{"id":289,"date":"2021-03-25T03:03:35","date_gmt":"2021-03-25T03:03:35","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=289"},"modified":"2021-11-17T23:51:30","modified_gmt":"2021-11-17T23:51:30","slug":"putting-it-together-parametric-equations-and-polar-coordinates","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/putting-it-together-parametric-equations-and-polar-coordinates\/","title":{"raw":"Putting It Together: Parametric Equations and Polar Coordinates","rendered":"Putting It Together: Parametric Equations and Polar Coordinates"},"content":{"raw":"<h3>Describing a Spiral<\/h3>\r\n<div id=\"fs-id1167794047296\" data-type=\"problem\">\r\n<p id=\"fs-id1167794047301\">Recall the <span class=\"no-emphasis\" data-type=\"term\">chambered nautilus<\/span> introduced in the chapter opener. This creature displays a spiral when half the outer shell is cut away. It is possible to describe a spiral using rectangular coordinates. Figure 1 below shows a spiral in rectangular coordinates. How can we describe this curve mathematically?<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_03_010\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"417\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234834\/CNX_Calc_Figure_11_03_010.jpg\" alt=\"A spiral starting at the origin and continually increasing its radius to a point P(x, y).\" width=\"417\" height=\"422\" data-media-type=\"image\/jpeg\" \/> Figure 1. How can we describe a spiral graph mathematically?[\/caption]<\/figure>\r\n<\/div>\r\nSolution:\r\n<div id=\"fs-id1167794332290\" data-type=\"solution\">\r\n<p id=\"fs-id1167794332292\">As the point <em data-effect=\"italics\">P<\/em> travels around the spiral in a counterclockwise direction, its distance <em data-effect=\"italics\">d<\/em> from the origin increases. Assume that the distance <em data-effect=\"italics\">d<\/em> is a constant multiple <em data-effect=\"italics\">k<\/em> of the angle [latex]\\theta [\/latex] that the line segment <em data-effect=\"italics\">OP<\/em> makes with the positive <em data-effect=\"italics\">x<\/em>-axis. Therefore [latex]d\\left(P,O\\right)=k\\theta [\/latex], where [latex]O[\/latex] is the origin. Now use the distance formula and some trigonometry:<\/p>\r\n\r\n<div id=\"fs-id1167794333799\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill d\\left(P,O\\right)&amp; =\\hfill &amp; k\\theta \\hfill \\\\ \\hfill \\sqrt{{\\left(x - 0\\right)}^{2}+{\\left(y - 0\\right)}^{2}}&amp; =\\hfill &amp; k\\text{arctan}\\left(\\frac{y}{x}\\right)\\hfill \\\\ \\hfill \\sqrt{{x}^{2}+{y}^{2}}&amp; =\\hfill &amp; k\\text{arctan}\\left(\\frac{y}{x}\\right)\\hfill \\\\ \\hfill \\text{arctan}\\left(\\frac{y}{x}\\right)&amp; =\\hfill &amp; \\frac{\\sqrt{{x}^{2}+{y}^{2}}}{k}\\hfill \\\\ \\hfill y&amp; =\\hfill &amp; x\\tan\\left(\\frac{\\sqrt{{x}^{2}+{y}^{2}}}{k}\\right).\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794214981\">Although this equation describes the spiral, it is not possible to solve it directly for either <em data-effect=\"italics\">x<\/em> or <em data-effect=\"italics\">y<\/em>. However, if we use polar coordinates, the equation becomes much simpler. In particular, [latex]d\\left(P,O\\right)=r[\/latex], and [latex]\\theta [\/latex] is the second coordinate. Therefore the equation for the spiral becomes [latex]r=k\\theta [\/latex]. Note that when [latex]\\theta =0[\/latex] we also have [latex]r=0[\/latex], so the spiral emanates from the origin. We can remove this restriction by adding a constant to the equation. Then the equation for the spiral becomes [latex]r=a+k\\theta [\/latex] for arbitrary constants [latex]a[\/latex] and [latex]k[\/latex]. This is referred to as an <span class=\"no-emphasis\" data-type=\"term\">Archimedean spiral<\/span>, after the Greek mathematician Archimedes.<\/p>\r\n<p id=\"fs-id1167794175932\">Another type of spiral is the logarithmic spiral, described by the function [latex]r=a\\cdot {b}^{\\theta }[\/latex]. A graph of the function [latex]r=1.2\\left({1.25}^{\\theta }\\right)[\/latex] is given in Figure 2. This spiral describes the shell shape of the chambered nautilus.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_03_011\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"858\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234837\/CNX_Calc_Figure_11_03_011.jpg\" alt=\"This figure has two figures. The first is a shell with many chambers that increase in size from the center out. The second is a spiral with equation r = 1.2(1.25\u03b8).\" width=\"858\" height=\"422\" data-media-type=\"image\/jpeg\" \/> Figure 2. A logarithmic spiral is similar to the shape of the chambered nautilus shell. (credit: modification of work by Jitze Couperus, Flickr)[\/caption]<\/figure>\r\n<\/div>","rendered":"<h3>Describing a Spiral<\/h3>\n<div id=\"fs-id1167794047296\" data-type=\"problem\">\n<p id=\"fs-id1167794047301\">Recall the <span class=\"no-emphasis\" data-type=\"term\">chambered nautilus<\/span> introduced in the chapter opener. This creature displays a spiral when half the outer shell is cut away. It is possible to describe a spiral using rectangular coordinates. Figure 1 below shows a spiral in rectangular coordinates. How can we describe this curve mathematically?<\/p>\n<figure id=\"CNX_Calc_Figure_11_03_010\"><figcaption><\/figcaption><div style=\"width: 427px\" 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\/11234834\/CNX_Calc_Figure_11_03_010.jpg\" alt=\"A spiral starting at the origin and continually increasing its radius to a point P(x, y).\" width=\"417\" height=\"422\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. How can we describe a spiral graph mathematically?<\/p>\n<\/div>\n<\/figure>\n<\/div>\n<p>Solution:<\/p>\n<div id=\"fs-id1167794332290\" data-type=\"solution\">\n<p id=\"fs-id1167794332292\">As the point <em data-effect=\"italics\">P<\/em> travels around the spiral in a counterclockwise direction, its distance <em data-effect=\"italics\">d<\/em> from the origin increases. Assume that the distance <em data-effect=\"italics\">d<\/em> is a constant multiple <em data-effect=\"italics\">k<\/em> of the angle [latex]\\theta[\/latex] that the line segment <em data-effect=\"italics\">OP<\/em> makes with the positive <em data-effect=\"italics\">x<\/em>-axis. Therefore [latex]d\\left(P,O\\right)=k\\theta[\/latex], where [latex]O[\/latex] is the origin. Now use the distance formula and some trigonometry:<\/p>\n<div id=\"fs-id1167794333799\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill d\\left(P,O\\right)& =\\hfill & k\\theta \\hfill \\\\ \\hfill \\sqrt{{\\left(x - 0\\right)}^{2}+{\\left(y - 0\\right)}^{2}}& =\\hfill & k\\text{arctan}\\left(\\frac{y}{x}\\right)\\hfill \\\\ \\hfill \\sqrt{{x}^{2}+{y}^{2}}& =\\hfill & k\\text{arctan}\\left(\\frac{y}{x}\\right)\\hfill \\\\ \\hfill \\text{arctan}\\left(\\frac{y}{x}\\right)& =\\hfill & \\frac{\\sqrt{{x}^{2}+{y}^{2}}}{k}\\hfill \\\\ \\hfill y& =\\hfill & x\\tan\\left(\\frac{\\sqrt{{x}^{2}+{y}^{2}}}{k}\\right).\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794214981\">Although this equation describes the spiral, it is not possible to solve it directly for either <em data-effect=\"italics\">x<\/em> or <em data-effect=\"italics\">y<\/em>. However, if we use polar coordinates, the equation becomes much simpler. In particular, [latex]d\\left(P,O\\right)=r[\/latex], and [latex]\\theta[\/latex] is the second coordinate. Therefore the equation for the spiral becomes [latex]r=k\\theta[\/latex]. Note that when [latex]\\theta =0[\/latex] we also have [latex]r=0[\/latex], so the spiral emanates from the origin. We can remove this restriction by adding a constant to the equation. Then the equation for the spiral becomes [latex]r=a+k\\theta[\/latex] for arbitrary constants [latex]a[\/latex] and [latex]k[\/latex]. This is referred to as an <span class=\"no-emphasis\" data-type=\"term\">Archimedean spiral<\/span>, after the Greek mathematician Archimedes.<\/p>\n<p id=\"fs-id1167794175932\">Another type of spiral is the logarithmic spiral, described by the function [latex]r=a\\cdot {b}^{\\theta }[\/latex]. A graph of the function [latex]r=1.2\\left({1.25}^{\\theta }\\right)[\/latex] is given in Figure 2. This spiral describes the shell shape of the chambered nautilus.<\/p>\n<figure id=\"CNX_Calc_Figure_11_03_011\"><figcaption><\/figcaption><div style=\"width: 868px\" 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\/11234837\/CNX_Calc_Figure_11_03_011.jpg\" alt=\"This figure has two figures. The first is a shell with many chambers that increase in size from the center out. The second is a spiral with equation r = 1.2(1.25\u03b8).\" width=\"858\" height=\"422\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. A logarithmic spiral is similar to the shape of the chambered nautilus shell. (credit: modification of work by Jitze Couperus, Flickr)<\/p>\n<\/div>\n<\/figure>\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-289\">\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 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":21,"template":"","meta":{"_candela_citation":"{\"1\":{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"}}","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-289","chapter","type-chapter","status-publish","hentry"],"part":162,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/289","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":6,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/289\/revisions"}],"predecessor-version":[{"id":1868,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/289\/revisions\/1868"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/162"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/289\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=289"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=289"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=289"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=289"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}