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Math 114 – Rimmer
16.1/16.2 Vector Fields
and Line Integrals
Section 16.2 Vector Fields
Velocity vector fields showing the wind speed and direction
Math 114 – Rimmer
16.1/16.2 Vector Fields
and Line Integrals
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Math 114 – Rimmer
16.1/16.2 Vector Fields
and Line Integrals
Math 114 – Rimmer
16.1/16.2 Vector Fields
and Line Integrals
( ), ,x y y x= −F
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Math 114 – Rimmer
16.1/16.2 Vector Fields
and Line Integrals
( ), ,sinx y y x=F ( ) ( ) ( )2 2, ln 1 , ln 1x y y y= + +F
Math 114 – Rimmer
16.1/16.2 Vector Fields
and Line Integrals
) ( ) , ,A x y y x= −F
) ( ) , 1,sinB x y y=F
) ( ) , 2, 1C x y x x= − +F
) ( )1
, ,D x y yx
=F
II
IV
I
III
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Math 114 – Rimmer
16.1/16.2 Vector Fields
and Line Integrals
( ), , , ,x y z y z x=F ( ), , , 2,x y z y x= −F ( ), , , ,4
y x zx y z
z z
−=F
Math 114 – Rimmer
16.1/16.2 Vector Fields
and Line Integrals
Physics applications
Fluid dynamics
Aerodynamics
Velocity Field
Dynamics
Thermodynamics
Force Field
Gravitational Field
Electric Field
Magnetic Field
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Math 114 – Rimmer
16.1/16.2 Vector Fields
and Line Integrals
Gradient Vector Field
( ),f x y
( ) ( ) ( ), , , ,x yf x y f x y f x y∇ =
( ) 2 3,f x y x y y= − ( ) 2 2, 2 , 3f x y xy x y∇ = −
contour map with of
with the gradient field
f
f∇ As we saw in 15.6, the gradient vectors
are perpendicular to the level curves
Gradient vectors are long where the
level curves are close and short where
they are far apart
Directional derivative
maximum valuef= ∇
tightly packed level curves steep surface longer gradient vectors⇒ ⇒
Math 114 – Rimmer
16.1/16.2 Vector Fields
and Line Integrals
f= ∇F
A vector field is called if it is the
gradient of some scalar function.
F conservative
If there exists a function such thatf , then is conservative.F
is called the for .f potential function F
In 17.3 we learn how to tell whether a vector
field is conservative and how to find when it is.f
( ) 2 3,f x y x y y= − 2 22 , 3xy x y= −F
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Math 114 – Rimmer
16.1/16.2 Vector Fields
and Line Integrals
Section 16.1 Line Integrals
( ) ( )
Parametric Curve
,x f t y g t= =
[ ]
and continuous
for in ,
f g
t a b
′ ′ Consists of a finite
number of smooth curves
Starts and ends at the same
point and doesn't cross itself
Starts and ends
at the same pt.
Orientation of the curve
Math 114 – Rimmer
16.1/16.2 Vector Fields
and Line Integrals
Line Integral
Definite Integral
Geometric Interpretation of a Line Integral
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Math 114 – Rimmer
16.1/16.2 Vector Fields
and Line Integrals( ) ( ), ,C
P x y dx Q x y dy+∫
( ) ( )
Parametrize the path
, x f t y g t a t b= = ≤ ≤
( ) ( ), dx f t dt dy g t dt′ ′= =
( ) ( ) ( )( ) ( )
( ) ( ) ( )( ) ( )
Substitute everything
, ,
, ,
P x y P f t g t U t
Q x y Q f t g t V t
= =
= =
( ) ( ) ( ) ( )b
a
U t f t dt V t g t dt′ ′= +∫
Work done by a vector field as its point of application
moves along from to .C A B
F
Calc II
C line
W d
→
= ⋅F
Now
the parametrization of as a vector
C
C curve
C
W d
→
= ⋅∫
r
F r
Math 114 – Rimmer
16.1/16.2 Vector Fields
and Line Integrals
( ) ( )
( ) ( )
Evaluate
on : line segments from 0,0 to 1,0
and from 1,0 to 1,1
C
ydx xdy
C
+∫
1 2C C C
ydx xdy ydx xdy ydx xdy+ = + + +∫ ∫ ∫
1
0: 0 1
0
x t yC t
dx dt dy
= =≤ ≤
= =
( )1
1
0
0 0 0C
ydx xdy dt+ = + =∫ ∫
2
1: 0 1
0
x y tC t
dx dy dt
= =≤ ≤
= =
( )2
1
0
0 1 1C
ydx xdy dt+ = + =∫ ∫
1C
ydx xdy+ =∫
1C
2C
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Math 114 – Rimmer
16.1/16.2 Vector Fields
and Line Integrals
( )2 2Evaluate 2 on the given closed curve .
C
x y dx xydy C+ −∫
1C
2C
2y x=
y x=
( ) ( ) ( )1 2
2 2 2 2 2 22 2 2C C C
x y dx xydy x y dx xydy x y dx xydy+ − = + − + + −∫ ∫ ∫
Math 114 – Rimmer
16.1/16.2 Vector Fields
and Line Integrals
( )2 2Evaluate 2 on the given closed curve .
C
x y dx xydy C+ −∫2
1 : 0 12
x t y tC t
dx dt dy tdt
= =≤ ≤
= =
( ) ( ) ( )1
1 1
2 2 2 4 4 2 4
0 0
2 4 3C
x y dx xydy t t t dt t t dt+ − = + − = −∫ ∫ ∫
2 : starts at 1 and ends at 01
2
x t y t
C tdx dt dy dt
t
= =
= =
( ) ( )2
0
2 2 2
1
2C
x y dx xydy t t t dt+ − = + −∫ ∫11 3
2
0 0
1
3 3
tt dt= − = − = −∫
( )2 2 4 12
15 3C
x y dx xydy−
+ − = −∫4 5 9
15 15
3
5
− − −=
−= =
1C
2C
2y x=
y x=
13 5
0
3 1 3 5 9 4
3 5 3 5 15 15
t t − −= − = − = =
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Math 114 – Rimmer
16.1/16.2 Vector Fields
and Line Integrals
( )
( ) ( ) ( ) ( )
2Find the work done by the force , 2 4 acting
along the piecewise smooth curve consisting of line segments
from 2, 2 to 0,0 and from 0,0 to 2,3 .
x y xy y= +
−
F i j
1C
2C
1 2C C C
Work dr dr dr= ⋅ = ⋅ + ⋅∫ ∫ ∫F F F
Math 114 – Rimmer
16.1/16.2 Vector Fields
and Line Integrals
( )
( ) ( ) ( ) ( )
2Find the work done by the force , 2 4 acting
along the piecewise smooth curve consisting of line segments
from 2,2 to 0,0 and from 0,0 to 2,3 .
x y xy y= +
−
F i j
1C
dr⋅∫ F
1 : 2 0x t y t
C tdx dt dy dt
= = −− ≤ ≤
= = −
2 2
1
1, 1
on : 2 ,4
r t t dr dt
C t t
= − ⇒ = −
−
i j
F
( )1
0
2 2
2
2 4C
dr t t dt−
⋅ = − −∫ ∫F
( )( )0
02 3
22
6 2 2 0 8 16t dt t−
−
= − = − = − − − = −∫
2
3
2: 0 2
3
2
x t y t
C t
dx dt dy dt
= =
≤ ≤
= =
2 2
2
3 3 1,
2 2
on : 3 ,9
r t t dr dt
C t t
= + ⇒ =i j
F
2 2
2
2 2 2
0
27 333
2 2C C
dr t t dt t dt
⋅ = + =
∫ ∫ ∫F
( )2
3
0
11 118 0 44
2 2t= = − =
216 44 8C
Work dr= ⋅ = − + =∫F
2C
dr⋅∫ F
