{"id":1121,"date":"2021-11-09T23:41:38","date_gmt":"2021-11-09T23:41:38","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=1121"},"modified":"2022-11-01T05:15:16","modified_gmt":"2022-11-01T05:15:16","slug":"conservative-vector-fields","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/conservative-vector-fields\/","title":{"raw":"Conservative Vector Fields","rendered":"Conservative Vector Fields"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul class=\"os-abstract\">\r\n \t<li><span class=\"os-abstract-content\">Explain how to find a potential function for a conservative vector field.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Use the Fundamental Theorem for Line Integrals to evaluate a line integral in a vector field.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Explain how to test a vector field to determine whether it is conservative.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 data-type=\"title\">Conservative Vector Fields and Potential Functions<\/h2>\r\n<p id=\"fs-id1167793404648\">As we have learned, the Fundamental Theorem for Line Integrals says that if\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative, then calculating [latex]\\displaystyle\\int_{C}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex] has two steps: first, find a potential function [latex]f[\/latex] for\u00a0[latex]{\\bf{F}}[\/latex]\u00a0and, second, calculate [latex]f(p_1)-f(P_0)[\/latex], where [latex]P_1[\/latex] is the endpoint of [latex]C[\/latex] and [latex]P_0[\/latex] is the starting point. To use this theorem for a conservative field\u00a0[latex]{\\bf{F}}[\/latex], we must be able to find a potential function [latex]f[\/latex] for\u00a0[latex]{\\bf{F}}[\/latex]. Therefore, we must answer the following question: Given a conservative vector field\u00a0[latex]{\\bf{F}}[\/latex], how do we find a function [latex]f[\/latex] such that [latex]\\nabla{f}={\\bf{F}}[\/latex]? Before giving a general method for finding a potential function, let\u2019s motivate the method with an example.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding a potential function<\/h3>\r\nFind a potential function for [latex]{\\bf{F}}(x,y)=\\langle2xy^3,3x^2y^2+\\cos{(y)}\\rangle[\/latex], thereby showing that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.\r\n\r\n[reveal-answer q=\"342587498\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"342587498\"]\r\n<p id=\"fs-id1167794097416\">Suppose that [latex]f(x, y)[\/latex] is a potential function for\u00a0[latex]{\\bf{F}}[\/latex]. Then, [latex]\\nabla{f}={\\bf{F}}[\/latex], and therefore<\/p>\r\n<p style=\"text-align: center;\">[latex]f_x=2xy^3\\text{ and }f_y=3x^2y^2+\\cos{y}[\/latex].<\/p>\r\n<p id=\"fs-id1167793416675\">Integrating the equation [latex]f_x=2xy^3[\/latex] with respect to [latex]x[\/latex] yields the equation<\/p>\r\n<p style=\"text-align: center;\">[latex]f(x,y)=x^2y^2+h(y)[\/latex].<\/p>\r\n<p id=\"fs-id1167793952315\">Notice that since we are integrating a two-variable function with respect to [latex]x[\/latex], we must add a constant of integration that is a constant with respect to [latex]x[\/latex], but may still be a function of [latex]y[\/latex]. The equation [latex]f(x,y)=x^2y^2+h(y)[\/latex] can be confirmed by taking the partial derivative with respect to [latex]x[\/latex]:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{\\partial{f}}{\\partial{x}}=\\frac{\\partial}{\\partial{x}}(x^2y^3)\\frac{\\partial}{\\partial{x}}(h(y))=2xy^3+0=2xy^3[\/latex].<\/p>\r\n<p id=\"fs-id1167793928372\">Since [latex]f[\/latex] is a potential function for\u00a0[latex]{\\bf{F}}[\/latex],<\/p>\r\n<p style=\"text-align: center;\">[latex]f_y=3x^2y^2+\\cos{(y)}[\/latex],<\/p>\r\n<p id=\"fs-id1167794126213\">and therefore<\/p>\r\n<p style=\"text-align: center;\">[latex]3x^2y^2+h^\\prime(y)=3x^2y^2+\\cos{(y)}[\/latex].<\/p>\r\n<p id=\"fs-id1167793931827\">This implies that [latex]h^\\prime(y)=\\cos{(y)}[\/latex], so [latex]h(y)=\\sin{y}+C[\/latex]. Therefore,\u00a0<em data-effect=\"italics\">any<\/em>\u00a0function of the form [latex]f(x,y)=x^2y^3+\\sin{(y)}+C[\/latex] is a potential function. Taking, in particular, [latex]C=0[\/latex] gives the potential function [latex]f(x,y)=x^2y^3+\\sin{(y)}[\/latex].<\/p>\r\n<p id=\"fs-id1167793505765\">To verify that [latex]f[\/latex] is a potential function, note that [latex]\\nabla{f}=\\langle2xy^3,3x^2y^2+\\cos{y}\\rangle={\\bf{F}}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind a potential function for [latex]{\\bf{F}}(x,y)=\\langle{e}^xy^3+y,3e^xy^2+x\\rangle[\/latex].\r\n\r\n[reveal-answer q=\"834728049\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"834728049\"]\r\n\r\n[latex]f(x,y)=e^xy^3+xy[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe logic of the previous example extends to finding the potential function for any conservative vector field in [latex]\\mathbb{R}^2[\/latex]. Thus, we have the following problem-solving strategy for finding potential functions:\r\n<div id=\"fs-id1167794333153\" class=\"problem-solving\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox examples\">\r\n<h3>Problem solving strategy: finding a potential function for a conservative vector field\u00a0[latex]{\\bf{F}}(X,Y)=&lt;{P}(X,Y),Q(X,Y)&gt;[\/latex]<\/h3>\r\n<ol id=\"fs-id1167793503117\" type=\"1\">\r\n \t<li>Integrate [latex]P[\/latex] with respect to [latex]x[\/latex]. This results in a function of the form [latex]g(x, y)+h(y)[\/latex], where [latex]h(y)[\/latex] is unknown.<\/li>\r\n \t<li>Take the partial derivative of [latex]g(x, y)+h(y)[\/latex] with respect to [latex]y[\/latex], which results in the function [latex]g_y(x, y)+h'(y)[\/latex].<\/li>\r\n \t<li>Use the equation [latex]g_y(x, y)+h'(y)=Q(x ,y)[\/latex] to find [latex]h'(y)[\/latex].<\/li>\r\n \t<li>Integrate [latex]h'(y)[\/latex] to find [latex]h(y)[\/latex].<\/li>\r\n \t<li>Any function of the form [latex]f(x,y)=g(x,y)+h(y)+C[\/latex], where [latex]C[\/latex] is a constant, is a potential function for\u00a0[latex]{\\bf{F}}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\nWe can adapt this strategy to find potential functions for vector fields in [latex]\\mathbb{R}^3[\/latex], as shown in the next example.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding a potential function in\u00a0[latex]\\mathbb{R}^3[\/latex]<\/h3>\r\nFind a potential function for [latex]{\\bf{F}}(x,y,z)=\\langle2xy,x^2+2yz^3,3y^2z^2+2z\\rangle[\/latex], thereby showing that [latex]{\\bf{F}}[\/latex] is conservative.\r\n\r\n[reveal-answer q=\"764523874\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"764523874\"]\r\n<p id=\"fs-id1167794042874\">Suppose that [latex]f[\/latex] is a potential function. Then, [latex]\\nabla{f}={\\bf{F}}[\/latex] and therefore [latex]f_x=2xy[\/latex]. Integrating this equation with respect to [latex]x[\/latex] yields the equation [latex]f(x, y, z)=x^{2}y+g(y, z)[\/latex] for some function [latex]g[\/latex]. Notice that, in this case, the constant of integration with respect to [latex]x[\/latex] is a function of [latex]y[\/latex] and\u00a0[latex]z[\/latex].<\/p>\r\n<p id=\"fs-id1167793631905\">Since [latex]f[\/latex] is a potential function,<\/p>\r\n<p style=\"text-align: center;\">[latex]x^{2}+2yz^{3}=f_y=x^{2}+g_y[\/latex].<\/p>\r\n<p id=\"fs-id1167793632719\">Therefore,<\/p>\r\n<p style=\"text-align: center;\">[latex]g_y=2yz^{3}[\/latex].<\/p>\r\n<p id=\"fs-id1167794210246\">Integrating this function with respect to [latex]y[\/latex] yields<\/p>\r\n<p style=\"text-align: center;\">[latex]g(y, z)=y^{2}z^{3}+h(z)[\/latex]<\/p>\r\n<p id=\"fs-id1167794136180\">for some function [latex]h(z)[\/latex] of [latex]z[\/latex] alone. (Notice that, because we know that [latex]g[\/latex] is a function of only [latex]y[\/latex] and [latex]z[\/latex], we do not need to write [latex]g(y, z)=y^{2}z^{3}+h(x,z)[\/latex].) Therefore,<\/p>\r\n<p style=\"text-align: center;\">[latex]f(z,y,z)=x^2+y+g(y,z)=x^2y+y^2z^3+h(z)[\/latex].<\/p>\r\n<p id=\"fs-id1167793397795\">To find [latex]f[\/latex], we now must only find [latex]h[\/latex] Since [latex]f[\/latex] is a potential function,<\/p>\r\n<p style=\"text-align: center;\">[latex]3y^2z^2+2z=g_z=3y^2z^2+h^\\prime(z)[\/latex].<\/p>\r\n<p id=\"fs-id1167793244255\">This implies that [latex]h^\\prime(z)=2z[\/latex], so [latex]h(z)=z^{2}+C[\/latex]. Letting [latex]C=0[\/latex] gives the potential function<\/p>\r\n<p style=\"text-align: center;\">[latex]f(x,y,z)=x^2y+y^2z^3+z^2[\/latex].<\/p>\r\n<p id=\"fs-id1167793498629\">To verify that [latex]f[\/latex] is a potential function, note that [latex]\\nabla{f}=\\langle2xy,x^2+2yz^3,3y^2z^2+2z\\rangle[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind a potential function for [latex]{\\bf{F}}(x,y,z)=\\langle12x^2,\\cos{y}\\cos{z},1-\\sin{y}\\sin{z}\\rangle[\/latex].\r\n\r\n[reveal-answer q=\"349856201\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"349856201\"]\r\n\r\n[latex]f(x,y,z)=4x^3+\\sin{y}\\cos{z}+z[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<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\"><\/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>\r\n<p id=\"fs-id1167793829179\">We can apply the process of finding a potential function to a\u00a0<span id=\"47dae7f0-fc5d-4d6c-a882-d19c748319d9_term258\" class=\"no-emphasis\" data-type=\"term\">gravitational force<\/span>. Recall that, if an object has unit mass and is located at the origin, then the gravitational force in [latex]\\mathbb{R}^2[\/latex] that the object exerts on another object of unit mass at the point [latex](x, y)[\/latex] is given by vector field<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{\\bf{F}}(x,y)=-G\\left\\langle\\frac{x}{(x^2+y^2)^{3\/2}},\\frac{y}{(x^2+y^2)^{3\/2}}\\right\\rangle}[\/latex],<\/p>\r\n<p id=\"fs-id1167793266873\">where [latex]G[\/latex] is the universal gravitational constant. In the next example, we build a potential function for\u00a0[latex]{\\bf{F}}[\/latex], thus confirming what we already know: that gravity is conservative.<\/p>\r\n\r\n<div id=\"fs-id1167793266888\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding a potential function<\/h3>\r\nFind a potential function [latex]f[\/latex] for [latex]{\\bf{F}}(x,y)=-G\\left\\langle\\frac{x}{(x^2+y^2)^{3\/2}},\\frac{y}{(x^2+y^2)^{3\/2}}\\right\\rangle[\/latex].\r\n\r\n[reveal-answer q=\"384572983\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"384572983\"]\r\n<p id=\"fs-id1167793398054\">Suppose that [latex]f[\/latex] is a potential function. Then, [latex]\\nabla{f}={\\bf{F}}[\/latex] and therefore<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{f_x=\\frac{-Gx}{(x^2+y^2)^{3\/2}}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793604079\">To integrate this function with respect to [latex]x[\/latex], we can use [latex]u[\/latex]-substitution. If [latex]u=x^{2}+y^{2}[\/latex] then [latex]\\frac{du}2=x \\ dx[\/latex], so<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\int\\frac{-Gx}{(x^2+y^2)^{3\/2}}&amp;=\\displaystyle\\int\\frac{-G}{2u^{3\/2}} \\ du \\\\\r\n&amp;=\\frac{G}{\\sqrt{u}}+h(y) \\\\\r\n&amp;=\\frac{G}{\\sqrt{x^2+y^2}}+h(y)\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167793505997\">for some function [latex]h(y)[\/latex]. Therefore,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{f(x,y)=\\frac{G}{\\sqrt{(x^2+y^2)}}+h(y)}[\/latex].<\/p>\r\n<p id=\"fs-id1167793432912\">Since [latex]f[\/latex] i<span style=\"font-size: 1rem; text-align: initial;\">s a potential function for\u00a0[latex]{\\bf{F}}[\/latex]<\/span><span style=\"font-size: 1rem; text-align: initial;\">,<\/span><\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{f_y=\\frac{-Gy}{(x^2+y^2)^{3\/2}}}[\/latex].<\/p>\r\n<p id=\"fs-id1167794097277\">Since [latex]f(x,y)=\\frac{G}{\\sqrt{x^2+y^2}}+h(y)[\/latex],\u00a0[latex]f_y[\/latex] also equals [latex]\\frac{-Gy}{(x^2+y^2)^{3\/2}}+h^\\prime(y)[\/latex].<\/p>\r\n<p id=\"fs-id1167794200726\">Therefore,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{-Gy}{(x^2+y^2)^{3\/2}}+h^\\prime(y)=\\frac{-Gy}{(x^2+y^2)^{3\/2}}}[\/latex],<\/p>\r\n<p id=\"fs-id1167793613305\">which implies that [latex]h^\\prime(y)=0[\/latex]. Thus, we can take [latex]h(y)[\/latex] to be any constant; in particular, we can let [latex]h(y)=0[\/latex]. The function<\/p>\r\n<p style=\"text-align: center;\">[latex]f(x,y)=\\frac{G}{\\sqrt{x^2+y^2}}[\/latex]<\/p>\r\n<p id=\"fs-id1167794127219\">is a potential function for the gravitational field\u00a0[latex]{\\bf{F}}[\/latex]. To confirm that [latex]f[\/latex] is a potential function, note that<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\nabla{f}&amp;=\\left\\langle-\\frac12\\frac{G}{(x^2+y^2)^{3\/2}}(2x),-\\frac12\\frac{G}{(x^2+y^2)^{3\/2}}(2y)\\right\\rangle \\\\\r\n&amp;=\\left\\langle\\frac{-Gx}{(x^2+y^2)^{3\/2}},\\frac{-Gy}{(x^2+y^2)^{3\/2}}\\right\\rangle \\\\\r\n&amp;={\\bf{F}}\r\n\\end{aligned}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind 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].\r\n\r\n[reveal-answer q=\"204873094\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"204873094\"]\r\n\r\n[latex]f(x,y,z)=\\frac{G}{\\sqrt{x^2+y^2+z^2}}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Testing a Vector Field<\/h2>\r\n<p id=\"fs-id1167793720011\">Until now, we have worked with vector fields that we know are conservative, but if we are not told that a vector field is conservative, we need to be able to test whether it is conservative. Recall that, if\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative, then\u00a0[latex]{\\bf{F}}[\/latex]\u00a0has the cross-partial property (see\u00a0The Cross-Partial Property of Conservative Vector Fields Theorem). That is, if [latex]{\\bf{F}}=\\langle{P},Q,R\\rangle[\/latex] is conservative, then [latex]P_y=Q_x[\/latex], [latex]P_z=R_x[\/latex], and [latex]Q_z=R_y[\/latex],\u00a0 So, if\u00a0[latex]{\\bf{F}}[\/latex]\u00a0has the cross-partial property, then is\u00a0[latex]{\\bf{F}}[\/latex]\u00a0conservative? If the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is open and simply connected, then the answer is yes.<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"example\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: the cross-partial test for conservative fields<\/h3>\r\n\r\n<hr \/>\r\n\r\nIf [latex]{\\bf{F}}=\\langle{P},Q,R\\rangle[\/latex] is a vector field on an open, simply connected region [latex]D[\/latex] and [latex]P_y=Q_x[\/latex], [latex]P_z=R_x[\/latex], and [latex]Q_z=R_y[\/latex] throughout [latex]D[\/latex], then\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793945273\">Although a proof of this theorem is beyond the scope of the text, we can discover its power with some examples. Later, we see why it is necessary for the region to be simply connected.<\/p>\r\n<p id=\"fs-id1167793945278\">Combining this theorem with the cross-partial property, we can determine whether a given vector field is conservative:<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"example\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: the cross-partial property of conservative fields<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]{\\bf{F}}=\\langle{P},Q,R\\rangle[\/latex] be a vector field on an open, simply connected region [latex]D[\/latex]. Then [latex]P_y=Q_x[\/latex], [latex]P_z=R_x[\/latex], and [latex]Q_z=R_y[\/latex] throughout [latex]D[\/latex] if and only if\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.\r\n\r\n<\/div>\r\nThe version of this theorem in [latex]\\mathbb{R}^2[\/latex] is also true. If [latex]{\\bf{F}}=\\langle{P},Q\\rangle[\/latex] is a vector field on an open, simply connected domain in [latex]\\mathbb{R}^2[\/latex], then\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative if and only if [latex]P_y=Q_x[\/latex].\r\n\r\n<\/div>\r\n<div data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: determining whether a vector field is conservative<\/h3>\r\nDetermine whether vector field [latex]{\\bf{F}}(x,y,z)=\\langle{x}y^2z,x^2yz,z^2\\rangle[\/latex] is conservative.\r\n\r\n[reveal-answer q=\"432875399\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"432875399\"]\r\n<p id=\"fs-id1167793441838\">Note that the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is all of [latex]\\mathbb{R}^2[\/latex] and [latex]\\mathbb{R}^3[\/latex] is simply connected. Therefore, we can use\u00a0Cross-Partial Property of Conservative Fields Theorem\u00a0to determine whether\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative. Let<\/p>\r\n<p style=\"text-align: center;\">[latex]P(x,y,z)=xy^2z,\\text{ }Q(x,y,z)=x^2yz,\\text{ and }R(x,y,z)=z^2[\/latex].<\/p>\r\n<p id=\"fs-id1167793718546\">Since [latex]Q_z=x^{2}y[\/latex] and [latex]R_y=0[\/latex], the vector field is not conservative.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: determining whether a vector field is conservative<\/h3>\r\nDetermine vector field [latex]{\\bf{F}}(x,y)=\\left\\langle{x}\\ln{(y)},\\frac{x^2}{2y}\\right\\rangle[\/latex] is conservative.\r\n\r\n[reveal-answer q=\"403750008\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"403750008\"]\r\n<p id=\"fs-id1167793940254\">Note that the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is the part of [latex]\\mathbb{R}^2[\/latex] in which [latex]y&gt;0[\/latex]. Thus, the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is part of a plane above the\u00a0<em data-effect=\"italics\">x<\/em>-axis, and this domain is simply connected (there are no holes in this region and this region is connected). Therefore, we can use\u00a0Cross-Partial Property of Conservative Fields Theorem\u00a0to determine whether\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative. Let<\/p>\r\n<p style=\"text-align: center;\">[latex]P(x,y)=x\\ln{(y)}\\text{ and\r\n}Q(x,y)=\\frac{x^2}{2y}[\/latex].<\/p>\r\n<p id=\"fs-id1167793925934\">Then [latex]P_y=\\frac{x}y=Q_x[\/latex] and thus\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nDetermine whether [latex]{\\bf{F}}(x,y)=\\langle\\sin{x}\\cos{y},\\cos{x}\\sin{y}\\rangle[\/latex] is conservative.\r\n\r\n[reveal-answer q=\"207340123\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"207340123\"]\r\n\r\nIt is conservative.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793939840\">When using\u00a0Cross-Partial Property of Conservative Fields Theorem, it is important to remember that a theorem is a tool, and like any tool, it can be applied only under the right conditions. In the case of\u00a0Cross-Partial Property of Conservative Fields Theorem, the theorem can be applied only if the domain of the vector field is simply connected.<\/p>\r\n<p id=\"fs-id1167793939853\">To see what can go wrong when misapplying the theorem, consider the vector field:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{\\bf{F}}(x,y)=\\frac{y}{x^2+y^2}{\\bf{i}}+\\frac{-x}{x^2+y^2}{\\bf{j}}}[\/latex].<\/p>\r\n<p id=\"fs-id1167794246756\">This vector field satisfies the cross-partial property, since<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{\\partial}{\\partial{y}}\\left(\\frac{y}{x^2+y^2}\\right)=\\frac{(x^2+y^2)-y(2y)}{(x^2+y^2)^2}=\\frac{x^2-y^2}{(x^2+y^2)^2}}[\/latex]<\/p>\r\n<p id=\"fs-id1167793373571\">and<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{\\partial}{\\partial{x}}\\left(\\frac{-x}{x^2+y^2}\\right)=\\frac{-(x^2+y^2)+x(2x)}{(x^2+y^2)^2}=\\frac{x^2-y^2}{(x^2+y^2)^2}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793464054\">Since\u00a0[latex]{\\bf{F}}[\/latex]\u00a0satisfies the cross-partial property, we might be tempted to conclude that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative. However,\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not conservative. To see this, let<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{\\bf{r}}(t)=\\langle\\cos{t},\\sin{t}\\rangle\\text{, }0\\leq{t}\\leq\\pi}[\/latex]<\/p>\r\n<p id=\"fs-id1167794005233\">be a parameterization of the upper half of a unit circle oriented counterclockwise (denote this [latex]C_1[\/latex]) and let<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{s(t)=\\langle\\cos{t},-\\sin{t}\\rangle\\text{, }0\\leq{t}\\leq\\pi}[\/latex]<\/p>\r\n<p id=\"fs-id1167793427791\">be a parameterization of the lower half of a unit circle oriented clockwise (denote this [latex]C_2[\/latex]). Notice that [latex]C_1[\/latex] and [latex]C_2[\/latex] have the same starting point and endpoint. Since [latex]\\sin^2t+\\cos^2t=1[\/latex],<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^\\prime(t)=\\langle\\sin{(t)},-\\cos{(t)}\\rangle\\cdot\\langle-\\sin{(t)},-=\\cos{(t)}\\rangle}=-1[\/latex]<\/p>\r\n<p id=\"fs-id1167793366626\">and<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\begin{aligned}\r\n{\\bf{F}}(s(t))\\cdot{s}^\\prime(t)&amp;=\\langle-\\sin{t},-\\cos{t}\\rangle\\cdot\\langle-\\sin{t},-\\cos{t}\\rangle \\\\\r\n&amp;=\\sin^2t+\\cos^2t \\\\\r\n&amp;=1\r\n\\end{aligned}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793929615\">Therefore,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\int_0^\\pi-1 \\ dt = -\\pi\\text{ and }\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\int_0^\\pi1 \\ dt=\\pi[\/latex].<\/p>\r\n<p id=\"fs-id1167793488586\">Thus, [latex]C_1[\/latex] and [latex]C_2[\/latex] have the same starting point and endpoint, but [latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}\\ne\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex]. Therefore,\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not independent of path and\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not conservative.<\/p>\r\n<p id=\"fs-id1167793312572\">To summarize:\u00a0[latex]{\\bf{F}}[\/latex]\u00a0satisfies the cross-partial property and yet\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not conservative. What went wrong? Does this contradict\u00a0Cross-Partial Property of Conservative Fields Theorem? The issue is that the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is all of [latex]\\mathbb{R}^2[\/latex] except for the origin. In other words, the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0has a hole at the origin, and therefore the domain is not simply connected. Since the domain is not simply connected,\u00a0Cross-Partial Property of Conservative Fields Theorem\u00a0does not apply to\u00a0[latex]{\\bf{F}}[\/latex].<\/p>\r\n<p id=\"fs-id1167794121462\">We close this section by looking at an example of the usefulness of the Fundamental Theorem for Line Integrals. Now that we can test whether a vector field is conservative, we can always decide whether the Fundamental Theorem for Line Integrals can be used to calculate a vector line integral. If we are asked to calculate an integral of the form [latex]\\displaystyle\\int_{C}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], then our first question should be: Is\u00a0[latex]{\\bf{F}}[\/latex]\u00a0conservative? If the answer is yes, then we should find a potential function and use the Fundamental Theorem for Line Integrals to calculate the integral. If the answer is no, then the Fundamental Theorem for Line Integrals can\u2019t help us and we have to use other methods, such as using the equation to compute vector line integral.<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: using the fundamental theorem for line integrals<\/h3>\r\nCalculate line integral [latex]\\displaystyle\\int_{C}{\\bf{F}}\\cdot{d}r[\/latex], where [latex]{\\bf{F}}(x,y,z)=\\langle2xe^yz+e^xz,x^2e^yz,x^2e^y+e^x\\rangle[\/latex] and [latex]C[\/latex] is any smooth curve that goes from the origin to [latex](1, 1, 1)[\/latex].\r\n\r\n[reveal-answer q=\"539424700\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"539424700\"]\r\n<div id=\"fs-id1167793414462\" class=\" ui-solution-visible\" data-type=\"solution\" aria-label=\"hide solution\" aria-expanded=\"true\"><section class=\"ui-body\" role=\"alert\">\r\n<h2 data-type=\"solution-title\"><span class=\"os-title-label\">Solution<\/span><\/h2>\r\n<div class=\"os-solution-container\">\r\n<p id=\"fs-id1167793414464\">Before trying to compute the integral, we need to determine whether\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative and whether the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is simply connected. The domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is all of [latex]\\mathbb{R}^3[\/latex], which is connected and has no holes. Therefore, the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is simply connected. Let<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{P(x,y,z)=2xe^yz+e^xz,\\text{ }Q(x,y,z)=x^2e^yz,\\text{and }R(x,y,z)=x^2e^y+e^x}[\/latex]<\/p>\r\n<p id=\"fs-id1167794122015\">so that [latex]{\\bf{F}}=\\langle{P},Q,R\\rangle[\/latex]. Since the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is simply connected, we can check the cross partials to determine whether\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative. Note that<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nP_y&amp;=2xe^yz=Q_x \\\\\r\nP_z&amp;=2xe^y+e^x=R_x \\\\\r\nQ_z&amp;=x^2e^y=R_y\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167793300319\">Therefore,\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.<\/p>\r\n<p id=\"fs-id1167793300328\">To evaluate [latex]\\displaystyle\\int_{C}{\\bf{F}}\\cdot{d}r[\/latex] using the Fundamental Theorem for Line Integrals, we need to find a potential function [latex]f[\/latex] for\u00a0[latex]{\\bf{F}}[\/latex]. Let [latex]f[\/latex] be a potential function for\u00a0[latex]{\\bf{F}}[\/latex]. Then, [latex]\\nabla{f}={\\bf{F}}[\/latex], and therefore [latex]f_x=2xe^yz+e^xz[\/latex]. Integrating this equation with respect to [latex]x[\/latex] gives [latex]f(x,y,z)=x^2e^yz+e^xz+h(y,z)[\/latex] for some function [latex]h[\/latex]. Differentiating this equation with respect to [latex]y[\/latex] gives [latex]x^2e^yz+h_y=Q=x^2e^yz[\/latex], which implies that [latex]h_y=0[\/latex]. Therefore, [latex]h[\/latex] is a function of [latex]z[\/latex] only, and [latex]f(x,y,z)=x^2e^yz+e^x+h(z)[\/latex]. To find [latex]h[\/latex], note that [latex]f_z=x^2e^y+e^x+h^\\prime(z)[\/latex]. Therefore, [latex]h^\\prime(z)=0[\/latex] and we can take [latex]h(z)=0[\/latex]. A potential function for\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is [latex]f(x,y,z)=x^2e^yz+e^xz[\/latex].<\/p>\r\n<p id=\"fs-id1167793287126\">Now that we have a potential function, we can use the Fundamental Theorem for Line Integrals to evaluate the integral. By the theorem,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\int_{C}{\\bf{F}}\\cdot{d}{\\bf{r}}&amp;=\\displaystyle\\int_C\\nabla{f}\\cdot{d}{\\bf{r}} \\\\\r\n&amp;=f(1,1,1)-f(0,0,0) \\\\\r\n&amp;=2e\r\n\\end{aligned}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1167794052588\" data-type=\"commentary\">\r\n<h2 id=\"38\" data-type=\"commentary-title\"><span class=\"os-title-label\">Analysis<\/span><\/h2>\r\n<p id=\"fs-id1167794052593\">Notice that if we hadn\u2019t recognized that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative, we would have had to parameterize [latex]C[\/latex] and use [latex]\\displaystyle\\int_{C}{\\bf{F}}\\cdot{\\bf{T}}ds=\\displaystyle\\int_{a}^{b}{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^{\\prime}(t)dt[\/latex]. Since curve [latex]C[\/latex] is unknown, using the Fundamental Theorem for Line Integrals is much simpler.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nCalculate 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].\r\n\r\n[reveal-answer q=\"984750429\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"984750429\"]\r\n\r\n[latex]-10\\pi[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=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\"><\/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>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: work done on a particle<\/h3>\r\nLet [latex]{\\bf{F}}(x,y)=\\langle2xy^2,2x^2y\\rangle[\/latex] be a force field. Suppose that a particle begins its motion at the origin and ends its movement at any point in a plane that is not on the\u00a0<em data-effect=\"italics\">x<\/em>-axis or the\u00a0<em data-effect=\"italics\">y<\/em>-axis. Furthermore, the particle\u2019s motion can be modeled with a smooth parameterization. Show that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0does positive work on the particle.\r\n\r\n[reveal-answer q=\"238416602\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"238416602\"]\r\n<div id=\"fs-id1167794143309\" class=\" ui-solution-visible\" data-type=\"solution\" aria-label=\"hide solution\" aria-expanded=\"true\"><section class=\"ui-body\" role=\"alert\">\r\n<h2 data-type=\"solution-title\"><span class=\"os-title-label\">Solution<\/span><\/h2>\r\n<div class=\"os-solution-container\">\r\n<p id=\"fs-id1167794143311\">We show that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0does positive work on the particle by showing that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative and then by using the Fundamental Theorem for Line Integrals.<\/p>\r\n<p id=\"fs-id1167794143324\">To show that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative, suppose [latex]f(x, y)[\/latex] were a potential function for\u00a0[latex]{\\bf{F}}[\/latex]. Then, [latex]\\nabla{f}={\\bf{F}}=\\langle2xy^2,2x^2y\\rangle[\/latex] and therefore [latex]f_x=2xy^{2}[\/latex] and [latex]f_y=2x^{2}y[\/latex]. Equation [latex]f_x=2xy^{2}[\/latex] implies that [latex]f(x, y)=x^{2}y^{2}+h(y)[\/latex]. Deriving both sides with respect to [latex]y[\/latex] yields [latex]f_y=2x^{2}y+h'(y)[\/latex]. Therefore, [latex]h'(y)[\/latex] and we can take [latex]h(y)=0[\/latex].<\/p>\r\n<p id=\"fs-id1167794209702\">If [latex]f(x, y)=x^{2}y^{2}[\/latex], then note that [latex]\\nabla{f}=\\langle2xy^2,2x^2y\\rangle={\\bf{F}}[\/latex], and therefore [latex]f[\/latex] is a potential function for\u00a0[latex]{\\bf{F}}[\/latex].<\/p>\r\n<p id=\"fs-id1167794024153\">Let [latex](a, b)[\/latex] be the point at which the particle stops is motion, and let [latex]C[\/latex] denote the curve that models the particle\u2019s motion. The work done by\u00a0[latex]{\\bf{F}}[\/latex]\u00a0on the particle is [latex]\\displaystyle\\int_{C}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex]. By the Fundamental Theorem for Line Integrals,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\int_{C}{\\bf{F}}\\cdot{d}{\\bf{r}}&amp;=\\displaystyle\\int_c\\nabla{f}\\cdot{d}{\\bf{r}} \\\\\r\n&amp;=f(a,b)-f(0,0) \\\\\r\n&amp;=a^2b^2\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167793219678\">Since [latex]a\\ne0[\/latex] and [latex]b\\ne0[\/latex], by assumption, [latex]a^2b^2&gt;0[\/latex]. Therefore, [latex]\\displaystyle\\int_{C}{\\bf{F}}\\cdot{d}{\\bf{r}}&gt;0[\/latex], and\u00a0[latex]{\\bf{F}}[\/latex]\u00a0does positive work on the particle.<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1167793421194\" data-type=\"commentary\">\r\n<h2 id=\"41\" data-type=\"commentary-title\"><span class=\"os-title-label\">Analysis<\/span><\/h2>\r\n<p id=\"fs-id1167793421199\">Notice that this problem would be much more difficult without using the Fundamental Theorem for Line Integrals. To apply the tools we have learned, we would need to give a curve parameterization and use [latex]\\displaystyle\\int_{C}{\\bf{F}}\\cdot{\\bf{T}}ds=\\displaystyle\\int_{a}^{b}{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^{\\prime}(t)dt[\/latex]. Since the path of motion [latex]C[\/latex] can be as exotic as we wish (as long as it is smooth), it can be very difficult to parameterize the motion of the particle.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nLet [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?\r\n\r\n[reveal-answer q=\"329893779\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"329893779\"]\r\n\r\nNegative.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Explain how to find a potential function for a conservative vector field.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Use the Fundamental Theorem for Line Integrals to evaluate a line integral in a vector field.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Explain how to test a vector field to determine whether it is conservative.<\/span><\/li>\n<\/ul>\n<\/div>\n<h2 data-type=\"title\">Conservative Vector Fields and Potential Functions<\/h2>\n<p id=\"fs-id1167793404648\">As we have learned, the Fundamental Theorem for Line Integrals says that if\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative, then calculating [latex]\\displaystyle\\int_{C}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex] has two steps: first, find a potential function [latex]f[\/latex] for\u00a0[latex]{\\bf{F}}[\/latex]\u00a0and, second, calculate [latex]f(p_1)-f(P_0)[\/latex], where [latex]P_1[\/latex] is the endpoint of [latex]C[\/latex] and [latex]P_0[\/latex] is the starting point. To use this theorem for a conservative field\u00a0[latex]{\\bf{F}}[\/latex], we must be able to find a potential function [latex]f[\/latex] for\u00a0[latex]{\\bf{F}}[\/latex]. Therefore, we must answer the following question: Given a conservative vector field\u00a0[latex]{\\bf{F}}[\/latex], how do we find a function [latex]f[\/latex] such that [latex]\\nabla{f}={\\bf{F}}[\/latex]? Before giving a general method for finding a potential function, let\u2019s motivate the method with an example.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: finding a potential function<\/h3>\n<p>Find a potential function for [latex]{\\bf{F}}(x,y)=\\langle2xy^3,3x^2y^2+\\cos{(y)}\\rangle[\/latex], thereby showing that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q342587498\">Show Solution<\/span><\/p>\n<div id=\"q342587498\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794097416\">Suppose that [latex]f(x, y)[\/latex] is a potential function for\u00a0[latex]{\\bf{F}}[\/latex]. Then, [latex]\\nabla{f}={\\bf{F}}[\/latex], and therefore<\/p>\n<p style=\"text-align: center;\">[latex]f_x=2xy^3\\text{ and }f_y=3x^2y^2+\\cos{y}[\/latex].<\/p>\n<p id=\"fs-id1167793416675\">Integrating the equation [latex]f_x=2xy^3[\/latex] with respect to [latex]x[\/latex] yields the equation<\/p>\n<p style=\"text-align: center;\">[latex]f(x,y)=x^2y^2+h(y)[\/latex].<\/p>\n<p id=\"fs-id1167793952315\">Notice that since we are integrating a two-variable function with respect to [latex]x[\/latex], we must add a constant of integration that is a constant with respect to [latex]x[\/latex], but may still be a function of [latex]y[\/latex]. The equation [latex]f(x,y)=x^2y^2+h(y)[\/latex] can be confirmed by taking the partial derivative with respect to [latex]x[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\partial{f}}{\\partial{x}}=\\frac{\\partial}{\\partial{x}}(x^2y^3)\\frac{\\partial}{\\partial{x}}(h(y))=2xy^3+0=2xy^3[\/latex].<\/p>\n<p id=\"fs-id1167793928372\">Since [latex]f[\/latex] is a potential function for\u00a0[latex]{\\bf{F}}[\/latex],<\/p>\n<p style=\"text-align: center;\">[latex]f_y=3x^2y^2+\\cos{(y)}[\/latex],<\/p>\n<p id=\"fs-id1167794126213\">and therefore<\/p>\n<p style=\"text-align: center;\">[latex]3x^2y^2+h^\\prime(y)=3x^2y^2+\\cos{(y)}[\/latex].<\/p>\n<p id=\"fs-id1167793931827\">This implies that [latex]h^\\prime(y)=\\cos{(y)}[\/latex], so [latex]h(y)=\\sin{y}+C[\/latex]. Therefore,\u00a0<em data-effect=\"italics\">any<\/em>\u00a0function of the form [latex]f(x,y)=x^2y^3+\\sin{(y)}+C[\/latex] is a potential function. Taking, in particular, [latex]C=0[\/latex] gives the potential function [latex]f(x,y)=x^2y^3+\\sin{(y)}[\/latex].<\/p>\n<p id=\"fs-id1167793505765\">To verify that [latex]f[\/latex] is a potential function, note that [latex]\\nabla{f}=\\langle2xy^3,3x^2y^2+\\cos{y}\\rangle={\\bf{F}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q834728049\">Show Solution<\/span><\/p>\n<div id=\"q834728049\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f(x,y)=e^xy^3+xy[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The logic of the previous example extends to finding the potential function for any conservative vector field in [latex]\\mathbb{R}^2[\/latex]. Thus, we have the following problem-solving strategy for finding potential functions:<\/p>\n<div id=\"fs-id1167794333153\" class=\"problem-solving\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox examples\">\n<h3>Problem solving strategy: finding a potential function for a conservative vector field\u00a0[latex]{\\bf{F}}(X,Y)=<{P}(X,Y),Q(X,Y)>[\/latex]<\/h3>\n<ol id=\"fs-id1167793503117\" type=\"1\">\n<li>Integrate [latex]P[\/latex] with respect to [latex]x[\/latex]. This results in a function of the form [latex]g(x, y)+h(y)[\/latex], where [latex]h(y)[\/latex] is unknown.<\/li>\n<li>Take the partial derivative of [latex]g(x, y)+h(y)[\/latex] with respect to [latex]y[\/latex], which results in the function [latex]g_y(x, y)+h'(y)[\/latex].<\/li>\n<li>Use the equation [latex]g_y(x, y)+h'(y)=Q(x ,y)[\/latex] to find [latex]h'(y)[\/latex].<\/li>\n<li>Integrate [latex]h'(y)[\/latex] to find [latex]h(y)[\/latex].<\/li>\n<li>Any function of the form [latex]f(x,y)=g(x,y)+h(y)+C[\/latex], where [latex]C[\/latex] is a constant, is a potential function for\u00a0[latex]{\\bf{F}}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<p>We can adapt this strategy to find potential functions for vector fields in [latex]\\mathbb{R}^3[\/latex], as shown in the next example.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: finding a potential function in\u00a0[latex]\\mathbb{R}^3[\/latex]<\/h3>\n<p>Find a potential function for [latex]{\\bf{F}}(x,y,z)=\\langle2xy,x^2+2yz^3,3y^2z^2+2z\\rangle[\/latex], thereby showing that [latex]{\\bf{F}}[\/latex] is conservative.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q764523874\">Show Solution<\/span><\/p>\n<div id=\"q764523874\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794042874\">Suppose that [latex]f[\/latex] is a potential function. Then, [latex]\\nabla{f}={\\bf{F}}[\/latex] and therefore [latex]f_x=2xy[\/latex]. Integrating this equation with respect to [latex]x[\/latex] yields the equation [latex]f(x, y, z)=x^{2}y+g(y, z)[\/latex] for some function [latex]g[\/latex]. Notice that, in this case, the constant of integration with respect to [latex]x[\/latex] is a function of [latex]y[\/latex] and\u00a0[latex]z[\/latex].<\/p>\n<p id=\"fs-id1167793631905\">Since [latex]f[\/latex] is a potential function,<\/p>\n<p style=\"text-align: center;\">[latex]x^{2}+2yz^{3}=f_y=x^{2}+g_y[\/latex].<\/p>\n<p id=\"fs-id1167793632719\">Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]g_y=2yz^{3}[\/latex].<\/p>\n<p id=\"fs-id1167794210246\">Integrating this function with respect to [latex]y[\/latex] yields<\/p>\n<p style=\"text-align: center;\">[latex]g(y, z)=y^{2}z^{3}+h(z)[\/latex]<\/p>\n<p id=\"fs-id1167794136180\">for some function [latex]h(z)[\/latex] of [latex]z[\/latex] alone. (Notice that, because we know that [latex]g[\/latex] is a function of only [latex]y[\/latex] and [latex]z[\/latex], we do not need to write [latex]g(y, z)=y^{2}z^{3}+h(x,z)[\/latex].) Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]f(z,y,z)=x^2+y+g(y,z)=x^2y+y^2z^3+h(z)[\/latex].<\/p>\n<p id=\"fs-id1167793397795\">To find [latex]f[\/latex], we now must only find [latex]h[\/latex] Since [latex]f[\/latex] is a potential function,<\/p>\n<p style=\"text-align: center;\">[latex]3y^2z^2+2z=g_z=3y^2z^2+h^\\prime(z)[\/latex].<\/p>\n<p id=\"fs-id1167793244255\">This implies that [latex]h^\\prime(z)=2z[\/latex], so [latex]h(z)=z^{2}+C[\/latex]. Letting [latex]C=0[\/latex] gives the potential function<\/p>\n<p style=\"text-align: center;\">[latex]f(x,y,z)=x^2y+y^2z^3+z^2[\/latex].<\/p>\n<p id=\"fs-id1167793498629\">To verify that [latex]f[\/latex] is a potential function, note that [latex]\\nabla{f}=\\langle2xy,x^2+2yz^3,3y^2z^2+2z\\rangle[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q349856201\">Show Solution<\/span><\/p>\n<div id=\"q349856201\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f(x,y,z)=4x^3+\\sin{y}\\cos{z}+z[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\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\"><\/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<p id=\"fs-id1167793829179\">We can apply the process of finding a potential function to a\u00a0<span id=\"47dae7f0-fc5d-4d6c-a882-d19c748319d9_term258\" class=\"no-emphasis\" data-type=\"term\">gravitational force<\/span>. Recall that, if an object has unit mass and is located at the origin, then the gravitational force in [latex]\\mathbb{R}^2[\/latex] that the object exerts on another object of unit mass at the point [latex](x, y)[\/latex] is given by vector field<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{\\bf{F}}(x,y)=-G\\left\\langle\\frac{x}{(x^2+y^2)^{3\/2}},\\frac{y}{(x^2+y^2)^{3\/2}}\\right\\rangle}[\/latex],<\/p>\n<p id=\"fs-id1167793266873\">where [latex]G[\/latex] is the universal gravitational constant. In the next example, we build a potential function for\u00a0[latex]{\\bf{F}}[\/latex], thus confirming what we already know: that gravity is conservative.<\/p>\n<div id=\"fs-id1167793266888\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: finding a potential function<\/h3>\n<p>Find a potential function [latex]f[\/latex] for [latex]{\\bf{F}}(x,y)=-G\\left\\langle\\frac{x}{(x^2+y^2)^{3\/2}},\\frac{y}{(x^2+y^2)^{3\/2}}\\right\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q384572983\">Show Solution<\/span><\/p>\n<div id=\"q384572983\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793398054\">Suppose that [latex]f[\/latex] is a potential function. Then, [latex]\\nabla{f}={\\bf{F}}[\/latex] and therefore<\/p>\n<p style=\"text-align: center;\">[latex]\\large{f_x=\\frac{-Gx}{(x^2+y^2)^{3\/2}}}[\/latex].<\/p>\n<p id=\"fs-id1167793604079\">To integrate this function with respect to [latex]x[\/latex], we can use [latex]u[\/latex]-substitution. If [latex]u=x^{2}+y^{2}[\/latex] then [latex]\\frac{du}2=x \\ dx[\/latex], so<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\int\\frac{-Gx}{(x^2+y^2)^{3\/2}}&=\\displaystyle\\int\\frac{-G}{2u^{3\/2}} \\ du \\\\  &=\\frac{G}{\\sqrt{u}}+h(y) \\\\  &=\\frac{G}{\\sqrt{x^2+y^2}}+h(y)  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167793505997\">for some function [latex]h(y)[\/latex]. Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{f(x,y)=\\frac{G}{\\sqrt{(x^2+y^2)}}+h(y)}[\/latex].<\/p>\n<p id=\"fs-id1167793432912\">Since [latex]f[\/latex] i<span style=\"font-size: 1rem; text-align: initial;\">s a potential function for\u00a0[latex]{\\bf{F}}[\/latex]<\/span><span style=\"font-size: 1rem; text-align: initial;\">,<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\large{f_y=\\frac{-Gy}{(x^2+y^2)^{3\/2}}}[\/latex].<\/p>\n<p id=\"fs-id1167794097277\">Since [latex]f(x,y)=\\frac{G}{\\sqrt{x^2+y^2}}+h(y)[\/latex],\u00a0[latex]f_y[\/latex] also equals [latex]\\frac{-Gy}{(x^2+y^2)^{3\/2}}+h^\\prime(y)[\/latex].<\/p>\n<p id=\"fs-id1167794200726\">Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac{-Gy}{(x^2+y^2)^{3\/2}}+h^\\prime(y)=\\frac{-Gy}{(x^2+y^2)^{3\/2}}}[\/latex],<\/p>\n<p id=\"fs-id1167793613305\">which implies that [latex]h^\\prime(y)=0[\/latex]. Thus, we can take [latex]h(y)[\/latex] to be any constant; in particular, we can let [latex]h(y)=0[\/latex]. The function<\/p>\n<p style=\"text-align: center;\">[latex]f(x,y)=\\frac{G}{\\sqrt{x^2+y^2}}[\/latex]<\/p>\n<p id=\"fs-id1167794127219\">is a potential function for the gravitational field\u00a0[latex]{\\bf{F}}[\/latex]. To confirm that [latex]f[\/latex] is a potential function, note that<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\nabla{f}&=\\left\\langle-\\frac12\\frac{G}{(x^2+y^2)^{3\/2}}(2x),-\\frac12\\frac{G}{(x^2+y^2)^{3\/2}}(2y)\\right\\rangle \\\\  &=\\left\\langle\\frac{-Gx}{(x^2+y^2)^{3\/2}},\\frac{-Gy}{(x^2+y^2)^{3\/2}}\\right\\rangle \\\\  &={\\bf{F}}  \\end{aligned}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q204873094\">Show Solution<\/span><\/p>\n<div id=\"q204873094\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f(x,y,z)=\\frac{G}{\\sqrt{x^2+y^2+z^2}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2 data-type=\"title\">Testing a Vector Field<\/h2>\n<p id=\"fs-id1167793720011\">Until now, we have worked with vector fields that we know are conservative, but if we are not told that a vector field is conservative, we need to be able to test whether it is conservative. Recall that, if\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative, then\u00a0[latex]{\\bf{F}}[\/latex]\u00a0has the cross-partial property (see\u00a0The Cross-Partial Property of Conservative Vector Fields Theorem). That is, if [latex]{\\bf{F}}=\\langle{P},Q,R\\rangle[\/latex] is conservative, then [latex]P_y=Q_x[\/latex], [latex]P_z=R_x[\/latex], and [latex]Q_z=R_y[\/latex],\u00a0 So, if\u00a0[latex]{\\bf{F}}[\/latex]\u00a0has the cross-partial property, then is\u00a0[latex]{\\bf{F}}[\/latex]\u00a0conservative? If the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is open and simply connected, then the answer is yes.<\/p>\n<\/div>\n<div data-type=\"example\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: the cross-partial test for conservative fields<\/h3>\n<hr \/>\n<p>If [latex]{\\bf{F}}=\\langle{P},Q,R\\rangle[\/latex] is a vector field on an open, simply connected region [latex]D[\/latex] and [latex]P_y=Q_x[\/latex], [latex]P_z=R_x[\/latex], and [latex]Q_z=R_y[\/latex] throughout [latex]D[\/latex], then\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.<\/p>\n<\/div>\n<p id=\"fs-id1167793945273\">Although a proof of this theorem is beyond the scope of the text, we can discover its power with some examples. Later, we see why it is necessary for the region to be simply connected.<\/p>\n<p id=\"fs-id1167793945278\">Combining this theorem with the cross-partial property, we can determine whether a given vector field is conservative:<\/p>\n<\/div>\n<div data-type=\"example\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: the cross-partial property of conservative fields<\/h3>\n<hr \/>\n<p>Let [latex]{\\bf{F}}=\\langle{P},Q,R\\rangle[\/latex] be a vector field on an open, simply connected region [latex]D[\/latex]. Then [latex]P_y=Q_x[\/latex], [latex]P_z=R_x[\/latex], and [latex]Q_z=R_y[\/latex] throughout [latex]D[\/latex] if and only if\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.<\/p>\n<\/div>\n<p>The version of this theorem in [latex]\\mathbb{R}^2[\/latex] is also true. If [latex]{\\bf{F}}=\\langle{P},Q\\rangle[\/latex] is a vector field on an open, simply connected domain in [latex]\\mathbb{R}^2[\/latex], then\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative if and only if [latex]P_y=Q_x[\/latex].<\/p>\n<\/div>\n<div data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: determining whether a vector field is conservative<\/h3>\n<p>Determine whether vector field [latex]{\\bf{F}}(x,y,z)=\\langle{x}y^2z,x^2yz,z^2\\rangle[\/latex] is conservative.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q432875399\">Show Solution<\/span><\/p>\n<div id=\"q432875399\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793441838\">Note that the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is all of [latex]\\mathbb{R}^2[\/latex] and [latex]\\mathbb{R}^3[\/latex] is simply connected. Therefore, we can use\u00a0Cross-Partial Property of Conservative Fields Theorem\u00a0to determine whether\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative. Let<\/p>\n<p style=\"text-align: center;\">[latex]P(x,y,z)=xy^2z,\\text{ }Q(x,y,z)=x^2yz,\\text{ and }R(x,y,z)=z^2[\/latex].<\/p>\n<p id=\"fs-id1167793718546\">Since [latex]Q_z=x^{2}y[\/latex] and [latex]R_y=0[\/latex], the vector field is not conservative.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: determining whether a vector field is conservative<\/h3>\n<p>Determine vector field [latex]{\\bf{F}}(x,y)=\\left\\langle{x}\\ln{(y)},\\frac{x^2}{2y}\\right\\rangle[\/latex] is conservative.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q403750008\">Show Solution<\/span><\/p>\n<div id=\"q403750008\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793940254\">Note that the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is the part of [latex]\\mathbb{R}^2[\/latex] in which [latex]y>0[\/latex]. Thus, the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is part of a plane above the\u00a0<em data-effect=\"italics\">x<\/em>-axis, and this domain is simply connected (there are no holes in this region and this region is connected). Therefore, we can use\u00a0Cross-Partial Property of Conservative Fields Theorem\u00a0to determine whether\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative. Let<\/p>\n<p style=\"text-align: center;\">[latex]P(x,y)=x\\ln{(y)}\\text{ and  }Q(x,y)=\\frac{x^2}{2y}[\/latex].<\/p>\n<p id=\"fs-id1167793925934\">Then [latex]P_y=\\frac{x}y=Q_x[\/latex] and thus\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q207340123\">Show Solution<\/span><\/p>\n<div id=\"q207340123\" class=\"hidden-answer\" style=\"display: none\">\n<p>It is conservative.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167793939840\">When using\u00a0Cross-Partial Property of Conservative Fields Theorem, it is important to remember that a theorem is a tool, and like any tool, it can be applied only under the right conditions. In the case of\u00a0Cross-Partial Property of Conservative Fields Theorem, the theorem can be applied only if the domain of the vector field is simply connected.<\/p>\n<p id=\"fs-id1167793939853\">To see what can go wrong when misapplying the theorem, consider the vector field:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{\\bf{F}}(x,y)=\\frac{y}{x^2+y^2}{\\bf{i}}+\\frac{-x}{x^2+y^2}{\\bf{j}}}[\/latex].<\/p>\n<p id=\"fs-id1167794246756\">This vector field satisfies the cross-partial property, since<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac{\\partial}{\\partial{y}}\\left(\\frac{y}{x^2+y^2}\\right)=\\frac{(x^2+y^2)-y(2y)}{(x^2+y^2)^2}=\\frac{x^2-y^2}{(x^2+y^2)^2}}[\/latex]<\/p>\n<p id=\"fs-id1167793373571\">and<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac{\\partial}{\\partial{x}}\\left(\\frac{-x}{x^2+y^2}\\right)=\\frac{-(x^2+y^2)+x(2x)}{(x^2+y^2)^2}=\\frac{x^2-y^2}{(x^2+y^2)^2}}[\/latex].<\/p>\n<p id=\"fs-id1167793464054\">Since\u00a0[latex]{\\bf{F}}[\/latex]\u00a0satisfies the cross-partial property, we might be tempted to conclude that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative. However,\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not conservative. To see this, let<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{\\bf{r}}(t)=\\langle\\cos{t},\\sin{t}\\rangle\\text{, }0\\leq{t}\\leq\\pi}[\/latex]<\/p>\n<p id=\"fs-id1167794005233\">be a parameterization of the upper half of a unit circle oriented counterclockwise (denote this [latex]C_1[\/latex]) and let<\/p>\n<p style=\"text-align: center;\">[latex]\\large{s(t)=\\langle\\cos{t},-\\sin{t}\\rangle\\text{, }0\\leq{t}\\leq\\pi}[\/latex]<\/p>\n<p id=\"fs-id1167793427791\">be a parameterization of the lower half of a unit circle oriented clockwise (denote this [latex]C_2[\/latex]). Notice that [latex]C_1[\/latex] and [latex]C_2[\/latex] have the same starting point and endpoint. Since [latex]\\sin^2t+\\cos^2t=1[\/latex],<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^\\prime(t)=\\langle\\sin{(t)},-\\cos{(t)}\\rangle\\cdot\\langle-\\sin{(t)},-=\\cos{(t)}\\rangle}=-1[\/latex]<\/p>\n<p id=\"fs-id1167793366626\">and<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\begin{aligned}  {\\bf{F}}(s(t))\\cdot{s}^\\prime(t)&=\\langle-\\sin{t},-\\cos{t}\\rangle\\cdot\\langle-\\sin{t},-\\cos{t}\\rangle \\\\  &=\\sin^2t+\\cos^2t \\\\  &=1  \\end{aligned}}[\/latex].<\/p>\n<p id=\"fs-id1167793929615\">Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\int_0^\\pi-1 \\ dt = -\\pi\\text{ and }\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\int_0^\\pi1 \\ dt=\\pi[\/latex].<\/p>\n<p id=\"fs-id1167793488586\">Thus, [latex]C_1[\/latex] and [latex]C_2[\/latex] have the same starting point and endpoint, but [latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}\\ne\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex]. Therefore,\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not independent of path and\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not conservative.<\/p>\n<p id=\"fs-id1167793312572\">To summarize:\u00a0[latex]{\\bf{F}}[\/latex]\u00a0satisfies the cross-partial property and yet\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not conservative. What went wrong? Does this contradict\u00a0Cross-Partial Property of Conservative Fields Theorem? The issue is that the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is all of [latex]\\mathbb{R}^2[\/latex] except for the origin. In other words, the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0has a hole at the origin, and therefore the domain is not simply connected. Since the domain is not simply connected,\u00a0Cross-Partial Property of Conservative Fields Theorem\u00a0does not apply to\u00a0[latex]{\\bf{F}}[\/latex].<\/p>\n<p id=\"fs-id1167794121462\">We close this section by looking at an example of the usefulness of the Fundamental Theorem for Line Integrals. Now that we can test whether a vector field is conservative, we can always decide whether the Fundamental Theorem for Line Integrals can be used to calculate a vector line integral. If we are asked to calculate an integral of the form [latex]\\displaystyle\\int_{C}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], then our first question should be: Is\u00a0[latex]{\\bf{F}}[\/latex]\u00a0conservative? If the answer is yes, then we should find a potential function and use the Fundamental Theorem for Line Integrals to calculate the integral. If the answer is no, then the Fundamental Theorem for Line Integrals can\u2019t help us and we have to use other methods, such as using the equation to compute vector line integral.<\/p>\n<\/div>\n<div data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: using the fundamental theorem for line integrals<\/h3>\n<p>Calculate line integral [latex]\\displaystyle\\int_{C}{\\bf{F}}\\cdot{d}r[\/latex], where [latex]{\\bf{F}}(x,y,z)=\\langle2xe^yz+e^xz,x^2e^yz,x^2e^y+e^x\\rangle[\/latex] and [latex]C[\/latex] is any smooth curve that goes from the origin to [latex](1, 1, 1)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q539424700\">Show Solution<\/span><\/p>\n<div id=\"q539424700\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167793414462\" class=\"ui-solution-visible\" data-type=\"solution\" aria-label=\"hide solution\" aria-expanded=\"true\">\n<section class=\"ui-body\" role=\"alert\">\n<h2 data-type=\"solution-title\"><span class=\"os-title-label\">Solution<\/span><\/h2>\n<div class=\"os-solution-container\">\n<p id=\"fs-id1167793414464\">Before trying to compute the integral, we need to determine whether\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative and whether the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is simply connected. The domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is all of [latex]\\mathbb{R}^3[\/latex], which is connected and has no holes. Therefore, the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is simply connected. Let<\/p>\n<p style=\"text-align: center;\">[latex]\\large{P(x,y,z)=2xe^yz+e^xz,\\text{ }Q(x,y,z)=x^2e^yz,\\text{and }R(x,y,z)=x^2e^y+e^x}[\/latex]<\/p>\n<p id=\"fs-id1167794122015\">so that [latex]{\\bf{F}}=\\langle{P},Q,R\\rangle[\/latex]. Since the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is simply connected, we can check the cross partials to determine whether\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative. Note that<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  P_y&=2xe^yz=Q_x \\\\  P_z&=2xe^y+e^x=R_x \\\\  Q_z&=x^2e^y=R_y  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167793300319\">Therefore,\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.<\/p>\n<p id=\"fs-id1167793300328\">To evaluate [latex]\\displaystyle\\int_{C}{\\bf{F}}\\cdot{d}r[\/latex] using the Fundamental Theorem for Line Integrals, we need to find a potential function [latex]f[\/latex] for\u00a0[latex]{\\bf{F}}[\/latex]. Let [latex]f[\/latex] be a potential function for\u00a0[latex]{\\bf{F}}[\/latex]. Then, [latex]\\nabla{f}={\\bf{F}}[\/latex], and therefore [latex]f_x=2xe^yz+e^xz[\/latex]. Integrating this equation with respect to [latex]x[\/latex] gives [latex]f(x,y,z)=x^2e^yz+e^xz+h(y,z)[\/latex] for some function [latex]h[\/latex]. Differentiating this equation with respect to [latex]y[\/latex] gives [latex]x^2e^yz+h_y=Q=x^2e^yz[\/latex], which implies that [latex]h_y=0[\/latex]. Therefore, [latex]h[\/latex] is a function of [latex]z[\/latex] only, and [latex]f(x,y,z)=x^2e^yz+e^x+h(z)[\/latex]. To find [latex]h[\/latex], note that [latex]f_z=x^2e^y+e^x+h^\\prime(z)[\/latex]. Therefore, [latex]h^\\prime(z)=0[\/latex] and we can take [latex]h(z)=0[\/latex]. A potential function for\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is [latex]f(x,y,z)=x^2e^yz+e^xz[\/latex].<\/p>\n<p id=\"fs-id1167793287126\">Now that we have a potential function, we can use the Fundamental Theorem for Line Integrals to evaluate the integral. By the theorem,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\int_{C}{\\bf{F}}\\cdot{d}{\\bf{r}}&=\\displaystyle\\int_C\\nabla{f}\\cdot{d}{\\bf{r}} \\\\  &=f(1,1,1)-f(0,0,0) \\\\  &=2e  \\end{aligned}[\/latex].<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1167794052588\" data-type=\"commentary\">\n<h2 id=\"38\" data-type=\"commentary-title\"><span class=\"os-title-label\">Analysis<\/span><\/h2>\n<p id=\"fs-id1167794052593\">Notice that if we hadn\u2019t recognized that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative, we would have had to parameterize [latex]C[\/latex] and use [latex]\\displaystyle\\int_{C}{\\bf{F}}\\cdot{\\bf{T}}ds=\\displaystyle\\int_{a}^{b}{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^{\\prime}(t)dt[\/latex]. Since curve [latex]C[\/latex] is unknown, using the Fundamental Theorem for Line Integrals is much simpler.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q984750429\">Show Solution<\/span><\/p>\n<div id=\"q984750429\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]-10\\pi[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=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\"><\/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<div class=\"textbox exercises\">\n<h3>Example: work done on a particle<\/h3>\n<p>Let [latex]{\\bf{F}}(x,y)=\\langle2xy^2,2x^2y\\rangle[\/latex] be a force field. Suppose that a particle begins its motion at the origin and ends its movement at any point in a plane that is not on the\u00a0<em data-effect=\"italics\">x<\/em>-axis or the\u00a0<em data-effect=\"italics\">y<\/em>-axis. Furthermore, the particle\u2019s motion can be modeled with a smooth parameterization. Show that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0does positive work on the particle.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q238416602\">Show Solution<\/span><\/p>\n<div id=\"q238416602\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794143309\" class=\"ui-solution-visible\" data-type=\"solution\" aria-label=\"hide solution\" aria-expanded=\"true\">\n<section class=\"ui-body\" role=\"alert\">\n<h2 data-type=\"solution-title\"><span class=\"os-title-label\">Solution<\/span><\/h2>\n<div class=\"os-solution-container\">\n<p id=\"fs-id1167794143311\">We show that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0does positive work on the particle by showing that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative and then by using the Fundamental Theorem for Line Integrals.<\/p>\n<p id=\"fs-id1167794143324\">To show that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative, suppose [latex]f(x, y)[\/latex] were a potential function for\u00a0[latex]{\\bf{F}}[\/latex]. Then, [latex]\\nabla{f}={\\bf{F}}=\\langle2xy^2,2x^2y\\rangle[\/latex] and therefore [latex]f_x=2xy^{2}[\/latex] and [latex]f_y=2x^{2}y[\/latex]. Equation [latex]f_x=2xy^{2}[\/latex] implies that [latex]f(x, y)=x^{2}y^{2}+h(y)[\/latex]. Deriving both sides with respect to [latex]y[\/latex] yields [latex]f_y=2x^{2}y+h'(y)[\/latex]. Therefore, [latex]h'(y)[\/latex] and we can take [latex]h(y)=0[\/latex].<\/p>\n<p id=\"fs-id1167794209702\">If [latex]f(x, y)=x^{2}y^{2}[\/latex], then note that [latex]\\nabla{f}=\\langle2xy^2,2x^2y\\rangle={\\bf{F}}[\/latex], and therefore [latex]f[\/latex] is a potential function for\u00a0[latex]{\\bf{F}}[\/latex].<\/p>\n<p id=\"fs-id1167794024153\">Let [latex](a, b)[\/latex] be the point at which the particle stops is motion, and let [latex]C[\/latex] denote the curve that models the particle\u2019s motion. The work done by\u00a0[latex]{\\bf{F}}[\/latex]\u00a0on the particle is [latex]\\displaystyle\\int_{C}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex]. By the Fundamental Theorem for Line Integrals,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\int_{C}{\\bf{F}}\\cdot{d}{\\bf{r}}&=\\displaystyle\\int_c\\nabla{f}\\cdot{d}{\\bf{r}} \\\\  &=f(a,b)-f(0,0) \\\\  &=a^2b^2  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167793219678\">Since [latex]a\\ne0[\/latex] and [latex]b\\ne0[\/latex], by assumption, [latex]a^2b^2>0[\/latex]. Therefore, [latex]\\displaystyle\\int_{C}{\\bf{F}}\\cdot{d}{\\bf{r}}>0[\/latex], and\u00a0[latex]{\\bf{F}}[\/latex]\u00a0does positive work on the particle.<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1167793421194\" data-type=\"commentary\">\n<h2 id=\"41\" data-type=\"commentary-title\"><span class=\"os-title-label\">Analysis<\/span><\/h2>\n<p id=\"fs-id1167793421199\">Notice that this problem would be much more difficult without using the Fundamental Theorem for Line Integrals. To apply the tools we have learned, we would need to give a curve parameterization and use [latex]\\displaystyle\\int_{C}{\\bf{F}}\\cdot{\\bf{T}}ds=\\displaystyle\\int_{a}^{b}{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^{\\prime}(t)dt[\/latex]. Since the path of motion [latex]C[\/latex] can be as exotic as we wish (as long as it is smooth), it can be very difficult to parameterize the motion of the particle.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q329893779\">Show Solution<\/span><\/p>\n<div id=\"q329893779\" class=\"hidden-answer\" style=\"display: none\">\n<p>Negative.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1121\">\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 6.29. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>CP 6.32. <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":428269,"menu_order":13,"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 6.29\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"CP 6.32\",\"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-1121","chapter","type-chapter","status-publish","hentry"],"part":24,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1121","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\/428269"}],"version-history":[{"count":115,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1121\/revisions"}],"predecessor-version":[{"id":6398,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1121\/revisions\/6398"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/24"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1121\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=1121"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=1121"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=1121"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=1121"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}