{"id":4088,"date":"2022-04-14T18:15:52","date_gmt":"2022-04-14T18:15:52","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-cylindrical-and-spherical-coordinates\/"},"modified":"2022-11-09T16:32:46","modified_gmt":"2022-11-09T16:32:46","slug":"skills-review-for-cylindrical-and-spherical-coordinates","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-cylindrical-and-spherical-coordinates\/","title":{"raw":"Skills Review for Cylindrical and Spherical Coordinates","rendered":"Skills Review for Cylindrical and Spherical Coordinates"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Identify reference angles for angles measured in both radians and degrees<\/li>\r\n \t<li>Convert points between rectangular and polar coordinates<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Cylindrical and Spherical Coordinates section, we will learn about two other coordinates systems, both extensions of the polar coordinate system. Here we will review how to evaluate the sine and cosine functions and convert points between rectangular and polar form.\r\n<h2>Find Reference Angles<\/h2>\r\n<strong><em>(See <a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-vectors-in-the-plane\/\" target=\"_blank\" rel=\"noopener\">Module 2, Skills Review for Vectors in the Plane<\/a>)<\/em><\/strong>\r\n<h2>Converting between Rectangular and Polar Coordinates<\/h2>\r\n<div id=\"fs-id1167793948751\" class=\"theorem\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Converting Points between Coordinate Systems<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167794024884\">Given a point [latex]P[\/latex] in the plane with Cartesian coordinates [latex]\\left(x,y\\right)[\/latex] and polar coordinates [latex]\\left(r,\\theta \\right)[\/latex], the following conversion formulas hold true:<\/p>\r\n\r\n<div id=\"fs-id1167794066096\" style=\"text-align: center;\" data-type=\"equation\">[latex]x=r\\cos\\theta \\:\\text{and}\\:y=r\\sin\\theta [\/latex],<\/div>\r\n&nbsp;\r\n<div id=\"fs-id1167793834620\" style=\"text-align: center;\" data-type=\"equation\">[latex]{r}^{2}={x}^{2}+{y}^{2}\\:\\text{and}\\:\\tan\\theta =\\frac{y}{x}[\/latex].<\/div>\r\n&nbsp;\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<\/div>\r\n<\/div>\r\n<\/div>\r\nAs we can note carefully above, the equation [latex]\\tan\\theta =\\frac{y}{x}[\/latex] is <em>not<\/em> expressed using the inverse tangent function.\u00a0 The reason for this is because of how the domain restriction of the tangent function leads to a restricted range of the inverse tangent function, reviewed below.\r\n<div id=\"fs-id1167793948751\" class=\"theorem\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Using the Inverse Tangent Function in The Coordinate Plane<\/h3>\r\n\r\n<hr \/>\r\n\r\nIf [latex] -\\frac{\\pi}{2} &lt; \\theta &lt; \\frac{\\pi}{2} [\/latex], and [latex] \\tan \\theta = \\frac{y}{x} [\/latex], then [latex] \\theta = \\tan^{-1} \\left( \\frac{y}{x} \\right) [\/latex]\r\n<p style=\"padding-left: 30px;\">That is, the inverse tangent function has a range of [latex] \\left(-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right) [\/latex], meaning that it always produces a positive angle in Quadrant I or a negative angle in Quadrant IV.<\/p>\r\nIf [latex] \\frac{\\pi}{2} &lt; \\theta &lt; \\frac{3\\pi}{2} [\/latex] and [latex] \\tan \\theta = \\frac{y}{x} [\/latex], then [latex] \\theta = \\tan^{-1} \\left( \\frac{y}{x} \\right) + \\pi [\/latex]\r\n<p style=\"padding-left: 30px;\">In other words, if the point [latex] \\left(x, y \\right)[\/latex] is in Quadrant II or III, the preceding rule means that you must add [latex] \\pi [\/latex] to the output of the inverse tangent function to produce an angle in the correct quadrant.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167794030722\" data-type=\"example\">\r\n<div id=\"fs-id1167794003809\" data-type=\"exercise\">\r\n<div id=\"fs-id1167794044637\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Converting between Rectangular and Polar Coordinates<\/h3>\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<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Converting between Rectangular and Polar Coordinates.\r\n\r\n<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\"><\/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<div id=\"fs-id1167793901804\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167793901807\" data-type=\"exercise\">\r\n<div id=\"fs-id1167793901809\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167793901809\" data-type=\"problem\">\r\n<p id=\"fs-id1167794066199\">Convert [latex]\\left(-8,-8\\right)[\/latex] into polar coordinates and [latex]\\left(4,\\frac{2\\pi }{3}\\right)[\/latex] into rectangular coordinates.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558897\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1167794064210\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167793262229\">Use both equations from the theorem. Make sure to check the quadrant when calculating [latex]\\theta [\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1167793805544\" data-type=\"solution\">\r\n<p id=\"fs-id1167793270874\" style=\"text-align: center;\">[latex]\\left(8\\sqrt{2},\\frac{5\\pi }{4}\\right)[\/latex] and [latex]\\left(-2,2\\sqrt{3}\\right)[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Identify reference angles for angles measured in both radians and degrees<\/li>\n<li>Convert points between rectangular and polar coordinates<\/li>\n<\/ul>\n<\/div>\n<p>In the Cylindrical and Spherical Coordinates section, we will learn about two other coordinates systems, both extensions of the polar coordinate system. Here we will review how to evaluate the sine and cosine functions and convert points between rectangular and polar form.<\/p>\n<h2>Find Reference Angles<\/h2>\n<p><strong><em>(See <a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-vectors-in-the-plane\/\" target=\"_blank\" rel=\"noopener\">Module 2, Skills Review for Vectors in the Plane<\/a>)<\/em><\/strong><\/p>\n<h2>Converting between Rectangular and Polar Coordinates<\/h2>\n<div id=\"fs-id1167793948751\" class=\"theorem\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Converting Points between Coordinate Systems<\/h3>\n<hr \/>\n<p id=\"fs-id1167794024884\">Given a point [latex]P[\/latex] in the plane with Cartesian coordinates [latex]\\left(x,y\\right)[\/latex] and polar coordinates [latex]\\left(r,\\theta \\right)[\/latex], the following conversion formulas hold true:<\/p>\n<div id=\"fs-id1167794066096\" style=\"text-align: center;\" data-type=\"equation\">[latex]x=r\\cos\\theta \\:\\text{and}\\:y=r\\sin\\theta[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1167793834620\" style=\"text-align: center;\" data-type=\"equation\">[latex]{r}^{2}={x}^{2}+{y}^{2}\\:\\text{and}\\:\\tan\\theta =\\frac{y}{x}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793913264\">These formulas can be used to convert from rectangular to polar or from polar to rectangular coordinates.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>As we can note carefully above, the equation [latex]\\tan\\theta =\\frac{y}{x}[\/latex] is <em>not<\/em> expressed using the inverse tangent function.\u00a0 The reason for this is because of how the domain restriction of the tangent function leads to a restricted range of the inverse tangent function, reviewed below.<\/p>\n<div id=\"fs-id1167793948751\" class=\"theorem\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Using the Inverse Tangent Function in The Coordinate Plane<\/h3>\n<hr \/>\n<p>If [latex]-\\frac{\\pi}{2} < \\theta < \\frac{\\pi}{2}[\/latex], and [latex]\\tan \\theta = \\frac{y}{x}[\/latex], then [latex]\\theta = \\tan^{-1} \\left( \\frac{y}{x} \\right)[\/latex]\n\n\n<p style=\"padding-left: 30px;\">That is, the inverse tangent function has a range of [latex]\\left(-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right)[\/latex], meaning that it always produces a positive angle in Quadrant I or a negative angle in Quadrant IV.<\/p>\n<p>If [latex]\\frac{\\pi}{2} < \\theta < \\frac{3\\pi}{2}[\/latex] and [latex]\\tan \\theta = \\frac{y}{x}[\/latex], then [latex]\\theta = \\tan^{-1} \\left( \\frac{y}{x} \\right) + \\pi[\/latex]\n\n\n<p style=\"padding-left: 30px;\">In other words, if the point [latex]\\left(x, y \\right)[\/latex] is in Quadrant II or III, the preceding rule means that you must add [latex]\\pi[\/latex] to the output of the inverse tangent function to produce an angle in the correct quadrant.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167794030722\" data-type=\"example\">\n<div id=\"fs-id1167794003809\" data-type=\"exercise\">\n<div id=\"fs-id1167794044637\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Converting between Rectangular and Polar Coordinates<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558899\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Converting between Rectangular and Polar Coordinates.<\/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\"><\/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<div id=\"fs-id1167793901804\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167793901807\" data-type=\"exercise\">\n<div id=\"fs-id1167793901809\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167793901809\" data-type=\"problem\">\n<p id=\"fs-id1167794066199\">Convert [latex]\\left(-8,-8\\right)[\/latex] into polar coordinates and [latex]\\left(4,\\frac{2\\pi }{3}\\right)[\/latex] into rectangular coordinates.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558897\">Hint<\/span><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794064210\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167793262229\">Use both equations from the theorem. Make sure to check the quadrant when calculating [latex]\\theta[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558898\">Show Solution<\/span><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167793805544\" data-type=\"solution\">\n<p id=\"fs-id1167793270874\" style=\"text-align: center;\">[latex]\\left(8\\sqrt{2},\\frac{5\\pi }{4}\\right)[\/latex] and [latex]\\left(-2,2\\sqrt{3}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-4088\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/calculus1\/\">https:\/\/courses.lumenlearning.com\/calculus1\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Calculus Volume 2. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/calculus2\/\">https:\/\/courses.lumenlearning.com\/calculus2\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/calculus1\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/calculus2\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-4088","chapter","type-chapter","status-publish","hentry"],"part":4109,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4088","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":6,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4088\/revisions"}],"predecessor-version":[{"id":4254,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4088\/revisions\/4254"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/4109"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4088\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=4088"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=4088"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=4088"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=4088"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}