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Determine gradient of a function f(x,y)

  1. Feb 17, 2009 #1
    1. The problem statement, all variables and given/known data
    View the curve below as a contour of f(x,y).
    (y-x)^2 + 2 = xy - 3

    Use gradf (2,3) to find a vector normal to the curve at (2,3).


    2. Relevant equations



    3. The attempt at a solution
    I am not sure how do I get the vector normal to the curve, is it using a cross product>?
     
  2. jcsd
  3. Feb 17, 2009 #2

    Dick

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    Re: gradients

    No. Express your curve as f(x,y)=C and use the gradient. That's what they told you to do.
     
  4. Feb 17, 2009 #3
    Re: gradients

    what's c here?
     
  5. Feb 18, 2009 #4

    HallsofIvy

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    Re: gradients

    The constant you get by putting x= 2, y= 3 into the function. Since it will drop out of the derivative, its value is not important and you can just leave it as "c". Since the gradient always points in the direction of fastest increase it is always normal to level curves.
     
  6. Feb 18, 2009 #5
    Re: gradients

    so the gradient will be the normal to the level curves??
     
  7. Feb 18, 2009 #6

    Dick

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    Re: gradients

    That's exactly what Halls said, isn't it? C doesn't matter. The gradient will depend only on x and y.
     
  8. Feb 18, 2009 #7
    Re: gradients

    well the gradient I have is:

    (-3y + 2x) i + (2y - 3x) j

    is this true?
     
  9. Feb 18, 2009 #8

    Dick

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    Re: gradients

    Sure. Now just evaluate it at (2,3).
     
  10. Feb 18, 2009 #9
    Re: gradients

    okay I get -5i , if then I am asked to find an equation for the tangent line to the curve at (2,3) and determine whether it's vertical , diagonal, or horizontal. How do I do this?
     
  11. Feb 18, 2009 #10

    Dick

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    Re: gradients

    The tangent is perpendicular to the normal, isn't it?
     
  12. Feb 18, 2009 #11
    Re: gradients

    so then I am assuming it's vertical?
     
  13. Feb 18, 2009 #12

    Dick

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    Re: gradients

    I wouldn't say your "assuming" anything. It is.
     
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