{"id":993,"date":"2025-06-20T17:27:12","date_gmt":"2025-06-20T17:27:12","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=993"},"modified":"2025-09-10T17:58:36","modified_gmt":"2025-09-10T17:58:36","slug":"calculus-with-parametric-curves-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/calculus-with-parametric-curves-learn-it-1\/","title":{"raw":"Calculus with Parametric Curves: Learn It 1","rendered":"Calculus with Parametric Curves: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Find derivatives and tangent lines for curves written in parametric form<\/li>\r\n \t<li>Calculate the area underneath a parametric curve<\/li>\r\n \t<li>Find the length of a parametric curve using the arc length formula<\/li>\r\n \t<li>Calculate the surface area when a parametric curve is rotated to create a 3D shape<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Derivatives of Parametric Equations<\/h2>\r\n<p class=\"whitespace-normal break-words\">When working with parametric equations, we often need to find the slope of the curve at any point. This is essential for understanding tangent lines, velocity, and rates of change.<\/p>\r\n<p class=\"whitespace-normal break-words\">Let's start with a concrete example.<\/p>\r\n<p class=\"whitespace-normal break-words\">Consider the parametric equations:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]x(t) = 2t + 3[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]y(t) = 3t - 4[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">where [latex]-2 \\le t \\le 3[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">This represents a line segment from [latex](-1, -10)[\/latex] to [latex](9, 5)[\/latex]. The graph of this curve appears in Figure 1.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_02_001\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"489\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234739\/CNX_Calc_Figure_11_02_001.jpg\" alt=\"A straight line from (\u22121, \u221210) to (9, 5). The point (\u22121, \u221210) is marked t = \u22122, the point (3, \u22124) is marked t = 0, and the point (9, 5) is marked t = 3. There are three equations marked: x(t) = 2t + 3, y(t) = 3t \u2013 4, and \u22122 \u2264 t \u2264 3\" width=\"489\" height=\"647\" data-media-type=\"image\/jpeg\" \/> Figure 1. Graph of the line segment described by the given parametric equations.[\/caption]<\/figure>\r\n<p class=\"whitespace-normal break-words\">We can find the slope by converting to rectangular form, starting with solving for [latex]t[\/latex] from the [latex]x[\/latex]-equation:<\/p>\r\n\r\n<div id=\"fs-id1167793820750\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill x\\left(t\\right)&amp; =\\hfill &amp; 2t+3\\hfill \\\\ \\hfill x - 3&amp; =\\hfill &amp; 2t\\hfill \\\\ \\hfill t&amp; =\\hfill &amp; \\frac{x - 3}{2}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167793897626\">Substituting this into [latex]y\\left(t\\right)[\/latex], we obtain<\/p>\r\n\r\n<div id=\"fs-id1167793783749\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill y\\left(t\\right)&amp; =\\hfill &amp; 3t - 4\\hfill \\\\ \\hfill y&amp; =\\hfill &amp; 3\\left(\\frac{x - 3}{2}\\right)-4\\hfill \\\\ \\hfill y&amp; =\\hfill &amp; \\frac{3x}{2}-\\frac{9}{2}-4\\hfill \\\\ \\hfill y&amp; =\\hfill &amp; \\frac{3x}{2}-\\frac{17}{2}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794062930\">The slope of this line is given by [latex]\\frac{dy}{dx}=\\frac{3}{2}[\/latex].<\/p>\r\nThere's a more direct approach -\u00a0 calculating [latex]{x}^{\\prime }\\left(t\\right)[\/latex] and [latex]{y}^{\\prime }\\left(t\\right)[\/latex]. This gives [latex]{x}^{\\prime }\\left(t\\right)=2[\/latex] and [latex]{y}^{\\prime }\\left(t\\right)=3[\/latex]. Notice that [latex]\\frac{dy}{dx}=\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}=\\frac{3}{2}[\/latex].\r\n<p class=\"whitespace-normal break-words\">Both methods give the same answer\u2014but the second method is often much more efficient!<\/p>\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 connectIt\">\r\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\r\n\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793838466\">This theorem can be proven using the Chain Rule. In particular, assume that the parameter <em data-effect=\"italics\">t<\/em> can be eliminated, yielding a differentiable function [latex]y=F\\left(x\\right)[\/latex]. Then [latex]y\\left(t\\right)=F\\left(x\\left(t\\right)\\right)[\/latex]. Differentiating both sides of this equation using the Chain Rule yields<\/p>\r\n\r\n<div id=\"fs-id1167794100709\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }\\left(t\\right)={F}^{\\prime }\\left(x\\left(t\\right)\\right){x}^{\\prime }\\left(t\\right)[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794068898\">so<\/p>\r\n\r\n<div id=\"fs-id1167793905994\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{F}^{\\prime }\\left(x\\left(t\\right)\\right)=\\frac{{y}^{\\prime }\\left(t\\right)}{{x}^{\\prime }\\left(t\\right)}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793848552\">But [latex]{F}^{\\prime }\\left(x\\left(t\\right)\\right)=\\frac{dy}{dx}[\/latex], which proves the theorem.<\/p>\r\n<p id=\"fs-id1167794069690\">[latex]_\\blacksquare[\/latex]<\/p>\r\n\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\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the example above.<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<\/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\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the example above.<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<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]311317[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Find derivatives and tangent lines for curves written in parametric form<\/li>\n<li>Calculate the area underneath a parametric curve<\/li>\n<li>Find the length of a parametric curve using the arc length formula<\/li>\n<li>Calculate the surface area when a parametric curve is rotated to create a 3D shape<\/li>\n<\/ul>\n<\/section>\n<h2>Derivatives of Parametric Equations<\/h2>\n<p class=\"whitespace-normal break-words\">When working with parametric equations, we often need to find the slope of the curve at any point. This is essential for understanding tangent lines, velocity, and rates of change.<\/p>\n<p class=\"whitespace-normal break-words\">Let&#8217;s start with a concrete example.<\/p>\n<p class=\"whitespace-normal break-words\">Consider the parametric equations:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]x(t) = 2t + 3[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]y(t) = 3t - 4[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">where [latex]-2 \\le t \\le 3[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">This represents a line segment from [latex](-1, -10)[\/latex] to [latex](9, 5)[\/latex]. The graph of this curve appears in Figure 1.<\/p>\n<figure id=\"CNX_Calc_Figure_11_02_001\"><figcaption><\/figcaption><figure style=\"width: 489px\" 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\/11234739\/CNX_Calc_Figure_11_02_001.jpg\" alt=\"A straight line from (\u22121, \u221210) to (9, 5). The point (\u22121, \u221210) is marked t = \u22122, the point (3, \u22124) is marked t = 0, and the point (9, 5) is marked t = 3. There are three equations marked: x(t) = 2t + 3, y(t) = 3t \u2013 4, and \u22122 \u2264 t \u2264 3\" width=\"489\" height=\"647\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 1. Graph of the line segment described by the given parametric equations.<\/figcaption><\/figure>\n<\/figure>\n<p class=\"whitespace-normal break-words\">We can find the slope by converting to rectangular form, starting with solving for [latex]t[\/latex] from the [latex]x[\/latex]-equation:<\/p>\n<div id=\"fs-id1167793820750\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill x\\left(t\\right)& =\\hfill & 2t+3\\hfill \\\\ \\hfill x - 3& =\\hfill & 2t\\hfill \\\\ \\hfill t& =\\hfill & \\frac{x - 3}{2}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793897626\">Substituting this into [latex]y\\left(t\\right)[\/latex], we obtain<\/p>\n<div id=\"fs-id1167793783749\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill y\\left(t\\right)& =\\hfill & 3t - 4\\hfill \\\\ \\hfill y& =\\hfill & 3\\left(\\frac{x - 3}{2}\\right)-4\\hfill \\\\ \\hfill y& =\\hfill & \\frac{3x}{2}-\\frac{9}{2}-4\\hfill \\\\ \\hfill y& =\\hfill & \\frac{3x}{2}-\\frac{17}{2}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794062930\">The slope of this line is given by [latex]\\frac{dy}{dx}=\\frac{3}{2}[\/latex].<\/p>\n<p>There&#8217;s a more direct approach &#8211;\u00a0 calculating [latex]{x}^{\\prime }\\left(t\\right)[\/latex] and [latex]{y}^{\\prime }\\left(t\\right)[\/latex]. This gives [latex]{x}^{\\prime }\\left(t\\right)=2[\/latex] and [latex]{y}^{\\prime }\\left(t\\right)=3[\/latex]. Notice that [latex]\\frac{dy}{dx}=\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}=\\frac{3}{2}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Both methods give the same answer\u2014but the second method is often much more efficient!<\/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 connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\n<p id=\"fs-id1167793838466\">This theorem can be proven using the Chain Rule. In particular, assume that the parameter <em data-effect=\"italics\">t<\/em> can be eliminated, yielding a differentiable function [latex]y=F\\left(x\\right)[\/latex]. Then [latex]y\\left(t\\right)=F\\left(x\\left(t\\right)\\right)[\/latex]. Differentiating both sides of this equation using the Chain Rule yields<\/p>\n<div id=\"fs-id1167794100709\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }\\left(t\\right)={F}^{\\prime }\\left(x\\left(t\\right)\\right){x}^{\\prime }\\left(t\\right)[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794068898\">so<\/p>\n<div id=\"fs-id1167793905994\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{F}^{\\prime }\\left(x\\left(t\\right)\\right)=\\frac{{y}^{\\prime }\\left(t\\right)}{{x}^{\\prime }\\left(t\\right)}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793848552\">But [latex]{F}^{\\prime }\\left(x\\left(t\\right)\\right)=\\frac{dy}{dx}[\/latex], which proves the theorem.<\/p>\n<p id=\"fs-id1167794069690\">[latex]_\\blacksquare[\/latex]<\/p>\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<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">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<\/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<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">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<\/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","protected":false},"author":15,"menu_order":10,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":675,"module-header":"- Select Header -","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\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/993"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/993\/revisions"}],"predecessor-version":[{"id":1671,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/993\/revisions\/1671"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/675"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/993\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=993"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=993"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=993"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=993"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}