Test the following identities of the vector analysis, for fields $F, G: mathbb{R}^3→ mathbb{R}^3$ and...
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Test the following identities of the vector analysis, for fields $F, G: mathbb{R}^3→ mathbb{R}^3$ and functions $f, g: mathbb{R}^3 → mathbb{R}$.
$text{a)}spacenabla cdot (nabla f times nabla g)=0.$
$text{b)}space nabla cdot (Ftimes G)=G cdot (nabla times F)-Fcdot (nabla times G).$
$text{c)}space text{div}(fF)=fmathrm{div}F+Fcdot nabla f.$
$text{d)}spacetext{rot}(fF)=ftext{rot}F+nabla f times F.$
I think I can try these properties by hand following the definitions but I notice that the work gets too long, my question is, can this be done easier with some properties instead of using the definitions? Thank you
calculus real-analysis multivariable-calculus vector-analysis
edited Nov 18 at 9:22
Fakemistake
1,635815
asked Nov 17 at 23:21
user424241
20819
add a comment |
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0
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Test the following identities of the vector analysis, for fields $F, G: mathbb{R}^3→ mathbb{R}^3$ and functions $f, g: mathbb{R}^3 → mathbb{R}$.
$text{a)}spacenabla cdot (nabla f times nabla g)=0.$
$text{b)}space nabla cdot (Ftimes G)=G cdot (nabla times F)-Fcdot (nabla times G).$
$text{c)}space text{div}(fF)=fmathrm{div}F+Fcdot nabla f.$
$text{d)}spacetext{rot}(fF)=ftext{rot}F+nabla f times F.$
I think I can try these properties by hand following the definitions but I notice that the work gets too long, my question is, can this be done easier with some properties instead of using the definitions? Thank you
calculus real-analysis multivariable-calculus vector-analysis
edited Nov 18 at 9:22
Fakemistake
1,635815
asked Nov 17 at 23:21
user424241
20819
2
Actually they themselves are very useful properties. I think for the most part you need to verify by direct calculation.
– Apocalypse
Nov 17 at 23:27
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Test the following identities of the vector analysis, for fields $F, G: mathbb{R}^3→ mathbb{R}^3$ and functions $f, g: mathbb{R}^3 → mathbb{R}$.
$text{a)}spacenabla cdot (nabla f times nabla g)=0.$
$text{b)}space nabla cdot (Ftimes G)=G cdot (nabla times F)-Fcdot (nabla times G).$
$text{c)}space text{div}(fF)=fmathrm{div}F+Fcdot nabla f.$
$text{d)}spacetext{rot}(fF)=ftext{rot}F+nabla f times F.$
I think I can try these properties by hand following the definitions but I notice that the work gets too long, my question is, can this be done easier with some properties instead of using the definitions? Thank you
calculus real-analysis multivariable-calculus vector-analysis
edited Nov 18 at 9:22
Fakemistake
1,635815
asked Nov 17 at 23:21
user424241
20819
Test the following identities of the vector analysis, for fields $F, G: mathbb{R}^3→ mathbb{R}^3$ and functions $f, g: mathbb{R}^3 → mathbb{R}$.
$text{a)}spacenabla cdot (nabla f times nabla g)=0.$
$text{b)}space nabla cdot (Ftimes G)=G cdot (nabla times F)-Fcdot (nabla times G).$
$text{c)}space text{div}(fF)=fmathrm{div}F+Fcdot nabla f.$
$text{d)}spacetext{rot}(fF)=ftext{rot}F+nabla f times F.$
I think I can try these properties by hand following the definitions but I notice that the work gets too long, my question is, can this be done easier with some properties instead of using the definitions? Thank you
calculus real-analysis multivariable-calculus vector-analysis
calculus real-analysis multivariable-calculus vector-analysis
edited Nov 18 at 9:22
Fakemistake
1,635815
asked Nov 17 at 23:21
user424241
20819
edited Nov 18 at 9:22
Fakemistake
1,635815
asked Nov 17 at 23:21
user424241
20819
edited Nov 18 at 9:22
Fakemistake
1,635815
edited Nov 18 at 9:22
Fakemistake
1,635815
edited Nov 18 at 9:22
Fakemistake
1,635815
1,635815
asked Nov 17 at 23:21
user424241
20819
asked Nov 17 at 23:21
user424241
20819
asked Nov 17 at 23:21
user424241
20819
20819
2
Actually they themselves are very useful properties. I think for the most part you need to verify by direct calculation.
– Apocalypse
Nov 17 at 23:27
add a comment |
2
Actually they themselves are very useful properties. I think for the most part you need to verify by direct calculation.
– Apocalypse
Nov 17 at 23:27
2
2
Actually they themselves are very useful properties. I think for the most part you need to verify by direct calculation.
– Apocalypse
Nov 17 at 23:27
Actually they themselves are very useful properties. I think for the most part you need to verify by direct calculation.
– Apocalypse
Nov 17 at 23:27
add a comment |
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Actually they themselves are very useful properties. I think for the most part you need to verify by direct calculation.
– Apocalypse
Nov 17 at 23:27