{"id":3856,"date":"2022-04-04T16:08:05","date_gmt":"2022-04-04T16:08:05","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=3856"},"modified":"2022-10-29T00:34:55","modified_gmt":"2022-10-29T00:34:55","slug":"derivatives-of-vector-valued-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/derivatives-of-vector-valued-functions\/","title":{"raw":"Derivatives of Vector-Valued Functions","rendered":"Derivatives of Vector-Valued Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Write an expression for the derivative of a vector-valued function<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1169739028079\" class=\" \">Now that we have seen what a vector-valued function is and how to take its limit, the next step is to learn how to differentiate a vector-valued function. The definition of the derivative of a vector-valued function is nearly identical to the definition of a real-valued function of one variable. However, because the range of a vector-valued function consists of vectors, the same is true for the range of the derivative of a vector-valued function.<\/p>\r\n\r\n<div id=\"fs-id1169738955085\" class=\"ui-has-child-title\" data-type=\"note\">\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\nA <strong><span id=\"term108\" data-type=\"term\">derivative of a vector-valued function<\/span><\/strong>\u00a0[latex]{\\bf{r}}\\,(t)[\/latex] is\r\n<div style=\"text-align: center;\">[latex]{\\bf{r}}'\\,(t)=\\displaystyle\\lim_{\\Delta{t}{\\to}0}\\frac{{\\bf{r}}(t+\\Delta{t})-{\\bf{r}}\\,(t)}{\\Delta{t}},[\/latex]<\/div>\r\n&nbsp;\r\n\r\nprovided the limit exists. If [latex]{\\bf{r}}'\\,(t)[\/latex] exists, then [latex]{\\bf{r}}[\/latex] is differentiable at [latex]t[\/latex]. If [latex]{\\bf{r}}'\\,(t)[\/latex] exists for all [latex]t[\/latex] in an open interval [latex](a,\\ b)[\/latex], then [latex]{\\bf{r}}[\/latex] is differentiable over the interval [latex](a,\\ b)[\/latex]. For the function to be differentiable over the closed interval [latex][a,\\ b][\/latex], the following two limits must exist as well:\r\n<div style=\"text-align: center;\">[latex]{\\bf{r}}'\\,(a)=\\displaystyle\\lim_{\\Delta{t}{\\to}0^{+}}\\frac{{\\bf{r}}\\,(a+\\Delta{t})-{\\bf{r}}\\,(a)}{\\Delta{t}}[\/latex] and [latex]{\\bf{r}}'\\,(b)=\\displaystyle\\lim_{\\Delta{t}{\\to}0^{-}}\\frac{{\\bf{r}}\\,(b+\\Delta{t})-{\\bf{r}}\\,(b)}{\\Delta{t}}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<p id=\"fs-id1169738885200\" class=\" \">Many of the rules for calculating derivatives of real-valued functions can be applied to calculating the derivatives of vector-valued functions as well. Recall that the derivative of a real-valued function can be interpreted as the slope of a tangent line or the instantaneous rate of change of the function. The derivative of a vector-valued function can be understood to be an instantaneous rate of change as well; for example, when the function represents the position of an object at a given point in time, the derivative represents its velocity at that same point in time.<\/p>\r\n<p id=\"fs-id1169738822521\" class=\" \">We now demonstrate taking the derivative of a vector-valued function.<\/p>\r\n\r\n<div id=\"fs-id1167793900960\" class=\"textbook exercises\">\r\n<h3>Example:\u00a0Finding the derivative of a vector-valued function<\/h3>\r\nUse the definition to calculate the derivative of the function\r\n<p style=\"text-align: center;\">[latex]{\\bf{r}}\\,(t)=(3t+4)\\,{\\bf{i}}+(t^{2}-4t+3){\\bf{j}}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1167793361764\" class=\"exercise\">[reveal-answer q=\"fs-id1167794055154\"]Show Solution[\/reveal-answer]<\/div>\r\n[hidden-answer a=\"fs-id1167794055154\"]\r\n<div class=\"exercise\">\r\n\r\nLet's use the equation from the above definition:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\bf{r}}'\\,(t) &amp; =\\hfill &amp; {\\displaystyle\\lim_{\\Delta{t}{\\to}0}\\frac{{\\bf{r}}(t+\\Delta{t})-{\\bf{r}}(t)}{\\Delta{t}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{\\Delta{t}{\\to}0}\\frac{\\big[(3(t+\\Delta{t})+4){\\bf{i}}+\\big({(t+\\Delta{t})}^{2}-4(t+\\Delta{t})+3\\big){\\bf{j}}\\big]-\\big[(3t+4){\\bf{i}}+(t^{2}-4t+3){\\bf{j}}\\big]}{\\Delta{t}}}\\hfill\\\\\\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{\\Delta{t}{\\to}0}\\frac{(3t+3\\Delta{t}+4){\\bf{i}}-(3t+4){\\bf{i}}+(t^{2}+2t\\Delta{t}+(\\Delta{t})^{2}-4t-4\\Delta{t}+3){\\bf{j}}-(t^{2}-4t+3){\\bf{j}}}{\\Delta{t}}}\\hfill\\\\\\hfill&amp; =\\hfill &amp; {\\displaystyle\\lim_{\\Delta{t}{\\to}0}\\frac{(3\\Delta{t}){\\bf{i}}+\\big(2t\\Delta{t}+(\\Delta{t})^{2}-4\\Delta{t}\\big){\\bf{j}}}{\\Delta{t}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{\\Delta{t}{\\to}0}{(3\\,{\\bf{i}}+(2t+{\\Delta}t-4)\\,{\\bf{j}}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {3\\,{\\bf{i}}+(2t-4)\\,{\\bf{j}}}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793957091\" class=\"textbook key-takeaways\">\r\n<h3>TRY IT<\/h3>\r\nUse 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<\/div>\r\nNotice that in the calculations in the example above, we could also obtain the answer by first calculating the derivative of each component function, then putting these derivatives back into the vector-valued function. This is always true for calculating the derivative of a vector-valued function, whether it is in two or three dimensions. We state this in the following theorem. The proof of this theorem follows directly from the definitions of the limit of a vector-valued function and the derivative of a vector-valued function.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Differentiation of Vector-Valued functions theorem<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]f[\/latex], [latex]g[\/latex], and [latex]h[\/latex] be differentiable functions of [latex]t[\/latex].\r\n<ul>\r\n \t<li style=\"text-align: left;\">If [latex]{\\bf{r}}\\,(t)=f\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}[\/latex], then [latex]{\\bf{r}}'\\,(t)=f'\\,(t)\\,{\\bf{i}}+g'\\,(t)\\,{\\bf{j}}[\/latex].<\/li>\r\n \t<li>If\u00a0[latex]{\\bf{r}}\\,(t)=f\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}+h\\,(t)\\,{\\bf{k}}[\/latex], then [latex]{\\bf{r}}'\\,(t)=f'\\,(t)\\,{\\bf{i}}+g'\\,(t)\\,{\\bf{j}}+h'\\,(t)\\,{\\bf{k}}[\/latex].<\/li>\r\n<\/ul>\r\n<\/div>\r\nSince we will be using derivatives of common single-variable functions throughout this course, we briefly recall the derivatives of common functions, along with frequently-encountered derivative rules below.\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Derivatives of common functions<\/h3>\r\n<ul>\r\n \t<li>[latex] \\frac{d}{dx} (x^n) = nx^{n-1} [\/latex]\u00a0 (Power Rule)<\/li>\r\n \t<li>[latex] \\frac{d}{dx} (\\sin x) = \\cos x [\/latex]<\/li>\r\n \t<li>[latex] \\frac{d}{dx} (\\cos x) = -\\sin x [\/latex]<\/li>\r\n \t<li>[latex] \\frac{d}{dx} (\\sec x) = \\sec x \\tan x [\/latex]<\/li>\r\n \t<li>[latex] \\frac{d}{dx} (\\csc x) = -\\csc x \\cot x [\/latex]<\/li>\r\n \t<li>[latex] \\frac{d}{dx} (\\tan x) = \\sec^2 x [\/latex]<\/li>\r\n \t<li>[latex] \\frac{d}{dx} (\\cot x) = -\\csc^2 x [\/latex]<\/li>\r\n \t<li>[latex] \\frac{d}{dx} (e^x) = e^x [\/latex]<\/li>\r\n \t<li>[latex] \\frac{d}{dx} (\\ln x) = \\frac{1}{x} [\/latex]<\/li>\r\n \t<li>[latex] \\frac{d}{dx} (\\arctan x) = \\frac{1}{1+x^2} [\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Derivative Rules<\/h3>\r\n<strong>Constant Multiple Rule<\/strong>: [latex] \\frac{d}{dx} (cf(x)) = cf'(x) [\/latex]\r\n\r\n<strong>Sum and Difference Rule<\/strong>: [latex] \\frac{d}{dx} f(x) \\pm g(x)) = f'(x) \\pm g'(x) [\/latex]\r\n\r\n<strong>Product Rule<\/strong>: [latex] \\frac{d}{dx} (f(x)g(x)) = g(x)f'(x) + f(x)g'(x) [\/latex]\r\n\r\n<strong>Quotient Rule<\/strong>: [latex] \\frac{d}{dx} \\left( \\frac{f(x)}{g(x)} \\right) = \\frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2} [\/latex]\r\n\r\n<strong>Chain Rule<\/strong>: [latex] \\frac{d}{dx} f(g(x) = f'(g(x)) g'(x) [\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0calculating the derivative of vector-valued functions<\/h3>\r\nUse Differentiation of Vector-Valued Functions to calculate the derivative of each of the following functions.\r\n<ol>\r\n \t<li style=\"text-align: left;\">[latex]{\\bf{r}}\\,(t)=(6t+8)\\,{\\bf{i}}+(4t^{2}+2t-3){\\bf{j}}[\/latex]<\/li>\r\n \t<li>[latex]{\\bf{r}}\\,(t)=3\\cos{t\\,{\\bf{i}}}+4\\sin{t\\,{\\bf{j}}}[\/latex]<\/li>\r\n \t<li>[latex]{\\bf{r}}\\,(t)=e^{t}\\sin{t\\,{\\bf{i}}}+e^{t}\\cos{t{\\bf{j}}}-e^{2t}\\,{\\bf{k}}[\/latex]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167793361764\" class=\"exercise\">[reveal-answer q=\"fs-id1167794055156\"]Show Solution[\/reveal-answer]<\/div>\r\n[hidden-answer a=\"fs-id1167794055156\"]\r\n<div class=\"exercise\">\r\n\r\nWe use Differentiation of Vector-Valued Functions and what we know about differentiating functions of one variable.\r\n<ol>\r\n \t<li style=\"text-align: left;\">The first component of [latex]{\\bf{r}}\\,(t)=(6t+8)\\,{\\bf{i}}+(4t^{2}+2t-3)\\,{\\bf{j}}[\/latex] is [latex]f\\,(t)=6t+8[\/latex]. The second component is [latex]g\\,(t)=4t^{2}+2t-3[\/latex]. We have [latex]f'\\,(t)=6[\/latex] and [latex]g'\\,(t)=8t+2[\/latex], so the theorem gives [latex]{\\bf{r}}'\\,(t)=6\\,{\\bf{i}}+(8t+2)\\,{\\bf{j}}[\/latex]<\/li>\r\n \t<li>The first component is [latex]f\\,(t)=3\\cos{t}[\/latex] and the second component is [latex]g\\,(t)=4\\sin{t}[\/latex]. We have [latex]f'\\,(t)=-3\\sin{t}[\/latex] and [latex]g'\\,(t)=4\\cos{t}[\/latex], so we obtain [latex]{\\bf{r}}'\\,(t)=-3\\sin{t\\,{\\bf{i}}}+4\\cos{t\\,{\\bf{j}}}[\/latex]<\/li>\r\n \t<li>The first component of [latex]{\\bf{r}}\\,(t)=e^{t}\\sin{t\\,{\\bf{i}}}+e^{t}\\cos{t{\\bf{j}}}-e^{2t}\\,{\\bf{k}}[\/latex] is [latex]f\\,(t)=e^{t}\\sin{t}[\/latex], the second component is [latex]g\\,(t)=e^{t}\\cos{t}[\/latex], and the third component is [latex]h\\,(t)=-e^{2t}[\/latex]. We have [latex]f'\\,(t)=e^{t}(\\sin{t}+\\cos{t}),\\ g'\\,(t)=e^{t}(\\cos{t}-\\sin{t})[\/latex], and [latex]h'\\,(t)=-2e^{2t}[\/latex], so the theorem gives [latex]{\\bf{r}}'\\,(t)=e^{t}(\\sin{t}+\\cos{t})\\,{\\bf{i}}+e^{t}(\\cos{t}-\\sin{t})\\,{\\bf{j}}-2e^{2t}\\,{\\bf{k}}[\/latex]<\/li>\r\n<\/ol>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1167793957091\" class=\"textbook key-takeaways\">\r\n<h3>TRY IT<\/h3>\r\nCalculate 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<\/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=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\"><\/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>We can extend to vector-valued functions the properties of the derivative. In particular, the <span id=\"term122\" class=\"no-emphasis\" data-type=\"term\">constant multiple rule<\/span>, the <span id=\"term123\" class=\"no-emphasis\" data-type=\"term\">sum and difference rules<\/span>, the <span id=\"term124\" class=\"no-emphasis\" data-type=\"term\">product rule<\/span>, and the <span id=\"term125\" class=\"no-emphasis\" data-type=\"term\">chain rule<\/span> all extend to vector-valued functions. However, in the case of the product rule, there are actually three extensions: (1) for a real-valued function multiplied by a vector-valued function, (2) for the dot product of two vector-valued functions, and (3) for the cross product of two vector-valued functions.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Differentiation of Vector-Valued functions theorem<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]\\bf{r}[\/latex] and [latex]\\bf{u}[\/latex] be differentiable vector-valued functions of [latex]t[\/latex], let [latex]f[\/latex] be a differentiable real-valued function of\r\n\r\n[latex]\\begin{alignat}{2}\r\n\r\n\\hspace{5cm}\\text{1.}&amp; &amp;\\quad \\frac{d}{dt}\\big[c{\\bf{r}}\\,(t)\\big]&amp;=c{\\bf{r}}'\\,(t) &amp;\\quad &amp;\\text{Scalar multiple}\\\\\r\n\r\n\\text{2.}&amp; &amp;\\quad \\frac{d}{dt}\\big[{\\bf{r}}\\,(t)\\pm{\\bf{u}}\\,(t)\\big]&amp;={\\bf{r}}'\\,(t)\\pm{\\bf{u}}'\\,(t) &amp;\\quad &amp;\\text{Sum and difference}\\\\\r\n\r\n\\text{3.}&amp; &amp;\\quad \\frac{d}{dt}\\big[f\\,(t)\\,{\\bf{u}}\\,(t)\\big]&amp;=f'\\,(t)\\,{\\bf{u}}\\,(t)+f\\,(t)\\,{\\bf{u}}'\\,(t) &amp;\\quad &amp;\\text{Scalar product}\\\\\r\n\r\n\\text{4.}&amp; &amp;\\quad \\frac{d}{dt}\\big[{\\bf{r}}\\,(t)\\cdot{\\bf{u}}\\,(t)\\big]&amp;={\\bf{r}}'\\,(t)\\cdot{\\bf{u}}\\,(t)+{\\bf{r}}\\,(t)\\cdot{\\bf{u}}'\\,(t) &amp;\\quad &amp;\\text{Dot product}\\\\\r\n\r\n\\text{5.}&amp; &amp;\\quad \\frac{d}{dt}\\big[{\\bf{r}}\\,(t)\\times{\\bf{u}}\\,(t)\\big] &amp;={\\bf{r}}'\\,(t)\\times{\\bf{u}}\\,(t)+{\\bf{r}}\\,(t)\\times{\\bf{u}}'\\,(t) &amp;\\quad &amp;\\text{Cross product}\\\\\r\n\r\n\\text{6.}&amp; &amp;\\quad \\frac{d}{dt}\\big[{\\bf{r}}(f\\,(t))\\big]&amp;={\\bf{r}}'\\,(f\\,(t))\\cdot{f}'\\,(t) &amp;\\quad &amp;\\text{Chain rule}\\\\\r\n\r\n\\text{7.}&amp; &amp;\\quad {\\bf{r}}\\,(t)\\cdot{\\bf{r}}\\,(t) &amp;=c, \\text{ then }{\\bf{r}}\\,(t)\\cdot{\\bf{r}}'\\,(t)=0.\r\n\r\n\\end{alignat}[\/latex]\r\n\r\n<\/div>\r\n<h3>Proof<\/h3>\r\nThe proofs of the first two properties follow directly from the definition of the derivative of a vector-valued function. The third property can be derived from the first two properties, along with the product rule from the Introduction to Derivatives. Let [latex]{\\bf{u}}\\,(t)=g\\,(t)\\,{\\bf{i}}+h\\,(t)\\,{\\bf{j}}[\/latex]. Then\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{d}{dt}\\big[f\\,(t)\\,{\\bf{u}}\\,(t)\\big]} &amp; =\\hfill &amp; {\\frac{d}{dt}\\big[f\\,(t)(g\\,(t)\\,{\\bf{i}}+h\\,(t)\\,{\\bf{j}})\\big]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{d}{dt}\\big[f\\,(t)\\,g\\,(t)\\,{\\bf{i}}+f\\,(t)\\,h\\,(t)\\,{\\bf{j}}\\big]}\\hfill\\\\\\hfill &amp; =\\hfill &amp; {\\frac{d}{dt}\\big[f\\,(t)\\,g\\,(t)\\big]\\,{\\bf{i}}+\\frac{d}{dt}\\big[f\\,(t)\\,h\\,(t)\\big]\\,{\\bf{j}}}\\hfill\\\\\\hfill&amp; =\\hfill &amp; {\\big(f'\\,(t)\\,g\\,(t)+f\\,(t)\\,g'\\,(t)\\big)\\,{\\bf{i}}+\\big(f'\\,(t)\\,h\\,(t)+f\\,(t)\\,h'\\,(t)\\big)\\,{\\bf{j}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {f'\\,(t)\\,{\\bf{u}}\\,(t)+f\\,(t)\\,{\\bf{u}}'\\,(t)}\\hfill \\\\ \\hfill \\end{array}[\/latex].<\/div>\r\n&nbsp;\r\n\r\nTo prove property 4. Let [latex]{\\bf{r}}\\,(t)=f_{1}\\,(t)\\,{\\bf{i}}+g_{1}\\,(t)\\,{\\bf{j}}[\/latex] and [latex]{\\bf{u}}\\,(t)=f_{2}\\,(t)\\,{\\bf{i}}+g_{2}\\,(t)\\,{\\bf{j}}[\/latex]. Then\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{d}{dt}\\big[{\\bf{r}}\\,(t)\\cdot{\\bf{u}}\\,(t)\\big]} &amp; =\\hfill &amp; {\\frac{d}{dt}\\big[f_{1}\\,(t)\\,f_{2}\\,(t)+g_{1}\\,(t)\\,g_{2}\\,(t)\\big]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {f_{1}'\\,(t)f_{2}\\,(t)+f_{1}\\,(t)\\,f_{2}'\\,(t)+g_{1}'\\,(t)\\,g_{2}\\,(t)+g_{1}\\,(t)\\,g_{2}'\\,(t)}\\hfill\\\\\\hfill &amp; =\\hfill &amp; {f_{1}'\\,(t)f_{2}\\,(t)+g_{1}'\\,(t)\\,g_{2}\\,(t)+f_{1}\\,(t)\\,f_{2}'\\,(t)+g_{1}\\,(t)\\,g_{2}'\\,(t)}\\hfill\\\\\\hfill&amp; =\\hfill &amp; {\\big(f_{1}'{\\bf{i}}+g_{1}'{\\bf{j}}\\big)\\cdot(f_{2}{\\bf{i}}+g_{2}{\\bf{j}})+(f_{1}{\\bf{i}}+g_{1}{\\bf{j}})\\cdot(f_{2}'{\\bf{i}}+g_{2}'{\\bf{j}})} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {{\\bf{r}}'\\,(t)\\cdot{\\bf{u}}\\,(t)+{\\bf{r}}\\,(t)\\cdot{\\bf{u}}'\\,(t)}\\hfill \\\\ \\hfill \\end{array}[\/latex].<\/div>\r\n&nbsp;\r\n\r\nThe proof of property 5 is similar to that of property 4. Property 6 can be proved using the chain rule. Last, property 7 follows from property 4:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{d}{dt}\\big[{\\bf{r}}\\,(t)\\cdot{\\bf{r}}\\,(t)\\big]} &amp; =\\hfill &amp; {\\frac{d}{dt}[c]}\\hfill \\\\ \\hfill {{\\bf{r}}'\\,(t)\\cdot{\\bf{r}}\\,(t)+{\\bf{r}}\\,(t)\\cdot{\\bf{r}}'\\,(t)} &amp; =\\hfill &amp; {0} \\hfill\\\\\\hfill {2{\\bf{r}}\\,(t)\\cdot{\\bf{r}}'\\,(t)} &amp; =\\hfill &amp; {0} \\hfill\\\\\\hfill {{\\bf{r}}\\,(t)\\cdot{\\bf{r}}'\\,(t)} &amp; =\\hfill &amp; {0}.\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\nNow for some examples using these properties.\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Using the properties of derivatives of vector-valued functions<\/h3>\r\nGiven the vector-valued functions\r\n<p style=\"text-align: center;\">[latex]{\\bf{r}}\\,(t)=(6t+8)\\,{\\bf{i}}+(4t^{2}+2t-3)\\,{\\bf{j}}+5t\\,{\\bf{k}}[\/latex]<\/p>\r\nand\r\n<p style=\"text-align: center;\">[latex]{\\bf{u}}\\,(t)=(t^{2}-3)\\,{\\bf{i}}+(2t+4)\\,{\\bf{j}}+(t^{3}-3t)\\,{\\bf{k}}[\/latex],<\/p>\r\ncalculate each of the following derivatives using the properties of the derivative of vector-valued functions.\r\n<ol>\r\n \t<li>[latex]\\frac{d}{dt}\\big[{\\bf{r}}\\,(t)\\cdot{\\bf{u}}\\,(t)\\big][\/latex]<\/li>\r\n \t<li>[latex]\\frac{d}{dt}\\big[{\\bf{u}}\\,(t)\\times{\\bf{u}}'\\,(t)\\big][\/latex]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167793361764\" class=\"exercise\">[reveal-answer q=\"fs-id1167794055158\"]Show Solution[\/reveal-answer]<\/div>\r\n[hidden-answer a=\"fs-id1167794055158\"]\r\n<div class=\"exercise\">\r\n<ol>\r\n \t<li>We have [latex]{\\bf{r}}'\\,(t)=6\\,{\\bf{i}}+(8t+2)\\,{\\bf{j}}+5\\,{\\bf{k}}[\/latex] and [latex]{\\bf{u}}'\\,(t)=2t\\,{\\bf{i}}+2\\,{\\bf{j}}+(3t^{2}-3)\\,{\\bf{k}}[\/latex]. Therefore, according to property 4:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{d}{dt}\\big[{\\bf{r}}\\,(t)\\cdot{\\bf{u}}\\,(t)\\big]} &amp; =\\hfill &amp; {{\\bf{r}}'\\,(t)\\cdot{\\bf{u}}\\,(t)+{\\bf{r}}\\cdot{\\bf{u}}'\\,(t)}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {(6\\,{\\bf{i}}+(8t+2)\\,{\\bf{j}}+5\\,{\\bf{k}})\\cdot\\big((t^{2}-3)\\,{\\bf{i}}+(2t+4)\\,{\\bf{j}}+(t^{3}-3t)\\,{\\bf{k}}\\big) \\\\ +\\big((6t+8)\\,{\\bf{i}}+(4t^{2}+2t-3)\\,{\\bf{j}}+5t\\,{\\bf{k}}\\big)\\cdot\\big(2t\\,{\\bf{i}}+2\\,{\\bf{j}}+(3t^{2}-3)\\,{\\bf{k}}\\big)}\\hfill\\\\\\hfill &amp; =\\hfill &amp; {6(t^{2}-3)+(8t+2)(2t+4)+5(t^{3}-3t)+2t(6t+8)+2(4t^{2}+2t-3)+5t(3t^{2}-3)}\\hfill\\\\\\hfill&amp; =\\hfill &amp; {20t^{3}+42t^{2}+26t-16}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;<\/li>\r\n \t<li>First, we need to adapt property 5 for this problem:\r\n<div style=\"text-align: center;\">[latex]\\frac{d}{dt}\\big[{\\bf{u}}\\,(t)\\times{\\bf{u}}'\\,(t)\\big]={\\bf{u}}'\\,(t)\\times{\\bf{u}}'\\,(t)+{\\bf{u}}\\,(t)\\times{\\bf{u}}''\\,(t).[\/latex]<\/div>\r\n&nbsp;\r\n\r\nRecall that the cross product of any vector with itself is zero. Furthermore, [latex]{\\bf{u}}''\\,(t)[\/latex] represents the second derivative of [latex]{\\bf{u}}\\,(t)[\/latex]:\r\n<div style=\"text-align: center;\">[latex]{\\bf{u}}''\\,(t)=\\frac{d}{dt}[{\\bf{u}}'\\,(t)]=\\frac{d}{dt}[2t\\,{\\bf{i}}+2\\,{\\bf{j}}+(3t^{2}-3)\\,{\\bf{k}}]=2\\,{\\bf{i}}+6t\\,{\\bf{k}}.[\/latex]<\/div>\r\n&nbsp;\r\n\r\nTherefore,\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{d}{dt}\\big[{\\bf{u}}\\,(t)\\cdot{\\bf{u}}'\\,(t)\\big]} &amp; =\\hfill &amp; {0+\\big((t^{2}-3)\\,{\\bf{I}}+(2t+4)\\,{\\bf{j}}+(t^{3}-3t)\\,{\\bf{k}}\\big)\\times(2\\,{\\bf{I}}+6t\\,{\\bf{k}})}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\begin{vmatrix}{\\bf{i}}&amp;{\\bf{j}}&amp;{\\bf{k}}\\\\t^{2}-3&amp;2t+4&amp;t^{3}-3t\\\\2&amp;0&amp;6t\\end{vmatrix}}\\hfill\\\\\\hfill &amp; =\\hfill &amp; {6t\\,(2t+4)\\,{\\bf{i}}-\\big(6t\\,(t^{2}-3)-2(t^{3}-3t)\\big)\\,{\\bf{j}}-2\\,(2t-4)\\,{\\bf{k}}}\\hfill\\\\\\hfill&amp; =\\hfill &amp; {(12t^{2}+24t)\\,{\\bf{i}}+(12t-4t^{3})\\,{\\bf{j}}-(4t+8)\\,{\\bf{k}}}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>TRY IT<\/h3>\r\nGiven 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<\/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=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\"><\/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>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Write an expression for the derivative of a vector-valued function<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1169739028079\" class=\"\">Now that we have seen what a vector-valued function is and how to take its limit, the next step is to learn how to differentiate a vector-valued function. The definition of the derivative of a vector-valued function is nearly identical to the definition of a real-valued function of one variable. However, because the range of a vector-valued function consists of vectors, the same is true for the range of the derivative of a vector-valued function.<\/p>\n<div id=\"fs-id1169738955085\" class=\"ui-has-child-title\" data-type=\"note\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p>A <strong><span id=\"term108\" data-type=\"term\">derivative of a vector-valued function<\/span><\/strong>\u00a0[latex]{\\bf{r}}\\,(t)[\/latex] is<\/p>\n<div style=\"text-align: center;\">[latex]{\\bf{r}}'\\,(t)=\\displaystyle\\lim_{\\Delta{t}{\\to}0}\\frac{{\\bf{r}}(t+\\Delta{t})-{\\bf{r}}\\,(t)}{\\Delta{t}},[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>provided the limit exists. If [latex]{\\bf{r}}'\\,(t)[\/latex] exists, then [latex]{\\bf{r}}[\/latex] is differentiable at [latex]t[\/latex]. If [latex]{\\bf{r}}'\\,(t)[\/latex] exists for all [latex]t[\/latex] in an open interval [latex](a,\\ b)[\/latex], then [latex]{\\bf{r}}[\/latex] is differentiable over the interval [latex](a,\\ b)[\/latex]. For the function to be differentiable over the closed interval [latex][a,\\ b][\/latex], the following two limits must exist as well:<\/p>\n<div style=\"text-align: center;\">[latex]{\\bf{r}}'\\,(a)=\\displaystyle\\lim_{\\Delta{t}{\\to}0^{+}}\\frac{{\\bf{r}}\\,(a+\\Delta{t})-{\\bf{r}}\\,(a)}{\\Delta{t}}[\/latex] and [latex]{\\bf{r}}'\\,(b)=\\displaystyle\\lim_{\\Delta{t}{\\to}0^{-}}\\frac{{\\bf{r}}\\,(b+\\Delta{t})-{\\bf{r}}\\,(b)}{\\Delta{t}}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p id=\"fs-id1169738885200\" class=\"\">Many of the rules for calculating derivatives of real-valued functions can be applied to calculating the derivatives of vector-valued functions as well. Recall that the derivative of a real-valued function can be interpreted as the slope of a tangent line or the instantaneous rate of change of the function. The derivative of a vector-valued function can be understood to be an instantaneous rate of change as well; for example, when the function represents the position of an object at a given point in time, the derivative represents its velocity at that same point in time.<\/p>\n<p id=\"fs-id1169738822521\" class=\"\">We now demonstrate taking the derivative of a vector-valued function.<\/p>\n<div id=\"fs-id1167793900960\" class=\"textbook exercises\">\n<h3>Example:\u00a0Finding the derivative of a vector-valued function<\/h3>\n<p>Use the definition to calculate the derivative of the function<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{r}}\\,(t)=(3t+4)\\,{\\bf{i}}+(t^{2}-4t+3){\\bf{j}}[\/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-id1167794055154\">Show Solution<\/span><\/div>\n<div id=\"qfs-id1167794055154\" class=\"hidden-answer\" style=\"display: none\">\n<div class=\"exercise\">\n<p>Let&#8217;s use the equation from the above definition:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\bf{r}}'\\,(t) & =\\hfill & {\\displaystyle\\lim_{\\Delta{t}{\\to}0}\\frac{{\\bf{r}}(t+\\Delta{t})-{\\bf{r}}(t)}{\\Delta{t}}} \\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{\\Delta{t}{\\to}0}\\frac{\\big[(3(t+\\Delta{t})+4){\\bf{i}}+\\big({(t+\\Delta{t})}^{2}-4(t+\\Delta{t})+3\\big){\\bf{j}}\\big]-\\big[(3t+4){\\bf{i}}+(t^{2}-4t+3){\\bf{j}}\\big]}{\\Delta{t}}}\\hfill\\\\\\hfill & =\\hfill & {\\displaystyle\\lim_{\\Delta{t}{\\to}0}\\frac{(3t+3\\Delta{t}+4){\\bf{i}}-(3t+4){\\bf{i}}+(t^{2}+2t\\Delta{t}+(\\Delta{t})^{2}-4t-4\\Delta{t}+3){\\bf{j}}-(t^{2}-4t+3){\\bf{j}}}{\\Delta{t}}}\\hfill\\\\\\hfill& =\\hfill & {\\displaystyle\\lim_{\\Delta{t}{\\to}0}\\frac{(3\\Delta{t}){\\bf{i}}+\\big(2t\\Delta{t}+(\\Delta{t})^{2}-4\\Delta{t}\\big){\\bf{j}}}{\\Delta{t}}} \\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{\\Delta{t}{\\to}0}{(3\\,{\\bf{i}}+(2t+{\\Delta}t-4)\\,{\\bf{j}}}} \\hfill \\\\ \\hfill & =\\hfill & {3\\,{\\bf{i}}+(2t-4)\\,{\\bf{j}}}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793957091\" class=\"textbook key-takeaways\">\n<h3>TRY IT<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794055155\">Show Solution<\/span><\/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<\/div>\n<p>Notice that in the calculations in the example above, we could also obtain the answer by first calculating the derivative of each component function, then putting these derivatives back into the vector-valued function. This is always true for calculating the derivative of a vector-valued function, whether it is in two or three dimensions. We state this in the following theorem. The proof of this theorem follows directly from the definitions of the limit of a vector-valued function and the derivative of a vector-valued function.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Differentiation of Vector-Valued functions theorem<\/h3>\n<hr \/>\n<p>Let [latex]f[\/latex], [latex]g[\/latex], and [latex]h[\/latex] be differentiable functions of [latex]t[\/latex].<\/p>\n<ul>\n<li style=\"text-align: left;\">If [latex]{\\bf{r}}\\,(t)=f\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}[\/latex], then [latex]{\\bf{r}}'\\,(t)=f'\\,(t)\\,{\\bf{i}}+g'\\,(t)\\,{\\bf{j}}[\/latex].<\/li>\n<li>If\u00a0[latex]{\\bf{r}}\\,(t)=f\\,(t)\\,{\\bf{i}}+g\\,(t)\\,{\\bf{j}}+h\\,(t)\\,{\\bf{k}}[\/latex], then [latex]{\\bf{r}}'\\,(t)=f'\\,(t)\\,{\\bf{i}}+g'\\,(t)\\,{\\bf{j}}+h'\\,(t)\\,{\\bf{k}}[\/latex].<\/li>\n<\/ul>\n<\/div>\n<p>Since we will be using derivatives of common single-variable functions throughout this course, we briefly recall the derivatives of common functions, along with frequently-encountered derivative rules below.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Derivatives of common functions<\/h3>\n<ul>\n<li>[latex]\\frac{d}{dx} (x^n) = nx^{n-1}[\/latex]\u00a0 (Power Rule)<\/li>\n<li>[latex]\\frac{d}{dx} (\\sin x) = \\cos x[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx} (\\cos x) = -\\sin x[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx} (\\sec x) = \\sec x \\tan x[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx} (\\csc x) = -\\csc x \\cot x[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx} (\\tan x) = \\sec^2 x[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx} (\\cot x) = -\\csc^2 x[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx} (e^x) = e^x[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx} (\\ln x) = \\frac{1}{x}[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx} (\\arctan x) = \\frac{1}{1+x^2}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall: Derivative Rules<\/h3>\n<p><strong>Constant Multiple Rule<\/strong>: [latex]\\frac{d}{dx} (cf(x)) = cf'(x)[\/latex]<\/p>\n<p><strong>Sum and Difference Rule<\/strong>: [latex]\\frac{d}{dx} f(x) \\pm g(x)) = f'(x) \\pm g'(x)[\/latex]<\/p>\n<p><strong>Product Rule<\/strong>: [latex]\\frac{d}{dx} (f(x)g(x)) = g(x)f'(x) + f(x)g'(x)[\/latex]<\/p>\n<p><strong>Quotient Rule<\/strong>: [latex]\\frac{d}{dx} \\left( \\frac{f(x)}{g(x)} \\right) = \\frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}[\/latex]<\/p>\n<p><strong>Chain Rule<\/strong>: [latex]\\frac{d}{dx} f(g(x) = f'(g(x)) g'(x)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0calculating the derivative of vector-valued functions<\/h3>\n<p>Use Differentiation of Vector-Valued Functions to calculate the derivative of each of the following functions.<\/p>\n<ol>\n<li style=\"text-align: left;\">[latex]{\\bf{r}}\\,(t)=(6t+8)\\,{\\bf{i}}+(4t^{2}+2t-3){\\bf{j}}[\/latex]<\/li>\n<li>[latex]{\\bf{r}}\\,(t)=3\\cos{t\\,{\\bf{i}}}+4\\sin{t\\,{\\bf{j}}}[\/latex]<\/li>\n<li>[latex]{\\bf{r}}\\,(t)=e^{t}\\sin{t\\,{\\bf{i}}}+e^{t}\\cos{t{\\bf{j}}}-e^{2t}\\,{\\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-id1167794055156\">Show Solution<\/span><\/div>\n<div id=\"qfs-id1167794055156\" class=\"hidden-answer\" style=\"display: none\">\n<div class=\"exercise\">\n<p>We use Differentiation of Vector-Valued Functions and what we know about differentiating functions of one variable.<\/p>\n<ol>\n<li style=\"text-align: left;\">The first component of [latex]{\\bf{r}}\\,(t)=(6t+8)\\,{\\bf{i}}+(4t^{2}+2t-3)\\,{\\bf{j}}[\/latex] is [latex]f\\,(t)=6t+8[\/latex]. The second component is [latex]g\\,(t)=4t^{2}+2t-3[\/latex]. We have [latex]f'\\,(t)=6[\/latex] and [latex]g'\\,(t)=8t+2[\/latex], so the theorem gives [latex]{\\bf{r}}'\\,(t)=6\\,{\\bf{i}}+(8t+2)\\,{\\bf{j}}[\/latex]<\/li>\n<li>The first component is [latex]f\\,(t)=3\\cos{t}[\/latex] and the second component is [latex]g\\,(t)=4\\sin{t}[\/latex]. We have [latex]f'\\,(t)=-3\\sin{t}[\/latex] and [latex]g'\\,(t)=4\\cos{t}[\/latex], so we obtain [latex]{\\bf{r}}'\\,(t)=-3\\sin{t\\,{\\bf{i}}}+4\\cos{t\\,{\\bf{j}}}[\/latex]<\/li>\n<li>The first component of [latex]{\\bf{r}}\\,(t)=e^{t}\\sin{t\\,{\\bf{i}}}+e^{t}\\cos{t{\\bf{j}}}-e^{2t}\\,{\\bf{k}}[\/latex] is [latex]f\\,(t)=e^{t}\\sin{t}[\/latex], the second component is [latex]g\\,(t)=e^{t}\\cos{t}[\/latex], and the third component is [latex]h\\,(t)=-e^{2t}[\/latex]. We have [latex]f'\\,(t)=e^{t}(\\sin{t}+\\cos{t}),\\ g'\\,(t)=e^{t}(\\cos{t}-\\sin{t})[\/latex], and [latex]h'\\,(t)=-2e^{2t}[\/latex], so the theorem gives [latex]{\\bf{r}}'\\,(t)=e^{t}(\\sin{t}+\\cos{t})\\,{\\bf{i}}+e^{t}(\\cos{t}-\\sin{t})\\,{\\bf{j}}-2e^{2t}\\,{\\bf{k}}[\/latex]<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793957091\" class=\"textbook key-takeaways\">\n<h3>TRY IT<\/h3>\n<p>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 id=\"fs-id1167793361764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794055157\">Show Solution<\/span><\/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<\/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=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\"><\/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<p>We can extend to vector-valued functions the properties of the derivative. In particular, the <span id=\"term122\" class=\"no-emphasis\" data-type=\"term\">constant multiple rule<\/span>, the <span id=\"term123\" class=\"no-emphasis\" data-type=\"term\">sum and difference rules<\/span>, the <span id=\"term124\" class=\"no-emphasis\" data-type=\"term\">product rule<\/span>, and the <span id=\"term125\" class=\"no-emphasis\" data-type=\"term\">chain rule<\/span> all extend to vector-valued functions. However, in the case of the product rule, there are actually three extensions: (1) for a real-valued function multiplied by a vector-valued function, (2) for the dot product of two vector-valued functions, and (3) for the cross product of two vector-valued functions.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Differentiation of Vector-Valued functions theorem<\/h3>\n<hr \/>\n<p>Let [latex]\\bf{r}[\/latex] and [latex]\\bf{u}[\/latex] be differentiable vector-valued functions of [latex]t[\/latex], let [latex]f[\/latex] be a differentiable real-valued function of<\/p>\n<p>[latex]\\begin{alignat}{2}    \\hspace{5cm}\\text{1.}& &\\quad \\frac{d}{dt}\\big[c{\\bf{r}}\\,(t)\\big]&=c{\\bf{r}}'\\,(t) &\\quad &\\text{Scalar multiple}\\\\    \\text{2.}& &\\quad \\frac{d}{dt}\\big[{\\bf{r}}\\,(t)\\pm{\\bf{u}}\\,(t)\\big]&={\\bf{r}}'\\,(t)\\pm{\\bf{u}}'\\,(t) &\\quad &\\text{Sum and difference}\\\\    \\text{3.}& &\\quad \\frac{d}{dt}\\big[f\\,(t)\\,{\\bf{u}}\\,(t)\\big]&=f'\\,(t)\\,{\\bf{u}}\\,(t)+f\\,(t)\\,{\\bf{u}}'\\,(t) &\\quad &\\text{Scalar product}\\\\    \\text{4.}& &\\quad \\frac{d}{dt}\\big[{\\bf{r}}\\,(t)\\cdot{\\bf{u}}\\,(t)\\big]&={\\bf{r}}'\\,(t)\\cdot{\\bf{u}}\\,(t)+{\\bf{r}}\\,(t)\\cdot{\\bf{u}}'\\,(t) &\\quad &\\text{Dot product}\\\\    \\text{5.}& &\\quad \\frac{d}{dt}\\big[{\\bf{r}}\\,(t)\\times{\\bf{u}}\\,(t)\\big] &={\\bf{r}}'\\,(t)\\times{\\bf{u}}\\,(t)+{\\bf{r}}\\,(t)\\times{\\bf{u}}'\\,(t) &\\quad &\\text{Cross product}\\\\    \\text{6.}& &\\quad \\frac{d}{dt}\\big[{\\bf{r}}(f\\,(t))\\big]&={\\bf{r}}'\\,(f\\,(t))\\cdot{f}'\\,(t) &\\quad &\\text{Chain rule}\\\\    \\text{7.}& &\\quad {\\bf{r}}\\,(t)\\cdot{\\bf{r}}\\,(t) &=c, \\text{ then }{\\bf{r}}\\,(t)\\cdot{\\bf{r}}'\\,(t)=0.    \\end{alignat}[\/latex]<\/p>\n<\/div>\n<h3>Proof<\/h3>\n<p>The proofs of the first two properties follow directly from the definition of the derivative of a vector-valued function. The third property can be derived from the first two properties, along with the product rule from the Introduction to Derivatives. Let [latex]{\\bf{u}}\\,(t)=g\\,(t)\\,{\\bf{i}}+h\\,(t)\\,{\\bf{j}}[\/latex]. Then<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{d}{dt}\\big[f\\,(t)\\,{\\bf{u}}\\,(t)\\big]} & =\\hfill & {\\frac{d}{dt}\\big[f\\,(t)(g\\,(t)\\,{\\bf{i}}+h\\,(t)\\,{\\bf{j}})\\big]}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{d}{dt}\\big[f\\,(t)\\,g\\,(t)\\,{\\bf{i}}+f\\,(t)\\,h\\,(t)\\,{\\bf{j}}\\big]}\\hfill\\\\\\hfill & =\\hfill & {\\frac{d}{dt}\\big[f\\,(t)\\,g\\,(t)\\big]\\,{\\bf{i}}+\\frac{d}{dt}\\big[f\\,(t)\\,h\\,(t)\\big]\\,{\\bf{j}}}\\hfill\\\\\\hfill& =\\hfill & {\\big(f'\\,(t)\\,g\\,(t)+f\\,(t)\\,g'\\,(t)\\big)\\,{\\bf{i}}+\\big(f'\\,(t)\\,h\\,(t)+f\\,(t)\\,h'\\,(t)\\big)\\,{\\bf{j}}} \\hfill \\\\ \\hfill & =\\hfill & {f'\\,(t)\\,{\\bf{u}}\\,(t)+f\\,(t)\\,{\\bf{u}}'\\,(t)}\\hfill \\\\ \\hfill \\end{array}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>To prove property 4. Let [latex]{\\bf{r}}\\,(t)=f_{1}\\,(t)\\,{\\bf{i}}+g_{1}\\,(t)\\,{\\bf{j}}[\/latex] and [latex]{\\bf{u}}\\,(t)=f_{2}\\,(t)\\,{\\bf{i}}+g_{2}\\,(t)\\,{\\bf{j}}[\/latex]. Then<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{d}{dt}\\big[{\\bf{r}}\\,(t)\\cdot{\\bf{u}}\\,(t)\\big]} & =\\hfill & {\\frac{d}{dt}\\big[f_{1}\\,(t)\\,f_{2}\\,(t)+g_{1}\\,(t)\\,g_{2}\\,(t)\\big]}\\hfill \\\\ \\hfill & =\\hfill & {f_{1}'\\,(t)f_{2}\\,(t)+f_{1}\\,(t)\\,f_{2}'\\,(t)+g_{1}'\\,(t)\\,g_{2}\\,(t)+g_{1}\\,(t)\\,g_{2}'\\,(t)}\\hfill\\\\\\hfill & =\\hfill & {f_{1}'\\,(t)f_{2}\\,(t)+g_{1}'\\,(t)\\,g_{2}\\,(t)+f_{1}\\,(t)\\,f_{2}'\\,(t)+g_{1}\\,(t)\\,g_{2}'\\,(t)}\\hfill\\\\\\hfill& =\\hfill & {\\big(f_{1}'{\\bf{i}}+g_{1}'{\\bf{j}}\\big)\\cdot(f_{2}{\\bf{i}}+g_{2}{\\bf{j}})+(f_{1}{\\bf{i}}+g_{1}{\\bf{j}})\\cdot(f_{2}'{\\bf{i}}+g_{2}'{\\bf{j}})} \\hfill \\\\ \\hfill & =\\hfill & {{\\bf{r}}'\\,(t)\\cdot{\\bf{u}}\\,(t)+{\\bf{r}}\\,(t)\\cdot{\\bf{u}}'\\,(t)}\\hfill \\\\ \\hfill \\end{array}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>The proof of property 5 is similar to that of property 4. Property 6 can be proved using the chain rule. Last, property 7 follows from property 4:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{d}{dt}\\big[{\\bf{r}}\\,(t)\\cdot{\\bf{r}}\\,(t)\\big]} & =\\hfill & {\\frac{d}{dt}[c]}\\hfill \\\\ \\hfill {{\\bf{r}}'\\,(t)\\cdot{\\bf{r}}\\,(t)+{\\bf{r}}\\,(t)\\cdot{\\bf{r}}'\\,(t)} & =\\hfill & {0} \\hfill\\\\\\hfill {2{\\bf{r}}\\,(t)\\cdot{\\bf{r}}'\\,(t)} & =\\hfill & {0} \\hfill\\\\\\hfill {{\\bf{r}}\\,(t)\\cdot{\\bf{r}}'\\,(t)} & =\\hfill & {0}.\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<p>Now for some examples using these properties.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Using the properties of derivatives of vector-valued functions<\/h3>\n<p>Given the vector-valued functions<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{r}}\\,(t)=(6t+8)\\,{\\bf{i}}+(4t^{2}+2t-3)\\,{\\bf{j}}+5t\\,{\\bf{k}}[\/latex]<\/p>\n<p>and<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{u}}\\,(t)=(t^{2}-3)\\,{\\bf{i}}+(2t+4)\\,{\\bf{j}}+(t^{3}-3t)\\,{\\bf{k}}[\/latex],<\/p>\n<p>calculate each of the following derivatives using the properties of the derivative of vector-valued functions.<\/p>\n<ol>\n<li>[latex]\\frac{d}{dt}\\big[{\\bf{r}}\\,(t)\\cdot{\\bf{u}}\\,(t)\\big][\/latex]<\/li>\n<li>[latex]\\frac{d}{dt}\\big[{\\bf{u}}\\,(t)\\times{\\bf{u}}'\\,(t)\\big][\/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-id1167794055158\">Show Solution<\/span><\/div>\n<div id=\"qfs-id1167794055158\" class=\"hidden-answer\" style=\"display: none\">\n<div class=\"exercise\">\n<ol>\n<li>We have [latex]{\\bf{r}}'\\,(t)=6\\,{\\bf{i}}+(8t+2)\\,{\\bf{j}}+5\\,{\\bf{k}}[\/latex] and [latex]{\\bf{u}}'\\,(t)=2t\\,{\\bf{i}}+2\\,{\\bf{j}}+(3t^{2}-3)\\,{\\bf{k}}[\/latex]. Therefore, according to property 4:\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{d}{dt}\\big[{\\bf{r}}\\,(t)\\cdot{\\bf{u}}\\,(t)\\big]} & =\\hfill & {{\\bf{r}}'\\,(t)\\cdot{\\bf{u}}\\,(t)+{\\bf{r}}\\cdot{\\bf{u}}'\\,(t)}\\hfill \\\\ \\hfill & =\\hfill & {(6\\,{\\bf{i}}+(8t+2)\\,{\\bf{j}}+5\\,{\\bf{k}})\\cdot\\big((t^{2}-3)\\,{\\bf{i}}+(2t+4)\\,{\\bf{j}}+(t^{3}-3t)\\,{\\bf{k}}\\big) \\\\ +\\big((6t+8)\\,{\\bf{i}}+(4t^{2}+2t-3)\\,{\\bf{j}}+5t\\,{\\bf{k}}\\big)\\cdot\\big(2t\\,{\\bf{i}}+2\\,{\\bf{j}}+(3t^{2}-3)\\,{\\bf{k}}\\big)}\\hfill\\\\\\hfill & =\\hfill & {6(t^{2}-3)+(8t+2)(2t+4)+5(t^{3}-3t)+2t(6t+8)+2(4t^{2}+2t-3)+5t(3t^{2}-3)}\\hfill\\\\\\hfill& =\\hfill & {20t^{3}+42t^{2}+26t-16}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/li>\n<li>First, we need to adapt property 5 for this problem:\n<div style=\"text-align: center;\">[latex]\\frac{d}{dt}\\big[{\\bf{u}}\\,(t)\\times{\\bf{u}}'\\,(t)\\big]={\\bf{u}}'\\,(t)\\times{\\bf{u}}'\\,(t)+{\\bf{u}}\\,(t)\\times{\\bf{u}}''\\,(t).[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Recall that the cross product of any vector with itself is zero. Furthermore, [latex]{\\bf{u}}''\\,(t)[\/latex] represents the second derivative of [latex]{\\bf{u}}\\,(t)[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]{\\bf{u}}''\\,(t)=\\frac{d}{dt}[{\\bf{u}}'\\,(t)]=\\frac{d}{dt}[2t\\,{\\bf{i}}+2\\,{\\bf{j}}+(3t^{2}-3)\\,{\\bf{k}}]=2\\,{\\bf{i}}+6t\\,{\\bf{k}}.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Therefore,<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{d}{dt}\\big[{\\bf{u}}\\,(t)\\cdot{\\bf{u}}'\\,(t)\\big]} & =\\hfill & {0+\\big((t^{2}-3)\\,{\\bf{I}}+(2t+4)\\,{\\bf{j}}+(t^{3}-3t)\\,{\\bf{k}}\\big)\\times(2\\,{\\bf{I}}+6t\\,{\\bf{k}})}\\hfill \\\\ \\hfill & =\\hfill & {\\begin{vmatrix}{\\bf{i}}&{\\bf{j}}&{\\bf{k}}\\\\t^{2}-3&2t+4&t^{3}-3t\\\\2&0&6t\\end{vmatrix}}\\hfill\\\\\\hfill & =\\hfill & {6t\\,(2t+4)\\,{\\bf{i}}-\\big(6t\\,(t^{2}-3)-2(t^{3}-3t)\\big)\\,{\\bf{j}}-2\\,(2t-4)\\,{\\bf{k}}}\\hfill\\\\\\hfill& =\\hfill & {(12t^{2}+24t)\\,{\\bf{i}}+(12t-4t^{3})\\,{\\bf{j}}-(4t+8)\\,{\\bf{k}}}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>TRY IT<\/h3>\n<p>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 id=\"fs-id1167793361764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794055159\">Show Solution<\/span><\/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<\/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=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\"><\/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\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-3856\">\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 3.5. <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><li>CP 3.6. <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":7,"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 3.5\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"CP 3.6\",\"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-3856","chapter","type-chapter","status-publish","hentry"],"part":21,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3856","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":13,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3856\/revisions"}],"predecessor-version":[{"id":4803,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3856\/revisions\/4803"}],"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\/3856\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=3856"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=3856"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=3856"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=3856"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}