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Thread: Directional derivatives question

  1. #1
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    Directional derivatives question

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

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

    Attempt:

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

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



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

    How would you find the direction in which f occurs?
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  2. #2
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    If $\displaystyle \nabla f \neq \mathbf 0$ at $\displaystyle P$, then among all possible directional derivatives of $\displaystyle f$ at $\displaystyle P$, the derivative in the direction of $\displaystyle \nabla f$ at $\displaystyle P$ has the largest value. The value of this largest directional derivative is $\displaystyle ||\nabla f||$ at $\displaystyle P$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    and your problem is solved.
    Last edited by NOX Andrew; Mar 18th 2011 at 09:06 AM.
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