{"id":614,"date":"2025-10-09T15:25:32","date_gmt":"2025-10-09T15:25:32","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=614"},"modified":"2025-11-20T16:02:59","modified_gmt":"2025-11-20T16:02:59","slug":"advanced-partial-derivatives-background-youll-need-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/advanced-partial-derivatives-background-youll-need-2\/","title":{"raw":"Advanced Partial Derivatives: Background You'll Need 2","rendered":"Advanced Partial Derivatives: Background You&#8217;ll Need 2"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Calculate the dot product of two given vectors<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>The Dot Product<\/h2>\r\n<p class=\"whitespace-normal break-words\">The dot product is a new way to \"multiply\" vectors, but unlike vector addition and scalar multiplication, the result is a scalar (a number), not a vector.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>dot product<\/h3>\r\n<p class=\"whitespace-normal break-words\">For vectors [latex]\\mathbf{u} = \\langle u_1, u_2, u_3 \\rangle[\/latex] and [latex]\\mathbf{v} = \\langle v_1, v_2, v_3 \\rangle[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{u} \\cdot \\mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">For 2D vectors [latex]\\mathbf{u} = \\langle u_1, u_2 \\rangle[\/latex] and [latex]\\mathbf{v} = \\langle v_1, v_2 \\rangle[\/latex]:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\mathbf{u} \\cdot \\mathbf{v} = u_1v_1 + u_2v_2[\/latex]<\/p>\r\n\r\n<\/section>\r\n<p id=\"fs-id1163724101618\">The dot product is also called the <strong>scalar product<\/strong> (because the result is a scalar) or the <strong>inner product<\/strong>.<\/p>\r\n\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Multiply corresponding components and add the results. For [latex]\\langle 2,3,1 \\rangle \\cdot \\langle 4,-1,5 \\rangle[\/latex]:\r\n<center>[latex](2)(4) + (3)(-1) + (1)(5) = 8 - 3 + 5 = 10[\/latex]<\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<ol id=\"fs-id1163724076122\" type=\"a\">\r\n \t<li>Find the dot product of [latex]\\mathbf{u} = \\langle 3,5,2 \\rangle[\/latex] and\u00a0[latex]\\mathbf{v} =\\langle -1,3,0 \\rangle[\/latex].<\/li>\r\n \t<li>Find the scalar product of [latex]\\mathbf{p} = 10\\mathbf{i} - 4\\mathbf{j} +7 \\mathbf{k}[\/latex] and\u00a0[latex]\\mathbf{q} = -2\\mathbf{i} + \\mathbf{j} + 6\\mathbf{k}[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"471238733\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"471238733\"]\r\n<ol id=\"fs-id1163723802236\" type=\"a\">\r\n \t<li>Substitute the vector components into the formula for the dot product:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1163723138109\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{u \\cdot v} &amp;= u_1v_1 + u_2v_2 + u_3v_3 \\\\\r\n&amp;= 3(-1)+5(3)+2(0)= -3+15+0 = 12.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center><\/div><\/li>\r\n \t<li>The calculation is the same if the vectors are written using standard unit vectors. We still have three components for each vector to substitute into the formula for the dot product:\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{p \\cdot q} &amp;= u_1v_1 + u_2v_2 + u_3v_3 \\\\\r\n&amp;= 10(-2)+(-4)(1)+7(6)= -20-4+42 = 18.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]26041[\/ohm_question]<\/section>\r\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Properties of the Dot Product<\/h3>\r\n<p class=\"whitespace-normal break-words\">The dot product follows familiar algebraic rules, making it easy to work with in calculations.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>properties of the dot product<\/h3>\r\n<p class=\"whitespace-normal break-words\">Let [latex]\\mathbf{u}[\/latex], [latex]\\mathbf{v}[\/latex], and [latex]\\mathbf{w}[\/latex] be vectors, and 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>Commutative<\/strong>: [latex]\\mathbf{u} \\cdot \\mathbf{v} = \\mathbf{v} \\cdot \\mathbf{u}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Distributive<\/strong>: [latex]\\mathbf{u} \\cdot (\\mathbf{v} + \\mathbf{w}) = \\mathbf{u} \\cdot \\mathbf{v} + \\mathbf{u} \\cdot \\mathbf{w}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Associative with scalars<\/strong>: [latex]c(\\mathbf{u} \\cdot \\mathbf{v}) = (c\\mathbf{u}) \\cdot \\mathbf{v} = \\mathbf{u} \\cdot (c\\mathbf{v})[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Magnitude relationship<\/strong>: [latex]\\mathbf{v} \\cdot \\mathbf{v} = ||\\mathbf{v}||^2[\/latex]<\/li>\r\n<\/ol>\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">From the definition and properties, we can make several key observations:<\/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]\\mathbf{0} \\cdot \\mathbf{v} = 0[\/latex] for any vector [latex]\\mathbf{v}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]\\mathbf{v} \\cdot \\mathbf{v} = 0[\/latex], then [latex]\\mathbf{v} = \\mathbf{0}[\/latex] (only the zero vector has dot product zero with itself)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">The dot product of a vector with itself gives the square of its magnitude<\/li>\r\n<\/ul>\r\n<section class=\"textbox recall\" aria-label=\"Recall\">Remember that [latex]||\\mathbf{v}|| = \\sqrt{v_1^2 + v_2^2 + v_3^2}[\/latex], so [latex]||\\mathbf{v}||^2 = v_1^2 + v_2^2 + v_3^2 = \\mathbf{v} \\cdot \\mathbf{v}[\/latex].<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1163724075617\">Let [latex]\\mathbf{a} = \\langle 1,2,-3 \\rangle[\/latex], [latex]\\mathbf{b} = \\langle 0,2,4 \\rangle[\/latex], and [latex]\\mathbf{c} = \\langle 5,-1,3 \\rangle[\/latex]. Find each of the following products.<\/p>\r\n\r\n<ol id=\"fs-id1163724078047\" type=\"a\">\r\n \t<li>[latex](\\mathbf{a} \\cdot \\mathbf{b})\\mathbf{c} [\/latex]<\/li>\r\n \t<li>[latex]\\mathbf{a}\\cdot (2\\mathbf{c} )[\/latex]<\/li>\r\n \t<li>[latex]||\\mathbf{b}||^2[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"235187363\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"235187363\"]\r\n<ol id=\"fs-id1163723139665\" type=\"a\">\r\n \t<li>Note that this expression asks for the scalar multiple of\u00a0 [latex]\\mathbf{c}[\/latex]\u00a0by [latex]\\mathbf{a} \\cdot \\mathbf{b}[\/latex]:<center>\r\n[latex]\r\n\\begin{align*}\r\n(\\mathbf{a} \\cdot \\mathbf{b})\\mathbf{c} &amp;= (\\langle 1,2,-3 \\rangle \\cdot \\langle 0,2,4 \\rangle) \\langle 5,-1,3 \\rangle\\\\\r\n&amp;= (1(0)+2(2)+(-3)(4))\\langle 5,-1,3 \\rangle\\\\\r\n&amp;= -8\\langle 5,-1,3 \\rangle\\\\\r\n&amp;= \\langle -40,8,-24 \\rangle.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center><\/li>\r\n \t<li>This expression is a dot product of vector\u00a0[latex]\\mathbf{a}[\/latex]\u00a0and scalar multiple 2[latex]\\mathbf{c}[\/latex]:\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{a}\\cdot (2\\mathbf{c} ) &amp;= 2(\\mathbf{a}\\cdot \\mathbf{c})\\\\\r\n&amp;= 2(\\langle 1,2,-3 \\rangle \\cdot\\langle 5,-1,3 \\rangle)\\\\\r\n&amp;= 2(1(5) +2(-1)+(-3)(3))\\\\\r\n&amp;= 2(-6) = -12.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center><\/li>\r\n \t<li>Simplifying this expression is a straightforward application of the dot product:\r\n<center>[latex]||\\mathbf{b}||^2 = \\mathbf{b} \\cdot \\mathbf{b} = \\langle 0,2,4 \\rangle\\cdot\\langle 0,2,4 \\rangle = 0^2+2^2 +4^2 = 0+4+16 = 20 [\/latex].<\/center><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li><span data-sheets-root=\"1\">Calculate the dot product of two given vectors<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>The Dot Product<\/h2>\n<p class=\"whitespace-normal break-words\">The dot product is a new way to &#8220;multiply&#8221; vectors, but unlike vector addition and scalar multiplication, the result is a scalar (a number), not a vector.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>dot product<\/h3>\n<p class=\"whitespace-normal break-words\">For vectors [latex]\\mathbf{u} = \\langle u_1, u_2, u_3 \\rangle[\/latex] and [latex]\\mathbf{v} = \\langle v_1, v_2, v_3 \\rangle[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{u} \\cdot \\mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">For 2D vectors [latex]\\mathbf{u} = \\langle u_1, u_2 \\rangle[\/latex] and [latex]\\mathbf{v} = \\langle v_1, v_2 \\rangle[\/latex]:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\mathbf{u} \\cdot \\mathbf{v} = u_1v_1 + u_2v_2[\/latex]<\/p>\n<\/section>\n<p id=\"fs-id1163724101618\">The dot product is also called the <strong>scalar product<\/strong> (because the result is a scalar) or the <strong>inner product<\/strong>.<\/p>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Multiply corresponding components and add the results. For [latex]\\langle 2,3,1 \\rangle \\cdot \\langle 4,-1,5 \\rangle[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex](2)(4) + (3)(-1) + (1)(5) = 8 - 3 + 5 = 10[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<ol id=\"fs-id1163724076122\" type=\"a\">\n<li>Find the dot product of [latex]\\mathbf{u} = \\langle 3,5,2 \\rangle[\/latex] and\u00a0[latex]\\mathbf{v} =\\langle -1,3,0 \\rangle[\/latex].<\/li>\n<li>Find the scalar product of [latex]\\mathbf{p} = 10\\mathbf{i} - 4\\mathbf{j} +7 \\mathbf{k}[\/latex] and\u00a0[latex]\\mathbf{q} = -2\\mathbf{i} + \\mathbf{j} + 6\\mathbf{k}[\/latex].<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q471238733\">Show Solution<\/button><\/p>\n<div id=\"q471238733\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1163723802236\" type=\"a\">\n<li>Substitute the vector components into the formula for the dot product:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1163723138109\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{u \\cdot v} &= u_1v_1 + u_2v_2 + u_3v_3 \\\\  &= 3(-1)+5(3)+2(0)= -3+15+0 = 12.\\\\  \\end{align*}[\/latex]<\/div>\n<\/div>\n<\/li>\n<li>The calculation is the same if the vectors are written using standard unit vectors. We still have three components for each vector to substitute into the formula for the dot product:\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{p \\cdot q} &= u_1v_1 + u_2v_2 + u_3v_3 \\\\  &= 10(-2)+(-4)(1)+7(6)= -20-4+42 = 18.\\\\  \\end{align*}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm26041\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=26041&theme=lumen&iframe_resize_id=ohm26041&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Properties of the Dot Product<\/h3>\n<p class=\"whitespace-normal break-words\">The dot product follows familiar algebraic rules, making it easy to work with in calculations.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>properties of the dot product<\/h3>\n<p class=\"whitespace-normal break-words\">Let [latex]\\mathbf{u}[\/latex], [latex]\\mathbf{v}[\/latex], and [latex]\\mathbf{w}[\/latex] be vectors, and 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>Commutative<\/strong>: [latex]\\mathbf{u} \\cdot \\mathbf{v} = \\mathbf{v} \\cdot \\mathbf{u}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Distributive<\/strong>: [latex]\\mathbf{u} \\cdot (\\mathbf{v} + \\mathbf{w}) = \\mathbf{u} \\cdot \\mathbf{v} + \\mathbf{u} \\cdot \\mathbf{w}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Associative with scalars<\/strong>: [latex]c(\\mathbf{u} \\cdot \\mathbf{v}) = (c\\mathbf{u}) \\cdot \\mathbf{v} = \\mathbf{u} \\cdot (c\\mathbf{v})[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Magnitude relationship<\/strong>: [latex]\\mathbf{v} \\cdot \\mathbf{v} = ||\\mathbf{v}||^2[\/latex]<\/li>\n<\/ol>\n<\/section>\n<p class=\"whitespace-normal break-words\">From the definition and properties, we can make several key observations:<\/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]\\mathbf{0} \\cdot \\mathbf{v} = 0[\/latex] for any vector [latex]\\mathbf{v}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">If [latex]\\mathbf{v} \\cdot \\mathbf{v} = 0[\/latex], then [latex]\\mathbf{v} = \\mathbf{0}[\/latex] (only the zero vector has dot product zero with itself)<\/li>\n<li class=\"whitespace-normal break-words\">The dot product of a vector with itself gives the square of its magnitude<\/li>\n<\/ul>\n<section class=\"textbox recall\" aria-label=\"Recall\">Remember that [latex]||\\mathbf{v}|| = \\sqrt{v_1^2 + v_2^2 + v_3^2}[\/latex], so [latex]||\\mathbf{v}||^2 = v_1^2 + v_2^2 + v_3^2 = \\mathbf{v} \\cdot \\mathbf{v}[\/latex].<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1163724075617\">Let [latex]\\mathbf{a} = \\langle 1,2,-3 \\rangle[\/latex], [latex]\\mathbf{b} = \\langle 0,2,4 \\rangle[\/latex], and [latex]\\mathbf{c} = \\langle 5,-1,3 \\rangle[\/latex]. Find each of the following products.<\/p>\n<ol id=\"fs-id1163724078047\" type=\"a\">\n<li>[latex](\\mathbf{a} \\cdot \\mathbf{b})\\mathbf{c}[\/latex]<\/li>\n<li>[latex]\\mathbf{a}\\cdot (2\\mathbf{c} )[\/latex]<\/li>\n<li>[latex]||\\mathbf{b}||^2[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q235187363\">Show Solution<\/button><\/p>\n<div id=\"q235187363\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1163723139665\" type=\"a\">\n<li>Note that this expression asks for the scalar multiple of\u00a0 [latex]\\mathbf{c}[\/latex]\u00a0by [latex]\\mathbf{a} \\cdot \\mathbf{b}[\/latex]:\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  (\\mathbf{a} \\cdot \\mathbf{b})\\mathbf{c} &= (\\langle 1,2,-3 \\rangle \\cdot \\langle 0,2,4 \\rangle) \\langle 5,-1,3 \\rangle\\\\  &= (1(0)+2(2)+(-3)(4))\\langle 5,-1,3 \\rangle\\\\  &= -8\\langle 5,-1,3 \\rangle\\\\  &= \\langle -40,8,-24 \\rangle.\\\\  \\end{align*}[\/latex]<\/div>\n<\/li>\n<li>This expression is a dot product of vector\u00a0[latex]\\mathbf{a}[\/latex]\u00a0and scalar multiple 2[latex]\\mathbf{c}[\/latex]:\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{a}\\cdot (2\\mathbf{c} ) &= 2(\\mathbf{a}\\cdot \\mathbf{c})\\\\  &= 2(\\langle 1,2,-3 \\rangle \\cdot\\langle 5,-1,3 \\rangle)\\\\  &= 2(1(5) +2(-1)+(-3)(3))\\\\  &= 2(-6) = -12.\\\\  \\end{align*}[\/latex]<\/div>\n<\/li>\n<li>Simplifying this expression is a straightforward application of the dot product:\n<div style=\"text-align: center;\">[latex]||\\mathbf{b}||^2 = \\mathbf{b} \\cdot \\mathbf{b} = \\langle 0,2,4 \\rangle\\cdot\\langle 0,2,4 \\rangle = 0^2+2^2 +4^2 = 0+4+16 = 20[\/latex].<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":3,"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-614","chapter","type-chapter","status-publish","hentry"],"part":428,"module-header":"background_you_need","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\/614","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":3,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/614\/revisions"}],"predecessor-version":[{"id":2647,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/614\/revisions\/2647"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/428"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/614\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=614"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=614"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=614"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=614"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}