{"id":3859,"date":"2022-04-04T16:09:39","date_gmt":"2022-04-04T16:09:39","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=3859"},"modified":"2022-10-29T00:36:10","modified_gmt":"2022-10-29T00:36:10","slug":"tangent-vectors-and-unit-tangent-vectors","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/tangent-vectors-and-unit-tangent-vectors\/","title":{"raw":"Tangent Vectors and Unit Tangent Vectors","rendered":"Tangent Vectors and Unit Tangent Vectors"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Find the tangent vector at a point for a given position vector<\/li>\r\n \t<li>Find the unit tangent vector at a point for a given position vector and explain its significance<\/li>\r\n<\/ul>\r\n<\/div>\r\nRecall from the Introduction to Derivatives that the derivative at a point can be interpreted as the slope of the tangent line to the graph at that point. In the case of a vector-valued function, the derivative provides a tangent vector to the curve represented by the function. Consider the vector-valued function [latex]{\\bf{r}}\\,(t)=\\cos{t\\,{\\bf{i}}}+\\sin{t\\,{\\bf{j}}}[\/latex]. the derivative of this function is [latex]-\\sin{t\\,{\\bf{i}}}+\\cos{t\\,{\\bf{j}}}[\/latex]. If we substitute the value [latex]t=\\frac{\\pi}{6}[\/latex] into both functions we get\r\n<p style=\"text-align: center;\">[latex]{\\bf{r}}\\,\\big(\\frac{\\pi}{6}\\big)=\\frac{\\sqrt{3}}{2}\\,{\\bf{i}}+\\frac{1}{2}\\,{\\bf{j}}[\/latex] and [latex]{\\bf{r}}'\\,\\big(\\frac{\\pi}{6}\\big)=-\\frac{1}{2}\\,{\\bf{i}}+\\frac{\\sqrt{3}}{2}\\,{\\bf{j}}[\/latex].<\/p>\r\nThe graph of this function appears in Figure 1, along with the vectors [latex]{\\bf{r}}\\,\\big(\\frac{\\pi}{6}\\big)[\/latex] and [latex]{\\bf{r}}'\\,\\big(\\frac{\\pi}{6}\\big)[\/latex].\r\n\r\n[caption id=\"attachment_907\" align=\"aligncenter\" width=\"342\"]<img class=\"size-full wp-image-907\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21155717\/3-2-1.jpeg\" alt=\"This figure is the graph of a circle represented by the vector-valued function r(t) = cost i + sint j. It is a circle centered at the origin with radius of 1, and counter-clockwise orientation. It has a vector from the origin pointing to the curve and labeled r(pi\/6). At the same point on the circle there is a tangent vector labeled \u201cr\u2019(pi\/6)\u201d.\" width=\"342\" height=\"346\" \/> Figure 1.\u00a0The tangent line at a point is calculated from the derivative of the vector-valued function [latex]{\\bf{r}}(t)[\/latex].[\/caption]Notice that the vector [latex]{\\bf{r}}'\\,\\big(\\frac{\\pi}{6}\\big)[\/latex] is tangent to the circle at the point corresponding to [latex]t=\\frac{\\pi}{6}[\/latex]. This is an example of a\u00a0<strong>tangent vector<\/strong> to the plane curve defined by [latex]{\\bf{r}}\\,(t)=\\cos{t\\,{\\bf{i}}}+\\sin{t\\,{\\bf{j}}}[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]C[\/latex] be a curve defined by a vector-valued function [latex]\\bf{r}[\/latex], and assume that [latex]{\\bf{r}}'\\,(t)[\/latex] exists when [latex]t=t_{0}[\/latex]. A tangent vector [latex]\\bf{v}[\/latex] at [latex]t=t_{0}[\/latex] is any vector such that, when the tail of the vector is placed at point [latex]{\\bf{r}}\\,(t_{0})[\/latex] on the graph, vector [latex]\\bf{v}[\/latex] is tangent to curve [latex]C[\/latex]. Vector [latex]{\\bf{r}}'\\,(t_{0})[\/latex] is an example of a tangent vector at point [latex]t=t_{0}[\/latex]. Furthermore, assume that [latex]{\\bf{r}}'\\,(t)\\neq{\\bf{0}}[\/latex] The\u00a0<strong>principal unit tangent vector<\/strong> at [latex]t[\/latex] is defined to be\r\n<div style=\"text-align: center;\">[latex]{\\bf{T}}\\,(t)=\\frac{{\\bf{r}}'\\,(t)}{\\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert},[\/latex]<\/div>\r\n&nbsp;\r\n\r\nprovided [latex]\\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert\\neq{0}.[\/latex]\r\n\r\n<\/div>\r\nThe unit tangent vector is exactly what it sounds like: a unit vector that is tangent to the curve. To calculate a unit tangent vector, first find the derivative [latex]{\\bf{r}}'\\,(t)[\/latex].\u00a0Second, calculate the magnitude of the derivative. The third step is to divide the derivative by its magnitude.\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Finding a Unit Tangent Vector<\/h3>\r\nFind the unit tangent vector for each of the following vector-valued functions:\r\n<ol>\r\n \t<li style=\"text-align: left;\">[latex]{\\bf{r}}\\,(t)=\\cos{t\\,{\\bf{i}}}+\\sin{t\\,{\\bf{j}}}[\/latex]<\/li>\r\n \t<li>[latex]{\\bf{u}}\\,(t)=(3t^{2}+2t)\\,{\\bf{i}}+(2-4t^{3})\\,{\\bf{j}}+(6t+5)\\,{\\bf{k}}[\/latex]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167793361764\" class=\"exercise\">[reveal-answer q=\"fs-id1167794055168\"]Show Solution[\/reveal-answer]<\/div>\r\n[hidden-answer a=\"fs-id1167794055168\"]\r\n\r\n1.\r\n\r\n[latex] \\begin{alignat}{2}\r\n\r\n\\hspace{2cm}&amp;\\text{First step:} &amp;\\quad {\\bf{r}}'(t)&amp;=-\\sin{t{\\bf{i}}}+\\cos{t{\\bf{j}}}\\\\\r\n\r\n&amp;\\text{Second step:} &amp;\\quad ||{\\bf{r}}'(t)||&amp;=\\sqrt{(-\\sin{t})^2+(\\cos{t})^2}=1\\\\\r\n\r\n&amp;\\text{Third step:} &amp;\\quad {\\bf{T}}(t)&amp;=\\frac{{\\bf{r}'(t)}}{||{\\bf{r}'(t)||}}=\\frac{-\\sin{t{\\bf{i}}+\\cos{t{\\bf{j}}}}}1=-\\sin{t{\\bf{i}}}+\\cos{t{\\bf{j}}}&amp;\\quad\\\\\r\n\r\n\\end{alignat}[\/latex]\r\n\r\n2.\r\n\r\n[latex] \\begin{alignat}{2} &amp;\\text{ First step:} &amp;\\quad {\\bf{u}}'(t)&amp;=(6t+2){\\bf{i}}-12t^2{\\bf{j}}+6{\\bf{k}}\\\\\r\n\r\n\\hspace{2cm}&amp;\\text{Second step:} &amp;\\quad ||{\\bf{u}}'(t)||&amp;=\\sqrt{(6t+2)^2+(-12t^2)^2+6^2}\\\\\r\n\r\n\\text{ }&amp; &amp;\\quad &amp;=\\sqrt{144t^4+36t^2+24t+40}\\\\\r\n\r\n\\text{ }&amp; &amp;\\quad &amp;=2\\sqrt{36t^4+9t^2+6t+10}\\\\\r\n\r\n&amp;\\text{Third step:} &amp;\\quad {\\bf{T}}(t)&amp;=\\frac{{\\bf{u}}'(t)}{||{\\bf{u}}'(t)||}=\\frac{(6t+2){\\bf{i}}-12t^2{\\bf{j}}+6{\\bf{k}}}{2\\sqrt{36t^4+9t^2+6t+10}}\\\\\r\n\r\n\\text{ }&amp; &amp;\\quad &amp;=\\frac{3t+1}{\\sqrt{36t^4+9t^2+6t+10}}{\\bf{i}}-\\frac{6t^2}{\\sqrt{36t^4+9t^2+6t+10}}{\\bf{j}}+\\frac{3}{\\sqrt{36t^4+9t^2+6t+10}}{\\bf{k}}&amp;\\quad\\\\\r\n\r\n\\end{alignat}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>TRY IT<\/h3>\r\nFind the unit tangent vector for the vector-valued function\r\n<p style=\"text-align: center;\">[latex]{\\bf{r}}\\,(t)=(t^{2}-3)\\,{\\bf{i}}+(2t+1)\\,{\\bf{j}}+(t-2)\\,{\\bf{k}}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1167793361764\" class=\"exercise\">[reveal-answer q=\"fs-id1167794055179\"]Show Solution[\/reveal-answer]<\/div>\r\n[hidden-answer a=\"fs-id1167794055179\"]\r\n<p style=\"text-align: center;\">[latex]{\\bf{T}}\\,(t)=\\frac{2t}{\\sqrt{4t^{2}+5}}\\,{\\bf{i}}+\\frac{2}{\\sqrt{4t^{2}+5}}{\\bf{j}}+\\frac{1}{\\sqrt{4t^{2}+5}}\\,{\\bf{k}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7949606&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=eMy52uPy9fM&amp;video_target=tpm-plugin-n62yty0d-eMy52uPy9fM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.7_transcript.html\">transcript for \u201cCP 3.7\u201d here (opens in new window).<\/a><\/center>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Find the tangent vector at a point for a given position vector<\/li>\n<li>Find the unit tangent vector at a point for a given position vector and explain its significance<\/li>\n<\/ul>\n<\/div>\n<p>Recall from the Introduction to Derivatives that the derivative at a point can be interpreted as the slope of the tangent line to the graph at that point. In the case of a vector-valued function, the derivative provides a tangent vector to the curve represented by the function. Consider the vector-valued function [latex]{\\bf{r}}\\,(t)=\\cos{t\\,{\\bf{i}}}+\\sin{t\\,{\\bf{j}}}[\/latex]. the derivative of this function is [latex]-\\sin{t\\,{\\bf{i}}}+\\cos{t\\,{\\bf{j}}}[\/latex]. If we substitute the value [latex]t=\\frac{\\pi}{6}[\/latex] into both functions we get<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{r}}\\,\\big(\\frac{\\pi}{6}\\big)=\\frac{\\sqrt{3}}{2}\\,{\\bf{i}}+\\frac{1}{2}\\,{\\bf{j}}[\/latex] and [latex]{\\bf{r}}'\\,\\big(\\frac{\\pi}{6}\\big)=-\\frac{1}{2}\\,{\\bf{i}}+\\frac{\\sqrt{3}}{2}\\,{\\bf{j}}[\/latex].<\/p>\n<p>The graph of this function appears in Figure 1, along with the vectors [latex]{\\bf{r}}\\,\\big(\\frac{\\pi}{6}\\big)[\/latex] and [latex]{\\bf{r}}'\\,\\big(\\frac{\\pi}{6}\\big)[\/latex].<\/p>\n<div id=\"attachment_907\" style=\"width: 352px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-907\" class=\"size-full wp-image-907\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21155717\/3-2-1.jpeg\" alt=\"This figure is the graph of a circle represented by the vector-valued function r(t) = cost i + sint j. It is a circle centered at the origin with radius of 1, and counter-clockwise orientation. It has a vector from the origin pointing to the curve and labeled r(pi\/6). At the same point on the circle there is a tangent vector labeled \u201cr\u2019(pi\/6)\u201d.\" width=\"342\" height=\"346\" \/><\/p>\n<p id=\"caption-attachment-907\" class=\"wp-caption-text\">Figure 1.\u00a0The tangent line at a point is calculated from the derivative of the vector-valued function [latex]{\\bf{r}}(t)[\/latex].<\/p>\n<\/div>\n<p>Notice that the vector [latex]{\\bf{r}}'\\,\\big(\\frac{\\pi}{6}\\big)[\/latex] is tangent to the circle at the point corresponding to [latex]t=\\frac{\\pi}{6}[\/latex]. This is an example of a\u00a0<strong>tangent vector<\/strong> to the plane curve defined by [latex]{\\bf{r}}\\,(t)=\\cos{t\\,{\\bf{i}}}+\\sin{t\\,{\\bf{j}}}[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p>Let [latex]C[\/latex] be a curve defined by a vector-valued function [latex]\\bf{r}[\/latex], and assume that [latex]{\\bf{r}}'\\,(t)[\/latex] exists when [latex]t=t_{0}[\/latex]. A tangent vector [latex]\\bf{v}[\/latex] at [latex]t=t_{0}[\/latex] is any vector such that, when the tail of the vector is placed at point [latex]{\\bf{r}}\\,(t_{0})[\/latex] on the graph, vector [latex]\\bf{v}[\/latex] is tangent to curve [latex]C[\/latex]. Vector [latex]{\\bf{r}}'\\,(t_{0})[\/latex] is an example of a tangent vector at point [latex]t=t_{0}[\/latex]. Furthermore, assume that [latex]{\\bf{r}}'\\,(t)\\neq{\\bf{0}}[\/latex] The\u00a0<strong>principal unit tangent vector<\/strong> at [latex]t[\/latex] is defined to be<\/p>\n<div style=\"text-align: center;\">[latex]{\\bf{T}}\\,(t)=\\frac{{\\bf{r}}'\\,(t)}{\\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert},[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>provided [latex]\\left\\Vert{\\bf{r}}'\\,(t)\\right\\Vert\\neq{0}.[\/latex]<\/p>\n<\/div>\n<p>The unit tangent vector is exactly what it sounds like: a unit vector that is tangent to the curve. To calculate a unit tangent vector, first find the derivative [latex]{\\bf{r}}'\\,(t)[\/latex].\u00a0Second, calculate the magnitude of the derivative. The third step is to divide the derivative by its magnitude.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Finding a Unit Tangent Vector<\/h3>\n<p>Find the unit tangent vector for each of the following vector-valued functions:<\/p>\n<ol>\n<li style=\"text-align: left;\">[latex]{\\bf{r}}\\,(t)=\\cos{t\\,{\\bf{i}}}+\\sin{t\\,{\\bf{j}}}[\/latex]<\/li>\n<li>[latex]{\\bf{u}}\\,(t)=(3t^{2}+2t)\\,{\\bf{i}}+(2-4t^{3})\\,{\\bf{j}}+(6t+5)\\,{\\bf{k}}[\/latex]<\/li>\n<\/ol>\n<div id=\"fs-id1167793361764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794055168\">Show Solution<\/span><\/div>\n<div id=\"qfs-id1167794055168\" class=\"hidden-answer\" style=\"display: none\">\n<p>1.<\/p>\n<p>[latex]\\begin{alignat}{2}    \\hspace{2cm}&\\text{First step:} &\\quad {\\bf{r}}'(t)&=-\\sin{t{\\bf{i}}}+\\cos{t{\\bf{j}}}\\\\    &\\text{Second step:} &\\quad ||{\\bf{r}}'(t)||&=\\sqrt{(-\\sin{t})^2+(\\cos{t})^2}=1\\\\    &\\text{Third step:} &\\quad {\\bf{T}}(t)&=\\frac{{\\bf{r}'(t)}}{||{\\bf{r}'(t)||}}=\\frac{-\\sin{t{\\bf{i}}+\\cos{t{\\bf{j}}}}}1=-\\sin{t{\\bf{i}}}+\\cos{t{\\bf{j}}}&\\quad\\\\    \\end{alignat}[\/latex]<\/p>\n<p>2.<\/p>\n<p>[latex]\\begin{alignat}{2} &\\text{ First step:} &\\quad {\\bf{u}}'(t)&=(6t+2){\\bf{i}}-12t^2{\\bf{j}}+6{\\bf{k}}\\\\    \\hspace{2cm}&\\text{Second step:} &\\quad ||{\\bf{u}}'(t)||&=\\sqrt{(6t+2)^2+(-12t^2)^2+6^2}\\\\    \\text{ }& &\\quad &=\\sqrt{144t^4+36t^2+24t+40}\\\\    \\text{ }& &\\quad &=2\\sqrt{36t^4+9t^2+6t+10}\\\\    &\\text{Third step:} &\\quad {\\bf{T}}(t)&=\\frac{{\\bf{u}}'(t)}{||{\\bf{u}}'(t)||}=\\frac{(6t+2){\\bf{i}}-12t^2{\\bf{j}}+6{\\bf{k}}}{2\\sqrt{36t^4+9t^2+6t+10}}\\\\    \\text{ }& &\\quad &=\\frac{3t+1}{\\sqrt{36t^4+9t^2+6t+10}}{\\bf{i}}-\\frac{6t^2}{\\sqrt{36t^4+9t^2+6t+10}}{\\bf{j}}+\\frac{3}{\\sqrt{36t^4+9t^2+6t+10}}{\\bf{k}}&\\quad\\\\    \\end{alignat}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>TRY IT<\/h3>\n<p>Find the unit tangent vector for the vector-valued function<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{r}}\\,(t)=(t^{2}-3)\\,{\\bf{i}}+(2t+1)\\,{\\bf{j}}+(t-2)\\,{\\bf{k}}[\/latex]<\/p>\n<div id=\"fs-id1167793361764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794055179\">Show Solution<\/span><\/div>\n<div id=\"qfs-id1167794055179\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]{\\bf{T}}\\,(t)=\\frac{2t}{\\sqrt{4t^{2}+5}}\\,{\\bf{i}}+\\frac{2}{\\sqrt{4t^{2}+5}}{\\bf{j}}+\\frac{1}{\\sqrt{4t^{2}+5}}\\,{\\bf{k}}[\/latex]<\/p>\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\" src=\"\/\/plugin.3playmedia.com\/show?mf=7949606&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=eMy52uPy9fM&amp;video_target=tpm-plugin-n62yty0d-eMy52uPy9fM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.7_transcript.html\">transcript for \u201cCP 3.7\u201d here (opens in new window).<\/a><\/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-3859\">\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>CP 3.7. <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 3. <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\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/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-3\/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":349141,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 3.7\",\"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-3859","chapter","type-chapter","status-publish","hentry"],"part":21,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3859","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":5,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3859\/revisions"}],"predecessor-version":[{"id":4071,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3859\/revisions\/4071"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/21"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3859\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=3859"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=3859"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=3859"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=3859"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}