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galipop
May15-04, 11:37 PM
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

fourier jr
May16-04, 12:21 AM
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

Divergent13
May17-04, 09:36 PM
Believe me, once you get there, Stokes and Divergence Thms will make MOST of your troubles disappear for your course :)

Secret weapons.

Ebolamonk3y
May17-04, 10:46 PM
heh. w00t!

galipop
May19-04, 07:40 AM
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

fourier jr
May19-04, 12:52 PM
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.

galipop
May22-04, 08:29 AM
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?

arildno
May22-04, 08:41 AM
First of all, you should try to use the LATEX format when writing maths.
I take it that you meant:
\nabla{xy}=x\nabla{y}+y\nabla{x}=

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}})

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 \frac{\partial{x}}{\partial{z}} ?

galipop
May22-04, 08:49 AM
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?

arildno
May22-04, 09:07 AM
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:
\frac{\partial{f}}{\partial{x}}=\lim_{\bigtriangle up{x}\rightarrow{0}}\frac{f(x+\bigtriangleup{x},y, z)-f(x,y,z)}{\bigtriangleup{x}}=
\lim_{\bigtriangleup{x}\rightarrow{0}}\frac{y-y}{\bigtriangleup{x}}=0

galipop
May22-04, 06:50 PM
\nabla{xy}=x(0,1,0) + y(1,0,0)
\nabla{xy}=xj + yi

How does that look?

fourier jr
May22-04, 11:23 PM
Ok Thanks. One of those books is available in my library, so I'll go check it out on the weekend.

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.

arildno
May23-04, 03:26 AM
\nabla{xy}=x(0,1,0) + y(1,0,0)
\nabla{xy}=xj + yi

How does that look?
That looks great.

galipop
May23-04, 04:51 AM
I grabbed the first book you mentioned. I grabbed this book too - 'Vector Calculus' by Marsden Tromba which has heaps of examples.

I also found a cool introductory website: http://oregonstate.edu/Dept/math/CalculusQuestStudyGuides/vcalc/vcalc.html

Cheers for all your help.

Pete

arildno
May23-04, 06:36 AM
Marsden&Tromba is an excellent book, IMO.

galipop
May23-04, 07:52 AM
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:

\nabla(r^-2)

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

Thanks again

arildno
May23-04, 08:01 AM
If you do not know about spherical coordinate systems, here's how you should proceed
in Cartesian coordinates.
Clearly, we have:
r(x,y,z)=\sqrt{x^{2}+y^{2}+z^{2}}

By the chain rule, we must have, for example:
\frac{\partial}{\partial{x}}\frac{1}{r^{2}}=-\frac{2}{r^{3}}\frac{\partial{r}}{\partial{x}}

TALewis
May23-04, 08:21 AM
I might be off, but here's how I would approach that:

Using cartesian coordinates:


\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*}


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.

galipop
May24-04, 06:28 AM
Anyone care to show me how to do this using the spherical coordinate system?

arildno
May24-04, 07:16 AM
OK:
What you need, is the expression for the gradient operator in spherical coordinates:

\nabla=\vec{i}_{r}\frac{\partial}{\partial{r}}+\ve c{i}_{\theta}\frac{\partial}{r\sin\phi\partial\the ta}+\vec{i}_{\phi}\frac{\partial}{r\partial\phi}

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

By applying now the gradient to a function f(r), we gain:
\nabla{f}=\frac{df}{dr}\vec{i}_{r}

mathwonk
Jul23-04, 08:43 PM
curiously, i believe essentially all books on several variable calculus, and on vector calculus are rather good, as is true of books on complex variables. The same is not true of books on real variables, elementary analysis, books on elementary calculus, or books on algebra.

i learned about gradients and several variables calculus from "calculus of several variables", by george fleming. There is also an excellent appendix there on calculus of one variable inclduding elementary topology of the real line, which served as a review course of high quality on that topic.


i also like spivak's calculus on manifolds very much. for one thing it is only 140 pages.