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EmmaTassone

  • one year ago

Help with multivariable calculus please.

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  1. EmmaTassone
    • one year ago
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    Being F a vector field with continous partial derivatives, If rot(F)=0 and Div(F)=0 is F constant?

  2. EmmaTassone
    • one year ago
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    I tried using Gauss and stokes theorem but i dont see anything

  3. anonymous
    • one year ago
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    rot(F) = curl(F)?

  4. EmmaTassone
    • one year ago
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    yep, sry xD here in spanish we call it rot

  5. EmmaTassone
    • one year ago
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    i think the solution should be find using those theorems

  6. anonymous
    • one year ago
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    if the curl(F)=0 => F is a conservative vector field

  7. EmmaTassone
    • one year ago
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    yep, and if Div(F)=0 => F=curl(G) being G another vector field

  8. anonymous
    • one year ago
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    it's been a while since i've dealt with this stuff so i'm a bit rusty. sorry

  9. IrishBoy123
    • one year ago
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    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

  10. Michele_Laino
    • one year ago
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    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: \[\Large {\mathbf{0}} = {\nabla ^2}{\mathbf{F}}\]

  11. anonymous
    • one year ago
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    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?

  12. IrishBoy123
    • one year ago
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    @pgpilot326 +1 or its just not actually compressing

  13. IrishBoy123
    • one year ago
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    and does F is constant mean? like 1,2,3 ?

  14. EmmaTassone
    • one year ago
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    yep

  15. IrishBoy123
    • one year ago
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    well we have an example so that is not true

  16. EmmaTassone
    • one year ago
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    yep thank you very much guys , i really appreciate the help

  17. IrishBoy123
    • one year ago
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    or think gravity! zero curl, zero divergence.

  18. Michele_Laino
    • one year ago
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    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 \( \large x, \; y, \; z \)

  19. EmmaTassone
    • one year ago
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    yes and it doesnt have to be necessarily constant, thanks! :)

  20. Michele_Laino
    • one year ago
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    :)

  21. IrishBoy123
    • one year ago
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    @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 \( \nabla \times\vec A = 0\)

  22. IrishBoy123
    • one year ago
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    because that nails it :-)

  23. Michele_Laino
    • one year ago
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    since: \[\Large \nabla \times {\mathbf{F}} = {\mathbf{0}}\]

  24. IrishBoy123
    • one year ago
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    indeed. the curl is zero so can we use the bac cab rule for anything? like 2 x 0 = 3 x 0 2 = 3

  25. Michele_Laino
    • one year ago
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    yes! I think so, since the vector equation above, it is an identity

  26. Michele_Laino
    • one year ago
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    furthermore, also this condition: \[\Large \nabla \cdot {\mathbf{F}} = 0\] holds

  27. IrishBoy123
    • one year ago
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    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.

  28. Michele_Laino
    • one year ago
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    I have developed, component by component the cndition: \[\Large {\mathbf{0}} = {\nabla ^2}{\mathbf{F}}\] using this other condition: \[\Large \nabla \times {\mathbf{F}} = {\mathbf{0}}\]

  29. Michele_Laino
    • one year ago
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    condition*

  30. IrishBoy123
    • one year ago
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    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.

  31. Michele_Laino
    • one year ago
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    thanks!

  32. IrishBoy123
    • one year ago
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    you need a medal :-)

  33. IrishBoy123
    • one year ago
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    or 2

  34. Michele_Laino
    • one year ago
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    lol!

  35. Michele_Laino
    • one year ago
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    no worries! I'm happy so! thanks! @IrishBoy123

  36. IrishBoy123
    • one year ago
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    good!

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