{"id":1157,"date":"2021-11-11T17:37:25","date_gmt":"2021-11-11T17:37:25","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/derivatives-of-parametric-equations\/"},"modified":"2022-10-20T23:21:01","modified_gmt":"2022-10-20T23:21:01","slug":"derivatives-of-parametric-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/derivatives-of-parametric-equations\/","title":{"raw":"Derivatives of Parametric Equations","rendered":"Derivatives of Parametric Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Determine derivatives and equations of tangents for parametric curves<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1167794096206\" data-depth=\"1\">\r\n<p id=\"fs-id1167793803941\">We start by asking how to calculate the slope of a line tangent to a parametric curve at a point. Consider the plane curve defined by the parametric equations<\/p>\r\n\r\n<div id=\"fs-id1167793790815\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)=2t+3,y\\left(t\\right)=3t - 4,-2\\le t\\le 3[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794039551\">The graph of this curve appears in Figure 1. It is a line segment starting at [latex]\\left(-1,-10\\right)[\/latex] and ending at [latex]\\left(9,5\\right)[\/latex].<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_02_001\"><figcaption><\/figcaption>[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 [latex]t[\/latex] = \u22122, the point (3, \u22124) is marked [latex]t[\/latex] = 0, and the point (9, 5) is marked [latex]t[\/latex] = 3. There are three equations marked: x(t) = 2t + 3, y(t) = 3t \u2013 4, and \u22122 \u2264 [latex]t[\/latex] \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 id=\"fs-id1167793826913\">We can eliminate the parameter by first solving the equation [latex]x\\left(t\\right)=2t+3[\/latex] for <em data-effect=\"italics\">t<\/em>:<\/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&nbsp;\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]. Next we calculate [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]. This is no coincidence, as outlined in the following theorem.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: Derivative of Parametric Equations<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793961378\">Consider the plane curve defined by the parametric equations [latex]x=x\\left(t\\right)[\/latex] and [latex]y=y\\left(t\\right)[\/latex]. Suppose that [latex]{x}^{\\prime }\\left(t\\right)[\/latex] and [latex]{y}^{\\prime }\\left(t\\right)[\/latex] exist, and assume that [latex]{x}^{\\prime }\\left(t\\right)\\ne 0[\/latex]. Then the derivative [latex]\\frac{dy}{dx}[\/latex] is given by<\/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&nbsp;\r\n\r\n<\/div>\r\n<section id=\"fs-id1167793912242\" data-depth=\"2\">\r\n<h4 data-type=\"title\">Proof<\/h4>\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<p id=\"fs-id1167794029408\">The theorem can be used to calculate derivatives of plane curves, as well as critical points. 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. The theorem gives a formula for the slope of a tangent line to a curve defined parametrically regardless of whether the curve can be described by a function [latex]y=f\\left(x\\right)[\/latex] or not.<\/p>\r\n\r\n<div id=\"fs-id1167793984213\" data-type=\"example\">\r\n<div id=\"fs-id1167794071769\" data-type=\"exercise\">\r\n<div id=\"fs-id1167794048755\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Derivative of a Parametric Curve<\/h3>\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\">[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 [latex]t[\/latex] = \u22123, the point (\u22123, \u22121) is marked [latex]t[\/latex] = 0, and the point (13, 7) is marked [latex]t[\/latex] = 4. On the graph there are also written three equations: x(t) = t2 \u2212 3, y(t) = 2t \u2212 1, and \u22123 \u2264 [latex]t[\/latex] \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\">[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 [latex]t[\/latex] = \u22122, the point (\u22121, 6) is marked [latex]t[\/latex] = \u22121, the point (3, 2) is marked [latex]t[\/latex] = 1, and the point (5, 6) is marked [latex]t[\/latex] = 2. On the graph there are also written three equations: x(t) = 2t + 1, y(t) = t3 \u2013 3t + 4, and \u22122 \u2264 [latex]t[\/latex] \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\" style=\"height: 72px;\" 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 style=\"height: 12px;\" valign=\"top\">\r\n<th style=\"height: 12px; width: 157.401px;\" data-valign=\"top\" data-align=\"left\">[latex]t[\/latex]<\/th>\r\n<th style=\"height: 12px; width: 166.42px;\" data-valign=\"top\" data-align=\"left\">[latex]x\\left(t\\right)[\/latex]<\/th>\r\n<th style=\"height: 12px; width: 166.449px;\" 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 style=\"height: 12px;\" valign=\"top\">\r\n<td style=\"height: 12px; width: 157.401px;\" data-valign=\"top\" data-align=\"left\">[latex]0[\/latex]<\/td>\r\n<td style=\"height: 12px; width: 166.42px;\" data-valign=\"top\" data-align=\"left\">[latex]5[\/latex]<\/td>\r\n<td style=\"height: 12px; width: 166.449px;\" data-valign=\"top\" data-align=\"left\">[latex]0[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\" valign=\"top\">\r\n<td style=\"height: 12px; width: 157.401px;\" data-valign=\"top\" data-align=\"left\">[latex]\\frac{\\pi }{2}[\/latex]<\/td>\r\n<td style=\"height: 12px; width: 166.42px;\" data-valign=\"top\" data-align=\"left\">[latex]0[\/latex]<\/td>\r\n<td style=\"height: 12px; width: 166.449px;\" data-valign=\"top\" data-align=\"left\">[latex]5[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\" valign=\"top\">\r\n<td style=\"height: 12px; width: 157.401px;\" data-valign=\"top\" data-align=\"left\">[latex]\\pi [\/latex]<\/td>\r\n<td style=\"height: 12px; width: 166.42px;\" data-valign=\"top\" data-align=\"left\">[latex]-5[\/latex]<\/td>\r\n<td style=\"height: 12px; width: 166.449px;\" data-valign=\"top\" data-align=\"left\">[latex]0[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\" valign=\"top\">\r\n<td style=\"height: 12px; width: 157.401px;\" data-valign=\"top\" data-align=\"left\">[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\r\n<td style=\"height: 12px; width: 166.42px;\" data-valign=\"top\" data-align=\"left\">[latex]0[\/latex]<\/td>\r\n<td style=\"height: 12px; width: 166.449px;\" data-valign=\"top\" data-align=\"left\">[latex]-5[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\" valign=\"top\">\r\n<td style=\"height: 12px; width: 157.401px;\" data-valign=\"top\" data-align=\"left\">[latex]2\\pi [\/latex]<\/td>\r\n<td style=\"height: 12px; width: 166.42px;\" data-valign=\"top\" data-align=\"left\">[latex]5[\/latex]<\/td>\r\n<td style=\"height: 12px; width: 166.449px;\" data-valign=\"top\" data-align=\"left\">[latex]0[\/latex]<\/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\">[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 [latex]t[\/latex] = 0, the point (0, 5) is marked [latex]t[\/latex] = \u03c0\/2, the point (\u22125, 0) is marked [latex]t[\/latex] = \u03c0, and the point (0, \u22125) is marked [latex]t[\/latex] = 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 [latex]t[\/latex] \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<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Finding the Derivative of a Parametric Curve.\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\"><\/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<div id=\"fs-id1167794031032\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167794031036\" data-type=\"exercise\">\r\n<div id=\"fs-id1167794031038\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167794031038\" data-type=\"problem\">\r\n<p id=\"fs-id1167794140364\">Calculate the derivative [latex]\\frac{dy}{dx}[\/latex] for the plane curve defined by the equations<\/p>\r\n\r\n<div id=\"fs-id1167794071108\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)={t}^{2}-4t,y\\left(t\\right)=2{t}^{3}-6t,-2\\le t\\le 3[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794090911\">and locate any critical points on its graph.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558897\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1167793971067\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167793971073\">Calculate [latex]{x}^{\\prime }\\left(t\\right)[\/latex] and [latex]{y}^{\\prime }\\left(t\\right)[\/latex] and use the theorem.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1167794090915\" data-type=\"solution\">\r\n<p id=\"fs-id1167794090918\">[latex]{x}^{\\prime }\\left(t\\right)=2t - 4[\/latex] and [latex]{y}^{\\prime }\\left(t\\right)=6{t}^{2}-6[\/latex], so [latex]\\frac{dy}{dx}=\\frac{6{t}^{2}-6}{2t - 4}=\\frac{3{t}^{2}-3}{t - 2}[\/latex]. <span data-type=\"newline\">\r\n<\/span>\r\nThis expression is undefined when [latex]t=2[\/latex] and equal to zero when [latex]t=\\pm 1[\/latex]. <span data-type=\"newline\">\r\n<\/span><\/p>\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"388\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234751\/CNX_Calc_Figure_11_02_005.jpg\" alt=\"A curve going from (12, \u22124) through the origin and (\u22124, 0) to (\u22123, 36) with arrows in that order. The point (12, \u22124) is marked [latex]t[\/latex] = \u22122 and the point (\u22123, 36) is marked [latex]t[\/latex] = 3. On the graph there are also written three equations: x(t) = t2 \u2013 4t, y(t) = 2t3 \u2013 6t, and \u22122 \u2264 [latex]t[\/latex] \u2264 3.\" width=\"388\" height=\"384\" data-media-type=\"image\/jpeg\" \/> Figure 5.[\/caption]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167794042393\" data-type=\"example\">\r\n<div id=\"fs-id1167794042395\" data-type=\"exercise\">\r\n<div id=\"fs-id1167794042397\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding a Tangent Line<\/h3>\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 [latex]t[\/latex] = \u22123, the point (\u22123, \u22121) is marked [latex]t[\/latex] = 0, and the point (13, 7) is marked [latex]t[\/latex] = 4. On the graph there are also written three equations: x(t) = t2 \u2212 3, y(t) = 2t \u2212 1, and \u22123 \u2264 [latex]t[\/latex] \u2264 4. At the point (1, 3), which is marked [latex]t[\/latex] = 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<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Finding a Tangent Line.\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\"><\/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<div id=\"fs-id1167794011458\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167794011461\" data-type=\"exercise\">\r\n<div id=\"fs-id1167794011463\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167794011463\" data-type=\"problem\">\r\n<p id=\"fs-id1167794011465\">Find the equation of the tangent line to the curve defined by the equations<\/p>\r\n\r\n<div id=\"fs-id1167794011469\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)={t}^{2}-4t,y\\left(t\\right)=2{t}^{3}-6t,-2\\le t\\le 3\\text{ when }t=5[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558894\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1167794050248\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167794050254\">Calculate [latex]{x}^{\\prime }\\left(t\\right)[\/latex] and [latex]{y}^{\\prime }\\left(t\\right)[\/latex] and use the theorem.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1167794050224\" data-type=\"solution\">\r\n<p id=\"fs-id1167794050227\">The equation of the tangent line is [latex]y=24x+100[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]5087[\/ohm_question]\r\n\r\n<\/div>\r\n<\/section><\/section><section id=\"fs-id1167794024788\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Second-Order Derivatives<\/h2>\r\n<p id=\"fs-id1167794032313\">Our next goal is to see how to take the second derivative of a function defined parametrically. The second derivative of a function [latex]y=f\\left(x\\right)[\/latex] is defined to be the derivative of the first derivative; that is,<\/p>\r\n\r\n<div id=\"fs-id1167794032334\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{d}^{2}y}{d{x}^{2}}=\\frac{d}{dx}\\left[\\frac{dy}{dx}\\right][\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794069389\">Since [latex]\\frac{dy}{dx}=\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}[\/latex], we can replace the [latex]y[\/latex] on both sides of this equation with [latex]\\frac{dy}{dx}[\/latex]. This gives us<\/p>\r\n\r\n<div id=\"fs-id1167794072159\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\frac{{d}^{2}y}{d{x}^{2}}=\\frac{d}{dx}\\left(\\frac{dy}{dx}\\right)=\\frac{\\left(\\frac{d}{dt}\\right)\\left(\\frac{dy}{dx}\\right)}{\\frac{dx}{dt}}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794044383\">If we know [latex]\\frac{dy}{dx}[\/latex] as a function of <em data-effect=\"italics\">t,<\/em> then this formula is straightforward to apply.<\/p>\r\n\r\n<div id=\"fs-id1167794044408\" data-type=\"example\">\r\n<div id=\"fs-id1167794044410\" data-type=\"exercise\">\r\n<div id=\"fs-id1167794044412\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding a Second Derivative<\/h3>\r\n<div id=\"fs-id1167794044412\" data-type=\"problem\">\r\n<p id=\"fs-id1167794044418\">Calculate the second derivative [latex]\\frac{{d}^{2}y}{d{x}^{2}}[\/latex] for the plane curve defined by the parametric equations [latex]x\\left(t\\right)={t}^{2}-3,y\\left(t\\right)=2t - 1,-3\\le t\\le 4[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n<div id=\"fs-id1167794063388\" data-type=\"solution\">\r\n\r\nFrom the example: Finding the Derivative of a Parametric Curve we know that [latex]\\frac{dy}{dx}=\\frac{2}{2t}=\\frac{1}{t}[\/latex]. Using our above equation, we obtain\r\n<div id=\"fs-id1167794063436\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{d}^{2}y}{d{x}^{2}}=\\frac{\\left(\\frac{d}{dt}\\right)\\left(\\frac{dy}{dx}\\right)}{\\frac{dx}{dt}}=\\frac{\\left(\\frac{d}{dt}\\right)\\left(\\frac{1}{t}\\right)}{2t}=\\frac{-{t}^{-2}}{2t}=-\\frac{1}{2{t}^{3}}[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Finding a Second Derivative.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/bDvDdC2Wpu8?controls=0&amp;start=583&amp;end=697&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/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\/7.2CalculusOfParametricCurves583to697_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<div id=\"fs-id1167793984344\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167793984347\" data-type=\"exercise\">\r\n<div id=\"fs-id1167793984349\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167793984349\" data-type=\"problem\">\r\n<p id=\"fs-id1167793984351\">Calculate the second derivative [latex]\\frac{{d}^{2}y}{d{x}^{2}}[\/latex] for the plane curve defined by the equations<\/p>\r\n\r\n<div id=\"fs-id1167793984378\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)={t}^{2}-4t,y\\left(t\\right)=2{t}^{3}-6t,-2\\le t\\le 3[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793984455\">and locate any critical points on its graph.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558892\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n<div id=\"fs-id1167794065386\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167794065393\">Start with the solution from the previous checkpoint, and use the above equation.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558891\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558891\"]\r\n<div id=\"fs-id1167793984460\" data-type=\"solution\">\r\n<p id=\"fs-id1167793984462\" style=\"text-align: left;\">[latex]\\frac{{d}^{2}y}{d{x}^{2}}=\\frac{3{t}^{2}-12t+3}{2{\\left(t - 2\\right)}^{3}}[\/latex]. Critical points [latex]\\left(5,4\\right),\\left(-3,-4\\right),\\text{and}\\left(-4,6\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1167794065404\" data-depth=\"1\"><\/section>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Determine derivatives and equations of tangents for parametric curves<\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1167794096206\" data-depth=\"1\">\n<p id=\"fs-id1167793803941\">We start by asking how to calculate the slope of a line tangent to a parametric curve at a point. Consider the plane curve defined by the parametric equations<\/p>\n<div id=\"fs-id1167793790815\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)=2t+3,y\\left(t\\right)=3t - 4,-2\\le t\\le 3[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794039551\">The graph of this curve appears in Figure 1. It is a line segment starting at [latex]\\left(-1,-10\\right)[\/latex] and ending at [latex]\\left(9,5\\right)[\/latex].<\/p>\n<figure id=\"CNX_Calc_Figure_11_02_001\"><figcaption><\/figcaption><div style=\"width: 499px\" 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 [latex]t[\/latex] = \u22122, the point (3, \u22124) is marked [latex]t[\/latex] = 0, and the point (9, 5) is marked [latex]t[\/latex] = 3. There are three equations marked: x(t) = 2t + 3, y(t) = 3t \u2013 4, and \u22122 \u2264 [latex]t[\/latex] \u2264 3\" width=\"489\" height=\"647\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. Graph of the line segment described by the given parametric equations.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1167793826913\">We can eliminate the parameter by first solving the equation [latex]x\\left(t\\right)=2t+3[\/latex] for <em data-effect=\"italics\">t<\/em>:<\/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>&nbsp;<\/p>\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]. Next we calculate [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]. This is no coincidence, as outlined in the following theorem.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: Derivative of Parametric Equations<\/h3>\n<hr \/>\n<p id=\"fs-id1167793961378\">Consider the plane curve defined by the parametric equations [latex]x=x\\left(t\\right)[\/latex] and [latex]y=y\\left(t\\right)[\/latex]. Suppose that [latex]{x}^{\\prime }\\left(t\\right)[\/latex] and [latex]{y}^{\\prime }\\left(t\\right)[\/latex] exist, and assume that [latex]{x}^{\\prime }\\left(t\\right)\\ne 0[\/latex]. Then the derivative [latex]\\frac{dy}{dx}[\/latex] is given by<\/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<p>&nbsp;<\/p>\n<\/div>\n<section id=\"fs-id1167793912242\" data-depth=\"2\">\n<h4 data-type=\"title\">Proof<\/h4>\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<p id=\"fs-id1167794029408\">The theorem can be used to calculate derivatives of plane curves, as well as critical points. 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. The theorem gives a formula for the slope of a tangent line to a curve defined parametrically regardless of whether the curve can be described by a function [latex]y=f\\left(x\\right)[\/latex] or not.<\/p>\n<div id=\"fs-id1167793984213\" data-type=\"example\">\n<div id=\"fs-id1167794071769\" data-type=\"exercise\">\n<div id=\"fs-id1167794048755\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Derivative of a Parametric Curve<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558899\">Show Solution<\/span><\/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<div style=\"width: 427px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/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 &#091;latex&#093;t&#091;\/latex&#093; = \u22123, the point (\u22123, \u22121) is marked &#091;latex&#093;t&#091;\/latex&#093; = 0, and the point (13, 7) is marked &#091;latex&#093;t&#091;\/latex&#093; = 4. On the graph there are also written three equations: x(t) = t2 \u2212 3, y(t) = 2t \u2212 1, and \u22123 \u2264 &#091;latex&#093;t&#091;\/latex&#093; \u2264 4.\" width=\"417\" height=\"347\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. Graph of the parabola described by parametric equations in part a.<\/p>\n<\/div>\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<div style=\"width: 389px\" 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 &#091;latex&#093;t&#091;\/latex&#093; = \u22122, the point (\u22121, 6) is marked &#091;latex&#093;t&#091;\/latex&#093; = \u22121, the point (3, 2) is marked &#091;latex&#093;t&#091;\/latex&#093; = 1, and the point (5, 6) is marked &#091;latex&#093;t&#091;\/latex&#093; = 2. On the graph there are also written three equations: x(t) = 2t + 1, y(t) = t3 \u2013 3t + 4, and \u22122 \u2264 &#091;latex&#093;t&#091;\/latex&#093; \u2264 2.\" width=\"379\" height=\"347\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. Graph of the curve described by parametric equations in part b.<\/p>\n<\/div>\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\" style=\"height: 72px;\" 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 style=\"height: 12px;\" valign=\"top\">\n<th style=\"height: 12px; width: 157.401px;\" data-valign=\"top\" data-align=\"left\">[latex]t[\/latex]<\/th>\n<th style=\"height: 12px; width: 166.42px;\" data-valign=\"top\" data-align=\"left\">[latex]x\\left(t\\right)[\/latex]<\/th>\n<th style=\"height: 12px; width: 166.449px;\" data-valign=\"top\" data-align=\"left\">[latex]y\\left(t\\right)[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height: 12px;\" valign=\"top\">\n<td style=\"height: 12px; width: 157.401px;\" data-valign=\"top\" data-align=\"left\">[latex]0[\/latex]<\/td>\n<td style=\"height: 12px; width: 166.42px;\" data-valign=\"top\" data-align=\"left\">[latex]5[\/latex]<\/td>\n<td style=\"height: 12px; width: 166.449px;\" data-valign=\"top\" data-align=\"left\">[latex]0[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\" valign=\"top\">\n<td style=\"height: 12px; width: 157.401px;\" data-valign=\"top\" data-align=\"left\">[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<td style=\"height: 12px; width: 166.42px;\" data-valign=\"top\" data-align=\"left\">[latex]0[\/latex]<\/td>\n<td style=\"height: 12px; width: 166.449px;\" data-valign=\"top\" data-align=\"left\">[latex]5[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\" valign=\"top\">\n<td style=\"height: 12px; width: 157.401px;\" data-valign=\"top\" data-align=\"left\">[latex]\\pi[\/latex]<\/td>\n<td style=\"height: 12px; width: 166.42px;\" data-valign=\"top\" data-align=\"left\">[latex]-5[\/latex]<\/td>\n<td style=\"height: 12px; width: 166.449px;\" data-valign=\"top\" data-align=\"left\">[latex]0[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\" valign=\"top\">\n<td style=\"height: 12px; width: 157.401px;\" data-valign=\"top\" data-align=\"left\">[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\n<td style=\"height: 12px; width: 166.42px;\" data-valign=\"top\" data-align=\"left\">[latex]0[\/latex]<\/td>\n<td style=\"height: 12px; width: 166.449px;\" data-valign=\"top\" data-align=\"left\">[latex]-5[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\" valign=\"top\">\n<td style=\"height: 12px; width: 157.401px;\" data-valign=\"top\" data-align=\"left\">[latex]2\\pi[\/latex]<\/td>\n<td style=\"height: 12px; width: 166.42px;\" data-valign=\"top\" data-align=\"left\">[latex]5[\/latex]<\/td>\n<td style=\"height: 12px; width: 166.449px;\" data-valign=\"top\" data-align=\"left\">[latex]0[\/latex]<\/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<div style=\"width: 500px\" 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 &#091;latex&#093;t&#091;\/latex&#093; = 0, the point (0, 5) is marked &#091;latex&#093;t&#091;\/latex&#093; = \u03c0\/2, the point (\u22125, 0) is marked &#091;latex&#093;t&#091;\/latex&#093; = \u03c0, and the point (0, \u22125) is marked &#091;latex&#093;t&#091;\/latex&#093; = 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 &#091;latex&#093;t&#091;\/latex&#093; \u2264 2\u03c0.\" width=\"490\" height=\"497\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. Graph of the curve described by parametric equations in part c.<\/p>\n<\/div>\n<\/figure>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Finding the Derivative of a Parametric Curve.<\/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\"><\/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 id=\"fs-id1167794031032\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167794031036\" data-type=\"exercise\">\n<div id=\"fs-id1167794031038\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167794031038\" data-type=\"problem\">\n<p id=\"fs-id1167794140364\">Calculate the derivative [latex]\\frac{dy}{dx}[\/latex] for the plane curve defined by the equations<\/p>\n<div id=\"fs-id1167794071108\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)={t}^{2}-4t,y\\left(t\\right)=2{t}^{3}-6t,-2\\le t\\le 3[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794090911\">and locate any critical points on its graph.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558897\">Hint<\/span><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167793971067\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167793971073\">Calculate [latex]{x}^{\\prime }\\left(t\\right)[\/latex] and [latex]{y}^{\\prime }\\left(t\\right)[\/latex] and use the theorem.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558898\">Show Solution<\/span><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794090915\" data-type=\"solution\">\n<p id=\"fs-id1167794090918\">[latex]{x}^{\\prime }\\left(t\\right)=2t - 4[\/latex] and [latex]{y}^{\\prime }\\left(t\\right)=6{t}^{2}-6[\/latex], so [latex]\\frac{dy}{dx}=\\frac{6{t}^{2}-6}{2t - 4}=\\frac{3{t}^{2}-3}{t - 2}[\/latex]. <span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThis expression is undefined when [latex]t=2[\/latex] and equal to zero when [latex]t=\\pm 1[\/latex]. <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div style=\"width: 398px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234751\/CNX_Calc_Figure_11_02_005.jpg\" alt=\"A curve going from (12, \u22124) through the origin and (\u22124, 0) to (\u22123, 36) with arrows in that order. The point (12, \u22124) is marked &#091;latex&#093;t&#091;\/latex&#093; = \u22122 and the point (\u22123, 36) is marked &#091;latex&#093;t&#091;\/latex&#093; = 3. On the graph there are also written three equations: x(t) = t2 \u2013 4t, y(t) = 2t3 \u2013 6t, and \u22122 \u2264 &#091;latex&#093;t&#091;\/latex&#093; \u2264 3.\" width=\"388\" height=\"384\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167794042393\" data-type=\"example\">\n<div id=\"fs-id1167794042395\" data-type=\"exercise\">\n<div id=\"fs-id1167794042397\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Finding a Tangent Line<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558896\">Show Solution<\/span><\/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><div style=\"width: 389px\" 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 &#091;latex&#093;t&#091;\/latex&#093; = \u22123, the point (\u22123, \u22121) is marked &#091;latex&#093;t&#091;\/latex&#093; = 0, and the point (13, 7) is marked &#091;latex&#093;t&#091;\/latex&#093; = 4. On the graph there are also written three equations: x(t) = t2 \u2212 3, y(t) = 2t \u2212 1, and \u22123 \u2264 &#091;latex&#093;t&#091;\/latex&#093; \u2264 4. At the point (1, 3), which is marked &#091;latex&#093;t&#091;\/latex&#093; = 2, there is a tangent line with equation y = x\/2 + 5\/2.\" width=\"379\" height=\"383\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. Tangent line to the parabola described by the given parametric equations when [latex]t=2[\/latex].<\/p>\n<\/div>\n<\/figure>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Finding a Tangent Line.<\/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\"><\/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 id=\"fs-id1167794011458\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167794011461\" data-type=\"exercise\">\n<div id=\"fs-id1167794011463\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167794011463\" data-type=\"problem\">\n<p id=\"fs-id1167794011465\">Find the equation of the tangent line to the curve defined by the equations<\/p>\n<div id=\"fs-id1167794011469\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)={t}^{2}-4t,y\\left(t\\right)=2{t}^{3}-6t,-2\\le t\\le 3\\text{ when }t=5[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558894\">Hint<\/span><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794050248\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167794050254\">Calculate [latex]{x}^{\\prime }\\left(t\\right)[\/latex] and [latex]{y}^{\\prime }\\left(t\\right)[\/latex] and use the theorem.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558895\">Show Solution<\/span><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794050224\" data-type=\"solution\">\n<p id=\"fs-id1167794050227\">The equation of the tangent line is [latex]y=24x+100[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm5087\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5087&theme=oea&iframe_resize_id=ohm5087&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/section>\n<\/section>\n<section id=\"fs-id1167794024788\" data-depth=\"1\">\n<h2 data-type=\"title\">Second-Order Derivatives<\/h2>\n<p id=\"fs-id1167794032313\">Our next goal is to see how to take the second derivative of a function defined parametrically. The second derivative of a function [latex]y=f\\left(x\\right)[\/latex] is defined to be the derivative of the first derivative; that is,<\/p>\n<div id=\"fs-id1167794032334\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{d}^{2}y}{d{x}^{2}}=\\frac{d}{dx}\\left[\\frac{dy}{dx}\\right][\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794069389\">Since [latex]\\frac{dy}{dx}=\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}[\/latex], we can replace the [latex]y[\/latex] on both sides of this equation with [latex]\\frac{dy}{dx}[\/latex]. This gives us<\/p>\n<div id=\"fs-id1167794072159\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\frac{{d}^{2}y}{d{x}^{2}}=\\frac{d}{dx}\\left(\\frac{dy}{dx}\\right)=\\frac{\\left(\\frac{d}{dt}\\right)\\left(\\frac{dy}{dx}\\right)}{\\frac{dx}{dt}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794044383\">If we know [latex]\\frac{dy}{dx}[\/latex] as a function of <em data-effect=\"italics\">t,<\/em> then this formula is straightforward to apply.<\/p>\n<div id=\"fs-id1167794044408\" data-type=\"example\">\n<div id=\"fs-id1167794044410\" data-type=\"exercise\">\n<div id=\"fs-id1167794044412\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Finding a Second Derivative<\/h3>\n<div id=\"fs-id1167794044412\" data-type=\"problem\">\n<p id=\"fs-id1167794044418\">Calculate the second derivative [latex]\\frac{{d}^{2}y}{d{x}^{2}}[\/latex] for the plane curve defined by the parametric equations [latex]x\\left(t\\right)={t}^{2}-3,y\\left(t\\right)=2t - 1,-3\\le t\\le 4[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558893\">Show Solution<\/span><\/p>\n<div id=\"q44558893\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794063388\" data-type=\"solution\">\n<p>From the example: Finding the Derivative of a Parametric Curve we know that [latex]\\frac{dy}{dx}=\\frac{2}{2t}=\\frac{1}{t}[\/latex]. Using our above equation, we obtain<\/p>\n<div id=\"fs-id1167794063436\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{d}^{2}y}{d{x}^{2}}=\\frac{\\left(\\frac{d}{dt}\\right)\\left(\\frac{dy}{dx}\\right)}{\\frac{dx}{dt}}=\\frac{\\left(\\frac{d}{dt}\\right)\\left(\\frac{1}{t}\\right)}{2t}=\\frac{-{t}^{-2}}{2t}=-\\frac{1}{2{t}^{3}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Finding a Second Derivative.<\/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=583&amp;end=697&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/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\/7.2CalculusOfParametricCurves583to697_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 id=\"fs-id1167793984344\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167793984347\" data-type=\"exercise\">\n<div id=\"fs-id1167793984349\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167793984349\" data-type=\"problem\">\n<p id=\"fs-id1167793984351\">Calculate the second derivative [latex]\\frac{{d}^{2}y}{d{x}^{2}}[\/latex] for the plane curve defined by the equations<\/p>\n<div id=\"fs-id1167793984378\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)={t}^{2}-4t,y\\left(t\\right)=2{t}^{3}-6t,-2\\le t\\le 3[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793984455\">and locate any critical points on its graph.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558892\">Hint<\/span><\/p>\n<div id=\"q44558892\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794065386\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167794065393\">Start with the solution from the previous checkpoint, and use the above equation.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558891\">Show Solution<\/span><\/p>\n<div id=\"q44558891\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167793984460\" data-type=\"solution\">\n<p id=\"fs-id1167793984462\" style=\"text-align: left;\">[latex]\\frac{{d}^{2}y}{d{x}^{2}}=\\frac{3{t}^{2}-12t+3}{2{\\left(t - 2\\right)}^{3}}[\/latex]. Critical points [latex]\\left(5,4\\right),\\left(-3,-4\\right),\\text{and}\\left(-4,6\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1167794065404\" data-depth=\"1\"><\/section>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1157\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>7.2 Calculus of Parametric Curves. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"7.2 Calculus of Parametric Curves\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1157","chapter","type-chapter","status-publish","hentry"],"part":1150,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1157","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":2,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1157\/revisions"}],"predecessor-version":[{"id":4697,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1157\/revisions\/4697"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/1150"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1157\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=1157"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=1157"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=1157"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=1157"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}