{"id":1571,"date":"2025-07-25T03:51:06","date_gmt":"2025-07-25T03:51:06","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=1571"},"modified":"2026-03-12T06:59:02","modified_gmt":"2026-03-12T06:59:02","slug":"graphing-in-polar-coordinates-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/graphing-in-polar-coordinates-fresh-take\/","title":{"raw":"Graphing in Polar Coordinates: Fresh Take","rendered":"Graphing in Polar Coordinates: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\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<\/section>\r\n<h2 data-type=\"title\">Polar Curves<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Polar curves work differently from regular graphs. Instead of plotting [latex]y[\/latex] against [latex]x[\/latex], you're plotting how the distance from the origin changes as you rotate around it. Think of it as drawing with a pen tied to a string\u2014the length of the string ([latex]r[\/latex]) changes as you spin around ([latex]\\theta[\/latex]).<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy:<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Make a table with [latex]\\theta[\/latex] and [latex]r = f(\\theta)[\/latex] values<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Choose strategic [latex]\\theta[\/latex] values (consider the function's period)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Calculate corresponding [latex]r[\/latex] values<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Plot each [latex](r, \\theta)[\/latex] point<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Connect the points to reveal the pattern<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\">Many create recognizable shapes that would be complicated to describe in rectangular coordinates. For example, [latex]r = 4\\sin\\theta[\/latex] creates a perfect circle, even though the equation looks nothing like a circle formula.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Common polar curve families:<\/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>Lines through origin:<\/strong> [latex]\\theta = K[\/latex] creates a straight line with slope [latex]\\tan K[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Circles:<\/strong> [latex]r = a[\/latex] makes a circle centered at origin; [latex]r = a\\cos\\theta + b\\sin\\theta[\/latex] makes circles passing through origin<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Rose curves:<\/strong> [latex]r = a\\sin(b\\theta)[\/latex] or [latex]r = a\\cos(b\\theta)[\/latex] create flower-like petals<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Cardioids:<\/strong> [latex]r = a(1 \u00b1 \\cos\\theta)[\/latex] or [latex]r = a(1 \u00b1 \\sin\\theta)[\/latex] form heart shapes<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Rose petal rule:<\/strong> For [latex]r = a\\sin(b\\theta)[\/latex], if [latex]b[\/latex] is even, you get [latex]2b[\/latex] petals; if [latex]b[\/latex] is odd, you get [latex]b[\/latex] petals.<\/p>\r\n<p class=\"whitespace-normal break-words\">To switch a polar curve to rectangular coordinates, multiply both sides by [latex]r[\/latex] when helpful, then substitute [latex]r^2 = x^2 + y^2[\/latex], [latex]x = r\\cos\\theta[\/latex], and [latex]y = r\\sin\\theta[\/latex]. Complete the square when you end up with circle equations. Many polar functions repeat their patterns, so you often only need to evaluate one full period to capture the entire curve.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox watchIt\" aria-label=\"Watch It\"><script type=\"text\/javascript\" src=\"https:\/\/www.youtube.com\/iframe_api \"><\/script>\r\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-chgedhah-OAEjSiFGdhQ\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/OAEjSiFGdhQ?enablejsapi=1 \" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\r\n\r\n<div id=\"3p-plugin-target-chgedhah-OAEjSiFGdhQ\" class=\"p3sdk-target\"><\/div>\r\n<p class=\"cc-media-iframe-container\"><script type='text\/javascript' src='\/\/plugin.3playmedia.com\/ajax.js?cc=1&cc_minimizable=1&cc_minimize_on_load=0&cc_multi_text_track=0&cc_overlay=1&cc_searchable=0&embed=ajax&mf=14661427&p3sdk_version=1.11.7&p=20361&player_type=youtube&plugin_skin=dark&target=3p-plugin-target-chgedhah-OAEjSiFGdhQ&vembed=0&video_id=OAEjSiFGdhQ&video_target=tpm-plugin-chgedhah-OAEjSiFGdhQ'><\/script><\/p>\r\nYou can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Precalculus\/Transcripts\/Sketching+polar+curves+from+cartesian+curves+(KristaKingMath)_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cSketching polar curves from cartesian curves (KristaKingMath)\u201d here (opens in new window).<\/a>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1167793248902\" data-type=\"problem\">\r\n<p id=\"fs-id1167793248904\">Create a graph of the curve defined by the function [latex]r=4+4\\cos\\theta [\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558891\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558891\"]\r\n<div id=\"fs-id1167793249108\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167793249052\">Follow the problem-solving strategy for creating a graph in polar coordinates.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558892\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n<div id=\"fs-id1167793249087\" data-type=\"solution\">\r\n<p id=\"fs-id1167793249088\"><span data-type=\"newline\">\u00a0<\/span><\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"717\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234820\/CNX_Calc_Figure_11_03_006.jpg\" alt=\"The graph of r = 4 + 4 cos\u03b8 is given. It vaguely looks look a heart tipped on its side with a rounded bottom instead of a pointed one. Specifically, the graph starts at the origin, moves into the second quadrant and increases to a rounded circle-like figure. The graph is symmetric about the x axis, so it continues its rounded circle-like figure, goes into the third quadrant, and comes to a point at the origin.\" width=\"717\" height=\"717\" data-media-type=\"image\/jpeg\" \/> Figure 6.[\/caption]\r\n\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThe name of this shape is a cardioid, which we will study further later in this section.\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Rewrite the equation [latex]r=\\sec\\theta \\tan\\theta [\/latex] in rectangular coordinates and identify its graph.[reveal-answer q=\"44558879\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558879\"]\r\n<div id=\"fs-id1167793219344\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167793219351\">Convert to sine and cosine, then multiply both sides by cosine.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558889\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558889\"]\r\n<div id=\"fs-id1167794072980\" data-type=\"solution\">\r\n<p id=\"fs-id1167794072982\">[latex]y={x}^{2}[\/latex], which is the equation of a parabola opening upward.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h2 data-type=\"title\">Symmetry in Polar Coordinates<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Symmetry in polar coordinates is your shortcut to graphing complex curves without plotting every single point. Just like even and odd functions in rectangular coordinates have predictable patterns, polar curves have three types of symmetry that can cut your work down dramatically.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The three types of polar symmetry:<\/strong><\/p>\r\n\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Polar axis symmetry<\/strong> (reflects across the [latex]x[\/latex]-axis): If [latex](r, \\theta)[\/latex] is on the curve, then [latex](r, -\\theta)[\/latex] is too. Test by replacing [latex]\\theta[\/latex] with [latex]-\\theta[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Pole symmetry<\/strong> (reflects through the origin): If [latex](r, \\theta)[\/latex] is on the curve, then [latex](r, \\pi + \\theta)[\/latex] is too. Test by replacing [latex]r[\/latex] with [latex]-r[\/latex] or [latex]\\theta[\/latex] with [latex]\\pi + \\theta[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Vertical line symmetry<\/strong> (reflects across the [latex]y[\/latex]-axis): If [latex](r, \\theta)[\/latex] is on the curve, then [latex](r, \\pi - \\theta)[\/latex] is too. Test by replacing [latex]\\theta[\/latex] with [latex]\\pi - \\theta[\/latex].<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">When you identify symmetries, you only need to plot points in one region, then reflect them to complete the entire graph. For a rose curve like [latex]r = 3\\sin(2\\theta)[\/latex] with all three symmetries, you can plot just the first quadrant and reflect to get all four petals.<\/p>\r\n<p class=\"whitespace-normal break-words\">For each symmetry test, substitute the transformation into your equation. If you get back the original equation (possibly after algebraic manipulation), that symmetry exists.<\/p>\r\n<p class=\"whitespace-normal break-words\">Remember that the same point can have different polar coordinates. A symmetry test might fail in one form but succeed when you account for equivalent representations.<\/p>\r\n<p class=\"whitespace-normal break-words\">Even if a curve doesn't pass a particular symmetry test algebraically, you can still check visually after plotting a few key points to see if patterns emerge.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1167794202050\" data-type=\"problem\">\r\n<p id=\"fs-id1167794202052\">Determine the symmetry of the graph determined by the equation [latex]r=2\\cos\\left(3\\theta \\right)[\/latex] and create a graph.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558839\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558839\"]\r\n<div id=\"fs-id1167794324547\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167794324554\">Use the theorem.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558849\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558849\"]\r\n<div id=\"fs-id1167794324525\" data-type=\"solution\">\r\n<p id=\"fs-id1167794324527\">Symmetric with respect to the polar axis.<span data-type=\"newline\">\r\n<\/span><\/p>\r\n\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"417\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234847\/CNX_Calc_Figure_11_03_015.jpg\" alt=\"A three-petaled rose is graphed with equation r = 2 cos(3\u03b8). Each petal starts at the origin and reaches a maximum distance from the origin of 2.\" width=\"417\" height=\"417\" data-media-type=\"image\/jpeg\" \/> Figure 14.[\/caption]\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Test polar equations for symmetry.<\/li>\n<li>Graph polar equations by plotting points.<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Polar Curves<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Polar curves work differently from regular graphs. Instead of plotting [latex]y[\/latex] against [latex]x[\/latex], you&#8217;re plotting how the distance from the origin changes as you rotate around it. Think of it as drawing with a pen tied to a string\u2014the length of the string ([latex]r[\/latex]) changes as you spin around ([latex]\\theta[\/latex]).<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy:<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Make a table with [latex]\\theta[\/latex] and [latex]r = f(\\theta)[\/latex] values<\/li>\n<li class=\"whitespace-normal break-words\">Choose strategic [latex]\\theta[\/latex] values (consider the function&#8217;s period)<\/li>\n<li class=\"whitespace-normal break-words\">Calculate corresponding [latex]r[\/latex] values<\/li>\n<li class=\"whitespace-normal break-words\">Plot each [latex](r, \\theta)[\/latex] point<\/li>\n<li class=\"whitespace-normal break-words\">Connect the points to reveal the pattern<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\">Many create recognizable shapes that would be complicated to describe in rectangular coordinates. For example, [latex]r = 4\\sin\\theta[\/latex] creates a perfect circle, even though the equation looks nothing like a circle formula.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Common polar curve families:<\/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>Lines through origin:<\/strong> [latex]\\theta = K[\/latex] creates a straight line with slope [latex]\\tan K[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Circles:<\/strong> [latex]r = a[\/latex] makes a circle centered at origin; [latex]r = a\\cos\\theta + b\\sin\\theta[\/latex] makes circles passing through origin<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Rose curves:<\/strong> [latex]r = a\\sin(b\\theta)[\/latex] or [latex]r = a\\cos(b\\theta)[\/latex] create flower-like petals<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Cardioids:<\/strong> [latex]r = a(1 \u00b1 \\cos\\theta)[\/latex] or [latex]r = a(1 \u00b1 \\sin\\theta)[\/latex] form heart shapes<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Rose petal rule:<\/strong> For [latex]r = a\\sin(b\\theta)[\/latex], if [latex]b[\/latex] is even, you get [latex]2b[\/latex] petals; if [latex]b[\/latex] is odd, you get [latex]b[\/latex] petals.<\/p>\n<p class=\"whitespace-normal break-words\">To switch a polar curve to rectangular coordinates, multiply both sides by [latex]r[\/latex] when helpful, then substitute [latex]r^2 = x^2 + y^2[\/latex], [latex]x = r\\cos\\theta[\/latex], and [latex]y = r\\sin\\theta[\/latex]. Complete the square when you end up with circle equations. Many polar functions repeat their patterns, so you often only need to evaluate one full period to capture the entire curve.<\/p>\n<\/div>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\"><script type=\"text\/javascript\" src=\"https:\/\/www.youtube.com\/iframe_api\"><\/script><\/p>\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-chgedhah-OAEjSiFGdhQ\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/OAEjSiFGdhQ?enablejsapi=1\" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<div id=\"3p-plugin-target-chgedhah-OAEjSiFGdhQ\" class=\"p3sdk-target\"><\/div>\n<p class=\"cc-media-iframe-container\"><script type=\"text\/javascript\" src=\"\/\/plugin.3playmedia.com\/ajax.js?cc=1&#38;cc_minimizable=1&#38;cc_minimize_on_load=0&#38;cc_multi_text_track=0&#38;cc_overlay=1&#38;cc_searchable=0&#38;embed=ajax&#38;mf=14661427&#38;p3sdk_version=1.11.7&#38;p=20361&#38;player_type=youtube&#38;plugin_skin=dark&#38;target=3p-plugin-target-chgedhah-OAEjSiFGdhQ&#38;vembed=0&#38;video_id=OAEjSiFGdhQ&#38;video_target=tpm-plugin-chgedhah-OAEjSiFGdhQ\"><\/script><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Precalculus\/Transcripts\/Sketching+polar+curves+from+cartesian+curves+(KristaKingMath)_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cSketching polar curves from cartesian curves (KristaKingMath)\u201d here (opens in new window).<\/a><\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167793248902\" data-type=\"problem\">\n<p id=\"fs-id1167793248904\">Create a graph of the curve defined by the function [latex]r=4+4\\cos\\theta[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558891\">Hint<\/button><\/p>\n<div id=\"q44558891\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167793249108\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167793249052\">Follow the problem-solving strategy for creating a graph in polar coordinates.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558892\">Show Solution<\/button><\/p>\n<div id=\"q44558892\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167793249087\" data-type=\"solution\">\n<p id=\"fs-id1167793249088\"><span data-type=\"newline\">\u00a0<\/span><\/p>\n<figure style=\"width: 717px\" 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\/11234820\/CNX_Calc_Figure_11_03_006.jpg\" alt=\"The graph of r = 4 + 4 cos\u03b8 is given. It vaguely looks look a heart tipped on its side with a rounded bottom instead of a pointed one. Specifically, the graph starts at the origin, moves into the second quadrant and increases to a rounded circle-like figure. The graph is symmetric about the x axis, so it continues its rounded circle-like figure, goes into the third quadrant, and comes to a point at the origin.\" width=\"717\" height=\"717\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 6.<\/figcaption><\/figure>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThe name of this shape is a cardioid, which we will study further later in this section.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Rewrite the equation [latex]r=\\sec\\theta \\tan\\theta[\/latex] in rectangular coordinates and identify its graph.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558879\">Hint<\/button><\/p>\n<div id=\"q44558879\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167793219344\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167793219351\">Convert to sine and cosine, then multiply both sides by cosine.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558889\">Show Solution<\/button><\/p>\n<div id=\"q44558889\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794072980\" data-type=\"solution\">\n<p id=\"fs-id1167794072982\">[latex]y={x}^{2}[\/latex], which is the equation of a parabola opening upward.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Symmetry in Polar Coordinates<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Symmetry in polar coordinates is your shortcut to graphing complex curves without plotting every single point. Just like even and odd functions in rectangular coordinates have predictable patterns, polar curves have three types of symmetry that can cut your work down dramatically.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The three types of polar symmetry:<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\"><strong>Polar axis symmetry<\/strong> (reflects across the [latex]x[\/latex]-axis): If [latex](r, \\theta)[\/latex] is on the curve, then [latex](r, -\\theta)[\/latex] is too. Test by replacing [latex]\\theta[\/latex] with [latex]-\\theta[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Pole symmetry<\/strong> (reflects through the origin): If [latex](r, \\theta)[\/latex] is on the curve, then [latex](r, \\pi + \\theta)[\/latex] is too. Test by replacing [latex]r[\/latex] with [latex]-r[\/latex] or [latex]\\theta[\/latex] with [latex]\\pi + \\theta[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Vertical line symmetry<\/strong> (reflects across the [latex]y[\/latex]-axis): If [latex](r, \\theta)[\/latex] is on the curve, then [latex](r, \\pi - \\theta)[\/latex] is too. Test by replacing [latex]\\theta[\/latex] with [latex]\\pi - \\theta[\/latex].<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">When you identify symmetries, you only need to plot points in one region, then reflect them to complete the entire graph. For a rose curve like [latex]r = 3\\sin(2\\theta)[\/latex] with all three symmetries, you can plot just the first quadrant and reflect to get all four petals.<\/p>\n<p class=\"whitespace-normal break-words\">For each symmetry test, substitute the transformation into your equation. If you get back the original equation (possibly after algebraic manipulation), that symmetry exists.<\/p>\n<p class=\"whitespace-normal break-words\">Remember that the same point can have different polar coordinates. A symmetry test might fail in one form but succeed when you account for equivalent representations.<\/p>\n<p class=\"whitespace-normal break-words\">Even if a curve doesn&#8217;t pass a particular symmetry test algebraically, you can still check visually after plotting a few key points to see if patterns emerge.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167794202050\" data-type=\"problem\">\n<p id=\"fs-id1167794202052\">Determine the symmetry of the graph determined by the equation [latex]r=2\\cos\\left(3\\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=\"q44558839\">Hint<\/button><\/p>\n<div id=\"q44558839\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794324547\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167794324554\">Use the theorem.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558849\">Show Solution<\/button><\/p>\n<div id=\"q44558849\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794324525\" data-type=\"solution\">\n<p id=\"fs-id1167794324527\">Symmetric with respect to the polar axis.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<figure style=\"width: 417px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234847\/CNX_Calc_Figure_11_03_015.jpg\" alt=\"A three-petaled rose is graphed with equation r = 2 cos(3\u03b8). Each petal starts at the origin and reaches a maximum distance from the origin of 2.\" width=\"417\" height=\"417\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 14.<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":67,"menu_order":18,"template":"","meta":{"_candela_citation":"[{\"type\":\"copyrighted_video\",\"description\":\"Sketching polar curves from cartesian curves (KristaKingMath)\",\"author\":\"Krista King\",\"organization\":\"Krista King Math\",\"url\":\"https:\/\/youtu.be\/OAEjSiFGdhQ\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube License\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":247,"module-header":"fresh_take","content_attributions":[{"type":"copyrighted_video","description":"Sketching polar curves from cartesian curves (KristaKingMath)","author":"Krista King","organization":"Krista King Math","url":"https:\/\/youtu.be\/OAEjSiFGdhQ","project":"","license":"arr","license_terms":"Standard YouTube License"}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"<script type='text\/javascript' src='https:\/\/www.youtube.com\/iframe_api'><\/script><script type='text\/javascript' src='\/\/plugin.3playmedia.com\/ajax.js?cc=1&cc_minimizable=1&cc_minimize_on_load=0&cc_multi_text_track=0&cc_overlay=1&cc_searchable=0&embed=ajax&mf=14661427&p3sdk_version=1.11.7&p=20361&player_type=youtube&plugin_skin=dark&target=3p-plugin-target-chgedhah-OAEjSiFGdhQ&vembed=0&video_id=OAEjSiFGdhQ&video_target=tpm-plugin-chgedhah-OAEjSiFGdhQ'><\/script>\n","media_targets":["tpm-plugin-chgedhah-OAEjSiFGdhQ"]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1571"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/67"}],"version-history":[{"count":8,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1571\/revisions"}],"predecessor-version":[{"id":5820,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1571\/revisions\/5820"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/247"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1571\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=1571"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=1571"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=1571"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=1571"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}