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Homework Help: Emergency help

  1. Aug 12, 2004 #1
    I encountered many problems while doing exercises in text books. :confused: And i have stated it down in a word document attached in this post. Hope someone can help and teach me how to solve those problems. Answers are given. I just don't know how to get those answer.

    Thanks a lot. :smile:

    p/s : Emergency = next week test. :bugeye:

    (2), (4), (5), (6) solved.. thanks
     

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    Last edited: Aug 13, 2004
  2. jcsd
  3. Aug 12, 2004 #2

    arildno

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    Welcome to PF!

    You must show some of your own work here; don't expect your homework to be done for you!
    Having said that, let's take a specific question, nr. 4:
    Now, what are your problems with this particular exercise?
    Make a detailed comment on this.
     
  4. Aug 12, 2004 #3
    r dot grad T should be a scalar ? i wonder y answer's a vector. I used formulas and can't get those answers. Just can't understand why. So seeking help. Thanks.
     
  5. Aug 12, 2004 #4

    arildno

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    The answer should be a vector, as it is given in the answer (i haven't checked if the answer given is correct)

    I hope you know about the grad "vector":
    [tex]grad=\vec{a}_{x}\frac{\partial}{\partial{x}}+\vec{a}_{y}\frac{\partial}{\partial{y}}+\vec{a}_{z}\frac{\partial}{\partial{z}}[/tex]

    For example, the divergence of a vector [tex]\vec{v}[/tex] is given by:
    [tex]div\vec{v}=grad\cdot\vec{v}[/tex]

    Are you familiar with this notation?
     
  6. Aug 12, 2004 #5
    actually i dunno what means (r dot grad) ? i know div and grad as well
    p/s: how you draw those symbols?
     
  7. Aug 12, 2004 #6

    arildno

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    You can click on the LATEX code to see how you write it .

    OK, so you know the "grad", which I'll henceforth write as [tex]\nabla[/tex]

    Let's first review how we get the scalar known as "divergence"
    ([tex]\nabla\cdot\vec{v}[/tex])
    Let [tex]\vec{v}=u\vec{a}_{x}+v\vec{a}_{y}+w\vec{a}_{z}[/tex]

    We then have that:
    [tex]\nabla\cdot\vec{v}=\vec{a}_{x}\cdot(\frac{\partial}{\partial{x}}\vec{v})+\vec{a}_{y}\cdot(\frac{\partial}{\partial{y}}\vec{v})+\vec{a}_{z}\cdot(\frac{\partial}{\partial{z}}\vec{v})[/tex]

    This simplifies to, in our case:
    [tex]\nabla\cdot\vec{v}=\frac{\partial{u}}{\partial{x}}+\frac{\partial{v}}{\partial{y}}+\frac{\partial{w}}{\partial{z}}[/tex]

    (Please comment if this doesn't make sense to you!)

    Now, we're ready to tackle [tex]\vec{v}\cdot\nabla[/tex]
    This is also a "dot" product (scalar product), and looks like:
    [tex]\vec{v}\cdot\nabla=u\vec{a}_{x}\cdot\nabla+v\vec{a}_{y}\cdot\nabla+w\vec{a}_{z}\cdot\nabla[/tex]

    Or, simplified:
    [tex]\vec{v}\cdot\nabla=u\frac{\partial}{\partial{x}}+v\frac{\partial}{\partial{y}}+w\frac{\partial}{\partial{z}}[/tex]

    This is a "scalar" operator which you then apply on T.
     
  8. Aug 12, 2004 #7

    arildno

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    The answer is correct, BTW
     
  9. Aug 12, 2004 #8
    oic thanks
     
  10. Aug 12, 2004 #9
    in q 5 : i found the surface integral = -2, but line integral = 7/6. it didn't match Stoke's theorem.. i wonder y....
     
  11. Aug 12, 2004 #10
    i never learn that inverse dot product b4.. hehe... thanks
     
  12. Aug 12, 2004 #11
    and q 2 i really have no idea
     
  13. Aug 12, 2004 #12

    arildno

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    OK, first:
    Have you checked that you get no.4 right?

    Secondly, try and group together a few questions you think "belong" to each other, which you would like to focus on.
     
  14. Aug 12, 2004 #13
    no.1 to 6 are about vectors. All about integrals.. I just wonder y can't get answers using formulas. No.2 is really dunno how to start also.
     
  15. Aug 12, 2004 #14
    no.5 curl = -x^2 in z direction.. then integral can't get 7/6
     
  16. Aug 12, 2004 #15

    arildno

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    OK, we'll look into 2 (but you did check 4, or what?).

    2.
    Now, you've been given equations for two surfaces.
    In general, if you have a surface given on the form S(x,y,z)=0 (or constant),
    you know that the normal on that surface at a point (x,y,z) is parallell to the gradient of S (evaluated on the same point).
    Post what you get here in some detail.
     
  17. Aug 12, 2004 #16
    can't get no.4 answer
     
  18. Aug 12, 2004 #17

    arildno

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    Post what you've done. In detail.
     
  19. Aug 12, 2004 #18
    1) since z=0, [tex]a_{r}[/tex] has no [tex]a_{x}[/tex] component , [tex]a_{\phi} = -\sin{\phi}{a}_{r} + \cos{\phi}{a}_{\phi}[/tex] ... and i can't t the answer for (a) and so on
     
    Last edited: Aug 12, 2004
  20. Aug 12, 2004 #19

    arildno

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    I meant on Q4; the one I started with.
     
  21. Aug 12, 2004 #20
    (r dot grad T ) = [tex](\vec{r}\cdot\nabla){T}={{x}a_{x}}\frac{\partial{2zy}}{\partial{x}}+{{y}a_{y}}\frac{\partial{xy^2}}{\partial{y}}+{{z}a_{z}}\frac{\partial{x^2yz}}{\partial{z}}[/tex]
    is it?
     
    Last edited: Aug 12, 2004
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