{"id":777,"date":"2025-10-09T15:38:08","date_gmt":"2025-10-09T15:38:08","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=777"},"modified":"2025-11-20T18:42:00","modified_gmt":"2025-11-20T18:42:00","slug":"advanced-vector-calculus-background-youll-need-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/advanced-vector-calculus-background-youll-need-3\/","title":{"raw":"Advanced Vector Calculus: Background You'll Need 3","rendered":"Advanced Vector Calculus: Background You&#8217;ll Need 3"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Create parametric equations for curves and use them to calculate arc length.<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Derivatives of Parametric Equations<\/h2>\r\nWhen working with parametric equations [latex]x = x(t)[\/latex] and [latex]y = y(t)[\/latex], you can find the slope of the curve without eliminating the parameter.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>theorem: derivative of parametric equations<\/h3>\r\n<p id=\"fs-id1167793961378\">For parametric equations [latex]x = x(t)[\/latex] and [latex]y = y(t)[\/latex]:<\/p>\r\n\r\n<div id=\"fs-id1167793881791\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\frac{dy}{dx}=\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}=\\frac{{y}^{\\prime }\\left(t\\right)}{{x}^{\\prime }\\left(t\\right)}[\/latex].<\/div>\r\n<div data-type=\"equation\">Important condition: [latex]x'(t) \\neq 0[\/latex]<\/div>\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">\r\n<p class=\"whitespace-normal break-words\"><strong>Why This Formula Works<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Think of the chain rule: As [latex]t[\/latex] changes, both [latex]x[\/latex] and [latex]y[\/latex] change. The ratio of their rates of change gives us the slope of the curve.<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">This is particularly useful when eliminating the parameter would result in a complicated expression!<\/p>\r\n\r\n<\/section>The parametric derivative formula works for <strong>any<\/strong> parametric curve\u2014even curves that loop or cross themselves and can't be written as [latex]y = f(x)[\/latex].\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">Recall that a critical point of a differentiable function [latex]y=f\\left(x\\right)[\/latex] is any point [latex]x={x}_{0}[\/latex] such that either [latex]{f}^{\\prime }\\left({x}_{0}\\right)=0[\/latex] or [latex]{f}^{\\prime }\\left({x}_{0}\\right)[\/latex] does not exist.<\/section>\r\n<p class=\"whitespace-normal break-words\">The parametric approach gives us the slope at any point along the curve, regardless of how complicated the path might be.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1167794048755\" data-type=\"problem\">\r\n<p id=\"fs-id1167793895721\">Calculate the derivative [latex]\\frac{dy}{dx}[\/latex] for each of the following parametrically defined plane curves, and locate any critical points on their respective graphs.<\/p>\r\n\r\n<ol id=\"fs-id1167793998146\" type=\"a\">\r\n \t<li>[latex]x\\left(t\\right)={t}^{2}-3,y\\left(t\\right)=2t - 1,-3\\le t\\le 4[\/latex]<\/li>\r\n \t<li>[latex]x\\left(t\\right)=2t+1,y\\left(t\\right)={t}^{3}-3t+4,-2\\le t\\le 5[\/latex]<\/li>\r\n \t<li>[latex]x\\left(t\\right)=5\\cos{t},y\\left(t\\right)=5\\sin{t},0\\le t\\le 2\\pi [\/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-id1167793983758\" data-type=\"solution\">\r\n<ol id=\"fs-id1167793847325\" type=\"a\">\r\n \t<li>To apply the theorem , first calculate [latex]{x}^{\\prime }\\left(t\\right)[\/latex] and [latex]{y}^{\\prime }\\left(t\\right)\\text{:}[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167794072479\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{x}^{\\prime }\\left(t\\right)=2t\\hfill \\\\ {y}^{\\prime }\\left(t\\right)=2.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nNext substitute these into the equation:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167794068421\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}\\frac{dy}{dx}=\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}\\hfill \\\\ \\frac{dy}{dx}=\\frac{2}{2t}\\hfill \\\\ \\frac{dy}{dx}=\\frac{1}{t}.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThis derivative is undefined when [latex]t=0[\/latex]. Calculating [latex]x\\left(0\\right)[\/latex] and [latex]y\\left(0\\right)[\/latex] gives [latex]x\\left(0\\right)={\\left(0\\right)}^{2}-3=-3[\/latex] and [latex]y\\left(0\\right)=2\\left(0\\right)-1=-1[\/latex], which corresponds to the point [latex]\\left(-3,-1\\right)[\/latex] on the graph. The graph of this curve is a parabola opening to the right, and the point [latex]\\left(-3,-1\\right)[\/latex] is its vertex as shown.<span data-type=\"newline\">\r\n<\/span>\r\n<figure id=\"CNX_Calc_Figure_11_02_002\">\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\/11234741\/CNX_Calc_Figure_11_02_002.jpg\" alt=\"A curved line going from (6, \u22127) through (\u22123, \u22121) to (13, 7) with arrow pointing in that order. The point (6, \u22127) is marked t = \u22123, the point (\u22123, \u22121) is marked t = 0, and the point (13, 7) is marked t = 4. On the graph there are also written three equations: x(t) = t2 \u2212 3, y(t) = 2t \u2212 1, and \u22123 \u2264 t \u2264 4.\" width=\"417\" height=\"347\" data-media-type=\"image\/jpeg\" \/> Figure 2. Graph of the parabola described by parametric equations in part a.[\/caption]<\/figure>\r\n<\/li>\r\n \t<li>To apply the theorem, first calculate [latex]{x}^{\\prime }\\left(t\\right)[\/latex] and [latex]{y}^{\\prime }\\left(t\\right)\\text{:}[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167794060790\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{x}^{\\prime }\\left(t\\right)=2\\hfill \\\\ {y}^{\\prime }\\left(t\\right)=3{t}^{2}-3.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nNext substitute these into the equation:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167794098234\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}\\frac{dy}{dx}=\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}\\hfill \\\\ \\frac{dy}{dx}=\\frac{3{t}^{2}-3}{2}.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThis derivative is zero when [latex]t=\\pm 1[\/latex]. When [latex]t=-1[\/latex] we have<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793870851\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(-1\\right)=2\\left(-1\\right)+1=-1\\text{and}y\\left(-1\\right)={\\left(-1\\right)}^{3}-3\\left(-1\\right)+4=-1+3+4=6[\/latex],<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nwhich corresponds to the point [latex]\\left(-1,6\\right)[\/latex] on the graph. When [latex]t=1[\/latex] we have<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167794137428\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(1\\right)=2\\left(1\\right)+1=3\\text{and}y\\left(1\\right)={\\left(1\\right)}^{3}-3\\left(1\\right)+4=1 - 3+4=2[\/latex],<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nwhich corresponds to the point [latex]\\left(3,2\\right)[\/latex] on the graph. The point [latex]\\left(3,2\\right)[\/latex] is a relative minimum and the point [latex]\\left(-1,6\\right)[\/latex] is a relative maximum, as seen in the following graph.<span data-type=\"newline\">\r\n<\/span>\r\n<figure id=\"CNX_Calc_Figure_11_02_003\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"379\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234744\/CNX_Calc_Figure_11_02_003.jpg\" alt=\"A vaguely sinusoidal curve going from (\u22123, 2) through (\u22121, 6) and (3, 2) to (5, 6). The point (\u22123, 2) is marked t = \u22122, the point (\u22121, 6) is marked t = \u22121, the point (3, 2) is marked t = 1, and the point (5, 6) is marked t = 2. On the graph there are also written three equations: x(t) = 2t + 1, y(t) = t3 \u2013 3t + 4, and \u22122 \u2264 t \u2264 2.\" width=\"379\" height=\"347\" data-media-type=\"image\/jpeg\" \/> Figure 3. Graph of the curve described by parametric equations in part b.[\/caption]<\/figure>\r\n<\/li>\r\n \t<li>To apply the theorem, first calculate [latex]{x}^{\\prime }\\left(t\\right)[\/latex] and [latex]{y}^{\\prime }\\left(t\\right)\\text{:}[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167794140205\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{x}^{\\prime }\\left(t\\right)=-5\\sin{t}\\hfill \\\\ {y}^{\\prime }\\left(t\\right)=5\\cos{t}.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nNext substitute these into the equation:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167794098876\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}\\frac{dy}{dx}=\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}\\hfill \\\\ \\frac{dy}{dx}=\\frac{5\\cos{t}}{-5\\sin{t}}\\hfill \\\\ \\frac{dy}{dx}=-\\cot{t}.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThis derivative is zero when [latex]\\cos{t}=0[\/latex] and is undefined when [latex]\\sin{t}=0[\/latex]. This gives [latex]t=0,\\frac{\\pi }{2},\\pi ,\\frac{3\\pi }{2},\\text{and}2\\pi [\/latex] as critical points for <em data-effect=\"italics\">t.<\/em> Substituting each of these into [latex]x\\left(t\\right)[\/latex] and [latex]y\\left(t\\right)[\/latex], we obtain<span data-type=\"newline\">\r\n<\/span>\r\n<table id=\"fs-id1167793999790\" class=\"unnumbered\" summary=\"This table has three columns and six rows. The first row is a header row, and it reads from left to right t, x(t), and y(t). Below the header row, in the first column, the values read 0, \u03c0\/2, \u03c0, 3\u03c0\/2, and 2\u03c0. In the second column, the values read 5, 0, \u22125, 0, and 5. In the third column, the values read 0, 5, 0, \u22125, and 0.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th data-valign=\"top\" data-align=\"left\">[latex]t[\/latex]<\/th>\r\n<th data-valign=\"top\" data-align=\"left\">[latex]x\\left(t\\right)[\/latex]<\/th>\r\n<th data-valign=\"top\" data-align=\"left\">[latex]y\\left(t\\right)[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td data-valign=\"top\" data-align=\"left\">0<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">5<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">0<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-valign=\"top\" data-align=\"left\">[latex]\\frac{\\pi }{2}[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">0<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">5<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-valign=\"top\" data-align=\"left\">[latex]\\pi [\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">\u22125<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">0<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-valign=\"top\" data-align=\"left\">[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">0<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">\u22125<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-valign=\"top\" data-align=\"left\">[latex]2\\pi [\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">5<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThese points correspond to the sides, top, and bottom of the circle that is represented by the parametric equations (Figure 4). On the left and right edges of the circle, the derivative is undefined, and on the top and bottom, the derivative equals zero.<span data-type=\"newline\">\r\n<\/span>\r\n<figure id=\"CNX_Calc_Figure_11_02_004\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"490\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234746\/CNX_Calc_Figure_11_02_004.jpg\" alt=\"A circle with radius 5 centered at the origin is graphed with arrow going counterclockwise. The point (5, 0) is marked t = 0, the point (0, 5) is marked t = \u03c0\/2, the point (\u22125, 0) is marked t = \u03c0, and the point (0, \u22125) is marked t = 3\u03c0\/2. On the graph there are also written three equations: x(t) = 5 cos(t), y(t) = 5 sin(t), and 0 \u2264 t \u2264 2\u03c0.\" width=\"490\" height=\"497\" data-media-type=\"image\/jpeg\" \/> Figure 4. Graph of the curve described by parametric equations in part c.[\/caption]<\/figure>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\nWatch the following video to see the worked solution to the example above.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/bDvDdC2Wpu8?controls=0&amp;start=71&amp;end=465&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/7.2CalculusOfParametricCurves71to465_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"7.2 Calculus of Parametric Curves\" here (opens in new window)<\/a>.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1167794042397\" data-type=\"problem\">\r\n<p id=\"fs-id1167794044541\">Find the equation of the tangent line to the curve defined by the equations<\/p>\r\n\r\n<div id=\"fs-id1167794044544\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)={t}^{2}-3,y\\left(t\\right)=2t - 1,-3\\le t\\le 4\\text{ when }t=2[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n<div id=\"fs-id1167794090791\" data-type=\"solution\">\r\n<p id=\"fs-id1167794090794\">First find the slope of the tangent line using the theorem, which means calculating [latex]{x}^{\\prime }\\left(t\\right)[\/latex] and [latex]{y}^{\\prime }\\left(t\\right)\\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1167794071078\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{x}^{\\prime }\\left(t\\right)=2t\\hfill \\\\ {y}^{\\prime }\\left(t\\right)=2.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794098817\">Next substitute these into the equation:<\/p>\r\n\r\n<div id=\"fs-id1167794098820\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}\\frac{dy}{dx}=\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}\\hfill \\\\ \\frac{dy}{dx}=\\frac{2}{2t}\\hfill \\\\ \\frac{dy}{dx}=\\frac{1}{t}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794068318\">When [latex]t=2[\/latex], [latex]\\frac{dy}{dx}=\\frac{1}{2}[\/latex], so this is the slope of the tangent line. Calculating [latex]x\\left(2\\right)[\/latex] and [latex]y\\left(2\\right)[\/latex] gives<\/p>\r\n\r\n<div id=\"fs-id1167794028572\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(2\\right)={\\left(2\\right)}^{2}-3=1\\text{ and }y\\left(2\\right)=2\\left(2\\right)-1=3[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794098767\">which corresponds to the point [latex]\\left(1,3\\right)[\/latex] on the graph (Figure 6). Now use the point-slope form of the equation of a line to find the equation of the tangent line:<\/p>\r\n\r\n<div id=\"fs-id1167794098790\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill y-{y}_{0}&amp; =\\hfill &amp; m\\left(x-{x}_{0}\\right)\\hfill \\\\ \\hfill y - 3&amp; =\\hfill &amp; \\frac{1}{2}\\left(x - 1\\right)\\hfill \\\\ \\hfill y - 3&amp; =\\hfill &amp; \\frac{1}{2}x-\\frac{1}{2}\\hfill \\\\ \\hfill y&amp; =\\hfill &amp; \\frac{1}{2}x+\\frac{5}{2}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<figure id=\"CNX_Calc_Figure_11_02_006\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"379\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234753\/CNX_Calc_Figure_11_02_006.jpg\" alt=\"A curved line going from (6, \u22127) through (\u22123, \u22121) to (13, 7) with arrow pointing in that order. The point (6, \u22127) is marked t = \u22123, the point (\u22123, \u22121) is marked t = 0, and the point (13, 7) is marked t = 4. On the graph there are also written three equations: x(t) = t2 \u2212 3, y(t) = 2t \u2212 1, and \u22123 \u2264 t \u2264 4. At the point (1, 3), which is marked t = 2, there is a tangent line with equation y = x\/2 + 5\/2.\" width=\"379\" height=\"383\" data-media-type=\"image\/jpeg\" \/> Figure 6. Tangent line to the parabola described by the given parametric equations when [latex]t=2[\/latex].[\/caption]<\/figure>\r\n<\/div>\r\nWatch the following video to see the worked solution to the example above.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/bDvDdC2Wpu8?controls=0&amp;start=468&amp;end=581&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/7.2CalculusOfParametricCurves468to581_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"7.2 Calculus of Parametric Curves\" here (opens in new window)<\/a>.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]311317[\/ohm_question]<\/section><section aria-label=\"Try It\">\r\n<h2 data-type=\"title\">Arc Length of a Parametric Curve<\/h2>\r\nArc length measures the actual distance traveled along a curve. If a particle moves from point A to point B along a parametric path, the arc length tells you exactly how far it traveled.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>theorem: arc length of a parametric curve<\/h3>\r\n<p id=\"fs-id1167794065184\">For a plane curve defined by [latex]x = x(t)[\/latex], [latex]y = y(t)[\/latex] where [latex]t_1 \\le t \\le t_2[\/latex], with [latex]x(t)[\/latex] and [latex]y(t)[\/latex] differentiable then:<\/p>\r\n\r\n<div id=\"fs-id1167794023197\" style=\"text-align: center;\" data-type=\"equation\">[latex]s={\\displaystyle\\int }_{{t}_{1}}^{{t}_{2}}\\sqrt{{\\left(\\frac{dx}{dt}\\right)}^{2}+{\\left(\\frac{dy}{dt}\\right)}^{2}}dt[\/latex].<\/div>\r\n<div data-type=\"equation\">This formula gives the exact <strong>arc length of the parametric curve<\/strong>.<\/div>\r\n<\/section>\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How to: Calculate Arc Length<\/strong>\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Find [latex]\\frac{dx}{dt}[\/latex] and [latex]\\frac{dy}{dt}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Square both derivatives<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Add the squared derivatives under a square root<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Integrate with respect to [latex]t[\/latex] over the given interval<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Use trigonometric identities when needed (like [latex]\\sin^2 t + \\cos^2 t = 1[\/latex])<\/li>\r\n<\/ol>\r\n<\/section>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1167794100419\" data-type=\"problem\">\r\n<p id=\"fs-id1167794100424\">Find the arc length of the semicircle defined by the equations<\/p>\r\n\r\n<div id=\"fs-id1167794100427\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)=3\\cos{t},y\\left(t\\right)=3\\sin{t},0\\le t\\le \\pi [\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558859\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558859\"]\r\n<div id=\"fs-id1167794100493\" data-type=\"solution\">\r\n<p id=\"fs-id1167794100495\">The values [latex]t=0[\/latex] to [latex]t=\\pi [\/latex] trace out the red curve in Figure 9. To determine its length, use the theorem:<\/p>\r\n\r\n<div id=\"fs-id1167794061688\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill s&amp; ={\\displaystyle\\int }_{{t}_{1}}^{{t}_{2}}\\sqrt{{\\left(\\frac{dx}{dt}\\right)}^{2}+{\\left(\\frac{dy}{dt}\\right)}^{2}}dt\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{\\pi }\\sqrt{{\\left(-3\\sin{t}\\right)}^{2}+{\\left(3\\cos{t}\\right)}^{2}}dt\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{\\pi }\\sqrt{9{\\sin}^{2}t+9{\\cos}^{2}t}dt\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{\\pi }\\sqrt{9\\left({\\sin}^{2}t+{\\cos}^{2}t\\right)}dt\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{\\pi }3dt={3t|}_{0}^{\\pi }=3\\pi .\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794051248\">Note that the formula for the arc length of a semicircle is [latex]\\pi r[\/latex] and the radius of this circle is 3. This is a great example of using calculus to derive a known formula of a geometric quantity.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_02_010\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"339\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234801\/CNX_Calc_Figure_11_02_009.jpg\" alt=\"A semicircle is drawn with radius 3. There is an arrow pointing counterclockwise. On the graph there are also written three equations: x(t) = 3 cos(t), y(t) = 3 sin(t), and 0 \u2264 t \u2264 \u03c0.\" width=\"339\" height=\"347\" data-media-type=\"image\/jpeg\" \/> Figure 10. The arc length of the semicircle is equal to its radius times [latex]\\pi[\/latex].[\/caption]<\/figure>\r\n<\/div>\r\nWatch the following video to see the worked solution to the example above.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/bDvDdC2Wpu8?controls=0&amp;start=893&amp;end=993&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/7.2CalculusOfParametricCurves893to993_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"7.2 Calculus of Parametric Curves\" here (opens in new window)<\/a>.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]311320[\/ohm_question]<\/section><\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li><span data-sheets-root=\"1\">Create parametric equations for curves and use them to calculate arc length.<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>Derivatives of Parametric Equations<\/h2>\n<p>When working with parametric equations [latex]x = x(t)[\/latex] and [latex]y = y(t)[\/latex], you can find the slope of the curve without eliminating the parameter.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>theorem: derivative of parametric equations<\/h3>\n<p id=\"fs-id1167793961378\">For parametric equations [latex]x = x(t)[\/latex] and [latex]y = y(t)[\/latex]:<\/p>\n<div id=\"fs-id1167793881791\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\frac{dy}{dx}=\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}=\\frac{{y}^{\\prime }\\left(t\\right)}{{x}^{\\prime }\\left(t\\right)}[\/latex].<\/div>\n<div data-type=\"equation\">Important condition: [latex]x'(t) \\neq 0[\/latex]<\/div>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">\n<p class=\"whitespace-normal break-words\"><strong>Why This Formula Works<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Think of the chain rule: As [latex]t[\/latex] changes, both [latex]x[\/latex] and [latex]y[\/latex] change. The ratio of their rates of change gives us the slope of the curve.<\/p>\n<p class=\"whitespace-pre-wrap break-words\">This is particularly useful when eliminating the parameter would result in a complicated expression!<\/p>\n<\/section>\n<p>The parametric derivative formula works for <strong>any<\/strong> parametric curve\u2014even curves that loop or cross themselves and can&#8217;t be written as [latex]y = f(x)[\/latex].<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">Recall that a critical point of a differentiable function [latex]y=f\\left(x\\right)[\/latex] is any point [latex]x={x}_{0}[\/latex] such that either [latex]{f}^{\\prime }\\left({x}_{0}\\right)=0[\/latex] or [latex]{f}^{\\prime }\\left({x}_{0}\\right)[\/latex] does not exist.<\/section>\n<p class=\"whitespace-normal break-words\">The parametric approach gives us the slope at any point along the curve, regardless of how complicated the path might be.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167794048755\" data-type=\"problem\">\n<p id=\"fs-id1167793895721\">Calculate the derivative [latex]\\frac{dy}{dx}[\/latex] for each of the following parametrically defined plane curves, and locate any critical points on their respective graphs.<\/p>\n<ol id=\"fs-id1167793998146\" type=\"a\">\n<li>[latex]x\\left(t\\right)={t}^{2}-3,y\\left(t\\right)=2t - 1,-3\\le t\\le 4[\/latex]<\/li>\n<li>[latex]x\\left(t\\right)=2t+1,y\\left(t\\right)={t}^{3}-3t+4,-2\\le t\\le 5[\/latex]<\/li>\n<li>[latex]x\\left(t\\right)=5\\cos{t},y\\left(t\\right)=5\\sin{t},0\\le t\\le 2\\pi[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558899\">Show Solution<\/button><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167793983758\" data-type=\"solution\">\n<ol id=\"fs-id1167793847325\" type=\"a\">\n<li>To apply the theorem , first calculate [latex]{x}^{\\prime }\\left(t\\right)[\/latex] and [latex]{y}^{\\prime }\\left(t\\right)\\text{:}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167794072479\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{x}^{\\prime }\\left(t\\right)=2t\\hfill \\\\ {y}^{\\prime }\\left(t\\right)=2.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nNext substitute these into the equation:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167794068421\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}\\frac{dy}{dx}=\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}\\hfill \\\\ \\frac{dy}{dx}=\\frac{2}{2t}\\hfill \\\\ \\frac{dy}{dx}=\\frac{1}{t}.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThis derivative is undefined when [latex]t=0[\/latex]. Calculating [latex]x\\left(0\\right)[\/latex] and [latex]y\\left(0\\right)[\/latex] gives [latex]x\\left(0\\right)={\\left(0\\right)}^{2}-3=-3[\/latex] and [latex]y\\left(0\\right)=2\\left(0\\right)-1=-1[\/latex], which corresponds to the point [latex]\\left(-3,-1\\right)[\/latex] on the graph. The graph of this curve is a parabola opening to the right, and the point [latex]\\left(-3,-1\\right)[\/latex] is its vertex as shown.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<figure id=\"CNX_Calc_Figure_11_02_002\">\n<figure style=\"width: 417px\" 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\/11234741\/CNX_Calc_Figure_11_02_002.jpg\" alt=\"A curved line going from (6, \u22127) through (\u22123, \u22121) to (13, 7) with arrow pointing in that order. The point (6, \u22127) is marked t = \u22123, the point (\u22123, \u22121) is marked t = 0, and the point (13, 7) is marked t = 4. On the graph there are also written three equations: x(t) = t2 \u2212 3, y(t) = 2t \u2212 1, and \u22123 \u2264 t \u2264 4.\" width=\"417\" height=\"347\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 2. Graph of the parabola described by parametric equations in part a.<\/figcaption><\/figure>\n<\/figure>\n<\/li>\n<li>To apply the theorem, first calculate [latex]{x}^{\\prime }\\left(t\\right)[\/latex] and [latex]{y}^{\\prime }\\left(t\\right)\\text{:}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167794060790\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{x}^{\\prime }\\left(t\\right)=2\\hfill \\\\ {y}^{\\prime }\\left(t\\right)=3{t}^{2}-3.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nNext substitute these into the equation:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167794098234\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}\\frac{dy}{dx}=\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}\\hfill \\\\ \\frac{dy}{dx}=\\frac{3{t}^{2}-3}{2}.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThis derivative is zero when [latex]t=\\pm 1[\/latex]. When [latex]t=-1[\/latex] we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793870851\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(-1\\right)=2\\left(-1\\right)+1=-1\\text{and}y\\left(-1\\right)={\\left(-1\\right)}^{3}-3\\left(-1\\right)+4=-1+3+4=6[\/latex],<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nwhich corresponds to the point [latex]\\left(-1,6\\right)[\/latex] on the graph. When [latex]t=1[\/latex] we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167794137428\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(1\\right)=2\\left(1\\right)+1=3\\text{and}y\\left(1\\right)={\\left(1\\right)}^{3}-3\\left(1\\right)+4=1 - 3+4=2[\/latex],<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nwhich corresponds to the point [latex]\\left(3,2\\right)[\/latex] on the graph. The point [latex]\\left(3,2\\right)[\/latex] is a relative minimum and the point [latex]\\left(-1,6\\right)[\/latex] is a relative maximum, as seen in the following graph.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<figure id=\"CNX_Calc_Figure_11_02_003\">\n<figure style=\"width: 379px\" 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\/11234744\/CNX_Calc_Figure_11_02_003.jpg\" alt=\"A vaguely sinusoidal curve going from (\u22123, 2) through (\u22121, 6) and (3, 2) to (5, 6). The point (\u22123, 2) is marked t = \u22122, the point (\u22121, 6) is marked t = \u22121, the point (3, 2) is marked t = 1, and the point (5, 6) is marked t = 2. On the graph there are also written three equations: x(t) = 2t + 1, y(t) = t3 \u2013 3t + 4, and \u22122 \u2264 t \u2264 2.\" width=\"379\" height=\"347\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 3. Graph of the curve described by parametric equations in part b.<\/figcaption><\/figure>\n<\/figure>\n<\/li>\n<li>To apply the theorem, first calculate [latex]{x}^{\\prime }\\left(t\\right)[\/latex] and [latex]{y}^{\\prime }\\left(t\\right)\\text{:}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167794140205\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{x}^{\\prime }\\left(t\\right)=-5\\sin{t}\\hfill \\\\ {y}^{\\prime }\\left(t\\right)=5\\cos{t}.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nNext substitute these into the equation:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167794098876\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}\\frac{dy}{dx}=\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}\\hfill \\\\ \\frac{dy}{dx}=\\frac{5\\cos{t}}{-5\\sin{t}}\\hfill \\\\ \\frac{dy}{dx}=-\\cot{t}.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThis derivative is zero when [latex]\\cos{t}=0[\/latex] and is undefined when [latex]\\sin{t}=0[\/latex]. This gives [latex]t=0,\\frac{\\pi }{2},\\pi ,\\frac{3\\pi }{2},\\text{and}2\\pi[\/latex] as critical points for <em data-effect=\"italics\">t.<\/em> Substituting each of these into [latex]x\\left(t\\right)[\/latex] and [latex]y\\left(t\\right)[\/latex], we obtain<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<table id=\"fs-id1167793999790\" class=\"unnumbered\" summary=\"This table has three columns and six rows. The first row is a header row, and it reads from left to right t, x(t), and y(t). Below the header row, in the first column, the values read 0, \u03c0\/2, \u03c0, 3\u03c0\/2, and 2\u03c0. In the second column, the values read 5, 0, \u22125, 0, and 5. In the third column, the values read 0, 5, 0, \u22125, and 0.\">\n<thead>\n<tr valign=\"top\">\n<th data-valign=\"top\" data-align=\"left\">[latex]t[\/latex]<\/th>\n<th data-valign=\"top\" data-align=\"left\">[latex]x\\left(t\\right)[\/latex]<\/th>\n<th data-valign=\"top\" data-align=\"left\">[latex]y\\left(t\\right)[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td data-valign=\"top\" data-align=\"left\">0<\/td>\n<td data-valign=\"top\" data-align=\"left\">5<\/td>\n<td data-valign=\"top\" data-align=\"left\">0<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-valign=\"top\" data-align=\"left\">[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">0<\/td>\n<td data-valign=\"top\" data-align=\"left\">5<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-valign=\"top\" data-align=\"left\">[latex]\\pi[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">\u22125<\/td>\n<td data-valign=\"top\" data-align=\"left\">0<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-valign=\"top\" data-align=\"left\">[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">0<\/td>\n<td data-valign=\"top\" data-align=\"left\">\u22125<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-valign=\"top\" data-align=\"left\">[latex]2\\pi[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">5<\/td>\n<td data-valign=\"top\" data-align=\"left\">0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThese points correspond to the sides, top, and bottom of the circle that is represented by the parametric equations (Figure 4). On the left and right edges of the circle, the derivative is undefined, and on the top and bottom, the derivative equals zero.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<figure id=\"CNX_Calc_Figure_11_02_004\">\n<figure style=\"width: 490px\" 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\/11234746\/CNX_Calc_Figure_11_02_004.jpg\" alt=\"A circle with radius 5 centered at the origin is graphed with arrow going counterclockwise. The point (5, 0) is marked t = 0, the point (0, 5) is marked t = \u03c0\/2, the point (\u22125, 0) is marked t = \u03c0, and the point (0, \u22125) is marked t = 3\u03c0\/2. On the graph there are also written three equations: x(t) = 5 cos(t), y(t) = 5 sin(t), and 0 \u2264 t \u2264 2\u03c0.\" width=\"490\" height=\"497\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 4. Graph of the curve described by parametric equations in part c.<\/figcaption><\/figure>\n<\/figure>\n<\/li>\n<\/ol>\n<\/div>\n<p>Watch the following video to see the worked solution to the example above.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/bDvDdC2Wpu8?controls=0&amp;start=71&amp;end=465&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/7.2CalculusOfParametricCurves71to465_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;7.2 Calculus of Parametric Curves&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167794042397\" data-type=\"problem\">\n<p id=\"fs-id1167794044541\">Find the equation of the tangent line to the curve defined by the equations<\/p>\n<div id=\"fs-id1167794044544\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)={t}^{2}-3,y\\left(t\\right)=2t - 1,-3\\le t\\le 4\\text{ when }t=2[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558896\">Show Solution<\/button><\/p>\n<div id=\"q44558896\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794090791\" data-type=\"solution\">\n<p id=\"fs-id1167794090794\">First find the slope of the tangent line using the theorem, which means calculating [latex]{x}^{\\prime }\\left(t\\right)[\/latex] and [latex]{y}^{\\prime }\\left(t\\right)\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1167794071078\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{x}^{\\prime }\\left(t\\right)=2t\\hfill \\\\ {y}^{\\prime }\\left(t\\right)=2.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794098817\">Next substitute these into the equation:<\/p>\n<div id=\"fs-id1167794098820\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}\\frac{dy}{dx}=\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}\\hfill \\\\ \\frac{dy}{dx}=\\frac{2}{2t}\\hfill \\\\ \\frac{dy}{dx}=\\frac{1}{t}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794068318\">When [latex]t=2[\/latex], [latex]\\frac{dy}{dx}=\\frac{1}{2}[\/latex], so this is the slope of the tangent line. Calculating [latex]x\\left(2\\right)[\/latex] and [latex]y\\left(2\\right)[\/latex] gives<\/p>\n<div id=\"fs-id1167794028572\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(2\\right)={\\left(2\\right)}^{2}-3=1\\text{ and }y\\left(2\\right)=2\\left(2\\right)-1=3[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794098767\">which corresponds to the point [latex]\\left(1,3\\right)[\/latex] on the graph (Figure 6). Now use the point-slope form of the equation of a line to find the equation of the tangent line:<\/p>\n<div id=\"fs-id1167794098790\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill y-{y}_{0}& =\\hfill & m\\left(x-{x}_{0}\\right)\\hfill \\\\ \\hfill y - 3& =\\hfill & \\frac{1}{2}\\left(x - 1\\right)\\hfill \\\\ \\hfill y - 3& =\\hfill & \\frac{1}{2}x-\\frac{1}{2}\\hfill \\\\ \\hfill y& =\\hfill & \\frac{1}{2}x+\\frac{5}{2}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<figure id=\"CNX_Calc_Figure_11_02_006\"><figcaption><\/figcaption><figure style=\"width: 379px\" 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\/11234753\/CNX_Calc_Figure_11_02_006.jpg\" alt=\"A curved line going from (6, \u22127) through (\u22123, \u22121) to (13, 7) with arrow pointing in that order. The point (6, \u22127) is marked t = \u22123, the point (\u22123, \u22121) is marked t = 0, and the point (13, 7) is marked t = 4. On the graph there are also written three equations: x(t) = t2 \u2212 3, y(t) = 2t \u2212 1, and \u22123 \u2264 t \u2264 4. At the point (1, 3), which is marked t = 2, there is a tangent line with equation y = x\/2 + 5\/2.\" width=\"379\" height=\"383\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 6. Tangent line to the parabola described by the given parametric equations when [latex]t=2[\/latex].<\/figcaption><\/figure>\n<\/figure>\n<\/div>\n<p>Watch the following video to see the worked solution to the example above.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/bDvDdC2Wpu8?controls=0&amp;start=468&amp;end=581&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/7.2CalculusOfParametricCurves468to581_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;7.2 Calculus of Parametric Curves&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm311317\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=311317&theme=lumen&iframe_resize_id=ohm311317&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section aria-label=\"Try It\">\n<h2 data-type=\"title\">Arc Length of a Parametric Curve<\/h2>\n<p>Arc length measures the actual distance traveled along a curve. If a particle moves from point A to point B along a parametric path, the arc length tells you exactly how far it traveled.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>theorem: arc length of a parametric curve<\/h3>\n<p id=\"fs-id1167794065184\">For a plane curve defined by [latex]x = x(t)[\/latex], [latex]y = y(t)[\/latex] where [latex]t_1 \\le t \\le t_2[\/latex], with [latex]x(t)[\/latex] and [latex]y(t)[\/latex] differentiable then:<\/p>\n<div id=\"fs-id1167794023197\" style=\"text-align: center;\" data-type=\"equation\">[latex]s={\\displaystyle\\int }_{{t}_{1}}^{{t}_{2}}\\sqrt{{\\left(\\frac{dx}{dt}\\right)}^{2}+{\\left(\\frac{dy}{dt}\\right)}^{2}}dt[\/latex].<\/div>\n<div data-type=\"equation\">This formula gives the exact <strong>arc length of the parametric curve<\/strong>.<\/div>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How to: Calculate Arc Length<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Find [latex]\\frac{dx}{dt}[\/latex] and [latex]\\frac{dy}{dt}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Square both derivatives<\/li>\n<li class=\"whitespace-normal break-words\">Add the squared derivatives under a square root<\/li>\n<li class=\"whitespace-normal break-words\">Integrate with respect to [latex]t[\/latex] over the given interval<\/li>\n<li class=\"whitespace-normal break-words\">Use trigonometric identities when needed (like [latex]\\sin^2 t + \\cos^2 t = 1[\/latex])<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167794100419\" data-type=\"problem\">\n<p id=\"fs-id1167794100424\">Find the arc length of the semicircle defined by the equations<\/p>\n<div id=\"fs-id1167794100427\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)=3\\cos{t},y\\left(t\\right)=3\\sin{t},0\\le t\\le \\pi[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558859\">Show Solution<\/button><\/p>\n<div id=\"q44558859\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794100493\" data-type=\"solution\">\n<p id=\"fs-id1167794100495\">The values [latex]t=0[\/latex] to [latex]t=\\pi[\/latex] trace out the red curve in Figure 9. To determine its length, use the theorem:<\/p>\n<div id=\"fs-id1167794061688\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill s& ={\\displaystyle\\int }_{{t}_{1}}^{{t}_{2}}\\sqrt{{\\left(\\frac{dx}{dt}\\right)}^{2}+{\\left(\\frac{dy}{dt}\\right)}^{2}}dt\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{\\pi }\\sqrt{{\\left(-3\\sin{t}\\right)}^{2}+{\\left(3\\cos{t}\\right)}^{2}}dt\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{\\pi }\\sqrt{9{\\sin}^{2}t+9{\\cos}^{2}t}dt\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{\\pi }\\sqrt{9\\left({\\sin}^{2}t+{\\cos}^{2}t\\right)}dt\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{\\pi }3dt={3t|}_{0}^{\\pi }=3\\pi .\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794051248\">Note that the formula for the arc length of a semicircle is [latex]\\pi r[\/latex] and the radius of this circle is 3. This is a great example of using calculus to derive a known formula of a geometric quantity.<\/p>\n<figure id=\"CNX_Calc_Figure_11_02_010\"><figcaption><\/figcaption><figure style=\"width: 339px\" 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\/11234801\/CNX_Calc_Figure_11_02_009.jpg\" alt=\"A semicircle is drawn with radius 3. There is an arrow pointing counterclockwise. On the graph there are also written three equations: x(t) = 3 cos(t), y(t) = 3 sin(t), and 0 \u2264 t \u2264 \u03c0.\" width=\"339\" height=\"347\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 10. The arc length of the semicircle is equal to its radius times [latex]\\pi[\/latex].<\/figcaption><\/figure>\n<\/figure>\n<\/div>\n<p>Watch the following video to see the worked solution to the example above.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/bDvDdC2Wpu8?controls=0&amp;start=893&amp;end=993&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/7.2CalculusOfParametricCurves893to993_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;7.2 Calculus of Parametric Curves&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm311320\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=311320&theme=lumen&iframe_resize_id=ohm311320&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/section>\n","protected":false},"author":15,"menu_order":4,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-777","chapter","type-chapter","status-publish","hentry"],"part":438,"module-header":"background_you_need","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\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/777","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/777\/revisions"}],"predecessor-version":[{"id":2684,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/777\/revisions\/2684"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/438"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/777\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=777"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=777"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=777"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=777"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}