{"id":861,"date":"2021-09-03T17:24:55","date_gmt":"2021-09-03T17:24:55","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=861"},"modified":"2022-10-29T01:11:19","modified_gmt":"2022-10-29T01:11:19","slug":"functions-of-more-than-two-variables","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/functions-of-more-than-two-variables\/","title":{"raw":"Functions of More Than Two Variables","rendered":"Functions of More Than Two Variables"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Recognize a function of three or more variables and identify its level surfaces.<\/li>\r\n<\/ul>\r\n<\/div>\r\nSo far, we have examined only functions of two variables. However, it is useful to take a brief look at functions of more than two variables. Two such examples are\r\n<p style=\"text-align: center;\">[latex]f\\,(x,\\ y,\\ z)=x^{2}-2xy+y^{2}+3yz-z^{2}+4x-2y+3x-6[\/latex] (a polynomial in three variables)<\/p>\r\nand\r\n<p style=\"text-align: center;\">[latex]g\\,(x,\\ y,\\ t)=(x^{2}-4xy+y^{2})\\sin{t}-(3x+5y)\\cos{t}[\/latex].<\/p>\r\nIn the first function, [latex](x,\\ y,\\ z)[\/latex] represents\u00a0a point in space, and the function [latex]f[\/latex]\u00a0maps each point in space to a fourth quantity, such as temperature or wind speed. In the second function, [latex](x,\\ y)[\/latex]\u00a0can represent a point in the plane, and [latex]t[\/latex]\u00a0can represent time. The function might map a point in the plane to a third quantity (for example, pressure) at a given time [latex]t[\/latex].\u00a0The method for finding the domain of a function of more than two variables is analogous to the method for functions of one or two variables.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Domains for Functions of Three Variables<\/h3>\r\nFind the domain of each of the following functions:\r\n<ol>\r\n \t<li>[latex]f\\,(x,\\ y,\\ z)=\\frac{3x-4y+2z}{\\sqrt{9-x^{2}-y^{2}-z^{2}}}[\/latex]<\/li>\r\n \t<li>[latex]g\\,(x,\\ y,\\ t)=\\frac{\\sqrt{2t-4}}{x^{2}-y^{2}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id2177793933114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id2177793933114\"]\r\n<ol>\r\n \t<li>For the function [latex]f\\,(x,\\ y,\\ z)=\\frac{3x-4y+2z}{\\sqrt{9-x^{2}-y^{2}-z^{2}}}[\/latex] to be defined\u00a0(and be a real value), two conditions must hold:\r\n<ol>\r\n \t<li>The denominator cannot be zero<\/li>\r\n \t<li>The radicand cannot be zero<\/li>\r\n<\/ol>\r\nCombining\u00a0these conditions leads to the inequality\r\n<div style=\"text-align: center;\">[latex]9-x^{2}-y^{2}-z^{2}\\,&gt;\\,0[\/latex].<\/div>\r\n&nbsp;\r\n\r\nMoving the variables\u00a0to the other side and reversing the inequality gives the domain as\r\n<div style=\"text-align: center;\">[latex]\\text{domain}\\ (f)=\\{(x,\\ y,\\ z)\\in\\mathbb{R}^{3}|x^{2}+y^{2}+z^{2}\\,&lt;\\,9\\}[\/latex],<\/div>\r\n&nbsp;\r\n\r\nwhich describes a ball of radius [latex]3[\/latex]\u00a0centered at the origin. (<em data-effect=\"italics\">Note<\/em>: The surface of the ball is not included in this domain.)<\/li>\r\n \t<li>For the function [latex]g\\,(x,\\ y,\\ t)=\\frac{\\sqrt{2t-4}}{x^{2}-y^{2}}[\/latex] to be defined\u00a0(and be a real value), two conditions must hold:\r\n<ol>\r\n \t<li>The radicand cannot be negative.<\/li>\r\n \t<li>The denominator cannot be zero.<\/li>\r\n<\/ol>\r\nSince the radicand cannot be negative, this implies [latex]2t-4\\,\\geq\\,0[\/latex], and therefore that [latex]t\\,\\geq\\,2[\/latex].\u00a0Since the denominator cannot be zero, [latex]x^{2}-y^{2}\\neq{0}[\/latex], or [latex]x^{2}\\neq{y}^{2}[\/latex], which can be rewritten as [latex]y\\neq\\pm{x}[\/latex],\u00a0which are the equations of two lines passing through the origin. Therefore, the domain of [latex]g[\/latex] is\r\n<div style=\"text-align: center;\">[latex]\\text{domain}\\ (g)=\\{(x,\\ y,\\ t)|y\\neq\\pm{x},\\,t\\,\\geq\\,2\\}[\/latex].<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the domain of the function [latex]h\\,(x,\\ y,\\ t)=(3t-6)\\,\\sqrt{y-4x^{2}+4}[\/latex].\r\n\r\n[reveal-answer q=\"fs-id2168793953124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id2168793953124\"]\r\n<p style=\"text-align: center;\">[latex]\\text{domain}\\,(h)=\\{(x,\\ y,\\ t)\\in\\mathbb{R}^{3}|y\\,\\geq\\,4x^{2}-4\\}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\nFunctions of two variables have level curves, which are shown as curves in the [latex]xy[\/latex]-plane.\u00a0However, when the function has three variables, the curves become surfaces, so we can define level surfaces for functions of three variables.\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\nGiven a function [latex]f\\,(x,\\ y,\\, z)[\/latex] and a number [latex]c[\/latex] in the range of [latex]f[\/latex]\u00a0a <span id=\"term156\" data-type=\"term\"><strong>level surface of a function of three variables<\/strong><\/span> is defined to be the set of points satisfying the equation [latex]f\\,(x,\\ y,\\ z)=c[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding a Level Surface<\/h3>\r\nFind the level surface for the function [latex]f\\,(x,\\ y,\\ z)=4x^{2}+9y^{2}-z^{2}[\/latex] corresponding to [latex]c=1[\/latex].\r\n\r\n[reveal-answer q=\"fs-id2168743953124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id2168743953124\"]\r\n<p style=\"text-align: left;\">The\u00a0level surface is defined by the equation [latex]4x^{2}+9y^{2}-z^{2}=1[\/latex].\u00a0This equation describes a hyperboloid of one sheet as shown in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"attachment_965\" align=\"aligncenter\" width=\"875\"]<img class=\"wp-image-965 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/27170622\/4-1-12.jpeg\" alt=\"This figure consists of four figures. The first is marked c = 0 and consists of a double cone (that is, two nappes) with their apex at the origin. The second is marked c = 1 and it looks remarkably similar to the first except that there is no apex at which the cones meet: instead, the two nappes are connected. Similarly, the next figure marked c = 2 has the two nappes connect, but this time their connection is larger (that is, the radius of their connection is greater). The final figure marked c = 3 also has the two nappes connect in an even larger fashion.\" width=\"875\" height=\"971\" \/> Figure 1.\u00a0A hyperboloid of one sheet with some of its level surfaces.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the equation of the level surface of the function\r\n<p style=\"text-align: center;\">[latex]g\\,(x,\\ y,\\ t)=x^{2}+y^{2}+z^{2}-2x+4y-6z[\/latex]<\/p>\r\ncorresponding to [latex]c=2[\/latex], and describe the surface, if possible.\r\n\r\n[reveal-answer q=\"fs-id2168793959124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id2168793959124\"]\r\n<p style=\"text-align: center;\">[latex](x-1)^{2}+(y+2)^{2}+(z-3)^{2}=16[\/latex] describes\u00a0a sphere of radius [latex]4[\/latex]\u00a0centered at the point [latex](1,-2, 3)[\/latex].<\/p>\r\n<p style=\"text-align: left;\">[\/hidden-answer]<\/p>\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=8186149&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=OoAPC9MiAhw&amp;video_target=tpm-plugin-yx2gcocw-OoAPC9MiAhw\" 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\/CP4.5_transcript.html\">transcript for \u201cCP 4.5\u201d here (opens in new window).<\/a><\/center>&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Recognize a function of three or more variables and identify its level surfaces.<\/li>\n<\/ul>\n<\/div>\n<p>So far, we have examined only functions of two variables. However, it is useful to take a brief look at functions of more than two variables. Two such examples are<\/p>\n<p style=\"text-align: center;\">[latex]f\\,(x,\\ y,\\ z)=x^{2}-2xy+y^{2}+3yz-z^{2}+4x-2y+3x-6[\/latex] (a polynomial in three variables)<\/p>\n<p>and<\/p>\n<p style=\"text-align: center;\">[latex]g\\,(x,\\ y,\\ t)=(x^{2}-4xy+y^{2})\\sin{t}-(3x+5y)\\cos{t}[\/latex].<\/p>\n<p>In the first function, [latex](x,\\ y,\\ z)[\/latex] represents\u00a0a point in space, and the function [latex]f[\/latex]\u00a0maps each point in space to a fourth quantity, such as temperature or wind speed. In the second function, [latex](x,\\ y)[\/latex]\u00a0can represent a point in the plane, and [latex]t[\/latex]\u00a0can represent time. The function might map a point in the plane to a third quantity (for example, pressure) at a given time [latex]t[\/latex].\u00a0The method for finding the domain of a function of more than two variables is analogous to the method for functions of one or two variables.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Domains for Functions of Three Variables<\/h3>\n<p>Find the domain of each of the following functions:<\/p>\n<ol>\n<li>[latex]f\\,(x,\\ y,\\ z)=\\frac{3x-4y+2z}{\\sqrt{9-x^{2}-y^{2}-z^{2}}}[\/latex]<\/li>\n<li>[latex]g\\,(x,\\ y,\\ t)=\\frac{\\sqrt{2t-4}}{x^{2}-y^{2}}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id2177793933114\">Show Solution<\/span><\/p>\n<div id=\"qfs-id2177793933114\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>For the function [latex]f\\,(x,\\ y,\\ z)=\\frac{3x-4y+2z}{\\sqrt{9-x^{2}-y^{2}-z^{2}}}[\/latex] to be defined\u00a0(and be a real value), two conditions must hold:\n<ol>\n<li>The denominator cannot be zero<\/li>\n<li>The radicand cannot be zero<\/li>\n<\/ol>\n<p>Combining\u00a0these conditions leads to the inequality<\/p>\n<div style=\"text-align: center;\">[latex]9-x^{2}-y^{2}-z^{2}\\,>\\,0[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>Moving the variables\u00a0to the other side and reversing the inequality gives the domain as<\/p>\n<div style=\"text-align: center;\">[latex]\\text{domain}\\ (f)=\\{(x,\\ y,\\ z)\\in\\mathbb{R}^{3}|x^{2}+y^{2}+z^{2}\\,<\\,9\\}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p>which describes a ball of radius [latex]3[\/latex]\u00a0centered at the origin. (<em data-effect=\"italics\">Note<\/em>: The surface of the ball is not included in this domain.)<\/li>\n<li>For the function [latex]g\\,(x,\\ y,\\ t)=\\frac{\\sqrt{2t-4}}{x^{2}-y^{2}}[\/latex] to be defined\u00a0(and be a real value), two conditions must hold:\n<ol>\n<li>The radicand cannot be negative.<\/li>\n<li>The denominator cannot be zero.<\/li>\n<\/ol>\n<p>Since the radicand cannot be negative, this implies [latex]2t-4\\,\\geq\\,0[\/latex], and therefore that [latex]t\\,\\geq\\,2[\/latex].\u00a0Since the denominator cannot be zero, [latex]x^{2}-y^{2}\\neq{0}[\/latex], or [latex]x^{2}\\neq{y}^{2}[\/latex], which can be rewritten as [latex]y\\neq\\pm{x}[\/latex],\u00a0which are the equations of two lines passing through the origin. Therefore, the domain of [latex]g[\/latex] is<\/p>\n<div style=\"text-align: center;\">[latex]\\text{domain}\\ (g)=\\{(x,\\ y,\\ t)|y\\neq\\pm{x},\\,t\\,\\geq\\,2\\}[\/latex].<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the domain of the function [latex]h\\,(x,\\ y,\\ t)=(3t-6)\\,\\sqrt{y-4x^{2}+4}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id2168793953124\">Show Solution<\/span><\/p>\n<div id=\"qfs-id2168793953124\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\text{domain}\\,(h)=\\{(x,\\ y,\\ t)\\in\\mathbb{R}^{3}|y\\,\\geq\\,4x^{2}-4\\}[\/latex]<\/p>\n<p style=\"text-align: left;\"><\/div>\n<\/div>\n<\/div>\n<p>Functions of two variables have level curves, which are shown as curves in the [latex]xy[\/latex]-plane.\u00a0However, when the function has three variables, the curves become surfaces, so we can define level surfaces for functions of three variables.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p>Given a function [latex]f\\,(x,\\ y,\\, z)[\/latex] and a number [latex]c[\/latex] in the range of [latex]f[\/latex]\u00a0a <span id=\"term156\" data-type=\"term\"><strong>level surface of a function of three variables<\/strong><\/span> is defined to be the set of points satisfying the equation [latex]f\\,(x,\\ y,\\ z)=c[\/latex].<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding a Level Surface<\/h3>\n<p>Find the level surface for the function [latex]f\\,(x,\\ y,\\ z)=4x^{2}+9y^{2}-z^{2}[\/latex] corresponding to [latex]c=1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id2168743953124\">Show Solution<\/span><\/p>\n<div id=\"qfs-id2168743953124\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">The\u00a0level surface is defined by the equation [latex]4x^{2}+9y^{2}-z^{2}=1[\/latex].\u00a0This equation describes a hyperboloid of one sheet as shown in the following figure.<\/p>\n<div id=\"attachment_965\" style=\"width: 885px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-965\" class=\"wp-image-965 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/27170622\/4-1-12.jpeg\" alt=\"This figure consists of four figures. The first is marked c = 0 and consists of a double cone (that is, two nappes) with their apex at the origin. The second is marked c = 1 and it looks remarkably similar to the first except that there is no apex at which the cones meet: instead, the two nappes are connected. Similarly, the next figure marked c = 2 has the two nappes connect, but this time their connection is larger (that is, the radius of their connection is greater). The final figure marked c = 3 also has the two nappes connect in an even larger fashion.\" width=\"875\" height=\"971\" \/><\/p>\n<p id=\"caption-attachment-965\" class=\"wp-caption-text\">Figure 1.\u00a0A hyperboloid of one sheet with some of its level surfaces.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the equation of the level surface of the function<\/p>\n<p style=\"text-align: center;\">[latex]g\\,(x,\\ y,\\ t)=x^{2}+y^{2}+z^{2}-2x+4y-6z[\/latex]<\/p>\n<p>corresponding to [latex]c=2[\/latex], and describe the surface, if possible.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id2168793959124\">Show Solution<\/span><\/p>\n<div id=\"qfs-id2168793959124\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex](x-1)^{2}+(y+2)^{2}+(z-3)^{2}=16[\/latex] describes\u00a0a sphere of radius [latex]4[\/latex]\u00a0centered at the point [latex](1,-2, 3)[\/latex].<\/p>\n<p style=\"text-align: left;\"><\/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=8186149&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=OoAPC9MiAhw&amp;video_target=tpm-plugin-yx2gcocw-OoAPC9MiAhw\" 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\/CP4.5_transcript.html\">transcript for \u201cCP 4.5\u201d here (opens in new window).<\/a><\/div>\n<p>&nbsp;<\/p>\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-861\">\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 4.5. <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><\/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":5,"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 4.5\",\"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-861","chapter","type-chapter","status-publish","hentry"],"part":22,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/861","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":70,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/861\/revisions"}],"predecessor-version":[{"id":5833,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/861\/revisions\/5833"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/22"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/861\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=861"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=861"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=861"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=861"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}