{"id":562,"date":"2025-10-09T15:22:25","date_gmt":"2025-10-09T15:22:25","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=562"},"modified":"2025-10-24T13:21:38","modified_gmt":"2025-10-24T13:21:38","slug":"calculus-of-vector-valued-functions-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/calculus-of-vector-valued-functions-fresh-take\/","title":{"raw":"Calculus of Vector-Valued Functions: Fresh Take","rendered":"Calculus of Vector-Valued Functions: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Find derivatives of vector-valued functions<\/li>\r\n \t<li>Find the tangent vector at a point for a given position vector<\/li>\r\n \t<li>Find and interpret unit tangent vectors<\/li>\r\n \t<li>Calculate the definite integral of a vector-valued function<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Derivatives of Vector-Valued Functions<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Differentiating vector-valued functions works just like differentiating regular functions\u2014you apply the definition or, more practically, <strong>differentiate each component separately<\/strong>.<\/p>\r\n<p class=\"whitespace-normal break-words\">The derivative of [latex]{\\bf{r}}(t)[\/latex] is defined as:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{\\bf{r}}'(t)=\\underset{\\Delta t\\to 0}{\\text{lim}}\\frac{{\\bf{r}}(t+\\Delta t)-{\\bf{r}}(t)}{\\Delta t}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This measures the instantaneous rate of change of the vector function. When [latex]{\\bf{r}}(t)[\/latex] represents an object's position at time [latex]t[\/latex], the derivative [latex]{\\bf{r}}'(t)[\/latex] represents its velocity at that moment.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Practical Approach: Component-by-Component<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Rather than wrestling with the limit definition, use this shortcut: <strong>differentiate each component function individually<\/strong>.<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">For plane curves [latex]{\\bf{r}}(t)=f(t){\\bf{i}}+g(t){\\bf{j}}[\/latex]: [latex]{\\bf{r}}'(t)=f'(t){\\bf{i}}+g'(t){\\bf{j}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">For space curves [latex]{\\bf{r}}(t)=f(t){\\bf{i}}+g(t){\\bf{j}}+h(t){\\bf{k}}[\/latex]: [latex]{\\bf{r}}'(t)=f'(t){\\bf{i}}+g'(t){\\bf{j}}+h'(t){\\bf{k}}[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">The vector structure stays intact\u2014only the component functions change. Apply your familiar derivative rules (power rule, product rule, chain rule, trig derivatives, etc.) to each component.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Derivative Properties for Vector-Valued Functions<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The derivative rules you know extend naturally to vector-valued functions:<\/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>Scalar multiple<\/strong>: [latex]\\frac{d}{dt}[c{\\bf{r}}(t)]=c{\\bf{r}}'(t)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Sum and difference<\/strong>: [latex]\\frac{d}{dt}[{\\bf{r}}(t)\\pm{\\bf{u}}(t)]={\\bf{r}}'(t)\\pm{\\bf{u}}'(t)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Scalar product<\/strong>: [latex]\\frac{d}{dt}[f(t){\\bf{u}}(t)]=f'(t){\\bf{u}}(t)+f(t){\\bf{u}}'(t)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Dot product<\/strong>: [latex]\\frac{d}{dt}[{\\bf{r}}(t)\\cdot{\\bf{u}}(t)]={\\bf{r}}'(t)\\cdot{\\bf{u}}(t)+{\\bf{r}}(t)\\cdot{\\bf{u}}'(t)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Cross product<\/strong>: [latex]\\frac{d}{dt}[{\\bf{r}}(t)\\times{\\bf{u}}(t)]={\\bf{r}}'(t)\\times{\\bf{u}}(t)+{\\bf{r}}(t)\\times{\\bf{u}}'(t)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Chain rule<\/strong>: [latex]\\frac{d}{dt}[{\\bf{r}}(f(t))]={\\bf{r}}'(f(t))\\cdot f'(t)[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Special Property Worth Noting<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">If [latex]{\\bf{r}}(t)\\cdot{\\bf{r}}(t)=c[\/latex] (a constant), then [latex]{\\bf{r}}(t)\\cdot{\\bf{r}}'(t)=0[\/latex]. This tells us that when a vector has constant magnitude, it's always perpendicular to its derivative\u2014a useful geometric insight.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Use the definition to calculate the derivative of the function [latex]{\\bf{r}}(t)=(2t^{2}+3){\\bf{i}}+(5t-6){\\bf{j}}.[\/latex]\r\n<div id=\"fs-id1167793361764\" class=\"exercise\">[reveal-answer q=\"fs-id1167794055155\"]Show Solution[\/reveal-answer]<\/div>\r\n[hidden-answer a=\"fs-id1167794055155\"]\r\n<p style=\"text-align: center;\">[latex]{\\bf{r}}'(t)=4t{\\bf{i}}+5{\\bf{j}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">Calculate the derivative of the function\r\n<p style=\"text-align: center;\">[latex]{\\bf{r}}(t)=(t\\ln{t}){\\bf{i}}+(5e^{t}){\\bf{j}}+(\\cos{t}-\\sin{t}){\\bf{k}}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1167793361764\" class=\"exercise\">[reveal-answer q=\"fs-id1167794055157\"]Show Solution[\/reveal-answer]<\/div>\r\n[hidden-answer a=\"fs-id1167794055157\"]\r\n<p style=\"text-align: center;\">[latex]{\\bf{r}}'\\,(t)=(1+\\ln{t}){\\bf{i}}+5e^{t}{\\bf{j}}-(\\sin{t}+\\cos{t}){\\bf{k}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7949600&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=und25wfMXIc&amp;video_target=tpm-plugin-9ecrji6c-und25wfMXIc\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.5_transcript.html\">transcript for \u201cCP 3.5\u201d here (opens in new window).<\/a><\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">Given the vector-valued functions [latex]{\\bf{r}}\\,(t)=\\cos{t\\,{\\bf{i}}}+\\sin{t\\,{\\bf{j}}}-e^{2t}\\,{\\bf{k}}[\/latex] and [latex]{\\bf{u}}\\,(t)=t\\,{\\bf{i}}+\\sin{t\\,{\\bf{j}}}+\\cos{t\\,{\\bf{k}}}[\/latex], calculate [latex]\\frac{d}{dt}\\big[{\\bf{r}}\\,(t)\\cdot{\\bf{r}}'\\,(t)\\big][\/latex] and [latex]\\frac{d}{dt}\\big[{\\bf{u}}\\,(t)\\times{\\bf{r}}\\,(t)\\big][\/latex].\r\n<div id=\"fs-id1167793361764\" class=\"exercise\">[reveal-answer q=\"fs-id1167794055159\"]Show Solution[\/reveal-answer]<\/div>\r\n[hidden-answer a=\"fs-id1167794055159\"]\r\n<p style=\"text-align: center;\">[latex]\\frac{d}{dt}\\big[{\\bf{r}}\\,(t)\\cdot{\\bf{r}}'\\,(t)\\big]=8e^{4t}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{d}{dt}\\big[{\\bf{u}}\\,(t)\\times{\\bf{r}}\\,(t)\\big]=-\\big(e^{2t}(\\cos{t}+2\\sin{t})+\\cos{2t}\\big)\\,{\\bf{i}}+\\big(e^{2t}(2t+1)-\\sin2t\\big)\\,{\\bf{j}}+\\big(t\\cos{t}+\\sin{t}-\\cos{2t}\\big)\\,{\\bf{k}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7949601&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=t9a9as9Zsio&amp;video_target=tpm-plugin-f5h4a6t9-t9a9as9Zsio\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.6_transcript.html\">transcript for \u201cCP 3.6\u201d here (opens in new window).<\/a><\/center><\/section>\r\n<h2>Tangent Vectors and Unit Tangent Vectors<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">In single-variable calculus, the derivative gives the slope of the tangent line. For vector-valued functions, the derivative gives you a <strong>tangent vector<\/strong> that points along the curve's direction.<\/p>\r\n<p class=\"whitespace-normal break-words\">A <strong>tangent vector<\/strong> at [latex]t = t_0[\/latex] is any vector that lies tangent to the curve when placed at point [latex]{\\bf{r}}(t_0)[\/latex]. The derivative [latex]{\\bf{r}}'(t_0)[\/latex] is the most natural tangent vector. It points in the direction the curve is heading as [latex]t[\/latex] increases.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Principal Unit Tangent Vector<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">While [latex]{\\bf{r}}'(t)[\/latex] gives direction, its magnitude can vary. The <strong>principal unit tangent vector<\/strong> [latex]{\\bf{T}}(t)[\/latex] always has magnitude [latex]1[\/latex]:<\/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<p class=\"whitespace-normal break-words\">This unit vector points in the direction of motion. It strips away information about speed\u2014it's purely directional.<\/p>\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] by differentiating each component<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Find the magnitude<\/strong>: Calculate [latex]\\left\\Vert{\\bf{r}}'(t)\\right\\Vert = \\sqrt{(f'(t))^2 + (g'(t))^2 + (h'(t))^2}[\/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<p class=\"whitespace-normal break-words\">The formula only works when [latex]\\left\\Vert{\\bf{r}}'(t)\\right\\Vert \\neq 0[\/latex]. If the derivative's magnitude is zero, the unit tangent vector is undefined. This typically happens at cusps or points where the motion stops.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Why This Matters<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The unit tangent vector always has length [latex]1[\/latex]. It points in the direction of travel along the curve. It's useful when you need to know which way the curve is heading. You don't have to worry about how fast it's moving. Think of it as \"direction of motion\" isolated from \"speed of motion.\"<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Find 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<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above Try It<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\" data-mce-fragment=\"1\"><\/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><\/section>\r\n<h2>Integrals of Vector-Valued Functions<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Integrating vector-valued functions works just like differentiating them. You handle each component separately using the integration techniques you already know. The Fundamental Theorem of Calculus applies here too.<\/p>\r\n<p class=\"whitespace-normal break-words\">Just as you differentiate each component individually, you integrate the same way.<\/p>\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\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Indefinite integral<\/strong>: [latex]\\int [f(t){\\bf{i}}+g(t){\\bf{j}}]dt=\\Bigg[\\int f(t)dt\\Bigg]{\\bf{i}}+\\Bigg[\\int g(t)dt\\Bigg]{\\bf{j}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Definite integral<\/strong>: [latex]\\int_{a}^{b} [f(t){\\bf{i}}+g(t){\\bf{j}}]dt=\\Bigg[\\int_{a}^{b} f(t)dt\\Bigg]{\\bf{i}}+\\Bigg[\\int_{a}^{b} g(t)dt\\Bigg]{\\bf{j}}[\/latex]<\/li>\r\n<\/ul>\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\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Indefinite integral<\/strong>: [latex]\\int [f(t){\\bf{i}}+g(t){\\bf{j}}+h(t){\\bf{k}}]dt=\\Bigg[\\int f(t)dt\\Bigg]{\\bf{i}}+\\Bigg[\\int g(t)dt\\Bigg]{\\bf{j}}+\\Bigg[\\int h(t)dt\\Bigg]{\\bf{k}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Definite integral<\/strong>: [latex]\\int_{a}^{b} [f(t){\\bf{i}}+g(t){\\bf{j}}+h(t){\\bf{k}}]dt=\\Bigg[\\int_{a}^{b} f(t)dt\\Bigg]{\\bf{i}}+\\Bigg[\\int_{a}^{b} g(t)dt\\Bigg]{\\bf{j}}+\\Bigg[\\int_{a}^{b} h(t)dt\\Bigg]{\\bf{k}}[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Handling Constants of Integration<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">When you integrate a vector-valued function, each component gets its own integration constant. You can combine these into a single constant vector [latex]{\\bf{C}}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">For example: [latex]\\int [f(t){\\bf{i}}+g(t){\\bf{j}}]dt=F(t){\\bf{i}}+G(t){\\bf{j}}+{\\bf{C}}[\/latex] where [latex]{\\bf{C}}=C_{1}{\\bf{i}}+C_{2}{\\bf{j}}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Break the vector function into components. Integrate each scalar function using your standard techniques. Reassemble the results into vector form. Use the same integration formulas from Calculus 1 for each component.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Why This Matters<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Antiderivatives of vector-valued functions appear constantly in applications. If a function represents velocity, its antiderivative gives position. If the function represents acceleration, the antiderivative gives velocity.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Calculate the following integral:\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_{1}^{3} \\ \\big[(2t+4)\\,{\\bf{i}}+(3t^{2}-4t)\\,{\\bf{j}}\\big]\\,dt[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1167793361764\" class=\"exercise\">[reveal-answer q=\"fs-id1167794055279\"]Show Solution[\/reveal-answer]<\/div>\r\n[hidden-answer a=\"fs-id1167794055279\"]\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_{1}^{3} \\ \\big[(2t+4)\\,{\\bf{i}}+(3t^{2}-4t)\\,{\\bf{j}}\\big]\\,dt=16\\,{\\bf{i}}+10\\,{\\bf{j}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7949613&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=VeLUdurbQN4&amp;video_target=tpm-plugin-4s0tfrxc-VeLUdurbQN4\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.8_transcript.html\">transcript for \u201cCP 3.8\u201d here (opens in new window).<\/a><\/center><\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Find derivatives of vector-valued functions<\/li>\n<li>Find the tangent vector at a point for a given position vector<\/li>\n<li>Find and interpret unit tangent vectors<\/li>\n<li>Calculate the definite integral of a vector-valued function<\/li>\n<\/ul>\n<\/section>\n<h2>Derivatives of Vector-Valued Functions<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Differentiating vector-valued functions works just like differentiating regular functions\u2014you apply the definition or, more practically, <strong>differentiate each component separately<\/strong>.<\/p>\n<p class=\"whitespace-normal break-words\">The derivative of [latex]{\\bf{r}}(t)[\/latex] is defined as:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{\\bf{r}}'(t)=\\underset{\\Delta t\\to 0}{\\text{lim}}\\frac{{\\bf{r}}(t+\\Delta t)-{\\bf{r}}(t)}{\\Delta t}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This measures the instantaneous rate of change of the vector function. When [latex]{\\bf{r}}(t)[\/latex] represents an object&#8217;s position at time [latex]t[\/latex], the derivative [latex]{\\bf{r}}'(t)[\/latex] represents its velocity at that moment.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Practical Approach: Component-by-Component<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Rather than wrestling with the limit definition, use this shortcut: <strong>differentiate each component function individually<\/strong>.<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">For plane curves [latex]{\\bf{r}}(t)=f(t){\\bf{i}}+g(t){\\bf{j}}[\/latex]: [latex]{\\bf{r}}'(t)=f'(t){\\bf{i}}+g'(t){\\bf{j}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">For space curves [latex]{\\bf{r}}(t)=f(t){\\bf{i}}+g(t){\\bf{j}}+h(t){\\bf{k}}[\/latex]: [latex]{\\bf{r}}'(t)=f'(t){\\bf{i}}+g'(t){\\bf{j}}+h'(t){\\bf{k}}[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">The vector structure stays intact\u2014only the component functions change. Apply your familiar derivative rules (power rule, product rule, chain rule, trig derivatives, etc.) to each component.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Derivative Properties for Vector-Valued Functions<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The derivative rules you know extend naturally to vector-valued functions:<\/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>Scalar multiple<\/strong>: [latex]\\frac{d}{dt}[c{\\bf{r}}(t)]=c{\\bf{r}}'(t)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Sum and difference<\/strong>: [latex]\\frac{d}{dt}[{\\bf{r}}(t)\\pm{\\bf{u}}(t)]={\\bf{r}}'(t)\\pm{\\bf{u}}'(t)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Scalar product<\/strong>: [latex]\\frac{d}{dt}[f(t){\\bf{u}}(t)]=f'(t){\\bf{u}}(t)+f(t){\\bf{u}}'(t)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Dot product<\/strong>: [latex]\\frac{d}{dt}[{\\bf{r}}(t)\\cdot{\\bf{u}}(t)]={\\bf{r}}'(t)\\cdot{\\bf{u}}(t)+{\\bf{r}}(t)\\cdot{\\bf{u}}'(t)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Cross product<\/strong>: [latex]\\frac{d}{dt}[{\\bf{r}}(t)\\times{\\bf{u}}(t)]={\\bf{r}}'(t)\\times{\\bf{u}}(t)+{\\bf{r}}(t)\\times{\\bf{u}}'(t)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Chain rule<\/strong>: [latex]\\frac{d}{dt}[{\\bf{r}}(f(t))]={\\bf{r}}'(f(t))\\cdot f'(t)[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Special Property Worth Noting<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">If [latex]{\\bf{r}}(t)\\cdot{\\bf{r}}(t)=c[\/latex] (a constant), then [latex]{\\bf{r}}(t)\\cdot{\\bf{r}}'(t)=0[\/latex]. This tells us that when a vector has constant magnitude, it&#8217;s always perpendicular to its derivative\u2014a useful geometric insight.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Use the definition to calculate the derivative of the function [latex]{\\bf{r}}(t)=(2t^{2}+3){\\bf{i}}+(5t-6){\\bf{j}}.[\/latex]<\/p>\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-id1167794055155\">Show Solution<\/button><\/div>\n<div id=\"qfs-id1167794055155\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]{\\bf{r}}'(t)=4t{\\bf{i}}+5{\\bf{j}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">Calculate the derivative of the function<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{r}}(t)=(t\\ln{t}){\\bf{i}}+(5e^{t}){\\bf{j}}+(\\cos{t}-\\sin{t}){\\bf{k}}[\/latex]<\/p>\n<div class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794055157\">Show Solution<\/button><\/div>\n<div id=\"qfs-id1167794055157\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]{\\bf{r}}'\\,(t)=(1+\\ln{t}){\\bf{i}}+5e^{t}{\\bf{j}}-(\\sin{t}+\\cos{t}){\\bf{k}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7949600&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=und25wfMXIc&amp;video_target=tpm-plugin-9ecrji6c-und25wfMXIc\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/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\/CP3.5_transcript.html\">transcript for \u201cCP 3.5\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Given the vector-valued functions [latex]{\\bf{r}}\\,(t)=\\cos{t\\,{\\bf{i}}}+\\sin{t\\,{\\bf{j}}}-e^{2t}\\,{\\bf{k}}[\/latex] and [latex]{\\bf{u}}\\,(t)=t\\,{\\bf{i}}+\\sin{t\\,{\\bf{j}}}+\\cos{t\\,{\\bf{k}}}[\/latex], calculate [latex]\\frac{d}{dt}\\big[{\\bf{r}}\\,(t)\\cdot{\\bf{r}}'\\,(t)\\big][\/latex] and [latex]\\frac{d}{dt}\\big[{\\bf{u}}\\,(t)\\times{\\bf{r}}\\,(t)\\big][\/latex].<\/p>\n<div class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794055159\">Show Solution<\/button><\/div>\n<div id=\"qfs-id1167794055159\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\frac{d}{dt}\\big[{\\bf{r}}\\,(t)\\cdot{\\bf{r}}'\\,(t)\\big]=8e^{4t}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{d}{dt}\\big[{\\bf{u}}\\,(t)\\times{\\bf{r}}\\,(t)\\big]=-\\big(e^{2t}(\\cos{t}+2\\sin{t})+\\cos{2t}\\big)\\,{\\bf{i}}+\\big(e^{2t}(2t+1)-\\sin2t\\big)\\,{\\bf{j}}+\\big(t\\cos{t}+\\sin{t}-\\cos{2t}\\big)\\,{\\bf{k}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7949601&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=t9a9as9Zsio&amp;video_target=tpm-plugin-f5h4a6t9-t9a9as9Zsio\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/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\/CP3.6_transcript.html\">transcript for \u201cCP 3.6\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<h2>Tangent Vectors and Unit Tangent Vectors<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">In single-variable calculus, the derivative gives the slope of the tangent line. For vector-valued functions, the derivative gives you a <strong>tangent vector<\/strong> that points along the curve&#8217;s direction.<\/p>\n<p class=\"whitespace-normal break-words\">A <strong>tangent vector<\/strong> at [latex]t = t_0[\/latex] is any vector that lies tangent to the curve when placed at point [latex]{\\bf{r}}(t_0)[\/latex]. The derivative [latex]{\\bf{r}}'(t_0)[\/latex] is the most natural tangent vector. It points in the direction the curve is heading as [latex]t[\/latex] increases.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Principal Unit Tangent Vector<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">While [latex]{\\bf{r}}'(t)[\/latex] gives direction, its magnitude can vary. The <strong>principal unit tangent vector<\/strong> [latex]{\\bf{T}}(t)[\/latex] always has magnitude [latex]1[\/latex]:<\/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<p class=\"whitespace-normal break-words\">This unit vector points in the direction of motion. It strips away information about speed\u2014it&#8217;s purely directional.<\/p>\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] by differentiating each component<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Find the magnitude<\/strong>: Calculate [latex]\\left\\Vert{\\bf{r}}'(t)\\right\\Vert = \\sqrt{(f'(t))^2 + (g'(t))^2 + (h'(t))^2}[\/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<p class=\"whitespace-normal break-words\">The formula only works when [latex]\\left\\Vert{\\bf{r}}'(t)\\right\\Vert \\neq 0[\/latex]. If the derivative&#8217;s magnitude is zero, the unit tangent vector is undefined. This typically happens at cusps or points where the motion stops.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Why This Matters<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The unit tangent vector always has length [latex]1[\/latex]. It points in the direction of travel along the curve. It&#8217;s useful when you need to know which way the curve is heading. You don&#8217;t have to worry about how fast it&#8217;s moving. Think of it as &#8220;direction of motion&#8221; isolated from &#8220;speed of motion.&#8221;<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">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 class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794055179\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">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\" data-mce-fragment=\"1\"><\/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<\/section>\n<h2>Integrals of Vector-Valued Functions<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Integrating vector-valued functions works just like differentiating them. You handle each component separately using the integration techniques you already know. The Fundamental Theorem of Calculus applies here too.<\/p>\n<p class=\"whitespace-normal break-words\">Just as you differentiate each component individually, you integrate the same way.<\/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<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Indefinite integral<\/strong>: [latex]\\int [f(t){\\bf{i}}+g(t){\\bf{j}}]dt=\\Bigg[\\int f(t)dt\\Bigg]{\\bf{i}}+\\Bigg[\\int g(t)dt\\Bigg]{\\bf{j}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Definite integral<\/strong>: [latex]\\int_{a}^{b} [f(t){\\bf{i}}+g(t){\\bf{j}}]dt=\\Bigg[\\int_{a}^{b} f(t)dt\\Bigg]{\\bf{i}}+\\Bigg[\\int_{a}^{b} g(t)dt\\Bigg]{\\bf{j}}[\/latex]<\/li>\n<\/ul>\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<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Indefinite integral<\/strong>: [latex]\\int [f(t){\\bf{i}}+g(t){\\bf{j}}+h(t){\\bf{k}}]dt=\\Bigg[\\int f(t)dt\\Bigg]{\\bf{i}}+\\Bigg[\\int g(t)dt\\Bigg]{\\bf{j}}+\\Bigg[\\int h(t)dt\\Bigg]{\\bf{k}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Definite integral<\/strong>: [latex]\\int_{a}^{b} [f(t){\\bf{i}}+g(t){\\bf{j}}+h(t){\\bf{k}}]dt=\\Bigg[\\int_{a}^{b} f(t)dt\\Bigg]{\\bf{i}}+\\Bigg[\\int_{a}^{b} g(t)dt\\Bigg]{\\bf{j}}+\\Bigg[\\int_{a}^{b} h(t)dt\\Bigg]{\\bf{k}}[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Handling Constants of Integration<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When you integrate a vector-valued function, each component gets its own integration constant. You can combine these into a single constant vector [latex]{\\bf{C}}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">For example: [latex]\\int [f(t){\\bf{i}}+g(t){\\bf{j}}]dt=F(t){\\bf{i}}+G(t){\\bf{j}}+{\\bf{C}}[\/latex] where [latex]{\\bf{C}}=C_{1}{\\bf{i}}+C_{2}{\\bf{j}}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Break the vector function into components. Integrate each scalar function using your standard techniques. Reassemble the results into vector form. Use the same integration formulas from Calculus 1 for each component.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Why This Matters<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Antiderivatives of vector-valued functions appear constantly in applications. If a function represents velocity, its antiderivative gives position. If the function represents acceleration, the antiderivative gives velocity.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Calculate the following integral:<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_{1}^{3} \\ \\big[(2t+4)\\,{\\bf{i}}+(3t^{2}-4t)\\,{\\bf{j}}\\big]\\,dt[\/latex]<\/p>\n<div class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794055279\">Show Solution<\/button><\/div>\n<div id=\"qfs-id1167794055279\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_{1}^{3} \\ \\big[(2t+4)\\,{\\bf{i}}+(3t^{2}-4t)\\,{\\bf{j}}\\big]\\,dt=16\\,{\\bf{i}}+10\\,{\\bf{j}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7949613&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=VeLUdurbQN4&amp;video_target=tpm-plugin-4s0tfrxc-VeLUdurbQN4\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/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\/CP3.8_transcript.html\">transcript for \u201cCP 3.8\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":15,"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-562","chapter","type-chapter","status-publish","hentry"],"part":424,"module-header":"fresh_take","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\/562","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":9,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/562\/revisions"}],"predecessor-version":[{"id":1755,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/562\/revisions\/1755"}],"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\/562\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=562"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=562"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=562"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=562"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}