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Line Integral Example - mistake or am I missing something?

  1. Jun 2, 2015 #1
    This is an example at the beginning of the section on the Fundamental Theorem for Line Integrals.

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


    Find the work done by the gravitational field

    [tex]\vec{F}(\vec{x}) = -\frac{mMG}{|\vec{x}|^3}\vec{x}[/tex]

    in moving a particle from the point (3,4,12) to (2,2,0) along a piece wise smooth curve

    Now, I think that there's a mistake in the solution given...but this textbook is pretty good, and plenty of times in the past I've thought it had made a mistake and really I was mistaken.

    So, I understand all the concepts (edit: I obviously didn't), all good there. It's here where I'm scratching my head.

    [tex]f(x,y,z) = \frac{mMG}{\sqrt{x^2+y^2+z^2}}[/tex]

    So, we need to get the scalar function of f, call it the potential function, we know that in a conservative vector field [itex]\vec{F} = \nabla f[/itex], no worries. However, shouldn't it be

    [tex]\vec{F}(\vec{x}) = -\frac{mMG}{|\vec{x}|^3}\vec{x} = -\frac{mMG|\vec{x}|}{|\vec{x}|^3}\vec{u} = -\frac{mMG}{|\vec{x}|^2}\vec{u}[/tex]

    And so converting the vector form of F into a scalar field from which we can compute the grad vector, doesn't

    [tex]|\vec{x}|^2 = x^2 + y^2 + z^2[/tex]

    not

    [tex]\sqrt{x^2+y^2+z^2}[/tex]

    so shouldn't it be

    [tex]f(x,y,z) = \frac{mMG}{|\vec{x}|^2} = \frac{mMG}{x^2+y^2+z^2}[/tex]

    how would one end up with

    [tex]f(x,y,z) = \frac{mMG}{\sqrt{x^2+y^2+z^2}}[/tex]

    ??

    edit: both the 6th and 7th editions have the same example...so I'm guessing that I'm missing something.
     
    Last edited: Jun 2, 2015
  2. jcsd
  3. Jun 2, 2015 #2

    DEvens

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    Can you work out the gradient of the f(x,y,z) you suggest? Does it give the right force?
     
  4. Jun 2, 2015 #3
    No, it gives N not N m.

    I'm still not sure what's going on here though. Is it the case that we're NOT trying to just find the magnitude of the gravity vectors on a given domain, but rather searching for some other related function, that will produce a grad vec that when dot producted with a path vector will result in the dimensions that we need in our answer?

    I still don't know how they got to their answer.

    edit: Oh wait, I think I see what's happening...f(x,y,z) is essentially an antiderivative ?? Thus, we need to be able to get to the grad vec from that f(x,y,z)...??

    edit2: yep, that worked, and it made sense...thanks :)
     
    Last edited: Jun 2, 2015
  5. Jun 2, 2015 #4

    Ray Vickson

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