{"id":996,"date":"2025-06-20T17:27:21","date_gmt":"2025-06-20T17:27:21","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=996"},"modified":"2025-09-10T18:05:54","modified_gmt":"2025-09-10T18:05:54","slug":"calculus-with-parametric-curves-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/calculus-with-parametric-curves-learn-it-4\/","title":{"raw":"Calculus with Parametric Curves: Learn It 4","rendered":"Calculus with Parametric Curves: Learn It 4"},"content":{"raw":"

Arc Length of a Parametric Curve<\/h2>\r\n

Besides finding areas, we often need to calculate the arc length of parametric curves. Arc length measures the actual distance along a curve\u2014if a particle travels from point A to point B along a curve, the arc length tells us exactly how far that particle traveled. To develop a formula for arc length, we start with an approximation by line segments as shown in the following graph.<\/p>\r\n\r\n

<\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"304\"]\"A Figure 9. Approximation of a curve by line segments.[\/caption]<\/figure>\r\n

For a plane curve defined by [latex]x = x(t)[\/latex], [latex]y = y(t)[\/latex] where [latex]a \\le t \\le b[\/latex], we'll approximate the curve using line segments.<\/p>\r\n

Start by partitioning [latex][a,b][\/latex] into [latex]n[\/latex] equal subintervals: [latex]t_0 = a < t_1 < t_2 < \\cdots < t_n = b[\/latex], where each subinterval has width [latex]\\Delta t = \\frac{b-a}{n}[\/latex].<\/p>\r\nUsing the distance formula, we can calculate the length of each line segment:\r\n

[latex]\\begin{array}{}\\\\ {d}_{1}=\\sqrt{{\\left(x\\left({t}_{1}\\right)-x\\left({t}_{0}\\right)\\right)}^{2}+{\\left(y\\left({t}_{1}\\right)-y\\left({t}_{0}\\right)\\right)}^{2}}\\hfill \\\\ {d}_{2}=\\sqrt{{\\left(x\\left({t}_{2}\\right)-x\\left({t}_{1}\\right)\\right)}^{2}+{\\left(y\\left({t}_{2}\\right)-y\\left({t}_{1}\\right)\\right)}^{2}}\\text{etc}.\\hfill \\end{array}[\/latex]<\/div>\r\nThe approximate arc length [latex]s_n[\/latex] is the sum of all these segments:\r\n
[latex]s\\approx \\displaystyle\\sum _{k=1}^{n}{s}_{k}=\\displaystyle\\sum _{k=1}^{n}\\sqrt{{\\left(x\\left({t}_{k}\\right)-x\\left({t}_{k - 1}\\right)\\right)}^{2}+{\\left(y\\left({t}_{k}\\right)-y\\left({t}_{k - 1}\\right)\\right)}^{2}}[\/latex].<\/div>\r\nIf [latex]x(t)[\/latex] and [latex]y(t)[\/latex] are differentiable, the Mean Value Theorem tells us that in each subinterval [latex][t_{k-1}, t_k][\/latex], there exist points [latex]\\hat{t}_k[\/latex] and [latex]\\tilde{t}_k[\/latex] where:\r\n
[latex]\\begin{array}{}\\\\ x\\left({t}_{k}\\right)-x\\left({t}_{k - 1}\\right)={x}^{\\prime }\\left(\\hat{t}_{k}\\right)\\left({t}_{k}-{t}_{k - 1}\\right)={x}^{\\prime }\\left(\\hat{t}_{k}\\right)\\Delta t\\hfill \\\\ y\\left({t}_{k}\\right)-y\\left({t}_{k - 1}\\right)={y}^{\\prime }\\left(\\tilde{t}_{k}\\right)\\left({t}_{k}-{t}_{k - 1}\\right)={y}^{\\prime }\\left(\\tilde{t}_{k}\\right)\\Delta t.\\hfill \\end{array}[\/latex]<\/div>\r\n

Substituting these expressions into our sum and factoring out [latex]\\Delta t[\/latex]:<\/p>\r\n\r\n

[latex]\\begin{array}{cc}\\hfill s& \\approx {\\displaystyle\\sum _{k=1}^{n}}{s}_{k}\\hfill \\\\ & ={\\displaystyle\\sum _{k=1}^{n}}\\sqrt{{\\left({x}^{\\prime}\\left(\\hat{t}_{k}\\right)\\Delta t\\right)}^{2}+{\\left({y}^{\\prime}\\left(\\tilde{t}_{k}\\right)\\Delta t\\right)}^{2}}\\hfill \\\\ & ={\\displaystyle\\sum _{k=1}^{n}}\\sqrt{{\\left({x}^{\\prime}\\left(\\hat{t}_{k}\\right)\\right)}^{2}{\\left(\\Delta t\\right)}^{2}+{\\left({y}^{\\prime}\\left(\\tilde{t}_{k}\\right)\\right)}^{2}{\\left(\\Delta t\\right)}^{2}}\\hfill \\\\ & =\\left({\\displaystyle\\sum _{k=1}^{n}}\\sqrt{{\\left({x}^{\\prime}\\left(\\hat{t}_{k}\\right)\\right)}^{2}+{\\left({y}^{\\prime}\\left(\\tilde{t}_{k}\\right)\\right)}^{2}}\\right)\\Delta t.\\hfill \\end{array}[\/latex]<\/div>\r\nThis sum is a Riemann sum approximating the arc length over the partition of [latex][a,b][\/latex]. As [latex]n \\to \\infty[\/latex] and assuming the derivatives are continuous, we get:\r\n
[latex]\\begin{array}{cc}\\hfill s& ={\\underset{n\\to\\infty}\\lim} {\\displaystyle\\sum _{k=1}^{n}} {s}_{k}\\hfill \\\\ & = {\\underset{n\\to\\infty}\\lim}\\left({\\displaystyle\\sum _{k=1}^{n}}\\sqrt{{\\left({x}^{\\prime}\\left(\\hat{t}_{k}\\right)\\right)}^{2}+{\\left({y}^{\\prime}\\left(\\tilde{t}_{k}\\right)\\right)}^{2}}\\right)\\Delta t\\hfill \\\\ & ={\\displaystyle\\int_{a}^{b}}\\sqrt{{\\left({x}^{\\prime }\\left(t\\right)\\right)}^{2}+{\\left({y}^{\\prime }\\left(t\\right)\\right)}^{2}}dt.\\hfill \\end{array}[\/latex]<\/div>\r\nAs the partition gets finer, [latex]\\hat{t}_k[\/latex] and [latex]\\tilde{t}_k[\/latex] both lie in the same shrinking interval of width [latex]\\Delta t[\/latex], so they converge to the same value.\r\n

We can summarize this method in the following theorem.<\/p>\r\n\r\n

\r\n

theorem: arc length of a parametric curve<\/h3>\r\n

For a plane curve defined by [latex]x = x(t)[\/latex], [latex]y = y(t)[\/latex] where [latex]t_1 \\le t \\le t_2[\/latex], with [latex]x(t)[\/latex] and [latex]y(t)[\/latex] differentiable then:<\/p>\r\n\r\n

[latex]s={\\displaystyle\\int }_{{t}_{1}}^{{t}_{2}}\\sqrt{{\\left(\\frac{dx}{dt}\\right)}^{2}+{\\left(\\frac{dy}{dt}\\right)}^{2}}dt[\/latex].<\/div>\r\n
This formula gives the exact arc length of the parametric curve<\/strong>.<\/div>\r\n<\/section>
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\r\n

Find the arc length of the semicircle defined by the equations<\/p>\r\n\r\n

[latex]x\\left(t\\right)=3\\cos{t},y\\left(t\\right)=3\\sin{t},0\\le t\\le \\pi [\/latex].<\/div>\r\n \r\n\r\n<\/div>\r\n[reveal-answer q=\"44558859\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558859\"]\r\n
\r\n

The values [latex]t=0[\/latex] to [latex]t=\\pi [\/latex] trace out the red curve in Figure 9. To determine its length, use the theorem:<\/p>\r\n\r\n

[latex]\\begin{array}{cc}\\hfill s& ={\\displaystyle\\int }_{{t}_{1}}^{{t}_{2}}\\sqrt{{\\left(\\frac{dx}{dt}\\right)}^{2}+{\\left(\\frac{dy}{dt}\\right)}^{2}}dt\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{\\pi }\\sqrt{{\\left(-3\\sin{t}\\right)}^{2}+{\\left(3\\cos{t}\\right)}^{2}}dt\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{\\pi }\\sqrt{9{\\sin}^{2}t+9{\\cos}^{2}t}dt\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{\\pi }\\sqrt{9\\left({\\sin}^{2}t+{\\cos}^{2}t\\right)}dt\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{\\pi }3dt={3t|}_{0}^{\\pi }=3\\pi .\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

Note that the formula for the arc length of a semicircle is [latex]\\pi r[\/latex] and the radius of this circle is 3. This is a great example of using calculus to derive a known formula of a geometric quantity.<\/p>\r\n\r\n

<\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"339\"]\"A Figure 10. The arc length of the semicircle is equal to its radius times [latex]\\pi[\/latex].[\/caption]<\/figure>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
Watch the following video to see the worked solution to the example above.