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Conservative Work

  1. Nov 15, 2006 #1
    I have a force vector that is [tex]F = (x^2 + y)i + (y^2 + x)j +ze^z k)[/tex] and I am supposed to find the work done from (1,0,0) to (1,0,1). The question gives a bunch of paths to integrate, but I used the curl and found that the force was conservative (hence path independent), so I was going to make a function of x,y,z based on F.

    [tex]\frac{\partial f}{\partial x} = x^2 + y[/tex]

    [tex]f(x,y,z) = (1/3)x^3 +xy+g(y,z)[/tex]

    [tex]\frac{\partial f}{\partial y}=x+\frac{\partial g}{\partial y} = y^2 +x[/tex]

    [tex]\frac{\partial g}{\partial y} = y^2[/tex]

    [tex]g(y,z)= (1/3)y^3 +h(z)[/tex]

    so now continue to build the function

    [tex]f(x,y,z)=(1/3)x^3 +xy+(1/3)y^3 + h(z)[/tex]

    [tex]\frac{\partial f}{\partial z} = 0 + \frac{\partial h}{\partial z} = ze^z[/tex]

    solve by parts to get

    [tex]h(z)=ze^z - \int e^z dz = e^z(z-1)[/tex]

    [tex]f(x,y,z)=(1/3)x^3 + xy+(1/3)y^3 + e^z(z-1)[/tex]

    Now I know that the last part is the part that I screwed up because [tex]f_z[/tex] is supposed to equal ze^z, and it doesn't. What I can't figure out is what I screwed up. Anyone see it?
  2. jcsd
  3. Nov 15, 2006 #2
    No I get the same answer as you, dx=dy=0, so the only term left in [tex]\int_{1,0,0}^{1,0,1}{F.dl}[/tex] is [tex]\int_{0}^{1}{z e^{z} dz}[/tex], which gets you [tex][e^z(z-1)]_{0}^{1} = -1[/tex]. Why should it be ze^z? df is an element of work done, not dF.
  4. Nov 15, 2006 #3
    Don't you have to regain F with its respective partials? Taking the partial of the function with respect to x gives [tex]x^2 + y[/tex], which is the x (or M) term; with respect to y gives [tex]y^2 + x[/tex], which is the y (or N) term; yet, with repsect to z it is [tex]e^z+ze^z-e^z[/tex] (errr... whoops I just realized it did work out - for some reason I assumed it didn't).

    It would give +1 though, since evaluating gives 0--1=1.
    Last edited: Nov 15, 2006
  5. Nov 16, 2006 #4


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    Since you know the force is conservative, you know the work done is independent of the path. Integrate F along the straight line from (1, 0, 0) to (0, 0, 1): x= 1, y= 0, z= t. dx= 0, dy= 0, dz= dt so the integral is
    [tex]\int_0^1 te^t dt[/tex].
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