{"id":1147,"date":"2025-10-15T14:44:05","date_gmt":"2025-10-15T14:44:05","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1147"},"modified":"2026-05-26T15:02:12","modified_gmt":"2026-05-26T15:02:12","slug":"directional-derivatives-and-the-gradient-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/directional-derivatives-and-the-gradient-learn-it-2\/","title":{"raw":"Directional Derivatives and the Gradient: Learn It 2","rendered":"Directional Derivatives and the Gradient: Learn It 2"},"content":{"raw":"

Gradient<\/h2>\r\n

Look back at the formula for the directional derivative: [latex]D_{\\bf{u}}f(x,y)=f_x(x,y)\\cos\\theta+f_y(x,y)\\sin\\theta[\/latex]. The right-hand side can be written as a dot product of two vectors. Define the first vector as [latex]\\nabla{f}(x,y)=f_x(x,y){\\bf{i}}+f_y(x,y){\\bf{j}}[\/latex] and the second as [latex]{\\bf{u}}=(\\cos\\theta){\\bf{i}}+(\\sin\\theta){\\bf{j}}[\/latex]. Then:<\/p>\r\n

[latex]D_{\\bf{u}}f(x,y)=\\nabla{f}(x,y)\\cdot{\\bf{u}}[\/latex]<\/p>\r\n\r\n

The symbol [latex]\\nabla[\/latex] is called nabla, and the vector [latex]\\nabla{f}[\/latex] is read \"del [latex]f[\/latex]\" or \"gradient of [latex]f[\/latex]\".<\/section>The first vector in this equation has a special name: the gradient<\/strong> of the function [latex]f[\/latex].\r\n\r\n
\r\n

gradient<\/h3>\r\nLet\u00a0[latex]z=f(x, y)[\/latex] be a function of\u00a0[latex]x[\/latex] and\u00a0[latex]x[\/latex] such that\u00a0[latex]f_x[\/latex] and\u00a0[latex]f_y[\/latex] exist. The vector\u00a0[latex]\\nabla{f}(x,y)[\/latex] is called the gradient<\/strong> of [latex]f[\/latex] and is defined as:\r\n

[latex]\\nabla{f}(x,y)=f_x(x,y){\\bf{i}}+f_y(x,y){\\bf{j}}.[\/latex]<\/p>\r\nThe vector\u00a0[latex]\\nabla{f}(x,y)[\/latex] is also written as \"grad\u00a0[latex]f[\/latex]\".\r\n\r\n<\/section>

Find the gradient\u00a0[latex]\\nabla{f}(x,y)[\/latex] of each of the following functions:\r\n
    \r\n \t
  1. [latex]f(x,y)=x^2-xy+3y^2[\/latex]<\/li>\r\n \t
  2. [latex]f(x,y)=\\sin{3x}\\cos{3y}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"131344589\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"131344589\"]\r\n\r\nFor both parts a. and b., we first calculate the partial derivatives\u00a0[latex]f_x[\/latex] and\u00a0[latex]f_y[\/latex], then use\u00a0the gradient of [latex]f[\/latex].\r\n
      \r\n \t
    1. [latex]\\hspace{1cm}\\begin{align}f_x(x,y)&=2x-y\\text{ and }f_x(x,y)=-x+6y,\\text{ so} \\\\\\nabla{f}(x,y)&=f_x(x,y){\\bf{i}}+f_y(x,y){\\bf{j}} \\\\\r\n\r\n&=(2x-y){\\bf{i}}+(-x+6y){\\bf{j}}.\r\n\r\n\\end{align}[\/latex]<\/li>\r\n \t
    2. [latex]\\hspace{1cm}\\begin{align}f_x(x,y)&=3\\cos{3x}\\cos{3y}\\text{ and }f_x(x,y)=-3\\sin{3x}\\sin{3y},\\text{ so} \\\\\r\n\r\n\\nabla{f}(x,y)&=f_x(x,y){\\bf{i}}+f_y(x,y){\\bf{j}} \\\\\r\n\r\n&=(3\\cos{3x}\\cos{3y}){\\bf{i}}-(3\\sin{3x}\\sin{3y}){\\bf{j}}.\r\n\r\n\\end{align}\r\n\r\n[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>
      [ohm_question hide_question_numbers=1]325844[\/ohm_question]<\/section>\r\n

      Properties of the Gradient<\/h3>\r\n

      The gradient has important properties that make it useful for understanding how functions behave. We've already seen one: the formula for the directional derivative. Let's dig deeper.<\/p>\r\n

      Recall that if the angle between two vectors [latex]{\\bf{a}}[\/latex] and [latex]{\\bf{b}}[\/latex] is [latex]\\varphi[\/latex], then [latex]{\\bf{a}}\\cdot{\\bf{b}}=|{\\bf{a}}||{\\bf{b}}|\\cos\\varphi[\/latex]. If the angle between [latex]\\nabla{f}(x_0,y_0)[\/latex] and [latex]{\\bf{u}}=(\\cos\\theta){\\bf{i}}+(\\sin\\theta){\\bf{j}}[\/latex] is [latex]\\varphi[\/latex], then:<\/p>\r\n

      [latex]D_{\\bf{u}}f(x_0,y_0)=\\nabla{f}(x_0,y_0)\\cdot{\\bf{u}}=|\\nabla{f}(x_0,y_0)||{\\bf{u}}|\\cos\\varphi=|\\nabla{f}(x_0,y_0)|\\cos\\varphi[\/latex]<\/p>\r\n

      The [latex]|{\\bf{u}}|[\/latex] term disappears because [latex]{\\bf{u}}[\/latex] is a unit vector. The directional derivative equals the magnitude of the gradient at [latex](x_0, y_0)[\/latex], multiplied by [latex]\\cos\\varphi[\/latex].<\/p>\r\n

      Since [latex]\\cos\\varphi[\/latex] ranges from [latex]-1[\/latex] to [latex]1[\/latex], we can determine when the directional derivative is maximized or minimized:<\/p>\r\n\r\n

        \r\n \t
      • If [latex]\\varphi=0[\/latex], then [latex]\\cos\\varphi=1[\/latex] and [latex]\\nabla{f}(x_0,y_0)[\/latex] and [latex]{\\bf{u}}[\/latex] point in the same direction. This maximizes [latex]D_{\\bf{u}}f(x_0,y_0)[\/latex].<\/li>\r\n \t
      • If [latex]\\varphi=\\pi[\/latex], then [latex]\\cos\\varphi=-1[\/latex] and [latex]\\nabla{f}(x_0,y_0)[\/latex] and [latex]{\\bf{u}}[\/latex] point in opposite directions. This minimizes [latex]D_{\\bf{u}}f(x_0,y_0)[\/latex].<\/li>\r\n \t
      • If [latex]\\nabla{f}(x_0,y_0)={\\bf{0}}[\/latex], then [latex]D_{\\bf{u}}f(x_0,y_0)=0[\/latex] for any unit vector [latex]{\\bf{u}}[\/latex].<\/li>\r\n<\/ul>\r\nThese cases are outlined in the following theorem.\r\n\r\n
        \r\n

        properties of the gradient<\/h3>\r\nSuppose the function\u00a0[latex]z=f(x, y)[\/latex] is differentiable at\u00a0[latex](x_0, y_0)[\/latex] (Figure 3).\r\n
          \r\n \t
        1. If [latex]\\nabla{f}(x_0,y_0)=0[\/latex], then [latex]D_{\\bf{u}}f(x_0,y_0)=0[\/latex] for any unit vector [latex]\\bf{u}[\/latex].<\/li>\r\n \t
        2. If [latex]\\nabla{f}(x_0,y_0)\\ne 0[\/latex], then [latex]D_{\\bf{u}}f(x_0,y_0)[\/latex] is maximized when [latex]\\bf{u}[\/latex] points in the same direction as [latex]\\nabla{f}(x_0,y_0)[\/latex]. The maximum value of [latex]D_{\\bf{u}}f(x_0,y_0)[\/latex] is [latex]\\|\\nabla{f}(x_0,y_0)\\|[\/latex].<\/li>\r\n \t
        3. If [latex]\\nabla{f}(x_0,y_0)\\ne 0[\/latex], then [latex]D_{\\bf{u}}f(x_0,y_0)[\/latex] is minimized when [latex]\\bf{u}[\/latex] points in the opposite direction from [latex]\\nabla{f}(x_0,y_0)[\/latex]. The minimum value of [latex]D_{\\bf{u}}f(x_0,y_0)[\/latex] is [latex]-\\|\\nabla{f}(x_0,y_0)\\|[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>\r\n\r\n[caption id=\"attachment_1260\" align=\"aligncenter\" width=\"607\"]\"An Figure 1. The gradient indicates the maximum and minimum values of the directional derivative at a point.[\/caption]\r\n\r\n
          Find the direction for which the directional derivative of [latex]f(x, y)=3x^{2}-4xy+2y^{2}[\/latex] at [latex](-2, 3)[\/latex] is a maximum. What is the maximum value?[reveal-answer q=\"721568432\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"721568432\"]The maximum value of the directional derivative occurs when [latex]\\nabla{f}[\/latex] and the unit vector point in the same direction. Therefore, we start by calculating\u00a0[latex]\\nabla{f}(x,y)[\/latex]:[latex]\\hspace{6cm}\\large{\\begin{align}f_x(x,y)&=6x-4y\\text{ and }f_y(x,y)=-4x+4y,\\text{ so} \\\\\r\n\r\n\\nabla{f}(x,y)&=f_x(x,y){\\bf{i}}+f_y(x,y){\\bf{j}}=(6x-4y){\\bf{i}}+(-4x+4y){\\bf{j}}.\r\n\r\n\\end{align}}[\/latex]\r\n\r\nNext, we evaluate the gradient at\u00a0[latex](-2,3)[\/latex]:\r\n

          [latex]\\large{\\nabla{f}(-2,3)=(6(-2)-4(3)){\\bf{i}}+(-4(-2)+3(3)){\\bf{j}}=-24{\\bf{i}}+20{\\bf{j}}}.[\/latex]<\/p>\r\nWe need to find a unit vector that points in the same direction as [latex]\\nabla{f}(-2,3)[\/latex],<\/span>\u00a0so the next step is to divide [latex]\\nabla{f}(-2,3)[\/latex] by its magnitude, which is\u00a0[latex]\\sqrt{(-24)^2+(20)^2}=\\sqrt{976}=4\\sqrt{61}[\/latex]. Therefore,\r\n

          [latex]\\large{\\frac{\\nabla{f}(-2,3)}{\\|\\nabla{f}(-2,3)\\|}=\\frac{-24}{4\\sqrt{61}}{\\bf{i}}+\\frac{20}{4\\sqrt{61}}{\\bf{j}}=\\frac{-6\\sqrt{61}}{61}{\\bf{i}}+\\frac{5\\sqrt{61}}{61}{\\bf{j}}}.[\/latex]<\/p>\r\nThis is the unit vector that points in the same direction as [latex]\\nabla{f}(-2,3)[\/latex]. To find the angle corresponding to this unit vector, we solve the equations\r\n

          [latex]\\large{\\cos\\theta=\\frac{-6\\sqrt{61}}{61}}[\/latex] and\u00a0[latex]\\large{\\sin\\theta=\\frac{5\\sqrt{61}}{61}}.[\/latex]<\/p>\r\nfor\u00a0[latex]\\theta[\/latex].\u00a0Since cosine is negative and sine is positive, the angle must be in the second quadrant. Therefore,\r\n\r\n[latex]\\theta=\\pi-\\arcsin{((5\\sqrt{61})\/61)}\\approx2.45\\text{ rad}.[\/latex]\r\n\r\nThe maximum value of the directional derivative at\u00a0[latex](-2, 3)[\/latex] is\u00a0[latex]\\|\\nabla{f}(-2,3)\\|=4\\sqrt{61}[\/latex] (see the following figure).\r\n\r\n[caption id=\"attachment_1262\" align=\"aligncenter\" width=\"523\"]\"An Figure 2.\u00a0The maximum value of the directional derivative at [latex]\\small{(-2,3)}[\/latex]\u00a0is in the direction of the gradient.[\/caption][\/hidden-answer]<\/section>\r\n

          Interpreting the Gradient<\/h3>\r\nFigure 3 shows part of the graph of [latex]f(x,y)=3+\\sin x\\sin y[\/latex]. Given a point [latex](a, b)[\/latex] in the domain of [latex]f[\/latex], the maximum value of the gradient at that point is [latex]|\\nabla{f}(a,b)|[\/latex]. If the surface represented a topographical map, this would be the rate of greatest ascent. In the opposite direction, it would be the rate of greatest descent.\r\n\r\n[caption id=\"attachment_1265\" align=\"aligncenter\" width=\"724\"]\"An Figure 3. A typical surface in [latex]\\mathbb{R}^{3}[\/latex].\u00a0Given a point on the surface, the directional derivative can be calculated using the gradient.[\/caption]\r\n

          When using a topographical map, the steepest slope is always in the direction where the contour lines are closest together (see Figure 4). This is analogous to the contour map of a function, assuming the level curves are obtained for equally spaced values throughout the range of that function.<\/p>\r\n\r\n[caption id=\"attachment_1266\" align=\"aligncenter\" width=\"267\"]\"Two Figure 4. Contour map for the function [latex]\\small{f(x,y)=x^{2}-y^{2}}[\/latex] using level values between [latex]\\small{-5}[\/latex] and [latex]\\small{5}[\/latex].[\/caption]","rendered":"

          Gradient<\/h2>\n

          Look back at the formula for the directional derivative: [latex]D_{\\bf{u}}f(x,y)=f_x(x,y)\\cos\\theta+f_y(x,y)\\sin\\theta[\/latex]. The right-hand side can be written as a dot product of two vectors. Define the first vector as [latex]\\nabla{f}(x,y)=f_x(x,y){\\bf{i}}+f_y(x,y){\\bf{j}}[\/latex] and the second as [latex]{\\bf{u}}=(\\cos\\theta){\\bf{i}}+(\\sin\\theta){\\bf{j}}[\/latex]. Then:<\/p>\n

          [latex]D_{\\bf{u}}f(x,y)=\\nabla{f}(x,y)\\cdot{\\bf{u}}[\/latex]<\/p>\n

          The symbol [latex]\\nabla[\/latex] is called nabla, and the vector [latex]\\nabla{f}[\/latex] is read “del [latex]f[\/latex]” or “gradient of [latex]f[\/latex]“.<\/section>\n

          The first vector in this equation has a special name: the gradient<\/strong> of the function [latex]f[\/latex].<\/p>\n

          \n

          gradient<\/h3>\n

          Let\u00a0[latex]z=f(x, y)[\/latex] be a function of\u00a0[latex]x[\/latex] and\u00a0[latex]x[\/latex] such that\u00a0[latex]f_x[\/latex] and\u00a0[latex]f_y[\/latex] exist. The vector\u00a0[latex]\\nabla{f}(x,y)[\/latex] is called the gradient<\/strong> of [latex]f[\/latex] and is defined as:<\/p>\n

          [latex]\\nabla{f}(x,y)=f_x(x,y){\\bf{i}}+f_y(x,y){\\bf{j}}.[\/latex]<\/p>\n

          The vector\u00a0[latex]\\nabla{f}(x,y)[\/latex] is also written as “grad\u00a0[latex]f[\/latex]“.<\/p>\n<\/section>\n

          Find the gradient\u00a0[latex]\\nabla{f}(x,y)[\/latex] of each of the following functions:<\/p>\n
            \n
          1. [latex]f(x,y)=x^2-xy+3y^2[\/latex]<\/li>\n
          2. [latex]f(x,y)=\\sin{3x}\\cos{3y}[\/latex]<\/li>\n<\/ol>\n