Efficiently Calculate Math Div Expressions with Partial Derivatives and Vectors

[cscott]

Is the following true?

<br /> \newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } }<br /> <br /> \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)

where

\vec{r} = (x, y, z)

 

[BrendanH]

no, it would just be the partial of r times f in each case

 

[cscott]

Like this?

<br /> \newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } }<br /> <br /> \nabla \cdot \vec{r}f(r) = \pd{(rf)}{x}{} + f\pd{(rf)}{y}{} + f\pd{(rf}{z}{}<br />

or just 3f?

 

[cscott]

My initial expression seems right to me now...

 

[Vid]

Is f a scalar field or a vector field?
Is it del.(r*f) or (del.r)(f)

 

[cscott]

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||

 

[Vid]

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.