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Calculate the gradient of the function

  1. Sep 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Calculate the gradient of:
    (a) V1=6xy-2xz+z
    (b) V2=10ρcos(phi)-ρz
    (c) V3=(2/r)cos(phi)


    2. Relevant equations



    3. The attempt at a solution
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    Attached Files:

  2. jcsd
  3. Sep 6, 2013 #2

    rude man

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    You have the notation wrong for gradient. You wrote div instead. You can't take the div of a scalar. V1, V2 and V3 are all scalars.

    You also have the wrong expressions for del F in cartesian, cylindrical and spherical coordinates. Look them up.
     
  4. Sep 6, 2013 #3
    so what you are saying is that instead of just del I should have used del f for notation?
    (∂f/∂x)x+(∂f/∂y)y+(∂f/∂z)z (where x, y, and z should be x(hat), y(hat), and z(hat))
     
  5. Sep 6, 2013 #4
    so if I am right which I probably am not then (a) 6yUx+(-0)+zUz the reason I put 0 is the the partial derivative of -2xz with respect to y is 0 correct since there is no y term.
     
  6. Sep 6, 2013 #5

    rude man

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    You did not write "just del". You wrote del-dot. The dot was wrong. You can write del f or you can write grad f. Same thing. (But del-dot-F is OK whereas del F or grad F are not.)

    Your expression for grad f is correct for Cartesian coordinates x,y,z. I use i, j , and k for the unit vectors. I also use bold type for all vectors.

    When you get to V2 and V3 the expressions for grad V are very different from Cartesian. V2 is using cylindrical and V3 is using spherical.
     
  7. Sep 6, 2013 #6

    rude man

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    Yes, ∂(-2xz)/∂y = 0.
    No, you are not following your formula in post 3 for V1.
    Don't keep changing your notation for the unit vectors. Use i j k why not.
     
  8. Sep 6, 2013 #7
    dang I felt it might have actually been right. was it simply this 6yi+k
     
  9. Sep 6, 2013 #8

    rude man

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    Was what 6y i + k ?
     
  10. Sep 6, 2013 #9
    yes but it isn't correct I can imagine.
     
  11. Sep 6, 2013 #10

    rude man

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    I'm getting dizzy. Correct for what?

    Tell you what, start over with your expression for grad V1 from your post 3, use i j k unit vectors, and let's see if we can't figure this one out. Remember to use V1, not f.
     
  12. Sep 6, 2013 #11
    ∇V1=(∂f/∂x)i+(∂f/∂y)j+(∂f/∂z)k= 6y i + k
     
  13. Sep 6, 2013 #12

    rude man

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    I said use V1, not f.
    OK, one term at a time. Try ∂(V1)/∂x again?
     
  14. Sep 6, 2013 #13
    ok ∂(V1)/∂x= 6y
     
  15. Sep 6, 2013 #14
    ∂(V1)/∂x = 6y i sorry I forgot the "i" in the previous post.
     
  16. Sep 6, 2013 #15

    rude man

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    What about ∂(-2xz)/∂x?
     
  17. Sep 6, 2013 #16
  18. Sep 6, 2013 #17

    rude man

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    Better.
    Now, can you do the rest of grad V1?
     
  19. Sep 6, 2013 #18
    (6y-2z)i + (6x)j + (1-2x)k
     
  20. Sep 6, 2013 #19
    for V2 (10cos(phi))-z)i-(10sin(phi))j-(ρ)k
     
  21. Sep 6, 2013 #20
    for V3 ((-2/r^2)cos(phi))i-((2sin(phi))/(r^2sin(theta)))k
     
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