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Gradient Vector is Orthogonal to the Level Curve

  1. May 5, 2015 #1
    1. The problem statement, all variables and given/known data
    Let f(x,y)=arctan(x/y) and u={(√2)/2,(√2)/2}
    d.) Verify that ∇fp is orthogonal to the level curve through P for P=(x,y)≠(0,0) where y=mx for m≠0 are level curves for f.

    2. Relevant equations


    3. The attempt at a solution
    ∇f={(y)/(x^2+y^2),(-y)/(x^2+y^2)}
    m=1/tan(k) where k=f(x,y) and tan(k)≠0
    I'm stuck and very confused. The homework is doomed and turned in, but I still really want to understand how to think about problems like this one. Any help is appreciated.
     
  2. jcsd
  3. May 5, 2015 #2

    Orodruin

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    What is the tangent vector of the level curve? What is the requirement to be orthogonal to the level curve?
     
  4. May 5, 2015 #3
    See that's the thing. I know this is a simple question, but I'm having a total blank. Thank you for your help and I understand if you can't show me more.
     
  5. May 5, 2015 #4

    HallsofIvy

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    Uhh, do know what a level curve is?
     
  6. May 5, 2015 #5
    I think I do. Its the function f (x,y) equal to some constant, say k. Maybe this is right, maybe not. But when k=f (x,y), k=arctan (x/y) or y=x/(tan (k)), right? So I believe it is the function set so that the set of x and y values must equal the value k. I don't know, thank you for your help.
     
  7. May 5, 2015 #6

    Orodruin

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    Yes, but you really do not need to know what it is to solve this. You have already been told that the line y = mx is a level curve. This is a straight line. What is the tangent vector of a straight line? What is a tangent vector?
     
  8. May 6, 2015 #7

    HallsofIvy

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    You also should know that the directional derivative for f(x,y), in direction of unit vector [itex]\vec{u}[/itex], is [itex]\vec{u}\cdot \nabla f[/itex]. In particular, that directional derivative is 0 if and only if [itex]\vec{u}[/itex] is perpendicular to [itex]\nabla f[/itex]. And, of course, since the function is constant on a level curve, the derivative along it (in its direction) is 0.
     
    Last edited: May 7, 2015
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