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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)
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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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1this looks scary, what is it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i've been at this for 5 hours now

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i just have to prove the left side equals the right side

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1took, taking, taught meself, and all that stuff, yeah

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea..figured since u helped me the other day you might be able to hand this

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1can we clean it up some with a b c? V = ui + vj + wk ∇= ai + bj + ck  = au i +bv j +cw k

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so a b and c are the partials correct?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1yes (au +bv +cw) <u,v,k> is the left side then

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so then it comes out to (au^2 + bv^2+ cw^2

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1not 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0gotcha...distribute it

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1edit:\[\frac{1}{2}*(V.V)=\frac{ u^2+v^2+w^2}{2}\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1this is scaled to our gradient for that first part on the right

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0isn't it 1/2 the gradient times V dot V?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1grad x V is: a b c u v w x = bwvc y = (awuc) z = avbu and to cross that with V again we get: u v w bwvc ucaw avbu x = v(avbu)  w(ucaw) y = (u(avbu)w(bwvc)) z = u(ucaw)  v(bwvc)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1yes, and V.V is just: u^2 + v^2 + w^2

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1u v w u v w  u^2 + v^2 + w^2

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i think i got all the parts ready :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[(au+bv+cw)(u\bar i+v\bar j+w\bar k)=...\] \[...\frac{u^2+v^2+w^2}{2}(a\bar i+b\bar j+c\bar k)((av^2buv wuc+aw^2)\bar i\]\[+(uav+bu^2+bw^2wvc)\bar j+ (u^2cauw  vbw + v^2c)\bar k\] this look about right so far? :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its hard to say...b/c i been doing it partials and u v and w

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but...the setup looks correct from what i can decifer

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1a b c just stand in for partials to make it cleam up

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then yes...it appears to be correct...it's only the right hand side of the equation right?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the first line is the lhs; the rest is a bit lengthy and had to split it up some

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1if we multiply it all by 2 we get rid of the fraction

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[<(2au^2+2buv+2cuw),(2auv+2bv^2+2cwv),(2auw+2bvw+2cw^2)>\] is our LHS written out as its vector component stuff

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1lets try to work up the RHS to match it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0alright...i'm gonna try and re copy this for my homework in long hand without the a b and c lol

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[<au^2,bv^2,cw^2>etc\] \[<(au^22av^2+2buv +2wuc2aw^2),\]\[(bv^2+2uav2bu^22bw^2+2wvc),\]\[(cw^22u^2c+2auw +2vbw 2v^2c)>\] with any luck is how the RHS is looking after combining it all together

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1as long as components wise we zero out, it should be proofed then au^2 2au2 2av^2+2buv2buv +2wuc  2cuw2aw^2 = 0 wu is V parts, and c is a partial

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0nevermind lol....partial of w u over z

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1au^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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0trying to type it out with the partials

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1V = <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 = bwcu y = cvaw z = aubv V x that is bwcu cvaw aubv v u w  x = wcv+aw^2 + au^2 bvu y = vau+bv^2 +bw^2cuw z = ubw+cu^2 + cv^2awv 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 >

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1now compare component parts: 2au^2+2bvu+2cwu = av^2 + au^2 + aw^22wcv+2aw^2 +2au^2 2bvu ? au^2 +2cwu = av^2 + aw^22wcv+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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol...meteorology major

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry i'm trying to copy it all down lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its hard to see the difference between u and v when their close tother

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the wolf gives me this for VxgxV lets see if its does good with the rest too \[<a u^2+b u v+c v wa w^2, a u vb v^2+c u wb w^2, c u^2c v^2+b u w+a v w>\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0alright i'm all caught up writing lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea...from what i got..this does not equal each other lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.03 terms per component on the left hand side to 5 terms per component in the rhs

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[<\frac a2u^2+\frac a2v^2+\frac a2w^2 , \frac b2u^2+\frac b2v^2+\frac b2w^2, \frac c2u^2+\frac c2v^2+\frac c2w^2>\]\[<a u^2b u vc v w+a w^2, a u v+b v^2c u w+b w^2, c u^2+c v^2b u wa v w>\] \[av^2+buv+cwv=( \frac{au^2+ av^2+ aw^2+2au^22b u v2c v w+2a w^2}{2})\] \[2av^2+2buv+2cwv= au^2+ av^2+ aw^2+2au^22b u v2c v w+2a w^2\] \[av^2+4buv+4cwv= au^2+ aw^2+2au^2+2a w^2\] yeah, is it spose to proof correct? or is this just some sort of manuvuer on their part?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the left side should equal the right side exactly

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1try with some random easy numbers maybe, or functions lol just to test

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1V = x,x^2,x^3 g = 1,2x,3x^2 at x=1

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1(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

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1stuff = 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> ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so the answer above is correct as it stands?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1when 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it would just be weird b/c i have 5 of these to prove...and all 4 have matched perfectly

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1<9,6,3> = 2*stuff /2  <9/2, 3, 3/2> = stuff but still; thats not a match

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0alright...i gotta work out some stuff with wolf then lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i just emailed my professor..this is what i got back

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yep. 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now...i'm really confused b/c i don't see where i'm taking any derivatives

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1are v, u, amd w function of x?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i mean...exactly the way the question is written...is what i have

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1so grad is the operator on V and not a specific vector ....

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1del(V.V) is not what I had in mind at the begining

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1but then what is V.del? del of what?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1or is V.del, just V' ?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[\frac{d}{dx}u+\frac{d}{dx}v+\frac{d}{dx}w=<u',v',w'>\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1well, with the right parts that is :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well del is just = to partial/partial x etc.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(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.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[V = u i + v j + w k \] \[∇= \frac{∂}{∂x} i + \frac{∂}{∂y} j +\frac{∂}{∂z} k\] \[(V∙∇)= \frac{∂u}{∂x} + \frac{∂v}{∂y} +\frac{∂w}{∂z}\] \[(V∙∇)V= \left(\frac{∂u}{∂x}u + \frac{∂v}{∂y}u +\frac{∂w}{∂z}u\right)i+\left(\frac{∂u}{∂x}v + \frac{∂v}{∂y}v +\frac{∂w}{∂z}v\right)j+\left(\frac{∂u}{∂x}w + \frac{∂v}{∂y}w +\frac{∂w}{∂z}w\right)k\]  \[∇(V∙V)= \left(\frac{∂}{∂x}u^2i + \frac{∂}{∂y}v^2j +\frac{∂}{∂z}w^2k\right)\] \[∇(V∙V)= \left(\frac{∂u}{∂x}2ui + \frac{∂v}{∂y}2vj +\frac{∂w}{∂z}2wk\right)\] \[∇(V∙V)/2= \left(\frac{∂u}{∂x}ui + \frac{∂v}{∂y}vj +\frac{∂w}{∂z}wk\right)\]  so far −V X (∇XV)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/users/sonofa_nh#/updates/4f258d94e4b0a2a9c266fb07

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[VxV=<0,0,0>\]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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea...u see the link i just sent you.....a guy figured it all out...i can't make sense of it though

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1yeah, james is smart that way

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well the other guy if u scroll down solved it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but i can' t understand it

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1ok, so we seem to be good up to that v x del x v part; right?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1do you understand the notations?\[du/dx=u_x\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1if so: \[del(x)V=\begin{pmatrix}x&y&z\\d_x&d_y&d_z\\u&v&w\end{pmatrix}\] \[x=(w_yv_z)\] \[y=(u_zw_x)\] \[z=(v_xu_y)\] Vx(that stuff): \begin{pmatrix}x&y&z\\u&v&w\\(w_yv_z)&(u_zw_x)&(v_xu_y) \end{pmatrix} \[x=(vv_xvu_ywu_z+ww_x)\] \[y=(uu_yuv_x+ww_ywv_z)\] \[z=(vw_yvv_zuu_z+uw_x)\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i can make it out, i just cant seem to get a good explanation on it tho
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