vector calculus problems

[PG1995]

vector calculus problems, q11, chap3

Hi

Could you please help me with this query? Thanks.

Regards
PG

 

[dougy83]

1) any vector crossed with itself is zero: A x A = |A||A|sin theta, where theta is the angle between the A and itself; theta is of course 0 and sin(0) = 0

2) A.(A X B) = | j|A||B|sin(theta) | |A| cos(phi), where j is the unit vector pointing out 90 degrees to the plane containing A and B. A is on the plane, so there is 90 degrees between j and A -- this is phi. cos(90deg) is 0

 

[PG1995]

Thank you, Doug. I think I get it now.

For the expression labelled "2", you are essentially saying this.

Regards
PG

 

[dougy83]

Yes, where the non-bold A & B represent the vector length.

Sorry I should have used latex, but the editor doesn't work in my browser (chrome).

 

[PG1995]

Thank you for the reply.

Sorry I should have used latex, but the editor doesn't work in my browser (chrome).

No, it was okay. I got your point and that's the important thing. And besides it's good I did it myself so that I don't forget it in future.

Regards
PG

 

[PG1995]

Hi

Could you please help me with this problem? Thank you.

Regards
PG

 

[steveB]

Q1: It does mean that, in the limit as those deltas go to zero. More specifically, higher order is referring to the fact that those terms have higher powers with Δx^2, Δy^2, Δz^2, ΔxΔy, ... etc. and then 3rd powers and on higher and higher. When Δx, Δy and Δz are getting small, those terms are getting smaller much faster.

Q2: The partials derivatives are not with respect to any direction but each is with respect to a specific direction. When combined in an appropriate way, they can be used to determine derivatives in any direction. I'm not sure if the calculation should be considered an "average", but maybe "first-order" estimation might be more descriptive. However, overall I think you are correct, aside from these minor wording issues.

Q3: Well, it is pointing in the right direction and it has a magnitude of unity. That pretty much defines a unit vector. So, one needs to prove only that. I think another one of your attachments, from another thread, actually made this explicit. Intuitively, it should be clear that it points in the correct direction. The magnitude being one is maybe not as obvious, but think about how you calculate magnitude of any vector and then remember the Pythagorean theorem in multi-dimensions. These are the same formulas, so the distance traveled by the vector is the same as the linear distance traversed by the variable s. Hence, the rate of change of the vector magnitude with respect to the distance traveled is one.

 

[PG1995]

Thanks a lot.

Re: Q3

Well, it is pointing in the right direction and it has a magnitude of unity. That pretty much defines a unit vector.

Sorry but I still don't see how you can say it's a unit vector. I see that the vector (dx/ds)i + (dy/ds)j + (dz/ds)k points in the direction from the P to Q; that's the way it was calculated in part (b) of the question. Further I understand that how to apply Pythagorean theorem in three dimensions but you can say I have doubts that it has magnitude of unity.

I think another one of your attachments, from another thread, actually made this explicit.

Could you please tell which one was it? Perhaps, it could help me here. Thanks.

Regards
PG

 

[steveB]

Looking at the other thread, I see I was mistaken, and, there, the gradient of position is shown to be a unit vector. Although related, this is not explicitly related to your question. So, lets forget about that.

Just look at it intuitively instead. The Δs is the small distance between two close points. The derivatives dx/ds, dy/ds and dz/ds represent the rate of change of x, y and z relative to changes in s, as you move along the line that connects the two close points. Let's say s changes by 1 micro-unit a the infinitesimal level. Then, the corresponding changes in x, y and z (i.e. dx, dy and dz) will be rectangular components of the offset in s (i.e. ds), and the vector dr=dx i + dy j + dz k will have length of 1 micro-unit also. If you draw this out, you will see that the line segment between s and s+ds is the diagonal of a box with lengths dx, dy and dz. Hence, the magnitude of the vector dr=dx i + dy j +dz k , must be equal to ds since distance across a box is given by the Pythagorean Theorem and this formula is identical to vector magnitude. Even though this is not a formal proof, you should be able to see now that the result is nothing mysterious, and is tied to the fact that magnitude and distance are essentially the same thing. In other words, it is an obvious fact shouded in vector notation sufficiently that it does not look so obvious. The form looks confusing because the distances (dx, dy, dz) are shown as rates of change (dx/ds, dy/ds, dz/ds). But, it is the same thing at the infinitesimal level.

 

[PG1995]

Really appreciate your effort to make it easier for me. But you would need to make it still easier!

Just look at it intuitively instead. The Δs is the small distance between two close points. The derivatives dx/ds, dy/ds and dz/ds represent the rate of change of x, y and z relative to changes in s, as you move along the line that connects the two close points.

Don't you think that the term "line" in this context is misnomer or confusing because, in my view, it's going to be curve unless we have a flat plane in three dimensions?

Let's say s changes by 1 micro-unit a the infinitesimal level. Then, the corresponding changes in x, y and z (i.e. dx, dy and dz) will be rectangular components of the offset in s (i.e. ds), and the vector dr=dx i + dy j + dz k will have length of 1 micro-unit also. If you draw this out, you will see that the line segment between s and s+ds is the diagonal of a box with lengths dx, dy and dz. Hence, the magnitude of the vector dr=dx i + dy j +dz k , must be equal to ds since distance across a box is given by the Pythagorean Theorem and this formula is identical to vector magnitude.

You said that Δs is the small distance between two close points. So, Δs or s changes by 1 micro-unit. Do you mean that the distance between the points gets little larger? And why are you sticking with 1 micro-unit? Why not 2 or 3 micro-unit? Wouldn't it be more like a triangle in three space like this rather than a box?

Kindly help me with the queries above so that I can see where I'm going wrong. Thank you.

Regards
PG

 

[steveB]

I'm not clear on what the point of confusion is, but I can try to answer the specific questions you mentioned.

First, when dealing with small displacemnts that ultimately will be reduced to zero length in the limit, it is ok to think of small line segments that make up a curve that the distance s is linked to.

I say 1 micro-unit to mean any small unit dispacement. I could have said one unit, but then you might think of a finite and large displacement. So, micro just means small here. If 1 micro-unit is small, then 2 or 3 micro-units is also small and perfectly valid. Perhaps I just create confusion by trying to link the math with physics, but sometimes it helps to think in terms of something physical.

Your picture with a triangle looks ok. I just like to think of a box (i mean a 3d box, like a shoe-box) defined by the sides lengths dx, dy and dz, and then the diagonal across the box is the distance. Choose whichever picture you are comfortable with.

 

[PG1995]

Thanks a lot, Steve.

But I'm sorry I'm still struggling with it. If you don't mind let me give it another try. My query is about the Q3. And here is what confuses me. Thanks.

Reference:
1: Pythagoras theorem in 3-dimensional space.


Regards
PG

 

[steveB]

Lets try a different point of view.

Consider three separate examples. Imagine that Δs separates two points on the x axis. Then two points on the y axis and finally ( you guessed it ) two points on the z-axis. Do this out and I hope you get the i. J and k unit vectors respectively. If you gave trouble with this part, let us know.

Once you can visualize this, you can use a symmetry argument to prove that you get a unit vector in any direction. Since there is no preferred direction, we are free to redefine coordinate directions and transform any problem into this new reference frame. This means that any direction will give a unit vector because we can define any direction to be the x direction. This symmetry is based on the Euclidean norm which is just the vector magnitude or the Pythagorean distance formula. Our notion of distance is independent of position or choice of the reference frame in Euclidean space.

 

[PG1995]

Thanks a lot, Steve.

I believe I get it now. I know you hear it all the time but you are a genius without any tint of arrogance and rudeness! Moreover, we have been getting along for quite some time it shows that you are a person of remarkable patience!

Best wishes
PG

 

[PG1995]

Hi

Could you please help me with these queries? Thanks a lot.

Regards
PG

 

[steveB]

Overall, I'd say you have it correct. Perhaps a mathematician might argue slightly about a word choice here and there, but it seems good enough for engineering/science purposes.

For Q1, the only geometrical picture I use is that one vector is projected onto the other vector, and then the length of one vector is multiplied by the length of the projection of the other vector.

 

[PG1995]

Hi

My present query is related to Q1 in this problem. Could you please help me with it? Thank you.

Regards
PG

 

[steveB]

I don't see anything conceptually wrong with writing it that way. Although, there is a lot of detail about the structure of the ε1 and ε2 left out. Still, those quantities will go to zero in the limit, and the products of ε1Δx and ε2Δy represent second order and higher terms, so there is no problem to informally write it that way.

 

[PG1995]

chap 4, q28

Hi

Could you please help me with this query? Thank you.

Regards
PG

 

[steveB]

The vector function in question is spherically symmetric and always points radially outward. Hence, there is no rotation or vortex in the field and curl is always zero in that case.

This is much easier to see in spherical coordinates (r,ψ,θ), rather than rectangular coordinates (x,y,z). Imagine expressing that vector in terms of spherical coordinates. The vector always points in the r-direction. This means that there is only one component to the vector, which is r f(r) in the r-direction. Curl involves taking the derivative of each component relative to the other coordinates other than that coordinate direction. Hence you always have derivatives of the form ∂Ar/∂θ, ∂Ar/∂ψ, ∂Aψ/∂r, ∂Aψ/∂θ, ∂Aθ/∂r, ∂Aθ/∂ψ, but never derivatives of the form ∂Ar/∂r, ∂Aψ/∂ψ, ∂Aθ/∂θ. In this case there is only an Ar type of component, but Ar=f(r) only, hence ∂Ar/∂θ and ∂Ar/∂ψ are both zero. So, when you evaluate the curl of such a vector in spherical coordinates, all partial derivatives are zero from the very start, and no work needs to be done to prove that the answer is zero.