{"id":975,"date":"2025-06-20T17:26:12","date_gmt":"2025-06-20T17:26:12","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=975"},"modified":"2025-08-29T14:28:24","modified_gmt":"2025-08-29T14:28:24","slug":"parametric-curves-and-their-applications-background-youll-need-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/parametric-curves-and-their-applications-background-youll-need-2\/","title":{"raw":"Parametric Curves and Their Applications: Background You'll Need 2","rendered":"Parametric Curves and Their Applications: Background You’ll Need 2"},"content":{"raw":"
Type your Learning Goals<\/em> text here\r\n\r\n<\/section>\r\n

Arc Lengths of Curves<\/h2>\r\n

Arc Length of the Curve [latex]y[\/latex] = [latex]f[\/latex]([latex]x[\/latex])<\/h3>\r\n

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 smooth<\/em><\/span>.<\/p>\r\n\r\n

\r\n

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]<\/p>\r\nWe start by using line segments to approximate the length of the curve.\r\n\r\nFor [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]\r\n\r\nThen, 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. The figure below depicts this construct for [latex]n=5.[\/latex]\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"408\"]\"This Figure 1. We can approximate the length of a curve by adding line segments.[\/caption]\r\n

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.<\/p>\r\nBecause 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 below Note that some (or all) [latex]\\text{\u0394}{y}_{i}[\/latex] may be negative.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"317\"]\"This Figure 2. A representative line segment approximates the curve over the interval [latex]\\left[{x}_{i-1},{x}_{i}\\right].[\/latex][\/caption]\r\n

By the Pythagorean theorem, the length of the line segment is:<\/p>\r\n

[latex]\\sqrt{{(\\text{\u0394}x)}^{2}+{(\\text{\u0394}{y}_{i})}^{2}}.[\/latex]<\/p>\r\nWe can also write this as:\r\n

[latex]\\text{\u0394}x\\sqrt{1+{((\\text{\u0394}{y}_{i})\\text{\/}(\\text{\u0394}x))}^{2}}.[\/latex]<\/p>\r\nNow, 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]\r\n

Then the length of the line segment is given by:<\/p>\r\n

[latex]\\text{\u0394}x\\sqrt{1+{\\left[{f}^{\\prime }({x}_{i}^{*})\\right]}^{2}}.[\/latex]<\/p>\r\nAdding up the lengths of all the line segments, we get:\r\n

[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

This is a Riemann sum. Taking the limit as [latex]n\\to \\infty ,[\/latex] we have:<\/p>\r\n\r\n

[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<\/section>\r\n

We summarize these findings in the following theorem.<\/p>\r\n\r\n

\r\n

arc length for [latex]y[\/latex] = [latex]f[\/latex]([latex]x[\/latex])<\/h3>\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
[latex]\\text{Arc Length}={\\displaystyle\\int }_{a}^{b}\\sqrt{1+{\\left[{f}^{\\prime }(x)\\right]}^{2}}dx.[\/latex]<\/div>\r\n<\/section>\r\n

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

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.\r\n\r\n[reveal-answer q=\"fs-id1167793949048\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793949048\"]\r\n

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

[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>\r\n

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

[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][\/hidden-answer]<\/div>\r\n<\/section>\r\n

Although it is nice to have a formula for calculating arc length, this particular theorem can generate expressions that are difficult to integrate. In some cases, we may have to use a computer or calculator to approximate the value of the integral.<\/p>\r\n\r\n

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]\r\n
[reveal-answer q=\"fs-id1167793372282\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793372282\"]\r\n

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

[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

Using a computer to approximate the value of this integral, we get<\/p>\r\n\r\n

[latex]{\\displaystyle\\int }_{1}^{3}\\sqrt{1+4{x}^{2}}dx\\approx 8.26815.[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/section>\r\n
\r\n

Arc Length of the Curve [latex]x[\/latex] = [latex]g[\/latex]([latex]y[\/latex])<\/h3>\r\n

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]<\/p>\r\nThe figure below shows a representative line segment.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"454\"]\"This Figure 3. A representative line segment over the interval [latex]\\left[{y}_{i-1},{y}_{i}\\right].[\/latex][\/caption]\r\n

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

\r\n

arc length for [latex]x[\/latex] = [latex]g[\/latex]([latex]y[\/latex])<\/h3>\r\n

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

[latex]\\text{Arc Length}={\\displaystyle\\int }_{c}^{d}\\sqrt{1+{\\left[{g}^{\\prime }(y)\\right]}^{2}}dy[\/latex]<\/div>\r\n<\/section>
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]\r\n\r\n[reveal-answer q=\"fs-id1167793967146\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793967146\"]\r\n

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

[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

Using a computer to approximate the value of this integral, we obtain<\/p>\r\n\r\n

[latex]{\\displaystyle\\int }_{1}^{2}\\sqrt{1+81{y}^{4}}dy\\approx 21.0277.[\/latex][\/hidden-answer]<\/div>\r\n<\/section>
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.\r\n\r\n[reveal-answer q=\"fs-id1167793956207\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793956207\"]\r\n

[latex]\\text{Arc Length}=3.15018[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

Area of a Surface of Revolution<\/h2>\r\n

The concepts used to find the arc length of a curve can be extended to find the surface area of a surface of revolution. Surface area is the total area of the outer layer of an object. For objects such as cubes or bricks, the surface area of the object is the sum of the areas of all its faces. For curved surfaces, the situation is a little more complex.<\/p>\r\n

As with arc length, we can conduct a similar development for functions of [latex]y[\/latex] to get a formula for the surface area of surfaces of revolution about the [latex]y\\text{-axis}.[\/latex] These findings are summarized in the following theorem.<\/p>\r\n\r\n

\r\n

surface area of a surface of revolution<\/h3>\r\n

Let [latex]f(x)[\/latex] be a nonnegative smooth function over the interval [latex]\\left[a,b\\right].[\/latex] Then, the surface area of the surface of revolution formed by revolving the graph of [latex]f(x)[\/latex] around the [latex]x[\/latex]-axis is given by:<\/p>\r\n\r\n

[latex]\\text{Surface Area}={\\displaystyle\\int }_{a}^{b}(2\\pi f(x)\\sqrt{1+{({f}^{\\prime }(x))}^{2}})dx.[\/latex]<\/div>\r\n \r\n

Similarly, let [latex]g(y)[\/latex] be a nonnegative smooth function over the interval [latex]\\left[c,d\\right].[\/latex] Then, the surface area of the surface of revolution formed by revolving the graph of [latex]g(y)[\/latex] around the [latex]y\\text{-axis}[\/latex] is given by:<\/p>\r\n\r\n

[latex]\\text{Surface Area}={\\displaystyle\\int }_{c}^{d}(2\\pi g(y)\\sqrt{1+{({g}^{\\prime }(y))}^{2}})dy.[\/latex]<\/div>\r\n<\/section>
Let [latex]f(x)=\\sqrt{x}[\/latex] over the interval [latex]\\left[1,4\\right].[\/latex] Find the surface area of the surface generated by revolving the graph of [latex]f(x)[\/latex] around the [latex]x\\text{-axis}.[\/latex] Round the answer to three decimal places.\r\n\r\n[reveal-answer q=\"fs-id1167793957347\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793957347\"]\r\n

The graph of [latex]f(x)[\/latex] and the surface of rotation are shown in the following figure.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"740\"]\"This Figure 10. (a) The graph of [latex]f(x).[\/latex] (b) The surface of revolution.[\/caption]\r\n

We have [latex]f(x)=\\sqrt{x}.[\/latex] Then, [latex]{f}^{\\prime }(x)=\\frac{1}{(2\\sqrt{x})}[\/latex] and [latex]{({f}^{\\prime }(x))}^{2}=\\frac{1}{(4x)}.[\/latex] Then,<\/p>\r\n\r\n

[latex]\\begin{array}{cc}\\hfill \\text{Surface Area}& ={\\displaystyle\\int }_{a}^{b}(2\\pi f(x)\\sqrt{1+{({f}^{\\prime }(x))}^{2}})dx\\hfill \\\\ & ={\\displaystyle\\int }_{1}^{4}(2\\pi \\sqrt{x}\\sqrt{1+\\frac{1}{4x}})dx\\hfill \\\\ & ={\\displaystyle\\int }_{1}^{4}(2\\pi \\sqrt{x+\\frac{1}{4}})dx.\\hfill \\end{array}[\/latex]<\/div>\r\n

Let [latex]u=x+\\frac{1}{4}.[\/latex] Then, [latex]du=dx.[\/latex] When [latex]x=1,[\/latex] [latex]u=\\frac{5}{4},[\/latex] and when [latex]x=4,[\/latex] [latex]u=\\frac{17}{4}.[\/latex] This gives us:<\/p>\r\n\r\n

[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int }_{0}^{1}(2\\pi \\sqrt{x+\\frac{1}{4}})dx& ={\\displaystyle\\int }_{5\\text{\/}4}^{17\\text{\/}4}2\\pi \\sqrt{u}du\\hfill \\\\ & =2\\pi {\\left[\\frac{2}{3}{u}^{3\\text{\/}2}\\right]|}_{5\\text{\/}4}^{17\\text{\/}4}=\\frac{\\pi }{6}\\left[17\\sqrt{17}-5\\sqrt{5}\\right]\\approx 30.846.\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/section>
Let [latex]f(x)=y=\\sqrt[3]{3x}.[\/latex] Consider the portion of the curve where [latex]0\\le y\\le 2.[\/latex] Find the surface area of the surface generated by revolving the graph of [latex]f(x)[\/latex] around the [latex]y\\text{-axis}.[\/latex]\r\n\r\n[reveal-answer q=\"fs-id1167794163006\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794163006\"]\r\n

Notice that we are revolving the curve around the [latex]y\\text{-axis},[\/latex] and the interval is in terms of [latex]y,[\/latex] so we want to rewrite the function as a function of [latex]y[\/latex]. We get [latex]x=g(y)=\\left(\\frac{1}{3}\\right){y}^{3}.[\/latex] The graph of [latex]g(y)[\/latex] and the surface of rotation are shown in the following figure.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"740\"]\"This Figure 11. (a) The graph of [latex]g(y).[\/latex] (b) The surface of revolution.[\/caption]\r\n

We have [latex]g(y)=\\left(\\frac{1}{3}\\right){y}^{3},[\/latex] so [latex]{g}^{\\prime }(y)={y}^{2}[\/latex] and [latex]{({g}^{\\prime }(y))}^{2}={y}^{4}.[\/latex] Then:<\/p>\r\n\r\n

[latex]\\begin{array}{cc}\\hfill \\text{Surface Area}& ={\\displaystyle\\int }_{c}^{d}(2\\pi g(y)\\sqrt{1+{({g}^{\\prime }(y))}^{2}})dy\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{2}(2\\pi (\\frac{1}{3}{y}^{3})\\sqrt{1+{y}^{4}})dy\\hfill \\\\ & =\\frac{2\\pi }{3}{\\displaystyle\\int }_{0}^{2}({y}^{3}\\sqrt{1+{y}^{4}})dy.\\hfill \\end{array}[\/latex]<\/div>\r\n

Let [latex]u={y}^{4}+1.[\/latex] Then [latex]du=4{y}^{3}dy.[\/latex] When [latex]y=0,[\/latex] [latex]u=1,[\/latex] and when [latex]y=2,[\/latex] [latex]u=17.[\/latex] Then:<\/p>\r\n\r\n

[latex]\\begin{array}{cc}\\hfill \\frac{2\\pi }{3}{\\displaystyle\\int }_{0}^{2}({y}^{3}\\sqrt{1+{y}^{4}})dy& =\\frac{2\\pi }{3}{\\displaystyle\\int }_{1}^{17}\\frac{1}{4}\\sqrt{u}du\\hfill \\\\ & =\\frac{\\pi }{6}{\\left[\\frac{2}{3}{u}^{3\\text{\/}2}\\right]|}_{1}^{17}=\\frac{\\pi }{9}\\left[{(17)}^{3\\text{\/}2}-1\\right]\\approx 24.118.\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/section>
[ohm_question hide_question_numbers=1]15198[\/ohm_question]\r\n\r\n<\/section><\/div>","rendered":"
Type your Learning Goals<\/em> text here<\/p>\n<\/section>\n

Arc Lengths of Curves<\/h2>\n

Arc Length of the Curve [latex]y[\/latex] = [latex]f[\/latex]([latex]x[\/latex])<\/h3>\n

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 smooth<\/em><\/span>.<\/p>\n

\n

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]<\/p>\n

We start by using line segments to approximate the length of the curve.<\/p>\n

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]<\/p>\n

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. The figure below depicts this construct for [latex]n=5.[\/latex]<\/p>\n

\"This
Figure 1. We can approximate the length of a curve by adding line segments.<\/figcaption><\/figure>\n

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.<\/p>\n

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 below Note that some (or all) [latex]\\text{\u0394}{y}_{i}[\/latex] may be negative.<\/p>\n

\"This
Figure 2. A representative line segment approximates the curve over the interval [latex]\\left[{x}_{i-1},{x}_{i}\\right].[\/latex]<\/figcaption><\/figure>\n

By the Pythagorean theorem, the length of the line segment is:<\/p>\n

[latex]\\sqrt{{(\\text{\u0394}x)}^{2}+{(\\text{\u0394}{y}_{i})}^{2}}.[\/latex]<\/p>\n

We can also write this as:<\/p>\n

[latex]\\text{\u0394}x\\sqrt{1+{((\\text{\u0394}{y}_{i})\\text{\/}(\\text{\u0394}x))}^{2}}.[\/latex]<\/p>\n

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]<\/p>\n

Then the length of the line segment is given by:<\/p>\n

[latex]\\text{\u0394}x\\sqrt{1+{\\left[{f}^{\\prime }({x}_{i}^{*})\\right]}^{2}}.[\/latex]<\/p>\n

Adding up the lengths of all the line segments, we get:<\/p>\n

[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

This is a Riemann sum. Taking the limit as [latex]n\\to \\infty ,[\/latex] we have:<\/p>\n

[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<\/section>\n

We summarize these findings in the following theorem.<\/p>\n

\n

arc length for [latex]y[\/latex] = [latex]f[\/latex]([latex]x[\/latex])<\/h3>\n

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

[latex]\\text{Arc Length}={\\displaystyle\\int }_{a}^{b}\\sqrt{1+{\\left[{f}^{\\prime }(x)\\right]}^{2}}dx.[\/latex]<\/div>\n<\/section>\n

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

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