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Math Help - Potential Function

  1. #1
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    May 2010
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    Potential Function

    F(x,y) is defined by the integral:

    \\<br />
\int\nolimits_{(2,\pi )}^{(x,y)} {[-2uv^{2}\sin (u^{2}v)]du+[\cos (u^{2}v)-u^{2}v\sin (u^{2}v)]dv}

    Express F(x,y) as a function of x and y, eliminating the integral sign.

    _______________________

    My approach:

    I found the curl of the function to be zero, which means it is path independent and that a potential function exists... I'm now working towards building f from F and finally calculating f(x,y)-f(2,pi)

    Which I did and I got:
    f(u,v)=v[\cos (u^{2}v)]

    which means the final value for the integral is:

    y\cos \left ({x^{2}y} \right )-\pi

    Does my answer seem correct to you?
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  2. #2
    Newbie
    Joined
    May 2010
    Posts
    20

    Confirmed!

    I just thought of a way to confirm that my answer is correct. Here's how:

    If you differentiate f(u,v)=v[\cos (u^{2}v)] with respect to 'u' you get the first integrand and if you differentiate with respect to 'v' you get the second integrand in F(x,y)...

    So f(u,v)=v[\cos (u^{2}v)] must been the potential function to begin with.

    Anybody wanna tap me on the shoulder?
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