Results 1 to 2 of 2

Math Help - Directional derivatives question

  1. #1
    Member
    Joined
    Apr 2010
    Posts
    156

    Directional derivatives question

    Find the max rate of change of f at the given point and the direction in which it occurs.

    f(x, y, z) = \sqrt{x^{2} + y^{2} + z^{2}}, (3, 6, -2)

    Attempt:

    f_x = 0.5(x^2 + y^2 +z^2)^{-0.5}2x
    =0.5(3^{2} + 6^{2} + 4)^{-0.5}6
    = 0.43

    f_y = 0.5(x^2 + y^2 +z^2)^{-0.5}2y
    =0.5(3^{2} + 6^{2} + 4)^{-0.5}12
    = 0.86



    Gradient vector at (x, y) = (f_x, f_y) = (0.43, 0.86)

    How would you find the direction in which f occurs?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Dec 2009
    Posts
    226
    If \nabla f \neq \mathbf 0 at P, then among all possible directional derivatives of f at P, the derivative in the direction of \nabla f at P has the largest value. The value of this largest directional derivative is ||\nabla f|| at P.

    To see why this is so, you should know the directional derivative of f in the direction of \mathbf u at (x_0,y_0,z_0) is denoted by \mathbf{D_u}f(x_0,y_0,z_0) and is defined (for our purposes) by

    \mathbf{D_u}f(x_0,y_0,z_0) = \nabla f \cdot \mathbf u

    You should also be familiar with the following definition of the dot product of two vectors \mathbf a and [/tex]\mathbf b[/tex]

    \mathbf a \cdot \mathbf b = ||a|| \, ||b|| \cos \theta

    where \theta is the angle between \mathbf a and \mathbf b.

    Now, we can re-write the definition of the directional derivative as

    \mathbf{D_u}f(x_0,y_0,z_0) = || \nabla f || \, || \mathbf u || \cos \theta

    Because \mathbf u is a unit vector, || \mathbf u || = 1 and

    \mathbf{D_u}f(x_0,y_0,z_0) = || \nabla f || \cos \theta

    From the above equation, \mathbf{D_u} is maximum when \theta = 0, which means \nabla f and \mathbf u are parallel.

    Returning to your problem, first find \nabla f at the given point (3,6,-2).

    \nabla f = f_x(3,6,-2) \mathbf i + f_y(3,6,-2) \mathbf j + f_z(3,6,-2) \mathbf k = \dfrac{3}{7} \mathbf i + \dfrac{6}{7} \mathbf j - \dfrac{2}{7} \mathbf k

    To find the maximum rate of change of f, find ||\nabla f||.

    ||\nabla f|| = \sqrt{\left(\dfrac{3}{7}\right)^2 + \left(\dfrac{6}{7}\right)^2 + \left(-\dfrac{2}{7}\right)^2} = 1

    To find the direction in which the maximum rate of change of f occurs, find the unit vector \mathbf u in the direction of \nabla f.

    In your case, \nabla f is already a unit vector because ||\nabla f|| = 1. In other words,

    \mathbf u = \nabla f = \dfrac{3}{7} \mathbf i + \dfrac{6}{7} \mathbf j - \dfrac{2}{7} \mathbf k

    and your problem is solved.
    Last edited by NOX Andrew; March 18th 2011 at 09:06 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Directional derivatives
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 6th 2011, 08:28 PM
  2. A question about partial derivatives and directional derivative.
    Posted in the Differential Geometry Forum
    Replies: 8
    Last Post: February 25th 2010, 04:05 PM
  3. Directional derivatives - quick question
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 26th 2009, 09:20 AM
  4. Directional Derivatives
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 7th 2009, 11:45 PM
  5. directional derivatives
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 5th 2008, 06:08 PM

Search Tags


/mathhelpforum @mathhelpforum