{"id":3267,"date":"2025-08-15T23:54:32","date_gmt":"2025-08-15T23:54:32","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=3267"},"modified":"2025-10-20T17:14:24","modified_gmt":"2025-10-20T17:14:24","slug":"polar-coordinates-apply-it","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/polar-coordinates-apply-it\/","title":{"raw":"Polar Coordinates: Apply It","rendered":"Polar Coordinates: Apply It"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Plot points using polar coordinates.<\/li>\r\n \t<li>Convert between polar coordinates and rectangular coordinates.<\/li>\r\n \t<li>Transform equations between polar and rectangular forms.<\/li>\r\n \t<li>Identify and graph polar equations by converting to rectangular equations.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<div data-test-render-count=\"1\">\r\n<div class=\"group relative pb-3\" data-is-streaming=\"false\">\r\n<div class=\"font-claude-response relative leading-[1.65rem] [&amp;_pre&gt;div]:bg-bg-000\/50 [&amp;_pre&gt;div]:border-0.5 [&amp;_pre&gt;div]:border-border-400 [&amp;_.ignore-pre-bg&gt;div]:bg-transparent [&amp;_.standard-markdown_:is(p,blockquote,h1,h2,h3,h4,h5,h6)]:pl-2 [&amp;_.standard-markdown_:is(p,blockquote,ul,ol,h1,h2,h3,h4,h5,h6)]:pr-8 [&amp;_.progressive-markdown_:is(p,blockquote,h1,h2,h3,h4,h5,h6)]:pl-2 [&amp;_.progressive-markdown_:is(p,blockquote,ul,ol,h1,h2,h3,h4,h5,h6)]:pr-8\">\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0 standard-markdown\">\r\n<h2 class=\"text-2xl font-bold mt-1 text-text-100\">GPS Navigation and Location Services<\/h2>\r\n<p class=\"whitespace-normal break-words\">Modern GPS navigation systems work behind the scenes with both rectangular and polar coordinate systems. While your phone displays locations using latitude and longitude (similar to rectangular coordinates), GPS satellites and navigation algorithms often convert between coordinate systems to calculate distances, directions, and optimal routes.<\/p>\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">\r\n<p class=\"whitespace-normal break-words\"><strong>Polar to Rectangular:<\/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-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]x = r\\cos\\theta[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]y = r\\sin\\theta[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Rectangular to Polar:<\/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-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]r = \\sqrt{x^2 + y^2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\tan\\theta = \\frac{y}{x}[\/latex]<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">A search and rescue helicopter is located 8 miles from base at a bearing of [latex]\\frac{\\pi}{3}[\/latex] radians (60\u00b0) from north. Convert this polar location [latex]\\left(8, \\frac{\\pi}{3}\\right)[\/latex] to rectangular coordinates for mapping software.[reveal-answer q=\"31591\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"31591\"][latex]\\begin{align} x &amp;= r\\cos\\theta \\\\ x &amp;= 8\\cos\\left(\\frac{\\pi}{3}\\right) \\\\ x &amp;= 8 \\cdot \\frac{1}{2} = 4 \\text{ miles} \\\\ y &amp;= r\\sin\\theta \\\\ y &amp;= 8\\sin\\left(\\frac{\\pi}{3}\\right) \\\\ y &amp;= 8 \\cdot \\frac{\\sqrt{3}}{2} = 4\\sqrt{3} \\approx 6.93 \\text{ miles} \\end{align}[\/latex]\r\n\r\nThe helicopter's position in rectangular coordinates is [latex](4, 4\\sqrt{3})[\/latex] or approximately [latex](4, 6.93)[\/latex] miles.[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">A drone is positioned 12 miles from its launch point at an angle of [latex]\\frac{3\\pi}{4}[\/latex] radians. Find the rectangular coordinates of the drone's location.<\/section><section class=\"textbox example\" aria-label=\"Example\">A delivery vehicle is at position [latex](6, 8)[\/latex] miles from the distribution center (using east and north as positive directions). What distance and bearing should the navigation system display?[reveal-answer q=\"583108\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"583108\"]First, find the distance [latex]r[\/latex]:\r\n\r\n[latex]\\begin{align} r &amp;= \\sqrt{x^2 + y^2} \\\\ r &amp;= \\sqrt{6^2 + 8^2} \\\\ r &amp;= \\sqrt{36 + 64} = \\sqrt{100} = 10 \\text{ miles} \\end{align}[\/latex]\r\n\r\nNext, find the angle [latex]\\theta[\/latex]:\r\n\r\n[latex]\\begin{align} \\tan\\theta &amp;= \\frac{y}{x} = \\frac{8}{6} = \\frac{4}{3} \\\\ \\theta &amp;= \\tan^{-1}\\left(\\frac{4}{3}\\right) \\\\ \\theta &amp;\\approx 0.927 \\text{ radians or } 53.1\u00b0 \\end{align}[\/latex]\r\n\r\nSince both [latex]x[\/latex] and [latex]y[\/latex] are positive, the point is in the first quadrant, so our angle is correct. The vehicle is 10 miles away at a bearing of approximately 53.1\u00b0 from east.[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">A GPS tracker shows a hiker at rectangular position [latex](-5, 12)[\/latex] miles from the trailhead. Find the distance and bearing (angle from the positive x-axis) in polar form.<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">When converting from rectangular to polar coordinates using [latex]\\tan^{-1}\\left(\\frac{y}{x}\\right)[\/latex], always check which quadrant your point is in. The inverse tangent function only returns angles in [latex]\\left(-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right)[\/latex], so you may need to add [latex]\\pi[\/latex] for points in Quadrants II or III.<\/section><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Plot points using polar coordinates.<\/li>\n<li>Convert between polar coordinates and rectangular coordinates.<\/li>\n<li>Transform equations between polar and rectangular forms.<\/li>\n<li>Identify and graph polar equations by converting to rectangular equations.<\/li>\n<\/ul>\n<\/section>\n<div data-test-render-count=\"1\">\n<div class=\"group relative pb-3\" data-is-streaming=\"false\">\n<div class=\"font-claude-response relative leading-[1.65rem] [&amp;_pre&gt;div]:bg-bg-000\/50 [&amp;_pre&gt;div]:border-0.5 [&amp;_pre&gt;div]:border-border-400 [&amp;_.ignore-pre-bg&gt;div]:bg-transparent [&amp;_.standard-markdown_:is(p,blockquote,h1,h2,h3,h4,h5,h6)]:pl-2 [&amp;_.standard-markdown_:is(p,blockquote,ul,ol,h1,h2,h3,h4,h5,h6)]:pr-8 [&amp;_.progressive-markdown_:is(p,blockquote,h1,h2,h3,h4,h5,h6)]:pl-2 [&amp;_.progressive-markdown_:is(p,blockquote,ul,ol,h1,h2,h3,h4,h5,h6)]:pr-8\">\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0 standard-markdown\">\n<h2 class=\"text-2xl font-bold mt-1 text-text-100\">GPS Navigation and Location Services<\/h2>\n<p class=\"whitespace-normal break-words\">Modern GPS navigation systems work behind the scenes with both rectangular and polar coordinate systems. While your phone displays locations using latitude and longitude (similar to rectangular coordinates), GPS satellites and navigation algorithms often convert between coordinate systems to calculate distances, directions, and optimal routes.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">\n<p class=\"whitespace-normal break-words\"><strong>Polar to Rectangular:<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]x = r\\cos\\theta[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]y = r\\sin\\theta[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Rectangular to Polar:<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]r = \\sqrt{x^2 + y^2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\tan\\theta = \\frac{y}{x}[\/latex]<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">A search and rescue helicopter is located 8 miles from base at a bearing of [latex]\\frac{\\pi}{3}[\/latex] radians (60\u00b0) from north. Convert this polar location [latex]\\left(8, \\frac{\\pi}{3}\\right)[\/latex] to rectangular coordinates for mapping software.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q31591\">Show Solution<\/button><\/p>\n<div id=\"q31591\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{align} x &= r\\cos\\theta \\\\ x &= 8\\cos\\left(\\frac{\\pi}{3}\\right) \\\\ x &= 8 \\cdot \\frac{1}{2} = 4 \\text{ miles} \\\\ y &= r\\sin\\theta \\\\ y &= 8\\sin\\left(\\frac{\\pi}{3}\\right) \\\\ y &= 8 \\cdot \\frac{\\sqrt{3}}{2} = 4\\sqrt{3} \\approx 6.93 \\text{ miles} \\end{align}[\/latex]<\/p>\n<p>The helicopter&#8217;s position in rectangular coordinates is [latex](4, 4\\sqrt{3})[\/latex] or approximately [latex](4, 6.93)[\/latex] miles.<\/p><\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">A drone is positioned 12 miles from its launch point at an angle of [latex]\\frac{3\\pi}{4}[\/latex] radians. Find the rectangular coordinates of the drone&#8217;s location.<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">A delivery vehicle is at position [latex](6, 8)[\/latex] miles from the distribution center (using east and north as positive directions). What distance and bearing should the navigation system display?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q583108\">Show Solution<\/button><\/p>\n<div id=\"q583108\" class=\"hidden-answer\" style=\"display: none\">First, find the distance [latex]r[\/latex]:<\/p>\n<p>[latex]\\begin{align} r &= \\sqrt{x^2 + y^2} \\\\ r &= \\sqrt{6^2 + 8^2} \\\\ r &= \\sqrt{36 + 64} = \\sqrt{100} = 10 \\text{ miles} \\end{align}[\/latex]<\/p>\n<p>Next, find the angle [latex]\\theta[\/latex]:<\/p>\n<p>[latex]\\begin{align} \\tan\\theta &= \\frac{y}{x} = \\frac{8}{6} = \\frac{4}{3} \\\\ \\theta &= \\tan^{-1}\\left(\\frac{4}{3}\\right) \\\\ \\theta &\\approx 0.927 \\text{ radians or } 53.1\u00b0 \\end{align}[\/latex]<\/p>\n<p>Since both [latex]x[\/latex] and [latex]y[\/latex] are positive, the point is in the first quadrant, so our angle is correct. The vehicle is 10 miles away at a bearing of approximately 53.1\u00b0 from east.<\/p><\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">A GPS tracker shows a hiker at rectangular position [latex](-5, 12)[\/latex] miles from the trailhead. Find the distance and bearing (angle from the positive x-axis) in polar form.<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">When converting from rectangular to polar coordinates using [latex]\\tan^{-1}\\left(\\frac{y}{x}\\right)[\/latex], always check which quadrant your point is in. The inverse tangent function only returns angles in [latex]\\left(-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right)[\/latex], so you may need to add [latex]\\pi[\/latex] for points in Quadrants II or III.<\/section>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"author":67,"menu_order":9,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":247,"module-header":"apply_it","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\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3267"}],"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":6,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3267\/revisions"}],"predecessor-version":[{"id":4761,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3267\/revisions\/4761"}],"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\/3267\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=3267"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=3267"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=3267"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=3267"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}