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Calculus Identities

  1. May 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose f is a continously differentiable real valued function on R^3 and F is a continously differentiable vector field

    Prove 1)##\oint (f \nabla g +g\nabla f) \cdot dr=0##

    2) ##\oint(f \nabla f)\cdot dr=0##


    2. Relevant equations

    ##\nabla f = f_z i+ f_y j+f_z k##
    Real valued function ##f(x,y,z)## and ##g(x,y,z)##

    3. The attempt at a solution

    1)

    ##f \nabla g =fg_x i +fg_y j+fg_z k##
    ##g \nabla f =gf_x i +gf_y j+gf_z k##

    ##\implies (f \nabla g + g \nabla f )\cdot dr##

    ##= (fg_x i +fg_y j+fg_z k+gf_x i +gf_y j+gf_z k)\cdot(dx i+dyj+dzk)##

    2)

    ##(f \nabla f)\cdot dr= (ff_xi+ff_yj+ff_zk)\cdot(dxi+dyj+dzk)##

    How do these work out to be 0?

    Thanks
     
  2. jcsd
  3. May 12, 2012 #2
    Any chance ## f \nabla g +g\nabla f ## and ## f \nabla f## might be conservative?
     
  4. May 12, 2012 #3

    sharks

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    Gold Member

    In my opinion, there are some missing information in your problem.

    Is ##g## also a continously differentiable real valued function in R^3?

    Shouldn't this be: [itex]\nabla f = f_x i+ f_y j+f_z k[/itex]?
     
  5. May 12, 2012 #4
    Yes, you are correct sharks, that is a typo. It should be as you have stated.
    ##...f_xi...##

    It does not mention anything about g but perhaps we take it that it is also a real valued function?

    I believe I left out the following important information
    C is a smooth, simple closed curve which lies on the surface of a paraboloid in R^3. I guess this means integrand is conservative, right?
    But I still not sure how it goes to 0, there must be additional lines
    Thanks
     
  6. May 12, 2012 #5

    sharks

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    Gold Member

    Do you mean ##f## or did the question involve ##\vec F##, in which case, has any information been given about the latter?
     
  7. May 13, 2012 #6
    Here is the proper question asked in full. Apologies again.

    Suppose that sigma and C satisfy the hypothesis of Stokes Theorem and that f and g have continous second order partial dervivatives. Prove each of the following

    ##\oint_C (f \nabla g) \cdot dr = \oint \oint_\sigma (\nabla f \times \nabla g)\cdot dS##

    ##\oint_C (f \nabla f) \cdot dr=0##

    ##\oint (f \nabla g +g \nabla f)\cdot dr=0##

    I am interested in the last 2 but maybe the first one allows me to complete the last 2?

    Thanks
     
  8. May 13, 2012 #7

    HallsofIvy

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    Science Advisor

    This makes no sense. You have "g" in the conclusion but not in the hypotheses and "F" in the hypotheses but not in the conclusion. What is the problem, really?

     
  9. May 13, 2012 #8
    The correct thread/question is post #6 and not #1. The is no 'F' involved, that was in another very similar question (#1 which I will ignore). Only f and g are involved.

    THanks
     
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