{"id":1507,"date":"2025-07-25T02:19:34","date_gmt":"2025-07-25T02:19:34","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=1507"},"modified":"2026-03-24T07:03:40","modified_gmt":"2026-03-24T07:03:40","slug":"arcs-and-sectors-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/arcs-and-sectors-fresh-take\/","title":{"raw":"Arcs and Sectors: Fresh Take","rendered":"Arcs and Sectors: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Find the length of a circular arc.<\/li>\r\n \t<li>Find the area of a sector of a circle.<\/li>\r\n \t<li>Use linear and angular speed to describe motion on a circular path.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Finding the Length of a Circular Arc<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\r\n<\/strong>\r\n\r\nThe length of a circular arc tells us how far along the circle\u2019s edge we travel when sweeping out an angle. The key is that the angle must be measured in radians, because radians directly connect an angle to the arc length. If [latex]\\theta[\/latex] is the central angle in radians and [latex]r[\/latex] is the circle\u2019s radius, then the arc length is given by [latex]s = r\\theta[\/latex]. This formula works because radians are defined as arc length divided by radius<strong style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">.<\/strong>\r\n\r\n<strong style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Quick Tips: Finding Arc Length<\/strong>\r\n<ol>\r\n \t<li>Use the formula: [latex]s = r\\theta[\/latex], where [latex]s[\/latex] is arc length, [latex]r[\/latex] is radius, and [latex]\\theta[\/latex] is in radians.<\/li>\r\n \t<li>Covert first if needed: If the angle is given in degrees, convert to radians before using the formula.<\/li>\r\n \t<li>Check the units: The arc length will be in the same unit as the radius (e.g., if [latex]r[\/latex] is in cm, [latex]s[\/latex] will be in cm).<\/li>\r\n \t<li>Fraction of the circle: You can also think of arc length as a fraction of the whole circumference:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]s = \\dfrac{\\theta}{2\\pi}\\cdot (2\\pi r)[\/latex].<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Real-world connection: Arc length is like the \u201cdistance walked\u201d along the edge of the circle \u2014 useful for wheels, gears, and circular tracks.<\/li>\r\n<\/ol>\r\n<\/div>\r\n&nbsp;\r\n<div><section class=\"textbox example\">A city bus navigates a roundabout of radius [latex]14[\/latex]m, turning through [latex]95^\\circ[\/latex]. Find the arc length [latex]s[\/latex].[reveal-answer q=\"725997\"]Hint[\/reveal-answer][hidden-answer a=\"725997\"][latex]\\\\\\\\[\/latex]Convert [latex]95^\\circ[\/latex] to radians.\r\n[latex]s = r\\theta[\/latex] with [latex]\\theta[\/latex] in radians\r\n[\/hidden-answer][reveal-answer q=\"702833\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"702833\"][latex]95^\\circ \\Rightarrow[\/latex] radians:\r\n[latex]\r\n\\begin{aligned}\r\n95^\\circ\\cdot \\dfrac{\\pi}{180} &amp;= \\dfrac{95\\pi}{180} \\\\\r\n&amp;= \\dfrac{19\\pi}{36} \\text { radians}\r\n\\end{aligned}\r\n[\/latex]\r\n[latex]\\begin{aligned}s &amp;= r\\theta \\\\ &amp;= 14 \\cdot \\dfrac{19\\pi}{36} \\\\&amp;= \\dfrac{133\\pi}{18} \\\\ &amp;\\approx 23.21 \\text { m}\\end{aligned}[\/latex][\/hidden-answer]<\/section><\/div>\r\n&nbsp;\r\n<div><section class=\"textbox example\">A robot vacuum hugs a circular coffee table, following an arc with radius [latex]0.70[\/latex] m and central angle [latex]1.9[\/latex] radians. Find [latex]s[\/latex].[latex]\\\\\\\\[\/latex][reveal-answer q=\"268488\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"268488\"]Angle is already in radians, and the radius is given.[latex]\\\\\\\\[\/latex]Remember: [latex]s = r\\theta[\/latex][latex]\\\\\\\\[\/latex][\/hidden-answer][reveal-answer q=\"939558\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"939558\"][latex]\\begin{aligned}\r\ns &amp;= 0.70 \\cdot 1.9 \\\\ &amp;= 1.33 \\end{aligned}\r\n[\/latex][\/hidden-answer]<\/section><\/div>\r\n&nbsp;\r\n<div><section class=\"textbox example\">In the planetarium, a laser dot sweeps along the dome at radius [latex]11[\/latex] m through angle [latex]\\dfrac{7\\pi}{20}[\/latex]. Find [latex]s[\/latex] in exact form.[reveal-answer q=\"828216\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"828216\"][latex]\\begin{aligned} s &amp;= r\\theta \\\\ &amp;= 11 \\cdot \\dfrac{7\\pi}{20}\\end{aligned}[\/latex]\r\nKeep [latex]\\pi[\/latex] for an exact answer.[\/hidden-answer][reveal-answer q=\"97179\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"97179\"][latex]s = \\dfrac{77\\pi}{20}[\/latex][\/hidden-answer]\u00a0<\/section><\/div>\r\n&nbsp;\r\n<div><section class=\"textbox watchIt\"><script type=\"text\/javascript\" src=\"https:\/\/www.youtube.com\/iframe_api \"><\/script>\r\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-efedhghd-zD4CsKIYEHo\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/zD4CsKIYEHo?enablejsapi=1 \" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\r\n\r\n<div id=\"3p-plugin-target-efedhghd-zD4CsKIYEHo\" class=\"p3sdk-target\"><\/div>\r\n<p class=\"cc-media-iframe-container\"><script type='text\/javascript' src='\/\/plugin.3playmedia.com\/ajax.js?cc=1&cc_minimizable=1&cc_minimize_on_load=0&cc_multi_text_track=0&cc_overlay=1&cc_searchable=0&embed=ajax&mf=13892889&p3sdk_version=1.11.7&p=20361&player_type=youtube&plugin_skin=dark&target=3p-plugin-target-efedhghd-zD4CsKIYEHo&vembed=0&video_id=zD4CsKIYEHo&video_target=tpm-plugin-efedhghd-zD4CsKIYEHo'><\/script><\/p>\r\nYou can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Precalculus\/Transcripts\/Arc+Length+and+Area+of+a+Sector+(Up+to+5_40)_transcript.txt\">transcript for \"Arc Length and Area of a Sector\" here (opens in new window).<\/a>\r\n\r\n(stop at 5:40)\r\n\r\n<\/section><\/div>\r\n<h2>Finding the Area of a Sector<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\r\n<\/strong>\r\n\r\nA sector of a circle is like a <strong data-start=\"195\" data-end=\"213\">slice of pizza<\/strong> \u2014 it\u2019s the region between two radii and the arc that connects them. The area of that slice depends on how big the angle is and how large the circle is. If the radius is [latex]r[\/latex] and the central angle is [latex]\\theta[\/latex] in radians, then the area of the sector is given by [latex]A = \\tfrac{1}{2}r^2\\theta[\/latex]. This formula works because radians directly connect angle size to the fraction of the circle\u2019s area.\r\n\r\n<strong style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Quick Tips: Finding Area of a Sector<\/strong>\r\n<ol>\r\n \t<li>Use the formula: [latex]A = \\tfrac{1}{2}r^2\\theta[\/latex], where [latex]\\theta[\/latex] is in radians.<\/li>\r\n \t<li>Convert if necessary: If the angle is in degrees, convert to radians before plugging it in.<\/li>\r\n \t<li>Check the units: The area will be in square units (e.g., [latex]\\text{cm}^2[\/latex]) if the radius is in cm.<\/li>\r\n \t<li>Fraction of the whole circle: The formula also comes from [latex]\\dfrac{\\theta}{2\\pi}\\cdot \\pi r^2[\/latex]; the fraction of the circle\u2019s total area.<\/li>\r\n \t<li>Think pizza or pie: A small angle gives a skinny slice, a big angle gives a bigger slice \u2014 the formula measures the \u201csize of the slice.\u201d<\/li>\r\n<\/ol>\r\n<\/div>\r\n&nbsp;\r\n<div><section class=\"textbox example\">A sprinkler waters a sector of radius [latex]6[\/latex] m across [latex]110^\\circ[\/latex]. Find the watered area.[reveal-answer q=\"710677\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"710677\"]You know the radius. Convert [latex]110^\\circ[\/latex] to radians, and then plug into the formula.\r\n[latex]A=\\dfrac{1}{2}r^2\\theta[\/latex][\/hidden-answer][reveal-answer q=\"256399\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"256399\"][latex]\\begin{aligned} A &amp;= \\dfrac{1}{2}(6)^2 \\cdot \\dfrac{11\\pi}{18} \\\\ &amp;= 18 \\cdot \\dfrac{11\\pi}{18} \\\\ &amp;= 11\\pi \\\\ &amp;\\approx 34.56 \\text { m}^2 \\end{aligned}[\/latex] [\/hidden-answer]<\/section><\/div>\r\n&nbsp;\r\n<div><section class=\"textbox example\">At a night market, a spotlight covers a wedge of radius [latex]12[\/latex] m with angle [latex]0.90[\/latex] radians. Find the illuminated area.[reveal-answer q=\"446876\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"446876\"]You know the radius and the radians.\r\nRemember: [latex]A = \\dfrac{1}{2} r^2\\theta[\/latex][\/hidden-answer][reveal-answer q=\"680566\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"680566\"][latex]\\begin{aligned} A &amp;= \\dfrac{1}{2} \\cdot (12)^2 \\cdot 0.90 \\\\ &amp;= \\dfrac{1}{2} \\cdot 144 \\cdot 0.90 \\\\ &amp;= 64.8 \\text { m}^2 \\end{aligned}[\/latex][\/hidden-answer]<\/section><\/div>\r\n&nbsp;\r\n<div><section class=\"textbox example\">A beadwork medallion shows a sector of radius [latex]8[\/latex] cm with angle [latex]\\dfrac{5\\pi}{6}[\/latex]. Find the sector area in exact form.[reveal-answer q=\"175607\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"175607\"][latex]\\begin{aligned} A &amp;= \\dfrac{1}{2} r^2\\theta \\\\ &amp;= \\dfrac{1}{2} \\cdot 64 \\cdot \\dfrac{5\\pi}{6} \\end{aligned}[\/latex]\r\nDo not change [latex]\\pi[\/latex] to a decimal.[\/hidden-answer][reveal-answer q=\"209703\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"209703\"][latex]\\begin{aligned} A &amp;= 32 \\cdot \\dfrac{5\\pi}{6} \\\\ &amp;= \\dfrac{80\\pi}{3} \\\\ &amp;\\approx 83.78 \\text { cm}^2 \\end{aligned}[\/latex][\/hidden-answer]<\/section><\/div>\r\n&nbsp;\r\n<div><section class=\"textbox watchIt\"><script type=\"text\/javascript\" src=\"https:\/\/www.youtube.com\/iframe_api \"><\/script>\r\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-eefbaeeb-zD4CsKIYEHo\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/zD4CsKIYEHo?enablejsapi=1 \" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\r\n\r\n<div id=\"3p-plugin-target-eefbaeeb-zD4CsKIYEHo\" class=\"p3sdk-target\"><\/div>\r\n<p class=\"cc-media-iframe-container\"><script type='text\/javascript' src='\/\/plugin.3playmedia.com\/ajax.js?cc=1&cc_minimizable=1&cc_minimize_on_load=0&cc_multi_text_track=0&cc_overlay=1&cc_searchable=0&embed=ajax&mf=13892889&p3sdk_version=1.11.7&p=20361&player_type=youtube&plugin_skin=dark&target=3p-plugin-target-eefbaeeb-zD4CsKIYEHo&vembed=0&video_id=zD4CsKIYEHo&video_target=tpm-plugin-eefbaeeb-zD4CsKIYEHo'><\/script><\/p>\r\nYou can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Precalculus\/Transcripts\/Arc+Length+and+Area+of+a+Sector+(After+5_40)_transcript.txt\">transcript for \"Arc Length and Area of a Sector\" here (opens in new window).<\/a>\r\n\r\n(start at 5:40)\r\n\r\n<\/section><\/div>\r\n<h2>Describing Motion on a Circular Path<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\r\n<\/strong>\r\n\r\nWhen an object moves along a circular path, we can describe its motion in two connected ways: <strong data-start=\"281\" data-end=\"298\">angular speed<\/strong> and <strong data-start=\"303\" data-end=\"319\">linear speed<\/strong>. Angular speed measures <strong data-start=\"344\" data-end=\"374\">how fast the angle changes<\/strong> (in radians per unit of time), while linear speed measures <strong data-start=\"434\" data-end=\"481\">how fast the distance along the arc changes<\/strong> (in distance per unit of time). The two are tied together by the radius:<br data-start=\"554\" data-end=\"557\" \/>[latex]v = r\\omega[\/latex], where [latex]v[\/latex] is linear speed, [latex]r[\/latex] is the radius, and [latex]\\omega[\/latex] is angular speed. This relationship lets us move between \u201cspinning speed\u201d (angular) and \u201ctraveling speed\u201d (linear).\r\n\r\n<strong style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Quick Tips: Finding Area of a Sector<\/strong>\r\n<ol>\r\n \t<li>Linear speed formula: [latex]\\omega = \\dfrac{\\theta}{t}[\/latex], where [latex]\\theta[\/latex] is in radians and [latex]t[\/latex] is time.<\/li>\r\n \t<li>Angular speed formula: [latex]v = \\dfrac{s}{t}[\/latex], where [latex]s[\/latex] is the arc length traveled.<\/li>\r\n \t<li>Connect them: Use [latex]s = r\\theta[\/latex] to link the two formulas, giving [latex]v = r\\omega[\/latex].<\/li>\r\n \t<li>Units Matter:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Linear speed: feet per second, meters per second, etc.<\/li>\r\n \t<li>Angular speed: radians per second (or per minute).<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Think Real-World: On a spinning wheel, all points share the same angular speed, but points farther from the center move faster linearly.<\/li>\r\n<\/ol>\r\n<\/div>\r\n&nbsp;\r\n<div><section class=\"textbox example\">A smartwatch second hand has length [latex]18[\/latex] mm. Find its angular speed [latex]\\omega[\/latex] rad\/s and tip speed [latex]v[\/latex] m\/s.\r\n[reveal-answer q=\"138196\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"138196\"][latex]\\omega = \\dfrac{2\\pi}{T}[\/latex] rad\/s where [latex]T[\/latex] is the period of one revolution.\r\n[latex]\\begin{aligned} r &amp;= 18 \\text { mm} \\\\\u00a0 &amp;= 0.018 \\text { m} \\end{aligned}[\/latex]\r\n[latex]v = r\\omega[\/latex][\/hidden-answer][reveal-answer q=\"6300\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"6300\"][latex]\\begin{aligned} \\omega &amp;= \\dfrac{2\\pi}{60} \\\\ &amp;= \\dfrac{\\pi}{30} \\\\ &amp;\\approx 0.1047 \\text { rad\/s} \\end{aligned}[\/latex]\r\n[latex]\\begin{aligned} v &amp;= 0.018 \\cdot 0.1047 \\\\ &amp;= 0.0018846 \\\\ &amp;\\approx 1.885 \\cdot 10^{-3} \\text { m\/s} \\end{aligned}[\/latex][\/hidden-answer]<\/section><\/div>\r\n&nbsp;\r\n<div><section class=\"textbox example\">A wind turbine (blade length [latex]27[\/latex] m spins at [latex]12[\/latex] RPM. Find [latex]\\omega[\/latex] rad\/s and the tip speed [latex]v[\/latex] m\/s.[reveal-answer q=\"109771\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"109771\"]RPM: revolutions per minute\r\n[latex] \\omega = \\text{RPM} \\cdot \\dfrac{2\\pi}{60} \\text { rad\/s}[\/latex]\r\n[latex]v=r\\omega[\/latex][\/hidden-answer][reveal-answer q=\"367369\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"367369\"][latex]\\begin{aligned} \\omega &amp;= \\dfrac{2\\pi}{5} \\text {rad\/s} \\\\ &amp;\\approx 1.257 \\end{aligned}[\/latex]\r\n[latex]\\begin{aligned} v &amp;= 27 \\cdot 1.257 \\\\ &amp;\\approx 33.93 \\text { m\/s} \\end{aligned}[\/latex][\/hidden-answer]<\/section><\/div>\r\n&nbsp;\r\n<div><section class=\"textbox example\">In an aim-trainer, a circular targe rotates with angular speed [latex]1.8[\/latex] rad\/s. A marker sits [latex]0.75[\/latex] m from the center. Find its linear speed [latex]v[\/latex].[reveal-answer q=\"416218\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"416218\"][latex]\\begin{aligned} v &amp;= r\\omega \\\\\u00a0 &amp;= 0.75 \\cdot 1.8 \\end{aligned}[\/latex][\/hidden-answer][reveal-answer q=\"938851\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"938851\"][latex]v = 1.35[\/latex] m\/s[\/hidden-answer]<\/section><\/div>\r\n&nbsp;\r\n\r\n<section aria-label=\"Watch It\"><section class=\"textbox watchIt\" aria-label=\"Watch It\"><script type=\"text\/javascript\" src=\"https:\/\/www.youtube.com\/iframe_api \"><\/script>\r\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-bfdbahga-bfWkgA5GSE0\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/bfWkgA5GSE0?enablejsapi=1 \" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\r\n\r\n<div id=\"3p-plugin-target-bfdbahga-bfWkgA5GSE0\" class=\"p3sdk-target\"><\/div>\r\n<p class=\"cc-media-iframe-container\"><script type='text\/javascript' src='\/\/plugin.3playmedia.com\/ajax.js?cc=1&cc_minimizable=1&cc_minimize_on_load=0&cc_multi_text_track=0&cc_overlay=1&cc_searchable=0&embed=ajax&mf=13892886&p3sdk_version=1.11.7&p=20361&player_type=youtube&plugin_skin=dark&target=3p-plugin-target-bfdbahga-bfWkgA5GSE0&vembed=0&video_id=bfWkgA5GSE0&video_target=tpm-plugin-bfdbahga-bfWkgA5GSE0'><\/script><\/p>\r\nYou can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Precalculus\/Transcripts\/Example+-+Determine+Angular+and+Linear+Velocity_transcript.txt\">transcript for \"Example: Determine Angular and Linear Velocity\" here (opens in new window).<\/a>\r\n\r\n<\/section><\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Find the length of a circular arc.<\/li>\n<li>Find the area of a sector of a circle.<\/li>\n<li>Use linear and angular speed to describe motion on a circular path.<\/li>\n<\/ul>\n<\/section>\n<h2>Finding the Length of a Circular Arc<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea<br \/>\n<\/strong><\/p>\n<p>The length of a circular arc tells us how far along the circle\u2019s edge we travel when sweeping out an angle. The key is that the angle must be measured in radians, because radians directly connect an angle to the arc length. If [latex]\\theta[\/latex] is the central angle in radians and [latex]r[\/latex] is the circle\u2019s radius, then the arc length is given by [latex]s = r\\theta[\/latex]. This formula works because radians are defined as arc length divided by radius<strong style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">.<\/strong><\/p>\n<p><strong style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Quick Tips: Finding Arc Length<\/strong><\/p>\n<ol>\n<li>Use the formula: [latex]s = r\\theta[\/latex], where [latex]s[\/latex] is arc length, [latex]r[\/latex] is radius, and [latex]\\theta[\/latex] is in radians.<\/li>\n<li>Covert first if needed: If the angle is given in degrees, convert to radians before using the formula.<\/li>\n<li>Check the units: The arc length will be in the same unit as the radius (e.g., if [latex]r[\/latex] is in cm, [latex]s[\/latex] will be in cm).<\/li>\n<li>Fraction of the circle: You can also think of arc length as a fraction of the whole circumference:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]s = \\dfrac{\\theta}{2\\pi}\\cdot (2\\pi r)[\/latex].<\/li>\n<\/ol>\n<\/li>\n<li>Real-world connection: Arc length is like the \u201cdistance walked\u201d along the edge of the circle \u2014 useful for wheels, gears, and circular tracks.<\/li>\n<\/ol>\n<\/div>\n<p>&nbsp;<\/p>\n<div>\n<section class=\"textbox example\">A city bus navigates a roundabout of radius [latex]14[\/latex]m, turning through [latex]95^\\circ[\/latex]. Find the arc length [latex]s[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q725997\">Hint<\/button><\/p>\n<div id=\"q725997\" class=\"hidden-answer\" style=\"display: none\">[latex]\\\\\\\\[\/latex]Convert [latex]95^\\circ[\/latex] to radians.<br \/>\n[latex]s = r\\theta[\/latex] with [latex]\\theta[\/latex] in radians\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q702833\">Show Answer<\/button><\/p>\n<div id=\"q702833\" class=\"hidden-answer\" style=\"display: none\">[latex]95^\\circ \\Rightarrow[\/latex] radians:<br \/>\n[latex]\\begin{aligned}  95^\\circ\\cdot \\dfrac{\\pi}{180} &= \\dfrac{95\\pi}{180} \\\\  &= \\dfrac{19\\pi}{36} \\text { radians}  \\end{aligned}[\/latex]<br \/>\n[latex]\\begin{aligned}s &= r\\theta \\\\ &= 14 \\cdot \\dfrac{19\\pi}{36} \\\\&= \\dfrac{133\\pi}{18} \\\\ &\\approx 23.21 \\text { m}\\end{aligned}[\/latex]<\/div>\n<\/div>\n<\/section>\n<\/div>\n<p>&nbsp;<\/p>\n<div>\n<section class=\"textbox example\">A robot vacuum hugs a circular coffee table, following an arc with radius [latex]0.70[\/latex] m and central angle [latex]1.9[\/latex] radians. Find [latex]s[\/latex].[latex]\\\\\\\\[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q268488\">Hint<\/button><\/p>\n<div id=\"q268488\" class=\"hidden-answer\" style=\"display: none\">Angle is already in radians, and the radius is given.[latex]\\\\\\\\[\/latex]Remember: [latex]s = r\\theta[\/latex][latex]\\\\\\\\[\/latex]<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q939558\">Show Answer<\/button><\/p>\n<div id=\"q939558\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{aligned}  s &= 0.70 \\cdot 1.9 \\\\ &= 1.33 \\end{aligned}[\/latex]<\/div>\n<\/div>\n<\/section>\n<\/div>\n<p>&nbsp;<\/p>\n<div>\n<section class=\"textbox example\">In the planetarium, a laser dot sweeps along the dome at radius [latex]11[\/latex] m through angle [latex]\\dfrac{7\\pi}{20}[\/latex]. Find [latex]s[\/latex] in exact form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q828216\">Hint<\/button><\/p>\n<div id=\"q828216\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{aligned} s &= r\\theta \\\\ &= 11 \\cdot \\dfrac{7\\pi}{20}\\end{aligned}[\/latex]<br \/>\nKeep [latex]\\pi[\/latex] for an exact answer.<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q97179\">Show Answer<\/button><\/p>\n<div id=\"q97179\" class=\"hidden-answer\" style=\"display: none\">[latex]s = \\dfrac{77\\pi}{20}[\/latex]<\/div>\n<\/div>\n<p>\u00a0<\/section>\n<\/div>\n<p>&nbsp;<\/p>\n<div>\n<section class=\"textbox watchIt\"><script type=\"text\/javascript\" src=\"https:\/\/www.youtube.com\/iframe_api\"><\/script><\/p>\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-efedhghd-zD4CsKIYEHo\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/zD4CsKIYEHo?enablejsapi=1\" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<div id=\"3p-plugin-target-efedhghd-zD4CsKIYEHo\" class=\"p3sdk-target\"><\/div>\n<p class=\"cc-media-iframe-container\"><script type=\"text\/javascript\" src=\"\/\/plugin.3playmedia.com\/ajax.js?cc=1&#38;cc_minimizable=1&#38;cc_minimize_on_load=0&#38;cc_multi_text_track=0&#38;cc_overlay=1&#38;cc_searchable=0&#38;embed=ajax&#38;mf=13892889&#38;p3sdk_version=1.11.7&#38;p=20361&#38;player_type=youtube&#38;plugin_skin=dark&#38;target=3p-plugin-target-efedhghd-zD4CsKIYEHo&#38;vembed=0&#38;video_id=zD4CsKIYEHo&#38;video_target=tpm-plugin-efedhghd-zD4CsKIYEHo\"><\/script><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Precalculus\/Transcripts\/Arc+Length+and+Area+of+a+Sector+(Up+to+5_40)_transcript.txt\">transcript for &#8220;Arc Length and Area of a Sector&#8221; here (opens in new window).<\/a><\/p>\n<p>(stop at 5:40)<\/p>\n<\/section>\n<\/div>\n<h2>Finding the Area of a Sector<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea<br \/>\n<\/strong><\/p>\n<p>A sector of a circle is like a <strong data-start=\"195\" data-end=\"213\">slice of pizza<\/strong> \u2014 it\u2019s the region between two radii and the arc that connects them. The area of that slice depends on how big the angle is and how large the circle is. If the radius is [latex]r[\/latex] and the central angle is [latex]\\theta[\/latex] in radians, then the area of the sector is given by [latex]A = \\tfrac{1}{2}r^2\\theta[\/latex]. This formula works because radians directly connect angle size to the fraction of the circle\u2019s area.<\/p>\n<p><strong style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Quick Tips: Finding Area of a Sector<\/strong><\/p>\n<ol>\n<li>Use the formula: [latex]A = \\tfrac{1}{2}r^2\\theta[\/latex], where [latex]\\theta[\/latex] is in radians.<\/li>\n<li>Convert if necessary: If the angle is in degrees, convert to radians before plugging it in.<\/li>\n<li>Check the units: The area will be in square units (e.g., [latex]\\text{cm}^2[\/latex]) if the radius is in cm.<\/li>\n<li>Fraction of the whole circle: The formula also comes from [latex]\\dfrac{\\theta}{2\\pi}\\cdot \\pi r^2[\/latex]; the fraction of the circle\u2019s total area.<\/li>\n<li>Think pizza or pie: A small angle gives a skinny slice, a big angle gives a bigger slice \u2014 the formula measures the \u201csize of the slice.\u201d<\/li>\n<\/ol>\n<\/div>\n<p>&nbsp;<\/p>\n<div>\n<section class=\"textbox example\">A sprinkler waters a sector of radius [latex]6[\/latex] m across [latex]110^\\circ[\/latex]. Find the watered area.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q710677\">Hint<\/button><\/p>\n<div id=\"q710677\" class=\"hidden-answer\" style=\"display: none\">You know the radius. Convert [latex]110^\\circ[\/latex] to radians, and then plug into the formula.<br \/>\n[latex]A=\\dfrac{1}{2}r^2\\theta[\/latex]<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q256399\">Show Answer<\/button><\/p>\n<div id=\"q256399\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{aligned} A &= \\dfrac{1}{2}(6)^2 \\cdot \\dfrac{11\\pi}{18} \\\\ &= 18 \\cdot \\dfrac{11\\pi}{18} \\\\ &= 11\\pi \\\\ &\\approx 34.56 \\text { m}^2 \\end{aligned}[\/latex] <\/div>\n<\/div>\n<\/section>\n<\/div>\n<p>&nbsp;<\/p>\n<div>\n<section class=\"textbox example\">At a night market, a spotlight covers a wedge of radius [latex]12[\/latex] m with angle [latex]0.90[\/latex] radians. Find the illuminated area.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q446876\">Hint<\/button><\/p>\n<div id=\"q446876\" class=\"hidden-answer\" style=\"display: none\">You know the radius and the radians.<br \/>\nRemember: [latex]A = \\dfrac{1}{2} r^2\\theta[\/latex]<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q680566\">Show Answer<\/button><\/p>\n<div id=\"q680566\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{aligned} A &= \\dfrac{1}{2} \\cdot (12)^2 \\cdot 0.90 \\\\ &= \\dfrac{1}{2} \\cdot 144 \\cdot 0.90 \\\\ &= 64.8 \\text { m}^2 \\end{aligned}[\/latex]<\/div>\n<\/div>\n<\/section>\n<\/div>\n<p>&nbsp;<\/p>\n<div>\n<section class=\"textbox example\">A beadwork medallion shows a sector of radius [latex]8[\/latex] cm with angle [latex]\\dfrac{5\\pi}{6}[\/latex]. Find the sector area in exact form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q175607\">Hint<\/button><\/p>\n<div id=\"q175607\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{aligned} A &= \\dfrac{1}{2} r^2\\theta \\\\ &= \\dfrac{1}{2} \\cdot 64 \\cdot \\dfrac{5\\pi}{6} \\end{aligned}[\/latex]<br \/>\nDo not change [latex]\\pi[\/latex] to a decimal.<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q209703\">Show Answer<\/button><\/p>\n<div id=\"q209703\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{aligned} A &= 32 \\cdot \\dfrac{5\\pi}{6} \\\\ &= \\dfrac{80\\pi}{3} \\\\ &\\approx 83.78 \\text { cm}^2 \\end{aligned}[\/latex]<\/div>\n<\/div>\n<\/section>\n<\/div>\n<p>&nbsp;<\/p>\n<div>\n<section class=\"textbox watchIt\"><script type=\"text\/javascript\" src=\"https:\/\/www.youtube.com\/iframe_api\"><\/script><\/p>\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-eefbaeeb-zD4CsKIYEHo\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/zD4CsKIYEHo?enablejsapi=1\" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<div id=\"3p-plugin-target-eefbaeeb-zD4CsKIYEHo\" class=\"p3sdk-target\"><\/div>\n<p class=\"cc-media-iframe-container\"><script type=\"text\/javascript\" src=\"\/\/plugin.3playmedia.com\/ajax.js?cc=1&#38;cc_minimizable=1&#38;cc_minimize_on_load=0&#38;cc_multi_text_track=0&#38;cc_overlay=1&#38;cc_searchable=0&#38;embed=ajax&#38;mf=13892889&#38;p3sdk_version=1.11.7&#38;p=20361&#38;player_type=youtube&#38;plugin_skin=dark&#38;target=3p-plugin-target-eefbaeeb-zD4CsKIYEHo&#38;vembed=0&#38;video_id=zD4CsKIYEHo&#38;video_target=tpm-plugin-eefbaeeb-zD4CsKIYEHo\"><\/script><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Precalculus\/Transcripts\/Arc+Length+and+Area+of+a+Sector+(After+5_40)_transcript.txt\">transcript for &#8220;Arc Length and Area of a Sector&#8221; here (opens in new window).<\/a><\/p>\n<p>(start at 5:40)<\/p>\n<\/section>\n<\/div>\n<h2>Describing Motion on a Circular Path<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea<br \/>\n<\/strong><\/p>\n<p>When an object moves along a circular path, we can describe its motion in two connected ways: <strong data-start=\"281\" data-end=\"298\">angular speed<\/strong> and <strong data-start=\"303\" data-end=\"319\">linear speed<\/strong>. Angular speed measures <strong data-start=\"344\" data-end=\"374\">how fast the angle changes<\/strong> (in radians per unit of time), while linear speed measures <strong data-start=\"434\" data-end=\"481\">how fast the distance along the arc changes<\/strong> (in distance per unit of time). The two are tied together by the radius:<br data-start=\"554\" data-end=\"557\" \/>[latex]v = r\\omega[\/latex], where [latex]v[\/latex] is linear speed, [latex]r[\/latex] is the radius, and [latex]\\omega[\/latex] is angular speed. This relationship lets us move between \u201cspinning speed\u201d (angular) and \u201ctraveling speed\u201d (linear).<\/p>\n<p><strong style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Quick Tips: Finding Area of a Sector<\/strong><\/p>\n<ol>\n<li>Linear speed formula: [latex]\\omega = \\dfrac{\\theta}{t}[\/latex], where [latex]\\theta[\/latex] is in radians and [latex]t[\/latex] is time.<\/li>\n<li>Angular speed formula: [latex]v = \\dfrac{s}{t}[\/latex], where [latex]s[\/latex] is the arc length traveled.<\/li>\n<li>Connect them: Use [latex]s = r\\theta[\/latex] to link the two formulas, giving [latex]v = r\\omega[\/latex].<\/li>\n<li>Units Matter:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Linear speed: feet per second, meters per second, etc.<\/li>\n<li>Angular speed: radians per second (or per minute).<\/li>\n<\/ol>\n<\/li>\n<li>Think Real-World: On a spinning wheel, all points share the same angular speed, but points farther from the center move faster linearly.<\/li>\n<\/ol>\n<\/div>\n<p>&nbsp;<\/p>\n<div>\n<section class=\"textbox example\">A smartwatch second hand has length [latex]18[\/latex] mm. Find its angular speed [latex]\\omega[\/latex] rad\/s and tip speed [latex]v[\/latex] m\/s.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q138196\">Hint<\/button><\/p>\n<div id=\"q138196\" class=\"hidden-answer\" style=\"display: none\">[latex]\\omega = \\dfrac{2\\pi}{T}[\/latex] rad\/s where [latex]T[\/latex] is the period of one revolution.<br \/>\n[latex]\\begin{aligned} r &= 18 \\text { mm} \\\\\u00a0 &= 0.018 \\text { m} \\end{aligned}[\/latex]<br \/>\n[latex]v = r\\omega[\/latex]<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q6300\">Show Answer<\/button><\/p>\n<div id=\"q6300\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{aligned} \\omega &= \\dfrac{2\\pi}{60} \\\\ &= \\dfrac{\\pi}{30} \\\\ &\\approx 0.1047 \\text { rad\/s} \\end{aligned}[\/latex]<br \/>\n[latex]\\begin{aligned} v &= 0.018 \\cdot 0.1047 \\\\ &= 0.0018846 \\\\ &\\approx 1.885 \\cdot 10^{-3} \\text { m\/s} \\end{aligned}[\/latex]<\/div>\n<\/div>\n<\/section>\n<\/div>\n<p>&nbsp;<\/p>\n<div>\n<section class=\"textbox example\">A wind turbine (blade length [latex]27[\/latex] m spins at [latex]12[\/latex] RPM. Find [latex]\\omega[\/latex] rad\/s and the tip speed [latex]v[\/latex] m\/s.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q109771\">Hint<\/button><\/p>\n<div id=\"q109771\" class=\"hidden-answer\" style=\"display: none\">RPM: revolutions per minute<br \/>\n[latex]\\omega = \\text{RPM} \\cdot \\dfrac{2\\pi}{60} \\text { rad\/s}[\/latex]<br \/>\n[latex]v=r\\omega[\/latex]<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q367369\">Show Answer<\/button><\/p>\n<div id=\"q367369\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{aligned} \\omega &= \\dfrac{2\\pi}{5} \\text {rad\/s} \\\\ &\\approx 1.257 \\end{aligned}[\/latex]<br \/>\n[latex]\\begin{aligned} v &= 27 \\cdot 1.257 \\\\ &\\approx 33.93 \\text { m\/s} \\end{aligned}[\/latex]<\/div>\n<\/div>\n<\/section>\n<\/div>\n<p>&nbsp;<\/p>\n<div>\n<section class=\"textbox example\">In an aim-trainer, a circular targe rotates with angular speed [latex]1.8[\/latex] rad\/s. A marker sits [latex]0.75[\/latex] m from the center. Find its linear speed [latex]v[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q416218\">Hint<\/button><\/p>\n<div id=\"q416218\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{aligned} v &= r\\omega \\\\\u00a0 &= 0.75 \\cdot 1.8 \\end{aligned}[\/latex]<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q938851\">Show Answer<\/button><\/p>\n<div id=\"q938851\" class=\"hidden-answer\" style=\"display: none\">[latex]v = 1.35[\/latex] m\/s<\/div>\n<\/div>\n<\/section>\n<\/div>\n<p>&nbsp;<\/p>\n<section aria-label=\"Watch It\">\n<section class=\"textbox watchIt\" aria-label=\"Watch It\"><script type=\"text\/javascript\" src=\"https:\/\/www.youtube.com\/iframe_api\"><\/script><\/p>\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-bfdbahga-bfWkgA5GSE0\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/bfWkgA5GSE0?enablejsapi=1\" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<div id=\"3p-plugin-target-bfdbahga-bfWkgA5GSE0\" class=\"p3sdk-target\"><\/div>\n<p class=\"cc-media-iframe-container\"><script type=\"text\/javascript\" src=\"\/\/plugin.3playmedia.com\/ajax.js?cc=1&#38;cc_minimizable=1&#38;cc_minimize_on_load=0&#38;cc_multi_text_track=0&#38;cc_overlay=1&#38;cc_searchable=0&#38;embed=ajax&#38;mf=13892886&#38;p3sdk_version=1.11.7&#38;p=20361&#38;player_type=youtube&#38;plugin_skin=dark&#38;target=3p-plugin-target-bfdbahga-bfWkgA5GSE0&#38;vembed=0&#38;video_id=bfWkgA5GSE0&#38;video_target=tpm-plugin-bfdbahga-bfWkgA5GSE0\"><\/script><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Precalculus\/Transcripts\/Example+-+Determine+Angular+and+Linear+Velocity_transcript.txt\">transcript for &#8220;Example: Determine Angular and Linear Velocity&#8221; here (opens in new window).<\/a><\/p>\n<\/section>\n<\/section>\n","protected":false},"author":67,"menu_order":18,"template":"","meta":{"_candela_citation":"[{\"type\":\"copyrighted_video\",\"description\":\"Arc Length and Area of a Sector\",\"author\":\"\",\"organization\":\"Mathispower4u\",\"url\":\"https:\/\/youtu.be\/zD4CsKIYEHo\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube License\"},{\"type\":\"copyrighted_video\",\"description\":\"Example: Determine Angular and Linear Velocity\",\"author\":\"\",\"organization\":\"Mathispower4u\",\"url\":\"https:\/\/youtu.be\/bfWkgA5GSE0\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube License\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":178,"module-header":"fresh_take","content_attributions":[{"type":"copyrighted_video","description":"Arc Length and Area of a Sector","author":"","organization":"Mathispower4u","url":"https:\/\/youtu.be\/zD4CsKIYEHo","project":"","license":"arr","license_terms":"Standard YouTube License"},{"type":"copyrighted_video","description":"Example: Determine Angular and Linear Velocity","author":"","organization":"Mathispower4u","url":"https:\/\/youtu.be\/bfWkgA5GSE0","project":"","license":"arr","license_terms":"Standard YouTube License"}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"<script type='text\/javascript' src='https:\/\/www.youtube.com\/iframe_api'><\/script><script type='text\/javascript' src='\/\/plugin.3playmedia.com\/ajax.js?cc=1&cc_minimizable=1&cc_minimize_on_load=0&cc_multi_text_track=0&cc_overlay=1&cc_searchable=0&embed=ajax&mf=13892889&p3sdk_version=1.11.7&p=20361&player_type=youtube&plugin_skin=dark&target=3p-plugin-target-efedhghd-zD4CsKIYEHo&vembed=0&video_id=zD4CsKIYEHo&video_target=tpm-plugin-efedhghd-zD4CsKIYEHo'><\/script>\n<script type='text\/javascript' src='https:\/\/www.youtube.com\/iframe_api'><\/script><script type='text\/javascript' src='\/\/plugin.3playmedia.com\/ajax.js?cc=1&cc_minimizable=1&cc_minimize_on_load=0&cc_multi_text_track=0&cc_overlay=1&cc_searchable=0&embed=ajax&mf=13892889&p3sdk_version=1.11.7&p=20361&player_type=youtube&plugin_skin=dark&target=3p-plugin-target-eefbaeeb-zD4CsKIYEHo&vembed=0&video_id=zD4CsKIYEHo&video_target=tpm-plugin-eefbaeeb-zD4CsKIYEHo'><\/script>\n<script type='text\/javascript' src='https:\/\/www.youtube.com\/iframe_api'><\/script><script type='text\/javascript' src='\/\/plugin.3playmedia.com\/ajax.js?cc=1&cc_minimizable=1&cc_minimize_on_load=0&cc_multi_text_track=0&cc_overlay=1&cc_searchable=0&embed=ajax&mf=13892886&p3sdk_version=1.11.7&p=20361&player_type=youtube&plugin_skin=dark&target=3p-plugin-target-bfdbahga-bfWkgA5GSE0&vembed=0&video_id=bfWkgA5GSE0&video_target=tpm-plugin-bfdbahga-bfWkgA5GSE0'><\/script>\n","media_targets":["tpm-plugin-efedhghd-zD4CsKIYEHo","tpm-plugin-eefbaeeb-zD4CsKIYEHo","tpm-plugin-bfdbahga-bfWkgA5GSE0"]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1507"}],"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":125,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1507\/revisions"}],"predecessor-version":[{"id":5972,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1507\/revisions\/5972"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/178"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1507\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=1507"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=1507"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=1507"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=1507"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}