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Homework Help: Divergence Question

  1. Jan 15, 2010 #1
    Given two vectors, A and B:

    [tex]A = (x\widehat{x} + 2y\widehat{y} + 3z\widehat{z})[/tex]
    [tex]B = (3y\widehat{x} - 2x\widehat{y})[/tex]

    I need to calculate [tex](B [dot] \nabla)A[/tex], as part of a problem. The answer should be:

    [tex]\widehat{x}(3y) + \widehat{y}( -4x)[/tex]


    I get:

    [tex](B [dot] \nabla)A = ((3y) \delta / \delta x - (2x)\delta / \delta y) (x\widehat{x} + 2y\widehat{y} + 3z\widehat{z})[/tex]
    [tex]=(0+0)(x\widehat{x} + 2y\widehat{y} + 3z\widehat{z})[/tex]
    [tex]=0[/tex]

    I don't understand what I'm doing wrong. Can someone help me out please? Thanks in advance!
     
  2. jcsd
  3. Jan 15, 2010 #2
    Hey Aristata!

    First what is [tex]B \cdot \nabla[/tex]? Above you have written it out slightly wrong (but this may have been just a latex typo.) It should look like:

    [tex]3y \frac{\partial}{\partial x} - 2x\frac{\partial}{\partial y}[/tex]

    I would suggest rewriting [tex]\vec{A}[/tex] as [tex](x,2y,3z)[/tex] so you would obtain:

    [tex]\left (3y \frac{\partial}{\partial x} - 2x\frac{\partial}{\partial y}\right ) (x,2y,3z)[/tex].

    Now ''multiply'' through the operator as if it were a scalar acting on a vector. Similar to:

    [tex]\lambda \vec{A} = (\lambda a_1, \lambda a_2, \lambda a_3)[/tex]

    Since [tex]\lambda[/tex] in your case is an operator, you have to perform the action of [tex]\lambda[/tex] on each of [tex]a_1, a_2, a_3[/tex]. In particular take the derivatives of those elements.

    Hope this wasn't too vague :).
     
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