conservative vector fieldswziller/math114f19/ch16-3-4.pdfis called conservative (or a gradient...
TRANSCRIPT
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16.3
Conservative Vector Fields
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Review:
Work C
dF r = '( )C
t dtF r C
Mdx Ndy Pdz if M N P F i j k
Outward flux across a simple closed
curve C in the plane is C
dsF n if C
Mdy Ndx M N F i j
is called conservative (or a gradient vector field) if
The function is called the of .
f
f potential
F F
F
a) if and only if is path independent:C
f dr
Fundamental theorem for line integrals :
F F1 2
= C C
dr dr F F
= if C is a path from to .
B
C A
dr dr A B F F
b) If , then ( ) ( ) C
f dr f B f A F F
is also called if represents a velocity vector field.C
d circulation F r F
=C
dsF T
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When is a vector field conservative? Question : F
Need: f f
M N fx y
F = i j i j or: and
f fM N
x y
a necessary condition: M N
y x
In 3 dimensions: f f f
M N P fx y z
F = i j k i j k
need: , and f f f
M N Px y z
necessary conditions: , , M N M P N P
y x z x z y
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Recall:
curl x y z
P N M P N M
y z z x x yM N P
i j k
F i + j k
: If and then curl 0.M N P f Theorem F = i j k F F
necessary conditions: , , M N M P N P
y x z x z y
This condition is "almost" sufficient as well.
But it is always a good test.
In 3 dimensions: f f f
M N P fx y z
F = i j k i j k
Another way to express this, which is easier to remember:
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Some terminology for curves C :A region R is called if every closed curve in R
can be shrunk to a point by curves staying in R.
simply connected Definition :
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.
Important examples :
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: If is defined in a connected and
simply connected region, then if and only if
M N
M Nf
y x
Theorem F = i j
F
: If is defined in a connected and
simply connected region, then if and only if
, , or ( ) 0
M N P
f
M N M P N P
y x z x z y
Theorem F = i j k
F
curl F
: To find the potential , integrate one at a time
, (and if in 3-space)
f
f f fM N P
x y z
Hint
A differential is called if
for some function .
Mdx Ndy Pdz exact
Mdx Ndy Pdz df f
Definition :
This is equivalent to saying that (can use same theorems)M N P f i j k
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2 4
so is a gradient field.
y y
y y
y x
M x e N y xe
M e N e
F
2, 2 y yf x y Mdx x e dx x xe G y
2,0
2,0
2,02 2
2,0
,
2
4 2 4 2 4
C
y
Work dr f x y
x xe y
F
Find the work done by the force
, 2 4 along the indicated curve.y yx y x e y xe
Example :
F i j
, ,yyf x y xe G y N x y
4y yxe G y y xe
4G y y
22G y y C
2 2, 2yf x y x xe y C
need: and f f
M Nx y
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2 2 2 22 3 2 2 3 2xy xz x y y x z z F i j j
2 2
2
2
2 3
2 2
3 2
M xy xz
N x y y
P x z z
2 22 3f Mdx xy xz dx 2 2 2 23
2x y x z ,G y z
2 22 2 2 ( , )y yx y y N f x y G y z or ( , ) 2yG y z y
2 2 2 2 2 232the potential is , ,f x y z x y x z y z C
So is a gradient field, . Now let's find .f f F F
Determine wether the given vector field is a gradient field.
If so, find a potential function.
Example :
2 2 2 2
curl 0 6 6 4 4
2 3 2 2 3 2
x y zxz xz xy xy
xy xz x y y x z z
i j k
F i + j k = 0
2 23 2 3 '( )zx z z P f x z H z
2 2 2 2 2 232
or ( , ) ( ) hence ( )G y z y H z f x y x z y H z
2 or '( ) 2 hence ( )H z z H z z
need: , and f f f
M N Px y z
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: is conservative if and only if for every closed curve C:
0C
dr
Theorem F
F
Indeed, if and goes from to and
then ( ) ( ) 0C
f C A B A B
dr f B f A
F
F
: To indicate that the line integral is over a closed curve,
we often write C C
dr dr
Note
F F
1 2
Conversely, assume 0 for any closed curve
and let and be two curves from to with
C
dr C
C C A B A B
F
1 2 1 2
Then 0
C C C C
dr dr dr
F F F
1 2
and hence
C C
dr dr F F
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2 2 2 2Compute the work performed by the force
as a particle travels around a circle of radius counter clockwise.
y x
x y x y
r
Example : F i j
parametrize the circle: cos( ), sin( )x r t y r t sin( ), cos( )dx r t dy r t
2 2 2 2 C C
y xdr dx dy
x y x y
F i j i j
2 2sin( ) cos( )
= ( sin( )) ( cos( )C
r t r tr t r t dt
r r
i j i j
2
2 2 2 2
2
0
1= sin ( ) cos ( )r t r t dt
r
2
0
= 1 2dt
is conservative since with arctan (check it!)y
f fx
Puzzle 2 : F F
is not defined at (0,0), and the plane minus the origin is not simply connected !Solution : F
(for any radius !)r
2 2 2 2
2 22 2 2 2 2 2
is conservative since
1 2x y x xx x y
x x y x y x y
Puzzle 1 : F
2 2 2 2
2 22 2 2 2 2 2
1 2 and
x y y yy x y
y x y x y x y
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Gravitational vector fieldExample :
2
GmM
rF
| r | | r |11 2 2
is the vector from the center of the sun to the planet
is the mass of the sun
is the mass of the planet
is the gravitational constant
6.674 10 G = (from 1798 )
M
m
G
N m kg
r
3/23 2 2 2
x y z
x y z
r i j kF
| r |
1/2
2 2 2 is conservative with potential . . ( , , ) f i e f x y z
x y z
Claim : F
| r |
1/2
2 2 2 Indeed: ( , , ) f x y z x y z
and similarly for and f f
y z
The work necessary to "escape" the force field from a point :pExample :
= ( ) ( ) | |
C p
dr dr f f pp
F F
3/22 2 2
3/22 2 2
hence 22
f xx x y z
x x y z
GmM
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Newton: force = mass x acceleration ''mExample : F r
What is the work performed? Question :
= '( )dr t dt F F r ' ' '
2m dt
r r1 '' '( )
2m t dt r r
since ' ' ' '' ' ' '' 2 '' ' r r r r r r r r
' ' ' hence =
2C
dr m dt
r r
F
If is also conservative, (sign is physics convention)f F F
work performed is gain in kinetic energy
2' '2 2
m mdt dt v v | v |
2 is the kinetic energy:2
mk | v |
is constant along any path ( )
(conservation of kinetic energy + potential energy)
k f tClaim : r
0
= ( ( )) ' ( ( )) ( (0))
t
C
dr k t dt k t k F r r r
since = ( ( )) ( (0)) and ( ( )) ( (0))C
dr fdr f t f dr k t k F r r F r r hence ( ( )) ( ( )) ( (0)) ( (0))k t f t k f r r r r
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16.4
Greens Theorem
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Closed Curve Line Integral C
Pdx Qdy
Closed Curve Orientation:
Counter-clockwise
Clockwise
C
Pdx Qdy
C
Pdx QdyC
Pdx Qdy
Green’s Theorem (in the plane = 2 dim.)
Suppose that is a simple piecewise smooth closed curve.C
C is the boundary of a region .R If , , , and are continuous on , then y xP Q P Q R
x yR C
Q P dA Pdx Qdy
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2 , 3P y Q xy
3 , 2x yQ y P y
2 3GThm
C R
y dx xydy ydA 2
0 1
sinr rdrd
2
2
0 1
sin d r dr
14
3
2
3
0
1
cos3
r
8 1
cos cos03
7
1 13
2Compute the closed line intgeral 3
where C is the indicated curve.
C
y dx xydyExample : x yR CQ P dA Pdx Qdy
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x yC R
Pdx Qdy Q P dA
If 1, then x yQ P
R
dA R
The area of the
interior region
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 1choose: , 2
2 2x yP y Q x Q P
We can use Green's theorem to compute areas:
1( )
2R
area R xdy ydx
stands for the boundary
of the region
R
R
2R
dA x yC R
Pdx Qdy Q P dA
1 1
2 2Pdx Qdy xdy ydx
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aa
b
b
2 2
2 2The parametrization of the ellipse 1:
x y
a b
cosx a t
siny b t
1Area
2C
xdy ydx 0 2t
2
0
c1
cos sin2
os sina t bb t a tdtdt t
sindx a tdt
cosdy b tdt
2
2 2
0
1cos sin
2ab t ab t dt
2
2 2
0
cos sin 2
abt t dt
2
02
abdt
2ab
2 ab
2 2
2 2Find the area enclosed by the ellipse 1.
x y
a b Example :
positively oriented