axis, and the pole.\r\n\r\nLet\u2019s find the zeros. It should be routine by now, but we will approach this equation a little differently by making the substitution [latex]u=2\\theta [\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;0=4\\cos 2\\theta \\\\ &amp;0=4\\cos u \\\\ &amp;0=\\cos u \\\\ &amp;{\\cos }^{-1}0=\\frac{\\pi }{2} \\\\ &amp;u=\\frac{\\pi }{2}&amp;&amp; \\text{Substitute }2\\theta \\text{ back in for }u. \\\\ &amp;2\\theta =\\frac{\\pi }{2} \\\\ &amp;\\theta =\\frac{\\pi }{4} \\end{align}[\/latex]<\/p>\r\nSo, the point [latex]\\left(0,\\frac{\\pi }{4}\\right)[\/latex] is a zero of the equation.\r\n\r\nNow let\u2019s find the maximum value. Since the maximum of [latex]\\cos u=1[\/latex] when [latex]u=0[\/latex], the maximum [latex]\\cos 2\\theta =1[\/latex] when [latex]2\\theta =0[\/latex]. Thus,\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}{r}^{2}=4\\cos \\left(0\\right)\\\\ {r}^{2}=4\\left(1\\right)=4\\\\ r=\\pm \\sqrt{4}=2\\end{gathered}[\/latex]<\/p>\r\nWe have a maximum at (2, 0). Since this graph is symmetric with respect to the pole, the line [latex]\\theta =\\frac{\\pi }{2}[\/latex], and the polar axis, we only need to plot points in the first quadrant.\r\n\r\nMake a table similar to the table below.\r\n<table id=\"Table_08_04_06\" summary=\"Two rows and six columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,2), (pi\/6, rad2), (pi\/4,0), (pi\/3, rad2), (pi\/2,0).\">\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]\\theta [\/latex] <\/strong><\/td>\r\n<td>0<\/td>\r\n<td>[latex]\\frac{\\pi }{6}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\pi }{4}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\pi }{3}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong> [latex]r[\/latex] <\/strong><\/td>\r\n<td>2<\/td>\r\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\r\n<td>0<\/td>\r\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nPlot the points on the graph, such as the one shown in 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\/27165537\/CNX_Precalc_Figure_08_04_0152.jpg\" alt=\"Graph of r^2 = 4cos(2theta). Horizontal lemniscate, along x-axis. Points on edge plotted are (2,0), (rad2, pi\/6), (rad2 7pi\/6).\" width=\"487\" height=\"210\" \/> <b>Figure 14.<\/b> Lemniscate[\/caption]\r\n<h4>Analysis of the Solution<\/h4>\r\nMaking a substitution such as [latex]u=2\\theta [\/latex] is a common practice in mathematics because it can make calculations simpler. However, we must not forget to replace the substitution term with the original term at the end, and then solve for the unknown.\r\n\r\nSome of the points on this graph may not show up using the Trace function on the TI-84 graphing calculator, and the calculator table may show an error for these same points of [latex]r[\/latex]. This is because there are no real square roots for these values of [latex]\\theta [\/latex]. In other words, the corresponding <em>r<\/em>-values of [latex]\\sqrt{4\\cos \\left(2\\theta \\right)}[\/latex]\r\nare complex numbers because there is a negative number under the radical.\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;\">Investigating Rose Curves<\/span>\r\n\r\n<\/div>\r\nThe next type of polar equation produces a petal-like shape called a rose curve. Although the graphs look complex, a simple polar equation generates the pattern.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Rose Curves<\/h3>\r\nThe formulas that generate the graph of a <strong>rose curve<\/strong> are given by [latex]r=a\\cos n\\theta [\/latex] and [latex]r=a\\sin n\\theta [\/latex] where [latex]a\\ne 0[\/latex]. If [latex]n[\/latex] is even, the curve has [latex]2n[\/latex] petals. If [latex]n[\/latex] is odd, the curve has [latex]n[\/latex] petals.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165539\/CNX_Precalc_Figure_08_04_016new2.jpg\" alt=\"Graph of two rose curves side by side. (A) is r=acos(ntheta), where n is even. Eight petals extending from origin, equally spaced. (B) is r=asin(ntheta) where n is odd. Three petals extending from the origin, equally spaced.\" width=\"731\" height=\"385\" \/> <b>Figure 15<\/b>[\/caption]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 8: Sketching the Graph of a Rose Curve (<em>n<\/em> Even)<\/h3>\r\nSketch the graph of [latex]r=2\\cos 4\\theta [\/latex].\r\n\r\n[reveal-answer q=\"667305\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"667305\"]\r\n\r\nTesting for symmetry, we find again that the symmetry tests do not tell the whole story. The graph is not only symmetric with respect to the polar axis, but also with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex] and the pole.\r\n\r\nNow we will find the zeros. First make the substitution [latex]u=4\\theta [\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}0=2\\cos 4\\theta \\\\ 0=\\cos 4\\theta \\\\ 0=\\cos u\\\\ {\\cos }^{-1}0=u\\\\ u=\\frac{\\pi }{2}\\\\ 4\\theta =\\frac{\\pi }{2}\\\\ \\theta =\\frac{\\pi }{8}\\end{gathered}[\/latex]<\/p>\r\nThe zero is [latex]\\theta =\\frac{\\pi }{8}[\/latex]. The point [latex]\\left(0,\\frac{\\pi }{8}\\right)[\/latex] is on the curve.\r\n\r\nNext, we find the maximum [latex]|r|[\/latex]. We know that the maximum value of [latex]\\cos u=1[\/latex] when [latex]\\theta =0[\/latex]. Thus,\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}r=2\\cos \\left(4\\cdot 0\\right) \\\\ r=2\\cos \\left(0\\right) \\\\ r=2\\left(1\\right)=2 \\end{gathered}[\/latex]<\/p>\r\nThe point [latex]\\left(2,0\\right)[\/latex] is on the curve.\r\n\r\nThe graph of the rose curve has unique properties, which are revealed in the table below.\r\n<table id=\"Table_08_04_07\" summary=\"Two rows and eight columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,2), (pi\/8, 0), (pi\/4, -2), (3pi\/8, 0), (pi\/2, 2), (5pi\/8, 0), (3pi\/4, -2).\">\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]\\theta [\/latex] <\/strong><\/td>\r\n<td>0<\/td>\r\n<td>[latex]\\frac{\\pi }{8}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\pi }{4}[\/latex]<\/td>\r\n<td>[latex]\\frac{3\\pi }{8}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{5\\pi }{8}[\/latex]<\/td>\r\n<td>[latex]\\frac{3\\pi }{4}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong> [latex]r[\/latex] <\/strong><\/td>\r\n<td>2<\/td>\r\n<td>0<\/td>\r\n<td>\u22122<\/td>\r\n<td>0<\/td>\r\n<td>2<\/td>\r\n<td>0<\/td>\r\n<td>\u22122<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAs [latex]r=0[\/latex] when [latex]\\theta =\\frac{\\pi }{8}[\/latex], it makes sense to divide values in the table by [latex]\\frac{\\pi }{8}[\/latex] units. A definite pattern emerges. Look at the range of <em>r<\/em>-values: 2, 0, \u22122, 0, 2, 0, \u22122, and so on. This represents the development of the curve one petal at a time. Starting at [latex]r=0[\/latex], each petal extends out a distance of [latex]r=2[\/latex], and then turns back to zero [latex]2n[\/latex] times for a total of eight petals. See the graph in Figure 16.\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\/27165542\/CNX_Precalc_Figure_08_04_0172.jpg\" alt=\"Sketch of rose curve r=2*cos(4 theta). Goes out distance of 2 for each petal 2n times (here 2*4=8 times).\" width=\"487\" height=\"317\" \/> <b>Figure 16.<\/b> Rose curve, [latex]n[\/latex] even[\/caption]\r\n<h4>Analysis of the Solution<\/h4>\r\nWhen these curves are drawn, it is best to plot the points in order, as in the table of Example 8's solution. This allows us to see how the graph hits a maximum (the tip of a petal), loops back crossing the pole, hits the opposite maximum, and loops back to the pole. The action is continuous until all the petals are drawn.\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\nSketch the graph of [latex]r=4\\sin \\left(2\\theta \\right)[\/latex].\r\n\r\n[reveal-answer q=\"661678\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"661678\"]\r\n\r\nThe graph is a rose curve, [latex]n[\/latex] even\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165600\/CNX_Precalc_Figure_08_04_0252.jpg\" alt=\"Graph of rose curve r=4 sin(2 theta). Even - four petals equally spaced, each of length 4.\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 9: Sketching the Graph of a Rose Curve (<em>n<\/em> Odd)<\/h3>\r\nSketch the graph of [latex]r=2\\sin \\left(5\\theta \\right)[\/latex].\r\n\r\n[reveal-answer q=\"482872\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"482872\"]\r\n\r\nThe graph of the equation shows symmetry with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex]. Next, find the zeros and maximum. We will want to make the substitution [latex]u=5\\theta [\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}0=2\\sin \\left(5\\theta \\right)\\\\ 0=\\sin u\\\\ {\\sin }^{-1}0=0\\\\ u=0\\\\ 5\\theta =0\\\\ \\theta =0\\end{gathered}[\/latex]<\/p>\r\nThe maximum value is calculated at the angle where [latex]\\sin \\theta [\/latex] is a maximum. Therefore,\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered} r=2\\sin \\left(5\\cdot \\frac{\\pi }{2}\\right) \\\\ r=2\\left(1\\right)=2 \\end{gathered}[\/latex]<\/p>\r\nThus, the maximum value of the polar equation is 2. This is the length of each petal. As the curve for [latex]n[\/latex] odd yields the same number of petals as [latex]n[\/latex], there will be five petals on the graph.\r\n\r\nCreate a table of values similar to the table below.\r\n<table id=\"Table_08_04_08\" summary=\"Two rows and seven columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,0), (pi\/6, 1), (pi\/3, -1.73), (pi\/2, 2), (2pi\/3, -1.73), (5pi\/6, 1), (pi, 0).\">\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]\\theta [\/latex] <\/strong><\/td>\r\n<td>0<\/td>\r\n<td>[latex]\\frac{\\pi }{6}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\pi }{3}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{2\\pi }{3}[\/latex]<\/td>\r\n<td>[latex]\\frac{5\\pi }{6}[\/latex]<\/td>\r\n<td>[latex]\\pi [\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong> [latex]r[\/latex] <\/strong><\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>\u22121.73<\/td>\r\n<td>2<\/td>\r\n<td>\u22121.73<\/td>\r\n<td>1<\/td>\r\n<td>0<\/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\/27165544\/CNX_Precalc_Figure_08_04_019.jpg\" alt=\"Graph of rose curve r=2sin(5theta). Five petals equally spaced around origin. Point (2, pi\/2) on edge is marked.\" width=\"487\" height=\"369\" \/> <b>Figure 17.<\/b> Rose curve, [latex]n[\/latex] odd[\/caption][\/hidden-answer]<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nSketch the graph of [latex]r=3\\cos \\left(3\\theta \\right)[\/latex].\r\n\r\n[reveal-answer q=\"214154\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"214154\"]\r\n\r\nRose curve, [latex]n[\/latex] odd\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165602\/CNX_Precalc_Figure_08_04_018.jpg\" alt=\"Graph of rose curve r=3cos(3theta). Three petals equally spaced from origin. \" \/>\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]149355[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Investigating the Archimedes\u2019 Spiral<\/h2>\r\nThe final polar equation we will discuss is the Archimedes\u2019 spiral, named for its discoverer, the Greek mathematician Archimedes (c. 287 BCE - c. 212 BCE), who is credited with numerous discoveries in the fields of geometry and mechanics.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Archimedes\u2019 Spiral<\/h3>\r\nThe formula that generates the graph of the <strong>Archimedes\u2019 spiral<\/strong> is given by [latex]r=\\theta [\/latex]\u00a0for [latex]\\theta \\ge 0[\/latex]. As [latex]\\theta [\/latex] increases, [latex]r[\/latex]\u00a0increases at a constant rate in an ever-widening, never-ending, spiraling path.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165546\/CNX_Precalc_Figure_08_04_020new.jpg\" alt=\"Two graphs side by side of Archimedes' spiral. (A) is r= theta, [0, 2pi]. (B) is r=theta, [0, 4pi]. Both start at origin and spiral out counterclockwise. The second has two spirals out while the first has one.\" width=\"731\" height=\"385\" \/> <b>Figure 18<\/b>[\/caption]<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given an Archimedes\u2019 spiral over [latex]\\left[0,2\\pi \\right][\/latex], sketch the graph.<\/h3>\r\n<ol>\r\n \t<li>Make a table of values for [latex]r[\/latex] and [latex]\\theta [\/latex] over the given domain.<\/li>\r\n \t<li>Plot the points and sketch the graph.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 10: Sketching the Graph of an Archimedes\u2019 Spiral<\/h3>\r\nSketch the graph of [latex]r=\\theta [\/latex] over [latex]\\left[0,2\\pi \\right][\/latex].\r\n\r\n[reveal-answer q=\"329167\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"329167\"]\r\n\r\nAs [latex]r[\/latex] is equal to [latex]\\theta [\/latex], the plot of the Archimedes\u2019 spiral begins at the pole at the point (0, 0). While the graph hints of symmetry, there is no formal symmetry with regard to passing the symmetry tests. Further, there is no maximum value, unless the domain is restricted.\r\n\r\nCreate a table such as the one below.\r\n<table id=\"Table_08_04_09\" summary=\"Two rows and seven columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (pi\/4, 0.785), (pi\/2, 1.57), (pi, 3.14), (3pi\/2, 4.71), (7pi\/4, 5.5), (2pi, 6.28).\">\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]\\theta [\/latex] <\/strong><\/td>\r\n<td>[latex]\\frac{\\pi }{4}[\/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]\\frac{7\\pi }{4}[\/latex]<\/td>\r\n<td>[latex]2\\pi [\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong> [latex]r[\/latex] <\/strong><\/td>\r\n<td>0.785<\/td>\r\n<td>1.57<\/td>\r\n<td>3.14<\/td>\r\n<td>4.71<\/td>\r\n<td>5.50<\/td>\r\n<td>6.28<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNotice that the <em>r<\/em>-values are just the decimal form of the angle measured in radians. We can see them on a graph in Figure 19.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"488\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165549\/CNX_Precalc_Figure_08_04_021F.jpg\" alt=\"Graph of Archimedes' spiral r=theta over [0,2pi]. Starts at origin and spirals out in one loop counterclockwise. Points (pi\/4, pi\/4), (pi\/2,pi\/2), (pi,pi), (5pi\/4, 5pi\/4), (7pi\/4, pi\/4), and (2pi, 2pi) are marked.\" width=\"488\" height=\"420\" \/> <b>Figure 19.<\/b> Archimedes\u2019 spiral[\/caption]\r\n<h4>Analysis of the Solution<\/h4>\r\nThe domain of this polar curve is [latex]\\left[0,2\\pi \\right][\/latex]. In general, however, the domain of this function is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]. Graphing the equation of the Archimedes\u2019 spiral is rather simple, although the image makes it seem like it would be complex.\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\nSketch the graph of [latex]r=-\\theta [\/latex] over the interval [latex]\\left[0,4\\pi \\right][\/latex].\r\n\r\n[reveal-answer q=\"937587\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"937587\"]\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165605\/CNX_Precalc_Figure_08_04_022.jpg\" alt=\"\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>It is easier to graph polar equations if we can test the equations for symmetry with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex], the polar axis, or the pole.<\/li>\r\n \t<li>There are three symmetry tests that indicate whether the graph of a polar equation will exhibit symmetry. If an equation fails a symmetry test, the graph may or may not exhibit symmetry.<\/li>\r\n \t<li>Polar equations may be graphed by making a table of values for [latex]\\theta [\/latex] and [latex]r[\/latex].<\/li>\r\n \t<li>The maximum value of a polar equation is found by substituting the value [latex]\\theta [\/latex] that leads to the maximum value of the trigonometric expression.<\/li>\r\n \t<li>The zeros of a polar equation are found by setting [latex]r=0[\/latex] and solving for [latex]\\theta [\/latex].<\/li>\r\n \t<li>Some formulas that produce the graph of a circle in polar coordinates are given by [latex]r=a\\cos \\theta [\/latex] and [latex]r=a\\sin \\theta [\/latex].<\/li>\r\n \t<li>The formulas that produce the graphs of a cardioid are given by [latex]r=a\\pm b\\cos \\theta [\/latex] and [latex]r=a\\pm b\\sin \\theta [\/latex], for [latex]a&gt;0,b&gt;0[\/latex], and [latex]\\frac{a}{b}=1[\/latex].<\/li>\r\n \t<li>The formulas that produce the graphs of a one-loop lima\u00e7on are given by [latex]r=a\\pm b\\cos \\theta [\/latex] and [latex]r=a\\pm b\\sin \\theta [\/latex] for [latex]1&lt;\\frac{a}{b}&lt;2[\/latex].<\/li>\r\n \t<li>The formulas that produce the graphs of an inner-loop lima\u00e7on are given by [latex]r=a\\pm b\\cos \\theta [\/latex] and [latex]r=a\\pm b\\sin \\theta [\/latex] for [latex]a&gt;0,b&gt;0[\/latex],\u00a0and [latex]a&lt;b[\/latex].<\/li>\r\n \t<li>The formulas that produce the graphs of a lemniscates are given by [latex]{r}^{2}={a}^{2}\\cos 2\\theta [\/latex] and [latex]{r}^{2}={a}^{2}\\sin 2\\theta [\/latex], where [latex]a\\ne 0[\/latex].<\/li>\r\n \t<li>The formulas that produce the graphs of rose curves are given by [latex]r=a\\cos n\\theta [\/latex] and [latex]r=a\\sin n\\theta [\/latex], where [latex]a\\ne 0[\/latex]; if [latex]n[\/latex] is even, there are [latex]2n[\/latex] petals, and if [latex]n[\/latex] is odd, there are [latex]n[\/latex] petals.<\/li>\r\n \t<li>The formula that produces the graph of an Archimedes\u2019 spiral is given by [latex]r=\\theta ,\\theta \\ge 0[\/latex].<\/li>\r\n<\/ul>\r\n<h2>Summary of Curves<\/h2>\r\nWe have explored a number of seemingly complex polar curves in this section. Figures 20 and 21\u00a0summarize the graphs and equations for each of these curves.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165552\/CNX_Precalc_Figure_08_04_023n.jpg\" alt=\"&quot;Four\" \/>\r\n\r\n<img style=\"border: 0px; vertical-align: middle; max-width: 100%;\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165555\/CNX_Precalc_Figure_08_04_024n.jpg\" alt=\"&quot;Four\" \/>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165137410447\" class=\"definition\">\r\n \t<dt>Archimedes\u2019 spiral<\/dt>\r\n \t<dd id=\"fs-id1165137749714\">a polar curve given by [latex]r=\\theta [\/latex]. When multiplied by a constant, the equation appears as [latex]r=a\\theta [\/latex]. As [latex]r=\\theta [\/latex], the curve continues to widen in a spiral path over the domain.<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135378768\" class=\"definition\">\r\n \t<dt>cardioid<\/dt>\r\n \t<dd id=\"fs-id1165135378771\">a member of the lima\u00e7on family of curves, named for its resemblance to a heart; its equation is given as [latex]r=a\\pm b\\cos \\theta [\/latex] and [latex]r=a\\pm b\\sin \\theta [\/latex], where [latex]\\frac{a}{b}=1[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135344897\" class=\"definition\">\r\n \t<dt>convex lima\u04abon<\/dt>\r\n \t<dd id=\"fs-id1165135344900\">a type of one-loop lima\u00e7on represented by [latex]r=a\\pm b\\cos \\theta [\/latex] and [latex]r=a\\pm b\\sin \\theta [\/latex] such that [latex]\\frac{a}{b}\\ge 2[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135680172\" class=\"definition\">\r\n \t<dt>dimpled lima\u04abon<\/dt>\r\n \t<dd id=\"fs-id1165135680175\">a type of one-loop lima\u00e7on represented by [latex]r=a\\pm b\\cos \\theta [\/latex] and [latex]r=a\\pm b\\sin \\theta [\/latex] such that [latex]1&lt;\\frac{a}{b}&lt;2[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165134042074\" class=\"definition\">\r\n \t<dt>inner-loop lima\u00e7on<\/dt>\r\n \t<dd id=\"fs-id1165135191979\">a polar curve similar to the cardioid, but with an inner loop; passes through the pole twice; represented by [latex]r=a\\pm b\\cos \\theta [\/latex] and [latex]\\text{ }r=a\\pm b\\sin \\theta \\text{ }[\/latex] where [latex]a&lt;b[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137547628\" class=\"definition\">\r\n \t<dt>lemniscate<\/dt>\r\n \t<dd id=\"fs-id1165137547632\">a polar curve resembling a figure 8 and given by the equation [latex]{r}^{2}={a}^{2}\\cos 2\\theta [\/latex] and [latex]{r}^{2}={a}^{2}\\sin 2\\theta ,a\\ne 0[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137911243\" class=\"definition\">\r\n \t<dt>one-loop lima\u04abon<\/dt>\r\n \t<dd id=\"fs-id1165137911246\">a polar curve represented by [latex]r=a\\pm b\\cos \\theta [\/latex] and [latex]r=a\\pm b\\sin \\theta [\/latex] such that [latex]a&gt;0,b&gt;0[\/latex], and [latex]\\frac{a}{b}&gt;1[\/latex]; may be dimpled or convex; does not pass through the pole<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135564114\" class=\"definition\">\r\n \t<dt>polar equation<\/dt>\r\n \t<dd id=\"fs-id1165134331101\">an equation describing a curve on the polar grid.<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137626838\" class=\"definition\">\r\n \t<dt>rose curve<\/dt>\r\n \t<dd id=\"fs-id1165135306471\">a polar equation resembling a flower, given by the equations [latex]r=a\\cos n\\theta [\/latex] and [latex]r=a\\sin n\\theta [\/latex]; when [latex]n[\/latex] is even there are [latex]2n[\/latex] petals, and the curve is highly symmetrical; when [latex]n[\/latex] is odd there are [latex]n[\/latex] petals.<\/dd>\r\n<\/dl>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Test polar equations for symmetry.<\/li>\n<li>Graph polar equations by plotting points.<\/li>\n<\/ul>\n<\/div>\n<div style=\"width: 741px\" 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\/27165505\/CNX_Precalc_Figure_08_04_0012.jpg\" alt=\"Illustration of the solar system with the sun at the center and orbits of the planets Mercury, Venus, Earth, and Mars shown.\" width=\"731\" height=\"411\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1.<\/b> Planets follow elliptical paths as they orbit around the Sun. (credit: modification of work by NASA\/JPL-Caltech)<\/p>\n<\/div>\n<p>The planets move through space in elliptical, periodic orbits about the sun, as shown in\u00a0Figure 1. They are in constant motion, so fixing an exact position of any planet is valid only for a moment. In other words, we can fix only a planet\u2019s <em>instantaneous <\/em>position. This is one application of <strong>polar coordinates<\/strong>, represented as [latex]\\left(r,\\theta \\right)[\/latex].\u00a0We interpret [latex]r[\/latex] as the distance from the sun and [latex]\\theta[\/latex] as the planet\u2019s angular bearing, or its direction from a fixed point on the sun. In this section, we will focus on the polar system and the graphs that are generated directly from polar coordinates.<\/p>\n<h2>Testing Polar Equations for Symmetry<\/h2>\n<p>Just as a rectangular equation such as [latex]y={x}^{2}[\/latex] describes the relationship between [latex]x[\/latex] and [latex]y[\/latex] on a Cartesian grid, a <strong>polar equation <\/strong>describes a relationship between [latex]r[\/latex] and [latex]\\theta[\/latex] on a polar grid. Recall that the coordinate pair [latex]\\left(r,\\theta \\right)[\/latex] indicates that we move counterclockwise from the polar axis (positive <em>x<\/em>-axis) by an angle of [latex]\\theta[\/latex], and extend a ray from the pole (origin) [latex]r[\/latex] units in the direction of [latex]\\theta[\/latex]. All points that satisfy the polar equation are on the graph.<\/p>\n<p>Symmetry is a property that helps us recognize and plot the graph of any equation. If an equation has a graph that is symmetric with respect to an axis, it means that if we folded the graph in half over that axis, the portion of the graph on one side would coincide with the portion on the other side. By performing three tests, we will see how to apply the properties of symmetry to polar equations. Further, we will use symmetry (in addition to plotting key points, zeros, and maximums of [latex]r[\/latex])<br \/>\nto determine the graph of a polar equation.<\/p>\n<p>In the first test, we consider symmetry with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex] (<em>y<\/em>-axis). We replace [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(-r,-\\theta \\right)[\/latex] to determine if the new equation is equivalent to the original equation. For example, suppose we are given the equation [latex]r=2\\sin \\theta[\/latex];<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}r&=2\\sin \\theta \\\\ -r&=2\\sin \\left(-\\theta \\right)&& \\text{Replace}\\left(r,\\theta \\right)\\text{with }\\left(-r,-\\theta \\right). \\\\ -r&=-2\\sin \\theta&& \\text{Identity: }\\sin \\left(-\\theta \\right)=-\\sin \\theta. \\\\ r&=2\\sin \\theta&& \\text{Multiply both sides by}-1. \\end{align}[\/latex]<\/div>\n<p>This equation exhibits symmetry with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex].<\/p>\n<p>In the second test, we consider symmetry with respect to the polar axis ( [latex]x[\/latex] -axis). We replace [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(r,-\\theta \\right)[\/latex] or [latex]\\left(-r,\\pi -\\theta \\right)[\/latex] to determine equivalency between the tested equation and the original. For example, suppose we are given the equation [latex]r=1 - 2\\cos \\theta[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}r&=1 - 2\\cos \\theta \\\\ r&=1 - 2\\cos \\left(-\\theta \\right)&& \\text{Replace }\\left(r,\\theta \\right)\\text{with}\\left(r,-\\theta \\right).\\\\ r&=1 - 2\\cos \\theta&& \\text{Even\/Odd identity} \\end{align}[\/latex]<\/div>\n<p>The graph of this equation exhibits symmetry with respect to the polar axis.<\/p>\n<p>In the third test, we consider symmetry with respect to the pole (origin). We replace [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(-r,\\theta \\right)[\/latex] to determine if the tested equation is equivalent to the original equation. For example, suppose we are given the equation [latex]r=2\\sin \\left(3\\theta \\right)[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{gathered}r=2\\sin \\left(3\\theta \\right)\\\\ -r=2\\sin \\left(3\\theta \\right)\\end{gathered}[\/latex]<\/div>\n<p>The equation has failed the <strong>symmetry test<\/strong>, but that does not mean that it is not symmetric with respect to the pole. Passing one or more of the symmetry tests verifies that symmetry will be exhibited in a graph. However, failing the symmetry tests does not necessarily indicate that a graph will not be symmetric about the line [latex]\\theta =\\frac{\\pi }{2}[\/latex], the polar axis, or the pole. In these instances, we can confirm that symmetry exists by plotting reflecting points across the apparent axis of symmetry or the pole. Testing for symmetry is a technique that simplifies the graphing of polar equations, but its application is not perfect.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Symmetry Tests<\/h3>\n<p>A <strong>polar equation<\/strong> describes a curve on the polar grid. The graph of a polar equation can be evaluated for three types of symmetry, as shown in Figure 2.<\/p>\n<div style=\"width: 933px\" 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\/27165507\/CNX_Precalc_Figure_08_04_002new2.jpg\" alt=\"3 graphs side by side. (A) shows a ray extending into Q 1 and its symmetric version in Q 2. (B) shows a ray extending into Q 1 and its symmetric version in Q 4. (C) shows a ray extending into Q 1 and its symmetric version in Q 3. See caption for more information.\" width=\"923\" height=\"336\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2.<\/b> (a) A graph is symmetric with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex] (y-axis) if replacing [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(-r,-\\theta \\right)[\/latex] yields an equivalent equation. (b) A graph is symmetric with respect to the polar axis (x-axis) if replacing [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(r,-\\theta \\right)[\/latex] or [latex]\\left(-r,\\mathrm{\\pi -}\\theta \\right)[\/latex] yields an equivalent equation. (c) A graph is symmetric with respect to the pole (origin) if replacing [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(-r,\\theta \\right)[\/latex] yields an equivalent equation.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a polar equation, test for symmetry.<\/h3>\n<ol>\n<li>Substitute the appropriate combination of components for [latex]\\left(r,\\theta \\right):[\/latex] [latex]\\left(-r,-\\theta \\right)[\/latex] for [latex]\\theta =\\frac{\\pi }{2}[\/latex] symmetry; [latex]\\left(r,-\\theta \\right)[\/latex] for polar axis symmetry; and [latex]\\left(-r,\\theta \\right)[\/latex] for symmetry with respect to the pole.<\/li>\n<li>If the resulting equations are equivalent in one or more of the tests, the graph produces the expected symmetry.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Testing a Polar Equation for Symmetry<\/h3>\n<p>Test the equation [latex]r=2\\sin \\theta[\/latex] for symmetry.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q995200\">Show Solution<\/span><\/p>\n<div id=\"q995200\" class=\"hidden-answer\" style=\"display: none\">\n<p>Test for each of the three types of symmetry.<\/p>\n<table id=\"eip-id3747579\" summary=\"Three rows and two columns. The first column contains the steps to test for a type of symmetry, and the second column gives an example. The first column, first row tests symmetry with respect to theta= pi\/2. Test: Replacing (r, theta) with (-r, -theta) yields the same result. Thus, the graph is symmetric with respect to the line pi\/2. The example is -r = 2sin(-theta). By the even-odd identity, -r = -2sin(theta). After multiplying by -1, r=2sin(theta), so it passes the test. The next test is symmetry with respect to the polar axis. Test: Replacing theta with -theta does not yield the same equation. Therefore, the graph fails the test and may or may not be symmetric with respect to the polar axis. Example: r=2sin(-theta). By the even-odd identity, r=-2sin(theta). We have then r=-2sin(theta) which does not equal 2 sin(theta), so it fails. Finally, there is symmetry with respect to the pole. Test: Replacing r with -r changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole. Example: -r = 2sin(theta). r=-2sin(theta) which does not equal 2sin(theta), so it fails the test.\">\n<tbody>\n<tr>\n<td>1) Replacing [latex]\\left(r,\\theta \\right)[\/latex] with [latex]\\left(-r,-\\theta \\right)[\/latex] yields the same result. Thus, the graph is symmetric with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex].<\/td>\n<td>[latex]\\begin{align}&-r=2\\sin \\left(-\\theta \\right) \\\\ &-r=-2\\sin \\theta&& \\text{Even-odd identity} \\\\ &r=2\\sin \\theta&& \\text{Multiply}\\text{by}-1 \\\\ &\\text{Passed} \\end{align}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>2) Replacing [latex]\\theta[\/latex] with [latex]-\\theta[\/latex] does not yield the same equation. Therefore, the graph fails the test and may or may not be symmetric with respect to the polar axis.<\/td>\n<td>[latex]\\begin{align}&r=2\\sin \\left(-\\theta \\right) \\\\ r&=-2\\sin \\theta && \\text{Even-odd identity} \\\\ &r=-2\\sin \\theta \\ne 2\\sin \\theta \\\\ &\\text{Failed} \\end{align}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>3) Replacing [latex]r[\/latex] with [latex]-r[\/latex] changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole.<\/td>\n<td>[latex]\\begin{align}&-r=2\\sin \\theta \\\\ &r=-2\\sin \\theta \\ne 2\\sin \\theta \\\\ &\\text{Failed} \\end{align}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h4>Analysis of the Solution<\/h4>\n<p>Using a graphing calculator, we can see that the equation [latex]r=2\\sin \\theta[\/latex] is a circle centered at [latex]\\left(0,1\\right)[\/latex] with radius [latex]r=1[\/latex] and is indeed symmetric to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex]. We can also see that the graph is not symmetric with the polar axis or the pole. See 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\/27165509\/CNX_Precalc_Figure_08_04_0032.jpg\" alt=\"Graph of the given circle on the polar coordinate grid. Center is at (0,1), and it has radius 1.\" width=\"487\" height=\"369\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Test the equation for symmetry: [latex]r=-2\\cos \\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q47168\">Show Solution<\/span><\/p>\n<div id=\"q47168\" class=\"hidden-answer\" style=\"display: none\">\n<p>The equation fails the symmetry test with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex] and with respect to the pole. It passes the polar axis symmetry test.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm169519\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169519&theme=oea&iframe_resize_id=ohm169519\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>\u00a0Graphing Polar Equations by Plotting Points<\/h2>\n<p>To graph in the rectangular coordinate system we construct a table of [latex]x[\/latex] and [latex]y[\/latex] values. To graph in the polar coordinate system we construct a table of [latex]\\theta[\/latex] and [latex]r[\/latex] values. We enter values of [latex]\\theta[\/latex]\u00a0into a <strong>polar equation<\/strong> and calculate [latex]r[\/latex]. However, using the properties of symmetry and finding key values of [latex]\\theta[\/latex] and [latex]r[\/latex] means fewer calculations will be needed.<\/p>\n<h2>Finding Zeros and Maxima<\/h2>\n<p>To find the zeros of a polar equation, we solve for the values of [latex]\\theta[\/latex] that result in [latex]r=0[\/latex]. Recall that, to find the zeros of polynomial functions, we set the equation equal to zero and then solve for [latex]x[\/latex]. We use the same process for polar equations. Set [latex]r=0[\/latex], and solve for [latex]\\theta[\/latex].<\/p>\n<p>For many of the forms we will encounter, the maximum value of a polar equation is found by substituting those values of [latex]\\theta[\/latex] into the equation that result in the maximum value of the trigonometric functions. Consider [latex]r=5\\cos \\theta[\/latex]; the maximum distance between the curve and the pole is 5 units. The maximum value of the cosine function is 1 when [latex]\\theta =0[\/latex], so our polar equation is [latex]5\\cos \\theta[\/latex], and the value [latex]\\theta =0[\/latex] will yield the maximum [latex]|r|[\/latex].<\/p>\n<p>Similarly, the maximum value of the sine function is 1 when [latex]\\theta =\\frac{\\pi }{2}[\/latex], and if our polar equation is [latex]r=5\\sin \\theta[\/latex], the value [latex]\\theta =\\frac{\\pi }{2}[\/latex] will yield the maximum [latex]|r|[\/latex]. We may find additional information by calculating values of [latex]r[\/latex] when [latex]\\theta =0[\/latex]. These points would be polar axis intercepts, which may be helpful in drawing the graph and identifying the curve of a polar equation.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 2: Finding Zeros and Maximum Values for a Polar Equation<\/h3>\n<p>Using the equation in Example 1, find the zeros and maximum [latex]|r|[\/latex] and, if necessary, the polar axis intercepts of [latex]r=2\\sin \\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q927984\">Show Solution<\/span><\/p>\n<div id=\"q927984\" class=\"hidden-answer\" style=\"display: none\">\n<p>To find the zeros, set [latex]r[\/latex] equal to zero and solve for [latex]\\theta[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&2\\sin \\theta =0 \\\\ &\\sin \\theta =0 \\\\ &\\theta ={\\sin }^{-1}0 \\\\ &\\theta =n\\pi&& \\text{where }n\\text{ is an integer} \\end{align}[\/latex]<\/p>\n<p>Substitute any one of the [latex]\\theta[\/latex] values into the equation. We will use [latex]0[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} &r=2\\sin \\left(0\\right) \\\\ &r=0 \\end{align}[\/latex]<\/p>\n<p>The points [latex]\\left(0,0\\right)[\/latex] and [latex]\\left(0,\\pm n\\pi \\right)[\/latex] are the zeros of the equation. They all coincide, so only one point is visible on the graph. This point is also the only polar axis intercept.<\/p>\n<p>To find the maximum value of the equation, look at the maximum value of the trigonometric function [latex]\\sin \\theta[\/latex], which occurs when [latex]\\theta =\\frac{\\pi }{2}\\pm 2k\\pi[\/latex] resulting in [latex]\\sin \\left(\\frac{\\pi }{2}\\right)=1[\/latex]. Substitute [latex]\\frac{\\pi }{2}[\/latex] for [latex]\\mathrm{\\theta .}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&r=2\\sin \\left(\\frac{\\pi }{2}\\right) \\\\ &r=2\\left(1\\right) \\\\ &r=2 \\end{align}[\/latex]<\/p>\n<h3>Analysis of the Solution<\/h3>\n<p>The point [latex]\\left(2,\\frac{\\pi }{2}\\right)[\/latex] will be the maximum value on the graph. Let\u2019s plot a few more points to verify the graph of a circle.<\/p>\n<table id=\"Table_08_04_01\" summary=\"Eight rows and 3 columns. First column is labeled theta, second column is labeled r=2sin(theta), and third column is labeled r. The table has ordered triples of these column values: (0, r=2sin(0)=0, 0), (pi\/6, r=2sin(pi\/6)=1, 1), (pi\/3, r=2sin(pi\/3) = approx. 1.73, 1.73), (pi\/2, r=2sin(pi\/2) = 2, 2), (2pi\/3, r=2sin(2pi\/3)=approx. 1.73, 1.73), (5pi\/6, r=2sin(5pi\/6)=1, 1), and (pi, r=2sin(pi)=0).\">\n<thead>\n<tr>\n<th>[latex]\\theta[\/latex]<\/th>\n<th>[latex]r=2\\sin \\theta[\/latex]<\/th>\n<th>[latex]r[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>0<\/td>\n<td>[latex]r=2\\sin \\left(0\\right)=0[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\frac{\\pi }{6}[\/latex]<\/td>\n<td>[latex]r=2\\sin \\left(\\frac{\\pi }{6}\\right)=1[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\frac{\\pi }{3}[\/latex]<\/td>\n<td>[latex]r=2\\sin \\left(\\frac{\\pi }{3}\\right)\\approx 1.73[\/latex]<\/td>\n<td>[latex]1.73[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<td>[latex]r=2\\sin \\left(\\frac{\\pi }{2}\\right)=2[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\frac{2\\pi }{3}[\/latex]<\/td>\n<td>[latex]r=2\\sin \\left(\\frac{2\\pi }{3}\\right)\\approx 1.73[\/latex]<\/td>\n<td>[latex]1.73[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\frac{5\\pi }{6}[\/latex]<\/td>\n<td>[latex]r=2\\sin \\left(\\frac{5\\pi }{6}\\right)=1[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\pi[\/latex]<\/td>\n<td>[latex]r=2\\sin \\left(\\pi \\right)=0[\/latex]<\/td>\n<td>[latex]0[\/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\/27165512\/CNX_Precalc_Figure_08_04_0042.jpg\" alt=\"Graph of circle on the polar coordinate grid. The center is at (0,1), and it has radius 1. Six points along the circumference are marked: (0,0), (1, pi\/6), (1.3, pi\/3), (2, pi\/2), (1.73, 2pi\/3), and (1, 5pi\/6).\" width=\"487\" height=\"369\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Without converting to Cartesian coordinates, test the given equation for symmetry and find the zeros and maximum values of [latex]|r|:[\/latex] [latex]r=3\\cos \\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q16075\">Show Solution<\/span><\/p>\n<div id=\"q16075\" class=\"hidden-answer\" style=\"display: none\">\n<p>Tests will reveal symmetry about the polar axis. The zero is [latex]\\left(0,\\frac{\\pi }{2}\\right)[\/latex], and the maximum value is [latex]\\left(3,0\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Graphing Circles and the 5 Classic Polar Curves<\/h2>\n<h2>Investigating Circles<\/h2>\n<p>Now we have seen the equation of a circle in the polar coordinate system. In the last two examples, the same equation was used to illustrate the properties of symmetry and demonstrate how to find the zeros, maximum values, and plotted points that produced the graphs. However, the circle is only one of many shapes in the set of polar curves.<\/p>\n<p>There are five classic polar curves<strong>: cardioids<\/strong>, <strong>lima\u04abons, lemniscates, rose curves<\/strong>, and <strong>Archimedes\u2019 spirals<\/strong>. We will briefly touch on the polar formulas for the circle before moving on to the classic curves and their variations.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Formulas for the Equation of a Circle<\/h3>\n<p>Some of the formulas that produce the graph of a circle in polar coordinates are given by [latex]r=a\\cos \\theta[\/latex] and [latex]r=a\\sin \\theta[\/latex], where [latex]a[\/latex] is the diameter of the circle or the distance from the pole to the farthest point on the circumference. The radius is [latex]\\frac{|a|}{2}[\/latex], or one-half the diameter. For [latex]r=a\\cos \\theta ,[\/latex] the center is [latex]\\left(\\frac{a}{2},0\\right)[\/latex]. For [latex]r=a\\sin \\theta[\/latex], the center is [latex]\\left(\\frac{a}{2},\\pi \\right)[\/latex]. Figure 5\u00a0shows the graphs of these four circles.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165514\/CNX_Precalc_Figure_08_04_005new2.jpg\" alt=\"&quot;Four\" \/><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Sketching the Graph of a Polar Equation for a Circle<\/h3>\n<p>Sketch the graph of [latex]r=4\\cos \\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q198958\">Show Solution<\/span><\/p>\n<div id=\"q198958\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, testing the equation for symmetry, we find that the graph is symmetric about the polar axis. Next, we find the <strong>zeros<\/strong> and maximum [latex]|r|[\/latex] for [latex]r=4\\cos \\theta[\/latex]. First, set [latex]r=0[\/latex], and solve for [latex]\\theta[\/latex] . Thus, a zero occurs at [latex]\\theta =\\frac{\\pi }{2}\\pm k\\pi[\/latex]. A key point to plot is [latex]\\left(0,\\text{ }\\text{ }\\frac{\\pi }{2}\\right)[\/latex].<\/p>\n<p>To find the maximum value of [latex]r[\/latex], note that the maximum value of the cosine function is 1 when [latex]\\theta =0\\pm 2k\\pi[\/latex]. Substitute [latex]\\theta =0[\/latex] into the equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}r=4\\cos \\theta \\\\ r=4\\cos \\left(0\\right)\\\\ r=4\\left(1\\right)=4\\end{gathered}[\/latex]<\/p>\n<p>The maximum value of the equation is 4. A key point to plot is [latex]\\left(4,0\\right)[\/latex].<\/p>\n<p>As [latex]r=4\\cos \\theta[\/latex] is symmetric with respect to the polar axis, we only need to calculate <em>r<\/em>-values for [latex]\\theta[\/latex] over the interval [latex]\\left[0,\\pi \\right][\/latex]. Points in the upper quadrant can then be reflected to the lower quadrant. Make a table of values similar to the table below. The graph is shown in Figure 6.<\/p>\n<table id=\"Table_08_04_02\" summary=\"Two rows and ten columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,4), (pi\/6, 3.46), (pi\/4, 2.83), (pi\/3,2), (pi\/2,0), (2pi\/3,-2), (3pi\/4,-2.83), (5pi\/6, -3.46), and (pi,4).\">\n<tbody>\n<tr>\n<td><strong> [latex]\\theta[\/latex] <\/strong><\/td>\n<td>0<\/td>\n<td>[latex]\\frac{\\pi }{6}[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{4}[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{3}[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<td>[latex]\\frac{2\\pi }{3}[\/latex]<\/td>\n<td>[latex]\\frac{3\\pi }{4}[\/latex]<\/td>\n<td>[latex]\\frac{5\\pi }{6}[\/latex]<\/td>\n<td>[latex]\\pi[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong> [latex]r[\/latex] <\/strong><\/td>\n<td>4<\/td>\n<td>3.46<\/td>\n<td>2.83<\/td>\n<td>2<\/td>\n<td>0<\/td>\n<td>\u22122<\/td>\n<td>\u22122.83<\/td>\n<td>\u22123.46<\/td>\n<td>4<\/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\/27165517\/CNX_Precalc_Figure_08_04_0062.jpg\" alt=\"Graph of 4=4cos(theta) in polar coordinates. Points (0, pi\/2), (-2, 2pi\/3), (4,0), and (2, pi\/3) are marked on the circumference.\" width=\"487\" height=\"364\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>Investigating Cardioids<\/h2>\n<p>While translating from polar coordinates to Cartesian coordinates may seem simpler in some instances, graphing the classic curves is actually less complicated in the polar system. The next curve is called a cardioid, as it resembles a heart. This shape is often included with the family of curves called lima\u00e7ons, but here we will discuss the cardioid on its own.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Formulas for a Cardioid<\/h3>\n<p>The formulas that produce the graphs of a <strong>cardioid<\/strong> are given by [latex]r=a\\pm b\\cos \\theta[\/latex] and [latex]r=a\\pm b\\sin \\theta[\/latex] where [latex]a>0,b>0[\/latex], and [latex]\\frac{a}{b}=1[\/latex]. The cardioid graph passes through the pole, as we can see in Figure 7.<\/p>\n<div style=\"width: 933px\" 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\/27165519\/CNX_Precalc_Figure_08_04_007new2.jpg\" alt=\"Graph of four cardioids. (A) is r = a + bcos(theta). Cardioid extending to the right. (B) is r=a-bcos(theta). Cardioid extending to the left. (C) is r=a+bsin(theta). Cardioid extending up. (D) is r=a-bsin(theta). Cardioid extending down.\" width=\"923\" height=\"290\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 7<\/b><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given the polar equation of a cardioid, sketch its graph.<\/h3>\n<ol>\n<li>Check equation for the three types of symmetry.<\/li>\n<li>Find the zeros. Set [latex]r=0[\/latex].<\/li>\n<li>Find the maximum value of the equation according to the maximum value of the trigonometric expression.<\/li>\n<li>Make a table of values for [latex]r[\/latex] and [latex]\\theta[\/latex].<\/li>\n<li>Plot the points and sketch the graph.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 4: Sketching the Graph of a Cardioid<\/h3>\n<p>Sketch the graph of [latex]r=2+2\\cos \\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q869763\">Show Solution<\/span><\/p>\n<div id=\"q869763\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, testing the equation for symmetry, we find that the graph of this equation will be symmetric about the polar axis. Next, we find the zeros and maximums. Setting [latex]r=0[\/latex], we have [latex]\\theta =\\pi +2k\\pi[\/latex]. The zero of the equation is located at [latex]\\left(0,\\pi \\right)[\/latex]. The graph passes through this point.<\/p>\n<p>The maximum value of [latex]r=2+2\\cos \\theta[\/latex] occurs when [latex]\\cos \\theta[\/latex] is a maximum, which is when [latex]\\cos \\theta =1[\/latex] or when [latex]\\theta =0[\/latex]. Substitute [latex]\\theta =0[\/latex] into the equation, and solve for [latex]r[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered} r=2+2\\cos \\left(0\\right) \\\\ r=2+2\\left(1\\right)=4 \\end{gathered}[\/latex]<\/p>\n<p>The point [latex]\\left(4,0\\right)[\/latex] is the maximum value on the graph.<\/p>\n<p>We found that the polar equation is symmetric with respect to the polar axis, but as it extends to all four quadrants, we need to plot values over the interval [latex]\\left[0,\\pi \\right][\/latex]. The upper portion of the graph is then reflected over the polar axis. Next, we make a table of values, as in the table below, and then we plot the points and draw the graph. See Figure 8.<\/p>\n<table id=\"Table_08_04_03\" summary=\"Two rows and six columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,4), (pi\/4, 3.41), (pi\/2, 2), (2pi\/3, 1), and (pi, 0).\">\n<tbody>\n<tr>\n<td><strong> [latex]\\theta[\/latex] <\/strong><\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{4}[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<td>[latex]\\frac{2\\pi }{3}[\/latex]<\/td>\n<td>[latex]\\pi[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong> [latex]r[\/latex] <\/strong><\/td>\n<td>4<\/td>\n<td>3.41<\/td>\n<td>2<\/td>\n<td>1<\/td>\n<td>0<\/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\/27165522\/CNX_Precalc_Figure_08_04_0082.jpg\" alt=\"Graph of r=2+2cos(theta). Cardioid extending to the right. Points on the edge (0,pi), (4,0),(3.4, pi\/4), (2,pi\/2), and (1, 2pi\/3) are shown.\" width=\"487\" height=\"369\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 8<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>Investigating Lima\u00e7ons<\/h2>\n<p>The word <em>lima\u00e7on<\/em> is Old French for &#8220;snail,&#8221; a name that describes the shape of the graph. As mentioned earlier, the cardioid is a member of the lima\u00e7on family, and we can see the similarities in the graphs. The other images in this category include the one-loop lima\u00e7on and the two-loop (or inner-loop) lima\u00e7on. <strong>One-loop lima\u00e7ons<\/strong> are sometimes referred to as <strong>dimpled lima\u00e7ons<\/strong> when [latex]1<\\frac{a}{b}<2[\/latex] and <strong>convex lima\u00e7ons<\/strong> when [latex]\\frac{a}{b}\\ge 2[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Formulas for One-Loop Lima\u00e7ons<\/h3>\n<p>The formulas that produce the graph of a dimpled <strong>one-loop lima\u00e7on<\/strong> are given by [latex]r=a\\pm b\\cos \\theta[\/latex] and [latex]r=a\\pm b\\sin \\theta[\/latex] where [latex]a>0,b>0,\\text{and 1<}\\frac{a}{b}<2[\/latex]. All four graphs are shown in Figure 9.\n\n\n\n<div style=\"width: 933px\" 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\/27165524\/CNX_Precalc_Figure_08_04_009new2.jpg\" alt=\"Four dimpled lima\u00e7ons side by side. (A) is r=a+bcos(theta). Extending to the right. (B) is r=a-bcos(theta). Extending to the left. (C) is r=a+bsin(theta). Extending up. (D) is r=a-bsin(theta). Extending down.\" width=\"923\" height=\"290\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 9.<\/b> Dimpled lima\u00e7ons<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a polar equation for a one-loop lima\u00e7on, sketch the graph.<\/h3>\n<ol>\n<li>Test the equation for symmetry. Remember that failing a symmetry test does not mean that the shape will not exhibit symmetry. Often the symmetry may reveal itself when the points are plotted.<\/li>\n<li>Find the zeros.<\/li>\n<li>Find the maximum values according to the trigonometric expression.<\/li>\n<li>Make a table.<\/li>\n<li>Plot the points and sketch the graph.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Sketching the Graph of a One-Loop Lima\u00e7on<\/h3>\n<p>Graph the equation [latex]r=4 - 3\\sin \\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q739204\">Show Solution<\/span><\/p>\n<div id=\"q739204\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, testing the equation for symmetry, we find that it fails all three symmetry tests, meaning that the graph may or may not exhibit symmetry, so we cannot use the symmetry to help us graph it. However, this equation has a graph that clearly displays symmetry with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex], yet it fails all the three symmetry tests. A graphing calculator will immediately illustrate the graph\u2019s reflective quality.<\/p>\n<p>Next, we find the zeros and maximum, and plot the reflecting points to verify any symmetry. Setting [latex]r=0[\/latex] results in [latex]\\theta[\/latex] being undefined. What does this mean? How could [latex]\\theta[\/latex] be undefined? The angle [latex]\\theta[\/latex] is undefined for any value of [latex]\\sin \\theta >1[\/latex]. Therefore, [latex]\\theta[\/latex] is undefined because there is no value of [latex]\\theta[\/latex] for which [latex]\\sin \\theta >1[\/latex]. Consequently, the graph does not pass through the pole. Perhaps the graph does cross the polar axis, but not at the pole. We can investigate other intercepts by calculating [latex]r[\/latex] when [latex]\\theta =0[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}r\\left(0\\right)=4 - 3\\sin \\left(0\\right) \\\\ r=4 - 3\\cdot 0=4 \\end{gathered}[\/latex]<\/p>\n<p>So, there is at least one polar axis intercept at [latex]\\left(4,0\\right)[\/latex].<\/p>\n<p>Next, as the maximum value of the sine function is 1 when [latex]\\theta =\\frac{\\pi }{2}[\/latex], we will substitute [latex]\\theta =\\frac{\\pi }{2}[\/latex]\u00a0into the equation and solve for [latex]r[\/latex]. Thus, [latex]r=1[\/latex].<\/p>\n<p>Make a table of the coordinates similar to the table below.<\/p>\n<table id=\"Table_08_04_04\" summary=\"Two rows and fourteen columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,4), (pi\/6, 2.5), (pi\/3,1.4), (pi\/2, 1), (2pi\/3, 1.4), (5pi\/6, 2.5), (pi,4), (7pi\/6, 5.5), (4pi\/3,6.6), (3pi\/2, 7), (5pi\/3, 6.6), (11pi\/6, 5.5), and (2pi, 4).\">\n<tbody>\n<tr>\n<td><strong> [latex]\\theta[\/latex] <\/strong><\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{6}[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{3}[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<td>[latex]\\frac{2\\pi }{3}[\/latex]<\/td>\n<td>[latex]\\frac{5\\pi }{6}[\/latex]<\/td>\n<td>[latex]\\pi[\/latex]<\/td>\n<td>[latex]\\frac{7\\pi }{6}[\/latex]<\/td>\n<td>[latex]\\frac{4\\pi }{3}[\/latex]<\/td>\n<td>[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\n<td>[latex]\\frac{5\\pi }{3}[\/latex]<\/td>\n<td>[latex]\\frac{11\\pi }{6}[\/latex]<\/td>\n<td>[latex]2\\pi[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong> [latex]r[\/latex] <\/strong><\/td>\n<td>4<\/td>\n<td>2.5<\/td>\n<td>1.4<\/td>\n<td>1<\/td>\n<td>1.4<\/td>\n<td>2.5<\/td>\n<td>4<\/td>\n<td>5.5<\/td>\n<td>6.6<\/td>\n<td>7<\/td>\n<td>6.6<\/td>\n<td>5.5<\/td>\n<td>4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The graph is shown in Figure 10.<\/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\/27165527\/CNX_Precalc_Figure_08_04_0102.jpg\" alt=\"Graph of the lima\u00e7on r=4-3sin(theta). Extending down. Points on the edge are shown: (1,pi\/2), (4,0), (4,pi), and (7, 3pi\/2).\" width=\"487\" height=\"423\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 10.<\/b> One-loop lima\u00e7on<\/p>\n<\/div>\n<h4>Analysis of the Solution<\/h4>\n<p>This is an example of a curve for which making a table of values is critical to producing an accurate graph. The symmetry tests fail; the zero is undefined. While it may be apparent that an equation involving [latex]\\sin \\theta[\/latex] is likely symmetric with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex], evaluating more points helps to verify that the graph is correct.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Sketch the graph of [latex]r=3 - 2\\cos \\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q875395\">Show Solution<\/span><\/p>\n<div id=\"q875395\" class=\"hidden-answer\" style=\"display: none\">\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165558\/CNX_Precalc_Figure_08_04_0112.jpg\" alt=\"Graph of the lima\u00e7on r=3-2cos(theta). Extending to the left.\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm149412\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=149412&theme=oea&iframe_resize_id=ohm149412\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Another type of lima\u00e7on, the <strong>inner-loop lima\u00e7on<\/strong>, is named for the loop formed inside the general lima\u00e7on shape. It was discovered by the German artist Albrecht <strong>D\u00fcrer<\/strong>(1471-1528), who revealed a method for drawing the inner-loop lima\u00e7on in his 1525 book <em>Underweysung der Messing<\/em>. A century later, the father of mathematician Blaise <strong>Pascal<\/strong>, \u00c9tienne Pascal(1588-1651), rediscovered it.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Formulas for Inner-Loop Lima\u00e7ons<\/h3>\n<p>The formulas that generate the <strong>inner-loop lima\u00e7ons<\/strong> are given by [latex]r=a\\pm b\\cos \\theta[\/latex] and [latex]r=a\\pm b\\sin \\theta[\/latex] where [latex]a>0,b>0[\/latex], and [latex]a<b[\/latex]. The graph of the inner-loop lima\u00e7on passes through the pole twice: once for the outer loop, and once for the inner loop. See Figure 11\u00a0for the graphs.\n\n\n\n<div style=\"width: 933px\" 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\/27165529\/CNX_Precalc_Figure_08_04_012new2.jpg\" alt=\"Graph of four inner loop lima\u00e7ons side by side. (A) is r=a+bcos(theta),a&lt;b. Extended to the right. (B) is a-bcos(theta), a&lt;b. Extends to the left. (C) is r=a+bsin(theta), a&lt;b. Extends up. (D) is r=a-bsin(theta), a&lt;b. Extends down.\" width=\"923\" height=\"290\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 11<\/b><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 6: Sketching the Graph of an Inner-Loop Lima\u00e7on<\/h3>\n<p>Sketch the graph of [latex]r=2+5\\text{cos}\\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q360339\">Show Solution<\/span><\/p>\n<div id=\"q360339\" class=\"hidden-answer\" style=\"display: none\">\n<p>Testing for symmetry, we find that the graph of the equation is symmetric about the polar axis. Next, finding the zeros reveals that when [latex]r=0,\\theta =1.98[\/latex].\u00a0The maximum [latex]|r|[\/latex] is found when [latex]\\cos \\theta =1[\/latex] or when [latex]\\theta =0[\/latex]. Thus, the maximum is found at the point (7, 0).<\/p>\n<p>Even though we have found symmetry, the zero, and the maximum, plotting more points will help to define the shape, and then a pattern will emerge.<\/p>\n<table id=\"Table_08_04_05\" summary=\"Two rows and fourteen columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,7), (pi\/6, 6.3.), (pi\/3,4.5), (pi\/2, 2), (2pi\/3, -.5), (5pi\/6, -2.3), (pi,-3), (7pi\/6, -2.3), (4pi\/3,-.5), (3pi\/2, 2), (5pi\/3, 4.5), (11pi\/6, 6.3), and (2pi, 7).\">\n<tbody>\n<tr>\n<td><strong> [latex]\\theta[\/latex] <\/strong><\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{6}[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{3}[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<td>[latex]\\frac{2\\pi }{3}[\/latex]<\/td>\n<td>[latex]\\frac{5\\pi }{6}[\/latex]<\/td>\n<td>[latex]\\pi[\/latex]<\/td>\n<td>[latex]\\frac{7\\pi }{6}[\/latex]<\/td>\n<td>[latex]\\frac{4\\pi }{3}[\/latex]<\/td>\n<td>[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\n<td>[latex]\\frac{5\\pi }{3}[\/latex]<\/td>\n<td>[latex]\\frac{11\\pi }{6}[\/latex]<\/td>\n<td>[latex]2\\pi[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong> [latex]r[\/latex] <\/strong><\/td>\n<td>7<\/td>\n<td>6.3<\/td>\n<td>4.5<\/td>\n<td>2<\/td>\n<td>\u22120.5<\/td>\n<td>\u22122.3<\/td>\n<td>\u22123<\/td>\n<td>\u22122.3<\/td>\n<td>\u22120.5<\/td>\n<td>2<\/td>\n<td>4.5<\/td>\n<td>6.3<\/td>\n<td>7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>As expected, the values begin to repeat after [latex]\\theta =\\pi[\/latex]. The graph is shown in Figure 12.<\/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\/27165532\/CNX_Precalc_Figure_08_04_0132.jpg\" alt=\"Graph of inner loop lima\u00e7on r=2+5cos(theta). Extends to the right. Points on edge plotted are (7,0), (4.5, pi\/3), (2, pi\/2), and (-3, pi).\" width=\"487\" height=\"423\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 12.<\/b> Inner-loop lima\u00e7on<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>Investigating Lemniscates<\/h2>\n<p>The lemniscate is a polar curve resembling the infinity symbol [latex]\\infty[\/latex] or a figure 8. Centered at the pole, a lemniscate is symmetrical by definition.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Formulas for Lemniscates<\/h3>\n<p>The formulas that generate the graph of a <strong>lemniscate<\/strong> are given by [latex]{r}^{2}={a}^{2}\\cos 2\\theta[\/latex] and [latex]{r}^{2}={a}^{2}\\sin 2\\theta[\/latex] where [latex]a\\ne 0[\/latex]. The formula [latex]{r}^{2}={a}^{2}\\sin 2\\theta[\/latex] is symmetric with respect to the pole. The formula [latex]{r}^{2}={a}^{2}\\cos 2\\theta[\/latex] is symmetric with respect to the pole, the line [latex]\\theta =\\frac{\\pi }{2}[\/latex], and the polar axis. See Figure 13\u00a0for the graphs.<\/p>\n<div style=\"width: 933px\" 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\/27165535\/CNX_Precalc_Figure_08_04_014new2.jpg\" alt=\"Four graphs of lemniscates side by side. (A) is r^2 = a^2 * cos(2theta). Horizonatal figure eight, on x-axis. (B) is r^2 = - a^2 * cos(2theta). Vertical figure eight, on y axis. (C) is r^2 = a^2 * sin(2theta). Diagonal figure eight on line y=x. (D) is r^2 = -a^2 *sin(2theta). Diagonal figure eight on line y=-x.\" width=\"923\" height=\"282\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 13<\/b><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 7: Sketching the Graph of a Lemniscate<\/h3>\n<p>Sketch the graph of [latex]{r}^{2}=4\\cos 2\\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q721985\">Show Solution<\/span><\/p>\n<div id=\"q721985\" class=\"hidden-answer\" style=\"display: none\">\n<p>The equation exhibits symmetry with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex], the polar axis, and the pole.<\/p>\n<p>Let\u2019s find the zeros. It should be routine by now, but we will approach this equation a little differently by making the substitution [latex]u=2\\theta[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&0=4\\cos 2\\theta \\\\ &0=4\\cos u \\\\ &0=\\cos u \\\\ &{\\cos }^{-1}0=\\frac{\\pi }{2} \\\\ &u=\\frac{\\pi }{2}&& \\text{Substitute }2\\theta \\text{ back in for }u. \\\\ &2\\theta =\\frac{\\pi }{2} \\\\ &\\theta =\\frac{\\pi }{4} \\end{align}[\/latex]<\/p>\n<p>So, the point [latex]\\left(0,\\frac{\\pi }{4}\\right)[\/latex] is a zero of the equation.<\/p>\n<p>Now let\u2019s find the maximum value. Since the maximum of [latex]\\cos u=1[\/latex] when [latex]u=0[\/latex], the maximum [latex]\\cos 2\\theta =1[\/latex] when [latex]2\\theta =0[\/latex]. Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}{r}^{2}=4\\cos \\left(0\\right)\\\\ {r}^{2}=4\\left(1\\right)=4\\\\ r=\\pm \\sqrt{4}=2\\end{gathered}[\/latex]<\/p>\n<p>We have a maximum at (2, 0). Since this graph is symmetric with respect to the pole, the line [latex]\\theta =\\frac{\\pi }{2}[\/latex], and the polar axis, we only need to plot points in the first quadrant.<\/p>\n<p>Make a table similar to the table below.<\/p>\n<table id=\"Table_08_04_06\" summary=\"Two rows and six columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,2), (pi\/6, rad2), (pi\/4,0), (pi\/3, rad2), (pi\/2,0).\">\n<tbody>\n<tr>\n<td><strong> [latex]\\theta[\/latex] <\/strong><\/td>\n<td>0<\/td>\n<td>[latex]\\frac{\\pi }{6}[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{4}[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{3}[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong> [latex]r[\/latex] <\/strong><\/td>\n<td>2<\/td>\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\n<td>0<\/td>\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\n<td>0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Plot the points on the graph, such as the one shown in Figure 14.<\/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\/27165537\/CNX_Precalc_Figure_08_04_0152.jpg\" alt=\"Graph of r^2 = 4cos(2theta). Horizontal lemniscate, along x-axis. Points on edge plotted are (2,0), (rad2, pi\/6), (rad2 7pi\/6).\" width=\"487\" height=\"210\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 14.<\/b> Lemniscate<\/p>\n<\/div>\n<h4>Analysis of the Solution<\/h4>\n<p>Making a substitution such as [latex]u=2\\theta[\/latex] is a common practice in mathematics because it can make calculations simpler. However, we must not forget to replace the substitution term with the original term at the end, and then solve for the unknown.<\/p>\n<p>Some of the points on this graph may not show up using the Trace function on the TI-84 graphing calculator, and the calculator table may show an error for these same points of [latex]r[\/latex]. This is because there are no real square roots for these values of [latex]\\theta[\/latex]. In other words, the corresponding <em>r<\/em>-values of [latex]\\sqrt{4\\cos \\left(2\\theta \\right)}[\/latex]<br \/>\nare complex numbers because there is a negative number under the radical.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<p><span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">Investigating Rose Curves<\/span><\/p>\n<\/div>\n<p>The next type of polar equation produces a petal-like shape called a rose curve. Although the graphs look complex, a simple polar equation generates the pattern.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Rose Curves<\/h3>\n<p>The formulas that generate the graph of a <strong>rose curve<\/strong> are given by [latex]r=a\\cos n\\theta[\/latex] and [latex]r=a\\sin n\\theta[\/latex] where [latex]a\\ne 0[\/latex]. If [latex]n[\/latex] is even, the curve has [latex]2n[\/latex] petals. If [latex]n[\/latex] is odd, the curve has [latex]n[\/latex] petals.<\/p>\n<div style=\"width: 741px\" 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\/27165539\/CNX_Precalc_Figure_08_04_016new2.jpg\" alt=\"Graph of two rose curves side by side. (A) is r=acos(ntheta), where n is even. Eight petals extending from origin, equally spaced. (B) is r=asin(ntheta) where n is odd. Three petals extending from the origin, equally spaced.\" width=\"731\" height=\"385\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 15<\/b><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 8: Sketching the Graph of a Rose Curve (<em>n<\/em> Even)<\/h3>\n<p>Sketch the graph of [latex]r=2\\cos 4\\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q667305\">Show Solution<\/span><\/p>\n<div id=\"q667305\" class=\"hidden-answer\" style=\"display: none\">\n<p>Testing for symmetry, we find again that the symmetry tests do not tell the whole story. The graph is not only symmetric with respect to the polar axis, but also with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex] and the pole.<\/p>\n<p>Now we will find the zeros. First make the substitution [latex]u=4\\theta[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}0=2\\cos 4\\theta \\\\ 0=\\cos 4\\theta \\\\ 0=\\cos u\\\\ {\\cos }^{-1}0=u\\\\ u=\\frac{\\pi }{2}\\\\ 4\\theta =\\frac{\\pi }{2}\\\\ \\theta =\\frac{\\pi }{8}\\end{gathered}[\/latex]<\/p>\n<p>The zero is [latex]\\theta =\\frac{\\pi }{8}[\/latex]. The point [latex]\\left(0,\\frac{\\pi }{8}\\right)[\/latex] is on the curve.<\/p>\n<p>Next, we find the maximum [latex]|r|[\/latex]. We know that the maximum value of [latex]\\cos u=1[\/latex] when [latex]\\theta =0[\/latex]. Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}r=2\\cos \\left(4\\cdot 0\\right) \\\\ r=2\\cos \\left(0\\right) \\\\ r=2\\left(1\\right)=2 \\end{gathered}[\/latex]<\/p>\n<p>The point [latex]\\left(2,0\\right)[\/latex] is on the curve.<\/p>\n<p>The graph of the rose curve has unique properties, which are revealed in the table below.<\/p>\n<table id=\"Table_08_04_07\" summary=\"Two rows and eight columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,2), (pi\/8, 0), (pi\/4, -2), (3pi\/8, 0), (pi\/2, 2), (5pi\/8, 0), (3pi\/4, -2).\">\n<tbody>\n<tr>\n<td><strong> [latex]\\theta[\/latex] <\/strong><\/td>\n<td>0<\/td>\n<td>[latex]\\frac{\\pi }{8}[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{4}[\/latex]<\/td>\n<td>[latex]\\frac{3\\pi }{8}[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<td>[latex]\\frac{5\\pi }{8}[\/latex]<\/td>\n<td>[latex]\\frac{3\\pi }{4}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong> [latex]r[\/latex] <\/strong><\/td>\n<td>2<\/td>\n<td>0<\/td>\n<td>\u22122<\/td>\n<td>0<\/td>\n<td>2<\/td>\n<td>0<\/td>\n<td>\u22122<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>As [latex]r=0[\/latex] when [latex]\\theta =\\frac{\\pi }{8}[\/latex], it makes sense to divide values in the table by [latex]\\frac{\\pi }{8}[\/latex] units. A definite pattern emerges. Look at the range of <em>r<\/em>-values: 2, 0, \u22122, 0, 2, 0, \u22122, and so on. This represents the development of the curve one petal at a time. Starting at [latex]r=0[\/latex], each petal extends out a distance of [latex]r=2[\/latex], and then turns back to zero [latex]2n[\/latex] times for a total of eight petals. See the graph in Figure 16.<\/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\/27165542\/CNX_Precalc_Figure_08_04_0172.jpg\" alt=\"Sketch of rose curve r=2*cos(4 theta). Goes out distance of 2 for each petal 2n times (here 2*4=8 times).\" width=\"487\" height=\"317\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 16.<\/b> Rose curve, [latex]n[\/latex] even<\/p>\n<\/div>\n<h4>Analysis of the Solution<\/h4>\n<p>When these curves are drawn, it is best to plot the points in order, as in the table of Example 8&#8217;s solution. This allows us to see how the graph hits a maximum (the tip of a petal), loops back crossing the pole, hits the opposite maximum, and loops back to the pole. The action is continuous until all the petals are drawn.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Sketch the graph of [latex]r=4\\sin \\left(2\\theta \\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q661678\">Show Solution<\/span><\/p>\n<div id=\"q661678\" class=\"hidden-answer\" style=\"display: none\">\n<p>The graph is a rose curve, [latex]n[\/latex] even<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165600\/CNX_Precalc_Figure_08_04_0252.jpg\" alt=\"Graph of rose curve r=4 sin(2 theta). Even - four petals equally spaced, each of length 4.\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 9: Sketching the Graph of a Rose Curve (<em>n<\/em> Odd)<\/h3>\n<p>Sketch the graph of [latex]r=2\\sin \\left(5\\theta \\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q482872\">Show Solution<\/span><\/p>\n<div id=\"q482872\" class=\"hidden-answer\" style=\"display: none\">\n<p>The graph of the equation shows symmetry with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex]. Next, find the zeros and maximum. We will want to make the substitution [latex]u=5\\theta[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}0=2\\sin \\left(5\\theta \\right)\\\\ 0=\\sin u\\\\ {\\sin }^{-1}0=0\\\\ u=0\\\\ 5\\theta =0\\\\ \\theta =0\\end{gathered}[\/latex]<\/p>\n<p>The maximum value is calculated at the angle where [latex]\\sin \\theta[\/latex] is a maximum. Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered} r=2\\sin \\left(5\\cdot \\frac{\\pi }{2}\\right) \\\\ r=2\\left(1\\right)=2 \\end{gathered}[\/latex]<\/p>\n<p>Thus, the maximum value of the polar equation is 2. This is the length of each petal. As the curve for [latex]n[\/latex] odd yields the same number of petals as [latex]n[\/latex], there will be five petals on the graph.<\/p>\n<p>Create a table of values similar to the table below.<\/p>\n<table id=\"Table_08_04_08\" summary=\"Two rows and seven columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,0), (pi\/6, 1), (pi\/3, -1.73), (pi\/2, 2), (2pi\/3, -1.73), (5pi\/6, 1), (pi, 0).\">\n<tbody>\n<tr>\n<td><strong> [latex]\\theta[\/latex] <\/strong><\/td>\n<td>0<\/td>\n<td>[latex]\\frac{\\pi }{6}[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{3}[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<td>[latex]\\frac{2\\pi }{3}[\/latex]<\/td>\n<td>[latex]\\frac{5\\pi }{6}[\/latex]<\/td>\n<td>[latex]\\pi[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong> [latex]r[\/latex] <\/strong><\/td>\n<td>0<\/td>\n<td>1<\/td>\n<td>\u22121.73<\/td>\n<td>2<\/td>\n<td>\u22121.73<\/td>\n<td>1<\/td>\n<td>0<\/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\/27165544\/CNX_Precalc_Figure_08_04_019.jpg\" alt=\"Graph of rose curve r=2sin(5theta). Five petals equally spaced around origin. Point (2, pi\/2) on edge is marked.\" width=\"487\" height=\"369\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 17.<\/b> Rose curve, [latex]n[\/latex] odd<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Sketch the graph of [latex]r=3\\cos \\left(3\\theta \\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q214154\">Show Solution<\/span><\/p>\n<div id=\"q214154\" class=\"hidden-answer\" style=\"display: none\">\n<p>Rose curve, [latex]n[\/latex] odd<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165602\/CNX_Precalc_Figure_08_04_018.jpg\" alt=\"Graph of rose curve r=3cos(3theta). Three petals equally spaced from origin.\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm149355\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=149355&theme=oea&iframe_resize_id=ohm149355\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Investigating the Archimedes\u2019 Spiral<\/h2>\n<p>The final polar equation we will discuss is the Archimedes\u2019 spiral, named for its discoverer, the Greek mathematician Archimedes (c. 287 BCE &#8211; c. 212 BCE), who is credited with numerous discoveries in the fields of geometry and mechanics.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Archimedes\u2019 Spiral<\/h3>\n<p>The formula that generates the graph of the <strong>Archimedes\u2019 spiral<\/strong> is given by [latex]r=\\theta[\/latex]\u00a0for [latex]\\theta \\ge 0[\/latex]. As [latex]\\theta[\/latex] increases, [latex]r[\/latex]\u00a0increases at a constant rate in an ever-widening, never-ending, spiraling path.<\/p>\n<div style=\"width: 741px\" 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\/27165546\/CNX_Precalc_Figure_08_04_020new.jpg\" alt=\"Two graphs side by side of Archimedes' spiral. (A) is r= theta, [0, 2pi]. (B) is r=theta, [0, 4pi]. Both start at origin and spiral out counterclockwise. The second has two spirals out while the first has one.\" width=\"731\" height=\"385\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 18<\/b><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given an Archimedes\u2019 spiral over [latex]\\left[0,2\\pi \\right][\/latex], sketch the graph.<\/h3>\n<ol>\n<li>Make a table of values for [latex]r[\/latex] and [latex]\\theta[\/latex] over the given domain.<\/li>\n<li>Plot the points and sketch the graph.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 10: Sketching the Graph of an Archimedes\u2019 Spiral<\/h3>\n<p>Sketch the graph of [latex]r=\\theta[\/latex] over [latex]\\left[0,2\\pi \\right][\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q329167\">Show Solution<\/span><\/p>\n<div id=\"q329167\" class=\"hidden-answer\" style=\"display: none\">\n<p>As [latex]r[\/latex] is equal to [latex]\\theta[\/latex], the plot of the Archimedes\u2019 spiral begins at the pole at the point (0, 0). While the graph hints of symmetry, there is no formal symmetry with regard to passing the symmetry tests. Further, there is no maximum value, unless the domain is restricted.<\/p>\n<p>Create a table such as the one below.<\/p>\n<table id=\"Table_08_04_09\" summary=\"Two rows and seven columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (pi\/4, 0.785), (pi\/2, 1.57), (pi, 3.14), (3pi\/2, 4.71), (7pi\/4, 5.5), (2pi, 6.28).\">\n<tbody>\n<tr>\n<td><strong> [latex]\\theta[\/latex] <\/strong><\/td>\n<td>[latex]\\frac{\\pi }{4}[\/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]\\frac{7\\pi }{4}[\/latex]<\/td>\n<td>[latex]2\\pi[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong> [latex]r[\/latex] <\/strong><\/td>\n<td>0.785<\/td>\n<td>1.57<\/td>\n<td>3.14<\/td>\n<td>4.71<\/td>\n<td>5.50<\/td>\n<td>6.28<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Notice that the <em>r<\/em>-values are just the decimal form of the angle measured in radians. We can see them on a graph in Figure 19.<\/p>\n<div style=\"width: 498px\" 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\/27165549\/CNX_Precalc_Figure_08_04_021F.jpg\" alt=\"Graph of Archimedes' spiral r=theta over &#091;0,2pi&#093;. Starts at origin and spirals out in one loop counterclockwise. Points (pi\/4, pi\/4), (pi\/2,pi\/2), (pi,pi), (5pi\/4, 5pi\/4), (7pi\/4, pi\/4), and (2pi, 2pi) are marked.\" width=\"488\" height=\"420\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 19.<\/b> Archimedes\u2019 spiral<\/p>\n<\/div>\n<h4>Analysis of the Solution<\/h4>\n<p>The domain of this polar curve is [latex]\\left[0,2\\pi \\right][\/latex]. In general, however, the domain of this function is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]. Graphing the equation of the Archimedes\u2019 spiral is rather simple, although the image makes it seem like it would be complex.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Sketch the graph of [latex]r=-\\theta[\/latex] over the interval [latex]\\left[0,4\\pi \\right][\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q937587\">Show Solution<\/span><\/p>\n<div id=\"q937587\" class=\"hidden-answer\" style=\"display: none\">\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165605\/CNX_Precalc_Figure_08_04_022.jpg\" alt=\"\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>It is easier to graph polar equations if we can test the equations for symmetry with respect to the line [latex]\\theta =\\frac{\\pi }{2}[\/latex], the polar axis, or the pole.<\/li>\n<li>There are three symmetry tests that indicate whether the graph of a polar equation will exhibit symmetry. If an equation fails a symmetry test, the graph may or may not exhibit symmetry.<\/li>\n<li>Polar equations may be graphed by making a table of values for [latex]\\theta[\/latex] and [latex]r[\/latex].<\/li>\n<li>The maximum value of a polar equation is found by substituting the value [latex]\\theta[\/latex] that leads to the maximum value of the trigonometric expression.<\/li>\n<li>The zeros of a polar equation are found by setting [latex]r=0[\/latex] and solving for [latex]\\theta[\/latex].<\/li>\n<li>Some formulas that produce the graph of a circle in polar coordinates are given by [latex]r=a\\cos \\theta[\/latex] and [latex]r=a\\sin \\theta[\/latex].<\/li>\n<li>The formulas that produce the graphs of a cardioid are given by [latex]r=a\\pm b\\cos \\theta[\/latex] and [latex]r=a\\pm b\\sin \\theta[\/latex], for [latex]a>0,b>0[\/latex], and [latex]\\frac{a}{b}=1[\/latex].<\/li>\n<li>The formulas that produce the graphs of a one-loop lima\u00e7on are given by [latex]r=a\\pm b\\cos \\theta[\/latex] and [latex]r=a\\pm b\\sin \\theta[\/latex] for [latex]1<\\frac{a}{b}<2[\/latex].<\/li>\n<li>The formulas that produce the graphs of an inner-loop lima\u00e7on are given by [latex]r=a\\pm b\\cos \\theta[\/latex] and [latex]r=a\\pm b\\sin \\theta[\/latex] for [latex]a>0,b>0[\/latex],\u00a0and [latex]a<b[\/latex].<\/li>\n<li>The formulas that produce the graphs of a lemniscates are given by [latex]{r}^{2}={a}^{2}\\cos 2\\theta[\/latex] and [latex]{r}^{2}={a}^{2}\\sin 2\\theta[\/latex], where [latex]a\\ne 0[\/latex].<\/li>\n<li>The formulas that produce the graphs of rose curves are given by [latex]r=a\\cos n\\theta[\/latex] and [latex]r=a\\sin n\\theta[\/latex], where [latex]a\\ne 0[\/latex]; if [latex]n[\/latex] is even, there are [latex]2n[\/latex] petals, and if [latex]n[\/latex] is odd, there are [latex]n[\/latex] petals.<\/li>\n<li>The formula that produces the graph of an Archimedes\u2019 spiral is given by [latex]r=\\theta ,\\theta \\ge 0[\/latex].<\/li>\n<\/ul>\n<h2>Summary of Curves<\/h2>\n<p>We have explored a number of seemingly complex polar curves in this section. Figures 20 and 21\u00a0summarize the graphs and equations for each of these curves.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165552\/CNX_Precalc_Figure_08_04_023n.jpg\" alt=\"&quot;Four\" \/><\/p>\n<p><img decoding=\"async\" style=\"border: 0px; vertical-align: middle; max-width: 100%;\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165555\/CNX_Precalc_Figure_08_04_024n.jpg\" alt=\"&quot;Four\" \/><\/p>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165137410447\" class=\"definition\">\n<dt>Archimedes\u2019 spiral<\/dt>\n<dd id=\"fs-id1165137749714\">a polar curve given by [latex]r=\\theta[\/latex]. When multiplied by a constant, the equation appears as [latex]r=a\\theta[\/latex]. As [latex]r=\\theta[\/latex], the curve continues to widen in a spiral path over the domain.<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135378768\" class=\"definition\">\n<dt>cardioid<\/dt>\n<dd id=\"fs-id1165135378771\">a member of the lima\u00e7on family of curves, named for its resemblance to a heart; its equation is given as [latex]r=a\\pm b\\cos \\theta[\/latex] and [latex]r=a\\pm b\\sin \\theta[\/latex], where [latex]\\frac{a}{b}=1[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135344897\" class=\"definition\">\n<dt>convex lima\u04abon<\/dt>\n<dd id=\"fs-id1165135344900\">a type of one-loop lima\u00e7on represented by [latex]r=a\\pm b\\cos \\theta[\/latex] and [latex]r=a\\pm b\\sin \\theta[\/latex] such that [latex]\\frac{a}{b}\\ge 2[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135680172\" class=\"definition\">\n<dt>dimpled lima\u04abon<\/dt>\n<dd id=\"fs-id1165135680175\">a type of one-loop lima\u00e7on represented by [latex]r=a\\pm b\\cos \\theta[\/latex] and [latex]r=a\\pm b\\sin \\theta[\/latex] such that [latex]1<\\frac{a}{b}<2[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134042074\" class=\"definition\">\n<dt>inner-loop lima\u00e7on<\/dt>\n<dd id=\"fs-id1165135191979\">a polar curve similar to the cardioid, but with an inner loop; passes through the pole twice; represented by [latex]r=a\\pm b\\cos \\theta[\/latex] and [latex]\\text{ }r=a\\pm b\\sin \\theta \\text{ }[\/latex] where [latex]a<b[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137547628\" class=\"definition\">\n<dt>lemniscate<\/dt>\n<dd id=\"fs-id1165137547632\">a polar curve resembling a figure 8 and given by the equation [latex]{r}^{2}={a}^{2}\\cos 2\\theta[\/latex] and [latex]{r}^{2}={a}^{2}\\sin 2\\theta ,a\\ne 0[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137911243\" class=\"definition\">\n<dt>one-loop lima\u04abon<\/dt>\n<dd id=\"fs-id1165137911246\">a polar curve represented by [latex]r=a\\pm b\\cos \\theta[\/latex] and [latex]r=a\\pm b\\sin \\theta[\/latex] such that [latex]a>0,b>0[\/latex], and [latex]\\frac{a}{b}>1[\/latex]; may be dimpled or convex; does not pass through the pole<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135564114\" class=\"definition\">\n<dt>polar equation<\/dt>\n<dd id=\"fs-id1165134331101\">an equation describing a curve on the polar grid.<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137626838\" class=\"definition\">\n<dt>rose curve<\/dt>\n<dd id=\"fs-id1165135306471\">a polar equation resembling a flower, given by the equations [latex]r=a\\cos n\\theta[\/latex] and [latex]r=a\\sin n\\theta[\/latex]; when [latex]n[\/latex] is even there are [latex]2n[\/latex] petals, and the curve is highly symmetrical; when [latex]n[\/latex] is odd there are [latex]n[\/latex] petals.<\/dd>\n<\/dl>\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-14414\">\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":4,"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-14414","chapter","type-chapter","status-publish","hentry"],"part":14256,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/14414","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":6,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/14414\/revisions"}],"predecessor-version":[{"id":15879,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/14414\/revisions\/15879"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/14256"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/14414\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=14414"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=14414"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=14414"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=14414"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}