Turn on suggestions

Auto-suggest helps you quickly narrow down your search results by suggesting possible matches as you type.

Showing results for

Auto-suggest helps you quickly narrow down your search results by suggesting possible matches as you type.

Showing results for

- Subscribe to RSS Feed
- Mark Topic as New
- Mark Topic as Read
- Float this Topic for Current User
- Bookmark
- Subscribe
- Mute
- Printer Friendly Page

Mar 08, 2021
04:25 AM

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Notify Moderator

Mar 08, 2021
04:25 AM

Partial derivatives operator

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:

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.

Solved! Go to Solution.

Labels:

1 ACCEPTED SOLUTION

Accepted Solutions

Mar 08, 2021
05:03 AM

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Notify Moderator

5 REPLIES 5

Mar 08, 2021
05:03 AM

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Notify Moderator

Mar 08, 2021
08:35 AM

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Notify Moderator

Mar 08, 2021
08:35 AM

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

,

is it possible?

Mar 08, 2021
04:05 PM

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Notify Moderator

Mar 08, 2021
04:05 PM

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:

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.

Success!

Luc

Mar 09, 2021
02:41 AM

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Notify Moderator

Mar 09, 2021
02:41 AM

in fact that is a mnemonic rule for calculating the rotor. It is also given by the vector product between the vector of the operators and the given vector.

which also makes no sense.

Obviously, your displayed result does not make sense as well.

Mar 08, 2021
07:14 PM

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Notify Moderator

Mar 08, 2021
07:14 PM

Are you aware that the del symbol is the functional gradient operator? In Mathcad 15 (not Prime) you get it with cntrl+shift+G, you can subscript which variables you want to take the derivative of. There were a few old posts . ..