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Directional Derivative Equal to Zero

  1. Mar 15, 2013 #1
    The problem states:

    "In what direction is the directional derivative of [tex]f(x,y) = \frac{x^2 - y^2}{x^2 + y^2}[/tex] at (1,1) equal to zero?"

    I know that ##D_uf = \nabla{f}\cdot{{\bf{u}}}##. I believe the problem simply is asking for me to determine what vector ##{\bf{u}}## will yield zero. Thus:
    [tex]\nabla{f} = \left\langle\frac{4xy^2}{{(x^2 + y^2)}^2},\frac{-4{x^2}y}{{(x^2 + y^2)}^2}\right\rangle[/tex]
    At the point (1,1), we get ##\nabla{f} = \left\langle{1},{-1}\right\rangle##. From here, I think that a vector taken with the dot product of ##\left\langle{1},{-1}\right\rangle## to give zero would be ##\left\langle{1},{1}\right\rangle##. I'm not sure if it's that simple though because I could also say that ##\left\langle{0},{0}\right\rangle## gives zero, too. Perhaps my justification is flawed? Much appreciated.
     
  2. jcsd
  3. Mar 16, 2013 #2

    LCKurtz

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    Assuming your derivative is right it looks good except why would you say <0,0> gives zero? Also besides <1,1>, what about its opposite?
     
  4. Mar 16, 2013 #3
    True. I guess I was thinking that the problem was seeking a single answer. As for the former, ##\left\langle{0},{0}\right\rangle \cdot \left\langle{1},{-1}\right\rangle = 0##, but I suppose that's not really even a "direction" at all... ><
     
  5. Mar 16, 2013 #4

    LCKurtz

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    While it's true that ##\left\langle{0},{0}\right\rangle \cdot \left\langle{1},{-1}\right\rangle = 0##, what point ##(x,y)## have you found where ##\nabla f(x,y) = \langle 0,0\rangle##?
     
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