{"id":588,"date":"2021-02-04T16:26:31","date_gmt":"2021-02-04T16:26:31","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=588"},"modified":"2022-03-16T22:45:47","modified_gmt":"2022-03-16T22:45:47","slug":"arc-lengths-of-curves","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/arc-lengths-of-curves\/","title":{"raw":"Arc Lengths of Curves","rendered":"Arc Lengths of Curves"},"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 curve, [latex]y=f(x),[\/latex] between two points<\/li>\r\n \t<li>Determine the length of a curve, [latex]x=g(y),[\/latex] between two points<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1167793393429\" class=\"bc-section section\">\r\n<h2>Arc Length of the Curve [latex]y[\/latex] = [latex]f[\/latex]([latex]x[\/latex])<\/h2>\r\n<p id=\"fs-id1167794212707\">In previous applications of integration, we required the function [latex]f(x)[\/latex] to be integrable, or at most continuous. However, for calculating arc length we have a more stringent requirement for [latex]f(x).[\/latex] Here, we require [latex]f(x)[\/latex] to be differentiable, and furthermore we require its derivative, [latex]{f}^{\\prime }(x),[\/latex] to be continuous. Functions like this, which have continuous derivatives, are called <span class=\"no-emphasis\"><em>smooth<\/em><\/span>. (This property comes up again in later chapters.)<\/p>\r\n<p id=\"fs-id1167793495508\">Let [latex]f(x)[\/latex] be a smooth function defined over [latex]\\left[a,b\\right].[\/latex] We want to calculate the length of the curve from the point [latex](a,f(a))[\/latex] to the point [latex](b,f(b)).[\/latex] We start by using line segments to approximate the length of the curve. For [latex]i=0,1,2\\text{,\u2026},n,[\/latex] let [latex]P=\\left\\{{x}_{i}\\right\\}[\/latex] be a regular partition of [latex]\\left[a,b\\right].[\/latex] Then, for [latex]i=1,2\\text{,\u2026},n,[\/latex] construct a line segment from the point [latex]({x}_{i-1},f({x}_{i-1}))[\/latex] to the point [latex]({x}_{i},f({x}_{i})).[\/latex] Although it might seem logical to use either horizontal or vertical line segments, we want our line segments to approximate the curve as closely as possible. Figure 1 depicts this construct for [latex]n=5.[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"408\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213108\/CNX_Calc_Figure_06_04_001.jpg\" alt=\"This figure is a graph in the first quadrant. The curve increases and decreases. It is divided into parts at the points a=xsub0, xsub1, xsub2, xsub3, xsub4, and xsub5=b. Also, there are line segments between the points on the curve.\" width=\"408\" height=\"276\" \/> Figure 1. We can approximate the length of a curve by adding line segments.[\/caption]\r\n<p id=\"fs-id1167794196520\">To help us find the length of each line segment, we look at the change in vertical distance as well as the change in horizontal distance over each interval. Because we have used a regular partition, the change in horizontal distance over each interval is given by [latex]\\text{\u0394}x.[\/latex] The change in vertical distance varies from interval to interval, though, so we use [latex]\\text{\u0394}{y}_{i}=f({x}_{i})-f({x}_{i-1})[\/latex] to represent the change in vertical distance over the interval [latex]\\left[{x}_{i-1},{x}_{i}\\right],[\/latex] as shown in Figure 2. Note that some (or all) [latex]\\text{\u0394}{y}_{i}[\/latex] may be negative.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"317\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213111\/CNX_Calc_Figure_06_04_002.jpg\" alt=\"This figure is a graph. It is a curve above the x-axis beginning at the point f(xsubi-1). The curve ends in the first quadrant at the point f(xsubi). Between the two points on the curve is a line segment. A right triangle is formed with this line segment as the hypotenuse, a horizontal segment with length delta x, and a vertical line segment with length delta y.\" width=\"317\" height=\"322\" \/> Figure 2. A representative line segment approximates the curve over the interval [latex]\\left[{x}_{i-1},{x}_{i}\\right].[\/latex][\/caption]\r\n<p id=\"fs-id1167794212591\">By the Pythagorean theorem, the length of the line segment is [latex]\\sqrt{{(\\text{\u0394}x)}^{2}+{(\\text{\u0394}{y}_{i})}^{2}}.[\/latex] We can also write this as [latex]\\text{\u0394}x\\sqrt{1+{((\\text{\u0394}{y}_{i})\\text{\/}(\\text{\u0394}x))}^{2}}.[\/latex] Now, by the Mean Value Theorem, there is a point [latex]{x}_{i}^{*}\\in \\left[{x}_{i-1},{x}_{i}\\right][\/latex] such that [latex]{f}^{\\prime }({x}_{i}^{*})=(\\text{\u0394}{y}_{i})\\text{\/}(\\text{\u0394}x).[\/latex] Then the length of the line segment is given by [latex]\\text{\u0394}x\\sqrt{1+{\\left[{f}^{\\prime }({x}_{i}^{*})\\right]}^{2}}.[\/latex] Adding up the lengths of all the line segments, we get<\/p>\r\n\r\n<div id=\"fs-id1167793279503\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Arc Length}\\approx \\underset{i=1}{\\overset{n}{\\text{\u2211}}}\\sqrt{1+{\\left[{f}^{\\prime }({x}_{i}^{*})\\right]}^{2}}\\text{\u0394}x.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793367181\">This is a Riemann sum. Taking the limit as [latex]n\\to \\infty ,[\/latex] we have<\/p>\r\n\r\n<div id=\"fs-id1167793829102\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Arc Length}=\\underset{n\\to \\infty }{\\text{lim}}\\underset{i=1}{\\overset{n}{\\text{\u2211}}}\\sqrt{1+{\\left[{f}^{\\prime }({x}_{i}^{*})\\right]}^{2}}\\text{\u0394}x={\\displaystyle\\int }_{a}^{b}\\sqrt{1+{\\left[{f}^{\\prime }(x)\\right]}^{2}}dx.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793470793\">We summarize these findings in the following theorem.<\/p>\r\n\r\n<div id=\"fs-id1167793385865\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Arc Length for [latex]y[\/latex] = [latex]f[\/latex]([latex]x[\/latex])<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]f(x)[\/latex] be a smooth function over the interval [latex]\\left[a,b\\right].[\/latex] Then the arc length of the portion of the graph of [latex]f(x)[\/latex] from the point [latex](a,f(a))[\/latex] to the point [latex](b,f(b))[\/latex] is given by\r\n<div id=\"fs-id1167794043266\" class=\"equation\" style=\"text-align: center;\">[latex]\\text{Arc Length}={\\displaystyle\\int }_{a}^{b}\\sqrt{1+{\\left[{f}^{\\prime }(x)\\right]}^{2}}dx.[\/latex]<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167794034181\">Note that we are integrating an expression involving [latex]{f}^{\\prime }(x),[\/latex] so we need to be sure [latex]{f}^{\\prime }(x)[\/latex] is integrable. This is why we require [latex]f(x)[\/latex] to be smooth. The following example shows how to apply the theorem.<\/p>\r\n\r\n<div id=\"fs-id1167793940076\" class=\"textbook exercises\">\r\n<h3>Example: Calculating the Arc Length of a Function of [latex]x[\/latex]<\/h3>\r\nLet [latex]f(x)=2{x}^{3\\text{\/}2}.[\/latex] Calculate the arc length of the graph of [latex]f(x)[\/latex] over the interval [latex]\\left[0,1\\right].[\/latex] Round the answer to three decimal places.\r\n<div id=\"fs-id1167794099909\" class=\"exercise\">[reveal-answer q=\"fs-id1167793949048\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793949048\"]\r\n<p id=\"fs-id1167793949048\">We have [latex]{f}^{\\prime }(x)=3{x}^{1\\text{\/}2},[\/latex] so [latex]{\\left[{f}^{\\prime }(x)\\right]}^{2}=9x.[\/latex] Then, the arc length is<\/p>\r\n\r\n<div id=\"fs-id1167794212298\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\text{Arc Length}&amp; ={\\displaystyle\\int }_{a}^{b}\\sqrt{1+{\\left[{f}^{\\prime }(x)\\right]}^{2}}dx\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{1}\\sqrt{1+9x}dx.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167793510526\">Substitute [latex]u=1+9x.[\/latex] Then, [latex]du=9dx.[\/latex] When [latex]x=0,[\/latex] then [latex]u=1,[\/latex] and when [latex]x=1,[\/latex] then [latex]u=10.[\/latex] Thus,<\/p>\r\n\r\n<div id=\"fs-id1167793870742\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\text{Arc Length}&amp; ={\\displaystyle\\int }_{0}^{1}\\sqrt{1+9x}dx\\hfill \\\\ &amp; =\\frac{1}{9}{\\displaystyle\\int }_{0}^{1}\\sqrt{1+9x}9dx=\\dfrac{1}{9}{\\displaystyle\\int }_{1}^{10}\\sqrt{u}du\\hfill \\\\ &amp; ={\\dfrac{1}{9}\u00b7\\frac{2}{3}{u}^{3\\text{\/}2}|}_{1}^{10}=\\frac{2}{27}\\left[10\\sqrt{10}-1\\right]\\approx 2.268\\text{ units}.\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793515332\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nLet [latex]f(x)=\\left(\\dfrac{4}{3}\\right){x}^{3\\text{\/}2}.[\/latex] Calculate the arc length of the graph of [latex]f(x)[\/latex] over the interval [latex]\\left[0,1\\right].[\/latex] Round the answer to three decimal places.\r\n<div id=\"fs-id1167793984361\" class=\"exercise\">[reveal-answer q=\"fs-id1167794173109\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794173109\"]\r\n<p id=\"fs-id1167794173109\">[latex]\\dfrac{1}{6}(5\\sqrt{5}-1)\\approx 1.697[\/latex]<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1167793367044\">Use the process from the previous example. Don\u2019t forget to change the limits of integration.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lm7ZbIPZZlc?controls=0&amp;start=338&amp;end=488&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.4ArcLengthOfACurveAndSurfaceArea338to488_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.4 Arc Length of a Curve and Surface Area\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<p id=\"fs-id1167793964932\">Although it is nice to have a formula for calculating arc length, this particular theorem can generate expressions that are difficult to integrate. We study some techniques for integration in <a class=\"target-chapter\" href=\"https:\/\/cnx.org\/contents\/HTmjSAcf@2.46:Z4WWhBaa@3\/Introduction\">Introduction to Techniques of Integration<\/a>\u00a0in the second volume of this text. In some cases, we may have to use a computer or calculator to approximate the value of the integral.<\/p>\r\n\r\n<div id=\"fs-id1167794071518\" class=\"textbook exercises\">\r\n<h3>Example: Using a Computer or Calculator to Determine the Arc Length of a Function of [latex]x[\/latex]<\/h3>\r\nLet [latex]f(x)={x}^{2}.[\/latex] Calculate the arc length of the graph of [latex]f(x)[\/latex] over the interval [latex]\\left[1,3\\right].[\/latex]\r\n<div id=\"fs-id1167794055619\" class=\"exercise\">[reveal-answer q=\"fs-id1167793372282\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793372282\"]\r\n<p id=\"fs-id1167793372282\">We have [latex]{f}^{\\prime }(x)=2x,[\/latex] so [latex]{\\left[{f}^{\\prime }(x)\\right]}^{2}=4{x}^{2}.[\/latex] Then the arc length is given by<\/p>\r\n\r\n<div id=\"fs-id1167794122133\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Arc Length}={\\displaystyle\\int }_{a}^{b}\\sqrt{1+{\\left[{f}^{\\prime }(x)\\right]}^{2}}dx={\\displaystyle\\int }_{1}^{3}\\sqrt{1+4{x}^{2}}dx.[\/latex]<\/div>\r\n<p id=\"fs-id1167793498289\">Using a computer to approximate the value of this integral, we get<\/p>\r\n\r\n<div id=\"fs-id1167794025333\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{3}\\sqrt{1+4{x}^{2}}dx\\approx 8.26815.[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167794290608\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nLet [latex]f(x)= \\sin x.[\/latex] Calculate the arc length of the graph of [latex]f(x)[\/latex] over the interval [latex]\\left[0,\\pi \\right].[\/latex] Use a computer or calculator to approximate the value of the integral.\r\n<div id=\"fs-id1167794038452\" class=\"exercise\">[reveal-answer q=\"fs-id1167794337534\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794337534\"]\r\n<p id=\"fs-id1167794337534\">[latex]\\text{Arc Length}\\approx 3.8202[\/latex]<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1167793977067\">Use the process from the previous example.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lm7ZbIPZZlc?controls=0&amp;start=590&amp;end=630&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.4ArcLengthOfACurveAndSurfaceArea590to630_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.4 Arc Length of a Curve and Surface Area\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1167794037114\" class=\"bc-section section\">\r\n<h2>Arc Length of the Curve [latex]x[\/latex] = [latex]g[\/latex]([latex]y[\/latex])<\/h2>\r\n<p id=\"fs-id1167794144167\">We have just seen how to approximate the length of a curve with line segments. If we want to find the arc length of the graph of a function of [latex]y,[\/latex] we can repeat the same process, except we partition the [latex]y\\text{-axis}[\/latex] instead of the [latex]x\\text{-axis}.[\/latex] Figure 3 shows a representative line segment.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"454\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213114\/CNX_Calc_Figure_06_04_003.jpg\" alt=\"This figure is a graph. It is a curve to the right of the y-axis beginning at the point g(ysubi-1). The curve ends in the first quadrant at the point g(ysubi). Between the two points on the curve is a line segment. A right triangle is formed with this line segment as the hypotenuse, a horizontal segment with length delta x, and a vertical line segment with length delta y.\" width=\"454\" height=\"322\" \/> Figure 3. A representative line segment over the interval [latex]\\left[{y}_{i-1},{y}_{i}\\right].[\/latex][\/caption]\r\n<p id=\"fs-id1167794140191\">Then the length of the line segment is [latex]\\sqrt{{(\\text{\u0394}y)}^{2}+{(\\text{\u0394}{x}_{i})}^{2}},[\/latex] which can also be written as [latex]\\text{\u0394}y\\sqrt{1+{((\\text{\u0394}{x}_{i})\\text{\/}(\\text{\u0394}y))}^{2}}.[\/latex] If we now follow the same development we did earlier, we get a formula for arc length of a function [latex]x=g(y).[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1167793363528\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Arc Length for [latex]x[\/latex] = [latex]g[\/latex]([latex]y[\/latex])<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793900934\">Let [latex]g(y)[\/latex] be a smooth function over an interval [latex]\\left[c,d\\right].[\/latex] Then, the arc length of the graph of [latex]g(y)[\/latex] from the point [latex](c,g(c))[\/latex] to the point [latex](d,g(d))[\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1167794071093\" class=\"equation\" style=\"text-align: center;\">[latex]\\text{Arc Length}={\\displaystyle\\int }_{c}^{d}\\sqrt{1+{\\left[{g}^{\\prime }(y)\\right]}^{2}}dy[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1167794043530\" class=\"textbook exercises\">\r\n<h3>Example: Calculating the Arc Length of a Function of [latex]y[\/latex]<\/h3>\r\nLet [latex]g(y)=3{y}^{3}.[\/latex] Calculate the arc length of the graph of [latex]g(y)[\/latex] over the interval [latex]\\left[1,2\\right].[\/latex]\r\n<div id=\"fs-id1167793514590\" class=\"exercise\">[reveal-answer q=\"fs-id1167793967146\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793967146\"]\r\n<p id=\"fs-id1167793967146\">We have [latex]{g}^{\\prime }(y)=9{y}^{2},[\/latex] so [latex]{\\left[{g}^{\\prime }(y)\\right]}^{2}=81{y}^{4}.[\/latex] Then the arc length is<\/p>\r\n\r\n<div id=\"fs-id1167793937228\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Arc Length}={\\displaystyle\\int }_{c}^{d}\\sqrt{1+{\\left[{g}^{\\prime }(y)\\right]}^{2}}dy={\\displaystyle\\int }_{1}^{2}\\sqrt{1+81{y}^{4}}dy.[\/latex]<\/div>\r\n<p id=\"fs-id1167794074176\">Using a computer to approximate the value of this integral, we obtain<\/p>\r\n\r\n<div id=\"fs-id1167793949762\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{2}\\sqrt{1+81{y}^{4}}dy\\approx 21.0277.[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167794062618\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nLet [latex]g(y)=\\frac{1}{y}.[\/latex] Calculate the arc length of the graph of [latex]g(y)[\/latex] over the interval [latex]\\left[1,4\\right].[\/latex] Use a computer or calculator to approximate the value of the integral.\r\n<div id=\"fs-id1167793950303\" class=\"exercise\">[reveal-answer q=\"fs-id1167793956207\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793956207\"]\r\n<p id=\"fs-id1167793956207\">[latex]\\text{Arc Length}=3.15018[\/latex]<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1167794070916\">Use the process from the previous example.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Determine the length of a curve, [latex]y=f(x),[\/latex] between two points<\/li>\n<li>Determine the length of a curve, [latex]x=g(y),[\/latex] between two points<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1167793393429\" class=\"bc-section section\">\n<h2>Arc Length of the Curve [latex]y[\/latex] = [latex]f[\/latex]([latex]x[\/latex])<\/h2>\n<p id=\"fs-id1167794212707\">In previous applications of integration, we required the function [latex]f(x)[\/latex] to be integrable, or at most continuous. However, for calculating arc length we have a more stringent requirement for [latex]f(x).[\/latex] Here, we require [latex]f(x)[\/latex] to be differentiable, and furthermore we require its derivative, [latex]{f}^{\\prime }(x),[\/latex] to be continuous. Functions like this, which have continuous derivatives, are called <span class=\"no-emphasis\"><em>smooth<\/em><\/span>. (This property comes up again in later chapters.)<\/p>\n<p id=\"fs-id1167793495508\">Let [latex]f(x)[\/latex] be a smooth function defined over [latex]\\left[a,b\\right].[\/latex] We want to calculate the length of the curve from the point [latex](a,f(a))[\/latex] to the point [latex](b,f(b)).[\/latex] We start by using line segments to approximate the length of the curve. For [latex]i=0,1,2\\text{,\u2026},n,[\/latex] let [latex]P=\\left\\{{x}_{i}\\right\\}[\/latex] be a regular partition of [latex]\\left[a,b\\right].[\/latex] Then, for [latex]i=1,2\\text{,\u2026},n,[\/latex] construct a line segment from the point [latex]({x}_{i-1},f({x}_{i-1}))[\/latex] to the point [latex]({x}_{i},f({x}_{i})).[\/latex] Although it might seem logical to use either horizontal or vertical line segments, we want our line segments to approximate the curve as closely as possible. Figure 1 depicts this construct for [latex]n=5.[\/latex]<\/p>\n<div style=\"width: 418px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213108\/CNX_Calc_Figure_06_04_001.jpg\" alt=\"This figure is a graph in the first quadrant. The curve increases and decreases. It is divided into parts at the points a=xsub0, xsub1, xsub2, xsub3, xsub4, and xsub5=b. Also, there are line segments between the points on the curve.\" width=\"408\" height=\"276\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. We can approximate the length of a curve by adding line segments.<\/p>\n<\/div>\n<p id=\"fs-id1167794196520\">To help us find the length of each line segment, we look at the change in vertical distance as well as the change in horizontal distance over each interval. Because we have used a regular partition, the change in horizontal distance over each interval is given by [latex]\\text{\u0394}x.[\/latex] The change in vertical distance varies from interval to interval, though, so we use [latex]\\text{\u0394}{y}_{i}=f({x}_{i})-f({x}_{i-1})[\/latex] to represent the change in vertical distance over the interval [latex]\\left[{x}_{i-1},{x}_{i}\\right],[\/latex] as shown in Figure 2. Note that some (or all) [latex]\\text{\u0394}{y}_{i}[\/latex] may be negative.<\/p>\n<div style=\"width: 327px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213111\/CNX_Calc_Figure_06_04_002.jpg\" alt=\"This figure is a graph. It is a curve above the x-axis beginning at the point f(xsubi-1). The curve ends in the first quadrant at the point f(xsubi). Between the two points on the curve is a line segment. A right triangle is formed with this line segment as the hypotenuse, a horizontal segment with length delta x, and a vertical line segment with length delta y.\" width=\"317\" height=\"322\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. A representative line segment approximates the curve over the interval [latex]\\left[{x}_{i-1},{x}_{i}\\right].[\/latex]<\/p>\n<\/div>\n<p id=\"fs-id1167794212591\">By the Pythagorean theorem, the length of the line segment is [latex]\\sqrt{{(\\text{\u0394}x)}^{2}+{(\\text{\u0394}{y}_{i})}^{2}}.[\/latex] We can also write this as [latex]\\text{\u0394}x\\sqrt{1+{((\\text{\u0394}{y}_{i})\\text{\/}(\\text{\u0394}x))}^{2}}.[\/latex] Now, by the Mean Value Theorem, there is a point [latex]{x}_{i}^{*}\\in \\left[{x}_{i-1},{x}_{i}\\right][\/latex] such that [latex]{f}^{\\prime }({x}_{i}^{*})=(\\text{\u0394}{y}_{i})\\text{\/}(\\text{\u0394}x).[\/latex] Then the length of the line segment is given by [latex]\\text{\u0394}x\\sqrt{1+{\\left[{f}^{\\prime }({x}_{i}^{*})\\right]}^{2}}.[\/latex] Adding up the lengths of all the line segments, we get<\/p>\n<div id=\"fs-id1167793279503\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Arc Length}\\approx \\underset{i=1}{\\overset{n}{\\text{\u2211}}}\\sqrt{1+{\\left[{f}^{\\prime }({x}_{i}^{*})\\right]}^{2}}\\text{\u0394}x.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793367181\">This is a Riemann sum. Taking the limit as [latex]n\\to \\infty ,[\/latex] we have<\/p>\n<div id=\"fs-id1167793829102\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Arc Length}=\\underset{n\\to \\infty }{\\text{lim}}\\underset{i=1}{\\overset{n}{\\text{\u2211}}}\\sqrt{1+{\\left[{f}^{\\prime }({x}_{i}^{*})\\right]}^{2}}\\text{\u0394}x={\\displaystyle\\int }_{a}^{b}\\sqrt{1+{\\left[{f}^{\\prime }(x)\\right]}^{2}}dx.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793470793\">We summarize these findings in the following theorem.<\/p>\n<div id=\"fs-id1167793385865\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Arc Length for [latex]y[\/latex] = [latex]f[\/latex]([latex]x[\/latex])<\/h3>\n<hr \/>\n<p>Let [latex]f(x)[\/latex] be a smooth function over the interval [latex]\\left[a,b\\right].[\/latex] Then the arc length of the portion of the graph of [latex]f(x)[\/latex] from the point [latex](a,f(a))[\/latex] to the point [latex](b,f(b))[\/latex] is given by<\/p>\n<div id=\"fs-id1167794043266\" class=\"equation\" style=\"text-align: center;\">[latex]\\text{Arc Length}={\\displaystyle\\int }_{a}^{b}\\sqrt{1+{\\left[{f}^{\\prime }(x)\\right]}^{2}}dx.[\/latex]<\/div>\n<\/div>\n<p id=\"fs-id1167794034181\">Note that we are integrating an expression involving [latex]{f}^{\\prime }(x),[\/latex] so we need to be sure [latex]{f}^{\\prime }(x)[\/latex] is integrable. This is why we require [latex]f(x)[\/latex] to be smooth. The following example shows how to apply the theorem.<\/p>\n<div id=\"fs-id1167793940076\" class=\"textbook exercises\">\n<h3>Example: Calculating the Arc Length of a Function of [latex]x[\/latex]<\/h3>\n<p>Let [latex]f(x)=2{x}^{3\\text{\/}2}.[\/latex] Calculate the arc length of the graph of [latex]f(x)[\/latex] over the interval [latex]\\left[0,1\\right].[\/latex] Round the answer to three decimal places.<\/p>\n<div id=\"fs-id1167794099909\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793949048\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793949048\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793949048\">We have [latex]{f}^{\\prime }(x)=3{x}^{1\\text{\/}2},[\/latex] so [latex]{\\left[{f}^{\\prime }(x)\\right]}^{2}=9x.[\/latex] Then, the arc length is<\/p>\n<div id=\"fs-id1167794212298\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\text{Arc Length}& ={\\displaystyle\\int }_{a}^{b}\\sqrt{1+{\\left[{f}^{\\prime }(x)\\right]}^{2}}dx\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{1}\\sqrt{1+9x}dx.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793510526\">Substitute [latex]u=1+9x.[\/latex] Then, [latex]du=9dx.[\/latex] When [latex]x=0,[\/latex] then [latex]u=1,[\/latex] and when [latex]x=1,[\/latex] then [latex]u=10.[\/latex] Thus,<\/p>\n<div id=\"fs-id1167793870742\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\text{Arc Length}& ={\\displaystyle\\int }_{0}^{1}\\sqrt{1+9x}dx\\hfill \\\\ & =\\frac{1}{9}{\\displaystyle\\int }_{0}^{1}\\sqrt{1+9x}9dx=\\dfrac{1}{9}{\\displaystyle\\int }_{1}^{10}\\sqrt{u}du\\hfill \\\\ & ={\\dfrac{1}{9}\u00b7\\frac{2}{3}{u}^{3\\text{\/}2}|}_{1}^{10}=\\frac{2}{27}\\left[10\\sqrt{10}-1\\right]\\approx 2.268\\text{ units}.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793515332\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Let [latex]f(x)=\\left(\\dfrac{4}{3}\\right){x}^{3\\text{\/}2}.[\/latex] Calculate the arc length of the graph of [latex]f(x)[\/latex] over the interval [latex]\\left[0,1\\right].[\/latex] Round the answer to three decimal places.<\/p>\n<div id=\"fs-id1167793984361\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794173109\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794173109\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794173109\">[latex]\\dfrac{1}{6}(5\\sqrt{5}-1)\\approx 1.697[\/latex]<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1167793367044\">Use the process from the previous example. Don\u2019t forget to change the limits of integration.<\/p>\n<\/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\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lm7ZbIPZZlc?controls=0&amp;start=338&amp;end=488&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.4ArcLengthOfACurveAndSurfaceArea338to488_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.4 Arc Length of a Curve and Surface Area&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<p id=\"fs-id1167793964932\">Although it is nice to have a formula for calculating arc length, this particular theorem can generate expressions that are difficult to integrate. We study some techniques for integration in <a class=\"target-chapter\" href=\"https:\/\/cnx.org\/contents\/HTmjSAcf@2.46:Z4WWhBaa@3\/Introduction\">Introduction to Techniques of Integration<\/a>\u00a0in the second volume of this text. In some cases, we may have to use a computer or calculator to approximate the value of the integral.<\/p>\n<div id=\"fs-id1167794071518\" class=\"textbook exercises\">\n<h3>Example: Using a Computer or Calculator to Determine the Arc Length of a Function of [latex]x[\/latex]<\/h3>\n<p>Let [latex]f(x)={x}^{2}.[\/latex] Calculate the arc length of the graph of [latex]f(x)[\/latex] over the interval [latex]\\left[1,3\\right].[\/latex]<\/p>\n<div id=\"fs-id1167794055619\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793372282\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793372282\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793372282\">We have [latex]{f}^{\\prime }(x)=2x,[\/latex] so [latex]{\\left[{f}^{\\prime }(x)\\right]}^{2}=4{x}^{2}.[\/latex] Then the arc length is given by<\/p>\n<div id=\"fs-id1167794122133\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Arc Length}={\\displaystyle\\int }_{a}^{b}\\sqrt{1+{\\left[{f}^{\\prime }(x)\\right]}^{2}}dx={\\displaystyle\\int }_{1}^{3}\\sqrt{1+4{x}^{2}}dx.[\/latex]<\/div>\n<p id=\"fs-id1167793498289\">Using a computer to approximate the value of this integral, we get<\/p>\n<div id=\"fs-id1167794025333\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{3}\\sqrt{1+4{x}^{2}}dx\\approx 8.26815.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167794290608\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Let [latex]f(x)= \\sin x.[\/latex] Calculate the arc length of the graph of [latex]f(x)[\/latex] over the interval [latex]\\left[0,\\pi \\right].[\/latex] Use a computer or calculator to approximate the value of the integral.<\/p>\n<div id=\"fs-id1167794038452\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794337534\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794337534\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794337534\">[latex]\\text{Arc Length}\\approx 3.8202[\/latex]<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1167793977067\">Use the process from the previous example.<\/p>\n<\/div>\n<\/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\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lm7ZbIPZZlc?controls=0&amp;start=590&amp;end=630&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.4ArcLengthOfACurveAndSurfaceArea590to630_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.4 Arc Length of a Curve and Surface Area&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1167794037114\" class=\"bc-section section\">\n<h2>Arc Length of the Curve [latex]x[\/latex] = [latex]g[\/latex]([latex]y[\/latex])<\/h2>\n<p id=\"fs-id1167794144167\">We have just seen how to approximate the length of a curve with line segments. If we want to find the arc length of the graph of a function of [latex]y,[\/latex] we can repeat the same process, except we partition the [latex]y\\text{-axis}[\/latex] instead of the [latex]x\\text{-axis}.[\/latex] Figure 3 shows a representative line segment.<\/p>\n<div style=\"width: 464px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213114\/CNX_Calc_Figure_06_04_003.jpg\" alt=\"This figure is a graph. It is a curve to the right of the y-axis beginning at the point g(ysubi-1). The curve ends in the first quadrant at the point g(ysubi). Between the two points on the curve is a line segment. A right triangle is formed with this line segment as the hypotenuse, a horizontal segment with length delta x, and a vertical line segment with length delta y.\" width=\"454\" height=\"322\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. A representative line segment over the interval [latex]\\left[{y}_{i-1},{y}_{i}\\right].[\/latex]<\/p>\n<\/div>\n<p id=\"fs-id1167794140191\">Then the length of the line segment is [latex]\\sqrt{{(\\text{\u0394}y)}^{2}+{(\\text{\u0394}{x}_{i})}^{2}},[\/latex] which can also be written as [latex]\\text{\u0394}y\\sqrt{1+{((\\text{\u0394}{x}_{i})\\text{\/}(\\text{\u0394}y))}^{2}}.[\/latex] If we now follow the same development we did earlier, we get a formula for arc length of a function [latex]x=g(y).[\/latex]<\/p>\n<div id=\"fs-id1167793363528\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Arc Length for [latex]x[\/latex] = [latex]g[\/latex]([latex]y[\/latex])<\/h3>\n<hr \/>\n<p id=\"fs-id1167793900934\">Let [latex]g(y)[\/latex] be a smooth function over an interval [latex]\\left[c,d\\right].[\/latex] Then, the arc length of the graph of [latex]g(y)[\/latex] from the point [latex](c,g(c))[\/latex] to the point [latex](d,g(d))[\/latex] is given by<\/p>\n<div id=\"fs-id1167794071093\" class=\"equation\" style=\"text-align: center;\">[latex]\\text{Arc Length}={\\displaystyle\\int }_{c}^{d}\\sqrt{1+{\\left[{g}^{\\prime }(y)\\right]}^{2}}dy[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1167794043530\" class=\"textbook exercises\">\n<h3>Example: Calculating the Arc Length of a Function of [latex]y[\/latex]<\/h3>\n<p>Let [latex]g(y)=3{y}^{3}.[\/latex] Calculate the arc length of the graph of [latex]g(y)[\/latex] over the interval [latex]\\left[1,2\\right].[\/latex]<\/p>\n<div id=\"fs-id1167793514590\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793967146\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793967146\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793967146\">We have [latex]{g}^{\\prime }(y)=9{y}^{2},[\/latex] so [latex]{\\left[{g}^{\\prime }(y)\\right]}^{2}=81{y}^{4}.[\/latex] Then the arc length is<\/p>\n<div id=\"fs-id1167793937228\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Arc Length}={\\displaystyle\\int }_{c}^{d}\\sqrt{1+{\\left[{g}^{\\prime }(y)\\right]}^{2}}dy={\\displaystyle\\int }_{1}^{2}\\sqrt{1+81{y}^{4}}dy.[\/latex]<\/div>\n<p id=\"fs-id1167794074176\">Using a computer to approximate the value of this integral, we obtain<\/p>\n<div id=\"fs-id1167793949762\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{2}\\sqrt{1+81{y}^{4}}dy\\approx 21.0277.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167794062618\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Let [latex]g(y)=\\frac{1}{y}.[\/latex] Calculate the arc length of the graph of [latex]g(y)[\/latex] over the interval [latex]\\left[1,4\\right].[\/latex] Use a computer or calculator to approximate the value of the integral.<\/p>\n<div id=\"fs-id1167793950303\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793956207\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793956207\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793956207\">[latex]\\text{Arc Length}=3.15018[\/latex]<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1167794070916\">Use the process from the previous example.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/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-588\">\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>2.4 Arc Length of a Curve and Surface Area. <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 1. <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\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/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-1\/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":17533,"menu_order":15,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"2.4 Arc Length of a Curve and Surface Area\",\"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-588","chapter","type-chapter","status-publish","hentry"],"part":65,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/588","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":22,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/588\/revisions"}],"predecessor-version":[{"id":4898,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/588\/revisions\/4898"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/65"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/588\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=588"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=588"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=588"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=588"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}