 Quote by understand.
My calculus book states that a vector field is conservative if and only if the curl of the vector field is the zero vector. And, as far as I can tell a conservative vector field is the same as a path-independent vector field.
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"Conservative" is really a physics term, from conservative force fields. A more mathematical term is "exact differential".
The thing is, I came across this: http://www.math.umn.edu/~nykamp/m2374/readings/pathindex/
The site shows a vector field where the curl is equal to the zero vector, yet the vector field is not conservative.
As far as I can tell, saying "F is conservative iff Curl(F) = 0" contradicts the claims of the site I posted.
What conditions must be met for a vector field to be conservative?
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"F is conservative iff Curl(F) =
0" is not true. What is true is that if Curl(F)= 0
on a simply connected set then F is conservative. In the example given on that site, F is not defined at (0, 0) so is not conservative on any set that includes (0,0). If you were to integrate around any curve that does NOT have (0,0) in its interior, you would get an integral of 0- conservative field. But if the path does include (0,0) in its interior, its interior is not "simply connected" so you cannot expect to get a 0 integral.
"Simply connected" means that any closed curve can be "shrunk" to a single point while staying inside the region. Since (0,0) is not in the region (F is not defined at (0,0) so Curl(F)= 0 is not true there), any closed curve that includes (0,0) in its interior cannot be "shrunk" to a single point without passing through (0,0) which is not in the region.