{"id":566,"date":"2025-10-09T15:22:13","date_gmt":"2025-10-09T15:22:13","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=566"},"modified":"2025-10-23T12:50:25","modified_gmt":"2025-10-23T12:50:25","slug":"arc-length-and-curvature-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/arc-length-and-curvature-learn-it-1\/","title":{"raw":"Arc Length and Curvature: Learn It 1","rendered":"Arc Length and Curvature: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\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 \t<li>Calculate and understand what curvature means for curves in space<\/li>\r\n \t<li>Understand what normal and binormal vectors tell us about a curve's behavior<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Arc Length for Vector Functions<\/h2>\r\n<p class=\"whitespace-normal break-words\">You've already seen that a vector-valued function describes a curve in either two or three dimensions. Recall from earlier courses that the arc length of a curve defined by parametric functions [latex]x=x(t)[\/latex], [latex]y=y(t)[\/latex], [latex]t_{1}\\leq t\\leq t_{2}[\/latex] is:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s=\\int_{t_{1}}^{t_{2}} \\sqrt{(x'(t))^{2}+(y'(t))^{2}}dt[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Similarly, 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:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s=\\int_{a}^{b} \\sqrt{(f'(t))^{2}+(g'(t))^{2}}dt[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">In three dimensions, if [latex]{\\bf{r}}(t)=f(t){\\bf{i}}+g(t){\\bf{j}}+h(t){\\bf{k}}[\/latex] over the interval [latex]a\\leq t\\leq b[\/latex], the arc length is:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s=\\int_{a}^{b} \\sqrt{(f'(t))^{2}+(g'(t))^{2}+(h'(t))^{2}}dt[\/latex]<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>arc-length formulas theorem<\/h3>\r\n<p class=\"whitespace-normal break-words\">Let [latex]C[\/latex] be a smooth curve defined by [latex]{\\bf{r}}(t)[\/latex], where [latex]t[\/latex] lies within the interval [latex][a, b][\/latex].<\/p>\r\n&nbsp;\r\n<p class=\"whitespace-normal break-words\"><strong>For plane curves<\/strong> [latex]{\\bf{r}}(t)=f(t){\\bf{i}}+g(t){\\bf{j}}[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s=\\int_{a}^{b} \\sqrt{[f'(t)]^{2}+[g'(t)]^{2}}dt=\\int_{a}^{b} \\left\\Vert{\\bf{r}}'(t)\\right\\Vert dt[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>For space curves<\/strong> [latex]{\\bf{r}}(t)=f(t){\\bf{i}}+g(t){\\bf{j}}+h(t){\\bf{k}}[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s=\\int_{a}^{b} \\sqrt{[f'(t)]^{2}+[g'(t)]^{2}+[h'(t)]^{2}}dt=\\int_{a}^{b} \\left\\Vert{\\bf{r}}'(t)\\right\\Vert dt[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">\r\n<p class=\"whitespace-normal break-words\">The two formulas are nearly identical\u2014they differ only in that a space curve has three component functions instead of two. Notice that arc length is simply the integral of the magnitude of the derivative vector [latex]\\left\\Vert{\\bf{r}}'(t)\\right\\Vert[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">These formulas are defined for <strong>smooth curves<\/strong>: curves where [latex]{\\bf{r}}(t)[\/latex] is differentiable with a non-zero derivative. This smoothness condition guarantees that the curve has no cusps or corners that could make the formula problematic.<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Calculate the arc length for each of the following vector-valued functions:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]{\\bf{r}}\\,(t)=(3t-2)\\,{\\bf{i}}+(4t+5)\\,{\\bf{j}},\\ 1\\,\\leq\\,t\\,\\leq\\,5[\/latex]<\/li>\r\n \t<li>[latex]{\\bf{r}}\\,(t)=\\langle{t}\\cos{t}, \\ t\\sin{t},\\ 2t\\rangle,\\ 0\\,\\leq\\,t\\,\\leq\\,2\\pi[\/latex]<\/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<\/section>\r\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Arc Length of a Helix<\/h3>\r\n<p class=\"whitespace-normal break-words\">Let's return to the helix introduced earlier in this chapter. A vector-valued function describing a helix can be written as:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{\\bf{r}}(t)=R\\cos\\Big(\\frac{2\\pi Nt}{h}\\Big){\\bf{i}}+R\\sin\\Big(\\frac{2\\pi Nt}{h}\\Big){\\bf{j}}+t{\\bf{k}}[\/latex], [latex]0\\leq t\\leq h[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">where [latex]R[\/latex] is the radius of the helix, [latex]h[\/latex] is the height (distance between two consecutive turns), and the helix completes [latex]N[\/latex] turns. Let's derive a formula for the arc length using our formula. First:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{\\bf{r}}'(t)=-\\frac{2\\pi NR}{h}\\sin\\Big(\\frac{2\\pi Nt}{h}\\Big){\\bf{i}}+\\frac{2\\pi NR}{h}\\cos\\Big(\\frac{2\\pi Nt}{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\n<p class=\"whitespace-normal break-words\">This formula gives the length of wire needed to form a helix with [latex]N[\/latex] turns, radius [latex]R[\/latex], and height [latex]h[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s=\\sqrt{4{\\pi}^{2}N^{2}R^{2}+h^{2}}[\/latex]<\/p>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Determine the length of a particle&#8217;s path in space by using the arc-length function<\/li>\n<li>Calculate and understand what curvature means for curves in space<\/li>\n<li>Understand what normal and binormal vectors tell us about a curve&#8217;s behavior<\/li>\n<\/ul>\n<\/section>\n<h2>Arc Length for Vector Functions<\/h2>\n<p class=\"whitespace-normal break-words\">You&#8217;ve already seen that a vector-valued function describes a curve in either two or three dimensions. Recall from earlier courses that the arc length of a curve defined by parametric functions [latex]x=x(t)[\/latex], [latex]y=y(t)[\/latex], [latex]t_{1}\\leq t\\leq t_{2}[\/latex] is:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s=\\int_{t_{1}}^{t_{2}} \\sqrt{(x'(t))^{2}+(y'(t))^{2}}dt[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Similarly, 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:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s=\\int_{a}^{b} \\sqrt{(f'(t))^{2}+(g'(t))^{2}}dt[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">In three dimensions, if [latex]{\\bf{r}}(t)=f(t){\\bf{i}}+g(t){\\bf{j}}+h(t){\\bf{k}}[\/latex] over the interval [latex]a\\leq t\\leq b[\/latex], the arc length is:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s=\\int_{a}^{b} \\sqrt{(f'(t))^{2}+(g'(t))^{2}+(h'(t))^{2}}dt[\/latex]<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>arc-length formulas theorem<\/h3>\n<p class=\"whitespace-normal break-words\">Let [latex]C[\/latex] be a smooth curve defined by [latex]{\\bf{r}}(t)[\/latex], where [latex]t[\/latex] lies within the interval [latex][a, b][\/latex].<\/p>\n<p>&nbsp;<\/p>\n<p class=\"whitespace-normal break-words\"><strong>For plane curves<\/strong> [latex]{\\bf{r}}(t)=f(t){\\bf{i}}+g(t){\\bf{j}}[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s=\\int_{a}^{b} \\sqrt{[f'(t)]^{2}+[g'(t)]^{2}}dt=\\int_{a}^{b} \\left\\Vert{\\bf{r}}'(t)\\right\\Vert dt[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>For space curves<\/strong> [latex]{\\bf{r}}(t)=f(t){\\bf{i}}+g(t){\\bf{j}}+h(t){\\bf{k}}[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s=\\int_{a}^{b} \\sqrt{[f'(t)]^{2}+[g'(t)]^{2}+[h'(t)]^{2}}dt=\\int_{a}^{b} \\left\\Vert{\\bf{r}}'(t)\\right\\Vert dt[\/latex]<\/p>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">\n<p class=\"whitespace-normal break-words\">The two formulas are nearly identical\u2014they differ only in that a space curve has three component functions instead of two. Notice that arc length is simply the integral of the magnitude of the derivative vector [latex]\\left\\Vert{\\bf{r}}'(t)\\right\\Vert[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">These formulas are defined for <strong>smooth curves<\/strong>: curves where [latex]{\\bf{r}}(t)[\/latex] is differentiable with a non-zero derivative. This smoothness condition guarantees that the curve has no cusps or corners that could make the formula problematic.<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Calculate the arc length for each of the following vector-valued functions:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]{\\bf{r}}\\,(t)=(3t-2)\\,{\\bf{i}}+(4t+5)\\,{\\bf{j}},\\ 1\\,\\leq\\,t\\,\\leq\\,5[\/latex]<\/li>\n<li>[latex]{\\bf{r}}\\,(t)=\\langle{t}\\cos{t}, \\ t\\sin{t},\\ 2t\\rangle,\\ 0\\,\\leq\\,t\\,\\leq\\,2\\pi[\/latex]<\/li>\n<\/ol>\n<div id=\"fs-id1167793461764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167795055165\">Show Solution<\/button><\/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<\/section>\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Arc Length of a Helix<\/h3>\n<p class=\"whitespace-normal break-words\">Let&#8217;s return to the helix introduced earlier in this chapter. A vector-valued function describing a helix can be written as:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{\\bf{r}}(t)=R\\cos\\Big(\\frac{2\\pi Nt}{h}\\Big){\\bf{i}}+R\\sin\\Big(\\frac{2\\pi Nt}{h}\\Big){\\bf{j}}+t{\\bf{k}}[\/latex], [latex]0\\leq t\\leq h[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">where [latex]R[\/latex] is the radius of the helix, [latex]h[\/latex] is the height (distance between two consecutive turns), and the helix completes [latex]N[\/latex] turns. Let&#8217;s derive a formula for the arc length using our formula. First:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{\\bf{r}}'(t)=-\\frac{2\\pi NR}{h}\\sin\\Big(\\frac{2\\pi Nt}{h}\\Big){\\bf{i}}+\\frac{2\\pi NR}{h}\\cos\\Big(\\frac{2\\pi Nt}{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 class=\"whitespace-normal break-words\">This formula gives the length of wire needed to form a helix with [latex]N[\/latex] turns, radius [latex]R[\/latex], and height [latex]h[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s=\\sqrt{4{\\pi}^{2}N^{2}R^{2}+h^{2}}[\/latex]<\/p>\n","protected":false},"author":15,"menu_order":22,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-566","chapter","type-chapter","status-publish","hentry"],"part":424,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/566","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/566\/revisions"}],"predecessor-version":[{"id":1722,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/566\/revisions\/1722"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/424"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/566\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=566"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=566"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=566"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=566"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}