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anonymous

  • 4 years ago

V = u i + v j + w k and ∇= ∂╱∂x i + ∂╱∂y j + ∂╱∂z k that: (V∙∇)V = (1╱2)∇(V∙V)−V X (∇XV)

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  1. amistre64
    • 4 years ago
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    this looks scary, what is it?

  2. anonymous
    • 4 years ago
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    i've been at this for 5 hours now

  3. anonymous
    • 4 years ago
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    i just have to prove the left side equals the right side

  4. anonymous
    • 4 years ago
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    u took calc 3 right?

  5. amistre64
    • 4 years ago
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    took, taking, taught meself, and all that stuff, yeah

  6. anonymous
    • 4 years ago
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    yea..figured since u helped me the other day you might be able to hand this

  7. amistre64
    • 4 years ago
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    can we clean it up some with a b c? V = ui + vj + wk ∇= ai + bj + ck ------------------ = au i +bv j +cw k

  8. anonymous
    • 4 years ago
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    so a b and c are the partials correct?

  9. amistre64
    • 4 years ago
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    yes (au +bv +cw) <u,v,k> is the left side then

  10. anonymous
    • 4 years ago
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    right

  11. anonymous
    • 4 years ago
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    so then it comes out to (au^2 + bv^2+ cw^2

  12. amistre64
    • 4 years ago
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    not quite, that dot is a scalar, not a vector so its applied to all the vector parts to stretch or shrink it to fit V.V = |V|^2 |V|^2 = u^2+v^2+w^2

  13. anonymous
    • 4 years ago
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    gotcha...distribute it

  14. amistre64
    • 4 years ago
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    edit:\[\frac{1}{2}*(V.V)=\frac{ u^2+v^2+w^2}{2}\]

  15. amistre64
    • 4 years ago
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    this is scaled to our gradient for that first part on the right

  16. anonymous
    • 4 years ago
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    isn't it 1/2 the gradient times V dot V?

  17. amistre64
    • 4 years ago
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    grad x V is: a b c u v w x = bw-vc y = -(aw-uc) z = av-bu and to cross that with V again we get: u v w bw-vc uc-aw av-bu x = v(av-bu) - w(uc-aw) y = -(u(av-bu)-w(bw-vc)) z = u(uc-aw) - v(bw-vc)

  18. amistre64
    • 4 years ago
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    yes, and V.V is just: u^2 + v^2 + w^2

  19. amistre64
    • 4 years ago
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    u v w u v w --------------- u^2 + v^2 + w^2

  20. amistre64
    • 4 years ago
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    i think i got all the parts ready :)

  21. amistre64
    • 4 years ago
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    \[(au+bv+cw)(u\bar i+v\bar j+w\bar k)=...\] \[...\frac{u^2+v^2+w^2}{2}(a\bar i+b\bar j+c\bar k)-((av^2-buv -wuc+aw^2)\bar i\]\[+(-uav+bu^2+bw^2-wvc)\bar j+ (u^2c-auw - vbw + v^2c)\bar k\] this look about right so far? :)

  22. anonymous
    • 4 years ago
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    its hard to say...b/c i been doing it partials and u v and w

  23. anonymous
    • 4 years ago
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    but...the setup looks correct from what i can decifer

  24. amistre64
    • 4 years ago
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    a b c just stand in for partials to make it cleam up

  25. anonymous
    • 4 years ago
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    then yes...it appears to be correct...it's only the right hand side of the equation right?

  26. amistre64
    • 4 years ago
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    the first line is the lhs; the rest is a bit lengthy and had to split it up some

  27. amistre64
    • 4 years ago
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    if we multiply it all by 2 we get rid of the fraction

  28. anonymous
    • 4 years ago
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    oh okay..yea

  29. amistre64
    • 4 years ago
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    \[<(2au^2+2buv+2cuw),(2auv+2bv^2+2cwv),(2auw+2bvw+2cw^2)>\] is our LHS written out as its vector component stuff

  30. amistre64
    • 4 years ago
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    lets try to work up the RHS to match it

  31. anonymous
    • 4 years ago
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    alright...i'm gonna try and re copy this for my homework in long hand without the a b and c lol

  32. amistre64
    • 4 years ago
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    \[<au^2,bv^2,cw^2>-etc\] \[<(au^2-2av^2+2buv +2wuc-2aw^2),\]\[(bv^2+2uav-2bu^2-2bw^2+2wvc),\]\[(cw^2-2u^2c+2auw +2vbw -2v^2c)>\] with any luck is how the RHS is looking after combining it all together

  33. anonymous
    • 4 years ago
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    what is wuc?

  34. amistre64
    • 4 years ago
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    as long as components wise we zero out, it should be proofed then au^2- 2au2 -2av^2+2buv-2buv +2wuc - 2cuw-2aw^2 = 0 wu is V parts, and c is a partial

  35. anonymous
    • 4 years ago
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    nevermind lol....partial of w u over z

  36. amistre64
    • 4 years ago
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    -au^2 -2av^2 -2aw^2 = 0 might need to rechk the math in all this, but this is the basic concept that im thinking of

  37. anonymous
    • 4 years ago
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    yea i'm getting lost

  38. anonymous
    • 4 years ago
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    trying to type it out with the partials

  39. amistre64
    • 4 years ago
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    V = <u,v,w> g = <a,b,c> (V∙g)V = (1╱2)g(V∙V)−V X (gXV) (V∙g) = au+bv+cw (au+bv+cw)V = < au^2+bvu+cwu, auv+bv^2+cwv, auw+bvw+cw^2> is our LHS ............................................ (1╱2)g(V∙V) (V∙V) = v^2+u^2+w^2 (1╱2)(v^2+u^2+w^2) = v^2/2 + u^2/2 + w^2/2 (1╱2)(V∙V)g = av^2/2 + au^2/2 + aw^2/2 , bv^2/2 + bu^2/2 + bw^2/2 , cv^2/2 +cu^2/2 + cw^2/2 .......................................................... − V X (gXV) gxV: a b c v u w ----- x = bw-cu y = cv-aw z = au-bv -V x that is bw-cu cv-aw au-bv -v -u -w --------------------- x = -wcv+aw^2 + au^2 -bvu y = -vau+bv^2 +bw^2-cuw z = -ubw+cu^2 + cv^2-awv ugh lol -------------------------- to get rid of the /2 *2 it all < 2au^2+2bvu+2cwu, 2auv+2bv^2+2cwv, 2auw+2bvw+2cw^2> = = <av^2 + au^2 + aw^2 , bv^2 + bu^2 + bw^2 , cv^2 +cu^2 + cw^2> + <-2wcv+2aw^2 +2au^2 -2bvu, -2vau+2bv^2 +2bw^2 -2cuw, -2ubw +2cu^2 + 2cv^2 -2awv >

  40. amistre64
    • 4 years ago
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    now compare component parts: 2au^2+2bvu+2cwu = av^2 + au^2 + aw^2-2wcv+2aw^2 +2au^2 -2bvu ? au^2 +2cwu = av^2 + aw^2-2wcv+2aw^2 +2au^2 ? u(au +2cw) = a(v^2 + w^2+2w^2) -2wcv +2au^2 ? yeah, where do you get these problems at anyhoos?

  41. anonymous
    • 4 years ago
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    lol...meteorology major

  42. anonymous
    • 4 years ago
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    sorry i'm trying to copy it all down lol

  43. anonymous
    • 4 years ago
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    its hard to see the difference between u and v when their close tother

  44. amistre64
    • 4 years ago
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    the wolf gives me this for VxgxV lets see if its does good with the rest too \[<-a u^2+b u v+c v w-a w^2, a u v-b v^2+c u w-b w^2, -c u^2-c v^2+b u w+a v w>\]

  45. anonymous
    • 4 years ago
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    alright i'm all caught up writing lol

  46. anonymous
    • 4 years ago
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    yea...from what i got..this does not equal each other lol

  47. anonymous
    • 4 years ago
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    3 terms per component on the left hand side to 5 terms per component in the rhs

  48. amistre64
    • 4 years ago
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    \[<\frac a2u^2+\frac a2v^2+\frac a2w^2 , \frac b2u^2+\frac b2v^2+\frac b2w^2, \frac c2u^2+\frac c2v^2+\frac c2w^2>\]\[<a u^2-b u v-c v w+a w^2, -a u v+b v^2-c u w+b w^2, c u^2+c v^2-b u w-a v w>\] \[av^2+buv+cwv=( \frac{au^2+ av^2+ aw^2+2au^2-2b u v-2c v w+2a w^2}{2})\] \[2av^2+2buv+2cwv= au^2+ av^2+ aw^2+2au^2-2b u v-2c v w+2a w^2\] \[av^2+4buv+4cwv= au^2+ aw^2+2au^2+2a w^2\] yeah, is it spose to proof correct? or is this just some sort of manuvuer on their part?

  49. anonymous
    • 4 years ago
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    the left side should equal the right side exactly

  50. amistre64
    • 4 years ago
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    try with some random easy numbers maybe, or functions lol just to test

  51. amistre64
    • 4 years ago
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    V = x,x^2,x^3 g = 1,2x,3x^2 at x=1

  52. amistre64
    • 4 years ago
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    V = 1,1,1 g = 1,2,3

  53. amistre64
    • 4 years ago
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    (v.g)v = (1+2+3)<1,1,1> = <6,6,6> lol, nice 1/2 g = (1/2, 1 , 3/2)(3) = <3/2, 3 , 9/2> <6,6,6> = <3/2, 3 , 9/2> - stuff <12,12,12> = <3, 6 , 9> - 2*stuff -3 -6 -9 ---------------- <9,6,3> = -2*stuff <-18,-12,-6> = stuff

  54. amistre64
    • 4 years ago
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    stuff = vxgxv 1,1,1 1,2,3 ----- x=1 y=-2 z=1 1,-2,1 1, 1, 1 ------- x=-3 y= 0 z= 3 <-18,-9,-3> = <-3,0,3> ??

  55. anonymous
    • 4 years ago
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    so the answer above is correct as it stands?

  56. amistre64
    • 4 years ago
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    when i plug in some random function for V and grad it to g; this is what I get; so I dont see how it is spose to be true as a proof according to what youve listed

  57. anonymous
    • 4 years ago
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    it would just be weird b/c i have 5 of these to prove...and all 4 have matched perfectly

  58. amistre64
    • 4 years ago
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    <9,6,3> = -2*stuff /-2 -------- <-9/2, -3, -3/2> = stuff but still; thats not a match

  59. anonymous
    • 4 years ago
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    alright...i gotta work out some stuff with wolf then lol

  60. amistre64
    • 4 years ago
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    good luck :)

  61. anonymous
    • 4 years ago
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    u still around?

  62. amistre64
    • 4 years ago
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    im around

  63. anonymous
    • 4 years ago
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    i just emailed my professor..this is what i got back

  64. anonymous
    • 4 years ago
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    yep. they are supposed to match. Come by my office tomorrow with what you have done, and I can look at see where you are going wrong. Every year students think the 1/2 is an error, but it is correct......usually people go wrong when taking the derivative of u^2.

  65. anonymous
    • 4 years ago
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    now...i'm really confused b/c i don't see where i'm taking any derivatives

  66. amistre64
    • 4 years ago
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    hmmm, ...

  67. amistre64
    • 4 years ago
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    are v, u, amd w function of x?

  68. amistre64
    • 4 years ago
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    gradV = V' right?

  69. anonymous
    • 4 years ago
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    i mean...exactly the way the question is written...is what i have

  70. anonymous
    • 4 years ago
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    del operator?

  71. anonymous
    • 4 years ago
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    is what she calls it

  72. amistre64
    • 4 years ago
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    so grad is the operator on V and not a specific vector ....

  73. amistre64
    • 4 years ago
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    del(V.V) is not what I had in mind at the begining

  74. amistre64
    • 4 years ago
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    but then what is V.del? del of what?

  75. amistre64
    • 4 years ago
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    or is V.del, just V' ?

  76. amistre64
    • 4 years ago
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    \[\frac{d}{dx}u+\frac{d}{dx}v+\frac{d}{dx}w=<u',v',w'>\]

  77. amistre64
    • 4 years ago
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    well, with the right parts that is :)

  78. anonymous
    • 4 years ago
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    well del is just = to partial/partial x etc.

  79. anonymous
    • 4 years ago
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    (V∙∇)V = (1╱2)∇(V∙V)−V X (∇XV) let's do it term by term....... Letting V=(u(x,y,z),v(x,y,z),w(x,y,z)) and start with the term 12∇(V⋅V). V⋅V=u2+v2+w2 and this is a scalar quantity. Then ∇(u2+v2+w2) is a vector quantity, and as above, is equal to: [∂∂x(u2+v2+w2)i+∂∂y(u2+v2+w2)j+∂∂z(u2+v2+w2)k] in calculating each component of this vector, we will use subscripts to indicate partial differentiation wrt the subscripted variable. We have [(2uux+2vvx+2wwx)i+(2uuy+2vvy+2wwy)j+(2uuz+2vvz+2wwz)k] So multiplying by the 1/2 will remove all these 2s meaning: 12∇(V⋅V)=[(uux+vvx+wwx)i+(uuy+vvy+wwy)j+(uuz+vvz+wwz)k] Next we want to deal with the term: V×(∇×V) First we calculate ∇×V=(wy−vz)i+(uz−wx)j+(vx−uy)k Next we calculate V×[(wy−vz)i+(uz−wx)j+(vx−uy)k] =[v(vx−uy)−w(uz−wx)]i+[w(wy−vz)−u(vx−uy)]j +[u(uz−wx)−v(wy−vz)]k and this equals V×(∇×V). Now we do the subtraction. In the i component we get: [uux+vvx+wwx−v(vx−uy)+w(uz−wx)]i which is equal to [uux+vuy+wuz]i Looking good? Let's do the j and k component now: [uuy+vvy+wwy−w(wy−vz)+u(vx−uy)]j which is equal to [uvx+vvy+wvz]j and at long last, the k component... [uuz+vvz+wwz−u(uz−wx)+v(wy−vz)]k which is equal to [uwx+vwy+wwz]k THEREFORE we have shown that the right hand side reduces to the vector: [uux+vuy+wuz]i+[uvx+vvy+wvz]j+[uwx+vwy+wwz]k Now let's stop here and think for a bit. V⋅∇=u∂∂x+v∂∂y+w∂∂z and this is an operator which acts on vectors to the right of it, if you like, so we get: (V⋅∇)V=[u∂∂xu+v∂∂yu+w∂∂zu]i+[u∂∂xv+v∂∂yv+w∂∂zv]j +[u∂∂xw+v∂∂yw+w∂∂zw]k Notice that this is exactly what we just proved, therefore the identity has been verified.

  80. amistre64
    • 4 years ago
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    \[V = u i + v j + w k \] \[∇= \frac{∂}{∂x} i + \frac{∂}{∂y} j +\frac{∂}{∂z} k\] \[(V∙∇)= \frac{∂u}{∂x} + \frac{∂v}{∂y} +\frac{∂w}{∂z}\] \[(V∙∇)V= \left(\frac{∂u}{∂x}u + \frac{∂v}{∂y}u +\frac{∂w}{∂z}u\right)i+\left(\frac{∂u}{∂x}v + \frac{∂v}{∂y}v +\frac{∂w}{∂z}v\right)j+\left(\frac{∂u}{∂x}w + \frac{∂v}{∂y}w +\frac{∂w}{∂z}w\right)k\] ----------------------------------- \[∇(V∙V)= \left(\frac{∂}{∂x}u^2i + \frac{∂}{∂y}v^2j +\frac{∂}{∂z}w^2k\right)\] \[∇(V∙V)= \left(\frac{∂u}{∂x}2ui + \frac{∂v}{∂y}2vj +\frac{∂w}{∂z}2wk\right)\] \[∇(V∙V)/2= \left(\frac{∂u}{∂x}ui + \frac{∂v}{∂y}vj +\frac{∂w}{∂z}wk\right)\] ------------------------------------ so far −V X (∇XV)

  81. anonymous
    • 4 years ago
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    http://openstudy.com/users/sonofa_nh#/updates/4f258d94e4b0a2a9c266fb07

  82. amistre64
    • 4 years ago
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    \[-VxV=<0,0,0>\]so i have to wonder about that part :) \[(∇xV)=\frac{∂}{∂y}w-\frac{∂}{∂z}v,\ \frac{∂}{∂z}w-\frac{∂}{∂x}u,\ \frac{∂}{∂x}u - \frac{∂}{∂y} v \] \[-Vx(∇xV)=(\frac{∂w}{∂y}-\frac{∂v}{∂z},,\ \frac{∂u}{∂x} - \frac{∂v}{∂y} \] \[\ x=\frac{∂}{∂x}(wu)+\ \frac{∂}{∂x}(uv) - \frac{∂}{∂y}(v^2)-\frac{∂}{∂z}(w^2)\] \[\ x=\frac{∂u}{∂x}w+\frac{∂w}{∂x}u+\ \frac{∂u}{∂x}v+\ \frac{∂v}{∂x}u - \frac{∂v}{∂y}2v-\frac{∂w}{∂z}2w\] if this is right, its a killer trying to keep track of it all

  83. anonymous
    • 4 years ago
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    yea...u see the link i just sent you.....a guy figured it all out...i can't make sense of it though

  84. amistre64
    • 4 years ago
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    yeah, james is smart that way

  85. anonymous
    • 4 years ago
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    well the other guy if u scroll down solved it

  86. anonymous
    • 4 years ago
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    but i can' t understand it

  87. amistre64
    • 4 years ago
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    ok, so we seem to be good up to that -v x del x v part; right?

  88. amistre64
    • 4 years ago
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    do you understand the notations?\[du/dx=u_x\]

  89. amistre64
    • 4 years ago
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    if so: \[del(x)V=\begin{pmatrix}x&y&z\\d_x&d_y&d_z\\u&v&w\end{pmatrix}\] \[x=(w_y-v_z)\] \[y=(u_z-w_x)\] \[z=(v_x-u_y)\] Vx(that stuff): \begin{pmatrix}x&y&z\\u&v&w\\(w_y-v_z)&(u_z-w_x)&(v_x-u_y) \end{pmatrix} \[x=(vv_x-vu_y-wu_z+ww_x)\] \[y=(uu_y-uv_x+ww_y-wv_z)\] \[z=(vw_y-vv_z-uu_z+uw_x)\]

  90. amistre64
    • 4 years ago
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    i can make it out, i just cant seem to get a good explanation on it tho

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spraguer (Moderator)
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