{"id":5467,"date":"2022-06-02T18:37:41","date_gmt":"2022-06-02T18:37:41","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=5467"},"modified":"2022-11-01T05:01:31","modified_gmt":"2022-11-01T05:01:31","slug":"gradient-fields","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/gradient-fields\/","title":{"raw":"Gradient Fields","rendered":"Gradient 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\">Identify a conservative field and its associated potential function.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 data-type=\"title\">Gradient Fields<\/h2>\r\n<p id=\"fs-id1167793939446\">In this section, we study a special kind of vector field called a gradient field or a\u00a0<strong><span id=\"4ea1e6ac-b330-4b3a-a84a-b899b4fc0de5_term235\" data-type=\"term\">conservative field<\/span><\/strong>. These vector fields are extremely important in physics because they can be used to model physical systems in which energy is conserved. Gravitational fields and electric fields associated with a static charge are examples of gradient fields.<\/p>\r\n<p id=\"fs-id1167793939454\">Recall that if [latex]f[\/latex] is a (scalar) function of [latex]x[\/latex] and [latex]y[\/latex], then the gradient of [latex]f[\/latex] is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\text{grad}f=\\nabla{f}=f_x(x,y){\\bf{i}}+f_y(x,y){\\bf{j}}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793697779\">We can see from the form in which the gradient is written that [latex]\\nabla{f}[\/latex]. is a vector field in [latex]\\mathbb{R}^2[\/latex]. Similarly, if [latex]f[\/latex] is a function of [latex]x[\/latex], [latex]y[\/latex], and [latex]z[\/latex], then the gradient of [latex]f[\/latex] is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\text{grad}f=\\nabla{f}=f_x(x,y,z){\\bf{i}}+f_y(x,y,z){\\bf{j}}+f_z(x,y,z){\\bf{k}}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793450778\">The gradient of a three-variable function is a vector field in [latex]\\mathbb{R}^3[\/latex].<\/p>\r\n<p id=\"fs-id1167793266860\">A gradient field is a vector field that can be written as the gradient of a function, and we have the following definition.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\n<div id=\"fs-id1167793266865\" class=\"ui-has-child-title\" data-type=\"note\"><section>\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-id1167793266869\">A vector field\u00a0[latex]\\bf{F}[\/latex]\u00a0in [latex]\\mathbb{R}^2[\/latex] or in [latex]\\mathbb{R}^3[\/latex] is a<strong>\u00a0<span id=\"4ea1e6ac-b330-4b3a-a84a-b899b4fc0de5_term236\" data-type=\"term\">gradient field<\/span>\u00a0<\/strong>if there exists a scalar function [latex]f[\/latex] such that [latex]\\nabla{f}={\\bf{F}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: sketching a gradient vector field<\/h3>\r\nUse technology to plot the gradient vector field of [latex]f(x, y)=x^{2}y^{2}[\/latex].\r\n\r\n[reveal-answer q=\"277324891\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"277324891\"]\r\n<p id=\"fs-id1167793543917\">The gradient of [latex]f[\/latex] is [latex]\\nabla{f}=\\langle2xy^2,2x^2y\\rangle[\/latex]. To sketch the vector field, use a computer algebra system such as Mathematica.\u00a0Figure 1\u00a0shows [latex]\\nabla{f}[\/latex].<\/p>\r\n\r\n[caption id=\"attachment_5237\" align=\"aligncenter\" width=\"667\"]<img class=\"size-full wp-image-5237\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31215639\/6.9.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/b91be2eedc7aabbebf23e61b81fb423f0410f70e&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A visual representation of the given gradient vector field in two dimensions. The arrows point up above the x axis and down below the x axis, and they point left on the left side of the y axis and to the right on the right side of the y axis. The further the arrows are from zero, the more vertical they are, and the closer the arrows are to zero, the more horizontal they are.&quot; id=&quot;31&quot;&gt;\" width=\"667\" height=\"395\" \/> Figure 1. The gradient vector field is [latex]\\nabla{f}[\/latex], where [latex]f(x, y)=x^{2}y^{2}[\/latex].[\/caption][\/hidden-answer]<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nUse technology to plot the gradient vector field of\u00a0[latex]f(x,y)=\\sin{x}\\cos{y}[\/latex].\r\n\r\n[reveal-answer q=\"485726668\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"485726668\"]\r\n\r\n[caption id=\"attachment_3309\" align=\"aligncenter\" width=\"717\"]<img class=\"wp-image-3309 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/19162738\/6-1-tryitans4.jpeg\" alt=\"\" width=\"717\" height=\"497\" \/> Figure 2. A virtual representation of the gradient vector field\u00a0[latex]f(x,y)=\\sin{x}\\cos{y}[\/latex].[\/caption][\/hidden-answer]<\/div>\r\nConsider the function [latex]f(x, y)=x^{2}y^{2}[\/latex] from\u00a0Example \"Sketching a Gradient Vector Field\".\u00a0Figure 4\u00a0shows the level curves of this function overlaid on the function\u2019s gradient vector field. The gradient vectors are perpendicular to the level curves, and the magnitudes of the vectors get larger as the level curves get closer together, because closely grouped level curves indicate the graph is steep, and the magnitude of the gradient vector is the largest value of the directional derivative. Therefore, you can see the local steepness of a graph by investigating the corresponding function\u2019s gradient field.\r\n\r\n[caption id=\"attachment_3310\" align=\"aligncenter\" width=\"667\"]<img class=\"wp-image-3310 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/19162800\/6-1-7.jpeg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/516e6d511466956acbd12fe82171045a9794197b&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A visual representation of the given gradient field. The arrows are flatter the closer they are to the x axis and more vertical the further they are from the x axis. The arrows point left to the left of the y axis, and they point to the right to the right of the y axis. They point up above the x axis and down below the x axis. Several level curves are drawn, each asymptotically approaching the axes. As the level curves get closer together, the magnitude of the gradient vectors increases.&quot; id=&quot;34&quot;&gt;\" width=\"667\" height=\"395\" \/> Figure 3. The gradient field of [latex]f(x, y)=x^{2}y^{2}[\/latex] and several level curves of [latex]f[\/latex]. Notice that as the level curves get closer together, the magnitude of the gradient vectors increases.[\/caption]\r\n<p id=\"fs-id1167793662272\">As we learned earlier, a vector field\u00a0[latex]{\\bf{F}}[\/latex] is a conservative vector field, or a gradient field if there exists a scalar function [latex]f[\/latex] such that [latex]\\nabla{f}={\\bf{F}}[\/latex]. In this situation, [latex]f[\/latex] is called a\u00a0<strong><span id=\"4ea1e6ac-b330-4b3a-a84a-b899b4fc0de5_term237\" data-type=\"term\">potential function<\/span><\/strong>\u00a0for [latex]{\\bf{F}}[\/latex]. Conservative vector fields arise in many applications, particularly in physics. The reason such fields are called\u00a0<em data-effect=\"italics\">conservative<\/em>\u00a0is that they model forces of physical systems in which energy is conserved. We study conservative vector fields in more detail later in this chapter.<\/p>\r\n<p id=\"fs-id1167793776652\">You might notice that, in some applications, a potential function [latex]f[\/latex] for\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is defined instead as a function such that [latex]-\\nabla{f}={\\bf{F}}[\/latex]. This is the case for certain contexts in physics, for example.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: verifying a potential function<\/h3>\r\n<p id=\"fs-id1167793465191\">Is [latex]f(x,y,z)=x^2yz-\\sin{(xy)}[\/latex] a potential function for vector field\u00a0[latex]{\\bf{F}}(x,y,z)=\\langle2xyz-y\\cos{(xy)},x^2z-x\\cos{(xy)},x^2y\\rangle[\/latex]?<\/p>\r\n[reveal-answer q=\"802452044\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"802452044\"]\r\n<p id=\"fs-id1167793465191\">We need to confirm whether [latex]\\nabla{f}={\\bf{F}}[\/latex]. We have<\/p>\r\n<p style=\"text-align: center;\">[latex]f_x=2xyz-y\\cos{(xy)},f_y=x^2z-x\\cos{(xy)},\\text{ and }f_z=x^2y[\/latex]<\/p>\r\nTherefore, [latex]\\nabla{f}={\\bf{F}}[\/latex] and [latex]f[\/latex] is a potential function for [latex]{\\bf{F}}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nIs [latex]f(x,y,z)=x^2\\cos{(yz)}+y^2z^2[\/latex] a potential function for [latex]{\\bf{F}}(x,y,z)=\\langle2x\\cos{(yz)},-x^2z\\sin{(yz)}+2yz^2,y^2\\rangle[\/latex]?\r\n\r\n[reveal-answer q=\"764528702\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"764528702\"]\r\n\r\nNo.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: verifying a potential function<\/h3>\r\nThe velocity of a fluid is modeled by field [latex]{\\bf{v}}(x,y)=\\langle{x}y,\\frac{x^2}2-y\\rangle[\/latex]. Verify that [latex]f(x,y)=\\frac{x^2y}2-\\frac{y^2}2[\/latex] is a potential function for\u00a0[latex]\\bf{v}[\/latex].\r\n\r\n[reveal-answer q=\"227097452\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"227097452\"]\r\n<p id=\"fs-id1167794121472\">To show that [latex]f[\/latex] is a potential function, we must show that [latex]\\nabla{f}=\\bf{v}[\/latex]. Note that [latex]f_x=xy[\/latex] and [latex]f_y=\\frac{x^2}2-y[\/latex]. Therefore, [latex]\\nabla{f}=\\langle{x}y,\\frac{x^2}2-y\\rangle[\/latex] and [latex]f[\/latex] is a potential function for [latex]{\\bf{v}}[\/latex] (Figure 4).<\/p>\r\n\r\n[caption id=\"attachment_3311\" align=\"aligncenter\" width=\"717\"]<img class=\"wp-image-3311 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/19162840\/6-1-8.jpeg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/0b7976b357e7d7bfaf5b84aa0ac2a8557b2e2906&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A visual representation of the given directional field in two dimensions. The arrows in quadrant 1point to the right. Closer to the y axis, they point down, but they quickly cuve and soon point up at at roughly 90-degree angle. The closer the arrows are to the x axis, the more vertical they are. Quadrant 2 is a reflection of quadrant 1. In quadrant 3, the arrows are more vertical the closer they are to the x and y axes. They point up and to the right. The further they are from the axes, the closer the arrows are to a 90-degree angle. Quadrant 4 is a reflection of quadrant 3.&quot; id=&quot;38&quot;&gt;\" width=\"717\" height=\"497\" \/> Figure 4. Velocity field [latex]{\\bf{v}}(x,y)[\/latex] has a potential function and is a conservative field.[\/caption][\/hidden-answer]<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nVerify that [latex]f(x, y)=x^{2}y^{2}+x[\/latex] is a potential function for velocity field [latex]{\\bf{v}}(x,y)=\\langle2xy^2+1,2x^2y\\rangle[\/latex].\r\n\r\n[reveal-answer q=\"324714857\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"324714857\"]\r\n\r\n[latex]\\nabla{f}=\\bf{v}[\/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=8250312&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=4aWU5lKTDUU&amp;video_target=tpm-plugin-mfb47tbe-4aWU5lKTDUU\" 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\/CP6.10_transcript.html\">transcript for \u201cCP 6.10\u201d here (opens in new window).<\/a><\/center>\r\n<p id=\"fs-id1167794165412\">If\u00a0[latex]\\bf{F}[\/latex]\u00a0is a conservative vector field, then there is at least one potential function [latex]f[\/latex] such that [latex]\\nabla{f}=\\bf{F}[\/latex]. But, could there be more than one potential function? If so, is there any relationship between two potential functions for the same vector field? Before answering these questions, let\u2019s recall some facts from single-variable calculus to guide our intuition. Recall that if [latex]k(x)[\/latex] is an integrable function, then [latex]k[\/latex] has infinitely many antiderivatives. Furthermore, if [latex]F[\/latex] and [latex]G[\/latex] are both antiderivatives of [latex]k[\/latex], then [latex]F[\/latex] and [latex]G[\/latex] differ only by a constant. That is, there is some number [latex]C[\/latex] such that [latex]F(x)=G(x)+C[\/latex].<\/p>\r\n<p id=\"fs-id1167793419791\">Now let [latex]\\bf{F}[\/latex]\u00a0be a conservative vector field and let [latex]f[\/latex] and\u00a0[latex]g[\/latex] be potential functions for [latex]\\bf{F}[\/latex]. Since the gradient is like a derivative, [latex]\\bf{F}[\/latex] being conservative means that [latex]\\bf{F}[\/latex] is \u201cintegrable\u201d with \u201cantiderivatives\u201d [latex]f[\/latex] and\u00a0[latex]g[\/latex]. Therefore, if the analogy with single-variable calculus is valid, we expect there is some constant [latex]C[\/latex] such that [latex]f(x)=g(x)+C[\/latex]. The next theorem says that this is indeed the case.<\/p>\r\n<p id=\"fs-id1167793551157\">To state the next theorem with precision, we need to assume the domain of the vector field is connected and open. To be connected means if [latex]P_1[\/latex] and [latex]P_2[\/latex] are any two points in the domain, then you can walk from [latex]P_1[\/latex] to [latex]P_2[\/latex] along a path that stays entirely inside the domain.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: uniqueness of potential functions<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet\u00a0[latex]\\bf{F}[\/latex]\u00a0be a conservative vector field on an open and connected domain and let [latex]f[\/latex] and [latex]g[\/latex] be functions such that [latex]\\nabla{f}=\\bf{F}[\/latex] and [latex]\\nabla{g}=\\bf{F}[\/latex]. Then, there is a constant [latex]C[\/latex] such that [latex]f=g+C[\/latex].\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Proof<\/h2>\r\n<p id=\"fs-id1167793249873\">Since [latex]f[\/latex] and\u00a0[latex]g[\/latex] are both potential functions for\u00a0[latex]\\bf{F}[\/latex], then [latex]\\nabla{(f-g)}=\\nabla{f}-\\nabla{g}={\\bf{F}}-{\\bf{F}}=0[\/latex]. Let [latex]h=f-g[\/latex] then we have [latex]\\nabla{h}=0[\/latex]. We would like to show that [latex]h[\/latex] is a constant function.<\/p>\r\n<p id=\"fs-id1167793720015\">Assume [latex]h[\/latex] is a function of [latex]x[\/latex] and [latex]y[\/latex] (the logic of this proof extends to any number of independent variables). Since [latex]\\nabla{h}=0[\/latex], we have [latex]h_x=0[\/latex] and [latex]h_y=0[\/latex]. The expression [latex]h_x=0[\/latex] implies that [latex]h[\/latex] is a constant function with respect to [latex]x[\/latex]<em data-effect=\"italics\">\u2014<\/em>that is, [latex]h(x, y)=k_1(y)[\/latex] for some function [latex]k_1[\/latex].\u00a0 Similarly, [latex]h_y=0[\/latex] implies [latex]h(x, y)=k_2(x)[\/latex] for some function [latex]k_2[\/latex]. Therefore, function [latex]h[\/latex] depends only on [latex]y[\/latex] and also depends only on [latex]x[\/latex]. Thus, [latex]h(x, y)=C[\/latex] for some constant [latex]C[\/latex] on the connected domain of\u00a0[latex]\\bf{F}[\/latex]. Note that we really do need connectedness at this point; if the domain of\u00a0[latex]\\bf{F}[\/latex]\u00a0came in two separate pieces, then [latex]k[\/latex] could be a constant [latex]C_1[\/latex] on one piece but could be a different constant [latex]C_2[\/latex] on the other piece. Since [latex]f-g=h=C[\/latex], we have that [latex]f=g+C[\/latex], as desired.<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\nConservative vector fields also have a special property called the\u00a0<span id=\"4ea1e6ac-b330-4b3a-a84a-b899b4fc0de5_term238\" class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">cross-partial property<\/em><\/span>. This property helps test whether a given vector field is conservative.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: the cross-partial property of conservative vector fields<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167794327434\">Let\u00a0<strong data-effect=\"bold\">F<\/strong>\u00a0be a vector field in two or three dimensions such that the component functions of\u00a0[latex]\\bf{F}[\/latex]\u00a0have continuous second-order mixed-partial derivatives on the domain of\u00a0[latex]\\bf{F}[\/latex].<\/p>\r\n<p id=\"fs-id1167794327453\">If [latex]{\\bf{F}}(x,y)=\\langle{P}(x,y),Q(x,y)\\rangle[\/latex] is a conservative vector field in [latex]\\mathbb{R}^2[\/latex], then [latex]\\frac{\\partial{P}}{\\partial{y}}=\\frac{\\partial{Q}}{\\partial{x}}[\/latex]. If [latex]{\\bf{F}}(x,y,z)=\\langle{P}(x,y,z),Q(x,y,z),R(x,y,z)\\rangle[\/latex] is a conservative vector field in [latex]\\mathbb{R}^3[\/latex], then<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{\\partial{P}}{\\partial{y}}=\\frac{\\partial{Q}}{\\partial{x}},\\frac{\\partial{Q}}{\\partial{z}}=\\frac{\\partial{R}}{\\partial{y}},\\text{ and }\\frac{\\partial{R}}{\\partial{x}}=\\frac{\\partial{P}}{\\partial{z}}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Proof<\/h2>\r\n<p id=\"fs-id1167793584400\">Since\u00a0[latex]\\bf{F}[\/latex]\u00a0is conservative, there is a function [latex]f(x, y)[\/latex] such that [latex]\\nabla{f}=\\bf{F}[\/latex]. Therefore, by the definition of the gradient, [latex]f_x=P[\/latex] and [latex]f_y=Q[\/latex]. By Clairaut\u2019s theorem, [latex]f_{xy}=f_{yx}[\/latex], but, [latex]f_{xy}=P_y[\/latex] and [latex]f_{yx}=Q_x[\/latex], and thus [latex]P_y=Q_x[\/latex].<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n<p id=\"fs-id1167793378938\"><span id=\"4ea1e6ac-b330-4b3a-a84a-b899b4fc0de5_term239\" class=\"no-emphasis\" data-type=\"term\">Clairaut\u2019s theorem<\/span>\u00a0gives a fast proof of the cross-partial property of conservative vector fields in [latex]\\mathbb{R}^3[\/latex], just as it did for vector fields in [latex]\\mathbb{R}^2[\/latex].<\/p>\r\n<p id=\"fs-id1167793541422\">The Cross-Partial Property of Conservative Vector Fields Theorem\u00a0shows that most vector fields are not conservative. The cross-partial property is difficult to satisfy in general, so most vector fields won\u2019t have equal cross-partials.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: showing a vector field is not conservative<\/h3>\r\nShow that rotational vector field [latex]{\\bf{F}}(x,y)=\\langle{y},-x\\rangle[\/latex] is not conservative.\r\n\r\n[reveal-answer q=\"234872346\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"234872346\"]\r\n\r\nLet [latex]P(x, y)=y[\/latex] and [latex]Q(x, y)=-x[\/latex]. If\u00a0[latex]\\bf{F}[\/latex]\u00a0is conservative, then the cross-partials would be equal\u2014that is, [latex]P_y[\/latex] would equal [latex]Q_x[\/latex]. Therefore, to show that\u00a0[latex]\\bf{F}[\/latex]\u00a0is not conservative, check that [latex]P_y\\ne{Q}_x[\/latex]. Since [latex]P_y=1[\/latex] and [latex]Q_x=-1[\/latex], the vector field is not conservative.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nShow that vector field [latex]{\\bf{F}}(x,y)x=y{\\bf{i}}-x^2y{\\bf{j}}[\/latex] is not conservative.\r\n\r\n[reveal-answer q=\"736457208\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"736457208\"]\r\n\r\n[latex]\\sqrt{2}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: showing a vector field is not conservative<\/h3>\r\nIs vector field [latex]{\\bf{F}}(x,y,z)=\\langle7,-2,x^3\\rangle[\/latex] conservative?\r\n\r\n[reveal-answer q=\"873425822\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"873425822\"]\r\n\r\nLet [latex]P(x, y, z)=7[\/latex], [latex]Q(x, y, z)=-2[\/latex], and [latex]R(x,y,z)=x^3[\/latex]. If\u00a0[latex]\\bf{F}[\/latex]\u00a0is conservative, then all three cross-partial equations will be satisfied\u2014that is, if\u00a0[latex]\\bf{F}[\/latex]\u00a0is conservative, then [latex]p_y[\/latex] would equal [latex]Q_x[\/latex], [latex]Q_z[\/latex], would equal [latex]R_y[\/latex], and [latex]R_x[\/latex] would equal [latex]P_z[\/latex]. Note that [latex]P_y=Q_x=R_y=Q_z=0[\/latex], so the first two necessary equalities hold. However, [latex]R_x=3x^2[\/latex] and [latex]P_z=0[\/latex] so [latex]R_x\\ne{P}_z[\/latex]. Therefore, [latex]\\bf{F}[\/latex] is not conservative.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nIs vector field [latex]{\\bf{G}}(x,y,z)=\\langle{y},x,xyz\\rangle[\/latex] conservative?\r\n\r\n[reveal-answer q=\"183428645\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"183428645\"]\r\n\r\nNo.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe conclude this section with a word of warning:\u00a0The Cross-Partial Property of Conservative Vector Fields Theorem\u00a0says that if\u00a0[latex]\\bf{F}[\/latex]\u00a0is conservative, then\u00a0[latex]\\bf{F}[\/latex]\u00a0has the cross-partial property. The theorem does\u00a0<em data-effect=\"italics\">not<\/em>\u00a0say that, if\u00a0[latex]\\bf{F}[\/latex]\u00a0has the cross-partial property, then\u00a0[latex]\\bf{F}[\/latex]\u00a0is conservative (the converse of an implication is not logically equivalent to the original implication). In other words,\u00a0The Cross-Partial Property of Conservative Vector Fields Theorem\u00a0can only help determine that a field is not conservative; it does not let you conclude that a vector field is conservative. For example, consider vector field [latex]{\\bf{F}}(x,y)=\\langle{x}^2y,\\frac{x^3}3\\rangle[\/latex]. This field has the cross-partial property, so it is natural to try to use\u00a0The Cross-Partial Property of Conservative Vector Fields Theorem\u00a0to conclude this vector field is conservative. However, this is a misapplication of the theorem. We learn later how to conclude that\u00a0[latex]\\bf{F}[\/latex]\u00a0is conservative.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Identify a conservative field and its associated potential function.<\/span><\/li>\n<\/ul>\n<\/div>\n<h2 data-type=\"title\">Gradient Fields<\/h2>\n<p id=\"fs-id1167793939446\">In this section, we study a special kind of vector field called a gradient field or a\u00a0<strong><span id=\"4ea1e6ac-b330-4b3a-a84a-b899b4fc0de5_term235\" data-type=\"term\">conservative field<\/span><\/strong>. These vector fields are extremely important in physics because they can be used to model physical systems in which energy is conserved. Gravitational fields and electric fields associated with a static charge are examples of gradient fields.<\/p>\n<p id=\"fs-id1167793939454\">Recall that if [latex]f[\/latex] is a (scalar) function of [latex]x[\/latex] and [latex]y[\/latex], then the gradient of [latex]f[\/latex] is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\text{grad}f=\\nabla{f}=f_x(x,y){\\bf{i}}+f_y(x,y){\\bf{j}}}[\/latex].<\/p>\n<p id=\"fs-id1167793697779\">We can see from the form in which the gradient is written that [latex]\\nabla{f}[\/latex]. is a vector field in [latex]\\mathbb{R}^2[\/latex]. Similarly, if [latex]f[\/latex] is a function of [latex]x[\/latex], [latex]y[\/latex], and [latex]z[\/latex], then the gradient of [latex]f[\/latex] is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\text{grad}f=\\nabla{f}=f_x(x,y,z){\\bf{i}}+f_y(x,y,z){\\bf{j}}+f_z(x,y,z){\\bf{k}}}[\/latex].<\/p>\n<p id=\"fs-id1167793450778\">The gradient of a three-variable function is a vector field in [latex]\\mathbb{R}^3[\/latex].<\/p>\n<p id=\"fs-id1167793266860\">A gradient field is a vector field that can be written as the gradient of a function, and we have the following definition.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<div id=\"fs-id1167793266865\" class=\"ui-has-child-title\" data-type=\"note\">\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-id1167793266869\">A vector field\u00a0[latex]\\bf{F}[\/latex]\u00a0in [latex]\\mathbb{R}^2[\/latex] or in [latex]\\mathbb{R}^3[\/latex] is a<strong>\u00a0<span id=\"4ea1e6ac-b330-4b3a-a84a-b899b4fc0de5_term236\" data-type=\"term\">gradient field<\/span>\u00a0<\/strong>if there exists a scalar function [latex]f[\/latex] such that [latex]\\nabla{f}={\\bf{F}}[\/latex].<\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: sketching a gradient vector field<\/h3>\n<p>Use technology to plot the gradient vector field of [latex]f(x, y)=x^{2}y^{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q277324891\">Show Solution<\/span><\/p>\n<div id=\"q277324891\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793543917\">The gradient of [latex]f[\/latex] is [latex]\\nabla{f}=\\langle2xy^2,2x^2y\\rangle[\/latex]. To sketch the vector field, use a computer algebra system such as Mathematica.\u00a0Figure 1\u00a0shows [latex]\\nabla{f}[\/latex].<\/p>\n<div id=\"attachment_5237\" style=\"width: 677px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5237\" class=\"size-full wp-image-5237\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31215639\/6.9.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/b91be2eedc7aabbebf23e61b81fb423f0410f70e&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A visual representation of the given gradient vector field in two dimensions. The arrows point up above the x axis and down below the x axis, and they point left on the left side of the y axis and to the right on the right side of the y axis. The further the arrows are from zero, the more vertical they are, and the closer the arrows are to zero, the more horizontal they are.&quot; id=&quot;31&quot;&gt;\" width=\"667\" height=\"395\" \/><\/p>\n<p id=\"caption-attachment-5237\" class=\"wp-caption-text\">Figure 1. The gradient vector field is [latex]\\nabla{f}[\/latex], where [latex]f(x, y)=x^{2}y^{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Use technology to plot the gradient vector field of\u00a0[latex]f(x,y)=\\sin{x}\\cos{y}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q485726668\">Show Solution<\/span><\/p>\n<div id=\"q485726668\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"attachment_3309\" style=\"width: 727px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-3309\" class=\"wp-image-3309 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/19162738\/6-1-tryitans4.jpeg\" alt=\"\" width=\"717\" height=\"497\" \/><\/p>\n<p id=\"caption-attachment-3309\" class=\"wp-caption-text\">Figure 2. A virtual representation of the gradient vector field\u00a0[latex]f(x,y)=\\sin{x}\\cos{y}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Consider the function [latex]f(x, y)=x^{2}y^{2}[\/latex] from\u00a0Example &#8220;Sketching a Gradient Vector Field&#8221;.\u00a0Figure 4\u00a0shows the level curves of this function overlaid on the function\u2019s gradient vector field. The gradient vectors are perpendicular to the level curves, and the magnitudes of the vectors get larger as the level curves get closer together, because closely grouped level curves indicate the graph is steep, and the magnitude of the gradient vector is the largest value of the directional derivative. Therefore, you can see the local steepness of a graph by investigating the corresponding function\u2019s gradient field.<\/p>\n<div id=\"attachment_3310\" style=\"width: 677px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-3310\" class=\"wp-image-3310 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/19162800\/6-1-7.jpeg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/516e6d511466956acbd12fe82171045a9794197b&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A visual representation of the given gradient field. The arrows are flatter the closer they are to the x axis and more vertical the further they are from the x axis. The arrows point left to the left of the y axis, and they point to the right to the right of the y axis. They point up above the x axis and down below the x axis. Several level curves are drawn, each asymptotically approaching the axes. As the level curves get closer together, the magnitude of the gradient vectors increases.&quot; id=&quot;34&quot;&gt;\" width=\"667\" height=\"395\" \/><\/p>\n<p id=\"caption-attachment-3310\" class=\"wp-caption-text\">Figure 3. The gradient field of [latex]f(x, y)=x^{2}y^{2}[\/latex] and several level curves of [latex]f[\/latex]. Notice that as the level curves get closer together, the magnitude of the gradient vectors increases.<\/p>\n<\/div>\n<p id=\"fs-id1167793662272\">As we learned earlier, a vector field\u00a0[latex]{\\bf{F}}[\/latex] is a conservative vector field, or a gradient field if there exists a scalar function [latex]f[\/latex] such that [latex]\\nabla{f}={\\bf{F}}[\/latex]. In this situation, [latex]f[\/latex] is called a\u00a0<strong><span id=\"4ea1e6ac-b330-4b3a-a84a-b899b4fc0de5_term237\" data-type=\"term\">potential function<\/span><\/strong>\u00a0for [latex]{\\bf{F}}[\/latex]. Conservative vector fields arise in many applications, particularly in physics. The reason such fields are called\u00a0<em data-effect=\"italics\">conservative<\/em>\u00a0is that they model forces of physical systems in which energy is conserved. We study conservative vector fields in more detail later in this chapter.<\/p>\n<p id=\"fs-id1167793776652\">You might notice that, in some applications, a potential function [latex]f[\/latex] for\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is defined instead as a function such that [latex]-\\nabla{f}={\\bf{F}}[\/latex]. This is the case for certain contexts in physics, for example.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: verifying a potential function<\/h3>\n<p id=\"fs-id1167793465191\">Is [latex]f(x,y,z)=x^2yz-\\sin{(xy)}[\/latex] a potential function for vector field\u00a0[latex]{\\bf{F}}(x,y,z)=\\langle2xyz-y\\cos{(xy)},x^2z-x\\cos{(xy)},x^2y\\rangle[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q802452044\">Show Solution<\/span><\/p>\n<div id=\"q802452044\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793465191\">We need to confirm whether [latex]\\nabla{f}={\\bf{F}}[\/latex]. We have<\/p>\n<p style=\"text-align: center;\">[latex]f_x=2xyz-y\\cos{(xy)},f_y=x^2z-x\\cos{(xy)},\\text{ and }f_z=x^2y[\/latex]<\/p>\n<p>Therefore, [latex]\\nabla{f}={\\bf{F}}[\/latex] and [latex]f[\/latex] is a potential function for [latex]{\\bf{F}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Is [latex]f(x,y,z)=x^2\\cos{(yz)}+y^2z^2[\/latex] a potential function for [latex]{\\bf{F}}(x,y,z)=\\langle2x\\cos{(yz)},-x^2z\\sin{(yz)}+2yz^2,y^2\\rangle[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q764528702\">Show Solution<\/span><\/p>\n<div id=\"q764528702\" class=\"hidden-answer\" style=\"display: none\">\n<p>No.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: verifying a potential function<\/h3>\n<p>The velocity of a fluid is modeled by field [latex]{\\bf{v}}(x,y)=\\langle{x}y,\\frac{x^2}2-y\\rangle[\/latex]. Verify that [latex]f(x,y)=\\frac{x^2y}2-\\frac{y^2}2[\/latex] is a potential function for\u00a0[latex]\\bf{v}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q227097452\">Show Solution<\/span><\/p>\n<div id=\"q227097452\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794121472\">To show that [latex]f[\/latex] is a potential function, we must show that [latex]\\nabla{f}=\\bf{v}[\/latex]. Note that [latex]f_x=xy[\/latex] and [latex]f_y=\\frac{x^2}2-y[\/latex]. Therefore, [latex]\\nabla{f}=\\langle{x}y,\\frac{x^2}2-y\\rangle[\/latex] and [latex]f[\/latex] is a potential function for [latex]{\\bf{v}}[\/latex] (Figure 4).<\/p>\n<div id=\"attachment_3311\" style=\"width: 727px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-3311\" class=\"wp-image-3311 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/19162840\/6-1-8.jpeg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/0b7976b357e7d7bfaf5b84aa0ac2a8557b2e2906&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A visual representation of the given directional field in two dimensions. The arrows in quadrant 1point to the right. Closer to the y axis, they point down, but they quickly cuve and soon point up at at roughly 90-degree angle. The closer the arrows are to the x axis, the more vertical they are. Quadrant 2 is a reflection of quadrant 1. In quadrant 3, the arrows are more vertical the closer they are to the x and y axes. They point up and to the right. The further they are from the axes, the closer the arrows are to a 90-degree angle. Quadrant 4 is a reflection of quadrant 3.&quot; id=&quot;38&quot;&gt;\" width=\"717\" height=\"497\" \/><\/p>\n<p id=\"caption-attachment-3311\" class=\"wp-caption-text\">Figure 4. Velocity field [latex]{\\bf{v}}(x,y)[\/latex] has a potential function and is a conservative field.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Verify that [latex]f(x, y)=x^{2}y^{2}+x[\/latex] is a potential function for velocity field [latex]{\\bf{v}}(x,y)=\\langle2xy^2+1,2x^2y\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q324714857\">Show Solution<\/span><\/p>\n<div id=\"q324714857\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\nabla{f}=\\bf{v}[\/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=8250312&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=4aWU5lKTDUU&amp;video_target=tpm-plugin-mfb47tbe-4aWU5lKTDUU\" 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\/CP6.10_transcript.html\">transcript for \u201cCP 6.10\u201d here (opens in new window).<\/a><\/div>\n<p id=\"fs-id1167794165412\">If\u00a0[latex]\\bf{F}[\/latex]\u00a0is a conservative vector field, then there is at least one potential function [latex]f[\/latex] such that [latex]\\nabla{f}=\\bf{F}[\/latex]. But, could there be more than one potential function? If so, is there any relationship between two potential functions for the same vector field? Before answering these questions, let\u2019s recall some facts from single-variable calculus to guide our intuition. Recall that if [latex]k(x)[\/latex] is an integrable function, then [latex]k[\/latex] has infinitely many antiderivatives. Furthermore, if [latex]F[\/latex] and [latex]G[\/latex] are both antiderivatives of [latex]k[\/latex], then [latex]F[\/latex] and [latex]G[\/latex] differ only by a constant. That is, there is some number [latex]C[\/latex] such that [latex]F(x)=G(x)+C[\/latex].<\/p>\n<p id=\"fs-id1167793419791\">Now let [latex]\\bf{F}[\/latex]\u00a0be a conservative vector field and let [latex]f[\/latex] and\u00a0[latex]g[\/latex] be potential functions for [latex]\\bf{F}[\/latex]. Since the gradient is like a derivative, [latex]\\bf{F}[\/latex] being conservative means that [latex]\\bf{F}[\/latex] is \u201cintegrable\u201d with \u201cantiderivatives\u201d [latex]f[\/latex] and\u00a0[latex]g[\/latex]. Therefore, if the analogy with single-variable calculus is valid, we expect there is some constant [latex]C[\/latex] such that [latex]f(x)=g(x)+C[\/latex]. The next theorem says that this is indeed the case.<\/p>\n<p id=\"fs-id1167793551157\">To state the next theorem with precision, we need to assume the domain of the vector field is connected and open. To be connected means if [latex]P_1[\/latex] and [latex]P_2[\/latex] are any two points in the domain, then you can walk from [latex]P_1[\/latex] to [latex]P_2[\/latex] along a path that stays entirely inside the domain.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: uniqueness of potential functions<\/h3>\n<hr \/>\n<p>Let\u00a0[latex]\\bf{F}[\/latex]\u00a0be a conservative vector field on an open and connected domain and let [latex]f[\/latex] and [latex]g[\/latex] be functions such that [latex]\\nabla{f}=\\bf{F}[\/latex] and [latex]\\nabla{g}=\\bf{F}[\/latex]. Then, there is a constant [latex]C[\/latex] such that [latex]f=g+C[\/latex].<\/p>\n<\/div>\n<h2 data-type=\"title\">Proof<\/h2>\n<p id=\"fs-id1167793249873\">Since [latex]f[\/latex] and\u00a0[latex]g[\/latex] are both potential functions for\u00a0[latex]\\bf{F}[\/latex], then [latex]\\nabla{(f-g)}=\\nabla{f}-\\nabla{g}={\\bf{F}}-{\\bf{F}}=0[\/latex]. Let [latex]h=f-g[\/latex] then we have [latex]\\nabla{h}=0[\/latex]. We would like to show that [latex]h[\/latex] is a constant function.<\/p>\n<p id=\"fs-id1167793720015\">Assume [latex]h[\/latex] is a function of [latex]x[\/latex] and [latex]y[\/latex] (the logic of this proof extends to any number of independent variables). Since [latex]\\nabla{h}=0[\/latex], we have [latex]h_x=0[\/latex] and [latex]h_y=0[\/latex]. The expression [latex]h_x=0[\/latex] implies that [latex]h[\/latex] is a constant function with respect to [latex]x[\/latex]<em data-effect=\"italics\">\u2014<\/em>that is, [latex]h(x, y)=k_1(y)[\/latex] for some function [latex]k_1[\/latex].\u00a0 Similarly, [latex]h_y=0[\/latex] implies [latex]h(x, y)=k_2(x)[\/latex] for some function [latex]k_2[\/latex]. Therefore, function [latex]h[\/latex] depends only on [latex]y[\/latex] and also depends only on [latex]x[\/latex]. Thus, [latex]h(x, y)=C[\/latex] for some constant [latex]C[\/latex] on the connected domain of\u00a0[latex]\\bf{F}[\/latex]. Note that we really do need connectedness at this point; if the domain of\u00a0[latex]\\bf{F}[\/latex]\u00a0came in two separate pieces, then [latex]k[\/latex] could be a constant [latex]C_1[\/latex] on one piece but could be a different constant [latex]C_2[\/latex] on the other piece. Since [latex]f-g=h=C[\/latex], we have that [latex]f=g+C[\/latex], as desired.<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<p>Conservative vector fields also have a special property called the\u00a0<span id=\"4ea1e6ac-b330-4b3a-a84a-b899b4fc0de5_term238\" class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">cross-partial property<\/em><\/span>. This property helps test whether a given vector field is conservative.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: the cross-partial property of conservative vector fields<\/h3>\n<hr \/>\n<p id=\"fs-id1167794327434\">Let\u00a0<strong data-effect=\"bold\">F<\/strong>\u00a0be a vector field in two or three dimensions such that the component functions of\u00a0[latex]\\bf{F}[\/latex]\u00a0have continuous second-order mixed-partial derivatives on the domain of\u00a0[latex]\\bf{F}[\/latex].<\/p>\n<p id=\"fs-id1167794327453\">If [latex]{\\bf{F}}(x,y)=\\langle{P}(x,y),Q(x,y)\\rangle[\/latex] is a conservative vector field in [latex]\\mathbb{R}^2[\/latex], then [latex]\\frac{\\partial{P}}{\\partial{y}}=\\frac{\\partial{Q}}{\\partial{x}}[\/latex]. If [latex]{\\bf{F}}(x,y,z)=\\langle{P}(x,y,z),Q(x,y,z),R(x,y,z)\\rangle[\/latex] is a conservative vector field in [latex]\\mathbb{R}^3[\/latex], then<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac{\\partial{P}}{\\partial{y}}=\\frac{\\partial{Q}}{\\partial{x}},\\frac{\\partial{Q}}{\\partial{z}}=\\frac{\\partial{R}}{\\partial{y}},\\text{ and }\\frac{\\partial{R}}{\\partial{x}}=\\frac{\\partial{P}}{\\partial{z}}}[\/latex].<\/p>\n<\/div>\n<h2 data-type=\"title\">Proof<\/h2>\n<p id=\"fs-id1167793584400\">Since\u00a0[latex]\\bf{F}[\/latex]\u00a0is conservative, there is a function [latex]f(x, y)[\/latex] such that [latex]\\nabla{f}=\\bf{F}[\/latex]. Therefore, by the definition of the gradient, [latex]f_x=P[\/latex] and [latex]f_y=Q[\/latex]. By Clairaut\u2019s theorem, [latex]f_{xy}=f_{yx}[\/latex], but, [latex]f_{xy}=P_y[\/latex] and [latex]f_{yx}=Q_x[\/latex], and thus [latex]P_y=Q_x[\/latex].<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<p id=\"fs-id1167793378938\"><span id=\"4ea1e6ac-b330-4b3a-a84a-b899b4fc0de5_term239\" class=\"no-emphasis\" data-type=\"term\">Clairaut\u2019s theorem<\/span>\u00a0gives a fast proof of the cross-partial property of conservative vector fields in [latex]\\mathbb{R}^3[\/latex], just as it did for vector fields in [latex]\\mathbb{R}^2[\/latex].<\/p>\n<p id=\"fs-id1167793541422\">The Cross-Partial Property of Conservative Vector Fields Theorem\u00a0shows that most vector fields are not conservative. The cross-partial property is difficult to satisfy in general, so most vector fields won\u2019t have equal cross-partials.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: showing a vector field is not conservative<\/h3>\n<p>Show that rotational vector field [latex]{\\bf{F}}(x,y)=\\langle{y},-x\\rangle[\/latex] is not conservative.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q234872346\">Show Solution<\/span><\/p>\n<div id=\"q234872346\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let [latex]P(x, y)=y[\/latex] and [latex]Q(x, y)=-x[\/latex]. If\u00a0[latex]\\bf{F}[\/latex]\u00a0is conservative, then the cross-partials would be equal\u2014that is, [latex]P_y[\/latex] would equal [latex]Q_x[\/latex]. Therefore, to show that\u00a0[latex]\\bf{F}[\/latex]\u00a0is not conservative, check that [latex]P_y\\ne{Q}_x[\/latex]. Since [latex]P_y=1[\/latex] and [latex]Q_x=-1[\/latex], the vector field is not conservative.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Show that vector field [latex]{\\bf{F}}(x,y)x=y{\\bf{i}}-x^2y{\\bf{j}}[\/latex] is not conservative.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q736457208\">Show Solution<\/span><\/p>\n<div id=\"q736457208\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\sqrt{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: showing a vector field is not conservative<\/h3>\n<p>Is vector field [latex]{\\bf{F}}(x,y,z)=\\langle7,-2,x^3\\rangle[\/latex] conservative?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q873425822\">Show Solution<\/span><\/p>\n<div id=\"q873425822\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let [latex]P(x, y, z)=7[\/latex], [latex]Q(x, y, z)=-2[\/latex], and [latex]R(x,y,z)=x^3[\/latex]. If\u00a0[latex]\\bf{F}[\/latex]\u00a0is conservative, then all three cross-partial equations will be satisfied\u2014that is, if\u00a0[latex]\\bf{F}[\/latex]\u00a0is conservative, then [latex]p_y[\/latex] would equal [latex]Q_x[\/latex], [latex]Q_z[\/latex], would equal [latex]R_y[\/latex], and [latex]R_x[\/latex] would equal [latex]P_z[\/latex]. Note that [latex]P_y=Q_x=R_y=Q_z=0[\/latex], so the first two necessary equalities hold. However, [latex]R_x=3x^2[\/latex] and [latex]P_z=0[\/latex] so [latex]R_x\\ne{P}_z[\/latex]. Therefore, [latex]\\bf{F}[\/latex] is not conservative.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Is vector field [latex]{\\bf{G}}(x,y,z)=\\langle{y},x,xyz\\rangle[\/latex] conservative?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q183428645\">Show Solution<\/span><\/p>\n<div id=\"q183428645\" class=\"hidden-answer\" style=\"display: none\">\n<p>No.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>We conclude this section with a word of warning:\u00a0The Cross-Partial Property of Conservative Vector Fields Theorem\u00a0says that if\u00a0[latex]\\bf{F}[\/latex]\u00a0is conservative, then\u00a0[latex]\\bf{F}[\/latex]\u00a0has the cross-partial property. The theorem does\u00a0<em data-effect=\"italics\">not<\/em>\u00a0say that, if\u00a0[latex]\\bf{F}[\/latex]\u00a0has the cross-partial property, then\u00a0[latex]\\bf{F}[\/latex]\u00a0is conservative (the converse of an implication is not logically equivalent to the original implication). In other words,\u00a0The Cross-Partial Property of Conservative Vector Fields Theorem\u00a0can only help determine that a field is not conservative; it does not let you conclude that a vector field is conservative. For example, consider vector field [latex]{\\bf{F}}(x,y)=\\langle{x}^2y,\\frac{x^3}3\\rangle[\/latex]. This field has the cross-partial property, so it is natural to try to use\u00a0The Cross-Partial Property of Conservative Vector Fields Theorem\u00a0to conclude this vector field is conservative. However, this is a misapplication of the theorem. We learn later how to conclude that\u00a0[latex]\\bf{F}[\/latex]\u00a0is conservative.<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-5467\">\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 6.10. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":4,"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 6.10\",\"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-5467","chapter","type-chapter","status-publish","hentry"],"part":24,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5467","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":20,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5467\/revisions"}],"predecessor-version":[{"id":6410,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5467\/revisions\/6410"}],"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\/5467\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=5467"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=5467"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=5467"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=5467"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}