Problem with vector field proof

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Homework Help Overview

The discussion revolves around proving a relationship between two vector fields, F and G, given that their difference is the gradient of a real-valued function μ. The problem is situated within the context of vector calculus, specifically focusing on line integrals and closed curves in the xy-plane.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the relationship between the vector fields and question the notation used for line integrals. There is discussion about the nature of the curve C and its closure, as well as the application of Green's theorem in the context of the problem.

Discussion Status

Participants are actively engaging with the problem, raising questions about the setup and notation. Some have suggested using Green's theorem to approach the proof, while others express uncertainty about how to proceed. There is no explicit consensus on the next steps, but various lines of reasoning are being explored.

Contextual Notes

There is mention of the curve C being closed, which is significant for the application of Green's theorem. The notation for the line integral has also been questioned, indicating potential confusion in the problem's presentation.

Roidin
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Suppose that F and G are vector fields and that F-G = ▽μ for some real-valued function μ(x,y). Prove that

∫ F.dx = ∫ G.dx for all piecewise smooth curves C in the xy-plane

I just need some help in getting started really. Thanks
 
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Roidin said:
Suppose that F and G are vector fields and that F-G = ▽μ for some real-valued function μ(x,y). Prove that

∫ F.dx = ∫ G.dx for all piecewise smooth curves C in the xy-plane

I just need some help in getting started really. Thanks

If ##dx## is a scalar, what does ##\vec F \cdot dx## mean? Have you stated the problem correctly? Is C a closed curve?
 
Thanks LCKurtz,

F.dx is the line integral of the vector field and yes C is closed.
 
Roidin said:
Thanks LCKurtz,

F.dx is the line integral of the vector field and yes C is closed.

That's unusual notation. And since it matters that C is closed, don't you think it would have been good to mention that?

Let's call ##\vec H = \vec F -\vec G##. So say ##\vec H = \langle h_1,h_2\rangle## and you are talking about$$
\oint_C \vec H \cdot d\vec r =\oint_C h_1\, dx + h_2\, dy$$So now if you use ##\vec H(x,y) =\nabla \mu(x,y)## what happens?
 
Last edited:
Hi I am doing a similar problem so sooner than start a new thread...
\oint_C \vec H \cdot d\vec r =\oint_C \frac{dμ}{dx}\, dx + \frac{dμ}{dy}\, dy?
 
gtfitzpatrick said:
Hi I am doing a similar problem so sooner than start a new thread...
\oint_C \vec H \cdot d\vec r =\oint_C \frac{dμ}{dx}\, dx + \frac{dμ}{dy}\, dy?

Yes. Now do you see how to argue that it is zero?
 
No. in a word.
I thought i might be able to work something out using greens theorem(as the curve is closed) but its not working out...
 
gtfitzpatrick said:
Hi I am doing a similar problem so sooner than start a new thread...
\oint_C \vec H \cdot d\vec r =\oint_C \frac{dμ}{dx}\, dx + \frac{dμ}{dy}\, dy?

LCKurtz said:
Yes. Now do you see how to argue that it is zero?

gtfitzpatrick said:
No. in a word.
I thought i might be able to work something out using greens theorem(as the curve is closed) but its not working out...

Green's theorem is what you want. What do you get and why don't you think it works?

[Edit]I have to go for a couple of hours. I'll check back and see if you have it by then.
 
Last edited:
\oint_C \vec H \cdot d\vec r =\oint_C \frac{dμ}{dx}\, dx + \frac{dμ}{dy}\, dy = \int\int\frac{d^2μ}{dxdy}-\frac{d^2μ}{dxdy} dxdy = 0 but this can only be true if \oint F.dr = \oint G.dr qed?
 
  • #10
gtfitzpatrick said:
\oint_C \vec H \cdot d\vec r =\oint_C \frac{dμ}{dx}\, dx + \frac{dμ}{dy}\, dy = \int\int\frac{d^2μ}{dxdy}-\frac{d^2μ}{dxdy} dxdy = 0 but this can only be true if \oint F.dr = \oint G.dr qed?

Yes. It would be more clear if the left side had started with$$
\oint_C \vec F\cdot d\vec R -\oint_C\vec G\cdot d\vec R =\oint_C (\vec F-\vec G)\cdot d\vec R=\oint_C\vec H\cdot d\vec R =\ ...$$
 

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