- #1

- 13

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is it just simple closed curve?

is F=-ysin(x)i+cos(x)j able to use the curl test?

- Thread starter mathwizeguy
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- #1

- 13

- 0

is it just simple closed curve?

is F=-ysin(x)i+cos(x)j able to use the curl test?

- #2

HallsofIvy

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[tex]\frac{\partial f}{\partial y}= \frac{\partial g}{\partial x}[/tex]

which the same as saying

[tex]curl \vec{f}= \nabla\times\vec{f}= \vec{0}[/tex] where

[tex]\vec{f}= f(x,y)\vec{i}+ g(x,y)\vec{j}[/tex].

It follows from that that the integral of f(x,y)dx+ g(x,y)dy around any closed path is 0 and that the integral from one point to another in the xy-plane is independent of the path. Which of those are you referring to?

- #3

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i was referring to the df/dy-dg/dx.

- #4

HallsofIvy

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The first obviously requires a closed path- it says so! The second does not.

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