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Vector Operations?

  1. Jan 18, 2013 #1
    1. The problem statement, all variables and given/known data

    $$ \overrightarrow {F}=3xz^{2}i+2xyj-x^{2}k $$ $$\phi =3x^{2}-yz $$ are given vector and scalar fields, respectively.

    a) $$\quad \operatorname{div}\left( \operatorname{grad}\phi \operatorname{div}\overrightarrow {F}\right) =\quad? $$

    b) $$\quad \operatorname{curl}\left( \phi F\right) =\quad? $$

    c) $$\quad \operatorname{div}\left( \phi F\right) =\quad? $$

    d) $$\quad \overrightarrow {\nabla }\cdot \left( \nabla \phi \times \overrightarrow {F}\right) =\quad? $$

    e) $$\quad \nabla \cdot \left( \overrightarrow {F}\nabla \phi \right) =\quad? $$

    I know the operations such as the Gradient, Divergence, Curl, and Laplacian. But I don't have an idea what can I do in this kind of problems?



    2. Relevant equations
    http://en.wikipedia.org/wiki/Vector_calculus_identities


    3. The attempt at a solution

    a)I found gradient of scalar field and divergence of vector field.Gradient of scalar field is a vector and divergence of vector fields is a scalar.So how can I take the divergence of this?
    No idea for others.I think I have to use some formula for calculate these..But which? I don't wanna a solution just I want to understand logic.Please help me to figure this:) Then I will try to do it myself.

    And also,I'd like to say that I'm sorry for my bad English.
     
  2. jcsd
  3. Jan 18, 2013 #2

    mfb

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    Staff: Mentor

    Vector*Scalar (better: written as scalar*vector) is a vector again, and has a divergence.

    Vector*Vector: Scalar product ("dot product")
    Vector x Vector: Cross-product
    Scalar*Vector or Vector*Scalar: Scalar multiplication of the vector
    Scalar*Scalar: Just like multiplication with real numbers

    You can evaluate these expressions step by step.
     
  4. Jan 18, 2013 #3
    @mfb; Is there any another way,because there are a lot of process with step by step? Now,I'm trying step by step.I will tell the result,but I don't know where can I control it?
     
  5. Jan 18, 2013 #4

    mfb

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    Staff: Mentor

    WolframAlpha and other computer algebra systems should be able to do that.
    There are many steps, but they are all easy, and you can even re-use some.

    For more complicated expressions, vector identities can be useful, but I think you cannot use them (in a meaningful way) here.
     
  6. Jan 18, 2013 #5
    OK.Here is my expression.

    = (18z^2-12+72x+2z-6z^2+2y-6yz) + ( 24-4z-3z^2-8y+2x-6xz) + (36xz+36x^2-9z^2-4y+2x-12xz-6yz-6xy)

    = 36x^2-6xy-12yz+6y+18xz am I right?
     
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