{"id":1149,"date":"2025-10-15T14:44:07","date_gmt":"2025-10-15T14:44:07","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1149"},"modified":"2025-10-27T21:42:54","modified_gmt":"2025-10-27T21:42:54","slug":"directional-derivatives-and-the-gradient-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/directional-derivatives-and-the-gradient-learn-it-4\/","title":{"raw":"Directional Derivatives and the Gradient: Learn It 4","rendered":"Directional Derivatives and the Gradient: Learn It 4"},"content":{"raw":"

Three-Dimensional Gradients and Directional Derivatives<\/h2>\r\n

The definition of a gradient extends naturally to functions of more than two variables.<\/p>\r\n\r\n

\r\n

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

[latex]\\nabla{f}(x,y,z)=f_x(x,y,z){\\bf{i}}+f_y(x,y,z){\\bf{j}}+f_z(x,y,z){\\bf{k}}.[\/latex]<\/p>\r\nThe notation [latex]\\nabla{f}(x,y,z)[\/latex] can also be written as [latex]\\text{grad }f(x,y,z)[\/latex].\r\n\r\n<\/section>\r\n

Calculating the gradient of a function in three variables follows the same process as in two variables. First, calculate the partial derivatives [latex]f_x[\/latex], [latex]f_y[\/latex], and [latex]f_z[\/latex], then assemble them into the gradient vector.<\/p>\r\n\r\n

Find the gradient [latex]\\nabla{f}(x,y,z)[\/latex] of each of the following functions:\r\n
    \r\n \t
  1. [latex]f(x,y,z)=5x^2-2xy+y^2-4yz+z^2+3xz[\/latex]<\/li>\r\n \t
  2. [latex]f(x,y,z)=e^{-2z}\\sin 2x\\cos 2y[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"268490354\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"268490354\"]\r\n\r\nFor both parts a. and b., we first calculate the partial derivatives [latex]f_x[\/latex],\u00a0[latex]f_y[\/latex], and\u00a0[latex]f_z[\/latex], then use\u00a0the Equation for [latex]\\nabla{f}(x,y,z)[\/latex].\r\n
      \r\n \t
    1. [latex]f_x(x,y,z) =10x-2y+3z,f_y(x,y,z)=-2x+2y-4z[\/latex] and [latex]f_z(x,y,z)=3x-4y+2z[\/latex] so,\r\n[latex]\\hspace{1cm}\\begin{align} \\nabla{f}(x,y,z)&=f_x(x,y,z){\\bf{i}}+f_y(x,y,z){\\bf{j}}+f_z(x,y,z){\\bf{k}} \\\\&=(10x-2y+3z){\\bf{i}}+(-2x+2y-4z){\\bf{j}}+(-4x+3y+2z){\\bf{k}}\\end{align}[\/latex]<\/li>\r\n \t
    2. [latex]f_x(x,y,z)=-2e^{-2z}\\cos2x\\cos 2y[\/latex], [latex]f_y(x,y,z)=-2e^{-2z}\\sin 2x\\sin2y[\/latex] and [latex] f_z(x,y,z)=-2e^{-2z}\\sin 2x\\cos 2y [\/latex] so,\r\n[latex]\\hspace{1cm}\\begin{align}\\nabla{f}(x,y,z)&=f_x(x,y,z){\\bf{i}}+f_y(x,y,z){\\bf{j}}+f_z(x,y,z){\\bf{k}} \\\\&=(2e^{-2z}\\cos2x\\cos 2y){\\bf{i}}+(-2e^{-2z}){\\bf{j}}+(-2e^{-2z}){\\bf{k}} \\\\&=2e^{-2z}(\\cos2x\\cos 2y{\\bf{i}}-\\sin 2x\\sin2y{\\bf{j}}-\\sin 2x\\cos 2y{\\bf{k}}).\\end{align}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

      Directional Derivatives in Three Dimensions<\/h3>\r\n

      The directional derivative also generalizes to functions of three variables. To specify a direction in three dimensions, we need a vector with three components. This vector must be a unit vector, and its components are called directional cosines<\/strong>.<\/p>\r\n

      Given a three-dimensional unit vector [latex]{\\bf{u}}[\/latex] in standard position (initial point at the origin), this vector forms three angles with the positive [latex]x[\/latex]-, [latex]y[\/latex]-, and [latex]z[\/latex]-axes. Let's call these angles [latex]\\alpha[\/latex], [latex]\\beta[\/latex], and [latex]\\gamma[\/latex]. The directional cosines are [latex]\\cos\\alpha[\/latex], [latex]\\cos\\beta[\/latex], and [latex]\\cos\\gamma[\/latex]\u2014these are the components of the unit vector [latex]{\\bf{u}}[\/latex]. Since [latex]{\\bf{u}}[\/latex] is a unit vector, we have [latex]\\cos^2\\alpha+\\cos^2\\beta+\\cos^2\\gamma=1[\/latex].<\/p>\r\n\r\n

      \r\n

      directional derivative in three dimensions<\/h3>\r\nSuppose [latex]w=f(x,y,z)[\/latex] is a function of three variables with a domain of [latex]D[\/latex], Let [latex](x_0,y_0,z_0)\\in{D}[\/latex] and let [latex]{\\bf{u}}=cos\\alpha{\\bf{i}}+\\cos\\beta{\\bf{j}}+\\cos\\gamma{\\bf{k}}[\/latex] be a unit vector.\r\n\r\n[latex]\\\\[\/latex]\r\n\r\nThen, the directional derivative of [latex]f[\/latex] in the direction of [latex]u[\/latex] is given by:\r\n

      [latex]D_{\\bf{u}}f(x_0,y_0,z_0)=\\displaystyle{\\lim_{t\\to0}}\\frac{f(x_0+t\\cos\\alpha,y_0+t\\cos\\beta,z_0+t\\cos\\gamma)-f(x_0,y_0,z_0)}t,[\/latex]<\/p>\r\nproviding the limit exists.\r\n\r\n<\/section>\r\n

      Just as in two dimensions, we can calculate the directional derivative using the gradient. This leads to a formula analogous to the dot product definition we saw earlier.<\/p>\r\n\r\n

      \r\n

      directional derivative of a function of three variables<\/h3>\r\nLet\u00a0[latex]f(x,y,z)[\/latex] be a differentiable function of three variables and let\u00a0[latex]{\\bf{u}}=\\cos\\alpha{\\bf{i}}+\\cos\\beta{\\bf{j}}+\\cos\\gamma{\\bf{k}}[\/latex] be a unit vector.\r\n\r\n[latex]\\\\[\/latex]\r\n\r\nThen, the directional derivative of\u00a0[latex]f[\/latex] in the direction of\u00a0[latex]{\\bf{u}}[\/latex] is given by\r\n

      [latex]\\begin{array}{ccl}D_{\\bf{u}}f(x,y,z)&=&\\nabla{f}(x,y,z)\\cdot{\\bf{u}} \\\\&=&f_x(x,y,z)\\cos\\alpha+f_y(x,y,z)\\cos\\beta+f_z(x,y,z)\\cos\\gamma.\\end{array}[\/latex]<\/p>\r\n\r\n<\/section>

      The three angles [latex]\\alpha[\/latex], [latex]\\beta[\/latex], and [latex]\\gamma[\/latex] determine the unit vector [latex]{\\bf{u}}[\/latex]. In practice, you can use any (nonunit) vector, then divide by its magnitude to obtain a unit vector in the desired direction.<\/section>
      Calculate\u00a0[latex]D_{\\bf{u}}f(1,-2,3)[\/latex] in the direction of\u00a0[latex]v=-{\\bf{i}}+2{\\bf{j}}+2{\\bf{k}}[\/latex] for the function\r\n

      [latex]f(x,y,z)=5x^2-2xy+y^2-4yz+z^2+3xz[\/latex].<\/p>\r\n[reveal-answer q=\"960004812\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"960004812\"]\r\n\r\nFirst, we find the magnitude of [latex]{\\bf{v}}[\/latex]:\r\n

      [latex]\\|{\\bf{v}}\\|=\\sqrt{(-1)^2+(2)^2+(2)^2}=3[\/latex]<\/p>\r\nTherefore\u00a0[latex]\\frac{{\\bf{v}}}{\\|{\\bf{v}}\\|}=\\frac{-{\\bf{i}}+2{\\bf{j}}+2{\\bf{k}}}3=-\\frac13{\\bf{i}}+\\frac23{\\bf{j}}+\\frac23{\\bf{k}}[\/latex] is a unit vector in the direction of\u00a0[latex]{\\bf{v}}[\/latex], so\u00a0[latex]\\cos\\alpha=-\\frac13[\/latex],\u00a0[latex]\\cos\\beta=\\frac23[\/latex], and\u00a0[latex]\\cos\\gamma=\\frac23[\/latex]. Next, we calculate the partial derivatives of\u00a0[latex]f[\/latex]:\r\n\r\n[latex]\\hspace{7cm}\\begin{align}\r\n\r\nf_x(x,y,z)&=10x-2y+3z \\\\\r\n\r\nf_y(x,y,z)&=-2x+2y-4z \\\\\r\n\r\nf_z(x,y,z)&=-4y+2z+3x,\r\n\r\n\\end{align}[\/latex]\r\n\r\nThen substitute them into the\u00a0Directional Derivative of a Function of Three Variables:\r\n\r\n[latex]\\hspace{5cm}\\begin{align}\r\n\r\nD_{\\bf{u}}f(x,y,z)&=10x-2y+3z \\\\\r\n\r\nf_y(x,y,z)&=f_x(x,y,z)\\cos\\alpha+f_y(x,y,z)\\cos\\beta+f_z(x,y,z)\\cos\\gamma \\\\\r\n\r\n&=(10x-2y+3z)(-\\frac13)+(-2x+2y-4z)(\\frac23)+(-4y+2z+3x)(\\frac23) \\\\\r\n\r\n&=-\\frac{10x}3+\\frac{2y}3-\\frac{3z}3-\\frac{4x}3+\\frac{4y}3-\\frac{8z}3-\\frac{8y}3+\\frac{4z}3+\\frac{6x}3 \\\\\r\n\r\n&=-\\frac{8x}3-\\frac{2y}3-\\frac{7z}3\r\n\r\n\\end{align}[\/latex]\r\n\r\nLast, to find\u00a0[latex]D_{\\bf{u}}f(1,-2,3)[\/latex], we substitute\u00a0[latex]x=1[\/latex],\u00a0[latex]y=-2[\/latex], and\u00a0[latex]z=3[\/latex]:\r\n\r\n[latex]\\hspace{5cm}\\begin{align}\r\n\r\nD_{\\bf{u}}f(1,-2,3)&=-\\frac{8(1)}3-\\frac{2(-2)}3-\\frac{7(3)}3 \\\\\r\n\r\n&=-\\frac83+\\frac43-\\frac{21}3 \\\\\r\n\r\n&=-\\frac{25}3.\r\n\r\n\\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

      Three-Dimensional Gradients and Directional Derivatives<\/h2>\n

      The definition of a gradient extends naturally to functions of more than two variables.<\/p>\n

      \n

      gradient of a function of three variables<\/h3>\n

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

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

      The notation [latex]\\nabla{f}(x,y,z)[\/latex] can also be written as [latex]\\text{grad }f(x,y,z)[\/latex].<\/p>\n<\/section>\n

      Calculating the gradient of a function in three variables follows the same process as in two variables. First, calculate the partial derivatives [latex]f_x[\/latex], [latex]f_y[\/latex], and [latex]f_z[\/latex], then assemble them into the gradient vector.<\/p>\n

      Find the gradient [latex]\\nabla{f}(x,y,z)[\/latex] of each of the following functions:<\/p>\n
        \n
      1. [latex]f(x,y,z)=5x^2-2xy+y^2-4yz+z^2+3xz[\/latex]<\/li>\n
      2. [latex]f(x,y,z)=e^{-2z}\\sin 2x\\cos 2y[\/latex]<\/li>\n<\/ol>\n