{"id":1016,"date":"2025-06-20T17:28:55","date_gmt":"2025-06-20T17:28:55","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1016"},"modified":"2025-09-29T13:01:54","modified_gmt":"2025-09-29T13:01:54","slug":"understanding-polar-coordinates-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/understanding-polar-coordinates-learn-it-1\/","title":{"raw":"Understanding Polar Coordinates: Learn It 1","rendered":"Understanding Polar Coordinates: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Plot points using polar coordinates (<em>r,\u03b8<\/em>)<\/li>\r\n \t<li>Switch back and forth between polar and rectangular (<em>x,y<\/em>) coordinates<\/li>\r\n \t<li>Draw polar curves from their equations<\/li>\r\n \t<li>Identify when polar curves have symmetry<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Polar Coordinates<\/h2>\r\n<p class=\"whitespace-normal break-words\">The rectangular coordinate system maps points to ordered pairs using perpendicular axes. The <strong>polar coordinate system<\/strong> offers an alternative approach that can be more useful in certain situations, particularly when dealing with circular or rotational patterns.<\/p>\r\n<p class=\"whitespace-normal break-words\">Instead of describing a point's location using horizontal and vertical distances, polar coordinates use:<\/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>Distance<\/strong> from the origin (called [latex]r[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Angle<\/strong> from the positive [latex]x[\/latex]-axis (called [latex]\\theta[\/latex])<\/li>\r\n<\/ul>\r\nThe rectangular coordinate system (or Cartesian plane) provides a means of mapping points to ordered pairs and ordered pairs to points. This is called a <em data-effect=\"italics\">one-to-one mapping<\/em> from points in the plane to ordered pairs. The polar coordinate system provides an alternative method of mapping points to ordered pairs. In this section we see that in some circumstances, polar coordinates can be more useful than rectangular coordinates.\r\n<h3>Defining Polar Coordinates<\/h3>\r\n<p id=\"fs-id1167794039551\">Consider any point [latex]P[\/latex] in the coordinate plane with rectangular coordinates [latex]\\left(x,y\\right)[\/latex].<\/p>\r\nTo find its polar coordinates, we measure the <strong>radius<\/strong> [latex]r[\/latex] as the distance from the origin to point [latex]P[\/latex], and the <strong>angle<\/strong> [latex]\\theta[\/latex] as the angle between the positive [latex]x[\/latex]-axis and the line segment from the origin to [latex]P[\/latex]. This creates a <strong>one-to-one mapping<\/strong> between points and the ordered pair [latex](r,\\theta )[\/latex], just as rectangular coordinates use [latex]\\left(x,y\\right)[\/latex].\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>polar coordinate system<\/h3>\r\nEach point in the plane corresponds to an ordered pair [latex](r,\\theta )[\/latex], where [latex]r[\/latex] is the distance from the origin and [latex]\\theta[\/latex] is the angle from the positive [latex]x[\/latex]-axis.\r\n\r\n<\/section>\r\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Converting Between Coordinate Systems<\/h3>\r\n<p class=\"whitespace-normal break-words\">The connection between rectangular and polar coordinates relies on right triangle relationships. When you draw a line from the origin to point [latex]P[\/latex], you create a right triangle where the hypotenuse equals [latex]r[\/latex], the horizontal leg equals [latex]x[\/latex], the vertical leg equals [latex]y[\/latex], and one angle equals [latex]\\theta[\/latex].<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_03_001\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"304\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234807\/CNX_Calc_Figure_11_03_001.jpg\" alt=\"A point P(x, y) is given in the first quadrant with lines drawn to indicate its x and y values. There is a line from the origin to P(x, y) marked r and this line make an angle \u03b8 with the x axis.\" width=\"304\" height=\"309\" data-media-type=\"image\/jpeg\" \/> Figure 1. An arbitrary point in the Cartesian plane.[\/caption]<\/figure>\r\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Recall: Right Triangle Trigonometry\r\n[latex]\\\\[\/latex]\r\n<\/strong>Given a right triangle with an acute angle of [latex]\\theta[\/latex]:\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]\\begin{align}&amp;\\sin \\left(\\theta \\right)=\\frac{\\text{opposite}}{\\text{hypotenuse}} \\\\ &amp;\\cos \\left(\\theta \\right)=\\frac{\\text{adjacent}}{\\text{hypotenuse}} \\\\ &amp;\\tan \\left(\\theta \\right)=\\frac{\\text{opposite}}{\\text{adjacent}} \\end{align}[\/latex]<\/p>\r\nA common mnemonic for remembering these relationships is <strong>SohCahToa<\/strong>: <strong>S<\/strong>ine is <strong>O<\/strong>pposite over <strong>H<\/strong>ypotenuse, <strong>C<\/strong>osine is <strong>A<\/strong>djacent over <strong>H<\/strong>ypotenuse, <strong>T<\/strong>angent is <strong>O<\/strong>pposite over <strong>A<\/strong>djacent.\r\n\r\nThe side lengths of the right triangle with legs [latex] a [\/latex] and [latex] b [\/latex] and hypotenuse\u00a0[latex] c [\/latex] are related through the Pythagorean Theorem:\u00a0[latex] a^2 + b^2 = c^2 [\/latex]\r\n\r\n<\/section>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\">Using the right triangle formed by point [latex]P[\/latex], the origin, and the coordinate axes, we can derive relationships between rectangular and polar coordinates.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-pre-wrap break-words\">From the basic trigonometric ratios, we get:<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167794187190\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\cos\\theta =\\frac{x}{r}\\:\\:\\text{so}\\:\\:x=r\\cos\\theta [\/latex]<\/div>\r\n<div id=\"fs-id1167794170718\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\sin\\theta =\\frac{y}{r}\\:\\:\\text{ so }\\:\\:y=r\\sin\\theta [\/latex].<\/div>\r\n<p id=\"fs-id1167793785830\">Additionally, the Pythagorean theorem and tangent ratio give us:<\/p>\r\n\r\n<div id=\"fs-id1167793821731\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{r}^{2}={x}^{2}+{y}^{2}\\text{ and }\\tan\\theta =\\frac{y}{x}[\/latex].<\/div>\r\nThese relationships allow us to convert any point [latex]\\left(x,y\\right)[\/latex] in rectangular coordinates to polar coordinates [latex]\\left(r,\\theta \\right)[\/latex]. The first coordinate [latex]r[\/latex] is called the <strong>radial coordinate<\/strong>, and the second coordinate [latex]\\theta[\/latex] is called the <strong>angular coordinate<\/strong>.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>converting points between coordinate systems<\/h3>\r\n<p id=\"fs-id1167794024884\">For a point [latex]P[\/latex] with rectangular coordinates [latex]\\left(x,y\\right)[\/latex] and polar coordinates [latex]\\left(r,\\theta \\right)[\/latex]:<\/p>\r\n\r\n<ul>\r\n \t<li><strong>Polar to Rectangular<\/strong>: [latex]x=r\\cos\\theta \\:\\text{and}\\:y=r\\sin\\theta [\/latex]<\/li>\r\n \t<li><strong>Rectangular to Polar<\/strong>: [latex]{r}^{2}={x}^{2}+{y}^{2}\\:\\text{and}\\:\\tan\\theta =\\frac{y}{x}[\/latex]<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1167793913264\">These formulas can be used to convert from rectangular to polar or from polar to rectangular coordinates.<\/p>\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">The equation [latex]\\tan\\theta =\\frac{y}{x}[\/latex] has infinitely many solutions for any point [latex]\\left(x,y\\right)[\/latex]. However, when we restrict solutions to values between [latex]0[\/latex] and [latex]2\\pi[\/latex], we can find a unique angle that places the point in the correct quadrant. This ensures that the corresponding value of [latex]r[\/latex] is positive.<\/p>\r\n<p class=\"whitespace-normal break-words\">Notice that we don't simply write [latex]\\theta = \\tan^{-1}\\left(\\frac{y}{x}\\right)[\/latex]. This is because the inverse tangent function has a restricted range that can lead to incorrect quadrant placement.<\/p>\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Recall: Using The Inverse Tangent Function in the coordinate plane<\/strong>\r\n<p class=\"whitespace-normal break-words\">The inverse tangent function has range [latex]\\left(-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right)[\/latex], so it only produces angles in Quadrants I and IV.<\/p>\r\n\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\"><strong>For Quadrants I and IV<\/strong>: If [latex]-\\frac{\\pi}{2} &lt; \\theta &lt; \\frac{\\pi}{2}[\/latex], then [latex]\\theta = \\tan^{-1}\\left(\\frac{y}{x}\\right)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>For Quadrants II and III<\/strong>: If [latex]\\frac{\\pi}{2} &lt; \\theta &lt; \\frac{3\\pi}{2}[\/latex], then [latex]\\theta = \\tan^{-1}\\left(\\frac{y}{x}\\right) + \\pi[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">When the point [latex] \\left(x, y \\right)[\/latex] is in Quadrant II or III, you must add [latex]\\pi[\/latex] to the inverse tangent result to get the correct angle.<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1167794044637\" data-type=\"problem\">\r\n<p id=\"fs-id1167793814959\">Convert each of the following points into polar coordinates.<\/p>\r\n\r\n<ol id=\"fs-id1167793906173\" type=\"a\">\r\n \t<li>[latex]\\left(1,1\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\left(-3,4\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\left(0,3\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\left(5\\sqrt{3},-5\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1167793266138\">Convert each of the following points into rectangular coordinates.<\/p>\r\n\r\n<ol id=\"fs-id1167793821570\" start=\"4\" type=\"a\">\r\n \t<li>[latex]\\left(3,\\frac{\\pi}{3}\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\left(2,\\frac{3\\pi}{2}\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\left(6,\\frac{-5\\pi}{6}\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<div id=\"fs-id1167794058406\" data-type=\"solution\">\r\n<ol id=\"fs-id1167794045804\" type=\"a\">\r\n \t<li>Use [latex]x=1[\/latex] and [latex]y=1[\/latex] in the theorem:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793849839\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill {r}^{2}&amp; =\\hfill &amp; {x}^{2}+{y}^{2}\\hfill \\\\ &amp; =\\hfill &amp; {1}^{2}+{1}^{2}\\hfill \\\\ \\hfill r&amp; =\\hfill &amp; \\sqrt{2}\\hfill \\end{array}\\hfill &amp; &amp; &amp; \\text{and}\\hfill &amp; &amp; &amp; \\begin{array}{ccc}\\hfill \\tan\\theta &amp; =\\hfill &amp; \\frac{y}{x}\\hfill \\\\ &amp; =\\hfill &amp; \\frac{1}{1}=1\\hfill \\\\ \\hfill \\theta &amp; =\\hfill &amp; \\frac{\\pi }{4}.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore this point can be represented as [latex]\\left(\\sqrt{2},\\frac{\\pi }{4}\\right)[\/latex] in polar coordinates.<\/li>\r\n \t<li>Use [latex]x=-3[\/latex] and [latex]y=4[\/latex] in the theorem:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793813480\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill {r}^{2}&amp; =\\hfill &amp; {x}^{2}+{y}^{2}\\hfill \\\\ &amp; =\\hfill &amp; {\\left(-3\\right)}^{2}+{\\left(4\\right)}^{2}\\hfill \\\\ \\hfill r&amp; =\\hfill &amp; 5\\hfill \\end{array}\\hfill &amp; &amp; &amp; \\text{and}\\hfill &amp; &amp; &amp; \\begin{array}{ccc}\\hfill \\tan\\theta &amp; =\\hfill &amp; \\frac{y}{x}\\hfill \\\\ &amp; =\\hfill &amp; -\\frac{4}{3}\\hfill \\\\ \\hfill \\theta &amp; =\\hfill &amp; -\\text{arctan}\\left(\\frac{4}{3}\\right)\\hfill \\\\ &amp; \\approx \\hfill &amp; 2.21.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore this point can be represented as [latex]\\left(5,2.21\\right)[\/latex] in polar coordinates.<\/li>\r\n \t<li>Use [latex]x=0[\/latex] and [latex]y=3[\/latex] in the theorem:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793819715\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill {r}^{2}&amp; =\\hfill &amp; {x}^{2}+{y}^{2}\\hfill \\\\ &amp; =\\hfill &amp; {\\left(3\\right)}^{2}+{\\left(0\\right)}^{2}\\hfill \\\\ &amp; =\\hfill &amp; 9+0\\hfill \\\\ \\hfill r&amp; =\\hfill &amp; 3\\hfill \\end{array}\\hfill &amp; &amp; &amp; \\text{and}\\hfill &amp; &amp; &amp; \\begin{array}{ccc}\\hfill \\tan\\theta &amp; =\\hfill &amp; \\frac{y}{x}\\hfill \\\\ &amp; =\\hfill &amp; \\frac{3}{0}.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nDirect application of the second equation leads to division by zero. Graphing the point [latex]\\left(0,3\\right)[\/latex] on the rectangular coordinate system reveals that the point is located on the positive <em data-effect=\"italics\">y<\/em>-axis. The angle between the positive <em data-effect=\"italics\">x<\/em>-axis and the positive <em data-effect=\"italics\">y<\/em>-axis is [latex]\\frac{\\pi }{2}[\/latex]. Therefore this point can be represented as [latex]\\left(3,\\frac{\\pi }{2}\\right)[\/latex] in polar coordinates.<\/li>\r\n \t<li>Use [latex]x=5\\sqrt{3}[\/latex] and [latex]y=-5[\/latex] in the theorem:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793878815\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill {r}^{2}&amp; =\\hfill &amp; {x}^{2}+{y}^{2}\\hfill \\\\ &amp; =\\hfill &amp; {\\left(5\\sqrt{3}\\right)}^{2}+{\\left(-5\\right)}^{2}\\hfill \\\\ &amp; =\\hfill &amp; 75+25\\hfill \\\\ \\hfill r&amp; =\\hfill &amp; 10\\hfill \\end{array}\\hfill &amp; &amp; &amp; \\text{and}\\hfill &amp; &amp; &amp; \\begin{array}{ccc}\\hfill \\tan\\theta &amp; =\\hfill &amp; \\frac{y}{x}\\hfill \\\\ &amp; =\\hfill &amp; \\frac{-5}{5\\sqrt{3}}=-\\frac{\\sqrt{3}}{3}\\hfill \\\\ \\hfill \\theta &amp; =\\hfill &amp; -\\frac{\\pi }{6}.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore this point can be represented as [latex]\\left(10,-\\frac{\\pi }{6}\\right)[\/latex] in polar coordinates.<\/li>\r\n \t<li>Use [latex]r=3[\/latex] and [latex]\\theta =\\frac{\\pi }{3}[\/latex] in the theorem:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167794047098\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill x&amp; =\\hfill &amp; r\\cos\\theta \\hfill \\\\ &amp; =\\hfill &amp; 3\\cos\\left(\\frac{\\pi }{3}\\right)\\hfill \\\\ &amp; =\\hfill &amp; 3\\left(\\frac{1}{2}\\right)=\\frac{3}{2}\\hfill \\end{array}\\hfill &amp; &amp; &amp; \\text{and}\\hfill &amp; &amp; &amp; \\begin{array}{ccc}\\hfill y&amp; =\\hfill &amp; r\\sin\\theta \\hfill \\\\ &amp; =\\hfill &amp; 3\\sin\\left(\\frac{\\pi }{3}\\right)\\hfill \\\\ &amp; =\\hfill &amp; 3\\left(\\frac{\\sqrt{3}}{2}\\right)=\\frac{3\\sqrt{3}}{2}.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore this point can be represented as [latex]\\left(\\frac{3}{2},\\frac{3\\sqrt{3}}{2}\\right)[\/latex] in rectangular coordinates.<\/li>\r\n \t<li>Use [latex]r=2[\/latex] and [latex]\\theta =\\frac{3\\pi }{2}[\/latex] in the theorem:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167794336338\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill x&amp; =\\hfill &amp; r\\cos\\theta \\hfill \\\\ &amp; =\\hfill &amp; 2\\cos\\left(\\frac{3\\pi }{2}\\right)\\hfill \\\\ &amp; =\\hfill &amp; 2\\left(0\\right)=0\\hfill \\end{array}\\hfill &amp; &amp; &amp; \\text{and}\\hfill &amp; &amp; &amp; \\begin{array}{ccc}\\hfill y&amp; =\\hfill &amp; r\\sin\\theta \\hfill \\\\ &amp; =\\hfill &amp; 2\\sin\\left(\\frac{3\\pi }{2}\\right)\\hfill \\\\ &amp; =\\hfill &amp; 2\\left(-1\\right)=-2.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore this point can be represented as [latex]\\left(0,-2\\right)[\/latex] in rectangular coordinates.<\/li>\r\n \t<li>Use [latex]r=6[\/latex] and [latex]\\theta =-\\frac{5\\pi }{6}[\/latex] in the theorem:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167794047981\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill x&amp; =\\hfill &amp; r\\cos\\theta \\hfill \\\\ &amp; =\\hfill &amp; 6\\cos\\left(-\\frac{5\\pi }{6}\\right)\\hfill \\\\ &amp; =\\hfill &amp; 6\\left(-\\frac{\\sqrt{3}}{2}\\right)\\hfill \\\\ &amp; =\\hfill &amp; -3\\sqrt{3}\\hfill \\end{array}\\hfill &amp; &amp; &amp; \\text{and}\\hfill &amp; &amp; &amp; \\begin{array}{ccc}\\hfill y&amp; =\\hfill &amp; r\\sin\\theta \\hfill \\\\ &amp; =\\hfill &amp; 6\\sin\\left(-\\frac{5\\pi }{6}\\right)\\hfill \\\\ &amp; =\\hfill &amp; 6\\left(-\\frac{1}{2}\\right)\\hfill \\\\ &amp; =\\hfill &amp; -3.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore this point can be represented as [latex]\\left(-3\\sqrt{3},-3\\right)[\/latex] in rectangular coordinates.<\/li>\r\n<\/ol>\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=72&amp;end=510&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\/7.3PolarCoordinates72to510_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]311321[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Plot points using polar coordinates (<em>r,\u03b8<\/em>)<\/li>\n<li>Switch back and forth between polar and rectangular (<em>x,y<\/em>) coordinates<\/li>\n<li>Draw polar curves from their equations<\/li>\n<li>Identify when polar curves have symmetry<\/li>\n<\/ul>\n<\/section>\n<h2 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Polar Coordinates<\/h2>\n<p class=\"whitespace-normal break-words\">The rectangular coordinate system maps points to ordered pairs using perpendicular axes. The <strong>polar coordinate system<\/strong> offers an alternative approach that can be more useful in certain situations, particularly when dealing with circular or rotational patterns.<\/p>\n<p class=\"whitespace-normal break-words\">Instead of describing a point&#8217;s location using horizontal and vertical distances, polar coordinates use:<\/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>Distance<\/strong> from the origin (called [latex]r[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Angle<\/strong> from the positive [latex]x[\/latex]-axis (called [latex]\\theta[\/latex])<\/li>\n<\/ul>\n<p>The rectangular coordinate system (or Cartesian plane) provides a means of mapping points to ordered pairs and ordered pairs to points. This is called a <em data-effect=\"italics\">one-to-one mapping<\/em> from points in the plane to ordered pairs. The polar coordinate system provides an alternative method of mapping points to ordered pairs. In this section we see that in some circumstances, polar coordinates can be more useful than rectangular coordinates.<\/p>\n<h3>Defining Polar Coordinates<\/h3>\n<p id=\"fs-id1167794039551\">Consider any point [latex]P[\/latex] in the coordinate plane with rectangular coordinates [latex]\\left(x,y\\right)[\/latex].<\/p>\n<p>To find its polar coordinates, we measure the <strong>radius<\/strong> [latex]r[\/latex] as the distance from the origin to point [latex]P[\/latex], and the <strong>angle<\/strong> [latex]\\theta[\/latex] as the angle between the positive [latex]x[\/latex]-axis and the line segment from the origin to [latex]P[\/latex]. This creates a <strong>one-to-one mapping<\/strong> between points and the ordered pair [latex](r,\\theta )[\/latex], just as rectangular coordinates use [latex]\\left(x,y\\right)[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>polar coordinate system<\/h3>\n<p>Each point in the plane corresponds to an ordered pair [latex](r,\\theta )[\/latex], where [latex]r[\/latex] is the distance from the origin and [latex]\\theta[\/latex] is the angle from the positive [latex]x[\/latex]-axis.<\/p>\n<\/section>\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Converting Between Coordinate Systems<\/h3>\n<p class=\"whitespace-normal break-words\">The connection between rectangular and polar coordinates relies on right triangle relationships. When you draw a line from the origin to point [latex]P[\/latex], you create a right triangle where the hypotenuse equals [latex]r[\/latex], the horizontal leg equals [latex]x[\/latex], the vertical leg equals [latex]y[\/latex], and one angle equals [latex]\\theta[\/latex].<\/p>\n<figure id=\"CNX_Calc_Figure_11_03_001\"><figcaption><\/figcaption><figure style=\"width: 304px\" 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\/11234807\/CNX_Calc_Figure_11_03_001.jpg\" alt=\"A point P(x, y) is given in the first quadrant with lines drawn to indicate its x and y values. There is a line from the origin to P(x, y) marked r and this line make an angle \u03b8 with the x axis.\" width=\"304\" height=\"309\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 1. An arbitrary point in the Cartesian plane.<\/figcaption><\/figure>\n<\/figure>\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Recall: Right Triangle Trigonometry<br \/>\n[latex]\\\\[\/latex]<br \/>\n<\/strong>Given a right triangle with an acute angle of [latex]\\theta[\/latex]:<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]\\begin{align}&\\sin \\left(\\theta \\right)=\\frac{\\text{opposite}}{\\text{hypotenuse}} \\\\ &\\cos \\left(\\theta \\right)=\\frac{\\text{adjacent}}{\\text{hypotenuse}} \\\\ &\\tan \\left(\\theta \\right)=\\frac{\\text{opposite}}{\\text{adjacent}} \\end{align}[\/latex]<\/p>\n<p>A common mnemonic for remembering these relationships is <strong>SohCahToa<\/strong>: <strong>S<\/strong>ine is <strong>O<\/strong>pposite over <strong>H<\/strong>ypotenuse, <strong>C<\/strong>osine is <strong>A<\/strong>djacent over <strong>H<\/strong>ypotenuse, <strong>T<\/strong>angent is <strong>O<\/strong>pposite over <strong>A<\/strong>djacent.<\/p>\n<p>The side lengths of the right triangle with legs [latex]a[\/latex] and [latex]b[\/latex] and hypotenuse\u00a0[latex]c[\/latex] are related through the Pythagorean Theorem:\u00a0[latex]a^2 + b^2 = c^2[\/latex]<\/p>\n<\/section>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\">Using the right triangle formed by point [latex]P[\/latex], the origin, and the coordinate axes, we can derive relationships between rectangular and polar coordinates.<\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-pre-wrap break-words\">From the basic trigonometric ratios, we get:<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1167794187190\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\cos\\theta =\\frac{x}{r}\\:\\:\\text{so}\\:\\:x=r\\cos\\theta[\/latex]<\/div>\n<div id=\"fs-id1167794170718\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\sin\\theta =\\frac{y}{r}\\:\\:\\text{ so }\\:\\:y=r\\sin\\theta[\/latex].<\/div>\n<p id=\"fs-id1167793785830\">Additionally, the Pythagorean theorem and tangent ratio give us:<\/p>\n<div id=\"fs-id1167793821731\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{r}^{2}={x}^{2}+{y}^{2}\\text{ and }\\tan\\theta =\\frac{y}{x}[\/latex].<\/div>\n<p>These relationships allow us to convert any point [latex]\\left(x,y\\right)[\/latex] in rectangular coordinates to polar coordinates [latex]\\left(r,\\theta \\right)[\/latex]. The first coordinate [latex]r[\/latex] is called the <strong>radial coordinate<\/strong>, and the second coordinate [latex]\\theta[\/latex] is called the <strong>angular coordinate<\/strong>.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>converting points between coordinate systems<\/h3>\n<p id=\"fs-id1167794024884\">For a point [latex]P[\/latex] with rectangular coordinates [latex]\\left(x,y\\right)[\/latex] and polar coordinates [latex]\\left(r,\\theta \\right)[\/latex]:<\/p>\n<ul>\n<li><strong>Polar to Rectangular<\/strong>: [latex]x=r\\cos\\theta \\:\\text{and}\\:y=r\\sin\\theta[\/latex]<\/li>\n<li><strong>Rectangular to Polar<\/strong>: [latex]{r}^{2}={x}^{2}+{y}^{2}\\:\\text{and}\\:\\tan\\theta =\\frac{y}{x}[\/latex]<\/li>\n<\/ul>\n<p id=\"fs-id1167793913264\">These formulas can be used to convert from rectangular to polar or from polar to rectangular coordinates.<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">The equation [latex]\\tan\\theta =\\frac{y}{x}[\/latex] has infinitely many solutions for any point [latex]\\left(x,y\\right)[\/latex]. However, when we restrict solutions to values between [latex]0[\/latex] and [latex]2\\pi[\/latex], we can find a unique angle that places the point in the correct quadrant. This ensures that the corresponding value of [latex]r[\/latex] is positive.<\/p>\n<p class=\"whitespace-normal break-words\">Notice that we don&#8217;t simply write [latex]\\theta = \\tan^{-1}\\left(\\frac{y}{x}\\right)[\/latex]. This is because the inverse tangent function has a restricted range that can lead to incorrect quadrant placement.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Recall: Using The Inverse Tangent Function in the coordinate plane<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The inverse tangent function has range [latex]\\left(-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right)[\/latex], so it only produces angles in Quadrants I and IV.<\/p>\n<ul>\n<li class=\"whitespace-normal break-words\"><strong>For Quadrants I and IV<\/strong>: If [latex]-\\frac{\\pi}{2} < \\theta < \\frac{\\pi}{2}[\/latex], then [latex]\\theta = \\tan^{-1}\\left(\\frac{y}{x}\\right)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>For Quadrants II and III<\/strong>: If [latex]\\frac{\\pi}{2} < \\theta < \\frac{3\\pi}{2}[\/latex], then [latex]\\theta = \\tan^{-1}\\left(\\frac{y}{x}\\right) + \\pi[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">When the point [latex]\\left(x, y \\right)[\/latex] is in Quadrant II or III, you must add [latex]\\pi[\/latex] to the inverse tangent result to get the correct angle.<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167794044637\" data-type=\"problem\">\n<p id=\"fs-id1167793814959\">Convert each of the following points into polar coordinates.<\/p>\n<ol id=\"fs-id1167793906173\" type=\"a\">\n<li>[latex]\\left(1,1\\right)[\/latex]<\/li>\n<li>[latex]\\left(-3,4\\right)[\/latex]<\/li>\n<li>[latex]\\left(0,3\\right)[\/latex]<\/li>\n<li>[latex]\\left(5\\sqrt{3},-5\\right)[\/latex]<\/li>\n<\/ol>\n<p id=\"fs-id1167793266138\">Convert each of the following points into rectangular coordinates.<\/p>\n<ol id=\"fs-id1167793821570\" start=\"4\" type=\"a\">\n<li>[latex]\\left(3,\\frac{\\pi}{3}\\right)[\/latex]<\/li>\n<li>[latex]\\left(2,\\frac{3\\pi}{2}\\right)[\/latex]<\/li>\n<li>[latex]\\left(6,\\frac{-5\\pi}{6}\\right)[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558899\">Show Solution<\/button><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794058406\" data-type=\"solution\">\n<ol id=\"fs-id1167794045804\" type=\"a\">\n<li>Use [latex]x=1[\/latex] and [latex]y=1[\/latex] in the theorem:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793849839\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill {r}^{2}& =\\hfill & {x}^{2}+{y}^{2}\\hfill \\\\ & =\\hfill & {1}^{2}+{1}^{2}\\hfill \\\\ \\hfill r& =\\hfill & \\sqrt{2}\\hfill \\end{array}\\hfill & & & \\text{and}\\hfill & & & \\begin{array}{ccc}\\hfill \\tan\\theta & =\\hfill & \\frac{y}{x}\\hfill \\\\ & =\\hfill & \\frac{1}{1}=1\\hfill \\\\ \\hfill \\theta & =\\hfill & \\frac{\\pi }{4}.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore this point can be represented as [latex]\\left(\\sqrt{2},\\frac{\\pi }{4}\\right)[\/latex] in polar coordinates.<\/li>\n<li>Use [latex]x=-3[\/latex] and [latex]y=4[\/latex] in the theorem:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793813480\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill {r}^{2}& =\\hfill & {x}^{2}+{y}^{2}\\hfill \\\\ & =\\hfill & {\\left(-3\\right)}^{2}+{\\left(4\\right)}^{2}\\hfill \\\\ \\hfill r& =\\hfill & 5\\hfill \\end{array}\\hfill & & & \\text{and}\\hfill & & & \\begin{array}{ccc}\\hfill \\tan\\theta & =\\hfill & \\frac{y}{x}\\hfill \\\\ & =\\hfill & -\\frac{4}{3}\\hfill \\\\ \\hfill \\theta & =\\hfill & -\\text{arctan}\\left(\\frac{4}{3}\\right)\\hfill \\\\ & \\approx \\hfill & 2.21.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore this point can be represented as [latex]\\left(5,2.21\\right)[\/latex] in polar coordinates.<\/li>\n<li>Use [latex]x=0[\/latex] and [latex]y=3[\/latex] in the theorem:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793819715\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill {r}^{2}& =\\hfill & {x}^{2}+{y}^{2}\\hfill \\\\ & =\\hfill & {\\left(3\\right)}^{2}+{\\left(0\\right)}^{2}\\hfill \\\\ & =\\hfill & 9+0\\hfill \\\\ \\hfill r& =\\hfill & 3\\hfill \\end{array}\\hfill & & & \\text{and}\\hfill & & & \\begin{array}{ccc}\\hfill \\tan\\theta & =\\hfill & \\frac{y}{x}\\hfill \\\\ & =\\hfill & \\frac{3}{0}.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nDirect application of the second equation leads to division by zero. Graphing the point [latex]\\left(0,3\\right)[\/latex] on the rectangular coordinate system reveals that the point is located on the positive <em data-effect=\"italics\">y<\/em>-axis. The angle between the positive <em data-effect=\"italics\">x<\/em>-axis and the positive <em data-effect=\"italics\">y<\/em>-axis is [latex]\\frac{\\pi }{2}[\/latex]. Therefore this point can be represented as [latex]\\left(3,\\frac{\\pi }{2}\\right)[\/latex] in polar coordinates.<\/li>\n<li>Use [latex]x=5\\sqrt{3}[\/latex] and [latex]y=-5[\/latex] in the theorem:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793878815\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill {r}^{2}& =\\hfill & {x}^{2}+{y}^{2}\\hfill \\\\ & =\\hfill & {\\left(5\\sqrt{3}\\right)}^{2}+{\\left(-5\\right)}^{2}\\hfill \\\\ & =\\hfill & 75+25\\hfill \\\\ \\hfill r& =\\hfill & 10\\hfill \\end{array}\\hfill & & & \\text{and}\\hfill & & & \\begin{array}{ccc}\\hfill \\tan\\theta & =\\hfill & \\frac{y}{x}\\hfill \\\\ & =\\hfill & \\frac{-5}{5\\sqrt{3}}=-\\frac{\\sqrt{3}}{3}\\hfill \\\\ \\hfill \\theta & =\\hfill & -\\frac{\\pi }{6}.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore this point can be represented as [latex]\\left(10,-\\frac{\\pi }{6}\\right)[\/latex] in polar coordinates.<\/li>\n<li>Use [latex]r=3[\/latex] and [latex]\\theta =\\frac{\\pi }{3}[\/latex] in the theorem:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167794047098\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill x& =\\hfill & r\\cos\\theta \\hfill \\\\ & =\\hfill & 3\\cos\\left(\\frac{\\pi }{3}\\right)\\hfill \\\\ & =\\hfill & 3\\left(\\frac{1}{2}\\right)=\\frac{3}{2}\\hfill \\end{array}\\hfill & & & \\text{and}\\hfill & & & \\begin{array}{ccc}\\hfill y& =\\hfill & r\\sin\\theta \\hfill \\\\ & =\\hfill & 3\\sin\\left(\\frac{\\pi }{3}\\right)\\hfill \\\\ & =\\hfill & 3\\left(\\frac{\\sqrt{3}}{2}\\right)=\\frac{3\\sqrt{3}}{2}.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore this point can be represented as [latex]\\left(\\frac{3}{2},\\frac{3\\sqrt{3}}{2}\\right)[\/latex] in rectangular coordinates.<\/li>\n<li>Use [latex]r=2[\/latex] and [latex]\\theta =\\frac{3\\pi }{2}[\/latex] in the theorem:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167794336338\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill x& =\\hfill & r\\cos\\theta \\hfill \\\\ & =\\hfill & 2\\cos\\left(\\frac{3\\pi }{2}\\right)\\hfill \\\\ & =\\hfill & 2\\left(0\\right)=0\\hfill \\end{array}\\hfill & & & \\text{and}\\hfill & & & \\begin{array}{ccc}\\hfill y& =\\hfill & r\\sin\\theta \\hfill \\\\ & =\\hfill & 2\\sin\\left(\\frac{3\\pi }{2}\\right)\\hfill \\\\ & =\\hfill & 2\\left(-1\\right)=-2.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore this point can be represented as [latex]\\left(0,-2\\right)[\/latex] in rectangular coordinates.<\/li>\n<li>Use [latex]r=6[\/latex] and [latex]\\theta =-\\frac{5\\pi }{6}[\/latex] in the theorem:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167794047981\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill x& =\\hfill & r\\cos\\theta \\hfill \\\\ & =\\hfill & 6\\cos\\left(-\\frac{5\\pi }{6}\\right)\\hfill \\\\ & =\\hfill & 6\\left(-\\frac{\\sqrt{3}}{2}\\right)\\hfill \\\\ & =\\hfill & -3\\sqrt{3}\\hfill \\end{array}\\hfill & & & \\text{and}\\hfill & & & \\begin{array}{ccc}\\hfill y& =\\hfill & r\\sin\\theta \\hfill \\\\ & =\\hfill & 6\\sin\\left(-\\frac{5\\pi }{6}\\right)\\hfill \\\\ & =\\hfill & 6\\left(-\\frac{1}{2}\\right)\\hfill \\\\ & =\\hfill & -3.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore this point can be represented as [latex]\\left(-3\\sqrt{3},-3\\right)[\/latex] in rectangular coordinates.<\/li>\n<\/ol>\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=72&amp;end=510&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\/7.3PolarCoordinates72to510_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=\"ohm311321\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=311321&theme=lumen&iframe_resize_id=ohm311321&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":5,"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\/1016"}],"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":11,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1016\/revisions"}],"predecessor-version":[{"id":2396,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1016\/revisions\/2396"}],"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\/1016\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1016"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1016"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1016"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1016"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}