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Vector Calculus

  1. May 15, 2004 #1
    Hi All,

    I've started to study vector calculus and am finding it a bit daunting. The notes the lecturer has supplied aren't too helpful. Does anyone know of some decent resources on the web that could help me out?

    Many Thanks,

    Pete
     
  2. jcsd
  3. May 16, 2004 #2
    I don't know about the web but I know of two awesome books on vector analysis. Maybe they're in your library?
    - Div, Grad, Curl & All That by Schey. Schey was a theoretical physicist and all the stuff is motivated by stuff that happens in electricity & magnetism, so it's pretty concrete
    - Vector Analysis by Seymour Lipschutz (sp?) <- that's a Schaum's book
     
  4. May 17, 2004 #3
    Believe me, once you get there, Stokes and Divergence Thms will make MOST of your troubles disappear for your course :)

    Secret weapons.
     
  5. May 17, 2004 #4
    heh. w00t!
     
  6. May 19, 2004 #5
    Ok Thanks. One of those books is available in my library, so I'll go check it out on the weekend.

    I'm working on some proof exercises and I was hoping someone could show me how to get started on this one:

    nabla x nabla f = 0

    Thanks

    Pete
     
  7. May 19, 2004 #6
    I think that's curl( grad( f ) ) right? if that's what it is, just do it. That function f has different components, so differentiate it componentwise with the nabla, then do the curl on what you get. So that would be the x derivative of the first component (call it f_1 or something), y derivative of the 2nd component, etc then do the curl on that vector.
     
  8. May 22, 2004 #7
    Thanks to everyone for the tips so far. I'm starting to get an idea about vector fields and vector functions.

    I'm looking at another problem now:

    Vector field F is defined by F = grad xy. Write out the components of F.

    Ok, to begin i use the following identity: grad xy = x.grad y + y.grad x.

    next
    grad y = (dy/dx, dy/dy, dy/dz)
    and
    grad x = (dx/dx, dx/dy, dx/dz)

    so grad xy = x.(dy/dx, dy/dy, dy/dz) + y.(dx/dx, dx/dy, dx/dz)

    then i multiply this out and the components should just drop out right?
     
  9. May 22, 2004 #8

    arildno

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    First of all, you should try to use the LATEX format when writing maths.
    I take it that you meant:
    [tex]\nabla{xy}=x\nabla{y}+y\nabla{x}=[/tex]

    [tex]x(\frac{\partial{y}}{\partial{x}},\frac{\partial{y}}{\partial{y}},\frac{\partial{y}}{\partial{z}})+y(\frac{\partial{x}}{\partial{x}},\frac{\partial{x}}{\partial{y}},\frac{\partial{x}}{\partial{z}})[/tex]

    If this is what you meant to write, then it is correct.
    But, what must the values of the partial derivatives be; for example,
    what is [tex]\frac{\partial{x}}{\partial{z}}[/tex] ?
     
  10. May 22, 2004 #9
    thanks for the quick reply.

    I'm not sure how to exvalute the partial derivatives for this example. Are you saying I don't need z?
     
  11. May 22, 2004 #10

    arildno

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    x,y,z are independent variables!
    What does this imply about partial derivatives.

    I'll get you started:
    Suppose you have a function f in formally 3 variables (x,y,z), and let f
    have the functional form f(x,y,z)=y.
    That is, the value of f remains constant for any changes in the values of x and z!

    Now let us look at the definition of the partial derivative with respect to x:
    [tex]\frac{\partial{f}}{\partial{x}}=\lim_{\bigtriangleup{x}\rightarrow{0}}\frac{f(x+\bigtriangleup{x},y,z)-f(x,y,z)}{\bigtriangleup{x}}=[/tex]
    [tex]\lim_{\bigtriangleup{x}\rightarrow{0}}\frac{y-y}{\bigtriangleup{x}}=0[/tex]
     
  12. May 22, 2004 #11
    [tex]\nabla{xy}=x(0,1,0) + y(1,0,0)[/tex]
    [tex]\nabla{xy}=xj + yi[/tex]

    How does that look?
     
  13. May 22, 2004 #12
    Which one? That Div Grad Curl book is really good. See if you can get that one somehow if it's not at the library.
     
  14. May 23, 2004 #13

    arildno

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    That looks great.
     
  15. May 23, 2004 #14
  16. May 23, 2004 #15

    arildno

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    Marsden&Tromba is an excellent book, IMO.
     
  17. May 23, 2004 #16
    Ok Arildno I have one more question :) I've almost finished my exercises now.

    Let r be the distance from the origin to the given point P. Using either Cartesian or spherical coordinate system, evaluate:

    [tex]\nabla(r^-2)[/tex]

    where do I go from here? I'm not even sure what topic this falls under.

    Thanks again
     
  18. May 23, 2004 #17

    arildno

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    If you do not know about spherical coordinate systems, here's how you should proceed
    in Cartesian coordinates.
    Clearly, we have:
    [tex]r(x,y,z)=\sqrt{x^{2}+y^{2}+z^{2}}[/tex]

    By the chain rule, we must have, for example:
    [tex]\frac{\partial}{\partial{x}}\frac{1}{r^{2}}=-\frac{2}{r^{3}}\frac{\partial{r}}{\partial{x}}[/tex]
     
  19. May 23, 2004 #18
    I might be off, but here's how I would approach that:

    Using cartesian coordinates:

    [tex]
    \begin{align*}
    r&=\sqrt{x^2+y^2}\\
    r^{-2}&=(x^2+y^2)^{-1}\\
    \nabla r^{-2}&=\nabla (x^2+y^2)^{-1}\\
    \mbox{let }f&=(x^2+y^2)^{-1}\\
    \nabla f&=\frac{\partial f}{\partial x}\mathbf{i} + \frac{\partial f}{\partial y}\mathbf{j}\\
    \frac{\partial f}{\partial x}&=-(x^2+y^2)^{-2}(2x)\\
    \frac{\partial f}{\partial y}&=-(x^2+y^2)^{-2}(2y)\\
    \therefore \nabla r^{-2}&=-\frac{2x}{(x^2+y^2)^2}\mathbf{i}-
    \frac{2y}{(x^2+y^2)^2}\mathbf{j}
    \end{align*}
    [/tex]

    Edit: Fixed a misplaced sign in final result.

    Also, if r is supposed to be the position vector in 3 dimensions, I forgot to include the z direction in the above. However, the result should be similar.
     
    Last edited: May 24, 2004
  20. May 24, 2004 #19
    Anyone care to show me how to do this using the spherical coordinate system?
     
  21. May 24, 2004 #20

    arildno

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    OK:
    What you need, is the expression for the gradient operator in spherical coordinates:

    [tex]\nabla=\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta}\frac{\partial}{r\sin\phi\partial\theta}+\vec{i}_{\phi}\frac{\partial}{r\partial\phi}[/tex]

    Here, we have the transformation rules:
    [tex]x=r\sin\phi\cos\theta,y=r\sin\phi\sin\theta,z=r\cos\phi[/tex]
    whereas the unit vectors have the form:
    [tex]\vec{i}_{r}=\sin\phi(\cos\theta\vec{i}+\sin\theta\vec{j})+\cos\phi\vec{k}[/tex]
    [tex]\vec{i}_{\phi}=\cos\phi(\cos\theta\vec{i}+\sin\theta\vec{j})-\sin\phi\vec{k}[/tex]
    [tex]\vec{i}_{\theta}=-\sin\theta\vec{i}+\cos\theta\vec{j}[/tex]

    By applying now the gradient to a function f(r), we gain:
    [tex]\nabla{f}=\frac{df}{dr}\vec{i}_{r}[/tex]
     
    Last edited: May 24, 2004
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