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linearOperator
 one year ago
Best ResponseYou've already chosen the best response.1Find the curl and divergence for the vector field: \(\overrightarrow{F}\) \(= \frac{ 4xz }{ y^2 } \overrightarrow{i}+(xz)\overrightarrow{j}7y^3 \overrightarrow{k}\)

linearOperator
 one year ago
Best ResponseYou've already chosen the best response.1can you help with question please

james342234
 one year ago
Best ResponseYou've already chosen the best response.0first send me a pic

linearOperator
 one year ago
Best ResponseYou've already chosen the best response.1you cant not see the vector field up there?

AravindG
 one year ago
Best ResponseYou've already chosen the best response.0@james342234 keep the talk related to math ..

TSwizzle
 one year ago
Best ResponseYou've already chosen the best response.2Omg James, stop! Leave the poor guy alone! Go away if you're not gonna help!

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1find the divergence by using formula ... what's so difficult?

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1http://upload.wikimedia.org/math/8/a/b/8abded95c326725e73cf446fdcd2f5f8.png

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1calculate those partial derivatives of those components

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1http://en.wikipedia.org/wiki/Divergence#Application_in_Cartesian_coordinates

linearOperator
 one year ago
Best ResponseYou've already chosen the best response.1ok i see thank youu!!

linearOperator
 one year ago
Best ResponseYou've already chosen the best response.1if it is zero it is said conservative right?

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1if the field is conservative ... the curl is zero. for divergence ... if the field is solenoidal (no source or sink)... then it is zero

linearOperator
 one year ago
Best ResponseYou've already chosen the best response.1great thank you!!! i am so greatful for helping me
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