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Derivatives of vectors

  1. Apr 11, 2014 #1
    What means:

    imagem.png

    ?

    This guy, ##\vec{\nabla}_{\hat{\phi}} \hat{r}##, for example, means:

    [tex]\\ \hat{\phi}\cdot\vec{\nabla}\hat{r} = \begin{bmatrix}
    \phi _1 \\
    \phi _2 \\
    \end{bmatrix}
    \begin{bmatrix}
    \frac{\partial r_1}{\partial x} & \frac{\partial r_1}{\partial y} \\
    \frac{\partial r_2}{\partial x} & \frac{\partial r_2}{\partial y} \\
    \end{bmatrix}[/tex]
    ?
     
  2. jcsd
  3. Apr 11, 2014 #2

    jedishrfu

    Staff: Mentor

  4. Apr 12, 2014 #3
  5. Apr 12, 2014 #4

    Mark44

    Staff: Mentor

    Your question was not clear. Were you asking what each of the equations in the first image means? Or were you asking what the following means?
    $$\\ \hat{\phi}\cdot\vec{\nabla}\hat{r} = \begin{bmatrix}
    \phi _1 \\
    \phi _2 \\
    \end{bmatrix}
    \begin{bmatrix}
    \frac{\partial r_1}{\partial x} & \frac{\partial r_1}{\partial y} \\
    \frac{\partial r_2}{\partial x} & \frac{\partial r_2}{\partial y} \\
    \end{bmatrix}$$
    BTW, that matrix product looks flaky to me. On the left side of your equation you have a dot product, which should result in a scalar, but on the right side, you have a 2X1 matrix multiplying a 2 X 2 matrix. That multiplication is not defined.

    If you want a clear, concise answer, limit your questions to one per post.
     
  6. Apr 12, 2014 #5
    Wrong my, the correct would be [tex]\begin{bmatrix} \phi _1 & \phi _2 \\ \end{bmatrix} \begin{bmatrix} \frac{\partial r_1}{\partial x} & \frac{\partial r_1}{\partial y} \\ \frac{\partial r_2}{\partial x} & \frac{\partial r_2}{\partial y} \\ \end{bmatrix}[/tex]
    Anyway, this multiplication is just a tentative of answer my own ask (my unique ask), that is: what means ##\vec{\nabla}_\hat{u} \vec{v}## ?
     
  7. Apr 12, 2014 #6

    Mark44

    Staff: Mentor

    Well, at least the matrix product is now defined. But another problem is that I don't see how this can be equal to the dot product you showed in the OP; namely,
    $$\hat{\phi}\cdot\vec{\nabla}\hat{r} $$
     
  8. Apr 12, 2014 #7
    You already understood that I'm lost and that all my equations no pass of speculation. I will not argue anymore about it.

    The question is clear:
     
  9. Apr 12, 2014 #8

    jedishrfu

    Staff: Mentor

    Its a directional derivative of the v vector function in the u direction
     
  10. Apr 13, 2014 #9
    ok, given, u = (u1, u2) , v = (v1, v2) and = (∂x, ∂y), how you'll operate u, v and for get the "directional derivative of the v vector function in the u direction" ?
     
  11. Apr 13, 2014 #10

    jedishrfu

    Staff: Mentor

  12. Apr 13, 2014 #11

    Mark44

    Staff: Mentor

  13. Apr 13, 2014 #12
    In other words, ##\frac{\partial \vec{f}}{\partial \vec{v}} \cdot \vec{u}## is the same thing that $$\\ \ \vec{\nabla}\vec{f}\cdot \vec{u} =
    \begin{bmatrix}
    \frac{\partial f_1}{\partial v_1} & \frac{\partial f_1}{\partial v_2} \\
    \frac{\partial f_2}{\partial v_1} & \frac{\partial f_2}{\partial v_2} \\
    \end{bmatrix}
    \begin{bmatrix}
    u_1 \\
    u_2 \\
    \end{bmatrix} = \vec{\nabla}_{\vec{u}}\vec{f}
    $$

    (assuming that v = (v1, v2) = r = (x, y))

    Like I say in the principle... correct?
     
    Last edited: Apr 13, 2014
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