{"id":1023,"date":"2025-10-15T13:18:55","date_gmt":"2025-10-15T13:18:55","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1023"},"modified":"2025-10-23T12:56:30","modified_gmt":"2025-10-23T12:56:30","slug":"arc-length-and-curvature-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/arc-length-and-curvature-learn-it-2\/","title":{"raw":"Arc Length and Curvature: Learn It 2","rendered":"Arc Length and Curvature: Learn It 2"},"content":{"raw":"

Arc-Length Parameterization<\/h2>\r\n

Now that we have a formula for the arc length of a curve, let's examine what an arc-length function is and how we can use it to reparameterize curves.<\/p>\r\n\r\n

The Arc-Length Function<\/h3>\r\n

If a vector-valued function represents the position of a particle in space as a function of time, then the arc-length function<\/strong> measures how far that particle travels as a function of time. The formula follows directly from the arc length formula:<\/p>\r\n

[latex]s(t)=\\int_{a}^{t} \\sqrt{(f'(u))^{2}+(g'(u))^{2}+(h'(u))^{2}}du[\/latex]<\/p>\r\n

For curves in two dimensions, only two terms appear under the square root. We use the independent variable [latex]u[\/latex] to distinguish between time and the variable of integration.<\/p>\r\n

Since [latex]s(t)[\/latex] measures distance traveled as a function of time, [latex]s'(t)[\/latex] measures the speed of the particle at any given time. Differentiating both sides:<\/p>\r\n

[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>\r\n

If [latex]{\\bf{r}}(t)[\/latex] defines a smooth curve, then arc length is always increasing, so [latex]s'(t)>0[\/latex] for [latex]t>a[\/latex].<\/p>\r\n

There's a special case worth noting: if [latex]{\\bf{r}}(t)[\/latex] is a curve where [latex]\\left\\Vert{\\bf{r}}'(t)\\right\\Vert=1[\/latex] for all [latex]t[\/latex], then:<\/p>\r\n

[latex]s(t)=\\int_{a}^{t} \\left\\Vert{\\bf{r}}'(u)\\right\\Vert du=\\int_{a}^{t} 1 du=t-a[\/latex]<\/p>\r\n

This means [latex]t[\/latex] represents the arc length as long as [latex]a=0[\/latex].<\/p>\r\n\r\n

\r\n

arc-length function<\/h3>\r\n

Let [latex]{\\bf{r}}(t)[\/latex] describe a smooth curve for [latex]t\\geq a[\/latex]. The arc-length function is:<\/p>\r\n

[latex]s(t)=\\int_{a}^{t} \\left\\Vert{\\bf{r}}'(u)\\right\\Vert du[\/latex]<\/p>\r\n

Additionally:<\/p>\r\n

[latex]\\frac{ds}{dt}=\\left\\Vert{\\bf{r}}'(t)\\right\\Vert>0[\/latex]<\/p>\r\n

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>\r\n\r\n<\/section>\r\n

Reparameterizing by Arc Length<\/h3>\r\n

A useful application of this theorem is finding an alternative parameterization of a curve called an arc-length parameterization<\/strong>. Recall that any vector-valued function can be reparameterized via a change of variables.<\/p>\r\n

For example, the function [latex]{\\bf{r}}(t)=\\langle 3\\cos t, 3\\sin t\\rangle[\/latex] for [latex]0\\leq t\\leq 2\\pi[\/latex] parameterizes a circle of radius [latex]3[\/latex]. Changing the parameter from [latex]t[\/latex] to [latex]4t[\/latex] gives [latex]{\\bf{r}}(t)=\\langle 3\\cos 4t, 3\\sin 4t\\rangle[\/latex]. This still defines a circle of radius [latex]3[\/latex], but now we only need [latex]0\\leq t\\leq\\frac{\\pi}{2}[\/latex] to traverse the circle once.<\/p>\r\n

To create an arc-length parameterization, we find the arc-length function [latex]s(t)[\/latex] and solve for [latex]t[\/latex] as a function of [latex]s[\/latex]. Then we substitute this expression for [latex]t[\/latex] back into the original function [latex]{\\bf{r}}(t)[\/latex]. The result is a vector-valued function written in terms of the parameter [latex]s[\/latex].<\/p>\r\n\r\n

\r\n

How to: Find an Arc-Length Parameterization<\/strong><\/p>\r\n\r\n

    \r\n \t
  1. 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
  2. Solve for [latex]t[\/latex]<\/strong>: Express [latex]t[\/latex] as a function of [latex]s[\/latex]<\/li>\r\n \t
  3. Substitute<\/strong>: Replace [latex]t[\/latex] in the original function [latex]{\\bf{r}}(t)[\/latex] with your expression in terms of [latex]s[\/latex]<\/li>\r\n \t
  4. Adjust the domain<\/strong>: Convert the restriction on [latex]t[\/latex] to a restriction on [latex]s[\/latex]<\/li>\r\n<\/ol>\r\n<\/section>
    One advantage of arc-length parameterization is that the distance traveled along the curve starting from [latex]s=0[\/latex] equals the parameter [latex]s[\/latex]. This makes calculations more intuitive and simplifies formulas in contexts like curvature and line integrals.\r\n\r\n<\/section>
    Find the arc-length parameterization for each of the following curves:\r\n
      \r\n \t
    1. \r\n
        a. [latex]{\\bf{r}}(t)=4\\cos{t{\\bf{i}}}+4\\sin{t{\\bf{j}}},\\ t\\geq 0[\/latex]<\/ol>\r\n
          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
          [reveal-answer q=\"fs-id1167795055166\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167795055166\"]\r\n
            \r\n \t
          1. First we find the arc-length function using the Arc Length Function equations:\r\n
            [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}\\ 4du=4t,}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n \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
            [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 \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
          2. The arc-length function is given by the Arc Length Function equations:\r\n
            [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}\\ 3du}\\hfill \\\\ \\hfill & =\\hfill & {3t-9.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n \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
            [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 \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<\/section>","rendered":"

            Arc-Length Parameterization<\/h2>\n

            Now that we have a formula for the arc length of a curve, let’s examine what an arc-length function is and how we can use it to reparameterize curves.<\/p>\n

            The Arc-Length Function<\/h3>\n

            If a vector-valued function represents the position of a particle in space as a function of time, then the arc-length function<\/strong> measures how far that particle travels as a function of time. The formula follows directly from the arc length formula:<\/p>\n

            [latex]s(t)=\\int_{a}^{t} \\sqrt{(f'(u))^{2}+(g'(u))^{2}+(h'(u))^{2}}du[\/latex]<\/p>\n

            For curves in two dimensions, only two terms appear under the square root. We use the independent variable [latex]u[\/latex] to distinguish between time and the variable of integration.<\/p>\n

            Since [latex]s(t)[\/latex] measures distance traveled as a function of time, [latex]s'(t)[\/latex] measures the speed of the particle at any given time. Differentiating both sides:<\/p>\n

            [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

            If [latex]{\\bf{r}}(t)[\/latex] defines a smooth curve, then arc length is always increasing, so [latex]s'(t)>0[\/latex] for [latex]t>a[\/latex].<\/p>\n

            There’s a special case worth noting: if [latex]{\\bf{r}}(t)[\/latex] is a curve where [latex]\\left\\Vert{\\bf{r}}'(t)\\right\\Vert=1[\/latex] for all [latex]t[\/latex], then:<\/p>\n

            [latex]s(t)=\\int_{a}^{t} \\left\\Vert{\\bf{r}}'(u)\\right\\Vert du=\\int_{a}^{t} 1 du=t-a[\/latex]<\/p>\n

            This means [latex]t[\/latex] represents the arc length as long as [latex]a=0[\/latex].<\/p>\n

            \n

            arc-length function<\/h3>\n

            Let [latex]{\\bf{r}}(t)[\/latex] describe a smooth curve for [latex]t\\geq a[\/latex]. The arc-length function is:<\/p>\n

            [latex]s(t)=\\int_{a}^{t} \\left\\Vert{\\bf{r}}'(u)\\right\\Vert du[\/latex]<\/p>\n

            Additionally:<\/p>\n

            [latex]\\frac{ds}{dt}=\\left\\Vert{\\bf{r}}'(t)\\right\\Vert>0[\/latex]<\/p>\n

            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<\/section>\n

            Reparameterizing by Arc Length<\/h3>\n

            A useful application of this theorem is finding an alternative parameterization of a curve called an arc-length parameterization<\/strong>. Recall that any vector-valued function can be reparameterized via a change of variables.<\/p>\n

            For example, the function [latex]{\\bf{r}}(t)=\\langle 3\\cos t, 3\\sin t\\rangle[\/latex] for [latex]0\\leq t\\leq 2\\pi[\/latex] parameterizes a circle of radius [latex]3[\/latex]. Changing the parameter from [latex]t[\/latex] to [latex]4t[\/latex] gives [latex]{\\bf{r}}(t)=\\langle 3\\cos 4t, 3\\sin 4t\\rangle[\/latex]. This still defines a circle of radius [latex]3[\/latex], but now we only need [latex]0\\leq t\\leq\\frac{\\pi}{2}[\/latex] to traverse the circle once.<\/p>\n

            To create an arc-length parameterization, we find the arc-length function [latex]s(t)[\/latex] and solve for [latex]t[\/latex] as a function of [latex]s[\/latex]. Then we substitute this expression for [latex]t[\/latex] back into the original function [latex]{\\bf{r}}(t)[\/latex]. The result is a vector-valued function written in terms of the parameter [latex]s[\/latex].<\/p>\n

            \n

            How to: Find an Arc-Length Parameterization<\/strong><\/p>\n

              \n
            1. Find the arc-length function<\/strong>: Calculate [latex]s(t)=\\int_{a}^{t} \\left\\Vert{\\bf{r}}'(u)\\right\\Vert du[\/latex]<\/li>\n
            2. Solve for [latex]t[\/latex]<\/strong>: Express [latex]t[\/latex] as a function of [latex]s[\/latex]<\/li>\n
            3. Substitute<\/strong>: Replace [latex]t[\/latex] in the original function [latex]{\\bf{r}}(t)[\/latex] with your expression in terms of [latex]s[\/latex]<\/li>\n
            4. Adjust the domain<\/strong>: Convert the restriction on [latex]t[\/latex] to a restriction on [latex]s[\/latex]<\/li>\n<\/ol>\n<\/section>\n
              One advantage of arc-length parameterization is that the distance traveled along the curve starting from [latex]s=0[\/latex] equals the parameter [latex]s[\/latex]. This makes calculations more intuitive and simplifies formulas in contexts like curvature and line integrals.<\/p>\n<\/section>\n
              Find the arc-length parameterization for each of the following curves:<\/p>\n
                \n
              1. \n
                  <\/ol>\n
                    <\/ol>\n<\/li>\n<\/ol>\n
                    \n