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Rotational in terms of vector calculus

  1. Jan 12, 2014 #1
    Hellow!

    I was noting that several definitions are, in actually, expressions of vector calculus, for example:

    Jacobian:
    [tex]\frac{d\vec{f}}{d\vec{r}}=\begin{bmatrix} \frac{df_1}{dx} & \frac{df_1}{dy} \\ \frac{df_2}{dx} & \frac{df_2}{dy} \\ \end{bmatrix}[/tex]
    Hessian:
    [tex]\frac{d^2f}{d\vec{r}^2} = \begin{bmatrix} \frac{d^2f}{dxdx} & \frac{d^2f}{dydx}\\ \frac{d^2f}{dxdy} & \frac{d^2f}{dydy}\\ \end{bmatrix}[/tex]
    Gradient:
    [tex]\frac{df}{d\vec{r}}=\begin{bmatrix} \frac{df}{dx} & \frac{df}{dy} \end{bmatrix}[/tex]
    Divergence:
    [tex]trace\left ( \frac{d\vec{f}}{d\vec{r}} \right ) = trace \left ( \begin{bmatrix} \frac{df_1}{dx} & \frac{df_1}{dy} \\ \frac{df_2}{dx} & \frac{df_2}{dy} \\ \end{bmatrix} \right )[/tex]
    Laplacian:
    [tex]trace\left ( \frac{d^2f}{d\vec{r}^2} \right ) = trace \left ( \begin{bmatrix} \frac{d^2f}{dxdx} & \frac{d^2f}{dydx}\\ \frac{d^2f}{dxdy} & \frac{d^2f}{dydy}\\ \end{bmatrix} \right )[/tex]

    However, still remained a doubt, is possible to express the rotational in terms of vector/matrix calculus, like above?
     
  2. jcsd
  3. Jan 14, 2014 #2
    OMG! I forgot that in english you do not speak "rotational" and yes "curl". My question is wrt curl ...
     
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