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Math Div expression help

  • Thread starter cscott
  • Start date
  • #1
782
1
Is the following true?

[tex]
\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } }

\nabla \cdot \vec{r}f(r) = \left ( \pd{f}{r}{} \pd{r}{x}{} \cdot x + f \right) + \left ( \pd{f}{r}{} \pd{r}{y}{} \cdot y + f \right) + \left( \pd{f}{r}{} \pd{r}{z}{} \cdot z + f \right)[/tex]

where

[tex]\vec{r} = (x, y, z)[/tex]
 
Last edited:

Answers and Replies

  • #2
63
0
no, it would just be the partial of r times f in each case
 
  • #3
782
1
Like this?

[tex]
\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } }

\nabla \cdot \vec{r}f(r) = \pd{(rf)}{x}{} + f\pd{(rf)}{y}{} + f\pd{(rf}{z}{}
[/tex]

or just 3f?
 
Last edited:
  • #4
782
1
My initial expression seems right to me now...
 
  • #5
Vid
401
0
Is f a scalar field or a vector field?
Is it del.(r*f) or (del.r)(f)
 
  • #6
782
1
I wrote it out just how they gave it to me but I assumed del.(rf) based on the question they ask after it.

f is scalar field, f(r), r = ||r||
 
  • #7
Vid
401
0
There's a type of product rule for a scalar field times a vector field.

del.(sF) = (del.F)*s + grad(s).F
where . is the vector dot product.
 

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