# Multivariable Calculus Intro

1. Jun 8, 2004

### Gza

I was wondering, if it wouldn't be too boring for any of you, if someone could post an intro to multivariable calculus, from partial differentiation up through the grad, curl, divergence of vector fields, preferrably with applications to physics. Maybe you can post a sticky or something. I'm trying to study up on some more advanced E&M, and having a tough time hacking through all that stuff. I know it's all in my text, but I can hardly find derivations, or even reasons for many of the concepts. I also like the idea of getting instant feedback on difficulties. Thanks in advance.

2. Jun 8, 2004

### arildno

This is a great idea, Gza!
Hopefully, the administrators will think the same..

3. Jun 10, 2004

### Gza

Doesn't seem to be the case. Oh well, i'm taking the class in two weeks anyway, no worries.

4. Jun 10, 2004

### master_coda

I'm sure people will be willing to help you with specific difficulties. Posting a complete overview of multivariable calculus is a little much.

5. Jun 10, 2004

### Tom Mattson

Staff Emeritus
Go ahead and post your questions as they arise. Right now, I'm teaching Calculus III (exactly what you describe). I don't have time to post a tutorial, but I can answer specific questions.

6. Jun 10, 2004

### Gza

In that case, I had a question on exactly why the gradient of a function points in the direction of greatest change. How is it that, say, the gradient of a function of two variables, which lies on a tangent plane somewhere on the surface of the function "knows" to point to the direction of greatest change. (excuse my anthropomizing) And also, the del operator was described to me as being a vector in itself. I find it hard to see why this is true, since it is only an operator. Thanks for the help.

7. Jun 10, 2004

### Tom Mattson

Staff Emeritus
Start with the directional derivative of f(x,y) in the direction of u=cos(θ)i+sin(θ)j:

Duf(x,y)=fx(x,y)cos(θ)+fy(x,y)sin(θ)

This can be restated in terms of the gradient as follows:

Since this is a dot product, we can rewrite it as:

where α=the angle between gradf and u. Noting that |u|=1, we have:

Since gradf is fixed once a surface z=f(x,y) is chosen, the only variable parameter is α. And since the maximum of cos(α) is 1, the maximum of Duf(x,y) is |gradf(x,y)|, and this occurs when u is parallel to gradf(x,y).

Yes, it's an operator: A vector operator. When it operates on a function f(x,y), then gradf(x,y) is a plane vector.

edit: fixed subscript bracket

Last edited: Jun 11, 2004
8. Jun 10, 2004

### arildno

Just a tiny comment about the del operator as a vector.
It is, in general not true that $$\nabla\times\vec{F}$$ is normal on $$\vec{F}$$

So, even if we have the identities
$$\nabla\times\nabla=\vec{0}, \nabla\cdot(\nabla\times\vec{F})=0$$

these do NOT follow from vector theorems about normals and parallells (as I have seen at times ), but from the independence of the order in which we differentiate.

9. Jun 12, 2004

### Gza

"Just a tiny comment about the del operator as a vector.
It is, in general not true that $$\nabla\times\vec{F}$$ is normal on $$\vec{F}$$"

I was interested in why the curl of a vector field would even in the first place be normal to the field?

10. Jun 12, 2004

### Tom Mattson

Staff Emeritus
Because the cross product AxB is perpendicular to both of the vectors A and B.

11. Jun 12, 2004

### arildno

You're right, of course, there's no reason for it.

It's not my fault that I have met several persons who persist in this idiotic fallacy

(Either they do so on basis of the cross-product theorem for vectors, or the fact that physically, the curl of a velocity field is parallell to the local angular velocity vector)

Last edited: Jun 12, 2004
12. Jun 12, 2004

### Tom Mattson

Staff Emeritus
I don't want you to think that Ardilno and I are telling you two different things, so just to clarify:

Some people look at $\nabla\times\vec{F}$ and say, "Well, $\nabla$ is a vector and $\vec{F}$ is a vector, and $\nabla\times\vec{F}$ is their cross product, so therefore they're perpendicular in general".

But they aren't. It shouldn't be too hard to find a counterexample.

edit: fixed tex bracket

13. Jun 12, 2004

### arildno

I whole-heartedly concur with Tom Mattson here;
my original post was given because
from my experience of assistant teaching on elementary maths/fluid mechanics that several students do, in fact, have these faulty notions..

14. Jun 12, 2004

### jdavel

Tom Mattson and arildno,

I think the reason people think they can do this is because there is some rule about substituting the del operator into vector identities. Sometimes you can do it. What are the restrictions that rule out the curlF.F case?

15. Jun 12, 2004

### Tom Mattson

Staff Emeritus
I've never been taught that you can substitute del into any vector identity. In all my courses that treated the del operator, we had to establish the identities all over again.

I think it's just a coincidence when you can.

I don't know of a general condition for curl F.F=0 other than the obvious one:

(&part;Fz/&part;y-&part;Fy/&part;z)Fx+...=0

Maybe Ardilno can state a more compact restriction.

Anyway, I worked out a counterexample to the statement:

F(x,y,z)=xyzi+xyzj+xyzk

Plug that into curl F.F, and you get:

x2(z-y)-y2(z-x)+z2(y-x).

16. Jun 12, 2004

### arildno

:uuh: :blush: :shame:

Last edited: Jun 12, 2004
17. Jul 22, 2004

### mathwonk

Re: why does the gradient point in the direction of greatest change?

That depends on what you think the gradient is. You could say that is a definition, but lets take it to be a vector whose x and y coordinates are the rates of change in the x and in the y direction.
Then why does the vector with those x and y components point in the direction of greatest change?

Imagine a stone thrown in the water and the wave of water rippling out in all directions. A particle of water has a velocity vector pointing straight away from the source of the pebble. if we want to measure the component of velocity in some other direction, we just project that velocity vector in the given direction. I.e. we take the component of the vector pointing in that direction.

For example, the particle is not moving at all in the direction perpendicular to the (radially pointing) velocity vector, since that projection is zero. If we pick two arbitrary perpendicular directions, called the x and y directions, and project thje velocity vector in those directions, then the two component vectors we get are called the velocity in the x and y directions respectively, or the partial derivatives in the x and y directions

But since our space is only two dimensional, in fact we can reason backwards as well, and from those two component vectors, we can reconstruct the original velocity vector by vector addition. So it appears as if the gradient is defiend in terms of the two partials, whereas really they are only devices for recovering it.

Note that each projection of the original velocity vector is shorter than the original velocity vector. Hence the vector sum of the two x and y velocity vectors, i.e. the original velocity vector is the longest velocity vector in any possible direction.

This shows that the only velocity vector with intrinsic meaning is the longest one, the one in the direction of greatest change. the partial derivatives in the x and y directions are artificial constructs which fortunately recover the intrinsic vector, the gradient, as their vector sum.

This attempted an intuitive explanation. For a mathematical proof, you could define the partial derivatives by limits as usual in the x and y directions, and then deduce that the directional derivative in any other direction v (also defined by an approapriate limit) is obtained by dotting the direction vector v with the gradient vector formed from the two partials. It then follows that the direction in which the change is greatest is the direction vector with largest dot product with the gradient. but this is the gradient itself.

does that make any sense?