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-MFra-21-Topaz II
21-Topaz II
March 8, 2021
Solved

Partial derivatives operator

  • March 8, 2021
  • 1 reply
  • 2690 views

Hi,

How can I modify the operator so that it works correctly and not as shown in the figure? In the image what I would like is marked in red:

symbolic.jpg

I used a letter linked to the sword symbol (but I could have used others) because when it is used as an argument of a function in a reference file, it is sure that, in the calling program (s) (in which I will be careful not to use that symbol, linked to the sword symbol), no conflicts arise.

Best answer by LucMeekes

LucMeekes_0-1615197698614.png

Success!

Luc

 

1 reply

L
LucMeekes23-Emerald IVAnswer
23-Emerald IV
March 8, 2021

LucMeekes_0-1615197698614.png

Success!

Luc

 

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    -MFra-21-Topaz IIAuthor
    21-Topaz II
    March 8, 2021

    I would need something more than previously said. For example, use those operators in the rotor calculation as in the figure:

    symbolic 1.jpg,

    is it possible?

    L
    LucMeekes23-Emerald IV
    23-Emerald IV
    March 8, 2021

    I'm afraid not. And I doubt if the operation, defined through the determinant, is mathematically correct.

    Anyway, if all items of the 3x3 matrix are just symbols, this is what you get:

    LucMeekes_0-1615237182162.png

    So it looks great, until you look closer. In the first term of i.x, the partial derivative to y is not applied to V.z, but simply multiplied with it.

    With the partial derivative 'operators' (in fact they're defined as functions) defined you're getting undefined items.

    LucMeekes_1-1615237356871.png

    Success!
    Luc

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