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Directiontal derivative

  1. Mar 19, 2004 #1
    OK, I having a small problem understanding how my text book came about an answer to an example problem.

    [tex]f(x,y)=4-x^2-\frac{1}{4}y^2[/tex]
    at P(1,2)

    This next step is the one that's bugging me:

    [tex]u^\rightarrow=\cos(\frac{\pi}{3})\imath+\sin(\frac{\pi}{3})\jmath[/tex]

    This is one of those instances where something magic happens because right now I have little to no clue where the [tex]\frac{\pi}{3}[/tex] came from.

    Thanks...
     
  2. jcsd
  3. Mar 19, 2004 #2

    jamesrc

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    I may be wrong, but isn't that just part of the given information? I mean, you're being asked to find the directional derivative, meaning the rate of change of the function at a point in a given direction. So you know the function z=f(x,y), you're given a point P(1,2), and you're given the unit vector of the direction you're interested in.
     
  4. Mar 19, 2004 #3
    Yeah your correct... The question is poorly written (or at least poorly formated). Thanks for showing me my stupid mistake.
     
  5. Mar 20, 2004 #4

    HallsofIvy

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    Specifically, [tex]u^\rightarrow=\cos(\frac{\pi}{3})\imath+\sin(\frac{\pi}{3})\jmath[/tex] is the unit vector point at an angle [tex]\pi/3[/tex] radians from the positive x-axis.
     
  6. Mar 20, 2004 #5
    (Psst. Use "\vec v" in LaTeX to display a vector... :smile:)
     
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