1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
     

    Attached Files:

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

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Emergency help
  1. Emergency Hw Help (Replies: 6)

  2. Emergency hw help (Replies: 2)

Loading...