{"id":998,"date":"2025-06-20T17:27:26","date_gmt":"2025-06-20T17:27:26","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=998"},"modified":"2025-09-09T19:19:55","modified_gmt":"2025-09-09T19:19:55","slug":"calculus-with-parametric-curves-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/calculus-with-parametric-curves-fresh-take\/","title":{"raw":"Calculus with Parametric Curves: Fresh Take","rendered":"Calculus with Parametric Curves: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Find derivatives and tangent lines for curves written in parametric form<\/li>\r\n \t<li>Calculate the area underneath a parametric curve<\/li>\r\n \t<li>Find the length of a parametric curve using the arc length formula<\/li>\r\n \t<li>Calculate the surface area when a parametric curve is rotated to create a 3D shape<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Derivatives of Parametric Equations<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">When working with parametric curves, you often need to find the slope at any point. Instead of the tedious process of eliminating the parameter, there's a direct formula that uses the chain rule.<\/p>\r\n<p class=\"whitespace-normal break-words\">For parametric equations [latex]x = x(t)[\/latex] and [latex]y = y(t)[\/latex], the slope is: [latex]\\frac{dy}{dx} = \\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}} = \\frac{y'(t)}{x'(t)}[\/latex]<\/p>\r\nThis formula works for ANY parametric curve\u2014even loops, cusps, and curves that cross themselves. You don't need to worry about whether the curve can be written as [latex]y = f(x)[\/latex].\r\n<p class=\"whitespace-normal break-words\"><strong>Critical Points:<\/strong> The derivative is zero when [latex]y'(t) = 0[\/latex] (horizontal tangent) and undefined when [latex]x'(t) = 0[\/latex] (vertical tangent). These give you the critical points of the parametric curve.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Finding Tangent Lines:<\/strong> Once you have the slope at parameter value [latex]t_0[\/latex], find the point [latex](x(t_0), y(t_0))[\/latex] and use point-slope form: [latex]y - y(t_0) = \\frac{dy}{dx}\\big|_{t=t_0}(x - x(t_0))[\/latex].<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\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\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 t = \u22122 and the point (\u22123, 36) is marked t = 3. On the graph there are also written three equations: x(t) = t2 \u2013 4t, y(t) = 2t3 \u2013 6t, and \u22122 \u2264 t \u2264 3.\" width=\"388\" height=\"384\" data-media-type=\"image\/jpeg\" \/> Figure 5.[\/caption]\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\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<\/section>\r\n<h2 data-type=\"title\">Second-Order Derivatives<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">For parametric equations, finding the second derivative follows naturally from the first derivative formula. Since we know [latex]\\frac{dy}{dx} = \\frac{y'(t)}{x'(t)}[\/latex], we can find the second derivative by treating this ratio as a function of [latex]t[\/latex] and applying the chain rule again.<\/p>\r\n<p class=\"whitespace-normal break-words\">The formula is: [latex]\\frac{d^2y}{dx^2} = \\frac{\\frac{d}{dt}\\left(\\frac{dy}{dx}\\right)}{\\frac{dx}{dt}}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy:<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Find the first derivative [latex]\\frac{dy}{dx}[\/latex] in terms of [latex]t[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Differentiate that expression with respect to [latex]t[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Divide by [latex]\\frac{dx}{dt}[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">The second derivative formula comes from applying the chain rule to [latex]\\frac{d}{dx}\\left(\\frac{dy}{dx}\\right)[\/latex]. Since both [latex]x[\/latex] and [latex]\\frac{dy}{dx}[\/latex] depend on [latex]t[\/latex], we need to account for how [latex]x[\/latex] changes with [latex]t[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">This approach is much more efficient than converting parametric equations to rectangular form first, especially when the elimination process would create complex expressions.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\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<\/section>\r\n<h2 data-type=\"title\">Integrals Involving Parametric Equations<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Finding the area under a parametric curve requires a different approach than regular integration. Instead of integrating with respect to [latex]x[\/latex], we integrate with respect to the parameter [latex]t[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">The formula comes from using rectangle approximations. For each small rectangle, the height is [latex]y(t)[\/latex] and the width is the change in [latex]x[\/latex] over a small time interval. As the intervals get smaller, this width approaches [latex]x'(t) \\cdot dt[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Area Formula:<\/strong> For a parametric curve [latex]x = x(t), y = y(t)[\/latex] where [latex]a \\leq t \\leq b[\/latex]: [latex]A = \\int_a^b y(t) \\cdot x'(t) , dt[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Key Requirements:<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">The curve must not cross itself<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]x(t)[\/latex] should increase as [latex]t[\/latex] increases from [latex]a[\/latex] to [latex]b[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]x(t)[\/latex] must be differentiable<\/li>\r\n<\/ul>\r\n<strong>Why This Works:<\/strong> The Mean Value Theorem guarantees that [latex]\\frac{x(t_i) - x(t_{i-1})}{t_i - t_{i-1}}[\/latex] equals [latex]x'(c)[\/latex] for some point [latex]c[\/latex] in the interval. As intervals shrink, this becomes [latex]x'(t)[\/latex].\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\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<\/section>\r\n<h2 data-type=\"title\">Arc Length of a Parametric Curve<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Arc length measures the actual distance along a curve\u2014if you walked along the path, this tells you how far you'd travel. For parametric curves, we can't use the usual arc length formula, so we need a different approach.<\/p>\r\n<p class=\"whitespace-normal break-words\">The key insight comes from approximating the curve with tiny line segments. Each segment has length [latex]\\sqrt{(\\Delta x)^2 + (\\Delta y)^2}[\/latex] using the distance formula. As we make the segments smaller and smaller, this approximation becomes exact.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Arc Length Formula:<\/strong> For a parametric curve [latex]x = x(t), y = y(t)[\/latex] from [latex]t_1[\/latex] to [latex]t_2[\/latex]: [latex]s = \\int_{t_1}^{t_2} \\sqrt{\\left(\\frac{dx}{dt}\\right)^2 + \\left(\\frac{dy}{dt}\\right)^2} dt[\/latex]<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\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<\/section>\r\n<h2 data-type=\"title\">Surface Area Generated by a Parametric Curve<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Think of a parametric curve as a flexible wire that you can bend into any shape you want. When you spin that wire around the x-axis, it sweeps out a 3D surface\u2014like how spinning a jump rope creates a cylinder-like shape in the air. Finding the surface area of that 3D shape is what we're after here.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The key insight:<\/strong> We're essentially wrapping tiny circumferences around each point on our curve and adding them all up. For each point [latex](x(t), y(t))[\/latex] on our parametric curve, we create a circle with circumference [latex]2\\pi y(t)[\/latex] when we rotate around the [latex]x[\/latex]-axis. But here's the catch\u2014we need to account for how stretched out our curve is as we move along it.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The key formula:<\/strong> [latex]S = 2\\pi \\int_a^b y(t) \\sqrt{[x'(t)]^2 + [y'(t)]^2} dt[\/latex]<\/p>\r\n<strong>Critical requirement:<\/strong> [latex]y(t) \\geq 0[\/latex] throughout your interval. This just means your curve stays on or above the x-axis\u2014which makes sense since negative radii don't exist in the real world.\r\n\r\nMake sure you're using the correct derivatives\u2014[latex]x'(t)[\/latex] and [latex]y'(t)[\/latex] are derivatives with respect to the parameter [latex]t[\/latex], not [latex]x[\/latex]. This is different from the regular surface area formula you might remember from earlier.\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\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 [latex]x[\/latex]-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<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Find derivatives and tangent lines for curves written in parametric form<\/li>\n<li>Calculate the area underneath a parametric curve<\/li>\n<li>Find the length of a parametric curve using the arc length formula<\/li>\n<li>Calculate the surface area when a parametric curve is rotated to create a 3D shape<\/li>\n<\/ul>\n<\/section>\n<h2>Derivatives of Parametric Equations<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When working with parametric curves, you often need to find the slope at any point. Instead of the tedious process of eliminating the parameter, there&#8217;s a direct formula that uses the chain rule.<\/p>\n<p class=\"whitespace-normal break-words\">For parametric equations [latex]x = x(t)[\/latex] and [latex]y = y(t)[\/latex], the slope is: [latex]\\frac{dy}{dx} = \\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}} = \\frac{y'(t)}{x'(t)}[\/latex]<\/p>\n<p>This formula works for ANY parametric curve\u2014even loops, cusps, and curves that cross themselves. You don&#8217;t need to worry about whether the curve can be written as [latex]y = f(x)[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Critical Points:<\/strong> The derivative is zero when [latex]y'(t) = 0[\/latex] (horizontal tangent) and undefined when [latex]x'(t) = 0[\/latex] (vertical tangent). These give you the critical points of the parametric curve.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Finding Tangent Lines:<\/strong> Once you have the slope at parameter value [latex]t_0[\/latex], find the point [latex](x(t_0), y(t_0))[\/latex] and use point-slope form: [latex]y - y(t_0) = \\frac{dy}{dx}\\big|_{t=t_0}(x - x(t_0))[\/latex].<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558897\">Hint<\/button><\/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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558898\">Show Solution<\/button><\/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<figure style=\"width: 388px\" 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 t = \u22122 and the point (\u22123, 36) is marked t = 3. On the graph there are also written three equations: x(t) = t2 \u2013 4t, y(t) = 2t3 \u2013 6t, and \u22122 \u2264 t \u2264 3.\" width=\"388\" height=\"384\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 5.<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558894\">Hint<\/button><\/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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558895\">Show Solution<\/button><\/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<\/section>\n<h2 data-type=\"title\">Second-Order Derivatives<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For parametric equations, finding the second derivative follows naturally from the first derivative formula. Since we know [latex]\\frac{dy}{dx} = \\frac{y'(t)}{x'(t)}[\/latex], we can find the second derivative by treating this ratio as a function of [latex]t[\/latex] and applying the chain rule again.<\/p>\n<p class=\"whitespace-normal break-words\">The formula is: [latex]\\frac{d^2y}{dx^2} = \\frac{\\frac{d}{dt}\\left(\\frac{dy}{dx}\\right)}{\\frac{dx}{dt}}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy:<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Find the first derivative [latex]\\frac{dy}{dx}[\/latex] in terms of [latex]t[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Differentiate that expression with respect to [latex]t[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Divide by [latex]\\frac{dx}{dt}[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">The second derivative formula comes from applying the chain rule to [latex]\\frac{d}{dx}\\left(\\frac{dy}{dx}\\right)[\/latex]. Since both [latex]x[\/latex] and [latex]\\frac{dy}{dx}[\/latex] depend on [latex]t[\/latex], we need to account for how [latex]x[\/latex] changes with [latex]t[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">This approach is much more efficient than converting parametric equations to rectangular form first, especially when the elimination process would create complex expressions.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558892\">Hint<\/button><\/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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558891\">Show Solution<\/button><\/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<\/section>\n<h2 data-type=\"title\">Integrals Involving Parametric Equations<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Finding the area under a parametric curve requires a different approach than regular integration. Instead of integrating with respect to [latex]x[\/latex], we integrate with respect to the parameter [latex]t[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">The formula comes from using rectangle approximations. For each small rectangle, the height is [latex]y(t)[\/latex] and the width is the change in [latex]x[\/latex] over a small time interval. As the intervals get smaller, this width approaches [latex]x'(t) \\cdot dt[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Area Formula:<\/strong> For a parametric curve [latex]x = x(t), y = y(t)[\/latex] where [latex]a \\leq t \\leq b[\/latex]: [latex]A = \\int_a^b y(t) \\cdot x'(t) , dt[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Key Requirements:<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">The curve must not cross itself<\/li>\n<li class=\"whitespace-normal break-words\">[latex]x(t)[\/latex] should increase as [latex]t[\/latex] increases from [latex]a[\/latex] to [latex]b[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]x(t)[\/latex] must be differentiable<\/li>\n<\/ul>\n<p><strong>Why This Works:<\/strong> The Mean Value Theorem guarantees that [latex]\\frac{x(t_i) - x(t_{i-1})}{t_i - t_{i-1}}[\/latex] equals [latex]x'(c)[\/latex] for some point [latex]c[\/latex] in the interval. As intervals shrink, this becomes [latex]x'(t)[\/latex].<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558879\">Hint<\/button><\/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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558869\">Show Solution<\/button><\/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<\/section>\n<h2 data-type=\"title\">Arc Length of a Parametric Curve<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Arc length measures the actual distance along a curve\u2014if you walked along the path, this tells you how far you&#8217;d travel. For parametric curves, we can&#8217;t use the usual arc length formula, so we need a different approach.<\/p>\n<p class=\"whitespace-normal break-words\">The key insight comes from approximating the curve with tiny line segments. Each segment has length [latex]\\sqrt{(\\Delta x)^2 + (\\Delta y)^2}[\/latex] using the distance formula. As we make the segments smaller and smaller, this approximation becomes exact.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Arc Length Formula:<\/strong> For a parametric curve [latex]x = x(t), y = y(t)[\/latex] from [latex]t_1[\/latex] to [latex]t_2[\/latex]: [latex]s = \\int_{t_1}^{t_2} \\sqrt{\\left(\\frac{dx}{dt}\\right)^2 + \\left(\\frac{dy}{dt}\\right)^2} dt[\/latex]<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558839\">Hint<\/button><\/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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558849\">Show Solution<\/button><\/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<\/section>\n<h2 data-type=\"title\">Surface Area Generated by a Parametric Curve<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Think of a parametric curve as a flexible wire that you can bend into any shape you want. When you spin that wire around the x-axis, it sweeps out a 3D surface\u2014like how spinning a jump rope creates a cylinder-like shape in the air. Finding the surface area of that 3D shape is what we&#8217;re after here.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The key insight:<\/strong> We&#8217;re essentially wrapping tiny circumferences around each point on our curve and adding them all up. For each point [latex](x(t), y(t))[\/latex] on our parametric curve, we create a circle with circumference [latex]2\\pi y(t)[\/latex] when we rotate around the [latex]x[\/latex]-axis. But here&#8217;s the catch\u2014we need to account for how stretched out our curve is as we move along it.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The key formula:<\/strong> [latex]S = 2\\pi \\int_a^b y(t) \\sqrt{[x'(t)]^2 + [y'(t)]^2} dt[\/latex]<\/p>\n<p><strong>Critical requirement:<\/strong> [latex]y(t) \\geq 0[\/latex] throughout your interval. This just means your curve stays on or above the x-axis\u2014which makes sense since negative radii don&#8217;t exist in the real world.<\/p>\n<p>Make sure you&#8217;re using the correct derivatives\u2014[latex]x'(t)[\/latex] and [latex]y'(t)[\/latex] are derivatives with respect to the parameter [latex]t[\/latex], not [latex]x[\/latex]. This is different from the regular surface area formula you might remember from earlier.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\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 [latex]x[\/latex]-axis.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558809\">Hint<\/button><\/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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558819\">Show Solution<\/button><\/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<\/section>\n","protected":false},"author":15,"menu_order":16,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":675,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/998"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/998\/revisions"}],"predecessor-version":[{"id":2251,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/998\/revisions\/2251"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/675"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/998\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=998"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=998"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=998"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=998"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}