16.
Conservative Vector
Fields
Review:
Work F dr = F Tds= F r '(t )dt Mdx Ndy Pdz
C C C C if F Mi Nj Pk
F dr is also called circulation if F represents a velocity vector field.
C
Outward flux across a simple closed
curve C in the plane is F nds Mdy Ndx if F Mi Nj
C C
F is called conservative (or a gradient vector field) if F f
The function f is called the potential of F.
Fundamental theorem for line integrals :
a) F f if and only if F dr is path independent: F dr = F dr
C1 C2
C
b) If F f , then F dr f ( B) f ( A)
C
B
F dr = F dr
C A
if C is a path from A to B.
Question : When is a vector field F conservative?
f f f f
Need: F = Mi Nj f i j or: M and N
x y x y
M N
a necessary condition:
y x
f f f
In 3 dimensions: F = Mi Nj Pk f i j k
x y z
f f f
need: M , N and P
x y z
M N M P N P
necessary conditions: , ,
y x z x z y
Recall: f f f
In 3 dimensions: F = Mi Nj Pk f i j k
x y z
M N M P N P
necessary conditions: , ,
y x z x z y
Another way to express this, which is easier to remember:
i j k
P N M P N M
curl F
x
y
z i + j k
y z z x x y
M N P
Theorem: If F = Mi Nj Pk and F f then curl F 0.
This condition is "almost" sufficient as well.
But it is always a good test.
Some terminology for curves C :
Definition : A region R is called simply connected if every closed curve in R
can be shrunk to a point by curves staying in R.
Important examples : The plane minus the origin is not simply connected.
3-space minus the origin is simply connected.
3-space minus the z-axis is not simply connected.
Theorem: If F = Mi Nj is defined in a connected and
M N
simply connected region, then F f if and only if
y x
Theorem: If F = Mi Nj Pk is defined in a connected and
simply connected region, then F f if and only if
M N M P N P
, , or curl(F) 0
y x z x z y
Hint: To find the potential f , integrate one at a time
f f f
M , N (and P if in 3-space)
x y z
Definition : A differential Mdx Ndy Pdz is called exact if
Mdx Ndy Pdz df for some function f .
This is equivalent to saying that Mi Nj Pk f (can use same theorems)
Example : Find the work done by the force
F x, y 2 x e y i 4 y xe y j along the indicated curve.
M 2 x e y N 4 y xe y
M y e y N x e y so F is a gradient field.
f f
need: M and N
x y
f x, y Mdx 2 x e y dx x 2 xe y G y
f y x, y xe y G y N x, y Work F dr f x, y
2,0
2,0
C
xe y G y 4 y xe y 2,0
x xe
2 y
2y 2
2,0
G y 4 y
4 2 4 2 4
G y 2 y2 C
f x, y x 2 xe y 2 y 2 C
Example : Determine wether the given vector field is a gradient field.
If so, find a potential function. M 2 xy 2 3xz 2
F 2 xy 3xz i 2 x y 2 y j 3x z 2 z j
2 2 2 2
N 2x2 y 2 y
P 3x 2 z 2 z
i j k
curl F
x
y
z 0 i + 6 xz 6 xz j 4 xy 4 xy k = 0
2 xy 2 3 xz 2 2 x 2 y 2 y 3x 2 z 2 z
f f f
So F is a gradient field, F f . Now let's find f . need: M , N and P
x y z
f Mdx 2 xy 2 3xz 2 dx x y 32 x z G y, z
2 2 2 2
2 x 2 y 2 y N f y 2 x 2 y G y ( y, z ) or G y ( y, z ) 2 y
or G ( y, z ) y 2 H ( z ) hence f x 2 y 2 32 x 2 z 2 y 2 H ( z )
3 x z 2 z P f z 3x z H '( z )
2 2
or H '( z ) 2 z hence H ( z ) z 2
the potential is f x, y, z x 2 y 2 32 x 2 z 2 y 2 z 2 C
Note : To indicate that the line integral is over a closed curve,
we often write F dr F dr
C C
Theorem: F is conservative if and only if for every closed curve C:
F dr 0
C
Indeed, if F f and C goes from A to B and A B
then F dr f ( B) f ( A) 0
C
Conversely, assume F dr 0 for any closed curve
C
C
and let C1 and C2 be two curves from A to B with A B
Then 0
C1 C2
F dr F dr F dr
C1 C2
and hence F dr F dr
C1 C2
y x
Example : Compute the work performed by the force F i j
x y
2 2
x y
2 2
as a particle travels around a circle of radius r counter clockwise.
parametrize the circle: x r cos(t ), y r sin(t ) dx r sin(t ), dy r cos(t )
y x
C F dr C x 2 y 2 x 2 y 2
i j dxi dyj
r sin(t ) r cos(t )
= i j ( r sin(t ))i (r cos(t ) jdt
C
2 2
r r
2 2
= r 2
1 2 2
r sin (t ) r 2
cos 2
(t ) dt = 1dt 2 (for any radius r !)
0 0
Puzzle 1 : F is conservative since
x 1 x y x 2 x x 2 y 2 y 1 x y y 2 y
2 2 2 2
x2 y 2
and 2 2
x x 2 y 2 x 2
y
2 2
x 2
y
2 2 y x y x 2
y
2 2
x 2
y
2 2
Solution : F is not defined at (0, 0), and the plane minus the origin is not simply connected !
y
Puzzle 2 : F is conservative since F f with f arctan (check it!)
x
r is the vector from the center of the sun to the planet
Example : Gravitational vector field
M is the mass of the sun
GmM r
F m is the mass of the planet
| r |2 | r | G is the gravitational constant
G = 6.674 1011 N m 2 kg 2 (from 1798)
r xi yj zk
F GmM
x y z 2 3/2
3
|r| 2 2
Claim : F is conservative with potential f i.e. f ( x, y, z )
x2 y 2 z 2
1/2
|r|
f ( x, y , z ) x 2 y 2 z 2
1/2
Indeed:
f x f f
hence 2x x y z
2 2 2 3/2
and similarly for and
x 2 x 2
y 2
z
2 3/2 y z
Example : The work necessary to "escape" the force field from a point p :
F dr = F dr f () f ( p)
C p
| p|
Example : Newton: force = mass x acceleration F mr ''
Question : What is the work performed?
1 r ' r ' '
F dr = F r '(t )dt m r '' r '(t )dt 2 m 2 dt
since r ' r ' ' r '' r ' r ' r '' 2r '' r '
r ' r ' ' dt
hence F dr =m 2
m
2 v v ' dt
m
2 | v |2
' dt
C
t
m
k | v |2 is the kinetic energy:
2
C
F dr = k (r (t )) ' dt k (r (t )) k (r (0))
0
work performed is gain in kinetic energy
If F is also conservative, F f (sign is physics convention)
Claim : k f is constant along any path r (t )
(conservation of kinetic energy + potential energy)
since F dr = fdr f (r (t )) f (r (0)) and F dr k (r (t )) k (r (0))
hence k (r (t )) f (r (t )) k (r (0)) f (r (0))
C
16.4
Greens Theorem
Closed Curve Line Integral Pdx Qdy
C
Closed Curve Orientation:
Counter-clockwise
Pdx Qdy
C
Clockwise
C
Pdx Qdy Pdx Qdy
C
Green’s Theorem (in the plane = 2 dim.)
Suppose that C is a simple piecewise smooth closed curve.
C is the boundary of a region R. If P, Q, Py , and Qx are continuous on R, then
Q
R
x Py dA Pdx Qdy
C
Example : Compute the closed line intgeral y 2 dx 3 xydy Q
R
x Py dA Pdx Qdy
C
C
where C is the indicated curve.
P y 2 , Q 3xy
Qx 3 y, Py 2 y
2
GThm
y 2 dx 3xydy ydA r sin rdrd
C R 0 1
2
r
2 3
cos 0
sin d r dr
2
0 1 3 1
8 1 7 14
cos cos 0 1 1
3 3 3
We can use Green's theorem to compute areas:
If Qx Py 1, then
Pdx Qdy Q
C R
x Py dA dA The area of the
R interior region R
If you have the parametrization of a closed curve and want to find the enclosed
area then you can use this consequence of Green's Theorem to set up the line integral.
1 1
choose: P y, Q x Qx Py 2
2 2
Pdx Qdy Q
C R
x Py dA 2dA
R
1 1 1 R stands for the boundary
Pdx Qdy
2
xdy ydx
2
area ( R)
2 R
xdy ydx
of the region R
x2 y 2
Example : Find the area enclosed by the ellipse 2 2 1.
a b
b
x2 y 2
The parametrization of the ellipse 2 2 1:
a a
a b
x a cos t dx a sin tdt
b
y b sin t dy b cos tdt
0 t 2
1
Area xdy ydx positively oriented
2C
2
1
2 a cos t b cos tdt b sin t a sin tdt
0
2 2
cos t sin 2 t dt
1 ab
ab cos t ab sin t dt
2 2 2
2 0 2 0
2
ab
ab ab
2 0 dt 2 2
