# Dependence of the path

• June 8th 2009, 09:05 AM
Apprentice123
Dependence of the path
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 ?

• June 8th 2009, 09:17 AM
Ruun
If $\nabla \times F = 0$ then F is a conservative field, and integral doesn't depend on the path, only on the initial and final position. Also the reciprocal is true
• June 8th 2009, 09:19 AM
Amer
f is independent of path in Q if

1) $\oint F.dr =0$ 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
• June 8th 2009, 09:34 AM
Apprentice123
Quote:

Originally Posted by Amer
f is independent of path in Q if

1) $\oint F.dr =0$ 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

$\int_S yzdx +xzdy + yx^2dz$

curl is different 0, then is dependent the path

Correct ?
• June 8th 2009, 10:15 AM
Apprentice123
Correct the example ?
• June 8th 2009, 10:18 AM
Amer
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

$F(x,y)=2xy^3i + (1+2x^2y^2)j$ is independent of path

the curl is zero or as the book solve it

$since f(x,y)=2xy^3$ and $g(x,y)=(1+3x^2y^2)$

$\frac{\partial g}{\partial x } = 6xy^2 = \frac{\partial f}{\partial y}$ 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
• June 8th 2009, 10:45 AM
Apprentice123
Quote:

Originally Posted by Amer
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

$F(x,y)=2xy^3i + (1+2x^2y^2)j$ is independent of path

the curl is zero or as the book solve it

$since f(x,y)=2xy^3$ and $g(x,y)=(1+3x^2y^2)$

$\frac{\partial g}{\partial x } = 6xy^2 = \frac{\partial f}{\partial y}$ 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

Ok. Thank you
• June 8th 2009, 12:01 PM
Ruun
If there exist a scalar function $f$ such that $F=\nabla f$ then $\nabla \times F = \nabla \times \nabla f = 0$

In Vector Calculos by Marsden & Tromba in the section 3.4 the Theorem: For every function $f \in C^2$ $\nabla \times \nabla f = 0$

The demonstration is just computing $\nabla \times \nabla f$ and use that $\frac{\partial f^2}{dxdy} = \frac{\partial f^2}{dydx}$ and so on.
• April 14th 2013, 01:39 PM
mathlover10
Re: Dependence of the path
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?