anonymous
  • anonymous
Come on Calculus Friends!! If you are given curl i.e. (del (cross) F) what is the method to figure out what F is, when (del (cross) F) is a vector quantity i.e. (#, #, #)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
there isn't an easy way to do it
anonymous
  • anonymous
you've got yourself a lovely set of 3 partial differential equations. \[\frac{\delta u_y}{\delta z}-\frac{\delta u_z}{\delta y}=f(x,y,z) \hat{x}\] etc.
anonymous
  • anonymous
can I show you the actual question and get your opinion on what to do from there?

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anonymous
  • anonymous
yeah, that'd be helpful, cause in general there's not really a set method of doing it that I'm aware of
anonymous
  • anonymous
Let F and G be vector fields such that (del) X F(0) = (-6.914, 5.159, 4.502), G(0) = (-8.196, -1.838, -2.2). Find the divergence of F X G at 0.
anonymous
  • anonymous
as far as i know the identity is shown as : (del)∗(FXG)=((del)crossF)dotG)−Fdot((del)crossG)
anonymous
  • anonymous
ah, okay, this is a bit of a trick, the divergence of a curl is always zero.
anonymous
  • anonymous
Ha! problem solved
anonymous
  • anonymous
oh wait, nvm, that wasnt in the question...
anonymous
  • anonymous
well the second part it kind of is no?
anonymous
  • anonymous
no, its the dot product with a curl, its different.
anonymous
  • anonymous
hmm..
anonymous
  • anonymous
so you are saying that F(dot)[del(cross)G) is not.. a divergence of a curl?
anonymous
  • anonymous
I guess, yeah that wouldn't make much sense, because then why would they have that identity in the first place
anonymous
  • anonymous
.. if it was just zero
anonymous
  • anonymous
yeah, g(0) is all constants though, shouldnt the curl of it be zero?
anonymous
  • anonymous
so all you'll have is the dot product of the two? I could be wrong, but from what I see you're definitly not given enough information to reverse engineer the field for f out of what you have
anonymous
  • anonymous
See, those are the types of things i'm still trying to grasp.. aha. it would be nice make things much easier
anonymous
  • anonymous
okay then that must be the best bet, I have possible answers, so i'll try and if I get it I'll let you know either way. thank you for your help!
anonymous
  • anonymous
k, best of luck

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