{"id":568,"date":"2025-10-09T15:22:07","date_gmt":"2025-10-09T15:22:07","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=568"},"modified":"2025-10-24T13:39:48","modified_gmt":"2025-10-24T13:39:48","slug":"arc-length-and-curvature-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/arc-length-and-curvature-fresh-take\/","title":{"raw":"Arc Length and Curvature: Fresh Take","rendered":"Arc Length and Curvature: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Determine the length of a particle's path in space by using the arc-length function<\/li>\r\n \t<li>Calculate and understand what curvature means for curves in space<\/li>\r\n \t<li>Understand what normal and binormal vectors tell us about a curve's behavior<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Arc Length for Vector Functions<\/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 distance along a curve. For vector-valued functions, the formula is straightforward: integrate the magnitude of the derivative vector.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Core Formula<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For any smooth curve defined by [latex]{\\bf{r}}(t)[\/latex] over [latex][a, b][\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s=\\int_{a}^{b} \\left\\Vert{\\bf{r}}'(t)\\right\\Vert dt[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Plane curves<\/strong> [latex]{\\bf{r}}(t)=f(t){\\bf{i}}+g(t){\\bf{j}}[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s=\\int_{a}^{b} \\sqrt{[f'(t)]^{2}+[g'(t)]^{2}}dt[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Space curves<\/strong> [latex]{\\bf{r}}(t)=f(t){\\bf{i}}+g(t){\\bf{j}}+h(t){\\bf{k}}[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s=\\int_{a}^{b} \\sqrt{[f'(t)]^{2}+[g'(t)]^{2}+[h'(t)]^{2}}dt[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">The only difference? Space curves include a third component.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Find [latex]{\\bf{r}}'(t)[\/latex] by differentiating each component<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Calculate [latex]\\left\\Vert{\\bf{r}}'(t)\\right\\Vert[\/latex] using the square root of the sum of squares<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Integrate [latex]\\left\\Vert{\\bf{r}}'(t)\\right\\Vert[\/latex] from [latex]a[\/latex] to [latex]b[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\">The curve must be <strong>smooth<\/strong>: [latex]{\\bf{r}}(t)[\/latex] must be differentiable with a non-zero derivative. This guarantees no cusps or corners that break the formula.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Special Case: Helices<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">A helix with radius [latex]R[\/latex], height [latex]h[\/latex], and [latex]N[\/latex] turns has arc length:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s=\\sqrt{4{\\pi}^{2}N^{2}R^{2}+h^{2}}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This tells you the length of wire needed to form the helix.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Calculate the arc length of the parameterized curve\r\n<p style=\"text-align: center;\">[latex]{\\bf{r}}\\,(t)=\\langle{2t^{2}}+1,\\ 2t^{2}-1,\\ t^{3}\\rangle,\\ 0\\,\\leq\\,t\\,\\leq\\,3.[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1167794933124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794933124\"]\r\n<div style=\"text-align: center;\">[latex]{\\bf{r}}'\\,(t)=\\langle{4t},\\ 4t,\\ 3t^{2}\\rangle[\/latex], so [latex]s=\\frac{1}{27}(113^{3\/2}-32^{3\/2})\\approx{37.785}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7949619&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=Gti0f6fEX6A&amp;video_target=tpm-plugin-e842wmnl-Gti0f6fEX6A\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.9_transcript.html\">transcript for \u201cCP 3.9\u201d here (opens in new window).<\/a><\/center><\/section>\r\n<h2>Arc-Length Parameterization<\/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 parameterization is a way to describe a curve where the parameter directly measures distance traveled along the curve. It's like using mile markers instead of time to track position on a road.<\/p>\r\n<p class=\"whitespace-normal break-words\">Regular parameterization can trace a curve at any speed. Arc-length parameterization always traces at unit speed, making distance calculations straightforward.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Arc-Length Function<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For a smooth curve [latex]{\\bf{r}}(t)[\/latex] where [latex]t \\geq a[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s(t)=\\int_{a}^{t} \\left\\Vert{\\bf{r}}'(u)\\right\\Vert du[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This function [latex]s(t)[\/latex] tells you the distance traveled from the starting point. Taking its derivative gives speed:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\frac{ds}{dt}=\\left\\Vert{\\bf{r}}'(t)\\right\\Vert[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Special Case: Unit Speed<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">When [latex]\\left\\Vert{\\bf{r}}'(t)\\right\\Vert=1[\/latex] for all [latex]t[\/latex], the parameter [latex]t[\/latex] already measures arc length. The curve traces at unit speed.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Creating an Arc-Length Parameterization<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Find the arc-length function<\/strong>: Calculate [latex]s(t)=\\int_{a}^{t} \\left\\Vert{\\bf{r}}'(u)\\right\\Vert du[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Solve for [latex]t[\/latex]<\/strong>: Express [latex]t[\/latex] as a function of [latex]s[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Substitute<\/strong>: Replace [latex]t[\/latex] in [latex]{\\bf{r}}(t)[\/latex] with your expression in terms of [latex]s[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Adjust the domain<\/strong>: Convert the restriction on [latex]t[\/latex] to a restriction on [latex]s[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\">Arc-length parameterization makes the parameter meaningful. The value of [latex]s[\/latex] tells you exactly how far you've traveled along the curve. This simplifies formulas for curvature and line integrals.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Find the arc-length function for the helix\r\n<p style=\"text-align: center;\">[latex]{\\bf{r}}\\,(t)=\\langle{3}\\cos{t},\\ 3\\sin{t},\\ 4t\\rangle{,}\\ t\\,\\geq\\,0.[\/latex]<\/p>\r\nThen, use the relationship between the arc length and the parameter [latex]t[\/latex] to find an arc-length parameterization of [latex]{\\bf{r}}(t)[\/latex].\r\n[reveal-answer q=\"fs-id1167794933134\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794933134\"]\r\n<div style=\"text-align: center;\">[latex]s=5t[\/latex], or [latex]t=\\frac{s}{5}[\/latex]. Substituting this into [latex]{\\bf{r}}\\,(t)=\\langle{3}\\cos{t},\\ 3\\sin{t},\\ 4t\\rangle[\/latex] gives [latex]{\\bf{r}}\\,(s)=\\Big\\langle3\\cos{\\big(\\frac{s}{5}\\big)},\\ 3\\sin{\\big(\\frac{s}{5}\\big)},\\ \\frac{4s}{5}\\Big\\rangle,\\ s\\geq{0}[\/latex].<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7994454&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=F7R34nEPVtM&amp;video_target=tpm-plugin-vfuem7m1-F7R34nEPVtM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.10_transcript.html\">transcript for \u201cCP 3.10\u201d here (opens in new window).<\/a><\/center><\/section>\r\n<h2>Curvature<\/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\">Curvature measures how sharply a curve bends. A straight line has zero curvature. A tight turn has high curvature. A gentle curve has low curvature.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Intuition<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Picture driving on a road. A gentle curve needs barely any steering. A sharp turn requires turning the wheel more. Curvature quantifies this difference. For a circle, smaller radius means greater curvature.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Definition<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For a smooth curve [latex]{\\bf{r}}(s)[\/latex] with arc-length parameter [latex]s[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\kappa=\\left\\Vert{\\bf{T}}'(s)\\right\\Vert[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">where [latex]{\\bf{T}}[\/latex] is the unit tangent vector.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Practical Formulas<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The definition requires arc-length parameterization. These alternatives work with any parameterization:<\/p>\r\n\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\"><strong>For any curve<\/strong> [latex]{\\bf{r}}(t)[\/latex]: [latex]\\kappa=\\frac{\\left\\Vert{\\bf{T}}'(t)\\right\\Vert}{\\left\\Vert{\\bf{r}}'(t)\\right\\Vert}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>For space curves<\/strong>: [latex]\\kappa=\\frac{\\left\\Vert{\\bf{r}}'(t)\\times{\\bf{r}}''(t)\\right\\Vert}{\\left\\Vert{\\bf{r}}'(t)\\right\\Vert^{3}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>For graphs<\/strong> [latex]y=f(x)[\/latex]: [latex]\\kappa=\\frac{|y''|}{[1+(y')^{2}]^{3\/2}}[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Choosing Your Formula<\/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-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Use the first when you already have [latex]{\\bf{T}}(t)[\/latex] calculated<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Use the cross product formula for space curves\u2014it only needs [latex]{\\bf{r}}'(t)[\/latex] and [latex]{\\bf{r}}''(t)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Use the [latex]y=f(x)[\/latex] formula for function graphs<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Circles<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For a circle of radius [latex]r[\/latex], curvature equals [latex]\\kappa = \\frac{1}{r}[\/latex]. Curvature is the reciprocal of radius.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox interact\" aria-label=\"Interact\">Visit\u00a0this <a href=\"http:\/\/www.openstax.org\/l\/20_spacecurve\" target=\"_blank\" rel=\"noopener nofollow\">website<\/a> for more information about the curvature of a space curve.<\/section><section class=\"textbox example\" aria-label=\"Example\">Find the curvature of the curve defined by the function\r\n<p style=\"text-align: center;\">[latex]y=3x^{2}-2x+4[\/latex]<\/p>\r\nat the point [latex]x=2[\/latex].\r\n\r\n[reveal-answer q=\"fs-id1167794933164\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794933164\"]\r\n<div style=\"text-align: center;\">[latex]\\large{\\kappa=\\frac{6}{101^{3\/2}}\\approx{0.0059}}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7949621&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=pUQ8QsPURLs&amp;video_target=tpm-plugin-n4ksl79f-pUQ8QsPURLs\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.11_transcript.html\">transcript for \u201cCP 3.11\u201d here (opens in new window).<\/a><\/center><\/section>\r\n<h2>The Normal and Binormal Vectors<\/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\">Beyond the tangent vector, two more vectors describe motion along a three-dimensional curve: the principal unit normal vector and the binormal vector. Together with the tangent, they form a moving frame of reference.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Principal Unit Normal Vector<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The principal unit normal vector [latex]{\\bf{N}}(t)[\/latex] points in the direction the curve is turning:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{\\bf{N}}(t)=\\frac{{\\bf{T}}'(t)}{\\left\\Vert{\\bf{T}}'(t)\\right\\Vert}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">It's always perpendicular to [latex]{\\bf{T}}(t)[\/latex] and has magnitude [latex]1[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Binormal Vector<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The binormal vector [latex]{\\bf{B}}(t)[\/latex] is perpendicular to both [latex]{\\bf{T}}[\/latex] and [latex]{\\bf{N}}[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{\\bf{B}}(t)={\\bf{T}}(t)\\times{\\bf{N}}(t)[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">It's automatically a unit vector.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Frenet Frame<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">These three vectors\u2014[latex]{\\bf{T}}[\/latex], [latex]{\\bf{N}}[\/latex], and [latex]{\\bf{B}}[\/latex]\u2014form a frame of reference that moves along the curve. They're mutually perpendicular and all have magnitude [latex]1[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Key Planes<\/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-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">The <strong>osculating plane<\/strong> contains [latex]{\\bf{T}}[\/latex] and [latex]{\\bf{N}}[\/latex]. The curve lies in this plane locally.<\/li>\r\n \t<li class=\"whitespace-normal break-words\">The <strong>normal plane<\/strong> contains [latex]{\\bf{N}}[\/latex] and [latex]{\\bf{B}}[\/latex]. It's perpendicular to the curve.<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>The Osculating Circle<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">At any point on a curve, you can fit a circle that matches the curve's curvature. This <strong>osculating circle<\/strong> has radius [latex]r = \\frac{1}{\\kappa}[\/latex], called the <strong>radius of curvature<\/strong>. It lies in the osculating plane, is tangent to the curve, and sits on the concave side.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Find [latex]{\\bf{T}}(t)[\/latex] first. Differentiate to get [latex]{\\bf{T}}'(t)[\/latex]. Normalize it to get [latex]{\\bf{N}}(t)[\/latex]. Take the cross product [latex]{\\bf{T}}\\times{\\bf{N}}[\/latex] to get [latex]{\\bf{B}}(t)[\/latex].<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Find the unit normal vector for the vector-valued function [latex]{\\bf{r}}\\,(t)=(t^{2}-3)\\,{\\bf{I}}+(4t+1)\\,{\\bf{j}}[\/latex] and evaluate it at [latex]t=2[\/latex].[reveal-answer q=\"fs-id1167794933264\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794933264\"]\r\n<div style=\"text-align: center;\">[latex]\\kappa=\\frac{6}{101^{3\/2}}\\approx{0.0059}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox interact\" aria-label=\"Interact\">For more information on osculating circles, see this <a href=\"http:\/\/www.openstax.org\/l\/20_OsculCircle1\" target=\"_blank\" rel=\"noopener nofollow\">demonstration<\/a> on curvature and torsion, this <a href=\"http:\/\/www.openstax.org\/l\/20_OsculCircle3\" target=\"_blank\" rel=\"noopener nofollow\">article<\/a> on osculating circles, and this <a href=\"http:\/\/www.openstax.org\/l\/20_OsculCircle2\" target=\"_blank\" rel=\"noopener nofollow\">discussion<\/a> of Serret formulas.<\/section><section class=\"textbox example\" aria-label=\"Example\">Find the equation of the osculating circle of the curve defined by the vector-valued function [latex]y=2x^{2}-4x+5[\/latex] at [latex]x=1[\/latex].[reveal-answer q=\"fs-id1167794934264\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794934264\"]\r\n<p style=\"text-align: left;\">[latex]\\kappa=\\frac{4}{[1+(4x-4)^{2}]^{3\/2}}[\/latex] At the point [latex]x=1[\/latex], the curvature is equal to [latex]4[\/latex]. Therefore, the radius of the osculating circle is [latex]\\frac{1}{4}[\/latex]. A graph of this functions appears next:<\/p>\r\n\r\n\r\n[caption id=\"attachment_921\" align=\"aligncenter\" width=\"362\"]<img class=\"size-full wp-image-921\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21161115\/3-3-tryitans.jpeg\" alt=\"This figure is the graph of the function y = 2x^2-4x+5. The curve is a parabola opening up with vertex at (1, 3).\" width=\"362\" height=\"347\" \/> Figure 9.[\/caption]\r\n\r\nThe vertex of this parabola is located at the point [latex](1,\\ 3)[\/latex]. Furthermore, the center of the osculating circle is directly above the vertex. Therefore, the coordinates of the center are [latex](1,\\ \\frac{13}{4})[\/latex]. The equation of the osculating circle is [latex](x-1)^{2}+(y-\\frac{13}{4})^{2}=\\frac{1}{16}[\/latex].\r\n<div style=\"text-align: center;\">[latex]{\\bf{v}}\\,(t)={\\bf{r}}'\\,(t)=(2t-3)\\,{\\bf{i}}+2\\,{\\bf{j}}+{\\bf{k}}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7949625&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=11CyZk8-Fa0&amp;video_target=tpm-plugin-h4q06w32-11CyZk8-Fa0\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.11_transcript.html\">transcript for \u201cCP 3.13\u201d here (opens in new window).<\/a><\/center><\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Determine the length of a particle&#8217;s path in space by using the arc-length function<\/li>\n<li>Calculate and understand what curvature means for curves in space<\/li>\n<li>Understand what normal and binormal vectors tell us about a curve&#8217;s behavior<\/li>\n<\/ul>\n<\/section>\n<h2>Arc Length for Vector Functions<\/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 distance along a curve. For vector-valued functions, the formula is straightforward: integrate the magnitude of the derivative vector.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Core Formula<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For any smooth curve defined by [latex]{\\bf{r}}(t)[\/latex] over [latex][a, b][\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s=\\int_{a}^{b} \\left\\Vert{\\bf{r}}'(t)\\right\\Vert dt[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Plane curves<\/strong> [latex]{\\bf{r}}(t)=f(t){\\bf{i}}+g(t){\\bf{j}}[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s=\\int_{a}^{b} \\sqrt{[f'(t)]^{2}+[g'(t)]^{2}}dt[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Space curves<\/strong> [latex]{\\bf{r}}(t)=f(t){\\bf{i}}+g(t){\\bf{j}}+h(t){\\bf{k}}[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s=\\int_{a}^{b} \\sqrt{[f'(t)]^{2}+[g'(t)]^{2}+[h'(t)]^{2}}dt[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">The only difference? Space curves include a third component.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Find [latex]{\\bf{r}}'(t)[\/latex] by differentiating each component<\/li>\n<li class=\"whitespace-normal break-words\">Calculate [latex]\\left\\Vert{\\bf{r}}'(t)\\right\\Vert[\/latex] using the square root of the sum of squares<\/li>\n<li class=\"whitespace-normal break-words\">Integrate [latex]\\left\\Vert{\\bf{r}}'(t)\\right\\Vert[\/latex] from [latex]a[\/latex] to [latex]b[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\">The curve must be <strong>smooth<\/strong>: [latex]{\\bf{r}}(t)[\/latex] must be differentiable with a non-zero derivative. This guarantees no cusps or corners that break the formula.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Special Case: Helices<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">A helix with radius [latex]R[\/latex], height [latex]h[\/latex], and [latex]N[\/latex] turns has arc length:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s=\\sqrt{4{\\pi}^{2}N^{2}R^{2}+h^{2}}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This tells you the length of wire needed to form the helix.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Calculate the arc length of the parameterized curve<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{r}}\\,(t)=\\langle{2t^{2}}+1,\\ 2t^{2}-1,\\ t^{3}\\rangle,\\ 0\\,\\leq\\,t\\,\\leq\\,3.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794933124\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167794933124\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">[latex]{\\bf{r}}'\\,(t)=\\langle{4t},\\ 4t,\\ 3t^{2}\\rangle[\/latex], so [latex]s=\\frac{1}{27}(113^{3\/2}-32^{3\/2})\\approx{37.785}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7949619&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=Gti0f6fEX6A&amp;video_target=tpm-plugin-e842wmnl-Gti0f6fEX6A\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.9_transcript.html\">transcript for \u201cCP 3.9\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<h2>Arc-Length Parameterization<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Arc-length parameterization is a way to describe a curve where the parameter directly measures distance traveled along the curve. It&#8217;s like using mile markers instead of time to track position on a road.<\/p>\n<p class=\"whitespace-normal break-words\">Regular parameterization can trace a curve at any speed. Arc-length parameterization always traces at unit speed, making distance calculations straightforward.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Arc-Length Function<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For a smooth curve [latex]{\\bf{r}}(t)[\/latex] where [latex]t \\geq a[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s(t)=\\int_{a}^{t} \\left\\Vert{\\bf{r}}'(u)\\right\\Vert du[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This function [latex]s(t)[\/latex] tells you the distance traveled from the starting point. Taking its derivative gives speed:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\frac{ds}{dt}=\\left\\Vert{\\bf{r}}'(t)\\right\\Vert[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Special Case: Unit Speed<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When [latex]\\left\\Vert{\\bf{r}}'(t)\\right\\Vert=1[\/latex] for all [latex]t[\/latex], the parameter [latex]t[\/latex] already measures arc length. The curve traces at unit speed.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Creating an Arc-Length Parameterization<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Find the arc-length function<\/strong>: Calculate [latex]s(t)=\\int_{a}^{t} \\left\\Vert{\\bf{r}}'(u)\\right\\Vert du[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Solve for [latex]t[\/latex]<\/strong>: Express [latex]t[\/latex] as a function of [latex]s[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Substitute<\/strong>: Replace [latex]t[\/latex] in [latex]{\\bf{r}}(t)[\/latex] with your expression in terms of [latex]s[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Adjust the domain<\/strong>: Convert the restriction on [latex]t[\/latex] to a restriction on [latex]s[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\">Arc-length parameterization makes the parameter meaningful. The value of [latex]s[\/latex] tells you exactly how far you&#8217;ve traveled along the curve. This simplifies formulas for curvature and line integrals.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Find the arc-length function for the helix<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{r}}\\,(t)=\\langle{3}\\cos{t},\\ 3\\sin{t},\\ 4t\\rangle{,}\\ t\\,\\geq\\,0.[\/latex]<\/p>\n<p>Then, use the relationship between the arc length and the parameter [latex]t[\/latex] to find an arc-length parameterization of [latex]{\\bf{r}}(t)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794933134\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167794933134\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">[latex]s=5t[\/latex], or [latex]t=\\frac{s}{5}[\/latex]. Substituting this into [latex]{\\bf{r}}\\,(t)=\\langle{3}\\cos{t},\\ 3\\sin{t},\\ 4t\\rangle[\/latex] gives [latex]{\\bf{r}}\\,(s)=\\Big\\langle3\\cos{\\big(\\frac{s}{5}\\big)},\\ 3\\sin{\\big(\\frac{s}{5}\\big)},\\ \\frac{4s}{5}\\Big\\rangle,\\ s\\geq{0}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7994454&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=F7R34nEPVtM&amp;video_target=tpm-plugin-vfuem7m1-F7R34nEPVtM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.10_transcript.html\">transcript for \u201cCP 3.10\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<h2>Curvature<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Curvature measures how sharply a curve bends. A straight line has zero curvature. A tight turn has high curvature. A gentle curve has low curvature.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Intuition<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Picture driving on a road. A gentle curve needs barely any steering. A sharp turn requires turning the wheel more. Curvature quantifies this difference. For a circle, smaller radius means greater curvature.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Definition<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For a smooth curve [latex]{\\bf{r}}(s)[\/latex] with arc-length parameter [latex]s[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\kappa=\\left\\Vert{\\bf{T}}'(s)\\right\\Vert[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">where [latex]{\\bf{T}}[\/latex] is the unit tangent vector.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Practical Formulas<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The definition requires arc-length parameterization. These alternatives work with any parameterization:<\/p>\n<ul>\n<li class=\"whitespace-normal break-words\"><strong>For any curve<\/strong> [latex]{\\bf{r}}(t)[\/latex]: [latex]\\kappa=\\frac{\\left\\Vert{\\bf{T}}'(t)\\right\\Vert}{\\left\\Vert{\\bf{r}}'(t)\\right\\Vert}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>For space curves<\/strong>: [latex]\\kappa=\\frac{\\left\\Vert{\\bf{r}}'(t)\\times{\\bf{r}}''(t)\\right\\Vert}{\\left\\Vert{\\bf{r}}'(t)\\right\\Vert^{3}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>For graphs<\/strong> [latex]y=f(x)[\/latex]: [latex]\\kappa=\\frac{|y''|}{[1+(y')^{2}]^{3\/2}}[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Choosing Your Formula<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Use the first when you already have [latex]{\\bf{T}}(t)[\/latex] calculated<\/li>\n<li class=\"whitespace-normal break-words\">Use the cross product formula for space curves\u2014it only needs [latex]{\\bf{r}}'(t)[\/latex] and [latex]{\\bf{r}}''(t)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Use the [latex]y=f(x)[\/latex] formula for function graphs<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Circles<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For a circle of radius [latex]r[\/latex], curvature equals [latex]\\kappa = \\frac{1}{r}[\/latex]. Curvature is the reciprocal of radius.<\/p>\n<\/div>\n<section class=\"textbox interact\" aria-label=\"Interact\">Visit\u00a0this <a href=\"http:\/\/www.openstax.org\/l\/20_spacecurve\" target=\"_blank\" rel=\"noopener nofollow\">website<\/a> for more information about the curvature of a space curve.<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the curvature of the curve defined by the function<\/p>\n<p style=\"text-align: center;\">[latex]y=3x^{2}-2x+4[\/latex]<\/p>\n<p>at the point [latex]x=2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794933164\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167794933164\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">[latex]\\large{\\kappa=\\frac{6}{101^{3\/2}}\\approx{0.0059}}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7949621&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=pUQ8QsPURLs&amp;video_target=tpm-plugin-n4ksl79f-pUQ8QsPURLs\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.11_transcript.html\">transcript for \u201cCP 3.11\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<h2>The Normal and Binormal Vectors<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Beyond the tangent vector, two more vectors describe motion along a three-dimensional curve: the principal unit normal vector and the binormal vector. Together with the tangent, they form a moving frame of reference.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Principal Unit Normal Vector<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The principal unit normal vector [latex]{\\bf{N}}(t)[\/latex] points in the direction the curve is turning:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{\\bf{N}}(t)=\\frac{{\\bf{T}}'(t)}{\\left\\Vert{\\bf{T}}'(t)\\right\\Vert}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">It&#8217;s always perpendicular to [latex]{\\bf{T}}(t)[\/latex] and has magnitude [latex]1[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Binormal Vector<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The binormal vector [latex]{\\bf{B}}(t)[\/latex] is perpendicular to both [latex]{\\bf{T}}[\/latex] and [latex]{\\bf{N}}[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{\\bf{B}}(t)={\\bf{T}}(t)\\times{\\bf{N}}(t)[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">It&#8217;s automatically a unit vector.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Frenet Frame<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">These three vectors\u2014[latex]{\\bf{T}}[\/latex], [latex]{\\bf{N}}[\/latex], and [latex]{\\bf{B}}[\/latex]\u2014form a frame of reference that moves along the curve. They&#8217;re mutually perpendicular and all have magnitude [latex]1[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Key Planes<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">The <strong>osculating plane<\/strong> contains [latex]{\\bf{T}}[\/latex] and [latex]{\\bf{N}}[\/latex]. The curve lies in this plane locally.<\/li>\n<li class=\"whitespace-normal break-words\">The <strong>normal plane<\/strong> contains [latex]{\\bf{N}}[\/latex] and [latex]{\\bf{B}}[\/latex]. It&#8217;s perpendicular to the curve.<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>The Osculating Circle<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">At any point on a curve, you can fit a circle that matches the curve&#8217;s curvature. This <strong>osculating circle<\/strong> has radius [latex]r = \\frac{1}{\\kappa}[\/latex], called the <strong>radius of curvature<\/strong>. It lies in the osculating plane, is tangent to the curve, and sits on the concave side.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Find [latex]{\\bf{T}}(t)[\/latex] first. Differentiate to get [latex]{\\bf{T}}'(t)[\/latex]. Normalize it to get [latex]{\\bf{N}}(t)[\/latex]. Take the cross product [latex]{\\bf{T}}\\times{\\bf{N}}[\/latex] to get [latex]{\\bf{B}}(t)[\/latex].<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Find the unit normal vector for the vector-valued function [latex]{\\bf{r}}\\,(t)=(t^{2}-3)\\,{\\bf{I}}+(4t+1)\\,{\\bf{j}}[\/latex] and evaluate it at [latex]t=2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794933264\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167794933264\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">[latex]\\kappa=\\frac{6}{101^{3\/2}}\\approx{0.0059}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox interact\" aria-label=\"Interact\">For more information on osculating circles, see this <a href=\"http:\/\/www.openstax.org\/l\/20_OsculCircle1\" target=\"_blank\" rel=\"noopener nofollow\">demonstration<\/a> on curvature and torsion, this <a href=\"http:\/\/www.openstax.org\/l\/20_OsculCircle3\" target=\"_blank\" rel=\"noopener nofollow\">article<\/a> on osculating circles, and this <a href=\"http:\/\/www.openstax.org\/l\/20_OsculCircle2\" target=\"_blank\" rel=\"noopener nofollow\">discussion<\/a> of Serret formulas.<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the equation of the osculating circle of the curve defined by the vector-valued function [latex]y=2x^{2}-4x+5[\/latex] at [latex]x=1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794934264\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167794934264\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">[latex]\\kappa=\\frac{4}{[1+(4x-4)^{2}]^{3\/2}}[\/latex] At the point [latex]x=1[\/latex], the curvature is equal to [latex]4[\/latex]. Therefore, the radius of the osculating circle is [latex]\\frac{1}{4}[\/latex]. A graph of this functions appears next:<\/p>\n<figure id=\"attachment_921\" aria-describedby=\"caption-attachment-921\" style=\"width: 362px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-921\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21161115\/3-3-tryitans.jpeg\" alt=\"This figure is the graph of the function y = 2x^2-4x+5. The curve is a parabola opening up with vertex at (1, 3).\" width=\"362\" height=\"347\" \/><figcaption id=\"caption-attachment-921\" class=\"wp-caption-text\">Figure 9.<\/figcaption><\/figure>\n<p>The vertex of this parabola is located at the point [latex](1,\\ 3)[\/latex]. Furthermore, the center of the osculating circle is directly above the vertex. Therefore, the coordinates of the center are [latex](1,\\ \\frac{13}{4})[\/latex]. The equation of the osculating circle is [latex](x-1)^{2}+(y-\\frac{13}{4})^{2}=\\frac{1}{16}[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]{\\bf{v}}\\,(t)={\\bf{r}}'\\,(t)=(2t-3)\\,{\\bf{i}}+2\\,{\\bf{j}}+{\\bf{k}}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7949625&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=11CyZk8-Fa0&amp;video_target=tpm-plugin-h4q06w32-11CyZk8-Fa0\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.11_transcript.html\">transcript for \u201cCP 3.13\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":27,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-568","chapter","type-chapter","status-publish","hentry"],"part":424,"module-header":"fresh_take","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/568","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":4,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/568\/revisions"}],"predecessor-version":[{"id":1757,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/568\/revisions\/1757"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/424"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/568\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=568"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=568"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=568"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=568"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}