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Thread: Directional derivatives

  1. #1
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    Directional derivatives

    Find a vector u for which Duf(1,1) =0

    Thank you.
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  2. #2
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    Quote Originally Posted by al1850 View Post
    Find a vector u for which Duf(1,1) =0

    Thank you.

    You ned to find a unit vector such that $\displaystyle {\rm{u}}.[(\nabla f (1,1)]=0$

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    You ned to find a unit vector such that $\displaystyle {\rm{u}}.[(\nabla f (1,1)]=0$

    RonL
    But how?
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  4. #4
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    well first for the definition of directional derivative:

    directional derivative of a scalar function, $\displaystyle f=f(x,y)$ in the direction of a unit vector,$\displaystyle \overrightarrow{u}=(u_1 , u_2) $ at a point $\displaystyle (x_0,y_0) $is given by:
    $\displaystyle D \overrightarrow {u} f(x_0,y_0) = \bigtriangledown f(x_0 , y_0) \cdot \overrightarrow {u} = u_1 \frac {\partial f}{\partial x} + u_2\frac {\partial f}{\partial y}
    $

    in your question:
    $\displaystyle
    f(x,y) = \frac{1}{x^2+y^2+1}
    $
    $\displaystyle
    \bigtriangledown f(x , y) = \frac {\partial f}{\partial x} +\frac {\partial f}{\partial y}
    $
    $\displaystyle
    = -\frac{2x}{(x^2+y^2+1)^2}\overrightarrow{i}-\frac{2y}{(x^2+y^2+1)^2}\overrightarrow{j}
    $

    at f(1,1):
    $\displaystyle
    = -\frac{2}{(1^2+1^2+1)^2}\overrightarrow{i}-\frac{2}{(1^2+1^2+1)^2}\overrightarrow{j}
    $
    $\displaystyle
    = -\frac{2}{9}\overrightarrow{i}--\frac{2}{9}\overrightarrow{j}
    $

    therefore from;
    $\displaystyle
    D \overrightarrow {u} f(x,y)
    $

    $\displaystyle
    D \overrightarrow {u} f(1,1)
    $
    $\displaystyle
    = (-\frac{2}{9}\overrightarrow{i}-\frac{2}{9}\overrightarrow{j})\cdot (\overrightarrow{u}) = 0
    $
    $\displaystyle
    = (-\frac{2}{9}\overrightarrow{i}-\frac{2}{9}\overrightarrow{j})\cdot (u_1\overrightarrow{i}+u_2\overrightarrow{j}) = 0
    $

    $\displaystyle
    -\frac{2}{9}u_1-\frac{2}{9} u_2 = 0
    $

    now to obtain a zero $\displaystyle u_1$ and $\displaystyle u_2$ must be equal in magnitude but opposite in sign:
    $\displaystyle
    \overrightarrow{u} = (n,-n)
    $

    n can take any value.

    Unit vector of $\displaystyle \overrightarrow{u}$
    = $\displaystyle \frac{\overrightarrow{u}}{|\overrightarrow{u}|}$
    = $\displaystyle \frac{u_1 \overrightarrow{i}+u_2\overrightarrow{j}}{|\sqrt{{ u_1}^2+{u_2}^2}|}$
    = $\displaystyle \frac{u_1 \overrightarrow{i}+u_2\overrightarrow{j}}{|\sqrt{n ^2+(-n)^2}|}$
    = $\displaystyle \frac{n \overrightarrow{i}-n\overrightarrow{j}}{|\sqrt{2}n|}$
    = $\displaystyle \frac{n }{|\sqrt{2}n|}\overrightarrow{i}-\frac{n }{|\sqrt{2}n|}\overrightarrow{j}$
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