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Math Help - Gradient of dot product

  1. #1
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    Gradient of dot product

    Removed this post because I thought it was solved, having to add it again. What is the gradient of acos(u1*u2) (u1 and u2 are the unit vectors)? Apparently the gradient of u1*u2 is u2-(u1*u2)u1 and I know the derivative of acos is -1/sqrt(1-x), but I'm slightly confused here. Thanks for any help.
    Last edited by aplrt; February 7th 2013 at 11:19 AM.
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  2. #2
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    Re: Gradient of dot product

    let f = u1.u2 = u11u21 + u12u22 = f(x,y)

    delf = df/dxi +df/dyj

    del acosf = [d(acosf)/dx]i + [d(acosf) /dy]j = -1/sqrt(1-f^2)[(df/dx)i +(df/dy)j]

    del acosf = -1/sqrt(1-f^2)delf

    delf = u21delu11 + u21delu22 + u11delu21 + u12delu22, or
    delf = delu1.u2 = u1.delu2 + u2.delu1 + u1x(delxu2) + u2x(delxu1) look it up, or google

    Same in 3 dimensions just more to write. If u1 and u2 are orthogonal unit vectors, go back and think about it, oh well, then u1.u2 =1 and delf=0, so probably not.
    Didn't feel like bolding del and all the vectors, ie, del = del and u1 = u1 etc. U11..are scalars, d is partial.
    Thanks from aplrt
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  3. #3
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    Re: Gradient of dot product

    if u1 & u2 are unit vectors, then u1.u2 = cosθ. Then
    acos(u1.u2) = θ

    but what is del(θ)? formally, in 3d,
    del(θ) = dθ/dx i+ dθ/dy j + dθ/dz k, where θ is angle between u1 & u2

    I have the feeling thereís a further simplification, but I donít see it.
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