{"id":3862,"date":"2022-04-04T16:25:18","date_gmt":"2022-04-04T16:25:18","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=3862"},"modified":"2022-10-29T00:51:03","modified_gmt":"2022-10-29T00:51:03","slug":"arc-length","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/arc-length\/","title":{"raw":"Arc Length","rendered":"Arc Length"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\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<\/ul>\r\n<\/div>\r\n<h2>Arc Length for Vector Functions<\/h2>\r\nWe have seen how a vector-valued function describes a curve in either two or three dimensions. Recall Alternative Formulas for Curvature, which states that the formula for the arc length of a curve defined by the parametric functions [latex]x=x\\,(t),\\ y=t\\,(t),\\ t_{1}\\,\\leq\\,t\\,\\leq\\,t_{2}[\/latex] is given by\r\n<p style=\"text-align: center;\">[latex]s=\\displaystyle\\int_{t_{1}}^{t^{2}}\\ \\sqrt{\\big(x'\\,(t)\\big)^{2}+\\big(y'\\,(t)\\big)^{2}}\\,dt[\/latex]<\/p>\r\nIn a similar fashion, if we define a smooth curve using a vector-valued function [latex]{\\bf{r}}\\,(t)=f\\,(t)\\,{\\bf{I}}+g\\,(t)\\,{\\bf{j}}[\/latex], where [latex]a\\,\\leq\\,t\\,\\leq\\,b[\/latex], the arc length is given by the formula\r\n<p style=\"text-align: center;\">[latex]s=\\displaystyle\\int_{a}^{b}\\ \\sqrt{\\big(f'\\,(t)\\big)^{2}+\\big(g'\\,(t)\\big)^{2}}\\,dt[\/latex]<\/p>\r\nIn three dimensions, if the vector-valued function is described by [latex]{\\bf{r}}\\,(t)=f\\,(t)\\,{\\bf{I}}+g\\,(t)\\,{\\bf{j}}+h\\,(t)\\,{\\bf{k}}[\/latex] over the same interval [latex]a\\,\\leq\\,t\\,\\leq\\,b[\/latex], the arc length is given by\r\n<p style=\"text-align: center;\">[latex]s=\\displaystyle\\int_{a}^{b}\\ \\sqrt{\\big(f'\\,(t)\\big)^{2}+\\big(g'\\,(t)\\big)^{2}+\\big(h'\\,(t)\\big)^{2}}\\,dt[\/latex]<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Arc-Length formulas Theorem<\/h3>\r\n\r\n<hr \/>\r\n\r\n<ul>\r\n \t<li><em>Plane curve<\/em>: Given a smooth curve [latex]C[\/latex] defined by the function [latex]{\\bf{r}}\\,(t)=g\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}[\/latex], where [latex]t[\/latex] lies within the interval [latex][a,\\ b][\/latex], the arc length of [latex]C[\/latex] over the interval is<\/li>\r\n<\/ul>\r\n<div style=\"text-align: center;\">[latex]s=\\displaystyle\\int_{a}^{b}\\ \\sqrt{\\big[f'\\,(t)\\big]^{2}+\\big[g'\\,(t)\\big]^{2}}\\,dt=\\displaystyle\\int_{a}^{b}\\ \\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert\\ dt[\/latex].<\/div>\r\n&nbsp;\r\n<ul>\r\n \t<li><em>Space curve<\/em>: Given a smooth curve [latex]C[\/latex] defined by the function [latex]{\\bf{r}}\\,(t)=f\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}+h\\,(t)\\,{\\bf{k}}[\/latex], where [latex]t[\/latex] lies within the interval [latex][a,\\ b][\/latex], the arc length of [latex]C[\/latex] over the interval is<\/li>\r\n<\/ul>\r\n<div style=\"text-align: center;\">[latex]s=\\displaystyle\\int_{a}^{b}\\ \\sqrt{\\big[f'\\,(t)\\big]^{2}+\\big[g'\\,(t)\\big]^{2}+\\big[h'\\,(t)\\big]^{2}}\\,dt=\\displaystyle\\int_{a}^{b}\\ \\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert\\ dt[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\nThe two formulas are very similar; they differ only in the fact that a space curve has three component functions instead of two. Note that the formulas are defined for smooth curves: curves where the vector-valued function [latex]{\\bf{r}}\\,(t)[\/latex] is differentiable with a non-zero derivative.\u00a0The smoothness condition guarantees that the curve has no cusps (or corners) that could make the formula problematic.\r\n<div id=\"fs-id1167793900968\" class=\"textbook exercises\">\r\n<h3>Example: finding the arc length<\/h3>\r\nCalculate the arc length for each of the following vector-valued functions:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">a. [latex]{\\bf{r}}\\,(t)=(3t-2)\\,{\\bf{i}}+(4t+5)\\,{\\bf{j}},\\ 1\\,\\leq\\,t\\,\\leq\\,5[\/latex]<\/ol>\r\n<ol style=\"list-style-type: lower-alpha;\">b. [latex]{\\bf{r}}\\,(t)=\\langle{t}\\cos{t}, \\ t\\sin{t},\\ 2t\\rangle,\\ 0\\,\\leq\\,t\\,\\leq\\,2\\pi[\/latex]<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167793461764\" class=\"exercise\">[reveal-answer q=\"fs-id1167795055165\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167795055165\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Using the Arc Length Formulas, [latex]{\\bf{r}}'\\,(t)=3\\,{\\bf{i}}+4\\,{\\bf{j}}[\/latex], so\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {s} &amp;=\\hfill&amp;{\\displaystyle\\int_{a}^{b}\\ \\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert\\ dt}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\int_{a}^{5}\\ \\sqrt{3^{2}+4^{2}}\\,dt} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\int_{1}^{5}\\ 5\\,dt=5t\\Big|^5_1=20.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;<\/li>\r\n \t<li>Using the Arc Length Formulas, [latex]{\\bf{r}}'\\,(t)=\\langle{\\cos{t}}-t\\sin{t},\\ \\sin{t}+t\\cos{t},\\ 2\\rangle[\/latex], so\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {s} &amp;=\\hfill&amp;{\\displaystyle\\int_{a}^{b}\\ \\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert\\ dt}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\int_{0}^{2\\pi}\\ \\sqrt{(\\cos{t}-t\\sin{t})^{2}+(\\sin{t}+t\\cos{t})^{2}+2^{2}}\\,dt} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\int_{0}^{2\\pi}\\ \\sqrt{(\\cos{^{2}t}-2t\\sin{t}\\cos{t}+t^{2}\\, \\sin{^{2}t})+(\\sin{^{2}t}+2t\\sin{t}\\cos{t}+t^{2}\\, \\cos{^{2}t})+4}\\,dt}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\int_{0}^{2\\pi}\\ \\sqrt{\\cos{^{2}t}+\\sin{^{2}t}+t^{2}\\,(\\cos{^{2}t}+\\sin{^{2}t})+4}\\,dt}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\int_{0}^{2\\pi}\\ \\sqrt{t^{2}+5}\\,dt}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nHere we can use a table integration formula\r\n<div style=\"text-align: center;\">[latex]\\displaystyle\\int_{}\\ \\sqrt{u^{2}+a^{2}}\\,du=\\frac{u}{2}\\,\\sqrt{u^{2}+a^{2}}+\\frac{a^{2}}{2}\\ln{\\Big|u+\\sqrt{u^{2}+a^{2}}\\Big|}+C[\/latex],<\/div>\r\nso we obtain\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int_{0}^{2\\pi}\\ \\sqrt{t^{2}+5}\\,dt} &amp;=\\hfill&amp;{\\frac{1}{2}\\big(t\\sqrt{t^{2}+5}+5\\,ln\\ \\Big|t+\\sqrt{t^{2}+5}\\Big|\\big)^{2\\pi}_{0}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{1}{2}\\,\\big(2\\pi\\sqrt{4{\\pi}^{2}+5}+5\\ln{(2\\pi+\\sqrt{4{\\pi}^{2}+5})\\big)}-\\frac{5}{2}\\ln{\\sqrt{5}}} \\hfill \\\\ \\hfill &amp; \\approx\\hfill &amp; {25.343.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1167793958097\" class=\"textbook key-takeaways\">\r\n<h3>try it<\/h3>\r\nCalculate 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<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<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\"><\/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>We now return to the helix introduced earlier in this chapter. A vector-valued function that describes a helix can be written in the form\r\n<p style=\"text-align: center;\">[latex]{\\bf{r}}\\,(t)=R\\cos{\\Big(\\frac{2\\pi\\,N\\,t}{h}\\Big)}\\,{\\bf{i}}+R\\sin{\\Big(\\frac{2\\pi\\,N\\,t}{h}\\Big)}\\,{\\bf{j}}+t\\,{\\bf{k}},\\ 0\\,\\leq\\,t\\,\\leq\\,h,[\/latex]<\/p>\r\nwhere [latex]R[\/latex]\u00a0represents the radius of the helix, [latex]h[\/latex]\u00a0represents the height (distance between two consecutive turns), and the helix completes [latex]N[\/latex]\u00a0turns. Let\u2019s derive a formula for the arc length of this helix using Arc-Length Formulas. First of all,\r\n<p style=\"text-align: center;\">[latex]{\\bf{r}}'\\,(t)=-\\frac{2\\pi\\,N\\,R}{h}\\sin{\\Big(\\frac{2\\pi\\,N\\,t}{h}\\Big)}\\,{\\bf{i}}+\\frac{2\\pi\\,N\\,R}{h}\\cos{\\Big(\\frac{2\\pi\\,N\\,t}{h}\\Big)}\\,{\\bf{j}}+{\\bf{k}}.[\/latex]<\/p>\r\nTherefore,\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {s} &amp;=\\hfill&amp;{\\displaystyle\\int_{a}^{b}\\ \\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert\\ dt}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\int_{0}^{h}\\ \\sqrt{\\Bigg(-\\frac{2\\pi\\,N\\,R}{h}\\sin{\\Big(\\frac{2\\pi\\,N\\,t}{h}\\Big)}\\Bigg)^{2}+\\Bigg(\\frac{2\\pi\\,N\\,R}{h}\\cos{\\Big(\\frac{2\\pi\\,N\\,t}{h}\\Big)}\\Bigg)^{2}+1^{2}}\\,dt} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\int_{0}^{h}\\ \\sqrt{\\frac{4{\\pi}^{2}N^{2}R^{2}}{h^{2}}\\,\\Bigg(\\sin{^{2}\\Big(\\frac{2\\pi{N}t}{h}\\Big)}+\\cos{^{2}\\Big(\\frac{2\\pi{N}t}{h}\\Big)\\Bigg)}+1}\\,dt}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\int_{0}^{h}\\ \\sqrt{\\frac{4{\\pi}^{2}N^{2}R^{2}}{h^{2}}+1}\\,dt}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\Big[t\\,\\sqrt{\\frac{4{\\pi}^{2}N^{2}R^{2}}{h^{2}}+1}\\Big]^h_0}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {h\\,\\sqrt{\\frac{4{\\pi}^{2}N^{2}R^{2}+h^{2}}{h^{2}}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\sqrt{4{\\pi}^{2}N^{2}R^{2}+h^{2}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\r\nThis gives a formula for the length of a wire needed to form a helix with [latex]N[\/latex]\u00a0turns that has radius [latex]R[\/latex]\u00a0and height [latex]h[\/latex]<em data-effect=\"italics\">.<\/em>\r\n<h2>Arc-Length Parameterization<\/h2>\r\n<p id=\"fs-id1169738022483\" class=\" \">We now have a formula for the arc length of a curve defined by a vector-valued function. Let\u2019s take this one step further and examine what an <strong><span id=\"term130\" data-type=\"term\">arc-length function<\/span><\/strong> is.<\/p>\r\n<p id=\"fs-id1169738186222\" class=\" \">If a vector-valued function represents the position of a particle in space as a function of time, then the arc-length function measures how far that particle travels as a function of time. The formula for the arc-length function follows directly from the formula for arc length:<\/p>\r\n<p style=\"text-align: center;\">[latex]s\\,(t)=\\displaystyle\\int_{a}^{t}\\ \\sqrt{\\big(f'\\,(u)\\big)^{2}+\\big(g'\\,(u)\\big)^{2}+\\big(h'\\,(u)\\big)^{2}}\\,du[\/latex]<\/p>\r\nIf the curve is in two dimensions, then only two terms appear under the square root inside the integral. The reason for using the independent variable [latex]u[\/latex]\u00a0is to distinguish between time and the variable of integration. Since [latex]s(t)[\/latex]\u00a0measures distance traveled as a function of time, [latex]s'(t)[\/latex]\u00a0measures the speed of the particle at any given time. Since we have a formula for [latex]s(t)[\/latex], we can differentiate both sides of the equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {s'\\,(t)} &amp;=\\hfill&amp;{\\frac{d}{dt}\\bigg[\\displaystyle\\int_{a}^{t}\\ \\sqrt{\\big(f'\\,(u)\\big)^{2}+\\big(g'\\,(u)\\big)^{2}+\\big(h'\\,(u)\\big)^{2}}\\,du\\bigg]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{d}{dt}\\bigg[\\displaystyle\\int_{a}^{t}\\ \\left\\Vert{\\bf{r}}'\\,(u)\\right\\Vert\\ du\\bigg]} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert\\,.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\r\nIf we assume that [latex]{\\bf{r}}\\,(t)[\/latex] defines a smooth curve, then the arc length is always increasing, so [latex]s'\\,(t)\\,&gt;\\,0[\/latex] for [latex]t\\,&gt;\\,a[\/latex]. Last, if [latex]{\\bf{r}}\\,(t)[\/latex] is a curve on which [latex]\\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert=1[\/latex] for all [latex]t[\/latex], then\r\n<p style=\"text-align: center;\">[latex]s\\,(t)=\\displaystyle\\int_{a}^{t}\\ \\left\\Vert{\\bf{r}}'\\,(u)\\right\\Vert\\ du=\\displaystyle\\int_{a}^{t}\\ 1\\,du=t-a,[\/latex]<\/p>\r\nwhich means that [latex]t[\/latex] represents the arc length as long as [latex]a=0[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Arc-Length function Theorem<\/h3>\r\nLet [latex]{\\bf{r}}\\,(t)[\/latex] describe a smooth curve for [latex]t\\,\\geq\\,a[\/latex]. Then the arc-length function is given by\r\n<div style=\"text-align: center;\">[latex]s\\,(t)=\\displaystyle\\int_{a}^{t}\\ \\left\\Vert{\\bf{r}}'\\,(u)\\right\\Vert\\ du[\/latex].<\/div>\r\n&nbsp;\r\n\r\nFurthermore, [latex]\\frac{ds}{dt}=\\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert\\,&gt;\\,0[\/latex]. If [latex]\\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert=1[\/latex] for all [latex]t\\,\\geq\\,a[\/latex], then the parameter [latex]t[\/latex] represents the arc length from the starting point at [latex]t=a[\/latex].\r\n\r\n<\/div>\r\nA useful application of this theorem is to find an alternative parameterization of a given curve, called an <strong><span id=\"term131\" data-type=\"term\">arc-length parameterization<\/span><\/strong>. Recall that any vector-valued function can be reparameterized via a change of variables. For example, if we have a function [latex]{\\bf{r}}\\,(t)=\\langle{3}\\cos{t},\\ 3\\sin{t}\\rangle,\\ 0\\,\\leq\\,t\\,\\leq\\,2\\pi[\/latex]\u00a0that parameterizes a circle of radius 3, we can change the parameter from [latex]t[\/latex] to [latex]4t[\/latex], obtaining a new parameterization [latex]{\\bf{r}}\\,(t)=\\langle{3}\\cos{4t},\\ 3\\sin{4t}\\rangle[\/latex]. The new parameterization still defines a circle of radius [latex]3[\/latex], but now we need only use the values [latex]0\\,\\leq\\,t\\,\\leq\\,\\frac{\\pi}{2}[\/latex] to traverse the circle once.\r\n\r\nSuppose that we find the arc-length function [latex]s(t)[\/latex] and are able to solve this function for [latex]t[\/latex] as a function of [latex]s[\/latex]. We can then reparameterize the original function [latex]{\\bf{r}}(t)[\/latex] by substituting the expression for [latex]t[\/latex] back into [latex]{\\bf{r}}(t)[\/latex]. The vector-valued function is now written in terms of the parameter [latex]s[\/latex]. Since the variable [latex]s[\/latex] represents the arc length, we call this an <em>arc-length parameterization<\/em> of the original function [latex]{\\bf{r}}(t)[\/latex]. One advantage of finding the arc-length parameterization is that the distance traveled along the curve starting from [latex]s=0[\/latex] is now equal to the parameter [latex]s[\/latex]. The arc-length parameterization also appears in the context of curvature (which we examine later in this section) and line integrals, which we study in the Introduction to Vector Calculus.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding an arc-length parameterization<\/h3>\r\nFind the arc-length parameterization for each of the following curves:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">a. [latex]{\\bf{r}}\\,(t)=4\\cos{t{\\bf{i}}}+4\\sin{t{\\bf{j}}},\\ t\\,\\geq\\,0[\/latex]<\/ol>\r\n<ol style=\"list-style-type: lower-alpha;\">b. [latex]{\\bf{r}}\\,(t)=\\langle{t+3},\\ 2t-4,\\ 2t\\rangle,\\ t\\,\\geq\\,3[\/latex]<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167793461764\" class=\"exercise\">[reveal-answer q=\"fs-id1167795055166\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167795055166\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>First we find the arc-length function using the Arc Length Function equations:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {s\\,(t)} &amp;=\\hfill&amp;{\\displaystyle\\int_{a}^{t}\\ \\left\\Vert{\\bf{r}}'\\,(u)\\right\\Vert\\ du}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\int_{0}^{t}\\ \\left\\Vert\\langle{-4}\\sin{u},\\ 4\\cos{u}\\rangle\\right\\Vert\\,du} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\int_{0}^{t}\\ \\sqrt{(-4\\sin{u})^{2}+(4\\cos{u})^{2}}\\,du}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\int_{0}^{t}\\ \\sqrt{16\\sin{^{2}u}+16\\cos{^{2}u}}\\,du}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\int_{0}^{t}\\ 4\\,du=4t,}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nwhich gives the relationship between the arc length [latex]s[\/latex] and the parameter [latex]t[\/latex] as [latex]s=4t[\/latex]; so, [latex]t=s\/4[\/latex]. Next we replace the variable [latex]t[\/latex] in the original function [latex]{\\bf{r}}\\,(t)=4\\cos{t{\\bf{i}}}+4\\sin{t{\\bf{j}}}[\/latex] with the expression [latex]s\/4[\/latex] to obtain\r\n<div style=\"text-align: center;\">[latex]{\\bf{r}}\\,(s)=4\\cos{\\big(\\frac{s}{4}\\big)}{\\bf{i}}+4\\sin{\\big(\\frac{s}{4}\\big)}{\\bf{j}}[\/latex].<\/div>\r\n&nbsp;\r\n\r\nThis is the arc-length parameterization of [latex]{\\bf{r}}(t)[\/latex]. Since the original restriction on [latex]t[\/latex] was given by [latex]t\\geq{0}[\/latex], the restriction on [latex]s[\/latex] becomes [latex]s\/4\\geq{0}[\/latex], or [latex]s\\geq{0}[\/latex].<\/li>\r\n \t<li>The arc-length function is given by the Arc Length Function equations:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {s\\,(t)} &amp;=\\hfill&amp;{\\displaystyle\\int_{a}^{t}\\ \\left\\Vert{\\bf{r}}'\\,(u)\\right\\Vert\\ du}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\int_{3}^{t}\\ \\left\\Vert\\langle{1},\\ 2,\\ 2\\rangle\\right\\Vert\\ du} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\int_{3}^{t}\\ \\sqrt{1^{2}+2^{2}+2^{2}}\\,du}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\int_{3}^{t}\\ 3\\,du}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {3t-9.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nTherefore, the relationship between the arc length [latex]s[\/latex] and the parameter [latex]t[\/latex] is [latex]s=3t-9[\/latex], so [latex]t=\\frac{s}{3}+3[\/latex]. Substituting this into the original function [latex]{\\bf{r}}\\,(t)=\\langle{t}+3,\\ 2t-4,\\ 2t\\rangle[\/latex] yields\r\n<div style=\"text-align: center;\">[latex]{\\bf{r}}\\,(s)=\\Big\\langle\\big(\\frac{s}{3}+3\\big)+3,\\ 2\\,\\big(\\frac{s}{3}+3\\big)-4,\\ 2\\,\\big(\\frac{s}{3}+3\\big)\\Big\\rangle=\\Big\\langle\\frac{s}{3}+6,\\ \\frac{2s}{3}+2,\\ \\frac{2s}{3}+6\\Big\\rangle[\/latex].<\/div>\r\n&nbsp;\r\n\r\nThis is an arc-length parameterization of [latex]{\\bf{r}}\\,(t)[\/latex]. The original restriction on the parameter [latex]t[\/latex] was [latex]t\\,\\geq\\,3[\/latex], so the restriction on [latex]s[\/latex] is [latex](s\/3)+3\\,\\geq\\,3[\/latex], or [latex]s\\,\\geq\\,0[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind 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<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<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\"><\/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>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Determine the length of a particle&#8217;s path in space by using the arc-length function.<\/li>\n<\/ul>\n<\/div>\n<h2>Arc Length for Vector Functions<\/h2>\n<p>We have seen how a vector-valued function describes a curve in either two or three dimensions. Recall Alternative Formulas for Curvature, which states that the formula for the arc length of a curve defined by the parametric functions [latex]x=x\\,(t),\\ y=t\\,(t),\\ t_{1}\\,\\leq\\,t\\,\\leq\\,t_{2}[\/latex] is given by<\/p>\n<p style=\"text-align: center;\">[latex]s=\\displaystyle\\int_{t_{1}}^{t^{2}}\\ \\sqrt{\\big(x'\\,(t)\\big)^{2}+\\big(y'\\,(t)\\big)^{2}}\\,dt[\/latex]<\/p>\n<p>In a similar fashion, if we define a smooth curve using a vector-valued function [latex]{\\bf{r}}\\,(t)=f\\,(t)\\,{\\bf{I}}+g\\,(t)\\,{\\bf{j}}[\/latex], where [latex]a\\,\\leq\\,t\\,\\leq\\,b[\/latex], the arc length is given by the formula<\/p>\n<p style=\"text-align: center;\">[latex]s=\\displaystyle\\int_{a}^{b}\\ \\sqrt{\\big(f'\\,(t)\\big)^{2}+\\big(g'\\,(t)\\big)^{2}}\\,dt[\/latex]<\/p>\n<p>In three dimensions, if the vector-valued function is described by [latex]{\\bf{r}}\\,(t)=f\\,(t)\\,{\\bf{I}}+g\\,(t)\\,{\\bf{j}}+h\\,(t)\\,{\\bf{k}}[\/latex] over the same interval [latex]a\\,\\leq\\,t\\,\\leq\\,b[\/latex], the arc length is given by<\/p>\n<p style=\"text-align: center;\">[latex]s=\\displaystyle\\int_{a}^{b}\\ \\sqrt{\\big(f'\\,(t)\\big)^{2}+\\big(g'\\,(t)\\big)^{2}+\\big(h'\\,(t)\\big)^{2}}\\,dt[\/latex]<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Arc-Length formulas Theorem<\/h3>\n<hr \/>\n<ul>\n<li><em>Plane curve<\/em>: Given a smooth curve [latex]C[\/latex] defined by the function [latex]{\\bf{r}}\\,(t)=g\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}[\/latex], where [latex]t[\/latex] lies within the interval [latex][a,\\ b][\/latex], the arc length of [latex]C[\/latex] over the interval is<\/li>\n<\/ul>\n<div style=\"text-align: center;\">[latex]s=\\displaystyle\\int_{a}^{b}\\ \\sqrt{\\big[f'\\,(t)\\big]^{2}+\\big[g'\\,(t)\\big]^{2}}\\,dt=\\displaystyle\\int_{a}^{b}\\ \\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert\\ dt[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<ul>\n<li><em>Space curve<\/em>: Given a smooth curve [latex]C[\/latex] defined by the function [latex]{\\bf{r}}\\,(t)=f\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}+h\\,(t)\\,{\\bf{k}}[\/latex], where [latex]t[\/latex] lies within the interval [latex][a,\\ b][\/latex], the arc length of [latex]C[\/latex] over the interval is<\/li>\n<\/ul>\n<div style=\"text-align: center;\">[latex]s=\\displaystyle\\int_{a}^{b}\\ \\sqrt{\\big[f'\\,(t)\\big]^{2}+\\big[g'\\,(t)\\big]^{2}+\\big[h'\\,(t)\\big]^{2}}\\,dt=\\displaystyle\\int_{a}^{b}\\ \\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert\\ dt[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p>The two formulas are very similar; they differ only in the fact that a space curve has three component functions instead of two. Note that the formulas are defined for smooth curves: curves where the vector-valued function [latex]{\\bf{r}}\\,(t)[\/latex] is differentiable with a non-zero derivative.\u00a0The smoothness condition guarantees that the curve has no cusps (or corners) that could make the formula problematic.<\/p>\n<div id=\"fs-id1167793900968\" class=\"textbook exercises\">\n<h3>Example: finding the arc length<\/h3>\n<p>Calculate the arc length for each of the following vector-valued functions:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">  <\/ol>\n<ol style=\"list-style-type: lower-alpha;\">     <\/ol>\n<\/li>\n<\/ol>\n<div id=\"fs-id1167793461764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167795055165\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167795055165\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Using the Arc Length Formulas, [latex]{\\bf{r}}'\\,(t)=3\\,{\\bf{i}}+4\\,{\\bf{j}}[\/latex], so\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {s} &=\\hfill&{\\displaystyle\\int_{a}^{b}\\ \\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert\\ dt}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\int_{a}^{5}\\ \\sqrt{3^{2}+4^{2}}\\,dt} \\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\int_{1}^{5}\\ 5\\,dt=5t\\Big|^5_1=20.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/li>\n<li>Using the Arc Length Formulas, [latex]{\\bf{r}}'\\,(t)=\\langle{\\cos{t}}-t\\sin{t},\\ \\sin{t}+t\\cos{t},\\ 2\\rangle[\/latex], so\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {s} &=\\hfill&{\\displaystyle\\int_{a}^{b}\\ \\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert\\ dt}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\int_{0}^{2\\pi}\\ \\sqrt{(\\cos{t}-t\\sin{t})^{2}+(\\sin{t}+t\\cos{t})^{2}+2^{2}}\\,dt} \\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\int_{0}^{2\\pi}\\ \\sqrt{(\\cos{^{2}t}-2t\\sin{t}\\cos{t}+t^{2}\\, \\sin{^{2}t})+(\\sin{^{2}t}+2t\\sin{t}\\cos{t}+t^{2}\\, \\cos{^{2}t})+4}\\,dt}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\int_{0}^{2\\pi}\\ \\sqrt{\\cos{^{2}t}+\\sin{^{2}t}+t^{2}\\,(\\cos{^{2}t}+\\sin{^{2}t})+4}\\,dt}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\int_{0}^{2\\pi}\\ \\sqrt{t^{2}+5}\\,dt}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Here we can use a table integration formula<\/p>\n<div style=\"text-align: center;\">[latex]\\displaystyle\\int_{}\\ \\sqrt{u^{2}+a^{2}}\\,du=\\frac{u}{2}\\,\\sqrt{u^{2}+a^{2}}+\\frac{a^{2}}{2}\\ln{\\Big|u+\\sqrt{u^{2}+a^{2}}\\Big|}+C[\/latex],<\/div>\n<p>so we obtain<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int_{0}^{2\\pi}\\ \\sqrt{t^{2}+5}\\,dt} &=\\hfill&{\\frac{1}{2}\\big(t\\sqrt{t^{2}+5}+5\\,ln\\ \\Big|t+\\sqrt{t^{2}+5}\\Big|\\big)^{2\\pi}_{0}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{1}{2}\\,\\big(2\\pi\\sqrt{4{\\pi}^{2}+5}+5\\ln{(2\\pi+\\sqrt{4{\\pi}^{2}+5})\\big)}-\\frac{5}{2}\\ln{\\sqrt{5}}} \\hfill \\\\ \\hfill & \\approx\\hfill & {25.343.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793958097\" class=\"textbook key-takeaways\">\n<h3>try it<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794933124\">Show Solution<\/span><\/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<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/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\"><\/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<p>We now return to the helix introduced earlier in this chapter. A vector-valued function that describes a helix can be written in the form<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{r}}\\,(t)=R\\cos{\\Big(\\frac{2\\pi\\,N\\,t}{h}\\Big)}\\,{\\bf{i}}+R\\sin{\\Big(\\frac{2\\pi\\,N\\,t}{h}\\Big)}\\,{\\bf{j}}+t\\,{\\bf{k}},\\ 0\\,\\leq\\,t\\,\\leq\\,h,[\/latex]<\/p>\n<p>where [latex]R[\/latex]\u00a0represents the radius of the helix, [latex]h[\/latex]\u00a0represents the height (distance between two consecutive turns), and the helix completes [latex]N[\/latex]\u00a0turns. Let\u2019s derive a formula for the arc length of this helix using Arc-Length Formulas. First of all,<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{r}}'\\,(t)=-\\frac{2\\pi\\,N\\,R}{h}\\sin{\\Big(\\frac{2\\pi\\,N\\,t}{h}\\Big)}\\,{\\bf{i}}+\\frac{2\\pi\\,N\\,R}{h}\\cos{\\Big(\\frac{2\\pi\\,N\\,t}{h}\\Big)}\\,{\\bf{j}}+{\\bf{k}}.[\/latex]<\/p>\n<p>Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {s} &=\\hfill&{\\displaystyle\\int_{a}^{b}\\ \\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert\\ dt}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\int_{0}^{h}\\ \\sqrt{\\Bigg(-\\frac{2\\pi\\,N\\,R}{h}\\sin{\\Big(\\frac{2\\pi\\,N\\,t}{h}\\Big)}\\Bigg)^{2}+\\Bigg(\\frac{2\\pi\\,N\\,R}{h}\\cos{\\Big(\\frac{2\\pi\\,N\\,t}{h}\\Big)}\\Bigg)^{2}+1^{2}}\\,dt} \\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\int_{0}^{h}\\ \\sqrt{\\frac{4{\\pi}^{2}N^{2}R^{2}}{h^{2}}\\,\\Bigg(\\sin{^{2}\\Big(\\frac{2\\pi{N}t}{h}\\Big)}+\\cos{^{2}\\Big(\\frac{2\\pi{N}t}{h}\\Big)\\Bigg)}+1}\\,dt}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\int_{0}^{h}\\ \\sqrt{\\frac{4{\\pi}^{2}N^{2}R^{2}}{h^{2}}+1}\\,dt}\\hfill \\\\ \\hfill & =\\hfill & {\\Big[t\\,\\sqrt{\\frac{4{\\pi}^{2}N^{2}R^{2}}{h^{2}}+1}\\Big]^h_0}\\hfill \\\\ \\hfill & =\\hfill & {h\\,\\sqrt{\\frac{4{\\pi}^{2}N^{2}R^{2}+h^{2}}{h^{2}}}}\\hfill \\\\ \\hfill & =\\hfill & {\\sqrt{4{\\pi}^{2}N^{2}R^{2}+h^{2}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\n<p>This gives a formula for the length of a wire needed to form a helix with [latex]N[\/latex]\u00a0turns that has radius [latex]R[\/latex]\u00a0and height [latex]h[\/latex]<em data-effect=\"italics\">.<\/em><\/p>\n<h2>Arc-Length Parameterization<\/h2>\n<p id=\"fs-id1169738022483\" class=\"\">We now have a formula for the arc length of a curve defined by a vector-valued function. Let\u2019s take this one step further and examine what an <strong><span id=\"term130\" data-type=\"term\">arc-length function<\/span><\/strong> is.<\/p>\n<p id=\"fs-id1169738186222\" class=\"\">If a vector-valued function represents the position of a particle in space as a function of time, then the arc-length function measures how far that particle travels as a function of time. The formula for the arc-length function follows directly from the formula for arc length:<\/p>\n<p style=\"text-align: center;\">[latex]s\\,(t)=\\displaystyle\\int_{a}^{t}\\ \\sqrt{\\big(f'\\,(u)\\big)^{2}+\\big(g'\\,(u)\\big)^{2}+\\big(h'\\,(u)\\big)^{2}}\\,du[\/latex]<\/p>\n<p>If the curve is in two dimensions, then only two terms appear under the square root inside the integral. The reason for using the independent variable [latex]u[\/latex]\u00a0is to distinguish between time and the variable of integration. Since [latex]s(t)[\/latex]\u00a0measures distance traveled as a function of time, [latex]s'(t)[\/latex]\u00a0measures the speed of the particle at any given time. Since we have a formula for [latex]s(t)[\/latex], we can differentiate both sides of the equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {s'\\,(t)} &=\\hfill&{\\frac{d}{dt}\\bigg[\\displaystyle\\int_{a}^{t}\\ \\sqrt{\\big(f'\\,(u)\\big)^{2}+\\big(g'\\,(u)\\big)^{2}+\\big(h'\\,(u)\\big)^{2}}\\,du\\bigg]}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{d}{dt}\\bigg[\\displaystyle\\int_{a}^{t}\\ \\left\\Vert{\\bf{r}}'\\,(u)\\right\\Vert\\ du\\bigg]} \\hfill \\\\ \\hfill & =\\hfill & {\\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert\\,.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\n<p>If we assume that [latex]{\\bf{r}}\\,(t)[\/latex] defines a smooth curve, then the arc length is always increasing, so [latex]s'\\,(t)\\,>\\,0[\/latex] for [latex]t\\,>\\,a[\/latex]. Last, if [latex]{\\bf{r}}\\,(t)[\/latex] is a curve on which [latex]\\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert=1[\/latex] for all [latex]t[\/latex], then<\/p>\n<p style=\"text-align: center;\">[latex]s\\,(t)=\\displaystyle\\int_{a}^{t}\\ \\left\\Vert{\\bf{r}}'\\,(u)\\right\\Vert\\ du=\\displaystyle\\int_{a}^{t}\\ 1\\,du=t-a,[\/latex]<\/p>\n<p>which means that [latex]t[\/latex] represents the arc length as long as [latex]a=0[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Arc-Length function Theorem<\/h3>\n<p>Let [latex]{\\bf{r}}\\,(t)[\/latex] describe a smooth curve for [latex]t\\,\\geq\\,a[\/latex]. Then the arc-length function is given by<\/p>\n<div style=\"text-align: center;\">[latex]s\\,(t)=\\displaystyle\\int_{a}^{t}\\ \\left\\Vert{\\bf{r}}'\\,(u)\\right\\Vert\\ du[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>Furthermore, [latex]\\frac{ds}{dt}=\\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert\\,>\\,0[\/latex]. If [latex]\\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert=1[\/latex] for all [latex]t\\,\\geq\\,a[\/latex], then the parameter [latex]t[\/latex] represents the arc length from the starting point at [latex]t=a[\/latex].<\/p>\n<\/div>\n<p>A useful application of this theorem is to find an alternative parameterization of a given curve, called an <strong><span id=\"term131\" data-type=\"term\">arc-length parameterization<\/span><\/strong>. Recall that any vector-valued function can be reparameterized via a change of variables. For example, if we have a function [latex]{\\bf{r}}\\,(t)=\\langle{3}\\cos{t},\\ 3\\sin{t}\\rangle,\\ 0\\,\\leq\\,t\\,\\leq\\,2\\pi[\/latex]\u00a0that parameterizes a circle of radius 3, we can change the parameter from [latex]t[\/latex] to [latex]4t[\/latex], obtaining a new parameterization [latex]{\\bf{r}}\\,(t)=\\langle{3}\\cos{4t},\\ 3\\sin{4t}\\rangle[\/latex]. The new parameterization still defines a circle of radius [latex]3[\/latex], but now we need only use the values [latex]0\\,\\leq\\,t\\,\\leq\\,\\frac{\\pi}{2}[\/latex] to traverse the circle once.<\/p>\n<p>Suppose that we find the arc-length function [latex]s(t)[\/latex] and are able to solve this function for [latex]t[\/latex] as a function of [latex]s[\/latex]. We can then reparameterize the original function [latex]{\\bf{r}}(t)[\/latex] by substituting the expression for [latex]t[\/latex] back into [latex]{\\bf{r}}(t)[\/latex]. The vector-valued function is now written in terms of the parameter [latex]s[\/latex]. Since the variable [latex]s[\/latex] represents the arc length, we call this an <em>arc-length parameterization<\/em> of the original function [latex]{\\bf{r}}(t)[\/latex]. One advantage of finding the arc-length parameterization is that the distance traveled along the curve starting from [latex]s=0[\/latex] is now equal to the parameter [latex]s[\/latex]. The arc-length parameterization also appears in the context of curvature (which we examine later in this section) and line integrals, which we study in the Introduction to Vector Calculus.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: finding an arc-length parameterization<\/h3>\n<p>Find the arc-length parameterization for each of the following curves:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">  <\/ol>\n<ol style=\"list-style-type: lower-alpha;\">    <\/ol>\n<\/li>\n<\/ol>\n<div id=\"fs-id1167793461764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167795055166\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167795055166\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>First we find the arc-length function using the Arc Length Function equations:\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {s\\,(t)} &=\\hfill&{\\displaystyle\\int_{a}^{t}\\ \\left\\Vert{\\bf{r}}'\\,(u)\\right\\Vert\\ du}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\int_{0}^{t}\\ \\left\\Vert\\langle{-4}\\sin{u},\\ 4\\cos{u}\\rangle\\right\\Vert\\,du} \\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\int_{0}^{t}\\ \\sqrt{(-4\\sin{u})^{2}+(4\\cos{u})^{2}}\\,du}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\int_{0}^{t}\\ \\sqrt{16\\sin{^{2}u}+16\\cos{^{2}u}}\\,du}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\int_{0}^{t}\\ 4\\,du=4t,}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>which gives the relationship between the arc length [latex]s[\/latex] and the parameter [latex]t[\/latex] as [latex]s=4t[\/latex]; so, [latex]t=s\/4[\/latex]. Next we replace the variable [latex]t[\/latex] in the original function [latex]{\\bf{r}}\\,(t)=4\\cos{t{\\bf{i}}}+4\\sin{t{\\bf{j}}}[\/latex] with the expression [latex]s\/4[\/latex] to obtain<\/p>\n<div style=\"text-align: center;\">[latex]{\\bf{r}}\\,(s)=4\\cos{\\big(\\frac{s}{4}\\big)}{\\bf{i}}+4\\sin{\\big(\\frac{s}{4}\\big)}{\\bf{j}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>This is the arc-length parameterization of [latex]{\\bf{r}}(t)[\/latex]. Since the original restriction on [latex]t[\/latex] was given by [latex]t\\geq{0}[\/latex], the restriction on [latex]s[\/latex] becomes [latex]s\/4\\geq{0}[\/latex], or [latex]s\\geq{0}[\/latex].<\/li>\n<li>The arc-length function is given by the Arc Length Function equations:\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {s\\,(t)} &=\\hfill&{\\displaystyle\\int_{a}^{t}\\ \\left\\Vert{\\bf{r}}'\\,(u)\\right\\Vert\\ du}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\int_{3}^{t}\\ \\left\\Vert\\langle{1},\\ 2,\\ 2\\rangle\\right\\Vert\\ du} \\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\int_{3}^{t}\\ \\sqrt{1^{2}+2^{2}+2^{2}}\\,du}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\int_{3}^{t}\\ 3\\,du}\\hfill \\\\ \\hfill & =\\hfill & {3t-9.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Therefore, the relationship between the arc length [latex]s[\/latex] and the parameter [latex]t[\/latex] is [latex]s=3t-9[\/latex], so [latex]t=\\frac{s}{3}+3[\/latex]. Substituting this into the original function [latex]{\\bf{r}}\\,(t)=\\langle{t}+3,\\ 2t-4,\\ 2t\\rangle[\/latex] yields<\/p>\n<div style=\"text-align: center;\">[latex]{\\bf{r}}\\,(s)=\\Big\\langle\\big(\\frac{s}{3}+3\\big)+3,\\ 2\\,\\big(\\frac{s}{3}+3\\big)-4,\\ 2\\,\\big(\\frac{s}{3}+3\\big)\\Big\\rangle=\\Big\\langle\\frac{s}{3}+6,\\ \\frac{2s}{3}+2,\\ \\frac{2s}{3}+6\\Big\\rangle[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>This is an arc-length parameterization of [latex]{\\bf{r}}\\,(t)[\/latex]. The original restriction on the parameter [latex]t[\/latex] was [latex]t\\,\\geq\\,3[\/latex], so the restriction on [latex]s[\/latex] is [latex](s\/3)+3\\,\\geq\\,3[\/latex], or [latex]s\\,\\geq\\,0[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794933134\">Show Solution<\/span><\/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<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/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\"><\/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\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-3862\">\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>CP 3.10. <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":12,"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\":\"CP 3.10\",\"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-3862","chapter","type-chapter","status-publish","hentry"],"part":21,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3862","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":6,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3862\/revisions"}],"predecessor-version":[{"id":4256,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3862\/revisions\/4256"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/21"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3862\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=3862"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=3862"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=3862"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=3862"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}