{"id":143,"date":"2022-02-22T21:42:02","date_gmt":"2022-02-23T02:42:02","guid":{"rendered":"https:\/\/pressbooks.library.ryerson.ca\/multivariatecalculus\/?post_type=chapter&p=143"},"modified":"2024-01-31T10:40:02","modified_gmt":"2024-01-31T15:40:02","slug":"unit-3-directional-derivative","status":"publish","type":"chapter","link":"https:\/\/pressbooks.library.torontomu.ca\/multivariatecalculus\/chapter\/unit-3-directional-derivative\/","title":{"raw":"Unit 3: Directional Derivative","rendered":"Unit 3: Directional Derivative"},"content":{"raw":"

The Concept<\/h2>\r\n

Directional derivatives<\/strong> look to extend the concept of partial derivative<\/strong>s by finding the tangent line parallel to more than the [latex]x[\/latex]-axis or [latex]y[\/latex]-axis.<\/p>\r\nWe start with the graph of a surface defined by the equation [latex]z = f(x,y)[\/latex]. Given a point [latex](x_0, y_0)[\/latex] in the domain of [latex]f(x,y)[\/latex], we choose a direction defined by a unit vector<\/strong>, [latex]\\vec{u}=\\langle a,b\\rangle[\/latex], where [latex]a^2+b^2=1[\/latex], to travel from that point. This direction vector<\/strong> can also be written as [latex]\\vec{u}= \\langle \\cos\\theta,\\,\\sin\\theta \\rangle[\/latex].\u00a0 Angle [latex]\\theta[\/latex] is measured counterclockwise on the [latex]xy[\/latex]-plane, starting at zero from the positive [latex]x[\/latex]-axis. The derivative along that direction (that is, the directional derivative) represents the traveling speed and it is defined as the dot product between the gradient vector<\/strong>, [latex]\u2207f = \\langle f_x, f_y \\rangle[\/latex], and direction vector, [latex]\\vec{u}[\/latex], i.e.,\r\n

[latex] D_u f(x_0,y_0) = \u2207f \\cdot \\vec{u}=f_x(x_0, y_0) \\, a+ f_y(x_0, y_0) \\,b [\/latex],<\/p>\r\nwhere [latex]a=\\cos\\theta[\/latex] and [latex]b=\\sin \\theta[\/latex].\r\n\r\nConsider two special cases of directional derivatives:\r\n

    \r\n \t
  1. When [latex] \\theta = 0 [\/latex], we travel in the direction that is parallel to positive [latex]x[\/latex]-axis, so the direction [latex]\\vec{u} = \\langle \\cos0, \\, \\sin0\\rangle = \\langle 1,\\,0\\rangle [\/latex] and the corresponding directional derivative is [latex]D_u f(x,y) = f_x(x_0,y_0)\\,1+ f_y(x_0,y_0)\\,0= f_x(x_0,y_0)[\/latex].<\/li>\r\n \t
  2. When [latex] \\theta = \\frac{\\pi}{2}[\/latex], we travel in the direction that is parallel to positive [latex]y[\/latex]-axis, so the direction [latex] \\vec{u} = \\langle \\cos \\frac{\\pi}{2} , \\, \\sin \\frac{\\pi}{2} \\rangle = \\langle 0,\\, 1 \\rangle[\/latex] and the directional derivative is [latex] D_u f(x_0,y_0) = f_x(x_0,y_0)\\,0+ f_y(x_0,y_0)\\,1= f_y(x_0,y_0)[\/latex].<\/li>\r\n<\/ol>\r\nThe concept of directional derivatives can be extended into high dimensions. For example, we consider the 3D gradient vector, [latex] \u2207f= \\langle f_x, f_y, f_z \\rangle[\/latex] and 3D direction vector, [latex]\\vec{u}=\\langle a,\\,b\\, c \\rangle[\/latex], where [latex]a^2+b^2+c^2=1[\/latex] because [latex]\\vec{u}[\/latex] is a unit vector. Thus the directional derivative of [latex] w = f(x,y,z)[\/latex] at point [latex](x_0, y_0, z_0)[\/latex] is<\/span>\r\n

    [latex] D_u f(x_0,y_0, z_0) = \u2207f \\cdot \\vec{u}=f_x(x_0, y_0, z_0)\\, a + f_y(x_0, y_0, z_0) \\,b + f_y(x_0, y_0, z_0) \\,c[\/latex].<\/span><\/p>\r\n\r\n

    The Plot<\/h2>\r\nNow, you should engage with the 3D plot below to understand directional derivatives[footnote]Made with GeoGebra, licensed Creative Commons CC BY-NC-SA 4.0.[\/footnote]. Follow the steps below to apply changes to the plot and observe the effects:\r\n
      \r\n \t
    1. There are two separate plots where the direction vector (i.e., the direction of the derivative, denoted by [latex]\\vec{u}[\/latex]) is defined by either an angle in radians or a vector.<\/li>\r\n \t
    2. Point (P) is adjusted with the [latex]x[\/latex] and [latex]y[\/latex] sliders.<\/li>\r\n \t
    3. [latex]\\vec{u}[\/latex] is selected by either the angle or vector and is indicated by the red arrow on the graph.<\/li>\r\n<\/ol>\r\n

      Directional Derivative (Defined by an angle)<\/h3>\r\n[h5p id=\"12\"]\r\n

      Direction Derivative (Defined by a vector)<\/h3>\r\n[h5p id=\"13\"]\r\n
      \r\n

      Self-Checking Questions<\/h2>\r\n<\/header>\r\n
      \r\n\r\nCheck your understanding by solving the following questions[footnote]Gilbert Strang, Edwin \u201cJed\u201d Herman,\u00a0 OpenStax,\u00a0Calculus Volume 3,\u00a0 Calculus Volumes 1, 2, and 3 are licensed under an Attribution-NonCommercial-Sharealike 4.0 International License (CC BY-NC-SA).[\/footnote]<\/span>:\r\n
        \r\n \t
      1. Find the gradient, [latex]\u2207f(x,y)[\/latex], of the function: [latex] f(x,y) = x^2 - xy + 3y^2[\/latex]<\/li>\r\n \t
      2. Find the directional derivative, [latex] D_u f[\/latex], of the function: [latex] f(x,y,z) = e^{-2z} sin(2x)cos(2y)[\/latex] at point (0,1).<\/li>\r\n<\/ol>\r\n[h5p id=\"28\"]<\/span>\r\n\r\n<\/div>\r\n<\/div>","rendered":"

        The Concept<\/h2>\n

        Directional derivatives<\/strong> look to extend the concept of partial derivative<\/strong>s by finding the tangent line parallel to more than the [latex]x[\/latex]-axis or [latex]y[\/latex]-axis.<\/p>\n

        We start with the graph of a surface defined by the equation [latex]z = f(x,y)[\/latex]. Given a point [latex](x_0, y_0)[\/latex] in the domain of [latex]f(x,y)[\/latex], we choose a direction defined by a unit vector<\/strong>, [latex]\\vec{u}=\\langle a,b\\rangle[\/latex], where [latex]a^2+b^2=1[\/latex], to travel from that point. This direction vector<\/strong> can also be written as [latex]\\vec{u}= \\langle \\cos\\theta,\\,\\sin\\theta \\rangle[\/latex].\u00a0 Angle [latex]\\theta[\/latex] is measured counterclockwise on the [latex]xy[\/latex]-plane, starting at zero from the positive [latex]x[\/latex]-axis. The derivative along that direction (that is, the directional derivative) represents the traveling speed and it is defined as the dot product between the gradient vector<\/strong>, [latex]\u2207f = \\langle f_x, f_y \\rangle[\/latex], and direction vector, [latex]\\vec{u}[\/latex], i.e.,<\/p>\n

        [latex]D_u f(x_0,y_0) = \u2207f \\cdot \\vec{u}=f_x(x_0, y_0) \\, a+ f_y(x_0, y_0) \\,b[\/latex],<\/p>\n

        where [latex]a=\\cos\\theta[\/latex] and [latex]b=\\sin \\theta[\/latex].<\/p>\n

        Consider two special cases of directional derivatives:<\/p>\n

          \n
        1. When [latex]\\theta = 0[\/latex], we travel in the direction that is parallel to positive [latex]x[\/latex]-axis, so the direction [latex]\\vec{u} = \\langle \\cos0, \\, \\sin0\\rangle = \\langle 1,\\,0\\rangle[\/latex] and the corresponding directional derivative is [latex]D_u f(x,y) = f_x(x_0,y_0)\\,1+ f_y(x_0,y_0)\\,0= f_x(x_0,y_0)[\/latex].<\/li>\n
        2. When [latex]\\theta = \\frac{\\pi}{2}[\/latex], we travel in the direction that is parallel to positive [latex]y[\/latex]-axis, so the direction [latex]\\vec{u} = \\langle \\cos \\frac{\\pi}{2} , \\, \\sin \\frac{\\pi}{2} \\rangle = \\langle 0,\\, 1 \\rangle[\/latex] and the directional derivative is [latex]D_u f(x_0,y_0) = f_x(x_0,y_0)\\,0+ f_y(x_0,y_0)\\,1= f_y(x_0,y_0)[\/latex].<\/li>\n<\/ol>\n

          The concept of directional derivatives can be extended into high dimensions. For example, we consider the 3D gradient vector, [latex]\u2207f= \\langle f_x, f_y, f_z \\rangle[\/latex] and 3D direction vector, [latex]\\vec{u}=\\langle a,\\,b\\, c \\rangle[\/latex], where [latex]a^2+b^2+c^2=1[\/latex] because [latex]\\vec{u}[\/latex] is a unit vector. Thus the directional derivative of [latex]w = f(x,y,z)[\/latex] at point [latex](x_0, y_0, z_0)[\/latex] is<\/span><\/p>\n

          [latex]D_u f(x_0,y_0, z_0) = \u2207f \\cdot \\vec{u}=f_x(x_0, y_0, z_0)\\, a + f_y(x_0, y_0, z_0) \\,b + f_y(x_0, y_0, z_0) \\,c[\/latex].<\/span><\/p>\n

          The Plot<\/h2>\n

          Now, you should engage with the 3D plot below to understand directional derivatives[1]<\/sup><\/a>. Follow the steps below to apply changes to the plot and observe the effects:<\/p>\n

            \n
          1. There are two separate plots where the direction vector (i.e., the direction of the derivative, denoted by [latex]\\vec{u}[\/latex]) is defined by either an angle in radians or a vector.<\/li>\n
          2. Point (P) is adjusted with the [latex]x[\/latex] and [latex]y[\/latex] sliders.<\/li>\n
          3. [latex]\\vec{u}[\/latex] is selected by either the angle or vector and is indicated by the red arrow on the graph.<\/li>\n<\/ol>\n

            Directional Derivative (Defined by an angle)<\/h3>\n
            \n
            <\/div>\n<\/div>\n

            Direction Derivative (Defined by a vector)<\/h3>\n
            \n
            <\/div>\n<\/div>\n
            \n
            \n

            Self-Checking Questions<\/h2>\n<\/header>\n
            \n

            Check your understanding by solving the following questions[2]<\/sup><\/a><\/span>:<\/p>\n

              \n
            1. Find the gradient, [latex]\u2207f(x,y)[\/latex], of the function: [latex]f(x,y) = x^2 - xy + 3y^2[\/latex]<\/li>\n
            2. Find the directional derivative, [latex]D_u f[\/latex], of the function: [latex]f(x,y,z) = e^{-2z} sin(2x)cos(2y)[\/latex] at point (0,1).<\/li>\n<\/ol>\n

              <\/p>\n

              \n