To check whether a vector field depends on the path just check if the rotation is equal to 0 ?
How can I check if the integral curvilinear depends on the path ?
f is independent of path in Q if
1)$\displaystyle \oint F.dr =0 $ for all simple closed curve in Q piece wise smooth
2)F is conservative (in Q) is gradient
3)curl F = o for all (x,y,z) in Q
if one is true then all is true
all the statement are equivalent
that what my professor said to us independent of path mean the curl of the vector equal zero
but I didn't see it in the book of calculus 3) the curl is zero
this from my professor so I do not know if it is true or not but 1) and 2)
but I think it is true take this example
show that this is independent of path
$\displaystyle F(x,y)=2xy^3i + (1+2x^2y^2)j $ is independent of path
the curl is zero or as the book solve it
$\displaystyle since f(x,y)=2xy^3$ and $\displaystyle g(x,y)=(1+3x^2y^2)$
$\displaystyle \frac{\partial g}{\partial x } = 6xy^2 = \frac{\partial f}{\partial y}$ so 6 holds for all (x,y)
see this
Conservative vector field - Wikipedia, the free encyclopedia
they said every conservative vector is irrotational irrotational mean the curl is zero
If there exist a scalar function $\displaystyle f$ such that $\displaystyle F=\nabla f$ then $\displaystyle \nabla \times F = \nabla \times \nabla f = 0$
In Vector Calculos by Marsden & Tromba in the section 3.4 the Theorem: For every function $\displaystyle f \in C^2$ $\displaystyle \nabla \times \nabla f = 0 $
The demonstration is just computing $\displaystyle \nabla \times \nabla f $ and use that $\displaystyle \frac{\partial f^2}{dxdy} = \frac{\partial f^2}{dydx}$ and so on.
so for example for a conservative force, divide the closed path from Po to P1, the work done on path c1 is equal to path c2 and since they are in opposite directions they sum to 0. is there any relationship to the physics sum of all paths integral? why are the conditions: positive oriented, piecewise smooth, simple closed curve necessary conditions for this and Green's Theorem?