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Finding a function in conservative field

  1. Mar 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the non-zero function h(x) for which:

    field F(x,y) = h(x) [xsiny + ycosy] i + h(x) [xcosy - ysiny] j

    is conservative.

    3. The attempt at a solution

    curlF=0
    d/dx [h(x) [xcosy - ysiny] ] - d/dy [h(x) [xsiny + y cos y] ] = 0

    xcosy = ysiny ???

    I have no idea!!!
     
    Last edited: Mar 22, 2009
  2. jcsd
  3. Mar 22, 2009 #2

    gabbagabbahey

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    That is only the z-component of curlF....all 3 components will have to be zero.

    You will need to use the product rule when computing the derivatives....
     
  4. Mar 22, 2009 #3
    Thanks for the reply but I'm not sure I understand... do u mean this? ...

    curlF= {d/dy[0]-d/dz[ h(x)[xcosy - ysiny]]} - {d/dx[0]-d/dz[ h(x)[xsiny + ysiny]]} + {d/dx[ h(x)[xcosy - ysiny]] - d/dy[ h(x)[xsiny + ycosy]]}

    = 0 - 0 + d/dx[ h(x)[xcosy - ysiny]] - d/dy[ h(x)[xsiny + ycosy]]}

    = [ h`(x)xcosy + h(x)cosy - h`(x)ysiny] - [ h(x)xcosy - h(x)ysiny + h(x)cosy ]

    = h`(x)xcosy + h(x)cosy - h`(x)ysiny - h(x)xcosy + h(x)ysiny - h(x)cosy

    = h`(x)xcosy - h`(x)ysiny - h(x)xcosy + h(x)ysiny

    = xcosy [h`(x) - h(x)] + ysiny [h(x) - h`(x)]

    = xcosy [h`(x) - h(x)] - ysiny [h`(x) - h(x)]

    = [h`(x) - h(x)] [xcosy - ysiny] = 0

    as h(x) is a non-zero vector then xcosy - ysiny = 0

    xcosy = ysiny


    And the no matter what I do I can't seem to get h(x).
     
  5. Mar 22, 2009 #4

    gabbagabbahey

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    No, you can't pick and choose x and y values such that xcosy = ysin y , you are looking to choose an h(x) that makes [h`(x) - h(x)] [xcosy - ysiny] = 0 for all x and y....The only way that can happen is if
    [h`(x) - h(x)]=0...right? What kind of non-trivial function accomplishes that?:wink:
     
  6. Mar 22, 2009 #5
    thanks! i got it.
     
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