VC.04 through VC.08 Highlights

Report
VC.04-VC.08 Redux
Vector Fields Acting on a Curve
VC.04: Find a surface with the given
gradient (Ex 1 From VC.04 Notes)
Example 1: Given f(x, y)    x  1,  y  1  , find a possible
equation for the surface f(x, y).
x2 y 2
f(x, y)  

 x  y?
2
2
VC.04: Gradient Fields
Example 2: Plot the gradient field for f(x, y) 
2
f  e x
 y2
xy
e
x2  y2
:
1  2x(x  y),1  2y(x  y)
How many points stand out to
you in this vector field?
This vector field has a source and
a sink.
What do the source and sink
correspond to on the surface?
Source  Local Minimum
Sink  Local Maximum
VC.04: Gradient Fields
Example 2: Finally, compare the gradient field and f(x, y ) 
xy
e
x2  y2
:
VC.04: Breaking Field Vectors
into Two Components
Let Field(x, y)  (m(x, y),n(x, y) be a vector field acting on the
curve (x(t),y(t)).
Forward/Backward Push:
Left/Right Push:
The push of the field
The push of the field
vector in the direction
of the tangent vector
to the curve:
vector in the direction
of the normal vector
to the curve:
 Field(x(t),y(t))  (x'(t),y'(t)) 

 (x'(t),y'(t))
(x'(t),y'(t))

(x'(t),y'(t))


 Field(x(t),y(t))  (y'(t),  x'(t)) 

 (y'(t),  x'(t))
(y'(t),

x'(t))

(y'(t),

x'(t))


"The flow of the vector field
"The flow of the vector field
ALONG the curve."
ACROSS the curve."

Let the vector field (3x2  3x)(y  3),(3y 2  3y)(x  3)

act on the ellipse (x(t), y(t))  ( 4,2)  (4cos(t),2sin(t)).
Use the push of the field vectors in the direction of the
tangent vectors to the curve to determine whether
the net flow ALONG the curve is clockwise or counterclockwise:
 Field(x(t),y(t))  (x'(t),y'(t)) 
Plot 
 (x'(t),y'(t))
 (x'(t),y'(t))  (x'(t),y'(t)) 
The net flow of the vector field
ALONG the curve is clockwise.

Let the vector field (3x2  3x)(y  3),(3y 2  3y)(x  3)

act on the ellipse (x(t), y(t))  ( 4,2)  (4cos(t),2sin(t)).
Use the push of the field vectors in the direction of the normal
vectors to the curve to determine whether the net flow ACROSS
the curve is "inside to outside" or "outside to inside."
 Field(x(t),y(t))  (y'(t),  x'(t)) 
Plot 
 (y'(t),  x'(t))
 (y'(t),  x'(t))  (y'(t),  x'(t)) 
The net flow of the vector
field ACROSS the curve is
"inside to outside."
VC.04: Flow Along and Flow Across
"The flow of the vector field
"The flow of the vector field
ALONG the curve."
ACROSS the curve."
 Field(x(t),y(t))  (x'(t),y'(t)) 
Plot 
 (x'(t),y'(t))
(x'(t),y'(t))

(x'(t),y'(t))


Counterclockwise:
 Field(x(t),y(t))  (y'(t),  x'(t)) 
Plot 
 (y'(t),  x'(t))
(y'(t),

x'(t))

(y'(t),

x'(t))


Inside to outside:
Clockwise:
Outside to Inside:
*Could be 0!
*Could be 0!
For this chapter, our only measurement tool
is to plot and then eyeball the results...
VC.05: Measuring the Flow of a Vector Field
ALONG a Curve
If C is a closed curve with a
counterclockwise parameterization:
If C is not closed:
Integral :
Integral :
b
 Field(x(t), y(t))  (x '(t), y '(t))dt
a
b

  m(x(t), y(t))x '(t)  n(x(t), y(t))y '(t)  dt
a

Ñ
 C m(x, y)dx  n(x, y)dy
These integrals are used to compute net
flow of the vector field along the closed
curve (clockwise or counterwise). They
can be modified to measure flow of the
vector field across the closed curve (inside
to outside or outside to inside) with relative
ease.
b
 Field(x(t), y(t))  (x '(t), y '(t))dt
a
b

  m(x(t), y(t))x '(t)  n(x(t), y(t))y '(t)  dt
a

 m(x, y)dx  n(x, y)dy
C
Without a closed curve, the integral is
measured whether the net flow along the
open curve is in the direction of
parameterization or against it, and whether
the net flow across the open curve is from
“above to below” or “below to above.”
These integrals are usually called line integrals or path
integrals.
VC.05: Measuring the Flow of a Vector Field
ALONG a Curve
Let C be a closed curve with a COUNTERCLOCKWISE parameterization :
Ñ
 C m(x, y)dx  n(x, y)dy  0 describes a net flow of the vector field
along the curve in the counterclockwise direction.
Ñ
 C m(x, y)dx  n(x, y)dy  0 describes a net flow of the vector field
along the curve in the clockwise direction.
Ñ
 C m(x, y)dx  n(x, y)dy can equal 0.
VC.05: Measuring the Flow of a Vector
Field Along Another Closed Curve
2
 x 1
2
Let Field(x, y)  (3y,  x ) be a vector field acting on the ellipse 
  y  1. Compute Ñ
 C m(x, y)dx  n(x, y)dy :
 2 
2
Component of Field
Vectors in the
Direction of the
Field and Curve :
Vectors on Curve :
Tangent Vectors
VC.05: Measuring the Flow of a Vector
Field Along Another Closed Curve
Let Field(x, y)  (3y,  x 2 ) be a vector field acting on the
2
 x 1
2
ellipse 
m(x, y)dx  n(x, y)dy :
  y  1. Compute Ñ

C
 2 
 Counterclockwise
Field(x, y)   m(x, y),n(x, y)  (3y, x ) E(t)  (2cos(t), sin(t))  (1, 0)
2
b
Ñ
 C m(x, y)dx  n(x, y)dy    m(x(t), y(t))x'(t)  n(x(t), y(t))y'(t)  dt
a
2

2
3
2
cos(t)

4
cos
(t)

4cos
(t)

6
si
n
(t)dt

0
  10 
Negative! The net flow of the vector field along the curve is
against the direction of the parameterization (clockwise).
VC.05: Measuring the Flow of a Vector Field
ACROSS a Curve
If C is a closed curve with a
counterclockwise parameterization:
Integral :
If C is not closed:
Integral :
b
b
 Field(x(t), y(t))  (y '(t),  x '(t))dt
a
  n(x(t), y(t))x '(t)  m(x(t), y(t))y '(t)  dt

Ñ
 n(x, y)dx  m(x, y)dy

a

a
b
b

 Field(x(t), y(t))  (y '(t),  x '(t))dt
C
With a closed curve, the integral
measures whether the net flow ACROSS
the closed curve is from “inside to outside”
or from “outside to inside.”
  n(x(t), y(t))x '(t)  m(x(t), y(t))y '(t)  dt
a
 n(x, y)dx  m(x, y)dy
C
Without a closed curve, the integral
measures whether the net flow ACROSS
the open curve is from above to below the
curve or from below to above.
VC.05: Measuring the Flow of a Vector
Field Across Another Closed Curve
Let Field(x, y)  (  x  cos(y),  y  sin(x)) be a vector field acting on
the circle x2  y2  1. Compute Ñ
 n(x, y)dx  m(x, y)dy :
C
 Counterclockwise
m(x, y),n(x, y)  (x  cos(y), y  sin(x)) c(t)  (cos(t), sin(t))
Ñ
 C n(x, y)dx  m(x, y)dy    n(x(t), y(t))x'(t)  m(x(t), y(t))y'(t)  dt
b
a
  2
Negative! The net flow of the vector field
across the curve is from outside to inside.
VC.05: The Gradient Test
A vector field, Field(x,y)   m(x, y),n(x, y  , is a gradient
field if and only if:
m(x, y) n(x, y)

y
x
(The Gradient Test)
Proof of "if" Part of Theorem:


If Field(x,y) is a gradient field, then Field(x,y)= fx , fy .
So fxy  fyx .
VC.05: The Flow of a Gradient Field Along
a Closed Curve
Let Field(x,y)= m(x, y),n(x, y)  be a gradient field, and let C be a
simple closed curve with a parameterization (x(t),y(t)) for a  t  b.
b
1)  Field(x(t), y(t))  (x'(t), y'(t))dt  0
a
2) Ñ
 m(x, y)dx  n(x, y)dy  0
C
3) The flow of a gradient field along a simple closed curve is 0.
Why is this intuitively true?
How do we know a closed curve can't be a trajectory
of a gradient field?
Is the flow of a gradient field ACROSS a closed curve 0?
VC.05: Path Independence: The Flow of a
Gradient Field Along an Open Curve
Let Field(x,y)= m(x, y),n(x, y)  be a gradient field, and let C1 and
C2 be different curves that share the same starting and ending point:
 m(x, y)dx  n(x, y)dy   m(x, y)dx  n(x, y)dy
C1
C2
A gradient field is said to be
path independent. The flow
of the vector field along any
two curves connecting two
points is the same...
VC.06: The Gauss-Green Formula
Let R be a region in the xy-plane whose boundary is parameterized
by (x(t),y(t)) for tlow  t  thigh . Then the following formula holds :
thigh

tlow
 n m 
m  x(t), y( t) x'(t)  n  x(t), y( t) y'(t) dt   

 dA
x y 
R 

With the proper interpretation, we can use this formula
to help us compute flow along/across measurements!!
The basic interpretation of Gauss-Green is that it is a
correspondence between a line integral of a closed
curve and a double integral of the interior region of
the closed curve. So for this to work correctly, we just
need to make sure the vector field has no
singularities in the interior region!
VC.06: Measuring the Flow of a Vector Field
ALONG a Closed Curve
Let C be a closed curve with a counterclockwise parameterization. Then the net
flow of the vector field ALONG the closed curve is measured by:
b
 Field(x(t), y(t))  (x '(t), y'(t))dt
a
b

  m(x(t), y(t))x '(t)  n(x(t), y(t))y'(t)  dt
a

Ñ
 C m(x, y)dx  n(x, y)dy
Let region R be the interior of C. If the vector field has no singularities in R, then
we can use Gauss-Green:
 n m 
  

 dx dy
x y 
R 

 rotField dx dy
R
Let rotField(x, y) 
n m

.
x y
VC.06: Summary: The Flow of A Vector Field
ALONG a Closed Curve:
Let C be a closed curve parameterized counterclockwise. Let
Field(x,y) be a vector field with no singularities on the interior
region R of C. Then:
Ñ
C m(x, y)dx  n(x, y)dy   rotField dx dy
R
This measures the net flow of the vector field ALONG the closed
curve.
We define the rotation of the vector field as:
n m
rotField(x, y) 

 D[n[x, y], x]  D[m[x, y], y]
x y
VC.06: Measuring the Flow of a Vector Field
ACROSS a Closed Curve
Let C be a closed curve with a counterclockwise parameterization. Then the net
flow of the vector field ACROSS the closed curve is measured by:
b
 Field(x(t), y(t))  (y '(t),  x '(t))dt
a
b

  n(x(t), y(t))x '(t)  m(x(t), y(t))y '(t)  dt
a

Ñ
 C n(x, y)dx  m(x, y)dy
Let region R be the interior of C. If the vector field has no singularities in R, then
we can use Gauss-Green:
 m n 
  

 dx dy
x y 
R 

 divField dx dy
R
Let divField(x, y) 
m n

.
x y
VC.06: Summary: The Flow of A Vector Field
ACROSS a Closed Curve:
Let C be a closed curve parameterized counterclockwise. Let
Field(x,y) be a vector field with no singularities on the interior
region R of C. Then:
Ñ
C n(x, y)dx  m(x, y)dy   divField dx dy
R
This measures the net flow of the vector field ACROSS the closed
curve.
We define the divergence of the vector field as:
m n
divField(x, y) 

 D[m[x, y], x]  D[n[x, y], y]
x y
VC.06: The Divergence Locates Sources and
Sinks
Let C be a closed curve with a counterclockwise parameterization with no
singularities on the interior of the curve. Then:
If divField(x,y)>0 for all points in C, then all these points
are sources and the net flow of the vector field across C
is from inside to outside.
If divField(x,y)<0 for all points in C, then all of these
points are sinks and the net flow of the vector field
across C is from outside to inside.
If divField(x, y)  0 for all points in C, then the net flow of
the vector field across C is 0.
VC.06: The Rotation Helps You Find
Clockwise/Counterclockwise Swirl
Let C be a closed curve with a counterclockwise parameterization with no
singularities on the interior of the curve. Then:
If rotField(x,y)>0 for all points in C, then all these points
add counterclockwise swirl and the net flow of the vector
field along C is from counterclockwise.
If rotField(x,y)<0 for all points in C, then all of these points
add clockwise swirl and the net flow of the vector field
across C is clockwise.
If rotField(x, y)  0 for all points in C, then these points have
no swirl, and the net flow of the vector field along C is 0
(irrotational).
VC.06: Avoiding Computation Altogether
Let Field(x, y)   7x  2, y  6  and let C be a closed curve given by

3
C(t)  (x(t), y(t))  sin (t), cos(t)  sin(t) for   t 
.
4
4
Is the flow of the vector field across the curve from inside to outside or


2
outside to inside?
m n
divField(x, y) 

 71  8
x y
Ñ
 C n(x, y)dx  m(x, y)dy   divField dx dy   8 dx dy
R
R
Since divField(x,y) is ALWAYS positive for all (x,y) and there are no
singularities for any (x,y), this integral is positive for any closed curve.
That is, for ANY closed curve, the net flow of the vector field across
the curve is from inside to outside.
VC.06: Find the Net Flow of a Vector Field
ACROSS Closed Curve


Let Field(x, y)  x 2  2xy,  y 2  x and let C be the rectangle bounded by
x  2, x  5, y  1,and y  4. Measure the flow of the vector field across
the curve.
m n
divField(x, y) 

 2x  4y
x y
Ñ
 C n(x, y)dx  m(x, y)dy   divField dx dy
R
4 5

   2x  4y  dx dy
1 2
 105
Negative. The net flow of the vector field across our closed curve
is from outside to inside.
VC.06: A Flow Along Measurement With a
Singularity
 y
x 
Let Field(x, y)   2
,
and let C be the curve described by
2
2
2 

x y x y 

1
1

C(t)   sin2 (t)  cos(t)  ,cos(t)+sin(t)+  for  t  2. Compute the
2
2
2

flow of the vector field along the curve.
n m
rotField(x, y) 

0
x y
Since rotField(x,y)=0, your field is a
gradient away from singularities.
 The only swirl can come from singularities!
There is a singularity at (0,0).
We can replace our curve with any curve that encapsulates the singularity:
C2 (t)   cos(t), sin(t) for 0  t  2
Note: Had there been no singularities in the curve, how would we
know that the net flow of the vector field ALONG the curve would be 0?
VC.06: A Flow Along Measurement With a
Singularity
 y
x 
Field(x, y)   2
, 2
and C(t)   cos(t), sin(t)  for 0  t  2
2
2 

x y x y 
Because of the singularity, we can't use
 rotField(x, y) dx dy. Instead, we will need to compute
Ñ
 m(x, y)dx  n(x, y)dy the old-fashioned way:
C
2
 Field(x(t), y(t))  (x'(t), y'(t)) dt
0
2



 sin(t)
cos(t)
,
0  cos2 (t)  sin2 (t) cos2 (t)  sin2 (t)   (  sin(t), cos(t)) dt
2

2
2
sin
(t)

cos
(t) dt

0
2

 1 dt
0
 2
So the net flow of the vector field along the
curve is counterclockwise!
VC.06: A Flow Along With Multiple
Singularities? No Problem!
 y

y 1
x
x
Let Field(x, y)   2
 2
, 2
 2
and let C be the
2
2
2
2 

x  (y  1) x  y
x  (y  1) 
x y
curve pictured below. Compute the flow of the vector field along the curve.
rotField(x, y) 
n m

0
x y
Since rotField(x,y)=0, your field is a
gradient away from singularities.
 The only swirl can come from singularities!
There are singularities at (0,0) and (0,1).
We can encapsulate the singularities with two little circles and
sum our results!
C1 (t)  0.5  cos(t), sin(t) for 0  t  2
C2 (t)  0.5  cos(t), sin(t)  (0,1) for 0  t  2
VC.06: Flow Along When rotField(x,y)=0
n m

 0. Here are some conclusions about the net flow
x y
of the vector field along various closed curves:
Let rotField(x, y) 
If C doesn't contain any singularities, then
Ñ
 m(x, y)dx  n(x, y)dy  0.
C
If C contains a singularity, then
Ñ
 m(x, y)dx  n(x, y)dy  Ñ
 m(x, y)dx  n(x, y)dy
C
C1
for any substitute curve C1 containing the same singularity (and no new extras).
If C contains n singularities, then
Ñ
 m(x, y)dx  n(x, y)dy  Ñ
 m(x, y)dx  n(x, y)dy  ...  Ñ
 m(x, y)dx  n(x, y)dy
C
C1
Cn
for little circles, C1 ,..., Cn , encapsulating each of these singularities.
VC.06: Flow Along When divField(x,y)=0
m n

 0. Here are some conclusions about the net flow
x y
of the vector field across various closed curves:
Let divField(x, y) 
If C doesn't contain any singularities, then
Ñ
 n(x, y)dx  m(x, y)dy  0.
C
If C contains a singularity, then
Ñ
 n(x, y)dx  m(x, y)dy  Ñ
 n(x, y)dx  m(x, y)dy
C
C1
for any substitute curve C1 containing the same singularity (and no new extras).
If C contains n singularities, then
Ñ
 n(x, y)dx  m(x, y)dy  Ñ
 n(x, y)dx  m(x, y)dy  ...  Ñ
 n(x, y)dx  m(x, y)dy
C
C1
Cn
for little circles, C1 ,..., Cn , encapsulating each of these singularities.
VC.07: The Area Conversion Factor:
If we let r andt tend to zero, we can use this to get
T1
T2
t
t
dA  r
T1
 T1

 r
 T1
 t

r drdt
T2
T2 

r  is called the Jacobian matrix, and
T2 
t 
T1
You can let A xy (r, t)  r
T1
T1
T2
t
t
r
T1
r is called the Jacobian determinant.
T2
T2
r . Think of the Jacobian determinant as an
T2
t
t
Area Conversion Factor that lets us compute an xy-space integral in rt-space:
Hence,  f(x, y)dA   f(x(r,t), y(r,t)) A xy (r,t) dr dt
Rxy
Rrt
VC.07: The Area Conversion Factor:
 f(x, y)dA  f(x(r,t), y(r,t)) A
Rxy
Rrt
xy
(r,t) dr dt
Let T(r, t) be a transformation from rt-space to xy-space.
That is, T(r,t)   T1 (r, t), T2 (r, t)   (x(r, t), y(r, t)).
T1
T2
t
t
Then A xy (r, t)  r
T1
r .
T2
Note: Since we derived A xy (r, t) as the magnitude of a cross product, we need
it to be positive. This is why we put absolute value bars into the formula above.
VC.07: : Fixing Example 2
Remember we wanted
 1 dA for a circle of radius 4 centered at (0,0).
R
T(r, t)  (x(r, t), y(r, t))  (r cos(t),r sin(t))
x y
A xy (r, t)  r r
x y
t t
cos(t)
sin(t)

r sin(t) r cos(t)
 1 dA
R

r

  1 r dr dt
0 0
2
4
r 
    dt
2 0
0 
2
2

 r cos2 (t)  r sin2 (t)
 r cos2 (t)  sin2 (t)
2 4

 8 dt
0
 16 
Phew! We did it!
VC.07: Mathematica-Aided Change of
Variables (Parallelogram Region)
Use Mathematica to compute  e y dA for R given by the parallelogram:
R
These lines are given by y  x  1, y  x  4 ,
y
1
13
1
x
, and y   x  6.
4
4
4
We can rewrite them as 1  y  x , 4  y  x ,
1
13
1
xy 
,and x  y  6.
4
4
4
1
13
Then we can let u  y  x and v  x  y for  4  u  1 and
 v  6!
4
4
But to compute our integral, we need the map from uv-space to xy-space...
That is, we have u(x, y) and v(x, y ), but we need x(u, v ) and y(u, v).
Let Mathematica do the work: Solve[{u  u[ x , y ], v  v[ x , y ]},{ x , y }]
4
1
x   (u  v ) and y  (u  4v )
5
5
VC.07: Mathematica-Aided Change of
Variables (Parallelogram Region)
Use Mathematica to compute  e y dA for R given by the parallelogram:
R
4
1
x   (u  v ) and y  (u  4v )
5
5
x
A xy (u, v)  u
x
v
y
4

u
 5
y
4
v
5
4
So we use
5
1
4
5 
5
4
5
VC.07: Mathematica-Aided Change of
Variables (Parallelogram Region)
Use Mathematica to compute  e y dA for R given by the parallelogram:
R
4
4
1
, x   (u  v), y  (u  4v),
5
5
5
13
for  4  u  1 and
v6:
4
A xy (u, v) 
 e
R
y
dA 
6
1
 e
13/ 4 4
(u  4v )/5
4
  du dv
5
 417.1 (from Mathematica)
VC.08: 3D Integrals:
 f(x, y, z) dx dy dz   f(x, y, z) V
R xyz
xyz
Ruvw
x
u
x
Vxyz (u, v, w) 
v
x
w
y
u
y
v
y
w
(u, v, w) du dv dw
z
u
z
v
z
w
In my opinion, a good way to think about
 f(x, y, z) dx dy dz is
R xyz
as a calculation of the mass of R xyz where f(x, y, z) is the density
of the solid at any given point (x, y, z).
VC.08: A 3D-Change of Variables with
an Integrand
Compute
 9y dx dy dz where R
R xyz
xyz
is the parallelepiped that is between
the planes z  3x and z  3x  2, y  x and y  x  4, and y  2x and
y  2x  3.
wv
x
3
All from Example 4:
u  z  3x 0  u  2
2v  w
y

v  yx 0 v  4
1
3
Vxyz (u, v, w) 
w  y  2x 0  w  3 z  u  w  v
3
3 4 2
 9y dx dy dz     9y(u, v, w) V
0 0 0
3 4 2
R xyz

(u, v, w) du dv dw
 2v  w  1
9
0 0 0  3  3 du dv dw
3 4 2

xyz
    2v  w  du dv dw
0 0 0
 132

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