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Conservative Vector Fields - Is this right?

  1. Mar 7, 2008 #1
    Conservative Vector Fields -- Is this right?

    1. The problem statement, all variables and given/known data

    G = <(1 + x)e[tex]^{x+y}[/tex], xe[tex]^{x+y}[/tex]+2z, -2y>
    Evaluate [tex]\int[/tex][tex]_{C}[/tex]G.dR
    where C is the path given by:

    x = (1 - t)e[tex]^{t}[/tex], y = t, z = 2t, 1=>t>=0
    2. Relevant equations



    3. The attempt at a solution

    First, i noticed that there is a scalar potential associated with several terms (not all) in
    the above vector field:

    [tex]\varphi[/tex] = xe[tex]^{x + y}[/tex]
    [tex]\nabla[/tex][tex]\varphi[/tex] = <(1 + x)e[tex]^{x + y}[/tex], xe[tex]^{x + y}[/tex],0>

    So, i then separated the initial vector field G into two parts:

    G=<0, 2z, -2y> + <(1 + x)e[tex]^{x + y}[/tex], xe[tex]^{x + y}[/tex],0>

    First, I compute some derivatives and then evaluate the first line integral:
    [tex]\frac{dc}{dt}[/tex] =
    {dx = -te[tex]^{t}[/tex]dt
    dy = 1dt
    dz = 2dt}


    so,

    [tex]\int[/tex][tex]_{C}[/tex]G[tex]_{1}[/tex] . [tex]\frac{dc}{dt}[/tex]dt =
    [tex]\int[/tex][tex]^{1}_{0}[/tex]<0, 2(2t), -2(t)> . <-te[tex]^{t}[/tex], 1, 2>
    =

    [tex]\int[/tex]0dt = 0

    I then computed the second part of G, keeping in mind that this portion of the vector field is conservative. Therefore, i just made a path from point t = 0 to t = 1 with a straight line r, parametrized like so:

    r = { x = 1-t, y = t, z = 2t}
    [tex]\frac{dr}{dt}[/tex] = {dx = -1, dy = 1, dz = 2}

    my second integral then becomes;

    [tex]\int[/tex][tex]^{1}_{0}[/tex]<(1 + 1 - t)e[tex]^{1 - t + t}[/tex], (1-t)e[tex]^{1-t + t}[/tex],0> . <-1, 1, 2> dt =


    [tex]\int[/tex][tex]^{1}_{0}[/tex]<(2-t)e, (1-t)e,0> . <-1, 1, 2> dt =

    So, is my final answer simply this integral evaluated, or am I doing this completely wrong? Sorry if this is a bad question, my book just has no example problems at all, and integrating the theory can be confusing. Any feedback would be appreciated greatly. Thank you
     
  2. jcsd
  3. Mar 7, 2008 #2

    Dick

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    That looks like a reasonable approach to me. I didn't check all of the details, though.
     
  4. Mar 7, 2008 #3
    The final answer that i got was -e, by computing this integral and adding the two portions (o & -e)
     
  5. Mar 7, 2008 #4
    Great, thanks. I don't want anyone to check the details, I just wanted to see if what I was doing was reasonable, or if i was doing it completely wrong.
     
  6. Mar 7, 2008 #5

    Dick

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    Homework Helper

    I get -e as well.
     
  7. Mar 7, 2008 #6
    Thanks a lot, I appreciate the help!
     
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