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Math Help - Dependence of the path

  1. #1
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    Dependence of the path

    To check whether a vector field depends on the path just check if the rotation is equal to 0 ?

    How can I check if the integral curvilinear depends on the path ?

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  2. #2
    Member Ruun's Avatar
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    If \nabla \times F = 0 then F is a conservative field, and integral doesn't depend on the path, only on the initial and final position. Also the reciprocal is true
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  3. #3
    MHF Contributor Amer's Avatar
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    f is independent of path in Q if

    1) \oint F.dr =0 for all simple closed curve in Q piece wise smooth

    2)F is conservative (in Q) is gradient

    3)curl F = o for all (x,y,z) in Q

    if one is true then all is true
    all the statement are equivalent
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  4. #4
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    Quote Originally Posted by Amer View Post
    f is independent of path in Q if

    1) \oint F.dr =0 for all simple closed curve in Q piece wise smooth

    2)F is conservative (in Q) is gradient

    3)curl F = o for all (x,y,z) in Q

    if one is true then all is true
    all the statement are equivalent

    \int_S yzdx +xzdy + yx^2dz

    curl is different 0, then is dependent the path

    Correct ?
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  5. #5
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    Correct the example ?
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  6. #6
    MHF Contributor Amer's Avatar
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    that what my professor said to us independent of path mean the curl of the vector equal zero

    but I didn't see it in the book of calculus 3) the curl is zero

    this from my professor so I do not know if it is true or not but 1) and 2)

    but I think it is true take this example

    show that this is independent of path

    F(x,y)=2xy^3i + (1+2x^2y^2)j is independent of path

    the curl is zero or as the book solve it

    since f(x,y)=2xy^3 and g(x,y)=(1+3x^2y^2)

    \frac{\partial g}{\partial x } = 6xy^2 = \frac{\partial f}{\partial y} so 6 holds for all (x,y)


    see this
    Conservative vector field - Wikipedia, the free encyclopedia

    they said every conservative vector is irrotational irrotational mean the curl is zero
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  7. #7
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    Quote Originally Posted by Amer View Post
    that what my professor said to us independent of path mean the curl of the vector equal zero

    but I didn't see it in the book of calculus 3) the curl is zero

    this from my professor so I do not know if it is true or not but 1) and 2)

    but I think it is true take this example

    show that this is independent of path

    F(x,y)=2xy^3i + (1+2x^2y^2)j is independent of path

    the curl is zero or as the book solve it

    since f(x,y)=2xy^3 and g(x,y)=(1+3x^2y^2)

    \frac{\partial g}{\partial x } = 6xy^2 = \frac{\partial f}{\partial y} so 6 holds for all (x,y)


    see this
    Conservative vector field - Wikipedia, the free encyclopedia

    they said every conservative vector is irrotational irrotational mean the curl is zero
    Ok. Thank you
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  8. #8
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    If there exist a scalar function f such that F=\nabla f then  \nabla \times F = \nabla \times \nabla f = 0

    In Vector Calculos by Marsden & Tromba in the section 3.4 the Theorem: For every function f \in C^2 \nabla \times \nabla f = 0

    The demonstration is just computing  \nabla \times \nabla f and use that \frac{\partial f^2}{dxdy} = \frac{\partial f^2}{dydx} and so on.
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  9. #9
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    Re: Dependence of the path

    so for example for a conservative force, divide the closed path from Po to P1, the work done on path c1 is equal to path c2 and since they are in opposite directions they sum to 0. is there any relationship to the physics sum of all paths integral? why are the conditions: positive oriented, piecewise smooth, simple closed curve necessary conditions for this and Green's Theorem?
    Last edited by mathlover10; April 14th 2013 at 01:47 PM.
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