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)

Example1: Given f(x,y) x 1, y 1 , find a possibleequation for the surfacef(x,y).

22 yxf(x,y) x y?2 2

VC.04: Gradient Fields

2 2x yExample2: Plot thegradient field x yf(x,y) :

efor

2 2x ye 1 2x(x y),f 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 sinkcorrespond to on the surface?

Source LocalMinimumSink LocalMaximum

VC.04: Gradient Fields

2 2x yExample2: Finally,compare thegra x yf(x,dient fielda yn ) :

ed

VC.04: Breaking Field Vectorsinto Two Components

Forward/Backward Push:

LetField(x,y) (m(x,y),n(x,y)be a vector field acting on the curve (x(t),y(t)).

The push of the fieldvector in the direction of the tangent vectorto the curve:

Field(x(t),y(t)) (x'(t),y'(t)) (x'(t),y'(t))(x'(t),y'(t)) (x'(t),y'(t))

"Theflow of the vector fieldALONG the curve."

Left/Right Push:

The push of the fieldvector in the direction of the normal vectorto the curve:

Field(x(t),y(t)) (y'(t), x'(t)) (y'(t), x'(t))(y'(t), x'(t)) (y'(t), x'(t))

"Theflow of the vector fieldACROSS the curve."

Use the push of the field vectors in the direction of the tangent vectors to the curve to determine whetherthe net flow ALONG the curve is clockwise or counterclockwise:

2 2Let the vector field (3x 3x)(y 3),(3y 3y)(x 3)act on the ellipse(x(t),y(t)) ( 4,2) (4cos(t),2sin(t)).

Field(x(t),y(t)) (x'(t),y'(t))Plot (x'(t),y'(t))(x'(t),y'(t)) (x'(t),y'(t))

Thenet flow of the vector fieldALONG the curve is clockwise.

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 "insideto outside" or"outside to inside."

2 2Let the vector field (3x 3x)(y 3),(3y 3y)(x 3)act on the ellipse(x(t),y(t)) ( 4,2) (4cos(t),2sin(t)).

Field(x(t),y(t)) (y'(t), x'(t))Plot (y'(t), x'(t))(y'(t), x'(t)) (y'(t), x'(t))

Thenet flow of the vector field ACROSS the curve is "insideto outside."

VC.04: Flow Along and Flow Across"Theflow of the vector fieldALONG the curve."

"Theflow of the vector fieldACROSS the curve."

Insideto outside:

Outsideto Inside:

Counterclockwise:

Clockwise:

Field(x(t),y(t)) (x'(t),y'(t))Plot (x'(t),y'(t))(x'(t),y'(t)) (x'(t),y'(t))

Field(x(t),y(t)) (y'(t), x'(t))Plot (y'(t), x'(t))(y'(t), x'(t)) (y'(t), x'(t))

For this chapter, our only measurement tool is to plot and then eyeball the results...

*Could be 0! *Could be 0!

If C is a closed curve with a counterclockwise parameterization:

If C is not closed:

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.

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.”

VC.05: Measuring the Flow of a Vector Field ALONG a Curve

b

ab

a

C

Integral :

Field(x(t),y(t)) (x'(t),y'(t))dt

m(x(t),y(t))x'(t) n(x(t),y(t))y'(t) dt

m(x,y)dx n(x,y)dy

b

ab

a

C

Integral :

Field(x(t),y(t)) (x'(t),y'(t))dt

m(x(t),y(t))x'(t) n(x(t),y(t))y'(t) dt

m(x,y)dx n(x,y)dyÑ

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:

Cm(x,y)dx n(x,y)dy 0describes a net flow of the vector fieldalong the curve in the counterclockwise direction.Ñ

Cm(x,y)dx n(x,y)dy 0describes a net flow of the vector fieldalong the curve in the clockwise direction.Ñ

Cm(x,y)dx n(x,y)dy can equal 0.Ñ

VC.05: Measuring the Flow of a Vector Field Along Another Closed Curve

22 2

Cx 1LetField(x,y) (3y, x ) be a vector field acting on the ellipse y 1.Compute m(x,y)dx n(x,y)dy :2 Ñ

FieldandCurve: VectorsonCurve:

Component of FieldVectorsintheDirection of the Tangent Vectors

2Field(x,y) m(x,y),n(x,y) (3y, x )

VC.05: Measuring the Flow of a Vector Field Along Another Closed Curve

2

22

C

LetField(x,y) (3y, x ) be a vector field acting on the x 1ellipse y 1.Compute m(x,y)dx n(x,y)dy :2 Ñ

Cm(x,y)dx n(x,y)dyÑ E(t) (2cos(t),sin(t)) (1,0)

b

am(x(t),y(t))x'(t) n(x(t),y(t))y'(t) dt

2 22

0

3cos(t) 4cos (t) 4cos d(t) 6si tn (t)

10Negative!Thenet flow of the vector field along the curve isagainst the direction of the parameterization(clockwise).

Counterclockwise

If C is a closed curve with a counterclockwise parameterization:

If C is not closed:

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.”

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 a Curve

b

ab

a

C

Integral :

Field(x(t),y(t)) (y'(t), x'(t))dt

n(x(t),y(t))x'(t) m(x(t),y(t))y'(t) dt

n(x,y)dx m(x,y)dy

b

ab

a

C

Integral :

Field(x(t),y(t)) (y'(t), x'(t))dt

n(x(t),y(t))x'(t) m(x(t),y(t))y'(t) dt

n(x,y)dx m(x,y)dyÑ

m(x,y),n(x,y) ( x cos(y), y sin(x))

VC.05: Measuring the Flow of a Vector Field Across Another Closed Curve

2 2C

LetField(x,y) ( x cos(y), y sin(x)) be a vector field acting on the circle x y 1.Compute n(x,y)dx m(x,y)dy :Ñ

C n(x,y)dx m(x,y)dyÑc(t) (cos(t),sin(t))

b

an(x(t),y(t))x'(t) m(x(t),y(t))y'(t) dt

2

Negative!The net flow of the vector field across the curve is fromoutside to inside.

Counterclockwise

VC.05: The Gradient Test

A vector field, Field(x,y) m(x,y),n(x,y , is a gradient fieldif and only if:

x y

xy yx

If Field(x,y) is a gradient field, then Field(x,y)= f ,f . So ff .

Proof of "if" Part of Theorem:

m(x,y) n(x,y)

y x(TheGradient Test)

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 curvewitha parameterization (x(t),y(t)) for a t b.

b

a1) Field(x(t),y(t)) (x'(t),y'(t))dt 0

C2) m(x,y)dx n(x,y)dy 0Ñ3) The flow of a gradient field alonga simpleclosed 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 ACROSSa closed curve 0?

VC.05: Path Independence: The Flow of a Gradient Field Along an Open Curve

2

1Let Field(x,y)= m(x,y),n(x,y) be a gradient field, and let andbe different curves that share the same starting and ending pC

Coint:

A gradient field is said to be path independent. The flow of the vector field alongany two curvesconnecting two points is the same...

21 CC

m(x,y)dx n(x,y)dy m(x,y)dx n(x,y)dy

VC.06: The Gauss-Green Formula

high

low

t

t R

n mdt dAxx'(t y'n x(t),y(m x(t),y( )t) y)t) (t

low high

LetR be a region in the xy-plane whoseboundary is parameterizedby(x(t),y(t)) for . Thenthefollowingformulahot :t st ld

With the proper interpretation, we can use this formula to helpus compute flowalong/across measurements!!The basic interpretation of Gauss-Green is that it is a correspondencebetween 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 singularitiesin the interior region!

VC.06: Measuring the Flow of a Vector Field ALONG a Closed Curve

b

ab

a

C

Field(x(t),y(t)) (x'(t),y'(t))dt

m(x(t),y(t))x'(t) n(x(t),y(t))y'(t) dt

m(x,y)dx n(x,y)dyÑ

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:

Let region R be the interior of C. If the vector field has no singularities in R, then we can use Gauss-Green:

R

n m dx dyx y

R

rotFielddx dy

n mLetrotField(x,y) .x y

VC.06: Summary: The Flow of A Vector Field ALONG a Closed Curve:

R

Cm(x,y)dx n(x,y)dy rotFielddx dyÑ

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:

This measures the net flow of the vector field ALONG the closed curve.

Wedefine the rotation of the vector field as:n mrotField(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

b

ab

a

C

Field(x(t),y(t)) (y'(t), x'(t))dt

n(x(t),y(t))x'(t) m(x(t),y(t))y'(t) dt

n(x,y)dx m(x,y)dyÑ

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:

Let region R be the interior of C. If the vector field has no singularities in R, then we can use Gauss-Green:

R

m n dx dyx y

R

divFielddx dy

m nLet divField(x,y) .x y

VC.06: Summary: The Flow of A Vector Field ACROSS a Closed Curve:

R

C n(x,y)dx m(x,y)dy divFielddx dyÑ

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:

This measures the net flow of the vector field ACROSS the closed curve.

Wedefine the divergence of the vector field as:m ndivField(x,y) D[m[x,y],x] D[n[x,y],y]x y

VC.06: The Divergence Locates Sources and Sinks

If divField(x,y)>0for all points in C, then all these pointsare sources and the net flow of the vector field across C is from inside to outside.

Let C be a closed curve with a counterclockwise parameterization with no singularities on the interior of the curve. Then:

If divField(x,y)<0for 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 ofthe vector field across C is 0.

VC.06: The Rotation Helps You Find Clockwise/Counterclockwise Swirl

If rotField(x,y)>0for all points in C, then all these pointsadd counterclockwise swirl and the net flow of the vector field along C is from counterclockwise.

Let C be a closed curve with a counterclockwise parameterization with no singularities on the interior of the curve. Then:

If rotField(x,y)<0for 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 haveno swirl, and the net flow of the vector field along C is 0(irrotational).

VC.06: Avoiding Computation Altogether

2

Let Field(x,y) 7x 2,y 6 and let C be a closed curvegiven by3C(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 outsideor outside to inside?

R R

C n(x,y)dx m(x,y)dy divFielddx dy 8dx dyÑSince divField(x,y) is ALWAYS positive for all (x,y) and thereare no singularities for any (x,y), this integral is positive for any closed curve.

m ndivField(x,y) 7 1 8x y

That is, for ANY closed curve, the net flow of the vector field acrossthe curve is from inside to outside.

VC.06: Find the Net Flow of a Vector Field ACROSS Closed Curve

2 2Let Field(x,y) x 2xy, y x and let C be the rectangleboundedby x 2, x 5,y 1,and y 4.Measure the flow of the vector field across thecurve.

m ndivField(x,y) 2x 4yx y

R

C n(x,y)dx m(x,y)dy divFielddx dyÑ

4 5

1 22x 4y dx dy

105Negative. The net flow of the vector field across our closed curveis from outside to inside.

VC.06: A Flow Along Measurement With a Singularity

2 2

2

2 2

1sin (t) cos(t)

y xLet Field(x,y) , and let C be the curve describedby x y x y

1C(t) ,cos(t)+sin(t)+ for t 2 .Compute the 2 2flow of the vector field along the

2 curve.

n mrotField(x,y) 0x y

Theonly swirl can come from singularities!There is a singularity at (0,0).

We can replace our curve with any curve that encapsulates the singularity: 2C (t) sin(t) for 0cos( ) tt 2,

Since rotField(x,y)=0,your field is a gradient away from singularities.

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

2 2 2 2y xField(x,y) , andC(t) sin(t) focos(t), r 0 t 2

x y x y

C

Because of the singularity, we can't use rotField(x,y)dx dy. Instead, we will need to compute m(x,y)dx n(x,y)dy theold-fashioned way:Ñ

2

0Field(x(t),y(t)) (x'(t),y'(t))dt

2

2 2 2 20

sin(t) cos(t), ( sin(t),cos(t))dtcos (t) sin (t) cos (t) sin (t)

2

2 2

0sin (t) cos (t)dt

2

01dt 2 So the net flow of the vector field along the

curve is counterclockwise!

VC.06: A Flow Along With Multiple Singularities? No Problem!

2 2 2 2 2 2 2 2

y y 1 x xLet Field(x,y) , and let C be the x y x (y 1) x y x (y 1)

curve picturedbelow.Compute the flow of the vector field along the curve.

n mrotField(x,y) 0x y

Theonly 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!

1 cos(t),C (t) 0.5 sin(t) for 0 t 2

Since rotField(x,y)=0, your field is a gradient away from singularities.

2C (t) 0.5 sin(t) (0,1) forcos( 0, tt 2)

VC.06: Flow Along When rotField(x,y)=0

n mLetrotField(x,y) 0. Here are some conclusions about the net flowx y

of the vector field along various closed curves:

CIf C doesn't contain any singularities, then m(x,y)dx n(x,y)dy 0.Ñ

1C C

1

If C contains asingularity, then m(x,y)dx n(x,y)dy m(x,y)dx n(x,y)dy

for any substitute curve C containing the same singularity(andno new extras).

Ñ Ñ

1 nC C C

1 n

If C contains nsingularities, then m(x,y)dx n(x,y)dy m(x,y)dx n(x,y)dy ... m(x,y)dx n(x,y)dy

for littlecircles,C ,...,C ,encapsulatingeach of these singularities.

Ñ Ñ Ñ

VC.06: Flow Along When divField(x,y)=0

m nLet divField(x,y) 0.Here are some conclusions about the net flowx y

of the vector field across various closed curves:

CIf C doesn't contain any singularities, then n(x,y)dx m(x,y)dy 0.Ñ

1C C

1

If C contains asingularity, then n(x,y)dx m(x,y)dy n(x,y)dx m(x,y)dy

for any substitute curve C containing the same singularity(andno new extras).

Ñ Ñ

1 nC C C

1 n

If C contains nsingularities, then n(x,y)dx m(x,y)dy n(x,y)dx m(x,y)dy ... n(x,y)dx m(x,y)dy

for littlecircles,C ,...,C ,encapsulatingeach of these singularities.

Ñ Ñ Ñ

VC.07: The Area Conversion Factor:

1 2

1 2

If we let and tend to zero, we can usrT Tr

e tt

dtT Tt

r d

his t

t

et

d r

o g

A

xy rt

xyR R

Hence, f(x,y)dA f(x(r,t),y(r,t)) A (r,t) dr dt

1 2

1 1

1 2

2 2 is called the Jacobian matrix, and is called the Jacobian determi

T T T T

T T T Tt t t t

r r n .r ar nt

x1 2

1 2

yYou can let A (r,t) . Think of the Jacobian determinant as an

Area Conversion Factor that lets us compute an xy-space integ

T

r

Tr r

al in rt-spat

:

Tt

c

T

e

VC.07: The Area Conversion Factor:

xy rt

xyR R

f(x,y)dA f(x(r,t),y(r,t)) A (r,t) dr dt

1 2

xy

1

1 2

2

Let T(r,t) be a transformation from rt-space to xy-space.That is, T(r,t) T (r,t),T (r,t) (x(r,t),y(r,t)).

ThenAT Tr(r,t Tt t

) .Tr

xyNote: Since we derived A (r,t) as the magnitude of a cross product,we needit to be positive. This is why we put absolute value bars into the formula above.

VC.07: : Fixing Example 2

RRemember we wanted 1dA for a circle of radius 4 centered at (0,0).

2 4

0 0r d t1 r d

R

1dA42 2

0 0

r dt2

2

08dt

T(r,t) (x(r,t),y(r,t)) (rcos(t),rsin(t))

xy

yxr rA (r,t) yxt t

cos(t) sin(t)rsin(t) rcos(t)

2 2rcos (t) rsin (t)

r 2 2r cos (t) sin (t) 16

Phew!We did it!

VC.07: Mathematica-Aided Change of Variables (Parallelogram Region)

y

RUseMathematica to compute e dA for R given by the parallelogram:These lines are given by ,

1 13 1y x y x 64 4 4

,

,and .

y x 1 y x 4

1 yWe xcan rewrite them as ,1 13 1x y x y 6

,

,and .

x

4 4

4 y

4

But to computeour integral, we need the map from uv-space to xy-space...v(x,yThat is, we have and ,but we need u(x,y) x(u,v) y(ua d) n ,v).

Solve[LetMathematica { [ , ], [ , ]},{ , }]do the work: u u x y v v x y x y

1T 13v x y v 64henwe can let and for andy 4 !u u 1 4x

and 1y4x (u 4v) )(u5 5v

VC.07: Mathematica-Aided Change of Variables (Parallelogram Region)

y

RUseMathematica to compute e dA for R given by the parallelogram:

and 1y4x (u 4v) )(u5 5v

xy

yxu uA (u,v) yxv v

4 15 54 45 5

45

So we use 45

VC.07: Mathematica-Aided Change of Variables (Parallelogram Region)

y

RUseMathematica to compute e dA for R given by the parallelogram:

xy1y (4A (u,v) , , ,5

for an

u 4v)513 v 6d :

4x (u

4

v)54 u 1

y

Re dA

6 1(u 4v)/5

13/ 4 4

4 dudv5e

417.1 (fromMathematica)

VC.08: 3D Integrals:

xyz uvw

xyzR R

f(x,y,z)dx dy dz f(x,y,z) V (u,v,w) dudv dw

xyz

yx zu

yx zw

V (u,v,u u

w) yx

w

v

w

zv v

xyzR

xyz

In my opinion, a good way to think about f(x,y,z)dx dy dz is

asacalculation of the mass of R 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

xyz

xyzR

Compute 9y dx dy dz whereR is the parallelepiped that is between

the planes z 3x andz 3x 2, y x and y x 4,and y 2x andy 2x 3.

0 u 20 v 40 w 3

3 4 2

xyz0 0 0

9y(u,v,w) V (u,v,w) dudv dw

u z 3xv y xw y 2x

w vx 32v wy 3

z u w v

xyz1V (u,v,w) 3

xyzR9y dx dy dz

All from Example4:

3 4 2

0 0 0

2v w 19 dudv dw3 3

3 4 2

0 0 02v w dudv dw 132