{"id":560,"date":"2025-10-09T15:22:32","date_gmt":"2025-10-09T15:22:32","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=560"},"modified":"2025-10-22T16:26:19","modified_gmt":"2025-10-22T16:26:19","slug":"calculus-of-vector-valued-functions-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/calculus-of-vector-valued-functions-learn-it-1\/","title":{"raw":"Calculus of Vector-Valued Functions: Learn It 1","rendered":"Calculus of Vector-Valued Functions: Learn It 1"},"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<p class=\"whitespace-normal break-words\">To study the calculus of vector-valued functions, we follow a path similar to real-valued functions: define the derivative, examine applications, then move on to integrals. The vector nature of these functions brings some interesting new ideas along the way.<\/p>\r\n<p class=\"whitespace-normal break-words\">Now that you know what a vector-valued function is and how to take its limit, the next step is learning to differentiate it. The definition closely parallels the derivative of a real-valued function, but because the range consists of vectors, the derivative is also a vector.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>derivative of a vector-valued function<\/h3>\r\n<p class=\"whitespace-normal break-words\">The <strong>derivative<\/strong> of a vector-valued function [latex]{\\bf{r}}(t)[\/latex] is<\/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\">provided the limit exists. If [latex]{\\bf{r}}'(t)[\/latex] exists, then [latex]{\\bf{r}}[\/latex] is <strong>differentiable at [latex]t[\/latex]<\/strong>.<\/p>\r\n[latex]\\\\[\/latex]\r\n<p class=\"whitespace-normal break-words\">For [latex]{\\bf{r}}[\/latex] to be differentiable over a closed interval [latex][a, b][\/latex], these one-sided limits must also exist:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{\\bf{r}}'(a)=\\underset{\\Delta t\\to 0^{+}}{\\text{lim}}\\frac{{\\bf{r}}(a+\\Delta t)-{\\bf{r}}(a)}{\\Delta t}[\/latex] and [latex]{\\bf{r}}'(b)=\\underset{\\Delta t\\to 0^{-}}{\\text{lim}}\\frac{{\\bf{r}}(b+\\Delta t)-{\\bf{r}}(b)}{\\Delta t}[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">Just as with real-valued functions, the derivative of a vector-valued function represents an instantaneous rate of change. When the function represents the position of an object at time [latex]t[\/latex], the derivative represents its velocity at that same moment.\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Use 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<\/section>Calculating derivatives using the limit definition works, but it's tedious. Fortunately, we can differentiate component by component\u2014just like we handled limits.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>differentiation of vector-valued functions theorem<\/h3>\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<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">Differentiate each component separately using the derivative rules you already know from Calculus 1. The vector structure doesn't change\u2014only the components do.\r\n\r\n<\/section>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.\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">\r\n<p class=\"whitespace-normal break-words\"><strong>Derivatives of Common Functions<\/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\">[latex]\\frac{d}{dx} (x^n) = nx^{n-1}[\/latex] (Power Rule)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx} (\\sin x) = \\cos x[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx} (\\cos x) = -\\sin x[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx} (\\tan x) = \\sec^2 x[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx} (\\sec x) = \\sec x \\tan x[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx} (\\csc x) = -\\csc x \\cot x[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx} (\\cot x) = -\\csc^2 x[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx} (e^x) = e^x[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx} (\\ln x) = \\frac{1}{x}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx} (\\arctan x) = \\frac{1}{1+x^2}[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Derivative Rules<\/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\"><strong>Constant Multiple<\/strong>: [latex]\\frac{d}{dx} (cf(x)) = cf'(x)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Sum and Difference<\/strong>: [latex]\\frac{d}{dx} (f(x) \\pm g(x)) = f'(x) \\pm g'(x)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Product Rule<\/strong>: [latex]\\frac{d}{dx} (f(x)g(x)) = g(x)f'(x) + f(x)g'(x)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><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]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Chain Rule<\/strong>: [latex]\\frac{d}{dx} f(g(x)) = f'(g(x)) g'(x)[\/latex]<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Use 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<\/section>\r\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Properties of Derivatives<\/h3>\r\n<p class=\"whitespace-normal break-words\">The derivative properties you learned in Calculus 1 extend to vector-valued functions. The constant multiple rule, sum and difference rules, and chain rule all work the same way. The product rule has three versions because vectors can be multiplied in three different ways: scalar multiplication, dot product, and cross product.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>differentiation rules for vector-valued functions<\/h3>\r\n<p class=\"whitespace-normal break-words\">Let [latex]{\\bf{r}}[\/latex] and [latex]{\\bf{u}}[\/latex] be differentiable vector-valued functions of [latex]t[\/latex], and let [latex]f[\/latex] be a differentiable real-valued function of [latex]t[\/latex]. Let [latex]c[\/latex] be a scalar.<\/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}\\big[c{\\bf{r}},(t)\\big]=c{\\bf{r}}',(t)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Sum and difference<\/strong>: [latex]\\frac{d}{dt}\\big[{\\bf{r}},(t)\\pm{\\bf{u}},(t)\\big]={\\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}\\big[f,(t),{\\bf{u}},(t)\\big]=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}\\big[{\\bf{r}},(t)\\cdot{\\bf{u}},(t)\\big]={\\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}\\big[{\\bf{r}},(t)\\times{\\bf{u}},(t)\\big]={\\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}\\big[{\\bf{r}}(f,(t))\\big]={\\bf{r}}',(f,(t))\\cdot{f}',(t)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]{\\bf{r}},(t)\\cdot{\\bf{r}},(t)=c[\/latex], then [latex]{\\bf{r}},(t)\\cdot{\\bf{r}}',(t)=0[\/latex]<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox connectIt\">\r\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\r\n\r\n\r\n<hr \/>\r\n\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\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Given 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<\/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<p class=\"whitespace-normal break-words\">To study the calculus of vector-valued functions, we follow a path similar to real-valued functions: define the derivative, examine applications, then move on to integrals. The vector nature of these functions brings some interesting new ideas along the way.<\/p>\n<p class=\"whitespace-normal break-words\">Now that you know what a vector-valued function is and how to take its limit, the next step is learning to differentiate it. The definition closely parallels the derivative of a real-valued function, but because the range consists of vectors, the derivative is also a vector.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>derivative of a vector-valued function<\/h3>\n<p class=\"whitespace-normal break-words\">The <strong>derivative<\/strong> of a vector-valued function [latex]{\\bf{r}}(t)[\/latex] is<\/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\">provided the limit exists. If [latex]{\\bf{r}}'(t)[\/latex] exists, then [latex]{\\bf{r}}[\/latex] is <strong>differentiable at [latex]t[\/latex]<\/strong>.<\/p>\n<p>[latex]\\\\[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">For [latex]{\\bf{r}}[\/latex] to be differentiable over a closed interval [latex][a, b][\/latex], these one-sided limits must also exist:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{\\bf{r}}'(a)=\\underset{\\Delta t\\to 0^{+}}{\\text{lim}}\\frac{{\\bf{r}}(a+\\Delta t)-{\\bf{r}}(a)}{\\Delta t}[\/latex] and [latex]{\\bf{r}}'(b)=\\underset{\\Delta t\\to 0^{-}}{\\text{lim}}\\frac{{\\bf{r}}(b+\\Delta t)-{\\bf{r}}(b)}{\\Delta t}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Just as with real-valued functions, the derivative of a vector-valued function represents an instantaneous rate of change. When the function represents the position of an object at time [latex]t[\/latex], the derivative represents its velocity at that same moment.<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794055154\">Show Solution<\/button><\/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<\/section>\n<p>Calculating derivatives using the limit definition works, but it&#8217;s tedious. Fortunately, we can differentiate component by component\u2014just like we handled limits.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>differentiation of vector-valued functions theorem<\/h3>\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<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Differentiate each component separately using the derivative rules you already know from Calculus 1. The vector structure doesn&#8217;t change\u2014only the components do.<\/p>\n<\/section>\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<section class=\"textbox recall\" aria-label=\"Recall\">\n<p class=\"whitespace-normal break-words\"><strong>Derivatives of Common Functions<\/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\">[latex]\\frac{d}{dx} (x^n) = nx^{n-1}[\/latex] (Power Rule)<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx} (\\sin x) = \\cos x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx} (\\cos x) = -\\sin x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx} (\\tan x) = \\sec^2 x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx} (\\sec x) = \\sec x \\tan x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx} (\\csc x) = -\\csc x \\cot x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx} (\\cot x) = -\\csc^2 x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx} (e^x) = e^x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx} (\\ln x) = \\frac{1}{x}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx} (\\arctan x) = \\frac{1}{1+x^2}[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Derivative Rules<\/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\"><strong>Constant Multiple<\/strong>: [latex]\\frac{d}{dx} (cf(x)) = cf'(x)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Sum and Difference<\/strong>: [latex]\\frac{d}{dx} (f(x) \\pm g(x)) = f'(x) \\pm g'(x)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Product Rule<\/strong>: [latex]\\frac{d}{dx} (f(x)g(x)) = g(x)f'(x) + f(x)g'(x)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><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]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Chain Rule<\/strong>: [latex]\\frac{d}{dx} f(g(x)) = f'(g(x)) g'(x)[\/latex]<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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 class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794055156\">Show Solution<\/button><\/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<\/section>\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Properties of Derivatives<\/h3>\n<p class=\"whitespace-normal break-words\">The derivative properties you learned in Calculus 1 extend to vector-valued functions. The constant multiple rule, sum and difference rules, and chain rule all work the same way. The product rule has three versions because vectors can be multiplied in three different ways: scalar multiplication, dot product, and cross product.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>differentiation rules for vector-valued functions<\/h3>\n<p class=\"whitespace-normal break-words\">Let [latex]{\\bf{r}}[\/latex] and [latex]{\\bf{u}}[\/latex] be differentiable vector-valued functions of [latex]t[\/latex], and let [latex]f[\/latex] be a differentiable real-valued function of [latex]t[\/latex]. Let [latex]c[\/latex] be a scalar.<\/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}\\big[c{\\bf{r}},(t)\\big]=c{\\bf{r}}',(t)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Sum and difference<\/strong>: [latex]\\frac{d}{dt}\\big[{\\bf{r}},(t)\\pm{\\bf{u}},(t)\\big]={\\bf{r}}',(t)\\pm{\\bf{u}}',(t)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Scalar product<\/strong>: [latex]\\frac{d}{dt}\\big[f,(t),{\\bf{u}},(t)\\big]=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}\\big[{\\bf{r}},(t)\\cdot{\\bf{u}},(t)\\big]={\\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}\\big[{\\bf{r}},(t)\\times{\\bf{u}},(t)\\big]={\\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}\\big[{\\bf{r}}(f,(t))\\big]={\\bf{r}}',(f,(t))\\cdot{f}',(t)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">If [latex]{\\bf{r}},(t)\\cdot{\\bf{r}},(t)=c[\/latex], then [latex]{\\bf{r}},(t)\\cdot{\\bf{r}}',(t)=0[\/latex]<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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 class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794055158\">Show Solution<\/button><\/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<\/section>\n","protected":false},"author":15,"menu_order":11,"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-560","chapter","type-chapter","status-publish","hentry"],"part":424,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/560","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":19,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/560\/revisions"}],"predecessor-version":[{"id":1700,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/560\/revisions\/1700"}],"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\/560\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=560"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=560"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=560"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=560"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}