Finally, we see that independence of path is equivalent to the vector field being conservative.
Theorem. Let F be a continuous vector field on, and C a curve in, an open, connected region. Then
R
C
F · dr
is independent of path if and only if F is conservative.
Proof of Theorem (for 2 dimensions). If F is conservative with potential function f then the Fundamen-
tal Theorem tells us that
R
C
F · dr depends only on the endpoints of C.
Conversely, suppose that
R
C
F · dr is independent of path. By the previous Theorem,
R
C
F · dr is in-
dependent of path for any curve C in D. Choose a point A and define f (x, y) =
R
C
F · dr where C is
any curve joining A to (x, y). The fucntion f is well-defined because
R
C
F · dr is independent of path.
We claim that f is a potential function for F.
Choose (x, y) ∈ D. Since D is open, there exists a point (x
1
, y) ∈ D
such that x
1
< x. Let C
1
be a path from A to (x
1
, y) and C
2
the line
segment thence to (x, y). Then
f (x, y) =
Z
C
1
F · dr +
Z
C
2
F · dr
Since x
1
is constant, the first integral is independent of x and so
∂ f
∂x
=
∂
∂x
Z
C
2
F · dr
Now let F =
P
Q
. Along the curve C
2
we have y constant, hence dy = 0. Therefore
∂ f
∂x
=
∂
∂x
Z
(x,y)
(x
1
,y)
P dx + Q dy =
∂
∂x
Z
(x,y)
(x
1
,y)
P dx =
d
dx
Z
x
x
1
P(t, y) dt = P(x, y)
by the Fundamental Theorem of Calculus.
A similar argument (choose (x, y
1
) ∈ D with y
1
< y) shows that Q(x, y) =
∂ f
∂y
. Putting this together
we see that ∇ f = F and so F is conservative.
The proof in three dimensions requires a similar third argument for
∂ f
∂z
.
Example Evaluate the line integrals
R
C
i
2y
x
· dr over the same
line and parabola as before.
For the first curve we have r(t) =
(
t
t
)
, whence
Z
C
1
2y dx + x dy =
Z
1
0
3t dt =
3
2
For the second curve we have r(t) =
t
t
2
, and so
Z
C
2
2y dx + x dy =
Z
1
0
2t
2
dt +
Z
1
0
2t
2
dt =
4
3
6=
3
2
Considering the strength of the arrows in the picture it should be clear why
R
C
2
<
R
C
1
. The fact that
these integrals give different values tells us that F =
2y
x
is not conservative.
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