{"id":1158,"date":"2021-11-11T17:37:25","date_gmt":"2021-11-11T17:37:25","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/geometric-calculations-of-parametric-curves\/"},"modified":"2022-10-20T23:23:36","modified_gmt":"2022-10-20T23:23:36","slug":"geometric-calculations-of-parametric-curves","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/geometric-calculations-of-parametric-curves\/","title":{"raw":"Geometric Calculations of Parametric Curves","rendered":"Geometric Calculations of Parametric Curves"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Find the area under a parametric curve<\/li>\r\n \t<li>Use the equation for arc length of a parametric curve<\/li>\r\n \t<li>Apply the formula for surface area to a volume generated by a parametric curve<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 data-type=\"title\">Integrals Involving Parametric Equations<\/h2>\r\n<p id=\"fs-id1167794065409\">Now that we have seen how to calculate the derivative of a plane curve, the next question is this: How do we find the area under a curve defined parametrically? Recall the cycloid defined by the equations [latex]x\\left(t\\right)=t-\\sin{t},y\\left(t\\right)=1-\\cos{t}[\/latex]. Suppose we want to find the area of the shaded region in the following graph.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_02_007\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234756\/CNX_Calc_Figure_11_02_007.jpg\" alt=\"A series of half circles drawn above the [latex]x[\/latex]-axis with x intercepts being multiples of 2\u03c0. The half circle between 0 and 2\u03c0 is highlighted. On the graph there are also written two equations: x(t) = [latex]t[\/latex] \u2013 sin(t) and y(t) = 1 \u2013 cos(t).\" width=\"487\" height=\"347\" data-media-type=\"image\/jpeg\" \/> Figure 1. Graph of a cycloid with the arch over [latex]\\left[0,2\\pi \\right][\/latex] highlighted.[\/caption]<\/figure>\r\n<p id=\"fs-id1167794030645\">To derive a formula for the area under the curve defined by the functions<\/p>\r\n\r\n<div id=\"fs-id1167794030648\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x=x\\left(t\\right),y=y\\left(t\\right),a\\le t\\le b[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794030704\">we assume that [latex]x\\left(t\\right)[\/latex] is differentiable and start with an equal partition of the interval [latex]a\\le t\\le b[\/latex]. Suppose [latex]{t}_{0}=a&lt;{t}_{1}&lt;{t}_{2}&lt;\\cdots &lt;{t}_{n}=b[\/latex] and consider the following graph.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_02_008\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"233\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234758\/CNX_Calc_Figure_11_02_012.jpg\" alt=\"A curved line is drawn in the first quadrant. Below it are a series of rectangles marked that begin at the [latex]x[\/latex]-axis and reach up to the curved line; the rectangle\u2019s height is determined by the location of the curved line at the leftmost point of the rectangle. These lines are noted as x(t0), x(t1), \u2026, x(tn).\" width=\"233\" height=\"245\" data-media-type=\"image\/jpeg\" \/> Figure 2. Approximating the area under a parametrically defined curve.[\/caption]<\/figure>\r\n<p id=\"fs-id1167794030800\">We use rectangles to approximate the area under the curve. The height of a typical rectangle in this parametrization is [latex]y\\left(x\\left({\\overline{t}}_{i}\\right)\\right)[\/latex] for some value [latex]{\\overline{t}}_{i}[\/latex] in the <em data-effect=\"italics\">i<\/em>th subinterval, and the width can be calculated as [latex]x\\left({t}_{i}\\right)-x\\left({t}_{i - 1}\\right)[\/latex]. Thus the area of the <em data-effect=\"italics\">i<\/em>th rectangle is given by<\/p>\r\n\r\n<div id=\"fs-id1167794065566\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{A}_{i}=y\\left(x\\left({\\overline{t}}_{i}\\right)\\right)\\left(x\\left({t}_{i}\\right)-x\\left({t}_{i - 1}\\right)\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794065651\">Then a Riemann sum for the area is<\/p>\r\n\r\n<div id=\"fs-id1167794065654\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{A}_{n}=\\displaystyle\\sum _{i=1}^{n}y\\left(x\\left({\\overline{t}}_{i}\\right)\\right)\\left(x\\left({t}_{i}\\right)-x\\left({t}_{i - 1}\\right)\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794064679\">Multiplying and dividing each area by [latex]{t}_{i}-{t}_{i - 1}[\/latex] gives<\/p>\r\n\r\n<div id=\"fs-id1167794064704\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{A}_{n}=\\displaystyle\\sum _{i=1}^{n}y\\left(x\\left({\\overline{t}}_{i}\\right)\\right)\\left(\\frac{x\\left({t}_{i}\\right)-x\\left({t}_{i - 1}\\right)}{{t}_{i}-{t}_{i - 1}}\\right)\\left({t}_{i}-{t}_{i - 1}\\right)=\\displaystyle\\sum _{i=1}^{n}y\\left(x\\left({\\overline{t}}_{i}\\right)\\right)\\left(\\frac{x\\left({t}_{i}\\right)-x\\left({t}_{i - 1}\\right)}{\\Delta t}\\right)\\Delta t[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794000276\">Taking the limit as [latex]n[\/latex] approaches infinity gives<\/p>\r\n\r\n<div id=\"fs-id1167794000283\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]A=\\underset{n\\to \\infty }{\\text{lim}}{A}_{n}={\\displaystyle\\int }_{a}^{b}y\\left(t\\right){x}^{\\prime }\\left(t\\right)dt[\/latex].<\/div>\r\n&nbsp;\r\n\r\nNote that as the value of [latex] n [\/latex] approaches [latex] \\infty [\/latex], the change in [latex] x [\/latex] over smaller and smaller time intervals can be rewritten as the instantaneous rate of change of [latex] x [\/latex] with respect to [latex] t [\/latex] at some value within the sub-interval of [latex] t [\/latex]. Recall that this fact is known as the Mean Value Theorem, and we briefly review it to clarify the proofs within this section.\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Mean Value Theorem<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1165042357018\">Let [latex]f[\/latex] be continuous over the closed interval [latex][a,b][\/latex] and differentiable over the open interval [latex](a,b)[\/latex]. Then, there exists at least one point [latex]c \\in (a,b)[\/latex] such that<\/p>\r\n\r\n<div id=\"fs-id1165043066505\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(c)=\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167794038312\">The preceding result leads to the following theorem.<\/p>\r\n\r\n<div id=\"fs-id1167794038315\" class=\"theorem\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">theorem: Area under a Parametric Curve<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167794038322\">Consider the non-self-intersecting plane curve defined by the parametric equations<\/p>\r\n\r\n<div id=\"fs-id1167794038325\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x=x\\left(t\\right),y=y\\left(t\\right),a\\le t\\le b[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794038378\">and assume that [latex]x\\left(t\\right)[\/latex] is differentiable. The area under this curve is given by<\/p>\r\n\r\n<div id=\"fs-id1167794038394\" style=\"text-align: center;\" data-type=\"equation\">[latex]A={\\displaystyle\\int }_{a}^{b}y\\left(t\\right){x}^{\\prime }\\left(t\\right)dt[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167794038444\" data-type=\"example\">\r\n<div id=\"fs-id1167794038447\" data-type=\"exercise\">\r\n<div id=\"fs-id1167794038449\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Area under a Parametric Curve<\/h3>\r\n<div id=\"fs-id1167794038449\" data-type=\"problem\">\r\n<p id=\"fs-id1167794038454\">Find the area under the curve of the cycloid defined by the equations<\/p>\r\n\r\n<div id=\"fs-id1167794038457\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)=t-\\sin{t},y\\left(t\\right)=1-\\cos{t},0\\le t\\le 2\\pi [\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558889\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558889\"]\r\n<div id=\"fs-id1167794070824\" data-type=\"solution\">\r\n<p id=\"fs-id1167794070826\">Using the theorem, we have<\/p>\r\n\r\n<div id=\"fs-id1167794070833\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill A&amp; ={\\displaystyle\\int }_{a}^{b}y\\left(t\\right){x}^{\\prime }\\left(t\\right)dt\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{2\\pi }\\left(1-\\cos{t}\\right)\\left(1-\\cos{t}\\right)dt\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{2\\pi }\\left(1 - 2\\cos{t}+{\\cos}^{2}t\\right)dt\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{2\\pi }\\left(1 - 2\\cos{t}+\\frac{1+\\cos2t}{2}\\right)dt\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{2\\pi }\\left(\\frac{3}{2}-2\\cos{t}+\\frac{\\cos2t}{2}\\right)dt\\hfill \\\\ &amp; ={\\frac{3t}{2}-2\\sin{t}+\\frac{\\sin2t}{4}|}_{0}^{2\\pi }\\hfill \\\\ &amp; =3\\pi .\\hfill \\end{array}[\/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 the Area under a Parametric Curve.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/bDvDdC2Wpu8?controls=0&amp;start=718&amp;end=855&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\/7.2CalculusOfParametricCurves718to855_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-id1167794074316\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167794074320\" data-type=\"exercise\">\r\n<div id=\"fs-id1167794074322\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167794074322\" data-type=\"problem\">\r\n<p id=\"fs-id1167794074324\">Find the area under the curve of the hypocycloid defined by the equations<\/p>\r\n\r\n<div id=\"fs-id1167794074327\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)=3\\cos{t}+\\cos3t,y\\left(t\\right)=3\\sin{t}-\\sin3t,0\\le t\\le \\pi [\/latex].<\/div>\r\n<div data-type=\"equation\" data-label=\"\"><\/div>\r\n<span style=\"font-size: 1rem; text-align: initial;\">[reveal-answer q=\"44558879\"]Hint[\/reveal-answer]<\/span>\r\n\r\n<\/div>\r\n[hidden-answer a=\"44558879\"]\r\n<div id=\"fs-id1167794074461\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167794074469\" style=\"text-align: left;\">Use the theorem, along with the identities [latex]\\sin\\alpha \\sin\\beta =\\frac{1}{2}\\left[\\cos\\left(\\alpha -\\beta \\right)-\\cos\\left(\\alpha +\\beta \\right)\\right][\/latex] and [latex]{\\sin}^{2}t=\\frac{1-\\cos2t}{2}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558869\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558869\"]\r\n<div id=\"fs-id1167794074410\" data-type=\"solution\">\r\n<p id=\"fs-id1167794074412\">[latex]A=3\\pi [\/latex] (Note that the integral formula actually yields a negative answer. This is due to the fact that [latex]x\\left(t\\right)[\/latex] is a decreasing function over the interval [latex]\\left[0,2\\pi \\right][\/latex]; that is, the curve is traced from right to left.)<\/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]169501[\/ohm_question]\r\n\r\n<\/div>\r\n<section id=\"fs-id1167794031305\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Arc Length of a Parametric Curve<\/h2>\r\n<p id=\"fs-id1167794031311\">In addition to finding the area under a parametric curve, we sometimes need to find the arc length of a parametric curve. In the case of a line segment, arc length is the same as the distance between the endpoints. If a particle travels from point <em data-effect=\"italics\">A<\/em> to point <em data-effect=\"italics\">B<\/em> along a curve, then the distance that particle travels is the arc length. To develop a formula for arc length, we start with an approximation by line segments as shown in the following graph.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_02_009\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"304\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234759\/CNX_Calc_Figure_11_02_008.jpg\" alt=\"A curved line in the first quadrant with points marked for x = 1, 2, 3, 4, and 5. These points have values roughly 2.1, 2.7, 3, 2.7, and 2.1, respectively. The points for x = 1 and 5 are marked A and B, respectively.\" width=\"304\" height=\"272\" data-media-type=\"image\/jpeg\" \/> Figure 3. Approximation of a curve by line segments.[\/caption]<\/figure>\r\n<p id=\"fs-id1167794031351\">Given a plane curve defined by the functions [latex]x=x\\left(t\\right),y=y\\left(t\\right),a\\le t\\le b[\/latex], we start by partitioning the interval [latex]\\left[a,b\\right][\/latex] into <em data-effect=\"italics\">n<\/em> equal subintervals: [latex]{t}_{0}=a&lt;{t}_{1}&lt;{t}_{2}&lt;\\cdots &lt;{t}_{n}=b[\/latex]. The width of each subinterval is given by [latex]\\Delta t=\\frac{\\left(b-a\\right)}{n}[\/latex]. We can calculate the length of each line segment:<\/p>\r\n\r\n<div id=\"fs-id1167794025695\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ {d}_{1}=\\sqrt{{\\left(x\\left({t}_{1}\\right)-x\\left({t}_{0}\\right)\\right)}^{2}+{\\left(y\\left({t}_{1}\\right)-y\\left({t}_{0}\\right)\\right)}^{2}}\\hfill \\\\ {d}_{2}=\\sqrt{{\\left(x\\left({t}_{2}\\right)-x\\left({t}_{1}\\right)\\right)}^{2}+{\\left(y\\left({t}_{2}\\right)-y\\left({t}_{1}\\right)\\right)}^{2}}\\text{etc}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794027132\">Then add these up. We let <em data-effect=\"italics\">s<\/em> denote the exact arc length and [latex]{s}_{n}[\/latex] denote the approximation by <em data-effect=\"italics\">n<\/em> line segments:<\/p>\r\n\r\n<div id=\"fs-id1167794027155\" style=\"text-align: center;\" data-type=\"equation\">[latex]s\\approx \\displaystyle\\sum _{k=1}^{n}{s}_{k}=\\displaystyle\\sum _{k=1}^{n}\\sqrt{{\\left(x\\left({t}_{k}\\right)-x\\left({t}_{k - 1}\\right)\\right)}^{2}+{\\left(y\\left({t}_{k}\\right)-y\\left({t}_{k - 1}\\right)\\right)}^{2}}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794043116\">If we assume that [latex]x\\left(t\\right)[\/latex] and [latex]y\\left(t\\right)[\/latex] are differentiable functions of <em data-effect=\"italics\">t,<\/em> then the Mean Value Theorem (Introduction to the Applications of Derivatives) applies, so in each subinterval [latex]\\left[{t}_{k - 1},{t}_{k}\\right][\/latex] there exist [latex]\\hat{t}_{k}[\/latex] and [latex]\\tilde{t}_{k}[\/latex] such that<\/p>\r\n\r\n<div id=\"fs-id1167793970821\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ x\\left({t}_{k}\\right)-x\\left({t}_{k - 1}\\right)={x}^{\\prime }\\left(\\hat{t}_{k}\\right)\\left({t}_{k}-{t}_{k - 1}\\right)={x}^{\\prime }\\left(\\hat{t}_{k}\\right)\\Delta t\\hfill \\\\ y\\left({t}_{k}\\right)-y\\left({t}_{k - 1}\\right)={y}^{\\prime }\\left(\\tilde{t}_{k}\\right)\\left({t}_{k}-{t}_{k - 1}\\right)={y}^{\\prime }\\left(\\tilde{t}_{k}\\right)\\Delta t.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794036796\">Therefore [latex]s[\/latex] becomes<\/p>\r\n\r\n<div id=\"fs-id1167793926605\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill s&amp; \\approx {\\displaystyle\\sum _{k=1}^{n}}{s}_{k}\\hfill \\\\ &amp; ={\\displaystyle\\sum _{k=1}^{n}}\\sqrt{{\\left({x}^{\\prime}\\left(\\hat{t}_{k}\\right)\\Delta t\\right)}^{2}+{\\left({y}^{\\prime}\\left(\\tilde{t}_{k}\\right)\\Delta t\\right)}^{2}}\\hfill \\\\ &amp; ={\\displaystyle\\sum _{k=1}^{n}}\\sqrt{{\\left({x}^{\\prime}\\left(\\hat{t}_{k}\\right)\\right)}^{2}{\\left(\\Delta t\\right)}^{2}+{\\left({y}^{\\prime}\\left(\\tilde{t}_{k}\\right)\\right)}^{2}{\\left(\\Delta t\\right)}^{2}}\\hfill \\\\ &amp; =\\left({\\displaystyle\\sum _{k=1}^{n}}\\sqrt{{\\left({x}^{\\prime}\\left(\\hat{t}_{k}\\right)\\right)}^{2}+{\\left({y}^{\\prime}\\left(\\tilde{t}_{k}\\right)\\right)}^{2}}\\right)\\Delta t.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794066389\">This is a Riemann sum that approximates the arc length over a partition of the interval [latex]\\left[a,b\\right][\/latex]. If we further assume that the derivatives are continuous and let the number of points in the partition increase without bound, the approximation approaches the exact arc length. This gives<\/p>\r\n\r\n<div id=\"fs-id1167793424451\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill s&amp; ={\\underset{n\\to\\infty}\\lim} {\\displaystyle\\sum _{k=1}^{n}} {s}_{k}\\hfill \\\\ &amp; = {\\underset{n\\to\\infty}\\lim}\\left({\\displaystyle\\sum _{k=1}^{n}}\\sqrt{{\\left({x}^{\\prime}\\left(\\hat{t}_{k}\\right)\\right)}^{2}+{\\left({y}^{\\prime}\\left(\\tilde{t}_{k}\\right)\\right)}^{2}}\\right)\\Delta t\\hfill \\\\ &amp; ={\\displaystyle\\int_{a}^{b}}\\sqrt{{\\left({x}^{\\prime }\\left(t\\right)\\right)}^{2}+{\\left({y}^{\\prime }\\left(t\\right)\\right)}^{2}}dt.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794065133\">When taking the limit, the values of [latex]\\hat{t}_{k}[\/latex] and [latex]\\tilde{t}_{k}[\/latex] are both contained within the same ever-shrinking interval of width [latex]\\Delta t[\/latex], so they must converge to the same value.<\/p>\r\n<p id=\"fs-id1167794065174\">We can summarize this method in the following theorem.<\/p>\r\n\r\n<div id=\"fs-id1167794065177\" class=\"theorem\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">theorem: Arc Length of a Parametric Curve<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167794065184\">Consider the plane curve defined by the parametric equations<\/p>\r\n\r\n<div id=\"fs-id1167794065188\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x=x\\left(t\\right),y=y\\left(t\\right),{t}_{1}\\le t\\le {t}_{2}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794023163\">and assume that [latex]x\\left(t\\right)[\/latex] and [latex]y\\left(t\\right)[\/latex] are differentiable functions of <em data-effect=\"italics\">t.<\/em> Then the arc length of this curve is given by<\/p>\r\n\r\n<div id=\"fs-id1167794023197\" style=\"text-align: center;\" data-type=\"equation\">[latex]s={\\displaystyle\\int }_{{t}_{1}}^{{t}_{2}}\\sqrt{{\\left(\\frac{dx}{dt}\\right)}^{2}+{\\left(\\frac{dy}{dt}\\right)}^{2}}dt[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167794023283\">At this point a side derivation leads to a previous formula for arc length. In particular, suppose the parameter can be eliminated, leading to a function [latex]y=F\\left(x\\right)[\/latex]. Then [latex]y\\left(t\\right)=F\\left(x\\left(t\\right)\\right)[\/latex] and the Chain Rule gives [latex]{y}^{\\prime }\\left(t\\right)={F}^{\\prime }\\left(x\\left(t\\right)\\right){x}^{\\prime }\\left(t\\right)[\/latex]. Substituting this into the theorem gives<\/p>\r\n\r\n<div id=\"fs-id1167794051442\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill s&amp; ={\\displaystyle\\int }_{{t}_{1}}^{{t}_{2}}\\sqrt{{\\left(\\frac{dx}{dt}\\right)}^{2}+{\\left(\\frac{dy}{dt}\\right)}^{2}}dt\\hfill \\\\ &amp; ={\\displaystyle\\int }_{{t}_{1}}^{{t}_{2}}\\sqrt{{\\left(\\frac{dx}{dt}\\right)}^{2}+{\\left({F}^{\\prime }\\left(x\\right)\\frac{dx}{dt}\\right)}^{2}}dt\\hfill \\\\ &amp; ={\\displaystyle\\int }_{{t}_{1}}^{{t}_{2}}\\sqrt{{\\left(\\frac{dx}{dt}\\right)}^{2}\\left(1+{\\left({F}^{\\prime }\\left(x\\right)\\right)}^{2}\\right)}dt\\hfill \\\\ &amp; ={\\displaystyle\\int }_{{t}_{1}}^{{t}_{2}}{x}^{\\prime }\\left(t\\right)\\sqrt{1+{\\left(\\frac{dy}{dx}\\right)}^{2}}dt.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794003192\">Here we have assumed that [latex]{x}^{\\prime }\\left(t\\right)&gt;0[\/latex], which is a reasonable assumption. The Chain Rule gives [latex]dx={x}^{\\prime }\\left(t\\right)dt[\/latex], and letting [latex]a=x\\left({t}_{1}\\right)[\/latex] and [latex]b=x\\left({t}_{2}\\right)[\/latex] we obtain the formula<\/p>\r\n\r\n<div id=\"fs-id1167794100345\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]s={\\displaystyle\\int }_{a}^{b}\\sqrt{1+{\\left(\\frac{dy}{dx}\\right)}^{2}}dx[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794100404\">which is the formula for arc length obtained in the Introduction to the Applications of Integration.<\/p>\r\n\r\n<div id=\"fs-id1167794100414\" data-type=\"example\">\r\n<div id=\"fs-id1167794100417\" data-type=\"exercise\">\r\n<div id=\"fs-id1167794100419\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Arc Length of a Parametric Curve<\/h3>\r\n<div id=\"fs-id1167794100419\" data-type=\"problem\">\r\n<p id=\"fs-id1167794100424\">Find the arc length of the semicircle defined by the equations<\/p>\r\n\r\n<div id=\"fs-id1167794100427\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)=3\\cos{t},y\\left(t\\right)=3\\sin{t},0\\le t\\le \\pi [\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558859\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558859\"]\r\n<div id=\"fs-id1167794100493\" data-type=\"solution\">\r\n<p id=\"fs-id1167794100495\">The values [latex]t=0[\/latex] to [latex]t=\\pi [\/latex] trace out the red curve in Figure 9. To determine its length, use the theorem:<\/p>\r\n\r\n<div id=\"fs-id1167794061688\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill s&amp; ={\\displaystyle\\int }_{{t}_{1}}^{{t}_{2}}\\sqrt{{\\left(\\frac{dx}{dt}\\right)}^{2}+{\\left(\\frac{dy}{dt}\\right)}^{2}}dt\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{\\pi }\\sqrt{{\\left(-3\\sin{t}\\right)}^{2}+{\\left(3\\cos{t}\\right)}^{2}}dt\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{\\pi }\\sqrt{9{\\sin}^{2}t+9{\\cos}^{2}t}dt\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{\\pi }\\sqrt{9\\left({\\sin}^{2}t+{\\cos}^{2}t\\right)}dt\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{\\pi }3dt={3t|}_{0}^{\\pi }=3\\pi .\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794051248\">Note that the formula for the arc length of a semicircle is [latex]\\pi r[\/latex] and the radius of this circle is 3. This is a great example of using calculus to derive a known formula of a geometric quantity.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_02_010\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"339\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234801\/CNX_Calc_Figure_11_02_009.jpg\" alt=\"A semicircle is drawn with radius 3. There is an arrow pointing counterclockwise. On the graph there are also written three equations: x(t) = 3 cos(t), y(t) = 3 sin(t), and 0 \u2264 [latex]t[\/latex] \u2264 \u03c0.\" width=\"339\" height=\"347\" data-media-type=\"image\/jpeg\" \/> Figure 4. The arc length of the semicircle is equal to its radius times [latex]\\pi[\/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 the Arc Length 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=893&amp;end=993&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\/7.2CalculusOfParametricCurves893to993_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"7.2 Calculus of Parametric Curves\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1167794051286\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167794051290\" data-type=\"exercise\">\r\n<div id=\"fs-id1167794051292\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167794051292\" data-type=\"problem\">\r\n<p id=\"fs-id1167794051294\">Find the arc length of the curve defined by the equations<\/p>\r\n\r\n<div id=\"fs-id1167794051297\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)=3{t}^{2},y\\left(t\\right)=2{t}^{3},1\\le t\\le 3[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558839\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558839\"]\r\n<div id=\"fs-id1167794027824\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167794027831\">Use the theorem.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558849\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558849\"]\r\n<div id=\"fs-id1167794051365\" data-type=\"solution\">\r\n<p id=\"fs-id1167794051368\" style=\"text-align: center;\">[latex]s=2\\left({10}^{\\frac{3}{2}}-{2}^{\\frac{3}{2}}\\right)\\approx 57.589[\/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]169505[\/ohm_question]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167794027842\">We now return to the problem posed at the beginning of the section about a baseball leaving a pitcher\u2019s hand. Ignoring the effect of air resistance (unless it is a curve ball!), the ball travels a parabolic path. Assuming the pitcher\u2019s hand is at the origin and the ball travels left to right in the direction of the positive <em data-effect=\"italics\">x<\/em>-axis, the parametric equations for this curve can be written as<\/p>\r\n\r\n<div id=\"fs-id1167794027857\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)=140t,y\\left(t\\right)=-16{t}^{2}+2t[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794027910\">where <em data-effect=\"italics\">t<\/em> represents time. We first calculate the distance the ball travels as a function of time. This distance is represented by the arc length. We can modify the arc length formula slightly. First rewrite the functions [latex]x\\left(t\\right)[\/latex] and [latex]y\\left(t\\right)[\/latex] using <em data-effect=\"italics\">v<\/em> as an independent variable, so as to eliminate any confusion with the parameter <em data-effect=\"italics\">t<\/em>:<\/p>\r\n\r\n<div id=\"fs-id1167794027956\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(v\\right)=140v,y\\left(v\\right)=-16{v}^{2}+2v[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794028011\">Then we write the arc length formula as follows:<\/p>\r\n\r\n<div id=\"fs-id1167794028015\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill s\\left(t\\right)&amp; ={\\displaystyle\\int }_{0}^{t}\\sqrt{{\\left(\\frac{dx}{dv}\\right)}^{2}+{\\left(\\frac{dy}{dv}\\right)}^{2}}dv\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{t}\\sqrt{{140}^{2}+{\\left(-32v+2\\right)}^{2}}dv.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793926178\">The variable <em data-effect=\"italics\">v<\/em> acts as a dummy variable that disappears after integration, leaving the arc length as a function of time <em data-effect=\"italics\">t.<\/em> To integrate this expression we can use a formula from Appendix A,<\/p>\r\n\r\n<div id=\"fs-id1167793926200\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\sqrt{{a}^{2}+{u}^{2}}du=\\frac{u}{2}\\sqrt{{a}^{2}+{u}^{2}}+\\frac{{a}^{2}}{2}\\text{ln}|u+\\sqrt{{a}^{2}+{u}^{2}}|+C[\/latex].<\/div>\r\n<p id=\"fs-id1167794094267\">We set [latex]a=140[\/latex] and [latex]u=-32v+2[\/latex]. This gives [latex]du=-32dv[\/latex], so [latex]dv=-\\frac{1}{32}du[\/latex]. Therefore<\/p>\r\n\r\n<div id=\"fs-id1167794094341\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int \\sqrt{{140}^{2}+{\\left(-32v+2\\right)}^{2}}dv}&amp; =-\\frac{1}{32}{\\displaystyle\\int \\sqrt{{a}^{2}+{u}^{2}}du}\\hfill \\\\ &amp; =-\\frac{1}{32}\\left[\\begin{array}{c}\\frac{\\left(-32v+2\\right)}{2}\\sqrt{{140}^{2}+{\\left(-32v+2\\right)}^{2}}\\hfill \\\\ +\\frac{{140}^{2}}{2}\\text{ln}|\\left(-32v+2\\right)+\\sqrt{{140}^{2}+{\\left(-32v+2\\right)}^{2}}|\\hfill \\end{array}\\right]+C\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794035510\">and<\/p>\r\n\r\n<div id=\"fs-id1167794035513\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill s\\left(t\\right)&amp; =-\\frac{1}{32}\\left[\\frac{\\left(-32t+2\\right)}{2}\\sqrt{{140}^{2}+{\\left(-32t+2\\right)}^{2}}+\\frac{{140}^{2}}{2}\\text{ln}|\\left(-32t+2\\right)+\\sqrt{{140}^{2}+{\\left(-32t+2\\right)}^{2}}|\\right]\\hfill \\\\ &amp; +\\frac{1}{32}\\left[\\sqrt{{140}^{2}+{2}^{2}}+\\frac{{140}^{2}}{2}\\text{ln}|2+\\sqrt{{140}^{2}+{2}^{2}}|\\right]\\hfill \\\\ &amp; =\\left(\\frac{t}{2}-\\frac{1}{32}\\right)\\sqrt{1024{t}^{2}-128t+19604}-\\frac{1225}{4}\\text{ln}|\\left(-32t+2\\right)+\\sqrt{1024{t}^{2}-128t+19604}|\\hfill \\\\ &amp; +\\frac{\\sqrt{19604}}{32}+\\frac{1225}{4}\\text{ln}\\left(2+\\sqrt{19604}\\right).\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794095955\">This function represents the distance traveled by the ball as a function of time. To calculate the speed, take the derivative of this function with respect to <em data-effect=\"italics\">t.<\/em> While this may seem like a daunting task, it is possible to obtain the answer directly from the Fundamental Theorem of Calculus:<\/p>\r\n\r\n<div id=\"fs-id1167794095966\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{d}{dx}{\\displaystyle\\int }_{a}^{x}f\\left(u\\right)du=f\\left(x\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794096023\">Therefore<\/p>\r\n\r\n<div id=\"fs-id1167794096026\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {s}^{\\prime }\\left(t\\right)&amp; =\\frac{d}{dt}\\left[s\\left(t\\right)\\right]\\hfill \\\\ &amp; =\\frac{d}{dt}\\left[{\\displaystyle\\int }_{0}^{t}\\sqrt{{140}^{2}+{\\left(-32v+2\\right)}^{2}}dv\\right]\\hfill \\\\ &amp; =\\sqrt{{140}^{2}+{\\left(-32t+2\\right)}^{2}}\\hfill \\\\ &amp; =\\sqrt{1024{t}^{2}-128t+19604}\\hfill \\\\ &amp; =2\\sqrt{256{t}^{2}-32t+4901}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794040883\">One third of a second after the ball leaves the pitcher\u2019s hand, the distance it travels is equal to<\/p>\r\n\r\n<div id=\"fs-id1167794040888\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill s\\left(\\frac{1}{3}\\right)&amp; =\\left(\\frac{\\frac{1}{3}}{2}-\\frac{1}{32}\\right)\\sqrt{1024{\\left(\\frac{1}{3}\\right)}^{2}-128\\left(\\frac{1}{3}\\right)+19604}\\hfill \\\\ &amp; -\\frac{1225}{4}\\text{ln}|\\left(-32\\left(\\frac{1}{3}\\right)+2\\right)+\\sqrt{1024{\\left(\\frac{1}{3}\\right)}^{2}-128\\left(\\frac{1}{3}\\right)+19604}|\\hfill \\\\ &amp; +\\frac{\\sqrt{19604}}{32}+\\frac{1225}{4}\\text{ln}\\left(2+\\sqrt{19604}\\right)\\hfill \\\\ &amp; \\approx 46.69\\text{feet}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794135263\">This value is just over three quarters of the way to home plate. The speed of the ball is<\/p>\r\n\r\n<div id=\"fs-id1167794135267\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{s}^{\\prime }\\left(\\frac{1}{3}\\right)=2\\sqrt{256{\\left(\\frac{1}{3}\\right)}^{2}-16\\left(\\frac{1}{3}\\right)+4901}\\approx 140.34\\text{ft\/s}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794135348\">This speed translates to approximately 95 mph\u2014a major-league fastball.<\/p>\r\n\r\n<\/section><section id=\"fs-id1167794135353\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Surface Area Generated by a Parametric Curve<\/h2>\r\n<p id=\"fs-id1167794135359\">Recall the problem of finding the surface area of a volume of revolution. In Curve Length and Surface Area, we derived a formula for finding the surface area of a volume generated by a function [latex]y=f\\left(x\\right)[\/latex] from [latex]x=a[\/latex] to [latex]x=b[\/latex], revolved around the <em data-effect=\"italics\">x<\/em>-axis:<\/p>\r\n\r\n<div id=\"fs-id1167794135414\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]S=2\\pi {\\displaystyle\\int }_{a}^{b}f\\left(x\\right)\\sqrt{1+{\\left({f}^{\\prime }\\left(x\\right)\\right)}^{2}}dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794096354\">We now consider a volume of revolution generated by revolving a parametrically defined curve [latex]x=x\\left(t\\right),y=y\\left(t\\right),a\\le t\\le b[\/latex] around the <em data-effect=\"italics\">x<\/em>-axis as shown in the following figure.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_02_011\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"341\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234803\/CNX_Calc_Figure_11_02_010.jpg\" alt=\"A curve is drawn in the first quadrant with endpoints marked [latex]t[\/latex] = a and [latex]t[\/latex] = b. On this curve, there is a point marked (x(t), y(t)). There is a circle with an arrow drawn around the [latex]x[\/latex]-axis that seems to indicate a rotation about the x axis, and there is a shape that accompanies that curve that seems to be what you would obtain if you rotated the curve about the x axis.\" width=\"341\" height=\"276\" data-media-type=\"image\/jpeg\" \/> Figure 5. A surface of revolution generated by a parametrically defined curve.[\/caption]<\/figure>\r\n<p id=\"fs-id1167794096433\">The analogous formula for a parametrically defined curve is<\/p>\r\n\r\n<div id=\"fs-id1167794096436\" style=\"text-align: center;\" data-type=\"equation\">[latex]S=2\\pi {\\displaystyle\\int }_{a}^{b}y\\left(t\\right)\\sqrt{{\\left({x}^{\\prime }\\left(t\\right)\\right)}^{2}+{\\left({y}^{\\prime }\\left(t\\right)\\right)}^{2}}dt[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794096527\">provided that [latex]y\\left(t\\right)[\/latex] is not negative on [latex]\\left[a,b\\right][\/latex].<\/p>\r\n\r\n<div id=\"fs-id1167794076111\" data-type=\"example\">\r\n<div id=\"fs-id1167794076114\" data-type=\"exercise\">\r\n<div id=\"fs-id1167794076116\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding Surface Area<\/h3>\r\n<div id=\"fs-id1167794076116\" data-type=\"problem\">\r\n<p id=\"fs-id1167794076121\">Find the surface area of a sphere of radius [latex]r[\/latex] centered at the origin.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558829\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558829\"]\r\n<div id=\"fs-id1167794076130\" data-type=\"solution\">\r\n<p id=\"fs-id1167794076133\">We start with the curve defined by the equations<\/p>\r\n\r\n<div id=\"fs-id1167794076136\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)=r\\cos{t},y\\left(t\\right)=r\\sin{t},0\\le t\\le \\pi [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794076201\">This generates an upper semicircle of radius <em data-effect=\"italics\">r<\/em> centered at the origin as shown in the following graph.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_02_012\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"272\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234805\/CNX_Calc_Figure_11_02_011.jpg\" alt=\"A semicircle is drawn with radius r. On the graph there are also written three equations: x(t) = r cos(t), y(t) = r sin(t), and 0 \u2264 [latex]t[\/latex] \u2264 \u03c0.\" width=\"272\" height=\"271\" data-media-type=\"image\/jpeg\" \/> Figure 6. A semicircle generated by parametric equations.[\/caption]<\/figure>\r\n<p id=\"fs-id1167794076232\">When this curve is revolved around the\u00a0[latex]x[\/latex]-axis, it generates a sphere of radius [latex]r[\/latex]. To calculate the surface area of the sphere, we use the above equation:<\/p>\r\n\r\n<div id=\"fs-id1167794076251\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill S&amp;=2\\pi {\\displaystyle\\int_{a}^{b}} y\\left(t\\right)\\sqrt{{\\left({x}^{\\prime }\\left(t\\right)\\right)}^{2}+{\\left({y}^{\\prime}\\left(t\\right)\\right)}^{2}}dt\\hfill \\\\ &amp;=2\\pi {\\displaystyle\\int_{0}^{\\pi}} r\\sin{t}\\sqrt{{\\left(-r\\sin{t}\\right)}^{2}+{\\left(r\\cos{t}\\right)}^{2}}dt\\hfill \\\\ &amp;=2\\pi {\\displaystyle\\int }_{0}^{\\pi }r\\sin{t}\\sqrt{{r}^{2}{\\sin}^{2}t+{r}^{2}{\\cos}^{2}t}dt\\hfill \\\\ &amp;=2\\pi {\\displaystyle\\int }_{0}^{\\pi }r\\sin{t}\\sqrt{{r}^{2}\\left({\\sin}^{2}t+{\\cos}^{2}t\\right)}dt\\hfill \\\\ &amp;=2\\pi {\\displaystyle\\int_{0}^{\\pi}{r}^{2}\\sin{t}dt}\\hfill \\\\ &amp;=2\\pi{r}^{2}\\left(-\\cos{t}|_{0}^{\\pi}\\right)\\hfill \\\\ &amp; =2\\pi {r}^{2}\\left(-\\cos\\pi +\\cos0\\right)\\hfill \\\\ &amp; =4\\pi {r}^{2}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794055273\">This is, in fact, the formula for the surface area of a sphere.<\/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>\r\nWatch the following video to see the worked solution to Example: Finding Surface Area.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/bDvDdC2Wpu8?controls=0&amp;start=1040&amp;end=1263&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\/7.2CalculusOfParametricCurves1040to1263_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-id1167794055279\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167794055283\" data-type=\"exercise\">\r\n<div id=\"fs-id1167794055286\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167794055286\" data-type=\"problem\">\r\n<p id=\"fs-id1167794055288\">Find the surface area generated when the plane curve defined by the equations<\/p>\r\n\r\n<div id=\"fs-id1167794055291\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)={t}^{3},y\\left(t\\right)={t}^{2},0\\le t\\le 1[\/latex]<\/div>\r\n<p id=\"fs-id1167794055352\">is revolved around the <em data-effect=\"italics\">x<\/em>-axis.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558809\"]Hint[\/reveal-answer]\r\n\r\n[hidden-answer a=\"44558809\"]\r\n<div id=\"fs-id1167794055397\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167794055405\">Use the above equation. When evaluating the integral, use a <em data-effect=\"italics\">u<\/em>-substitution.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558819\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"44558819\"]\r\n<div id=\"fs-id1167794055361\" data-type=\"solution\">\r\n<p id=\"fs-id1167794055363\">[latex]A=\\frac{\\pi \\left(494\\sqrt{13}+128\\right)}{1215}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1167794048157\" class=\"section-exercises\" data-depth=\"1\">\r\n<div id=\"fs-id1167794046049\" data-type=\"exercise\"><\/div>\r\n<\/section>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Find the area under a parametric curve<\/li>\n<li>Use the equation for arc length of a parametric curve<\/li>\n<li>Apply the formula for surface area to a volume generated by a parametric curve<\/li>\n<\/ul>\n<\/div>\n<h2 data-type=\"title\">Integrals Involving Parametric Equations<\/h2>\n<p id=\"fs-id1167794065409\">Now that we have seen how to calculate the derivative of a plane curve, the next question is this: How do we find the area under a curve defined parametrically? Recall the cycloid defined by the equations [latex]x\\left(t\\right)=t-\\sin{t},y\\left(t\\right)=1-\\cos{t}[\/latex]. Suppose we want to find the area of the shaded region in the following graph.<\/p>\n<figure id=\"CNX_Calc_Figure_11_02_007\"><figcaption><\/figcaption><div style=\"width: 497px\" 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\/11234756\/CNX_Calc_Figure_11_02_007.jpg\" alt=\"A series of half circles drawn above the [latex]x[\/latex]-axis with x intercepts being multiples of 2\u03c0. The half circle between 0 and 2\u03c0 is highlighted. On the graph there are also written two equations: x(t) = [latex]t[\/latex] \u2013 sin(t) and y(t) = 1 \u2013 cos(t).\" width=\"487\" height=\"347\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. Graph of a cycloid with the arch over [latex]\\left[0,2\\pi \\right][\/latex] highlighted.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1167794030645\">To derive a formula for the area under the curve defined by the functions<\/p>\n<div id=\"fs-id1167794030648\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x=x\\left(t\\right),y=y\\left(t\\right),a\\le t\\le b[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794030704\">we assume that [latex]x\\left(t\\right)[\/latex] is differentiable and start with an equal partition of the interval [latex]a\\le t\\le b[\/latex]. Suppose [latex]{t}_{0}=a<{t}_{1}<{t}_{2}<\\cdots <{t}_{n}=b[\/latex] and consider the following graph.<\/p>\n<figure id=\"CNX_Calc_Figure_11_02_008\"><figcaption><\/figcaption><div style=\"width: 243px\" 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\/11234758\/CNX_Calc_Figure_11_02_012.jpg\" alt=\"A curved line is drawn in the first quadrant. Below it are a series of rectangles marked that begin at the [latex]x[\/latex]-axis and reach up to the curved line; the rectangle\u2019s height is determined by the location of the curved line at the leftmost point of the rectangle. These lines are noted as x(t0), x(t1), \u2026, x(tn).\" width=\"233\" height=\"245\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. Approximating the area under a parametrically defined curve.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1167794030800\">We use rectangles to approximate the area under the curve. The height of a typical rectangle in this parametrization is [latex]y\\left(x\\left({\\overline{t}}_{i}\\right)\\right)[\/latex] for some value [latex]{\\overline{t}}_{i}[\/latex] in the <em data-effect=\"italics\">i<\/em>th subinterval, and the width can be calculated as [latex]x\\left({t}_{i}\\right)-x\\left({t}_{i - 1}\\right)[\/latex]. Thus the area of the <em data-effect=\"italics\">i<\/em>th rectangle is given by<\/p>\n<div id=\"fs-id1167794065566\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{A}_{i}=y\\left(x\\left({\\overline{t}}_{i}\\right)\\right)\\left(x\\left({t}_{i}\\right)-x\\left({t}_{i - 1}\\right)\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794065651\">Then a Riemann sum for the area is<\/p>\n<div id=\"fs-id1167794065654\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{A}_{n}=\\displaystyle\\sum _{i=1}^{n}y\\left(x\\left({\\overline{t}}_{i}\\right)\\right)\\left(x\\left({t}_{i}\\right)-x\\left({t}_{i - 1}\\right)\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794064679\">Multiplying and dividing each area by [latex]{t}_{i}-{t}_{i - 1}[\/latex] gives<\/p>\n<div id=\"fs-id1167794064704\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{A}_{n}=\\displaystyle\\sum _{i=1}^{n}y\\left(x\\left({\\overline{t}}_{i}\\right)\\right)\\left(\\frac{x\\left({t}_{i}\\right)-x\\left({t}_{i - 1}\\right)}{{t}_{i}-{t}_{i - 1}}\\right)\\left({t}_{i}-{t}_{i - 1}\\right)=\\displaystyle\\sum _{i=1}^{n}y\\left(x\\left({\\overline{t}}_{i}\\right)\\right)\\left(\\frac{x\\left({t}_{i}\\right)-x\\left({t}_{i - 1}\\right)}{\\Delta t}\\right)\\Delta t[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794000276\">Taking the limit as [latex]n[\/latex] approaches infinity gives<\/p>\n<div id=\"fs-id1167794000283\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]A=\\underset{n\\to \\infty }{\\text{lim}}{A}_{n}={\\displaystyle\\int }_{a}^{b}y\\left(t\\right){x}^{\\prime }\\left(t\\right)dt[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>Note that as the value of [latex]n[\/latex] approaches [latex]\\infty[\/latex], the change in [latex]x[\/latex] over smaller and smaller time intervals can be rewritten as the instantaneous rate of change of [latex]x[\/latex] with respect to [latex]t[\/latex] at some value within the sub-interval of [latex]t[\/latex]. Recall that this fact is known as the Mean Value Theorem, and we briefly review it to clarify the proofs within this section.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Mean Value Theorem<\/h3>\n<hr \/>\n<p id=\"fs-id1165042357018\">Let [latex]f[\/latex] be continuous over the closed interval [latex][a,b][\/latex] and differentiable over the open interval [latex](a,b)[\/latex]. Then, there exists at least one point [latex]c \\in (a,b)[\/latex] such that<\/p>\n<div id=\"fs-id1165043066505\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(c)=\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\n<\/div>\n<p id=\"fs-id1167794038312\">The preceding result leads to the following theorem.<\/p>\n<div id=\"fs-id1167794038315\" class=\"theorem\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">theorem: Area under a Parametric Curve<\/h3>\n<hr \/>\n<p id=\"fs-id1167794038322\">Consider the non-self-intersecting plane curve defined by the parametric equations<\/p>\n<div id=\"fs-id1167794038325\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x=x\\left(t\\right),y=y\\left(t\\right),a\\le t\\le b[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794038378\">and assume that [latex]x\\left(t\\right)[\/latex] is differentiable. The area under this curve is given by<\/p>\n<div id=\"fs-id1167794038394\" style=\"text-align: center;\" data-type=\"equation\">[latex]A={\\displaystyle\\int }_{a}^{b}y\\left(t\\right){x}^{\\prime }\\left(t\\right)dt[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167794038444\" data-type=\"example\">\n<div id=\"fs-id1167794038447\" data-type=\"exercise\">\n<div id=\"fs-id1167794038449\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Area under a Parametric Curve<\/h3>\n<div id=\"fs-id1167794038449\" data-type=\"problem\">\n<p id=\"fs-id1167794038454\">Find the area under the curve of the cycloid defined by the equations<\/p>\n<div id=\"fs-id1167794038457\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)=t-\\sin{t},y\\left(t\\right)=1-\\cos{t},0\\le t\\le 2\\pi[\/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=\"q44558889\">Show Solution<\/span><\/p>\n<div id=\"q44558889\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794070824\" data-type=\"solution\">\n<p id=\"fs-id1167794070826\">Using the theorem, we have<\/p>\n<div id=\"fs-id1167794070833\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill A& ={\\displaystyle\\int }_{a}^{b}y\\left(t\\right){x}^{\\prime }\\left(t\\right)dt\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{2\\pi }\\left(1-\\cos{t}\\right)\\left(1-\\cos{t}\\right)dt\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{2\\pi }\\left(1 - 2\\cos{t}+{\\cos}^{2}t\\right)dt\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{2\\pi }\\left(1 - 2\\cos{t}+\\frac{1+\\cos2t}{2}\\right)dt\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{2\\pi }\\left(\\frac{3}{2}-2\\cos{t}+\\frac{\\cos2t}{2}\\right)dt\\hfill \\\\ & ={\\frac{3t}{2}-2\\sin{t}+\\frac{\\sin2t}{4}|}_{0}^{2\\pi }\\hfill \\\\ & =3\\pi .\\hfill \\end{array}[\/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 the Area under 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=718&amp;end=855&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\/7.2CalculusOfParametricCurves718to855_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-id1167794074316\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167794074320\" data-type=\"exercise\">\n<div id=\"fs-id1167794074322\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167794074322\" data-type=\"problem\">\n<p id=\"fs-id1167794074324\">Find the area under the curve of the hypocycloid defined by the equations<\/p>\n<div id=\"fs-id1167794074327\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)=3\\cos{t}+\\cos3t,y\\left(t\\right)=3\\sin{t}-\\sin3t,0\\le t\\le \\pi[\/latex].<\/div>\n<div data-type=\"equation\" data-label=\"\"><\/div>\n<p><span style=\"font-size: 1rem; text-align: initial;\"><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558879\">Hint<\/span><\/span><\/p>\n<\/div>\n<div id=\"q44558879\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794074461\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167794074469\" style=\"text-align: left;\">Use the theorem, along with the identities [latex]\\sin\\alpha \\sin\\beta =\\frac{1}{2}\\left[\\cos\\left(\\alpha -\\beta \\right)-\\cos\\left(\\alpha +\\beta \\right)\\right][\/latex] and [latex]{\\sin}^{2}t=\\frac{1-\\cos2t}{2}[\/latex].<\/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=\"q44558869\">Show Solution<\/span><\/p>\n<div id=\"q44558869\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794074410\" data-type=\"solution\">\n<p id=\"fs-id1167794074412\">[latex]A=3\\pi[\/latex] (Note that the integral formula actually yields a negative answer. This is due to the fact that [latex]x\\left(t\\right)[\/latex] is a decreasing function over the interval [latex]\\left[0,2\\pi \\right][\/latex]; that is, the curve is traced from right to left.)<\/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=\"ohm169501\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169501&theme=oea&iframe_resize_id=ohm169501&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<section id=\"fs-id1167794031305\" data-depth=\"1\">\n<h2 data-type=\"title\">Arc Length of a Parametric Curve<\/h2>\n<p id=\"fs-id1167794031311\">In addition to finding the area under a parametric curve, we sometimes need to find the arc length of a parametric curve. In the case of a line segment, arc length is the same as the distance between the endpoints. If a particle travels from point <em data-effect=\"italics\">A<\/em> to point <em data-effect=\"italics\">B<\/em> along a curve, then the distance that particle travels is the arc length. To develop a formula for arc length, we start with an approximation by line segments as shown in the following graph.<\/p>\n<figure id=\"CNX_Calc_Figure_11_02_009\"><figcaption><\/figcaption><div style=\"width: 314px\" 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\/11234759\/CNX_Calc_Figure_11_02_008.jpg\" alt=\"A curved line in the first quadrant with points marked for x = 1, 2, 3, 4, and 5. These points have values roughly 2.1, 2.7, 3, 2.7, and 2.1, respectively. The points for x = 1 and 5 are marked A and B, respectively.\" width=\"304\" height=\"272\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. Approximation of a curve by line segments.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1167794031351\">Given a plane curve defined by the functions [latex]x=x\\left(t\\right),y=y\\left(t\\right),a\\le t\\le b[\/latex], we start by partitioning the interval [latex]\\left[a,b\\right][\/latex] into <em data-effect=\"italics\">n<\/em> equal subintervals: [latex]{t}_{0}=a<{t}_{1}<{t}_{2}<\\cdots <{t}_{n}=b[\/latex]. The width of each subinterval is given by [latex]\\Delta t=\\frac{\\left(b-a\\right)}{n}[\/latex]. We can calculate the length of each line segment:<\/p>\n<div id=\"fs-id1167794025695\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ {d}_{1}=\\sqrt{{\\left(x\\left({t}_{1}\\right)-x\\left({t}_{0}\\right)\\right)}^{2}+{\\left(y\\left({t}_{1}\\right)-y\\left({t}_{0}\\right)\\right)}^{2}}\\hfill \\\\ {d}_{2}=\\sqrt{{\\left(x\\left({t}_{2}\\right)-x\\left({t}_{1}\\right)\\right)}^{2}+{\\left(y\\left({t}_{2}\\right)-y\\left({t}_{1}\\right)\\right)}^{2}}\\text{etc}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794027132\">Then add these up. We let <em data-effect=\"italics\">s<\/em> denote the exact arc length and [latex]{s}_{n}[\/latex] denote the approximation by <em data-effect=\"italics\">n<\/em> line segments:<\/p>\n<div id=\"fs-id1167794027155\" style=\"text-align: center;\" data-type=\"equation\">[latex]s\\approx \\displaystyle\\sum _{k=1}^{n}{s}_{k}=\\displaystyle\\sum _{k=1}^{n}\\sqrt{{\\left(x\\left({t}_{k}\\right)-x\\left({t}_{k - 1}\\right)\\right)}^{2}+{\\left(y\\left({t}_{k}\\right)-y\\left({t}_{k - 1}\\right)\\right)}^{2}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794043116\">If we assume that [latex]x\\left(t\\right)[\/latex] and [latex]y\\left(t\\right)[\/latex] are differentiable functions of <em data-effect=\"italics\">t,<\/em> then the Mean Value Theorem (Introduction to the Applications of Derivatives) applies, so in each subinterval [latex]\\left[{t}_{k - 1},{t}_{k}\\right][\/latex] there exist [latex]\\hat{t}_{k}[\/latex] and [latex]\\tilde{t}_{k}[\/latex] such that<\/p>\n<div id=\"fs-id1167793970821\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ x\\left({t}_{k}\\right)-x\\left({t}_{k - 1}\\right)={x}^{\\prime }\\left(\\hat{t}_{k}\\right)\\left({t}_{k}-{t}_{k - 1}\\right)={x}^{\\prime }\\left(\\hat{t}_{k}\\right)\\Delta t\\hfill \\\\ y\\left({t}_{k}\\right)-y\\left({t}_{k - 1}\\right)={y}^{\\prime }\\left(\\tilde{t}_{k}\\right)\\left({t}_{k}-{t}_{k - 1}\\right)={y}^{\\prime }\\left(\\tilde{t}_{k}\\right)\\Delta t.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794036796\">Therefore [latex]s[\/latex] becomes<\/p>\n<div id=\"fs-id1167793926605\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill s& \\approx {\\displaystyle\\sum _{k=1}^{n}}{s}_{k}\\hfill \\\\ & ={\\displaystyle\\sum _{k=1}^{n}}\\sqrt{{\\left({x}^{\\prime}\\left(\\hat{t}_{k}\\right)\\Delta t\\right)}^{2}+{\\left({y}^{\\prime}\\left(\\tilde{t}_{k}\\right)\\Delta t\\right)}^{2}}\\hfill \\\\ & ={\\displaystyle\\sum _{k=1}^{n}}\\sqrt{{\\left({x}^{\\prime}\\left(\\hat{t}_{k}\\right)\\right)}^{2}{\\left(\\Delta t\\right)}^{2}+{\\left({y}^{\\prime}\\left(\\tilde{t}_{k}\\right)\\right)}^{2}{\\left(\\Delta t\\right)}^{2}}\\hfill \\\\ & =\\left({\\displaystyle\\sum _{k=1}^{n}}\\sqrt{{\\left({x}^{\\prime}\\left(\\hat{t}_{k}\\right)\\right)}^{2}+{\\left({y}^{\\prime}\\left(\\tilde{t}_{k}\\right)\\right)}^{2}}\\right)\\Delta t.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794066389\">This is a Riemann sum that approximates the arc length over a partition of the interval [latex]\\left[a,b\\right][\/latex]. If we further assume that the derivatives are continuous and let the number of points in the partition increase without bound, the approximation approaches the exact arc length. This gives<\/p>\n<div id=\"fs-id1167793424451\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill s& ={\\underset{n\\to\\infty}\\lim} {\\displaystyle\\sum _{k=1}^{n}} {s}_{k}\\hfill \\\\ & = {\\underset{n\\to\\infty}\\lim}\\left({\\displaystyle\\sum _{k=1}^{n}}\\sqrt{{\\left({x}^{\\prime}\\left(\\hat{t}_{k}\\right)\\right)}^{2}+{\\left({y}^{\\prime}\\left(\\tilde{t}_{k}\\right)\\right)}^{2}}\\right)\\Delta t\\hfill \\\\ & ={\\displaystyle\\int_{a}^{b}}\\sqrt{{\\left({x}^{\\prime }\\left(t\\right)\\right)}^{2}+{\\left({y}^{\\prime }\\left(t\\right)\\right)}^{2}}dt.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794065133\">When taking the limit, the values of [latex]\\hat{t}_{k}[\/latex] and [latex]\\tilde{t}_{k}[\/latex] are both contained within the same ever-shrinking interval of width [latex]\\Delta t[\/latex], so they must converge to the same value.<\/p>\n<p id=\"fs-id1167794065174\">We can summarize this method in the following theorem.<\/p>\n<div id=\"fs-id1167794065177\" class=\"theorem\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">theorem: Arc Length of a Parametric Curve<\/h3>\n<hr \/>\n<p id=\"fs-id1167794065184\">Consider the plane curve defined by the parametric equations<\/p>\n<div id=\"fs-id1167794065188\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x=x\\left(t\\right),y=y\\left(t\\right),{t}_{1}\\le t\\le {t}_{2}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794023163\">and assume that [latex]x\\left(t\\right)[\/latex] and [latex]y\\left(t\\right)[\/latex] are differentiable functions of <em data-effect=\"italics\">t.<\/em> Then the arc length of this curve is given by<\/p>\n<div id=\"fs-id1167794023197\" style=\"text-align: center;\" data-type=\"equation\">[latex]s={\\displaystyle\\int }_{{t}_{1}}^{{t}_{2}}\\sqrt{{\\left(\\frac{dx}{dt}\\right)}^{2}+{\\left(\\frac{dy}{dt}\\right)}^{2}}dt[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167794023283\">At this point a side derivation leads to a previous formula for arc length. In particular, suppose the parameter can be eliminated, leading to a function [latex]y=F\\left(x\\right)[\/latex]. Then [latex]y\\left(t\\right)=F\\left(x\\left(t\\right)\\right)[\/latex] and the Chain Rule gives [latex]{y}^{\\prime }\\left(t\\right)={F}^{\\prime }\\left(x\\left(t\\right)\\right){x}^{\\prime }\\left(t\\right)[\/latex]. Substituting this into the theorem gives<\/p>\n<div id=\"fs-id1167794051442\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill s& ={\\displaystyle\\int }_{{t}_{1}}^{{t}_{2}}\\sqrt{{\\left(\\frac{dx}{dt}\\right)}^{2}+{\\left(\\frac{dy}{dt}\\right)}^{2}}dt\\hfill \\\\ & ={\\displaystyle\\int }_{{t}_{1}}^{{t}_{2}}\\sqrt{{\\left(\\frac{dx}{dt}\\right)}^{2}+{\\left({F}^{\\prime }\\left(x\\right)\\frac{dx}{dt}\\right)}^{2}}dt\\hfill \\\\ & ={\\displaystyle\\int }_{{t}_{1}}^{{t}_{2}}\\sqrt{{\\left(\\frac{dx}{dt}\\right)}^{2}\\left(1+{\\left({F}^{\\prime }\\left(x\\right)\\right)}^{2}\\right)}dt\\hfill \\\\ & ={\\displaystyle\\int }_{{t}_{1}}^{{t}_{2}}{x}^{\\prime }\\left(t\\right)\\sqrt{1+{\\left(\\frac{dy}{dx}\\right)}^{2}}dt.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794003192\">Here we have assumed that [latex]{x}^{\\prime }\\left(t\\right)>0[\/latex], which is a reasonable assumption. The Chain Rule gives [latex]dx={x}^{\\prime }\\left(t\\right)dt[\/latex], and letting [latex]a=x\\left({t}_{1}\\right)[\/latex] and [latex]b=x\\left({t}_{2}\\right)[\/latex] we obtain the formula<\/p>\n<div id=\"fs-id1167794100345\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]s={\\displaystyle\\int }_{a}^{b}\\sqrt{1+{\\left(\\frac{dy}{dx}\\right)}^{2}}dx[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794100404\">which is the formula for arc length obtained in the Introduction to the Applications of Integration.<\/p>\n<div id=\"fs-id1167794100414\" data-type=\"example\">\n<div id=\"fs-id1167794100417\" data-type=\"exercise\">\n<div id=\"fs-id1167794100419\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Arc Length of a Parametric Curve<\/h3>\n<div id=\"fs-id1167794100419\" data-type=\"problem\">\n<p id=\"fs-id1167794100424\">Find the arc length of the semicircle defined by the equations<\/p>\n<div id=\"fs-id1167794100427\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)=3\\cos{t},y\\left(t\\right)=3\\sin{t},0\\le t\\le \\pi[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558859\">Show Solution<\/span><\/p>\n<div id=\"q44558859\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794100493\" data-type=\"solution\">\n<p id=\"fs-id1167794100495\">The values [latex]t=0[\/latex] to [latex]t=\\pi[\/latex] trace out the red curve in Figure 9. To determine its length, use the theorem:<\/p>\n<div id=\"fs-id1167794061688\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill s& ={\\displaystyle\\int }_{{t}_{1}}^{{t}_{2}}\\sqrt{{\\left(\\frac{dx}{dt}\\right)}^{2}+{\\left(\\frac{dy}{dt}\\right)}^{2}}dt\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{\\pi }\\sqrt{{\\left(-3\\sin{t}\\right)}^{2}+{\\left(3\\cos{t}\\right)}^{2}}dt\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{\\pi }\\sqrt{9{\\sin}^{2}t+9{\\cos}^{2}t}dt\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{\\pi }\\sqrt{9\\left({\\sin}^{2}t+{\\cos}^{2}t\\right)}dt\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{\\pi }3dt={3t|}_{0}^{\\pi }=3\\pi .\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794051248\">Note that the formula for the arc length of a semicircle is [latex]\\pi r[\/latex] and the radius of this circle is 3. This is a great example of using calculus to derive a known formula of a geometric quantity.<\/p>\n<figure id=\"CNX_Calc_Figure_11_02_010\"><figcaption><\/figcaption><div style=\"width: 349px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234801\/CNX_Calc_Figure_11_02_009.jpg\" alt=\"A semicircle is drawn with radius 3. There is an arrow pointing counterclockwise. On the graph there are also written three equations: x(t) = 3 cos(t), y(t) = 3 sin(t), and 0 \u2264 &#091;latex&#093;t&#091;\/latex&#093; \u2264 \u03c0.\" width=\"339\" height=\"347\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. The arc length of the semicircle is equal to its radius times [latex]\\pi[\/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 the Arc Length 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=893&amp;end=993&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\/7.2CalculusOfParametricCurves893to993_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;7.2 Calculus of Parametric Curves&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1167794051286\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167794051290\" data-type=\"exercise\">\n<div id=\"fs-id1167794051292\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167794051292\" data-type=\"problem\">\n<p id=\"fs-id1167794051294\">Find the arc length of the curve defined by the equations<\/p>\n<div id=\"fs-id1167794051297\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)=3{t}^{2},y\\left(t\\right)=2{t}^{3},1\\le t\\le 3[\/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=\"q44558839\">Hint<\/span><\/p>\n<div id=\"q44558839\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794027824\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167794027831\">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=\"q44558849\">Show Solution<\/span><\/p>\n<div id=\"q44558849\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794051365\" data-type=\"solution\">\n<p id=\"fs-id1167794051368\" style=\"text-align: center;\">[latex]s=2\\left({10}^{\\frac{3}{2}}-{2}^{\\frac{3}{2}}\\right)\\approx 57.589[\/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=\"ohm169505\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169505&theme=oea&iframe_resize_id=ohm169505&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p id=\"fs-id1167794027842\">We now return to the problem posed at the beginning of the section about a baseball leaving a pitcher\u2019s hand. Ignoring the effect of air resistance (unless it is a curve ball!), the ball travels a parabolic path. Assuming the pitcher\u2019s hand is at the origin and the ball travels left to right in the direction of the positive <em data-effect=\"italics\">x<\/em>-axis, the parametric equations for this curve can be written as<\/p>\n<div id=\"fs-id1167794027857\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)=140t,y\\left(t\\right)=-16{t}^{2}+2t[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794027910\">where <em data-effect=\"italics\">t<\/em> represents time. We first calculate the distance the ball travels as a function of time. This distance is represented by the arc length. We can modify the arc length formula slightly. First rewrite the functions [latex]x\\left(t\\right)[\/latex] and [latex]y\\left(t\\right)[\/latex] using <em data-effect=\"italics\">v<\/em> as an independent variable, so as to eliminate any confusion with the parameter <em data-effect=\"italics\">t<\/em>:<\/p>\n<div id=\"fs-id1167794027956\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(v\\right)=140v,y\\left(v\\right)=-16{v}^{2}+2v[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794028011\">Then we write the arc length formula as follows:<\/p>\n<div id=\"fs-id1167794028015\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill s\\left(t\\right)& ={\\displaystyle\\int }_{0}^{t}\\sqrt{{\\left(\\frac{dx}{dv}\\right)}^{2}+{\\left(\\frac{dy}{dv}\\right)}^{2}}dv\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{t}\\sqrt{{140}^{2}+{\\left(-32v+2\\right)}^{2}}dv.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793926178\">The variable <em data-effect=\"italics\">v<\/em> acts as a dummy variable that disappears after integration, leaving the arc length as a function of time <em data-effect=\"italics\">t.<\/em> To integrate this expression we can use a formula from Appendix A,<\/p>\n<div id=\"fs-id1167793926200\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\sqrt{{a}^{2}+{u}^{2}}du=\\frac{u}{2}\\sqrt{{a}^{2}+{u}^{2}}+\\frac{{a}^{2}}{2}\\text{ln}|u+\\sqrt{{a}^{2}+{u}^{2}}|+C[\/latex].<\/div>\n<p id=\"fs-id1167794094267\">We set [latex]a=140[\/latex] and [latex]u=-32v+2[\/latex]. This gives [latex]du=-32dv[\/latex], so [latex]dv=-\\frac{1}{32}du[\/latex]. Therefore<\/p>\n<div id=\"fs-id1167794094341\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int \\sqrt{{140}^{2}+{\\left(-32v+2\\right)}^{2}}dv}& =-\\frac{1}{32}{\\displaystyle\\int \\sqrt{{a}^{2}+{u}^{2}}du}\\hfill \\\\ & =-\\frac{1}{32}\\left[\\begin{array}{c}\\frac{\\left(-32v+2\\right)}{2}\\sqrt{{140}^{2}+{\\left(-32v+2\\right)}^{2}}\\hfill \\\\ +\\frac{{140}^{2}}{2}\\text{ln}|\\left(-32v+2\\right)+\\sqrt{{140}^{2}+{\\left(-32v+2\\right)}^{2}}|\\hfill \\end{array}\\right]+C\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794035510\">and<\/p>\n<div id=\"fs-id1167794035513\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill s\\left(t\\right)& =-\\frac{1}{32}\\left[\\frac{\\left(-32t+2\\right)}{2}\\sqrt{{140}^{2}+{\\left(-32t+2\\right)}^{2}}+\\frac{{140}^{2}}{2}\\text{ln}|\\left(-32t+2\\right)+\\sqrt{{140}^{2}+{\\left(-32t+2\\right)}^{2}}|\\right]\\hfill \\\\ & +\\frac{1}{32}\\left[\\sqrt{{140}^{2}+{2}^{2}}+\\frac{{140}^{2}}{2}\\text{ln}|2+\\sqrt{{140}^{2}+{2}^{2}}|\\right]\\hfill \\\\ & =\\left(\\frac{t}{2}-\\frac{1}{32}\\right)\\sqrt{1024{t}^{2}-128t+19604}-\\frac{1225}{4}\\text{ln}|\\left(-32t+2\\right)+\\sqrt{1024{t}^{2}-128t+19604}|\\hfill \\\\ & +\\frac{\\sqrt{19604}}{32}+\\frac{1225}{4}\\text{ln}\\left(2+\\sqrt{19604}\\right).\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794095955\">This function represents the distance traveled by the ball as a function of time. To calculate the speed, take the derivative of this function with respect to <em data-effect=\"italics\">t.<\/em> While this may seem like a daunting task, it is possible to obtain the answer directly from the Fundamental Theorem of Calculus:<\/p>\n<div id=\"fs-id1167794095966\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{d}{dx}{\\displaystyle\\int }_{a}^{x}f\\left(u\\right)du=f\\left(x\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794096023\">Therefore<\/p>\n<div id=\"fs-id1167794096026\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {s}^{\\prime }\\left(t\\right)& =\\frac{d}{dt}\\left[s\\left(t\\right)\\right]\\hfill \\\\ & =\\frac{d}{dt}\\left[{\\displaystyle\\int }_{0}^{t}\\sqrt{{140}^{2}+{\\left(-32v+2\\right)}^{2}}dv\\right]\\hfill \\\\ & =\\sqrt{{140}^{2}+{\\left(-32t+2\\right)}^{2}}\\hfill \\\\ & =\\sqrt{1024{t}^{2}-128t+19604}\\hfill \\\\ & =2\\sqrt{256{t}^{2}-32t+4901}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794040883\">One third of a second after the ball leaves the pitcher\u2019s hand, the distance it travels is equal to<\/p>\n<div id=\"fs-id1167794040888\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill s\\left(\\frac{1}{3}\\right)& =\\left(\\frac{\\frac{1}{3}}{2}-\\frac{1}{32}\\right)\\sqrt{1024{\\left(\\frac{1}{3}\\right)}^{2}-128\\left(\\frac{1}{3}\\right)+19604}\\hfill \\\\ & -\\frac{1225}{4}\\text{ln}|\\left(-32\\left(\\frac{1}{3}\\right)+2\\right)+\\sqrt{1024{\\left(\\frac{1}{3}\\right)}^{2}-128\\left(\\frac{1}{3}\\right)+19604}|\\hfill \\\\ & +\\frac{\\sqrt{19604}}{32}+\\frac{1225}{4}\\text{ln}\\left(2+\\sqrt{19604}\\right)\\hfill \\\\ & \\approx 46.69\\text{feet}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794135263\">This value is just over three quarters of the way to home plate. The speed of the ball is<\/p>\n<div id=\"fs-id1167794135267\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{s}^{\\prime }\\left(\\frac{1}{3}\\right)=2\\sqrt{256{\\left(\\frac{1}{3}\\right)}^{2}-16\\left(\\frac{1}{3}\\right)+4901}\\approx 140.34\\text{ft\/s}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794135348\">This speed translates to approximately 95 mph\u2014a major-league fastball.<\/p>\n<\/section>\n<section id=\"fs-id1167794135353\" data-depth=\"1\">\n<h2 data-type=\"title\">Surface Area Generated by a Parametric Curve<\/h2>\n<p id=\"fs-id1167794135359\">Recall the problem of finding the surface area of a volume of revolution. In Curve Length and Surface Area, we derived a formula for finding the surface area of a volume generated by a function [latex]y=f\\left(x\\right)[\/latex] from [latex]x=a[\/latex] to [latex]x=b[\/latex], revolved around the <em data-effect=\"italics\">x<\/em>-axis:<\/p>\n<div id=\"fs-id1167794135414\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]S=2\\pi {\\displaystyle\\int }_{a}^{b}f\\left(x\\right)\\sqrt{1+{\\left({f}^{\\prime }\\left(x\\right)\\right)}^{2}}dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794096354\">We now consider a volume of revolution generated by revolving a parametrically defined curve [latex]x=x\\left(t\\right),y=y\\left(t\\right),a\\le t\\le b[\/latex] around the <em data-effect=\"italics\">x<\/em>-axis as shown in the following figure.<\/p>\n<figure id=\"CNX_Calc_Figure_11_02_011\"><figcaption><\/figcaption><div style=\"width: 351px\" 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\/11234803\/CNX_Calc_Figure_11_02_010.jpg\" alt=\"A curve is drawn in the first quadrant with endpoints marked [latex]t[\/latex] = a and [latex]t[\/latex] = b. On this curve, there is a point marked (x(t), y(t)). There is a circle with an arrow drawn around the [latex]x[\/latex]-axis that seems to indicate a rotation about the x axis, and there is a shape that accompanies that curve that seems to be what you would obtain if you rotated the curve about the x axis.\" width=\"341\" height=\"276\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5. A surface of revolution generated by a parametrically defined curve.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1167794096433\">The analogous formula for a parametrically defined curve is<\/p>\n<div id=\"fs-id1167794096436\" style=\"text-align: center;\" data-type=\"equation\">[latex]S=2\\pi {\\displaystyle\\int }_{a}^{b}y\\left(t\\right)\\sqrt{{\\left({x}^{\\prime }\\left(t\\right)\\right)}^{2}+{\\left({y}^{\\prime }\\left(t\\right)\\right)}^{2}}dt[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794096527\">provided that [latex]y\\left(t\\right)[\/latex] is not negative on [latex]\\left[a,b\\right][\/latex].<\/p>\n<div id=\"fs-id1167794076111\" data-type=\"example\">\n<div id=\"fs-id1167794076114\" data-type=\"exercise\">\n<div id=\"fs-id1167794076116\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Finding Surface Area<\/h3>\n<div id=\"fs-id1167794076116\" data-type=\"problem\">\n<p id=\"fs-id1167794076121\">Find the surface area of a sphere of radius [latex]r[\/latex] centered at the origin.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558829\">Show Solution<\/span><\/p>\n<div id=\"q44558829\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794076130\" data-type=\"solution\">\n<p id=\"fs-id1167794076133\">We start with the curve defined by the equations<\/p>\n<div id=\"fs-id1167794076136\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)=r\\cos{t},y\\left(t\\right)=r\\sin{t},0\\le t\\le \\pi[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794076201\">This generates an upper semicircle of radius <em data-effect=\"italics\">r<\/em> centered at the origin as shown in the following graph.<\/p>\n<figure id=\"CNX_Calc_Figure_11_02_012\"><figcaption><\/figcaption><div style=\"width: 282px\" 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\/11234805\/CNX_Calc_Figure_11_02_011.jpg\" alt=\"A semicircle is drawn with radius r. On the graph there are also written three equations: x(t) = r cos(t), y(t) = r sin(t), and 0 \u2264 &#091;latex&#093;t&#091;\/latex&#093; \u2264 \u03c0.\" width=\"272\" height=\"271\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. A semicircle generated by parametric equations.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1167794076232\">When this curve is revolved around the\u00a0[latex]x[\/latex]-axis, it generates a sphere of radius [latex]r[\/latex]. To calculate the surface area of the sphere, we use the above equation:<\/p>\n<div id=\"fs-id1167794076251\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill S&=2\\pi {\\displaystyle\\int_{a}^{b}} y\\left(t\\right)\\sqrt{{\\left({x}^{\\prime }\\left(t\\right)\\right)}^{2}+{\\left({y}^{\\prime}\\left(t\\right)\\right)}^{2}}dt\\hfill \\\\ &=2\\pi {\\displaystyle\\int_{0}^{\\pi}} r\\sin{t}\\sqrt{{\\left(-r\\sin{t}\\right)}^{2}+{\\left(r\\cos{t}\\right)}^{2}}dt\\hfill \\\\ &=2\\pi {\\displaystyle\\int }_{0}^{\\pi }r\\sin{t}\\sqrt{{r}^{2}{\\sin}^{2}t+{r}^{2}{\\cos}^{2}t}dt\\hfill \\\\ &=2\\pi {\\displaystyle\\int }_{0}^{\\pi }r\\sin{t}\\sqrt{{r}^{2}\\left({\\sin}^{2}t+{\\cos}^{2}t\\right)}dt\\hfill \\\\ &=2\\pi {\\displaystyle\\int_{0}^{\\pi}{r}^{2}\\sin{t}dt}\\hfill \\\\ &=2\\pi{r}^{2}\\left(-\\cos{t}|_{0}^{\\pi}\\right)\\hfill \\\\ & =2\\pi {r}^{2}\\left(-\\cos\\pi +\\cos0\\right)\\hfill \\\\ & =4\\pi {r}^{2}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794055273\">This is, in fact, the formula for the surface area of a sphere.<\/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 Surface Area.<\/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=1040&amp;end=1263&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\/7.2CalculusOfParametricCurves1040to1263_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-id1167794055279\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167794055283\" data-type=\"exercise\">\n<div id=\"fs-id1167794055286\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167794055286\" data-type=\"problem\">\n<p id=\"fs-id1167794055288\">Find the surface area generated when the plane curve defined by the equations<\/p>\n<div id=\"fs-id1167794055291\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x\\left(t\\right)={t}^{3},y\\left(t\\right)={t}^{2},0\\le t\\le 1[\/latex]<\/div>\n<p id=\"fs-id1167794055352\">is revolved around the <em data-effect=\"italics\">x<\/em>-axis.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558809\">Hint<\/span><\/p>\n<div id=\"q44558809\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794055397\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167794055405\">Use the above equation. When evaluating the integral, use a <em data-effect=\"italics\">u<\/em>-substitution.<\/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=\"q44558819\">Show Solution<\/span><\/p>\n<div id=\"q44558819\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794055361\" data-type=\"solution\">\n<p id=\"fs-id1167794055363\">[latex]A=\\frac{\\pi \\left(494\\sqrt{13}+128\\right)}{1215}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1167794048157\" class=\"section-exercises\" data-depth=\"1\">\n<div id=\"fs-id1167794046049\" data-type=\"exercise\"><\/div>\n<\/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-1158\">\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":8,"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-1158","chapter","type-chapter","status-publish","hentry"],"part":1150,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1158","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\/1158\/revisions"}],"predecessor-version":[{"id":6418,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1158\/revisions\/6418"}],"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\/1158\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=1158"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=1158"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=1158"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=1158"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}