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Math Help - Directional derivative problem

  1. #1
    MHF Contributor arbolis's Avatar
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    Directional derivative problem

    I'd like you to check my answer. This is the most common exercise about directional derivatives...
    Let f(x,y,z)=xyz, \bold u =( \cos (\alpha) \sin (\beta) , \sin (\alpha) \sin (\beta) , \cos (\beta)   ) and \bold x = (1,0,1).
    Find the directional derivative of f in the direction of \bold u, in the point \bold x.
    Here's my work :
    \frac{\partial f}{\partial \bold u} (\bold x)= \nabla f( \bold x) \cdot \bold u = \sin (\alpha) \sin (\beta).

    I think it is right, but I'm unsure.
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  2. #2
    Super Member Random Variable's Avatar
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    It look correct to me.

     \nabla f(x,y.z) = (yz, xz, xy)

     \nabla f(1,0,1) = (0,1,0)

     \bold{u} is already a unit vector so it doesn't need to be normalized

     D_{\bold{u}}f(1,0,1) = \nabla f(1,0,1) \cdot \bold{u} = \sin \alpha \sin \beta
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Random Variable View Post
    It look correct to me.

     \nabla f(x,y.z) = (yz, xz, xy)

     \nabla f(1,0,1) = (0,1,0)

     \bold{u} is already a unit vector so it doesn't need to be normalized

     D_{\bold{u}}f(1,0,1) = \nabla f(1,0,1) \cdot \bold{u} = \sin \alpha \sin \beta
    Ok thanks. Yeah I've realized I had to check out if \bold u was normalized and I did the calculations after I posted here. Thanks for pointing it out.
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