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Math Help - [SOLVED] finding the gradient vector

  1. #1
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    [SOLVED] finding the gradient vector

    Well this is the question-

    df/dx( 2,-3) = 7 and df/dy (2,-3)=11

    i) Find the equation of tangent I get z-f(2,-3)=7(x-2)+11(y+3)
    ii) Find the directional derivative Duf(2,-3) in the direction of u=5i-2j
    I get 7(5/root29) -11(2/root29) then just simpfly

    now my real question is this

    Find a non-zero vector v such that Dvf(2,-3)=0

    What is the proper working out for this i can just make up any vector provided when I take the gradient vectors dot product i get zero..

    can someone show me the proper way ?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    For any vector a i + b j

    a perpendicular vector is then simply -b i + a j

    Of course there are others
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  3. #3
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    so if i write any vector but provided their dot products equate to 0 its still true ??

    I like that -bi +aj method ??

    soo what i did still makes it right ?
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  4. #4
    MHF Contributor Calculus26's Avatar
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    Yes any perpindicular vector will do-- but I find when I have an automatic

    I use it.

    In 2 space all perpindicular vectors will be a scalar multiple of -b i + a j
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  5. #5
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    Thanks I will use that way it seems more efficient
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