{"id":757,"date":"2025-10-09T15:36:58","date_gmt":"2025-10-09T15:36:58","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=757"},"modified":"2025-11-13T20:12:40","modified_gmt":"2025-11-13T20:12:40","slug":"conservative-vector-fields-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/conservative-vector-fields-fresh-take\/","title":{"raw":"Conservative Vector Fields: Fresh Take","rendered":"Conservative Vector Fields: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Identify simple and closed curves in various regions<\/li>\r\n \t<li>Find potential functions for conservative fields<\/li>\r\n \t<li>Apply the Fundamental Theorem for Line Integrals<\/li>\r\n \t<li>Test a vector field to determine whether it is conservative<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Curves and Regions<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">When working with line integrals and their theorems, the geometry of curves and regions matters. Think of it this way: the shape and structure of the path or area you're integrating over determines which theorems you can apply.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Curves: Closed vs. Simple<\/strong><\/p>\r\n\r\n<ul>\r\n \t<li class=\"font-claude-response-body whitespace-normal break-words\">A <strong>closed curve<\/strong> is like a loop\u2014it starts and ends at the same point. Mathematically, if [latex]\\mathbf{r}(t)[\/latex] parameterizes your curve on [latex]a\\leq t\\leq b[\/latex], then [latex]\\mathbf{r}(a)=\\mathbf{r}(b)[\/latex].<\/li>\r\n \t<li class=\"font-claude-response-body whitespace-normal break-words\">A <strong>simple curve<\/strong> doesn't cross itself. The parameterization [latex]\\mathbf{r}(t)[\/latex] is one-to-one on [latex](a,b)[\/latex], meaning no self-intersections occur (though the endpoints can still meet if it's also closed).<\/li>\r\n \t<li class=\"font-claude-response-body whitespace-normal break-words\">A <strong>simple closed curve<\/strong> combines both properties\u2014it loops back to its starting point without crossing itself. Picture a circle or an ellipse.<\/li>\r\n<\/ul>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Regions: Connected vs. Simply Connected<\/strong><\/p>\r\n\r\n<ul>\r\n \t<li class=\"font-claude-response-body whitespace-normal break-words\">A <strong>connected region<\/strong> is one where you can draw a path between any two points without leaving the region. No matter which two points you pick in [latex]D[\/latex], there's a way to connect them while staying inside.<\/li>\r\n \t<li class=\"font-claude-response-body whitespace-normal break-words\">A <strong>simply connected region<\/strong> takes this further\u2014it's connected <em>and<\/em> has no holes. If you draw any closed loop inside the region, you can continuously shrink it down to a point without ever leaving the region. In two dimensions, \"no holes\" is the key idea.<\/li>\r\n<\/ul>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Why This Matters<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Many theorems relate integrals over a region to integrals around its boundary. These theorems require specific geometric conditions\u2014usually that your region is connected or simply connected. Understanding these distinctions helps you identify when you can apply powerful tools like Green's Theorem or the Fundamental Theorem for Line Integrals.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Is the curve given by parameterization [latex]{\\bf{r}}(t)=\\langle2\\cos{t},3\\sin{t}\\rangle[\/latex], [latex]0\\leq{t}\\leq6\\pi[\/latex] a simple closed curve?[reveal-answer q=\"814582267\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"814582267\"]Yes.[\/hidden-answer]<\/section>\r\n<h2 data-type=\"title\">Fundamental Theorem for Line Integrals<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The Fundamental Theorem of Calculus tells us that [latex]\\int_a^b f(x),dx = F(b) - F(a)[\/latex] when [latex]F[\/latex] is an antiderivative of [latex]f[\/latex]. The Fundamental Theorem for Line Integrals extends this idea to vector fields: if you're integrating a gradient field [latex]\\nabla f[\/latex] along a curve [latex]C[\/latex], the answer depends only on the values of [latex]f[\/latex] at the endpoints.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>The Core Formula<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">For a conservative vector field [latex]\\mathbf{F} = \\nabla f[\/latex] along a curve [latex]C[\/latex] from point [latex]\\mathbf{r}(a)[\/latex] to [latex]\\mathbf{r}(b)[\/latex]:<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\int_C \\mathbf{F} \\cdot d\\mathbf{r} = f(\\mathbf{r}(b)) - f(\\mathbf{r}(a))[\/latex]<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">This works because the gradient [latex]\\nabla f[\/latex] acts like a derivative, and [latex]f[\/latex] acts like an antiderivative. The path you take doesn't matter\u2014only where you start and where you end.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Why This Matters<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">This theorem transforms difficult line integrals into simple evaluations. Instead of parameterizing the curve and computing [latex]\\int_a^b \\mathbf{F}(\\mathbf{r}(t)) \\cdot \\mathbf{r}'(t)\\,dt[\/latex], you just find a potential function [latex]f[\/latex] and plug in the endpoints.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Two Key Consequences<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">First, if [latex]\\mathbf{F}[\/latex] is conservative and [latex]C[\/latex] is a closed curve (starts and ends at the same point), then [latex]\\oint_C \\mathbf{F} \\cdot d\\mathbf{r} = 0[\/latex]. This makes physical sense: if gravity does work on an object that returns to its starting point, the net work is zero.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Second, conservative vector fields are <strong>independent of path<\/strong>. The line integral [latex]\\int_C \\mathbf{F} \\cdot d\\mathbf{r}[\/latex] gives the same value for any two paths with the same endpoints. Think of three hikers climbing a mountain via different routes\u2014gravity does the same total work on each, because they all start and end at the same elevation.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>The Critical Requirement<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">This theorem only applies to conservative vector fields\u2014those with a potential function. Not every continuous vector field is conservative, so you need to verify this before using the shortcut.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Given that [latex]f(x,y)=(x-1)^2y+(y+1)^2x[\/latex] is a potential function for [latex]{\\bf{F}}=\\langle2xy-2y+(y+1)^2,(x-1)^2+2yx+2x\\rangle[\/latex], calculate integral [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], where [latex]C[\/latex] is the lower half of the unit circle oriented counterclockwise.<img class=\"aligncenter size-full wp-image-5252\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31223444\/6.26.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/9f28a5ea7739065108e57a778883047758bdf7f7&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A vector field in two dimensions. The arrows near the origin are the shortest, and the arrows in the upper right and lower left corners of quadrants 1 and 3 are the shortest. The arrows go up and to the left in quadrants 1 and 3. In quadrant 2, the arrows stretch up and to the right for values greater than x=-1. The closer the arrows are to y=1, the more horizontal they become. For values less than x=-1, the arrows point up and form a curve to the left. The closer the arrows are to y=1, the more horizontal they become. Above y=1, it looks like the arrows are shifting from vertical, going down to horizontal. In quadrant 4, the arrows go up and to the right fairly regularly, but they tend to be curving to the right the larger the x value becomes. For y values less than -1, the arrows shift from pointing up to pointing down, following x=1. The lower half of the unit circle with center at the origin is drawn in quadrants 3 and 4.&quot; id=&quot;15&quot;&gt;\" width=\"642\" height=\"459\" \/>[reveal-answer q=\"964357240\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"964357240\"][latex]2[\/latex].[\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8250317&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=nfZUeb79cSM&amp;video_target=tpm-plugin-srmwxh3x-nfZUeb79cSM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP6.26_transcript.html\">transcript for \u201cCP 6.26\u201d here (opens in new window).<\/a><\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">Show that [latex]{\\bf{F}}(x,y)=\\langle{x}y,x^2y^2\\rangle[\/latex] is not path independent by considering the line segment from [latex](0, 0)[\/latex] to [latex](2, 2)[\/latex] and the piece of the graph of [latex]y=\\frac{x^2}2[\/latex] that goes from [latex](0, 0)[\/latex] to\u00a0[latex](2, 2)[\/latex].[reveal-answer q=\"483725098\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"483725098\"]If [latex]C_1[\/latex] and [latex]C_2[\/latex] represent the two curves, then [latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}\\ne\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex].[\/hidden-answer]<\/section>\r\n<h2 data-type=\"title\">Conservative Vector Fields and Potential Functions<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">To use the Fundamental Theorem for Line Integrals, you need a potential function [latex]f[\/latex] such that [latex]\\nabla f = \\mathbf{F}[\/latex]. Finding this function is systematic\u2014think of it as reverse differentiation, working backward from the gradient to reconstruct the original function.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>The Strategy for [latex]\\mathbb{R}^2[\/latex]<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Given [latex]\\mathbf{F}(x,y) = \\langle P(x,y), Q(x,y) \\rangle[\/latex], you're looking for [latex]f[\/latex] where [latex]f_x = P[\/latex] and [latex]f_y = Q[\/latex]. Here's the process:<\/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>Integrate [latex]P[\/latex] with respect to [latex]x[\/latex]<\/strong> to get [latex]g(x,y) + h(y)[\/latex]. The \"constant\" of integration is actually a function [latex]h(y)[\/latex] because you're treating [latex]y[\/latex] as constant during this step.<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Differentiate your result with respect to [latex]y[\/latex]<\/strong> to get [latex]g_y(x,y) + h'(y)[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Set this equal to [latex]Q(x,y)[\/latex]<\/strong> and solve for [latex]h'(y)[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Integrate [latex]h'(y)[\/latex]<\/strong> to find [latex]h(y)[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Combine the pieces<\/strong>: [latex]f(x,y) = g(x,y) + h(y) + C[\/latex].<\/li>\r\n<\/ol>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Extending to [latex]\\mathbb{R}^3[\/latex]<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">For three variables, the process is similar but requires more steps. After integrating with respect to [latex]x[\/latex], your \"constant\" is [latex]g(y,z)[\/latex]. You then use the condition [latex]f_y = Q[\/latex] to find information about [latex]g[\/latex], and finally use [latex]f_z = R[\/latex] to pin down any remaining unknowns.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Key Insight<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Each integration step introduces an unknown function of the remaining variables. You systematically determine these unknowns by matching partial derivatives to the components of [latex]\\mathbf{F}[\/latex]. The process confirms that [latex]\\mathbf{F}[\/latex] is conservative while constructing the potential function you need.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Find a potential function for [latex]{\\bf{F}}(x,y)=\\langle{e}^xy^3+y,3e^xy^2+x\\rangle[\/latex].[reveal-answer q=\"834728049\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"834728049\"][latex]f(x,y)=e^xy^3+xy[\/latex].[\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">Find a potential function for [latex]{\\bf{F}}(x,y,z)=\\langle12x^2,\\cos{y}\\cos{z},1-\\sin{y}\\sin{z}\\rangle[\/latex].[reveal-answer q=\"349856201\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"349856201\"][latex]f(x,y,z)=4x^3+\\sin{y}\\cos{z}+z[\/latex][\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8250319&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=mkJ7BURhD20&amp;video_target=tpm-plugin-k3pv79x7-mkJ7BURhD20\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP6.29_transcript.html\">transcript for \u201cCP 6.29\u201d here (opens in new window).<\/a><\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">Find a potential function [latex]f[\/latex] for the three-dimensional gravitational force\u00a0[latex]{\\bf{F}}(x,y,z)=\\left\\langle\\frac{-Gx}{(x^2+y^2+z^2)^{3\/2}},\\frac{-Gy}{(x^2+y^2+z^2)^{3\/2}},\\frac{-Gz}{(x^2+y^2+z^2)^{3\/2}}\\right\\rangle[\/latex].[reveal-answer q=\"204873094\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"204873094\"][latex]f(x,y,z)=\\frac{G}{\\sqrt{x^2+y^2+z^2}}[\/latex].[\/hidden-answer]<\/section>\r\n<h2 data-type=\"title\">Testing a Vector Field<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">When you encounter a vector field [latex]\\mathbf{F}[\/latex], your first question should be: Is it conservative? The answer determines your entire approach to calculating line integrals. If [latex]\\mathbf{F}[\/latex] is conservative, you can use the Fundamental Theorem for Line Integrals\u2014a huge shortcut. If not, you'll need other methods.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>The Cross-Partial Test<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">For [latex]\\mathbf{F} = \\langle P, Q, R \\rangle[\/latex] on an open, simply connected region, check whether the cross-partials match:<\/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]P_y = Q_x[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]P_z = R_x[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]Q_z = R_y[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">If all three conditions hold, then [latex]\\mathbf{F}[\/latex] is conservative. In [latex]\\mathbb{R}^2[\/latex], you only need to verify [latex]P_y = Q_x[\/latex].<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">This is an \"if and only if\" test: [latex]\\mathbf{F}[\/latex] is conservative if and only if the cross-partials match (assuming the domain is open and simply connected).<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Why \"Simply Connected\" Matters<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The domain must be simply connected\u2014no holes. A vector field can satisfy the cross-partial property yet fail to be conservative if its domain has holes. The classic example is [latex]\\mathbf{F}(x,y) = \\left\\langle \\frac{y}{x^2+y^2}, \\frac{-x}{x^2+y^2} \\right\\rangle[\/latex], which has a hole at the origin. Even though [latex]P_y = Q_x[\/latex], the field isn't conservative because line integrals around the origin give different values depending on the path.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">When calculating [latex]\\int_C \\mathbf{F} \\cdot d\\mathbf{r}[\/latex]:<\/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\">Check if the domain is simply connected<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Apply the cross-partial test<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If conservative: find a potential function and use the Fundamental Theorem<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If not conservative: parameterize [latex]C[\/latex] and compute directly<\/li>\r\n<\/ol>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Recognizing conservative fields saves enormous effort, especially when the curve [latex]C[\/latex] is complicated or unspecified.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Determine whether [latex]{\\bf{F}}(x,y)=\\langle\\sin{x}\\cos{y},\\cos{x}\\sin{y}\\rangle[\/latex] is conservative.[reveal-answer q=\"207340123\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"207340123\"]It is conservative.[\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">Calculate integral [latex]\\displaystyle\\int_{C}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], where [latex]{\\bf{F}}(x,y)=\\langle\\sin{x}\\sin{y},5-\\cos{x}\\cos{y}\\rangle[\/latex] and [latex]C[\/latex] is a semicircle with starting point [latex](0,\\pi)[\/latex] and endpoint [latex](0,-\\pi)[\/latex].[reveal-answer q=\"984750429\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"984750429\"][latex]-10\\pi[\/latex].[\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8250318&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=0ASftQKk-Co&amp;video_target=tpm-plugin-n97z90vj-0ASftQKk-Co\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP6.32_transcript.html\">transcript for \u201cCP 6.32\u201d here (opens in new window).<\/a><\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">Let [latex]{\\bf{F}}(x,y)=\\langle4x^3y^4,4x^4y^3\\rangle[\/latex], and suppose that a particle moves from point [latex](4, 4)[\/latex] to [latex](1, 1)[\/latex] along any smooth curve. Is the work done by\u00a0[latex]{\\bf{F}}[\/latex]\u00a0on the particle positive, negative, or zero?[reveal-answer q=\"329893779\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"329893779\"]Negative.[\/hidden-answer]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Identify simple and closed curves in various regions<\/li>\n<li>Find potential functions for conservative fields<\/li>\n<li>Apply the Fundamental Theorem for Line Integrals<\/li>\n<li>Test a vector field to determine whether it is conservative<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Curves and Regions<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">When working with line integrals and their theorems, the geometry of curves and regions matters. Think of it this way: the shape and structure of the path or area you&#8217;re integrating over determines which theorems you can apply.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Curves: Closed vs. Simple<\/strong><\/p>\n<ul>\n<li class=\"font-claude-response-body whitespace-normal break-words\">A <strong>closed curve<\/strong> is like a loop\u2014it starts and ends at the same point. Mathematically, if [latex]\\mathbf{r}(t)[\/latex] parameterizes your curve on [latex]a\\leq t\\leq b[\/latex], then [latex]\\mathbf{r}(a)=\\mathbf{r}(b)[\/latex].<\/li>\n<li class=\"font-claude-response-body whitespace-normal break-words\">A <strong>simple curve<\/strong> doesn&#8217;t cross itself. The parameterization [latex]\\mathbf{r}(t)[\/latex] is one-to-one on [latex](a,b)[\/latex], meaning no self-intersections occur (though the endpoints can still meet if it&#8217;s also closed).<\/li>\n<li class=\"font-claude-response-body whitespace-normal break-words\">A <strong>simple closed curve<\/strong> combines both properties\u2014it loops back to its starting point without crossing itself. Picture a circle or an ellipse.<\/li>\n<\/ul>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Regions: Connected vs. Simply Connected<\/strong><\/p>\n<ul>\n<li class=\"font-claude-response-body whitespace-normal break-words\">A <strong>connected region<\/strong> is one where you can draw a path between any two points without leaving the region. No matter which two points you pick in [latex]D[\/latex], there&#8217;s a way to connect them while staying inside.<\/li>\n<li class=\"font-claude-response-body whitespace-normal break-words\">A <strong>simply connected region<\/strong> takes this further\u2014it&#8217;s connected <em>and<\/em> has no holes. If you draw any closed loop inside the region, you can continuously shrink it down to a point without ever leaving the region. In two dimensions, &#8220;no holes&#8221; is the key idea.<\/li>\n<\/ul>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Why This Matters<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Many theorems relate integrals over a region to integrals around its boundary. These theorems require specific geometric conditions\u2014usually that your region is connected or simply connected. Understanding these distinctions helps you identify when you can apply powerful tools like Green&#8217;s Theorem or the Fundamental Theorem for Line Integrals.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Is the curve given by parameterization [latex]{\\bf{r}}(t)=\\langle2\\cos{t},3\\sin{t}\\rangle[\/latex], [latex]0\\leq{t}\\leq6\\pi[\/latex] a simple closed curve?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q814582267\">Show Solution<\/button><\/p>\n<div id=\"q814582267\" class=\"hidden-answer\" style=\"display: none\">Yes.<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Fundamental Theorem for Line Integrals<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The Fundamental Theorem of Calculus tells us that [latex]\\int_a^b f(x),dx = F(b) - F(a)[\/latex] when [latex]F[\/latex] is an antiderivative of [latex]f[\/latex]. The Fundamental Theorem for Line Integrals extends this idea to vector fields: if you&#8217;re integrating a gradient field [latex]\\nabla f[\/latex] along a curve [latex]C[\/latex], the answer depends only on the values of [latex]f[\/latex] at the endpoints.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>The Core Formula<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">For a conservative vector field [latex]\\mathbf{F} = \\nabla f[\/latex] along a curve [latex]C[\/latex] from point [latex]\\mathbf{r}(a)[\/latex] to [latex]\\mathbf{r}(b)[\/latex]:<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\int_C \\mathbf{F} \\cdot d\\mathbf{r} = f(\\mathbf{r}(b)) - f(\\mathbf{r}(a))[\/latex]<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">This works because the gradient [latex]\\nabla f[\/latex] acts like a derivative, and [latex]f[\/latex] acts like an antiderivative. The path you take doesn&#8217;t matter\u2014only where you start and where you end.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Why This Matters<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">This theorem transforms difficult line integrals into simple evaluations. Instead of parameterizing the curve and computing [latex]\\int_a^b \\mathbf{F}(\\mathbf{r}(t)) \\cdot \\mathbf{r}'(t)\\,dt[\/latex], you just find a potential function [latex]f[\/latex] and plug in the endpoints.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Two Key Consequences<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">First, if [latex]\\mathbf{F}[\/latex] is conservative and [latex]C[\/latex] is a closed curve (starts and ends at the same point), then [latex]\\oint_C \\mathbf{F} \\cdot d\\mathbf{r} = 0[\/latex]. This makes physical sense: if gravity does work on an object that returns to its starting point, the net work is zero.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Second, conservative vector fields are <strong>independent of path<\/strong>. The line integral [latex]\\int_C \\mathbf{F} \\cdot d\\mathbf{r}[\/latex] gives the same value for any two paths with the same endpoints. Think of three hikers climbing a mountain via different routes\u2014gravity does the same total work on each, because they all start and end at the same elevation.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>The Critical Requirement<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">This theorem only applies to conservative vector fields\u2014those with a potential function. Not every continuous vector field is conservative, so you need to verify this before using the shortcut.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Given that [latex]f(x,y)=(x-1)^2y+(y+1)^2x[\/latex] is a potential function for [latex]{\\bf{F}}=\\langle2xy-2y+(y+1)^2,(x-1)^2+2yx+2x\\rangle[\/latex], calculate integral [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], where [latex]C[\/latex] is the lower half of the unit circle oriented counterclockwise.<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-5252\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31223444\/6.26.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/9f28a5ea7739065108e57a778883047758bdf7f7&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A vector field in two dimensions. The arrows near the origin are the shortest, and the arrows in the upper right and lower left corners of quadrants 1 and 3 are the shortest. The arrows go up and to the left in quadrants 1 and 3. In quadrant 2, the arrows stretch up and to the right for values greater than x=-1. The closer the arrows are to y=1, the more horizontal they become. For values less than x=-1, the arrows point up and form a curve to the left. The closer the arrows are to y=1, the more horizontal they become. Above y=1, it looks like the arrows are shifting from vertical, going down to horizontal. In quadrant 4, the arrows go up and to the right fairly regularly, but they tend to be curving to the right the larger the x value becomes. For y values less than -1, the arrows shift from pointing up to pointing down, following x=1. The lower half of the unit circle with center at the origin is drawn in quadrants 3 and 4.&quot; id=&quot;15&quot;&gt;\" width=\"642\" height=\"459\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q964357240\">Show Solution<\/button><\/p>\n<div id=\"q964357240\" class=\"hidden-answer\" style=\"display: none\">[latex]2[\/latex].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8250317&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=nfZUeb79cSM&amp;video_target=tpm-plugin-srmwxh3x-nfZUeb79cSM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP6.26_transcript.html\">transcript for \u201cCP 6.26\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Show that [latex]{\\bf{F}}(x,y)=\\langle{x}y,x^2y^2\\rangle[\/latex] is not path independent by considering the line segment from [latex](0, 0)[\/latex] to [latex](2, 2)[\/latex] and the piece of the graph of [latex]y=\\frac{x^2}2[\/latex] that goes from [latex](0, 0)[\/latex] to\u00a0[latex](2, 2)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q483725098\">Show Solution<\/button><\/p>\n<div id=\"q483725098\" class=\"hidden-answer\" style=\"display: none\">If [latex]C_1[\/latex] and [latex]C_2[\/latex] represent the two curves, then [latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}\\ne\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex].<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Conservative Vector Fields and Potential Functions<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">To use the Fundamental Theorem for Line Integrals, you need a potential function [latex]f[\/latex] such that [latex]\\nabla f = \\mathbf{F}[\/latex]. Finding this function is systematic\u2014think of it as reverse differentiation, working backward from the gradient to reconstruct the original function.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>The Strategy for [latex]\\mathbb{R}^2[\/latex]<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Given [latex]\\mathbf{F}(x,y) = \\langle P(x,y), Q(x,y) \\rangle[\/latex], you&#8217;re looking for [latex]f[\/latex] where [latex]f_x = P[\/latex] and [latex]f_y = Q[\/latex]. Here&#8217;s the process:<\/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>Integrate [latex]P[\/latex] with respect to [latex]x[\/latex]<\/strong> to get [latex]g(x,y) + h(y)[\/latex]. The &#8220;constant&#8221; of integration is actually a function [latex]h(y)[\/latex] because you&#8217;re treating [latex]y[\/latex] as constant during this step.<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Differentiate your result with respect to [latex]y[\/latex]<\/strong> to get [latex]g_y(x,y) + h'(y)[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Set this equal to [latex]Q(x,y)[\/latex]<\/strong> and solve for [latex]h'(y)[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Integrate [latex]h'(y)[\/latex]<\/strong> to find [latex]h(y)[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Combine the pieces<\/strong>: [latex]f(x,y) = g(x,y) + h(y) + C[\/latex].<\/li>\n<\/ol>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Extending to [latex]\\mathbb{R}^3[\/latex]<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">For three variables, the process is similar but requires more steps. After integrating with respect to [latex]x[\/latex], your &#8220;constant&#8221; is [latex]g(y,z)[\/latex]. You then use the condition [latex]f_y = Q[\/latex] to find information about [latex]g[\/latex], and finally use [latex]f_z = R[\/latex] to pin down any remaining unknowns.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Key Insight<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Each integration step introduces an unknown function of the remaining variables. You systematically determine these unknowns by matching partial derivatives to the components of [latex]\\mathbf{F}[\/latex]. The process confirms that [latex]\\mathbf{F}[\/latex] is conservative while constructing the potential function you need.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Find a potential function for [latex]{\\bf{F}}(x,y)=\\langle{e}^xy^3+y,3e^xy^2+x\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q834728049\">Show Solution<\/button><\/p>\n<div id=\"q834728049\" class=\"hidden-answer\" style=\"display: none\">[latex]f(x,y)=e^xy^3+xy[\/latex].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find a potential function for [latex]{\\bf{F}}(x,y,z)=\\langle12x^2,\\cos{y}\\cos{z},1-\\sin{y}\\sin{z}\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q349856201\">Show Solution<\/button><\/p>\n<div id=\"q349856201\" class=\"hidden-answer\" style=\"display: none\">[latex]f(x,y,z)=4x^3+\\sin{y}\\cos{z}+z[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8250319&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=mkJ7BURhD20&amp;video_target=tpm-plugin-k3pv79x7-mkJ7BURhD20\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP6.29_transcript.html\">transcript for \u201cCP 6.29\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find a potential function [latex]f[\/latex] for the three-dimensional gravitational force\u00a0[latex]{\\bf{F}}(x,y,z)=\\left\\langle\\frac{-Gx}{(x^2+y^2+z^2)^{3\/2}},\\frac{-Gy}{(x^2+y^2+z^2)^{3\/2}},\\frac{-Gz}{(x^2+y^2+z^2)^{3\/2}}\\right\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q204873094\">Show Solution<\/button><\/p>\n<div id=\"q204873094\" class=\"hidden-answer\" style=\"display: none\">[latex]f(x,y,z)=\\frac{G}{\\sqrt{x^2+y^2+z^2}}[\/latex].<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Testing a Vector Field<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">When you encounter a vector field [latex]\\mathbf{F}[\/latex], your first question should be: Is it conservative? The answer determines your entire approach to calculating line integrals. If [latex]\\mathbf{F}[\/latex] is conservative, you can use the Fundamental Theorem for Line Integrals\u2014a huge shortcut. If not, you&#8217;ll need other methods.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>The Cross-Partial Test<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">For [latex]\\mathbf{F} = \\langle P, Q, R \\rangle[\/latex] on an open, simply connected region, check whether the cross-partials match:<\/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]P_y = Q_x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]P_z = R_x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]Q_z = R_y[\/latex]<\/li>\n<\/ul>\n<p class=\"font-claude-response-body whitespace-normal break-words\">If all three conditions hold, then [latex]\\mathbf{F}[\/latex] is conservative. In [latex]\\mathbb{R}^2[\/latex], you only need to verify [latex]P_y = Q_x[\/latex].<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">This is an &#8220;if and only if&#8221; test: [latex]\\mathbf{F}[\/latex] is conservative if and only if the cross-partials match (assuming the domain is open and simply connected).<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Why &#8220;Simply Connected&#8221; Matters<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The domain must be simply connected\u2014no holes. A vector field can satisfy the cross-partial property yet fail to be conservative if its domain has holes. The classic example is [latex]\\mathbf{F}(x,y) = \\left\\langle \\frac{y}{x^2+y^2}, \\frac{-x}{x^2+y^2} \\right\\rangle[\/latex], which has a hole at the origin. Even though [latex]P_y = Q_x[\/latex], the field isn&#8217;t conservative because line integrals around the origin give different values depending on the path.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">When calculating [latex]\\int_C \\mathbf{F} \\cdot d\\mathbf{r}[\/latex]:<\/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\">Check if the domain is simply connected<\/li>\n<li class=\"whitespace-normal break-words\">Apply the cross-partial test<\/li>\n<li class=\"whitespace-normal break-words\">If conservative: find a potential function and use the Fundamental Theorem<\/li>\n<li class=\"whitespace-normal break-words\">If not conservative: parameterize [latex]C[\/latex] and compute directly<\/li>\n<\/ol>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Recognizing conservative fields saves enormous effort, especially when the curve [latex]C[\/latex] is complicated or unspecified.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Determine whether [latex]{\\bf{F}}(x,y)=\\langle\\sin{x}\\cos{y},\\cos{x}\\sin{y}\\rangle[\/latex] is conservative.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q207340123\">Show Solution<\/button><\/p>\n<div id=\"q207340123\" class=\"hidden-answer\" style=\"display: none\">It is conservative.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Calculate integral [latex]\\displaystyle\\int_{C}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], where [latex]{\\bf{F}}(x,y)=\\langle\\sin{x}\\sin{y},5-\\cos{x}\\cos{y}\\rangle[\/latex] and [latex]C[\/latex] is a semicircle with starting point [latex](0,\\pi)[\/latex] and endpoint [latex](0,-\\pi)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q984750429\">Show Solution<\/button><\/p>\n<div id=\"q984750429\" class=\"hidden-answer\" style=\"display: none\">[latex]-10\\pi[\/latex].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8250318&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=0ASftQKk-Co&amp;video_target=tpm-plugin-n97z90vj-0ASftQKk-Co\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP6.32_transcript.html\">transcript for \u201cCP 6.32\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Let [latex]{\\bf{F}}(x,y)=\\langle4x^3y^4,4x^4y^3\\rangle[\/latex], and suppose that a particle moves from point [latex](4, 4)[\/latex] to [latex](1, 1)[\/latex] along any smooth curve. Is the work done by\u00a0[latex]{\\bf{F}}[\/latex]\u00a0on the particle positive, negative, or zero?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q329893779\">Show Solution<\/button><\/p>\n<div id=\"q329893779\" class=\"hidden-answer\" style=\"display: none\">Negative.<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":21,"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-757","chapter","type-chapter","status-publish","hentry"],"part":436,"module-header":"fresh_take","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/757","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":8,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/757\/revisions"}],"predecessor-version":[{"id":2232,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/757\/revisions\/2232"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/436"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/757\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=757"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=757"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=757"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=757"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}