anonymous
  • anonymous
V = u i + v j + w k and ∇= ∂╱∂x i + ∂╱∂y j + ∂╱∂z k that: (V∙∇)V = (1╱2)∇(V∙V)−V X (∇XV)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
this looks scary, what is it?
anonymous
  • anonymous
i've been at this for 5 hours now
anonymous
  • anonymous
i just have to prove the left side equals the right side

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
u took calc 3 right?
amistre64
  • amistre64
took, taking, taught meself, and all that stuff, yeah
anonymous
  • anonymous
yea..figured since u helped me the other day you might be able to hand this
amistre64
  • amistre64
can we clean it up some with a b c? V = ui + vj + wk ∇= ai + bj + ck ------------------ = au i +bv j +cw k
anonymous
  • anonymous
so a b and c are the partials correct?
amistre64
  • amistre64
yes (au +bv +cw) is the left side then
anonymous
  • anonymous
right
anonymous
  • anonymous
so then it comes out to (au^2 + bv^2+ cw^2
amistre64
  • amistre64
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
anonymous
  • anonymous
gotcha...distribute it
amistre64
  • amistre64
edit:\[\frac{1}{2}*(V.V)=\frac{ u^2+v^2+w^2}{2}\]
amistre64
  • amistre64
this is scaled to our gradient for that first part on the right
anonymous
  • anonymous
isn't it 1/2 the gradient times V dot V?
amistre64
  • amistre64
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)
amistre64
  • amistre64
yes, and V.V is just: u^2 + v^2 + w^2
amistre64
  • amistre64
u v w u v w --------------- u^2 + v^2 + w^2
amistre64
  • amistre64
i think i got all the parts ready :)
amistre64
  • amistre64
\[(au+bv+cw)(u\bar i+v\bar j+w\bar k)=...\] \[...\frac{u^2+v^2+w^2}{2}(a\bar i+b\bar j+c\bar k)-((av^2-buv -wuc+aw^2)\bar i\]\[+(-uav+bu^2+bw^2-wvc)\bar j+ (u^2c-auw - vbw + v^2c)\bar k\] this look about right so far? :)
anonymous
  • anonymous
its hard to say...b/c i been doing it partials and u v and w
anonymous
  • anonymous
but...the setup looks correct from what i can decifer
amistre64
  • amistre64
a b c just stand in for partials to make it cleam up
anonymous
  • anonymous
then yes...it appears to be correct...it's only the right hand side of the equation right?
amistre64
  • amistre64
the first line is the lhs; the rest is a bit lengthy and had to split it up some
amistre64
  • amistre64
if we multiply it all by 2 we get rid of the fraction
anonymous
  • anonymous
oh okay..yea
amistre64
  • amistre64
\[<(2au^2+2buv+2cuw),(2auv+2bv^2+2cwv),(2auw+2bvw+2cw^2)>\] is our LHS written out as its vector component stuff
amistre64
  • amistre64
lets try to work up the RHS to match it
anonymous
  • anonymous
alright...i'm gonna try and re copy this for my homework in long hand without the a b and c lol
amistre64
  • amistre64
\[-etc\] \[<(au^2-2av^2+2buv +2wuc-2aw^2),\]\[(bv^2+2uav-2bu^2-2bw^2+2wvc),\]\[(cw^2-2u^2c+2auw +2vbw -2v^2c)>\] with any luck is how the RHS is looking after combining it all together
anonymous
  • anonymous
what is wuc?
amistre64
  • amistre64
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
anonymous
  • anonymous
nevermind lol....partial of w u over z
amistre64
  • amistre64
-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
anonymous
  • anonymous
yea i'm getting lost
anonymous
  • anonymous
trying to type it out with the partials
amistre64
  • amistre64
V = g = (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> = = + <-2wcv+2aw^2 +2au^2 -2bvu, -2vau+2bv^2 +2bw^2 -2cuw, -2ubw +2cu^2 + 2cv^2 -2awv >
amistre64
  • amistre64
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?
anonymous
  • anonymous
lol...meteorology major
anonymous
  • anonymous
sorry i'm trying to copy it all down lol
anonymous
  • anonymous
its hard to see the difference between u and v when their close tother
amistre64
  • amistre64
the wolf gives me this for VxgxV lets see if its does good with the rest too \[<-a u^2+b u v+c v w-a w^2, a u v-b v^2+c u w-b w^2, -c u^2-c v^2+b u w+a v w>\]
anonymous
  • anonymous
alright i'm all caught up writing lol
anonymous
  • anonymous
yea...from what i got..this does not equal each other lol
anonymous
  • anonymous
3 terms per component on the left hand side to 5 terms per component in the rhs
anonymous
  • anonymous
the left side should equal the right side exactly
amistre64
  • amistre64
try with some random easy numbers maybe, or functions lol just to test
amistre64
  • amistre64
V = x,x^2,x^3 g = 1,2x,3x^2 at x=1
amistre64
  • amistre64
V = 1,1,1 g = 1,2,3
amistre64
  • amistre64
(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
  • amistre64
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> ??
anonymous
  • anonymous
so the answer above is correct as it stands?
amistre64
  • amistre64
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
anonymous
  • anonymous
it would just be weird b/c i have 5 of these to prove...and all 4 have matched perfectly
amistre64
  • amistre64
<9,6,3> = -2*stuff /-2 -------- <-9/2, -3, -3/2> = stuff but still; thats not a match
anonymous
  • anonymous
alright...i gotta work out some stuff with wolf then lol
amistre64
  • amistre64
good luck :)
anonymous
  • anonymous
u still around?
amistre64
  • amistre64
im around
anonymous
  • anonymous
i just emailed my professor..this is what i got back
anonymous
  • anonymous
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.
anonymous
  • anonymous
now...i'm really confused b/c i don't see where i'm taking any derivatives
amistre64
  • amistre64
hmmm, ...
amistre64
  • amistre64
are v, u, amd w function of x?
amistre64
  • amistre64
gradV = V' right?
anonymous
  • anonymous
i mean...exactly the way the question is written...is what i have
anonymous
  • anonymous
del operator?
anonymous
  • anonymous
is what she calls it
amistre64
  • amistre64
so grad is the operator on V and not a specific vector ....
amistre64
  • amistre64
del(V.V) is not what I had in mind at the begining
amistre64
  • amistre64
but then what is V.del? del of what?
amistre64
  • amistre64
or is V.del, just V' ?
amistre64
  • amistre64
\[\frac{d}{dx}u+\frac{d}{dx}v+\frac{d}{dx}w=\]
amistre64
  • amistre64
well, with the right parts that is :)
anonymous
  • anonymous
well del is just = to partial/partial x etc.
anonymous
  • anonymous
(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
  • amistre64
\[V = u i + v j + w k \] \[∇= \frac{∂}{∂x} i + \frac{∂}{∂y} j +\frac{∂}{∂z} k\] \[(V∙∇)= \frac{∂u}{∂x} + \frac{∂v}{∂y} +\frac{∂w}{∂z}\] \[(V∙∇)V= \left(\frac{∂u}{∂x}u + \frac{∂v}{∂y}u +\frac{∂w}{∂z}u\right)i+\left(\frac{∂u}{∂x}v + \frac{∂v}{∂y}v +\frac{∂w}{∂z}v\right)j+\left(\frac{∂u}{∂x}w + \frac{∂v}{∂y}w +\frac{∂w}{∂z}w\right)k\] ----------------------------------- \[∇(V∙V)= \left(\frac{∂}{∂x}u^2i + \frac{∂}{∂y}v^2j +\frac{∂}{∂z}w^2k\right)\] \[∇(V∙V)= \left(\frac{∂u}{∂x}2ui + \frac{∂v}{∂y}2vj +\frac{∂w}{∂z}2wk\right)\] \[∇(V∙V)/2= \left(\frac{∂u}{∂x}ui + \frac{∂v}{∂y}vj +\frac{∂w}{∂z}wk\right)\] ------------------------------------ so far −V X (∇XV)
amistre64
  • amistre64
\[-VxV=<0,0,0>\]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
  • anonymous
yea...u see the link i just sent you.....a guy figured it all out...i can't make sense of it though
amistre64
  • amistre64
yeah, james is smart that way
anonymous
  • anonymous
well the other guy if u scroll down solved it
anonymous
  • anonymous
but i can' t understand it
amistre64
  • amistre64
ok, so we seem to be good up to that -v x del x v part; right?
amistre64
  • amistre64
do you understand the notations?\[du/dx=u_x\]
amistre64
  • amistre64
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} \[x=(vv_x-vu_y-wu_z+ww_x)\] \[y=(uu_y-uv_x+ww_y-wv_z)\] \[z=(vw_y-vv_z-uu_z+uw_x)\]
amistre64
  • amistre64
i can make it out, i just cant seem to get a good explanation on it tho

Looking for something else?

Not the answer you are looking for? Search for more explanations.