Help with multivariable calculus please.

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Help with multivariable calculus please.

Mathematics
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Being F a vector field with continous partial derivatives, If rot(F)=0 and Div(F)=0 is F constant?
I tried using Gauss and stokes theorem but i dont see anything
rot(F) = curl(F)?

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yep, sry xD here in spanish we call it rot
i think the solution should be find using those theorems
if the curl(F)=0 => F is a conservative vector field
yep, and if Div(F)=0 => F=curl(G) being G another vector field
it's been a while since i've dealt with this stuff so i'm a bit rusty. sorry
1) curl F = 0 is not the definitive test for a conservative field: http://mathinsight.org/path_dependent_zero_curl 2) i've had a look about for examples and \((2x,2y,−4z)\) has zero curl and divergence
hint: if we apply this vector identity: \[\Large \nabla \times \left( {\nabla \times {\mathbf{F}}} \right) = \left( {\nabla \cdot {\mathbf{F}}} \right)\nabla - {\nabla ^2}{\mathbf{F}}\] using your condition, we have: \[\Large {\mathbf{0}} = {\nabla ^2}{\mathbf{F}}\]
curl is tendancy to rotate and divergence is compressibility. if they're both 0, doesn't that just mean that an incompressible fluid is flowing in a way that it tends not to rotate?
@pgpilot326 +1 or its just not actually compressing
and does F is constant mean? like 1,2,3 ?
yep
well we have an example so that is not true
yep thank you very much guys , i really appreciate the help
or think gravity! zero curl, zero divergence.
we can show, that, from the condition: \[\Large {\mathbf{0}} = {\nabla ^2}{\mathbf{F}}\] the general field \( \large {\mathbf{F}}\) depends linearly on the cartesian coordinates \( \large x, \; y, \; z \)
yes and it doesnt have to be necessarily constant, thanks! :)
:)
@Michele_Laino interesting why did you set the triple product to zero ? and what does \(\nabla \times ( \nabla \times\vec A)\) actually mean if we already know that \( \nabla \times\vec A = 0\)
because that nails it :-)
since: \[\Large \nabla \times {\mathbf{F}} = {\mathbf{0}}\]
indeed. the curl is zero so can we use the bac cab rule for anything? like 2 x 0 = 3 x 0 2 = 3
yes! I think so, since the vector equation above, it is an identity
furthermore, also this condition: \[\Large \nabla \cdot {\mathbf{F}} = 0\] holds
aaahhh! i think i am getting you. so, ok, ....the gravitational field, inverse square but no curl no divergence.... i hate to think how you fit that into cartesian to get a linear.
I have developed, component by component the cndition: \[\Large {\mathbf{0}} = {\nabla ^2}{\mathbf{F}}\] using this other condition: \[\Large \nabla \times {\mathbf{F}} = {\mathbf{0}}\]
condition*
yep, i get that bit the triple product is zero because AxB is zero and ..... if div F is zero, the laplacian must also be zero so we have a linear in x,y,z i get the reasoning, sure. it's very good.
thanks!
you need a medal :-)
or 2
lol!
no worries! I'm happy so! thanks! @IrishBoy123
good!

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