{"id":1019,"date":"2025-06-20T17:29:04","date_gmt":"2025-06-20T17:29:04","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1019"},"modified":"2025-09-10T18:19:06","modified_gmt":"2025-09-10T18:19:06","slug":"understanding-polar-coordinates-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/understanding-polar-coordinates-learn-it-4\/","title":{"raw":"Understanding Polar Coordinates: Learn It 4","rendered":"Understanding Polar Coordinates: Learn It 4"},"content":{"raw":"<h2 data-type=\"title\">Symmetry in Polar Coordinates<\/h2>\r\n<p class=\"whitespace-normal break-words\">Just as we analyze symmetry in rectangular coordinates to understand function behavior, we can examine symmetry in polar curves to reveal important properties and simplify graphing.<\/p>\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">\r\n<p class=\"whitespace-normal break-words\"><strong>Symmetry in Rectangular Coordinates<\/strong>:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Even functions<\/strong>: [latex]f(-x) = f(x)[\/latex] creates [latex]y[\/latex]-axis symmetry<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Odd functions<\/strong>: [latex]f(-x) = -f(x)[\/latex] creates origin symmetry<\/li>\r\n<\/ul>\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">Polar curves exhibit three main types of symmetry, each corresponding to reflection across a different line or point.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>symmetry in polar curves and equations<\/h3>\r\n<p id=\"fs-id1167794226656\">For a curve [latex]r=f\\left(\\theta \\right)[\/latex] in polar coordinates, we can test for three types of symmetry:<\/p>\r\n\r\n<ol style=\"list-style-type: decimal;\">\r\n \t<li><strong>Polar Axis Symmetry (horizontal line symmetry)<\/strong>\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">If point [latex]\\left(r,\\theta \\right)[\/latex] is on the graph, then [latex]\\left(r,-\\theta \\right)[\/latex] \u00a0is also on the graph<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Test: Replace [latex]\\theta[\/latex] with [latex]-\\theta[\/latex] in the equation<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li><strong>Pole Symmetry (origin symmetry)<\/strong>\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">If point [latex]\\left(r,\\theta \\right)[\/latex] is on the graph, then [latex](r,\\pi + \\theta)[\/latex] is also on the graph<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Test: Replace [latex]r[\/latex] with [latex]-r[\/latex], or replace [latex]\\theta[\/latex] with [latex]\\pi + \\theta[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li><strong>Vertical Line Symmetry<\/strong> (line [latex]\\theta = \\frac{\\pi}{2}[\/latex])<\/li>\r\n<\/ol>\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">If point [latex]\\left(r,\\theta \\right)[\/latex] is on the graph, then [latex](r,\\pi - \\theta)[\/latex] is also on the graph<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Test: Replace [latex]\\theta[\/latex] with [latex]\\pi - \\theta[\/latex] in the equation<\/li>\r\n<\/ul>\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">Understanding these symmetries helps you graph polar curves more efficiently. If you can establish that a curve has certain symmetries, you only need to plot points in one region and then reflect them to complete the entire graph.<\/p>\r\n<p id=\"fs-id1167794215128\">The following table shows examples of each type of symmetry.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"796\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234840\/CNX_Calc_Figure_11_03_018.jpg\" alt=\"This table has three rows and two columns. The first row reads \" width=\"796\" height=\"1002\" data-media-type=\"image\/jpeg\" \/> Figure 11.[\/caption]\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1167794226675\" data-type=\"problem\">\r\n<p id=\"fs-id1167794226680\">Find the symmetry of the rose defined by the equation [latex]r=3\\sin\\left(2\\theta \\right)[\/latex] and create a graph.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558859\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558859\"]\r\n<div id=\"fs-id1167794226707\" data-type=\"solution\">\r\n<p id=\"fs-id1167794226709\">Suppose the point [latex]\\left(r,\\theta \\right)[\/latex] is on the graph of [latex]r=3\\sin\\left(2\\theta \\right)[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1167793238191\" type=\"i\">\r\n \t<li>To test for symmetry about the polar axis, first try replacing [latex]\\theta [\/latex] with [latex]-\\theta [\/latex]. This gives [latex]r=3\\sin\\left(2\\left(-\\theta \\right)\\right)=-3\\sin\\left(2\\theta \\right)[\/latex]. Since this changes the original equation, this test is not satisfied. However, returning to the original equation and replacing [latex]r[\/latex] with [latex]-r[\/latex] and [latex]\\theta [\/latex] with [latex]\\pi -\\theta [\/latex] yields<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167794332476\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ -r=3\\sin\\left(2\\left(\\pi -\\theta \\right)\\right)\\hfill \\\\ -r=3\\sin\\left(2\\pi -2\\theta \\right)\\hfill \\\\ -r=3\\sin\\left(-2\\theta \\right)\\hfill \\\\ -r=-3\\sin2\\theta .\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nMultiplying both sides of this equation by [latex]-1[\/latex] gives [latex]r=3\\sin2\\theta [\/latex], which is the original equation. This demonstrates that the graph is symmetric with respect to the polar axis.<\/li>\r\n \t<li>To test for symmetry with respect to the pole, first replace [latex]r[\/latex] with [latex]-r[\/latex], which yields [latex]-r=3\\sin\\left(2\\theta \\right)[\/latex]. Multiplying both sides by \u22121 gives [latex]r=-3\\sin\\left(2\\theta \\right)[\/latex], which does not agree with the original equation. Therefore the equation does not pass the test for this symmetry. However, returning to the original equation and replacing [latex]\\theta [\/latex] with [latex]\\theta +\\pi [\/latex] gives<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793252278\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill r&amp; =3\\sin\\left(2\\left(\\theta +\\pi \\right)\\right)\\hfill \\\\ &amp; =3\\sin\\left(2\\theta +2\\pi \\right)\\hfill \\\\ &amp; =3\\left(\\sin2\\theta \\cos2\\pi +\\cos2\\theta \\sin2\\pi \\right)\\hfill \\\\ &amp; =3\\sin2\\theta .\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince this agrees with the original equation, the graph is symmetric about the pole.<\/li>\r\n \t<li>To test for symmetry with respect to the vertical line [latex]\\theta =\\frac{\\pi }{2}[\/latex], first replace both [latex]r[\/latex] with [latex]-r[\/latex] and [latex]\\theta [\/latex] with [latex]-\\theta [\/latex]. <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167794136442\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ -r=3\\sin\\left(2\\left(-\\theta \\right)\\right)\\hfill \\\\ -r=3\\sin\\left(-2\\theta \\right)\\hfill \\\\ -r=-3\\sin2\\theta .\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nMultiplying both sides of this equation by [latex]-1[\/latex] gives [latex]r=3\\sin2\\theta [\/latex], which is the original equation. Therefore the graph is symmetric about the vertical line [latex]\\theta =\\frac{\\pi }{2}[\/latex].<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1167793260184\">This graph has symmetry with respect to the polar axis, the origin, and the vertical line going through the pole. To graph the function, tabulate values of [latex]\\theta [\/latex] between 0 and [latex]\\frac{\\pi}{2}[\/latex] and then reflect the resulting graph.<\/p>\r\n\r\n<table id=\"fs-id1167793260210\" class=\"unnumbered\" summary=\"This table has two columns and six rows. The first row is a header row, and it reads from left to right \u03b8 and r. Below the header row, in the first column, the values read 0, \u03c0\/6, \u03c0\/4, \u03c0\/3, and \u03c0\/2. In the second column, the values read 0, (3 times the square root of 3) all divided by 2, which is approximately equal to 2.6, 3, (3 times the square root of 3) all divided by 2, which is approximately equal to 2.6, and 0.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th data-valign=\"top\" data-align=\"center\">[latex]\\theta [\/latex]<\/th>\r\n<th data-valign=\"top\" data-align=\"center\">[latex]r[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td data-valign=\"top\" data-align=\"center\">[latex]0[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"center\">[latex]0[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-valign=\"top\" data-align=\"center\">[latex]\\frac{\\pi }{6}[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"center\">[latex]\\frac{3\\sqrt{3}}{2}\\approx 2.6[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-valign=\"top\" data-align=\"center\">[latex]\\frac{\\pi }{4}[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"center\">[latex]3[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-valign=\"top\" data-align=\"center\">[latex]\\frac{\\pi }{3}[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"center\">[latex]\\frac{3\\sqrt{3}}{2}\\approx 2.6[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-valign=\"top\" data-align=\"center\">[latex]\\frac{\\pi }{2}[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"center\">[latex]0[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1167794201964\">This gives one petal of the rose, as shown in the following graph.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_03_013\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"417\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234842\/CNX_Calc_Figure_11_03_013.jpg\" alt=\"A single petal is graphed with equation r = 3 sin(2\u03b8) for 0 \u2264 \u03b8 \u2264 \u03c0\/2. It starts at the origin and reaches a maximum distance from the origin of 3.\" width=\"417\" height=\"417\" data-media-type=\"image\/jpeg\" \/> Figure 12. The graph of the equation between [latex]\\theta =0[\/latex] and [latex]\\theta =\\frac{\\pi}{2}[\/latex].[\/caption]<\/figure>\r\n<p id=\"fs-id1167794202015\">Reflecting this image into the other three quadrants gives the entire graph as shown.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_03_014\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"417\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234845\/CNX_Calc_Figure_11_03_014.jpg\" alt=\"A four-petaled rose is graphed with equation r = 3 sin(2\u03b8). Each petal starts at the origin and reaches a maximum distance from the origin of 3.\" width=\"417\" height=\"417\" data-media-type=\"image\/jpeg\" \/> Figure 13. The entire graph of the equation is called a four-petaled rose.[\/caption]<\/figure>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the example above.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/m_wIuLZn03U?controls=0&amp;start=1580&amp;end=1886&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/7.3PolarCoordinates1580to1886_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"7.3 Polar Coordinates\" here (opens in new window)<\/a>.\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]311324[\/ohm_question]<\/section>","rendered":"<h2 data-type=\"title\">Symmetry in Polar Coordinates<\/h2>\n<p class=\"whitespace-normal break-words\">Just as we analyze symmetry in rectangular coordinates to understand function behavior, we can examine symmetry in polar curves to reveal important properties and simplify graphing.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">\n<p class=\"whitespace-normal break-words\"><strong>Symmetry in Rectangular Coordinates<\/strong>:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Even functions<\/strong>: [latex]f(-x) = f(x)[\/latex] creates [latex]y[\/latex]-axis symmetry<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Odd functions<\/strong>: [latex]f(-x) = -f(x)[\/latex] creates origin symmetry<\/li>\n<\/ul>\n<\/section>\n<p class=\"whitespace-normal break-words\">Polar curves exhibit three main types of symmetry, each corresponding to reflection across a different line or point.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>symmetry in polar curves and equations<\/h3>\n<p id=\"fs-id1167794226656\">For a curve [latex]r=f\\left(\\theta \\right)[\/latex] in polar coordinates, we can test for three types of symmetry:<\/p>\n<ol style=\"list-style-type: decimal;\">\n<li><strong>Polar Axis Symmetry (horizontal line symmetry)<\/strong>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">If point [latex]\\left(r,\\theta \\right)[\/latex] is on the graph, then [latex]\\left(r,-\\theta \\right)[\/latex] \u00a0is also on the graph<\/li>\n<li class=\"whitespace-normal break-words\">Test: Replace [latex]\\theta[\/latex] with [latex]-\\theta[\/latex] in the equation<\/li>\n<\/ul>\n<\/li>\n<li><strong>Pole Symmetry (origin symmetry)<\/strong>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">If point [latex]\\left(r,\\theta \\right)[\/latex] is on the graph, then [latex](r,\\pi + \\theta)[\/latex] is also on the graph<\/li>\n<li class=\"whitespace-normal break-words\">Test: Replace [latex]r[\/latex] with [latex]-r[\/latex], or replace [latex]\\theta[\/latex] with [latex]\\pi + \\theta[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li><strong>Vertical Line Symmetry<\/strong> (line [latex]\\theta = \\frac{\\pi}{2}[\/latex])<\/li>\n<\/ol>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">If point [latex]\\left(r,\\theta \\right)[\/latex] is on the graph, then [latex](r,\\pi - \\theta)[\/latex] is also on the graph<\/li>\n<li class=\"whitespace-normal break-words\">Test: Replace [latex]\\theta[\/latex] with [latex]\\pi - \\theta[\/latex] in the equation<\/li>\n<\/ul>\n<\/section>\n<p class=\"whitespace-normal break-words\">Understanding these symmetries helps you graph polar curves more efficiently. If you can establish that a curve has certain symmetries, you only need to plot points in one region and then reflect them to complete the entire graph.<\/p>\n<p id=\"fs-id1167794215128\">The following table shows examples of each type of symmetry.<\/p>\n<figure style=\"width: 796px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234840\/CNX_Calc_Figure_11_03_018.jpg\" alt=\"This table has three rows and two columns. The first row reads\" width=\"796\" height=\"1002\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 11.<\/figcaption><\/figure>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167794226675\" data-type=\"problem\">\n<p id=\"fs-id1167794226680\">Find the symmetry of the rose defined by the equation [latex]r=3\\sin\\left(2\\theta \\right)[\/latex] and create a graph.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558859\">Show Solution<\/button><\/p>\n<div id=\"q44558859\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794226707\" data-type=\"solution\">\n<p id=\"fs-id1167794226709\">Suppose the point [latex]\\left(r,\\theta \\right)[\/latex] is on the graph of [latex]r=3\\sin\\left(2\\theta \\right)[\/latex].<\/p>\n<ol id=\"fs-id1167793238191\" type=\"i\">\n<li>To test for symmetry about the polar axis, first try replacing [latex]\\theta[\/latex] with [latex]-\\theta[\/latex]. This gives [latex]r=3\\sin\\left(2\\left(-\\theta \\right)\\right)=-3\\sin\\left(2\\theta \\right)[\/latex]. Since this changes the original equation, this test is not satisfied. However, returning to the original equation and replacing [latex]r[\/latex] with [latex]-r[\/latex] and [latex]\\theta[\/latex] with [latex]\\pi -\\theta[\/latex] yields<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167794332476\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ -r=3\\sin\\left(2\\left(\\pi -\\theta \\right)\\right)\\hfill \\\\ -r=3\\sin\\left(2\\pi -2\\theta \\right)\\hfill \\\\ -r=3\\sin\\left(-2\\theta \\right)\\hfill \\\\ -r=-3\\sin2\\theta .\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nMultiplying both sides of this equation by [latex]-1[\/latex] gives [latex]r=3\\sin2\\theta[\/latex], which is the original equation. This demonstrates that the graph is symmetric with respect to the polar axis.<\/li>\n<li>To test for symmetry with respect to the pole, first replace [latex]r[\/latex] with [latex]-r[\/latex], which yields [latex]-r=3\\sin\\left(2\\theta \\right)[\/latex]. Multiplying both sides by \u22121 gives [latex]r=-3\\sin\\left(2\\theta \\right)[\/latex], which does not agree with the original equation. Therefore the equation does not pass the test for this symmetry. However, returning to the original equation and replacing [latex]\\theta[\/latex] with [latex]\\theta +\\pi[\/latex] gives<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793252278\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill r& =3\\sin\\left(2\\left(\\theta +\\pi \\right)\\right)\\hfill \\\\ & =3\\sin\\left(2\\theta +2\\pi \\right)\\hfill \\\\ & =3\\left(\\sin2\\theta \\cos2\\pi +\\cos2\\theta \\sin2\\pi \\right)\\hfill \\\\ & =3\\sin2\\theta .\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince this agrees with the original equation, the graph is symmetric about the pole.<\/li>\n<li>To test for symmetry with respect to the vertical line [latex]\\theta =\\frac{\\pi }{2}[\/latex], first replace both [latex]r[\/latex] with [latex]-r[\/latex] and [latex]\\theta[\/latex] with [latex]-\\theta[\/latex]. <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167794136442\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ -r=3\\sin\\left(2\\left(-\\theta \\right)\\right)\\hfill \\\\ -r=3\\sin\\left(-2\\theta \\right)\\hfill \\\\ -r=-3\\sin2\\theta .\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nMultiplying both sides of this equation by [latex]-1[\/latex] gives [latex]r=3\\sin2\\theta[\/latex], which is the original equation. Therefore the graph is symmetric about the vertical line [latex]\\theta =\\frac{\\pi }{2}[\/latex].<\/li>\n<\/ol>\n<p id=\"fs-id1167793260184\">This graph has symmetry with respect to the polar axis, the origin, and the vertical line going through the pole. To graph the function, tabulate values of [latex]\\theta[\/latex] between 0 and [latex]\\frac{\\pi}{2}[\/latex] and then reflect the resulting graph.<\/p>\n<table id=\"fs-id1167793260210\" class=\"unnumbered\" summary=\"This table has two columns and six rows. The first row is a header row, and it reads from left to right \u03b8 and r. Below the header row, in the first column, the values read 0, \u03c0\/6, \u03c0\/4, \u03c0\/3, and \u03c0\/2. In the second column, the values read 0, (3 times the square root of 3) all divided by 2, which is approximately equal to 2.6, 3, (3 times the square root of 3) all divided by 2, which is approximately equal to 2.6, and 0.\" data-label=\"\">\n<thead>\n<tr valign=\"top\">\n<th data-valign=\"top\" data-align=\"center\">[latex]\\theta[\/latex]<\/th>\n<th data-valign=\"top\" data-align=\"center\">[latex]r[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td data-valign=\"top\" data-align=\"center\">[latex]0[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"center\">[latex]0[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-valign=\"top\" data-align=\"center\">[latex]\\frac{\\pi }{6}[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"center\">[latex]\\frac{3\\sqrt{3}}{2}\\approx 2.6[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-valign=\"top\" data-align=\"center\">[latex]\\frac{\\pi }{4}[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"center\">[latex]3[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-valign=\"top\" data-align=\"center\">[latex]\\frac{\\pi }{3}[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"center\">[latex]\\frac{3\\sqrt{3}}{2}\\approx 2.6[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-valign=\"top\" data-align=\"center\">[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"center\">[latex]0[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1167794201964\">This gives one petal of the rose, as shown in the following graph.<\/p>\n<figure id=\"CNX_Calc_Figure_11_03_013\"><figcaption><\/figcaption><figure style=\"width: 417px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234842\/CNX_Calc_Figure_11_03_013.jpg\" alt=\"A single petal is graphed with equation r = 3 sin(2\u03b8) for 0 \u2264 \u03b8 \u2264 \u03c0\/2. It starts at the origin and reaches a maximum distance from the origin of 3.\" width=\"417\" height=\"417\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 12. The graph of the equation between [latex]\\theta =0[\/latex] and [latex]\\theta =\\frac{\\pi}{2}[\/latex].<\/figcaption><\/figure>\n<\/figure>\n<p id=\"fs-id1167794202015\">Reflecting this image into the other three quadrants gives the entire graph as shown.<\/p>\n<figure id=\"CNX_Calc_Figure_11_03_014\"><figcaption><\/figcaption><figure style=\"width: 417px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234845\/CNX_Calc_Figure_11_03_014.jpg\" alt=\"A four-petaled rose is graphed with equation r = 3 sin(2\u03b8). Each petal starts at the origin and reaches a maximum distance from the origin of 3.\" width=\"417\" height=\"417\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 13. The entire graph of the equation is called a four-petaled rose.<\/figcaption><\/figure>\n<\/figure>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the example above.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/m_wIuLZn03U?controls=0&amp;start=1580&amp;end=1886&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/7.3PolarCoordinates1580to1886_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;7.3 Polar Coordinates&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm311324\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=311324&theme=lumen&iframe_resize_id=ohm311324&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":8,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":677,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1019"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1019\/revisions"}],"predecessor-version":[{"id":2307,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1019\/revisions\/2307"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/677"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1019\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1019"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1019"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1019"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1019"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}