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Vector calc

  1. Dec 26, 2008 #1
    1(a) For the vector function q(x,y,z)=(xy-z2)i +(yz+x2)j +(xz-y2)k calculate

    (i)∇ . q (ii) ∇ . q (iii) ∇ x (∇ x q)

    verify that the identity ∇ x (∇ x q) = ∇(∇ . q - ∇2q

    i have answers of y+z+x,(-3y,-3z,1x) and (3,-1,3) respectivly for the first part but cant fathom the second part, mainly because subbing back in (which i presume is how this is shown) the part where i sub in y+z+x for (∇ . q) what do i now do with ∇ . (y+z+x). the fact that i end up with this leads me to beleive that subbing in is wrong but if it is i really dont know where to go, help please!
     
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  3. Dec 26, 2008 #2

    Dick

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    There's nothing wrong with 'subbing in'. But you want to compute the gradient of (y+z+x). Not the divergence.
     
  4. Dec 26, 2008 #3
    the gradient being d/dx, d/dy or d/dz and why?
     
  5. Dec 26, 2008 #4

    Dick

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    The gradient of a scalar function u is (du/dx,du/dy,du/dz). (y+z+x) is a scalar function. Look up the definition of gradient. Both sides of your identity are vectors. What's the laplacian of q?
     
  6. Dec 26, 2008 #5
    so then that gets me down to ∇2q=(-2,2,-2) if i'm not mistaken is this the same as doing ∇(∇ . q) giving me ∇ . (y+z+x)=(-2,2,-2) which gives me (1,1,1)=(-2,2,-2) ? (something tells me i'm missing a scalar here)
     
  7. Dec 26, 2008 #6
    i assume that if i have made a mistake that's not going to be clear enough so i'll show my working ∇x(∇xq)=∇(∇.q)-∇2q

    (3,-1,3)=∇.(y+z+x)-∇2q
    (3,-1,3)=(1,1,1)-∇2q
    (2,-2,2)=∇2q
    (2,-2,2)=∇.(d/dx,d/dy,d/dz).q=(y+z+x).∇
    (2,-2,2)=(1,1,1)
     
  8. Dec 26, 2008 #7

    Dick

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    You can't take ∇ . (y+z+x). y+z+x is a scalar function. The problem says you should take it's gradient. Like I said before.
     
  9. Dec 26, 2008 #8

    Dick

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    To find the laplacian of q, take the second derivative of q with respect to x (getting a vector), the second derivative wrt y and the second derivative wrt z and add all three vectors. That's what the ∇^2 is supposed to convey.
     
  10. Dec 26, 2008 #9
    so im left with (3,-1,3)=(2,-2,2)+ ∇(∇.q)

    so (3,-1,3)=(2,-2,2)+∇(y+z+x)

    so ∇(y+z+x) is just the gradient of (y+z+x) and is just (1,1,1) which makes both sides equal nice on. i am right in thinking this ∇(y+z+x) is dot prod not cros rite?
     
  11. Dec 26, 2008 #10

    Dick

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    Yes, it's just gradient. ∇(∇ . q)=∇(y+z+x)=(1,1,1). As you said.
     
  12. Dec 26, 2008 #11
    orite, so now we have that cleared up i've managed to get almost instantly stuck on the second part of the question. Consider the vector fieldq=yzcos(xy)i+xz(cos(xy)j+(sin(xy)+2z)k show that ∇xq=0

    i tried to solve this like i did part 2 of question 1 ((∇x q)) by using the determinant, however i end up with =zcos(xy)+yxz cos(xy) -2 instead of zero?
     
  13. Dec 26, 2008 #12

    Dick

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    How did you wind up with a scalar instead of a vector?? E.g. the x component of the curl should be d/dy(sin(xy)+2z)-d/dz(xzcos(xy)). What's that? What are the other two components?
     
  14. Dec 26, 2008 #13
    i ended up with erm det{(i,j,k),(d/dx,d/dy,d/dz),(yzcos(xy),xzcos(xy),sin(xy)+2z

    giving me i((d/dy*sin(xy) +2z)-(d/dz*xzcos(xy)) -j((d/dx(sin(xy)+2z)-(d/dz*yzcos(xy))+k((d/dx*xzcos(xy))-(d/dy*yzcos(xy)

    giving me (xcos(xy)-xcos(xy))i - (ycos(xy)-ycos(xy))j + ((zcos(xy)-yzxsin(xy))-(zcos(xy)-(yzxsin(xy))=0

    ok so maybe i didnt get that but i do now and thats what matters sorry silly mistake:P i'll be back in 10 minutes when the next question trips me up (joke i hope) thanks for all your help very kind of you :)
     
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