{"id":955,"date":"2025-10-14T16:38:40","date_gmt":"2025-10-14T16:38:40","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=955"},"modified":"2025-10-22T16:32:54","modified_gmt":"2025-10-22T16:32:54","slug":"955","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/955\/","title":{"raw":"Calculus of Vector-Valued Functions: Learn It 2","rendered":"Calculus of Vector-Valued Functions: Learn It 2"},"content":{"raw":"<h2>Tangent Vectors and Unit Tangent Vectors<\/h2>\r\nRecall in single-variable calculus, the derivative at a point gives the slope of the tangent line to the graph at that point. For vector-valued functions, the derivative provides a tangent vector to the curve.\r\n<p class=\"whitespace-normal break-words\">Consider the vector-valued function [latex]{\\bf{r}}(t)=\\cos t{\\bf{i}}+\\sin t{\\bf{j}}[\/latex]. The derivative is [latex]{\\bf{r}}'(t)=-\\sin t{\\bf{i}}+\\cos t{\\bf{j}}[\/latex]. If we substitute [latex]t=\\frac{\\pi}{6}[\/latex] into both functions, we get:<\/p>\r\n<p class=\"whitespace-normal break-words\" 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\n<p class=\"whitespace-normal break-words\">Figure 1 shows the graph of this function along with the vectors [latex]{\\bf{r}}\\big(\\frac{\\pi}{6}\\big)[\/latex] and [latex]{\\bf{r}}'\\big(\\frac{\\pi}{6}\\big)[\/latex].<\/p>\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 [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 a <strong>tangent vector<\/strong> to the curve defined by [latex]{\\bf{r}}(t)[\/latex].\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>tangent vector and principal unit tangent vector<\/h3>\r\n<p class=\"whitespace-normal break-words\">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].<\/p>\r\n[latex]\\\\[\/latex]\r\n<p class=\"whitespace-normal break-words\">A <strong>tangent vector<\/strong> [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]. [latex]\\\\[\/latex]<em>The derivative [latex]{\\bf{r}}'(t_{0})[\/latex] is an example of a tangent vector at point [latex]t=t_{0}[\/latex].<\/em><\/p>\r\n[latex]\\\\[\/latex]\r\n<p class=\"whitespace-normal break-words\">The <strong>principal unit tangent vector<\/strong> at [latex]t[\/latex] is defined as:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{\\bf{T}}(t)=\\frac{{\\bf{r}}'(t)}{\\left\\Vert{\\bf{r}}'(t)\\right\\Vert}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">provided [latex]\\left\\Vert{\\bf{r}}'(t)\\right\\Vert\\neq 0[\/latex].<\/p>\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">The unit tangent vector is exactly what it sounds like: a unit vector that is tangent to the curve. It points in the direction of motion along the curve and always has magnitude [latex]1[\/latex].<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\">\r\n<p class=\"whitespace-normal break-words\"><strong>How to: Find the Unit Tangent Vector<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Find the derivative<\/strong>: Calculate [latex]{\\bf{r}}'(t)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Find the magnitude<\/strong>: Calculate [latex]\\left\\Vert{\\bf{r}}'(t)\\right\\Vert[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Divide<\/strong>: Compute [latex]{\\bf{T}}(t)=\\frac{{\\bf{r}}'(t)}{\\left\\Vert{\\bf{r}}'(t)\\right\\Vert}[\/latex]<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find 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<\/section>","rendered":"<h2>Tangent Vectors and Unit Tangent Vectors<\/h2>\n<p>Recall in single-variable calculus, the derivative at a point gives the slope of the tangent line to the graph at that point. For vector-valued functions, the derivative provides a tangent vector to the curve.<\/p>\n<p class=\"whitespace-normal break-words\">Consider the vector-valued function [latex]{\\bf{r}}(t)=\\cos t{\\bf{i}}+\\sin t{\\bf{j}}[\/latex]. The derivative is [latex]{\\bf{r}}'(t)=-\\sin t{\\bf{i}}+\\cos t{\\bf{j}}[\/latex]. If we substitute [latex]t=\\frac{\\pi}{6}[\/latex] into both functions, we get:<\/p>\n<p class=\"whitespace-normal break-words\" 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 class=\"whitespace-normal break-words\">Figure 1 shows the graph of this function 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<figure id=\"attachment_907\" aria-describedby=\"caption-attachment-907\" style=\"width: 342px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption 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].<\/figcaption><\/figure>\n<p>Notice that [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 a <strong>tangent vector<\/strong> to the curve defined by [latex]{\\bf{r}}(t)[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>tangent vector and principal unit tangent vector<\/h3>\n<p class=\"whitespace-normal break-words\">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].<\/p>\n<p>[latex]\\\\[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">A <strong>tangent vector<\/strong> [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]. [latex]\\\\[\/latex]<em>The derivative [latex]{\\bf{r}}'(t_{0})[\/latex] is an example of a tangent vector at point [latex]t=t_{0}[\/latex].<\/em><\/p>\n<p>[latex]\\\\[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">The <strong>principal unit tangent vector<\/strong> at [latex]t[\/latex] is defined as:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{\\bf{T}}(t)=\\frac{{\\bf{r}}'(t)}{\\left\\Vert{\\bf{r}}'(t)\\right\\Vert}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">provided [latex]\\left\\Vert{\\bf{r}}'(t)\\right\\Vert\\neq 0[\/latex].<\/p>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">The unit tangent vector is exactly what it sounds like: a unit vector that is tangent to the curve. It points in the direction of motion along the curve and always has magnitude [latex]1[\/latex].<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\n<p class=\"whitespace-normal break-words\"><strong>How to: Find the Unit Tangent Vector<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Find the derivative<\/strong>: Calculate [latex]{\\bf{r}}'(t)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Find the magnitude<\/strong>: Calculate [latex]\\left\\Vert{\\bf{r}}'(t)\\right\\Vert[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Divide<\/strong>: Compute [latex]{\\bf{T}}(t)=\\frac{{\\bf{r}}'(t)}{\\left\\Vert{\\bf{r}}'(t)\\right\\Vert}[\/latex]<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794055168\">Show Solution<\/button><\/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<\/section>\n","protected":false},"author":15,"menu_order":12,"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-955","chapter","type-chapter","status-publish","hentry"],"part":424,"module-header":"- Select Header -","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\/955","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":8,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/955\/revisions"}],"predecessor-version":[{"id":1704,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/955\/revisions\/1704"}],"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\/955\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=955"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=955"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=955"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=955"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}