3 vector analysis questions

[freeziepleasy]

Hello.

This summer I have been self-studying vector analysis, using the book "Vector Analysis" by L. Marder. I have a few questions I am a bit confused about.

1) "Tangential Line Integral of a Vector"
Evaluate the tangential line integrals of the vector functions:
ii) G = 2xyi + (x^2 + z)j + (y+z^2)k
along the curve C whose parametric equations are x = acost, y=asint, z=bt, where a and b are positive constants and t increases from 0 to 2pi.
answer: (8/3)(pi^3)(b^3)

The author started off by doing
G=(df/dx, df/dy, df/dz) <-where the d is partial derivative
Then he started solving for f:
f=x^2y + g(y,z). (3. cool
-When he is solving for f, is he basically making a new function to take the integral from? If so, why? Why can't he just take the integral of the function as itself without any modification?

Then, he did:
*d is partial derivative
"df/dy = x^2 + dg/dy = x^2 + z
so that
dg/dy = z
and on integrating with respect to y (and noting that g is a functino of y and z only)
g= yz + h(z)
where h(z) is to be determined. We thus have by (3. cool
f= (x^2)(y) + yz + h(z)"

- For "df/dy = x^2 + dg/dy = x^2 + z", how did he get the notation dg/dy?
- I can see why dg/dy = z from that equation, but how did he know that dg/dy equals z in the first place to be confident enough to put dg/dy in there?
- When he plugs g= yz + h(z) back into f=x^2y + g(y,z), why did he leave x^2 out of the equation (x^2 from x^2 + dg/dy = x^2 +z)?

-Also, instead of integrating from 0 to 2pi, he integrated from initial and terminal coordinates A=(1,0,0) and B=(1,0,(2pi)(b)). Why did he use those coordinates instead of t? Is it because he "modified" the equation above like the way he did so that t would not work in this case?
-Where did he get those coordinates?

He didn't finish on explaining how he got to the solution (because he assumes the reader can finish the rest using F = (df/dx,df/dy, df/dz)), but if anyone can finish the problem while explaining, it would be very helpful.

2) "Surfaces"
Consider a pair of neighboring points P(u,v) and P'(u+du, v+dv) where du, dv are differentials of the curvilinear coordinates u,v. When du and dv are small, the vector PP' is approximated by the displacement dr=(dx,dy,dz)....
dr=(dr/du)(du) + (dr/dv)(dv) <- (dr/du) and (dr/dv) are partial derivatives

The length of this vector is denoted by ds, so that
d(s^2)=dr.dr
...
....
d(s^2)=(dr/du)^2(du^2)+2(dr/du)(dr/dv)(du)(dv)+(dr/dv)^2(dv)^2
d(s^2)=g11(u,v)(du)^2 + 2g12(u,v)dudv + g22(u,v)(dv)^2

The quantities g11,g22,g12, and g21 (defined to be equal to g12) are the elements of a symmetric matrix called metric tensor and determine the geometrical properties of the surface near any point.
coordinate lines meet orthogonally if and only if g12=0.

(g11 = g sub 11)
-Why do g11,g22,g12, and g21 equal g12?
-What is the meaning of metric tensor, and what does he mean by saying "determine the geometrical properties of the surface near any point"?

3) "Cylindrical Polar and Spherical Polar Coordinates"
Cylindrical Polar Coordinates: R, psi, z
with x = Rcos(psi), y=Rsin(psi), z=z
-Why is x = to cos(psi) and y = to sin(psi)? Is it arbitrary? Can you switch so that x = Rsin(psi) and y = Rcos(psi) without changing how the cylinder would look like?

To anyone who answers, thank you very much.  

[VorpalNeko]

freeziepleasy

1) "Tangential Line Integral of a Vector"
Evaluate the tangential line integrals of the vector functions:
ii) G = 2xyi + (x^2 + z)j + (y+z^2)k
along the curve C whose parametric equations are x = acost, y=asint, z=bt, where a and b are positive constants and t increases from 0 to 2pi.
answer: (8/3)(pi^3)(b^3)

OK.

freeziepleasy

The author started off by doing
G=(df/dx, df/dy, df/dz) <-where the d is partial derivative
Then he started solving for f:
f=x^2y + g(y,z).
-When he is solving for f, is he basically making a new function to take the integral from? If so, why? Why can't he just take the integral of the function as itself without any modification?

But he is. You will see this pattern again and again: the integral of a differential of a form over a region is equal to the integral of the original form over the boundary of that region (subject to some criteria such as orientability, which aren't important right now). You've already seen two particular cases of it.

The fundamental theorem of calculus. Say you're integrating dF over an interval I = [a,b], which has as a boundary the two points ∂I = {a,b}. Then:
[1] Int_I[ dF ] = Int_{∂I}[ F ] = F(b) - F(a).
You've just turned an integral over an interval to an integral over two isolated points, by looking for the function that dF is a differential of, which is equivalent to taking antiderivative.

The fundamental theorem of line integrals. Say you're integrating a curve γ. Then its boundary is the endpoints, say, {p,q}, if it has any, and the theorem reads
[2] Int_γ[ ∇F·dr ] = Int_{∂γ}[ F ] = F(q) - F(p),
which is what the author is doing here. (Note: dF is the dual of ∇F, so the correspondence is valid.)

Something you will see later are Green's theorem, which connects area integrals with line integrals over the boundary of that area, the divergence theorem, connected volume integrals to surface integrals over the boundary of that volume, and Stokes' theorem, connecting surface integrals to its boundary (a curve). Confusedly, the general pattern is also has a formal theorem also called Stokes' theorem.

freeziepleasy

-Also, instead of integrating from 0 to 2pi, he integrated from initial and terminal coordinates A=(1,0,0) and B=(1,0,(2pi)(b)). Why did he use those coordinates instead of t?

Well, yeah. That's the boundary of the curve, as per the fundamental theorem of line integrals (aka gradient theorem).

freeziepleasy

He didn't finish on explaining how he got to the solution (because he assumes the reader can finish the rest using F = (df/dx,df/dy, df/dz)), but if anyone can finish the problem while explaining, it would be very helpful.

Once you understand the fundamental theorem, the only part left is to find an F such that ∇F = G. This is the proper analogue of finding an antiderivative you should already be familiar with. Is that the part you need help on, or do you see that F = x²y+yz+z³/3 works?

freeziepleasy

d(s^2)=g11(u,v)(du)^2 + 2g12(u,v)dudv + g22(u,v)(dv)^2

The quantities g11,g22,g12, and g21 (defined to be equal to g12) are the elements of a symmetric matrix called metric tensor and determine the geometrical properties of the surface near any point.
coordinate lines meet orthogonally if and only if g12=0.

-Why do g11,g22,g12, and g21 equal g12?

They don't; only g21 = g12. For the 2-dimensional case, that just the same as saying g is symmetric.

freeziepleasy

-What is the meaning of metric tensor, and what does he mean by saying "determine the geometrical properties of the surface near any point"?

It measures distances differential distances. I'm not going to give you a full description, particularly since you've a book in front of you, but as an example, consider the plane in Cartesian coordinates: ds² = dx² + dy², which reads:
[3] g11 = g22 = 1, g12 = g21 = 0.
This is just the Pythagorean theorem in differential form: if x-coordinate lapses by dx and y-coordinate lapses by dy, then you've gone a distance ds. If you want to find the length of a curve, you integrate ds over that curve. If you substitute {x = r cos θ, y = r sin θ), you get ds² = dr² + r²dθ², so that (with x^1 = r, x^2 = θ):
[4] g11 = 1, g22 = r², g12 = g21 = 0.
Note: If you see something in the form dudv, this is shorthand for (1/2)dudv + (1/2)dvdu in the metric tensor proper.

freeziepleasy

3) "Cylindrical Polar and Spherical Polar Coordinates"
Cylindrical Polar Coordinates: R, psi, z
with x = Rcos(psi), y=Rsin(psi), z=z
-Why is x = to cos(psi) and y = to sin(psi)? Is it arbitrary? Can you switch so that x = Rsin(psi) and y = Rcos(psi) without changing how the cylinder would look like?

Yes, it's arbitrary, but it is also conventional--in exactly the same way that we label positive angles in the plane as counterclockwise from the positive x axis. It natural in the sense of meshing with the orientation of the typical coordinate choices, but it doesn't have any deeper meaning that than.  

[freeziepleasy]

VorpalNeko

freeziepleasy

1) "Tangential Line Integral of a Vector"
Evaluate the tangential line integrals of the vector functions:
ii) G = 2xyi + (x^2 + z)j + (y+z^2)k
along the curve C whose parametric equations are x = acost, y=asint, z=bt, where a and b are positive constants and t increases from 0 to 2pi.
answer: (8/3)(pi^3)(b^3)

OK.

freeziepleasy

The author started off by doing
G=(df/dx, df/dy, df/dz) <-where the d is partial derivative
Then he started solving for f:
f=x^2y + g(y,z).
-When he is solving for f, is he basically making a new function to take the integral from? If so, why? Why can't he just take the integral of the function as itself without any modification?

But he is. You will see this pattern again and again: the integral of a differential of a form over a region is equal to the integral of the original form over the boundary of that region (subject to some criteria such as orientability, which aren't important right now). You've already seen two particular cases of it.

The fundamental theorem of calculus. Say you're integrating dF over an interval I = [a,b], which has as a boundary the two points ∂I = {a,b}. Then:
[1] Int_I[ dF ] = Int_{∂I}[ F ] = F(b) - F(a).
You've just turned an integral over an interval to an integral over two isolated points, by looking for the function that dF is a differential of, which is equivalent to taking antiderivative.

The fundamental theorem of line integrals. Say you're integrating a curve γ. Then its boundary is the endpoints, say, {p,q}, if it has any, and the theorem reads
[2] Int_γ[ ∇F·dr ] = Int_{∂γ}[ F ] = F(q) - F(p),
which is what the author is doing here. (Note: dF is the dual of ∇F, so the correspondence is valid.)

Something you will see later are Green's theorem, which connects area integrals with line integrals over the boundary of that area, the divergence theorem, connected volume integrals to surface integrals over the boundary of that volume, and Stokes' theorem, connecting surface integrals to its boundary (a curve). Confusedly, the general pattern is also has a formal theorem also called Stokes' theorem.

Thank you.
I looked over the fundamental theorem of calculus over the internet to see if I forgot any aspects of it. I related that to what he was trying to do. I wasn't aware he was doing this. Thank you.

freeziepleasy

-Also, instead of integrating from 0 to 2pi, he integrated from initial and terminal coordinates A=(1,0,0) and B=(1,0,(2pi)(b)). Why did he use those coordinates instead of t?

Quote:

Well, yeah. That's the boundary of the curve, as per the fundamental theorem of line integrals (aka gradient theorem).

I think I worded my question incorrectly.
How did he find the coordinate points, and why would he use that instead of the t min. and t max. (0 to 2pi)? He did a previous problem with F=xyi+(y^2)j+yzk where he used parameters 0 to 2pi. I was wondering why he didn't take the same approach with this equation as well.

His next chapter is about gradients. He hasn't introduced in this chapter yet, so I am not sure how it is expected for us to know those coordinates (by memory?)...

freeziepleasy

He didn't finish on explaining how he got to the solution (because he assumes the reader can finish the rest using F = (df/dx,df/dy, df/dz)), but if anyone can finish the problem while explaining, it would be very helpful.

Quote:

Once you understand the fundamental theorem, the only part left is to find an F such that ∇F = G. This is the proper analogue of finding an antiderivative you should already be familiar with. Is that the part you need help on, or do you see that F = x²y+yz+z³/3 works?

I think I made a mistake when I typed that. He was talking about another problem. Sorry.

I kind of understand F = x²y+yz+z³/3.
What I don't understand:
I got for ....
Integral of Gxdx = (x^2)y
Integral of Gydy = (x^2)y + yz
Integral of Gzdz = (yz) + 1/3(z^3)
So.....
G = (x^2)y + (x^2)y + (yz) + (yz) + 1/3(z^3)
Not what he or you got.
My answer differs from both of yours in that I have 1 extra (x^2)y and 1 extra (yz), and I was wondering why you guys took that away.

Sorry if I used any notation of math that strays away from the thread topic on Abbreviations. I didn't have time yet to take a look at it, and I had thought a lot about your reply... I wanted to write back my questions before I read that topic.

Also, sorry if my questions may seem elementary. I am not a mathematics or physics major.

Thank you very much.
I will take another look again at your answer to my other questions.

Edit:
In case your wondering, this (to me) is a very condensed, small book. It is not as descriptive as a textbook, nor does it provide many math problems like a textbook, so anything said in there that I do not understand...I must look up online. He does not define the terms he uses, so I must look it up online. Textbooks are not offered in the public library catalog.  

[VorpalNeko]

freeziepleasy

How did he find the coordinate points, ...

By plugging the t_min and t_max into the equations defining the curve over which the integration occurs. The boundary of the curve consists of the two endpoints in space, rather than just numbers, so it's inappropriate to use t (which is just a dummy variable that parametrizes the curve).

freeziepleasy

... and why would he use that instead of the t min. and t max. (0 to 2pi)? He did a previous problem with F=xyi+(y^2)j+yzk where he used parameters 0 to 2pi. I was wondering why he didn't take the same approach with this equation as well.

Because it's easier to use the fundamental theorem in many cases, including this one. It's simply a much more powerful method.

We could use something more direct, if you really prefer, by turning it into an ordinary one-dimensional calculus problem. The path is defined as
[1] r = r(t) = , t in [0,2pi],
with the differential
[2] dr = <-a.sin t, a.cos t, b>dt,
so that taking the dot product with G = <2xy, x²+z, y+z²> gives
[3] G·dr = 2xy(-a.sin t) + (x²+z)(a.cos t) + (y+z²)(b) = b³t² + ab sin t + a cos t [ bt + 3a²cos²t - 2a² ]
after substitution of x,y,z by their values along r. Integrating with respect to t over [0,2pi] gives
[4] (8/3)(pi.b)³,
where it's useful to remember that odd powers of sine and cosine always drop out over [0,2pi].

freeziepleasy

His next chapter is about gradients. He hasn't introduced in this chapter yet, so I am not sure how it is expected for us to know those coordinates (by memory?)...

That's strange. He's giving a solution in terms of gradients to a problem before gradients are introduced? Eh... well, while not wrong, it's a dubious pedagogical practice.

freeziepleasy

What I don't understand:
I got for ....
Integral of Gxdx = (x^2)y
Integral of Gydy = (x^2)y + yz
Integral of Gzdz = (yz) + 1/3(z^3)
So.....
G = (x^2)y + (x^2)y + (yz) + (yz) + 1/3(z^3)
Not what he or you got.

Systems of partial differential equations such as this one are a bit more complicated than just integration, because on term also contributes derivatives to the other components, rather than just its own. That's why he did it step by step.

freeziepleasy

Integral of Gxdx = (x^2)y

Alright, so you see that this correctly fixes the x-component of G, yes?
This means that if G = ∇F, then F = x²y + (some function of y and z only). Any F in that form correctly reproduces the x-component of G, though not necessarily the other components. The trouble starts with the fact that the x²y term also contributes to the y-component of the gradient: the F,y = ∂F/∂y term (I'll use subscript-comma to denote partial derivatives from now on, because they're easier to type).

So what does it get us? Well, for F = x²y + G(y,z), where G is the aforementioned "some function of y and z only", we have
[5] ∇F = x²yi + (x² + G,y)j + (G,z)k.
Comparing that with the vector field G, the missing on the y-component is z; hence: G,y = z. Integrating that means a yz term, which would correctly fix the y-component term. But again, this term would contribute a derivative to the z-component, so we once again need to be careful, like so: F = x²y + yz + H(z), where H(z) is some as yet unknown function of z only.
[6] ∇F = x²yi + (x² + z)j + (y + H,z)k.
The missing term is H,z = z², so that H = z³/3. Putting it all together gives
[7] F = x²y + yz + z³/3.  

[freeziepleasy]

@VorpalNeko: Thank you very much! I understand it now.