{"id":5498,"date":"2022-06-02T18:53:52","date_gmt":"2022-06-02T18:53:52","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&p=5498"},"modified":"2022-11-01T05:22:35","modified_gmt":"2022-11-01T05:22:35","slug":"divergence","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/divergence\/","title":{"raw":"Divergence","rendered":"Divergence"},"content":{"raw":"
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Learning Objectives<\/h3>\r\n
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  • Determine divergence from the formula for a given vector field.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n

    Divergence is an operation on a vector field that tells us how the field behaves toward or away from a point. Locally, the divergence of a vector field\u00a0[latex]{\\bf{F}}[\/latex]\u00a0in [latex]\\mathbb{R}^2[\/latex] or [latex]\\mathbb{R}^3[\/latex] at a particular point [latex]P[\/latex] is a measure of the \u201coutflowing-ness\u201d of the vector field at [latex]P[\/latex]. If\u00a0[latex]{\\bf{F}}[\/latex]\u00a0represents the velocity of a fluid, then the divergence of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0at [latex]P[\/latex] measures the net rate of change with respect to time of the amount of fluid flowing away from [latex]P[\/latex] (the tendency of the fluid to flow \u201cout of\u201d [latex]P[\/latex]). In particular, if the amount of fluid flowing into\u00a0P<\/em>\u00a0is the same as the amount flowing out, then the divergence at [latex]P[\/latex] is zero.<\/p>\r\n\r\n

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    definition<\/h3>\r\n\r\n
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    If [latex]{\\bf{F}}\\langle{P},Q,R\\rangle[\/latex] is a vector field in [latex]\\mathbb{R}^3[\/latex] and [latex]P_x[\/latex], [latex]Q_y[\/latex], and [latex]R_z[\/latex] all exist, then the\u00a0divergence<\/span><\/strong>\u00a0of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is defined by<\/p>\r\n

    [latex]\\large{\\text{div }{\\bf{F}}=P_x+Q_y+R_z=\\frac{\\partial{P}}{\\partial{x}}+\\frac{\\partial{Q}}{\\partial{y}}+\\frac{\\partial{R}}{\\partial{z}}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n

    Note the divergence of a vector field is not a vector field, but a scalar function. In terms of the gradient operator [latex]\\nabla=\\left\\langle\\frac{\\partial}{\\partial{x}},\\frac{\\partial}{\\partial{y}},\\frac{\\partial}{\\partial{z}}\\right\\rangle[\/latex], divergence can be written symbolically as the dot product<\/p>\r\n

    [latex]\\large{\\text{div }{\\bf{F}}=\\nabla\\cdot{\\bf{F}}}[\/latex].<\/p>\r\nNote this is merely helpful notation, because the dot product of a vector of operators and a vector of functions is not meaningfully defined given our current definition of dot product.\r\n

    If [latex]{\\bf{F}}\\langle{P},Q\\rangle[\/latex] is a vector field in [latex]\\mathbb{R}^2[\/latex], and [latex]P_x[\/latex] and [latex]Q_y[\/latex] both exist, then the divergence of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is defined similarly as<\/p>\r\n

    [latex]\\large{\\text{div }{\\bf{F}}=P_x+Q_y=\\frac{\\partial{P}}{\\partial{x}}+\\frac{\\partial{Q}}{\\partial{y}}=\\nabla\\cdot{\\bf{F}}}[\/latex].<\/p>\r\n

    To illustrate this point, consider the two vector fields in\u00a0Figure 1. At any particular point, the amount flowing in is the same as the amount flowing out, so at every point the \u201coutflowing-ness\u201d of the field is zero. Therefore, we expect the divergence of both fields to be zero, and this is indeed the case, as<\/p>\r\n

    [latex]\\large{\\text{div }(\\langle1,2\\rangle)=\\frac{\\partial}{\\partial{x}}(1)+\\frac{\\partial}{\\partial{y}}(2)=0\\text{ and div}(\\langle-y,x\\rangle)=\\frac{\\partial}{\\partial{x}}(-y)+\\frac{\\partial}{\\partial{y}}(x)=0}[\/latex].<\/p>\r\n\r\n<\/div>\r\n

    [caption id=\"attachment_5325\" align=\"aligncenter\" width=\"794\"]\"<img Figure 1. (a) Vector field [latex]\\langle1,2\\rangle[\/latex] has zero divergence. (b) Vector field [latex]\\langle-y,x\\rangle[\/latex] also has zero divergence.[\/caption]<\/div>\r\n
    <\/div>\r\n<\/div>\r\n
    By contrast, consider radial vector field [latex]{\\bf{R}}(x,y)=\\langle-x,-y\\rangle[\/latex] in\u00a0Figure 2. At any given point, more fluid is flowing in than is flowing out, and therefore the \u201coutgoingness\u201d of the field is negative. We expect the divergence of this field to be negative, and this is indeed the case, as [latex]\\text{div }({\\bf{R}})=\\frac{\\partial}{\\partial{x}}(-x)+\\frac{\\partial}{\\partial{y}}(-y)=-2[\/latex].<\/div>\r\n
    <\/div>\r\n
    \r\n\r\n[caption id=\"attachment_5327\" align=\"aligncenter\" width=\"717\"]\"<img Figure 2. This vector field has negative divergence.[\/caption]\r\n\r\n<\/div>\r\n
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    To get a global sense of what divergence is telling us, suppose that a vector field in [latex]\\mathbb{R}^2[\/latex] represents the velocity of a fluid. Imagine taking an elastic circle (a circle with a shape that can be changed by the vector field) and dropping it into a fluid. If the circle maintains its exact area as it flows through the fluid, then the divergence is zero. This would occur for both vector fields in\u00a0Figure 1. On the other hand, if the circle\u2019s shape is distorted so that its area shrinks or expands, then the divergence is not zero. Imagine dropping such an elastic circle into the radial vector field in\u00a0Figure 2\u00a0so that the center of the circle lands at point [latex](3, 3)[\/latex]. The circle would flow toward the origin, and as it did so the front of the circle would travel more slowly than the back, causing the circle to \u201cscrunch\u201d and lose area. This is how you can see a negative divergence.<\/div>\r\n
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    Example: calculating divergence at a Point<\/h3>\r\nIf [latex]{\\bf{F}}(x,y,z)=e^x{\\bf{i}}+yz{\\bf{j}}-y^2{\\bf{k}}[\/latex], then find the divergence of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0at [latex](0, 2, -1)[\/latex].\r\n\r\n[reveal-answer q=\"943504823\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"943504823\"]\r\n

    The divergence of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is<\/p>\r\n

    [latex]\\large{\\frac{\\partial}{\\partial{x}}\\left(e^x\\right)+\\frac{\\partial}{\\partial{y}}(yz)-\\frac{\\partial}{\\partial{z}}\\left(yz^2\\right)=e^x+z-2yz}[\/latex].<\/p>\r\n

    Therefore, the divergence at [latex](0, 2, -1)[\/latex] is [latex]e^0-1+4=4[\/latex]. If\u00a0[latex]{\\bf{F}}[\/latex]\u00a0represents the velocity of a fluid, then more fluid is flowing out than flowing in at point [latex](0, 2, -1)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

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    try it<\/h3>\r\nFind [latex]\\text{div }{\\bf{F}}[\/latex] for [latex]{\\bf{F}}(x,y,z)=\\langle{x}y,5-z^2y,x^2+y^2\\rangle[\/latex].\r\n\r\n[reveal-answer q=\"294776295\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"294776295\"]\r\n\r\n[latex]y-z^{2}[\/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