vector calculus 17. 2 vector calculus here, we define two operations that: can be performed on...

VECTOR CALCULUSVECTOR CALCULUS

17

2

VECTOR CALCULUS

Here, we define two operations

that:

Can be performed on vector fields.

Play a basic role in the applications of vector calculus to fluid flow, electricity, and magnetism.

3

VECTOR CALCULUS

Each operation resembles differentiation.

However, one produces a vector field

whereas the other produces a scalar field.

4

17.5 Curl and Divergence

VECTOR CALCULUS

In this section, we will learn about:

The operations of curl and divergence

and how they can be used to obtain

vector forms of Green’s Theorem.

5

CURL

Suppose:

F = P i + Q j + R k is a vector field on .

The partial derivatives of P, Q, and R all exist

3

6

CURL

Then, the curl of F is the vector field on

defined by:

curl

R Q P R Q P

y z z x x y

F

i j k

Equation 13

7

CURL

As a memory aid, let’s rewrite Equation 1

using operator notation.

We introduce the vector differential operator (“del”) as:

x y z

i j k

8

CURL

It has meaning when it operates on a scalar

function to produce the gradient of f :

f f ff

x y z

f f f

x y z

i j k

i j k

9

CURL

If we think of as a vector with

components ∂/∂x, ∂/∂y, and ∂/∂z, we can

also consider the formal cross product of

with the vector field F as follows.

10

CURL

curl

x y z

P Q R

R Q P R Q P

y z z x x y

F

i j k

i j k

F

11

CURL

Thus, the easiest way to remember

Definition 1 is by means of the symbolic

expression

curl F F

Equation 2

12

CURL

If F(x, y, z) = xz i + xyz j – y2 k

find curl F.

Using Equation 2, we have the following result.

Example 1

13

CURL Example 1

2

2

2

curl

2 0 0

2

x y z

xz xyz y

y xyzy z

y xz xyz xzx z x y

y xy x yz

y x x yz

i j k

F F

i

j k

i j k

i j k

14

CURL

Most computer algebra systems (CAS)

have commands that compute the curl and

divergence of vector fields.

If you have access to a CAS, use these commands to check the answers to the examples and exercises in this section.

15

CURL

Recall that the gradient of a function f of

three variables is a vector field on .

So, we can compute its curl.

The following theorem says that the curl of a gradient vector field is 0.

3

16

GRADIENT VECTOR FIELDS

If f is a function of three variables that has

continuous second-order partial derivatives,

then

curl f 0

Theorem 3

17


By Clairaut’s Theorem,

2 2 2 2

2 2

curl

0 0 0

f fx y z

f f f

x y z

f f f f

y z z y z x x z

f f

x y y x

i j k

i j

k

i j k 0

Proof

18


Notice the similarity to what we know

from Section 12.4:

a x a = 0 for every three-dimensional (3-D)

vector a.

19

CONSERVATIVE VECTOR FIELDS

A conservative vector field is one for which

So, Theorem 3 can be rephrased as:

If F is conservative, then curl F = 0.

This gives us a way of verifying that a vector field is not conservative.

fF

20


Show that the vector field

F(x, y, z) = xz i + xyz j – y2 k

is not conservative.

In Example 1, we showed that: curl F = –y(2 + x) i + x j + yz

k

This shows that curl F ≠ 0.

So, by Theorem 3, F is not conservative.

Example 2

21


The converse of Theorem 3 is not true in

general.

The following theorem, though, says that

it is true if F is defined everywhere.

More generally, it is true if the domain is simply-connected—that is, “has no hole.”

22


Theorem 4 is the 3-D version of

Theorem 6 in Section 16.3

Its proof requires Stokes’ Theorem and is sketched at the end of Section 16.8

23


If F is a vector field defined on all of

whose component functions have continuous

partial derivatives and curl F = 0, then

F is a conservative vector field.

3Theorem 4

24


a. Show that

F(x, y, z) = y2z3 i + 2xyz3 j + 3xy2z2 k

is a conservative vector field.

b. Find a function f such that .

Example 3

fF

25


As curl F = 0 and the domain of F is , F is a conservative vector field by Theorem 4.

Example 3 a

2 3 3 2 2

2 2 2 2 2 2

3 3

curl

2 3

6 6 3 3

2 2

x y z

y z xyz xy z

xyz xyz y z y z

yz yz

i j k

F F

i j

k

03

26


The technique for finding f was given in

Section 17.3

We have:

fx(x, y, z) = y2z3

fy(x, y, z) = 2xyz3

fz(x, y, z) = 3xy2z2

E. g. 3 b—Eqns. 5-7

27


Integrating Equation 5 with respect to x,

we obtain:

f(x, y, z) = xy2z3 + g(y, z)

E. g. 3 b—Eqn. 8

28


Differentiating Equation 8 with respect to y,

we get:

fy(x, y, z) = 2xyz3 + gy(y, z)

So, comparison with Equation 6 gives: gy(y, z)

= 0

Thus, g(y, z) = h(z) and fz(x, y, z) = 3xy2z2 +

h’(z)

Example 3 b

29


Then, Equation 7 gives:

h’(z) = 0

Therefore, f(x, y, z) = xy2z3 + K

Example 3 b

30

CURL

The reason for the name curl is that

the curl vector is associated with rotations.

One connection is explained in Exercise 37.

Another occurs when F represents the velocity field in fluid flow (Example 3 in Section 17.1).

31

CURL

Particles near (x, y, z) in the fluid tend

to rotate about the axis that points in

the direction of curl F(x, y, z).

The length of this curl vector is a measure of how quickly the particles move around the axis.

Fig. 17.5.1, p. 1100

32

F = 0 (IRROTATIONAL CURL)

If curl F = 0 at a point P, the fluid is free

from rotations at P.

F is called irrotational at P.

That is, there is no whirlpool or eddy at P.

33

F = 0 & F ≠ 0

If curl F = 0, a tiny paddle wheel moves with

the fluid but doesn’t rotate about its axis.

If curl F ≠ 0, the paddle wheel rotates about

its axis.

We give a more detailed explanation in Section 16.8 as a consequence of Stokes’ Theorem.

34

DIVERGENCE

If F = P i + Q j + R k is a vector field on

and ∂P/∂x, ∂Q/∂y, and ∂R/∂z exist,

the divergence of F is the function of three

variables defined by:

Equation 9

div P Q R

x y z

F

3

35

CURL F VS. DIV F

Observe that:

Curl F is a vector field.

Div F is a scalar field.

36

DIVERGENCE

In terms of the gradient operator

the divergence of F can be written

symbolically as the dot product of and F:

x y z

i j k

div F F

Equation 10

37

DIVERGENCE

If F(x, y, z) = xz i + xyz j – y2 k

find div F.

By the definition of divergence (Equation 9 or 10) we have:

2

div

xz xyz yx y z

z xz

F F

Example 4

38

DIVERGENCE

If F is a vector field on , then curl F is

also a vector field on .

As such, we can compute its divergence.

The next theorem shows that the result is 0.

33

39

DIVERGENCE

If F = P i + Q j + R k is a vector field on

and P, Q, and R have continuous second-

order partial derivatives, then

div curl F = 0

Theorem 113

40

DIVERGENCE

By the definitions of divergence and curl,

The terms cancel in pairs by Clairaut’s Theorem.

Proof

2 2 2 2 2 2

div curl

0

R Q P R Q P

x y z y z x z x y

R Q P R Q P

x y x z y z y x z x z y

F

F

41

DIVERGENCE

Note the analogy with the scalar triple

product:

a . (a x b) = 0

42

DIVERGENCE

Show that the vector field

F(x, y, z) = xz i + xyz j – y2 k

can’t be written as the curl of another vector

field, that is, F ≠ curl G

In Example 4, we showed that div F = z + xz

and therefore div F ≠ 0.

Example 5

43

DIVERGENCE

If it were true that F = curl G, then Theorem 11 would give:

div F = div curl G = 0

This contradicts div F ≠ 0.

Thus, F is not the curl of another vector field.

Example 5

44

DIVERGENCE

Again, the reason for the name divergence

can be understood in the context of fluid flow.

If F(x, y, z) is the velocity of a fluid (or gas), div F(x, y, z) represents the net rate of change (with respect to time) of the mass of fluid (or gas) flowing from the point (x, y, z) per unit volume.

45

INCOMPRESSIBLE DIVERGENCE

In other words, div F(x, y, z) measures

the tendency of the fluid to diverge from

the point (x, y, z).

If div F = 0, F is said to be incompressible.

46


Another differential operator occurs when

we compute the divergence of a gradient

vector field .

If f is a function of three variables, we have:

f

2 2 2

2 2 2

div f f

f f f

x y z

47

LAPLACE OPERATOR

This expression occurs so often that

we abbreviate it as .

The operator is called

the Laplace operator due to its relation to

Laplace’s equation

2 f

2

2 2 22

2 2 20

f f ff

x y z

48

LAPLACE OPERATOR

We can also apply the Laplace operator

to a vector field

F = P i + Q j + R k

in terms of its components:

2 2 2 2P Q R F i j k

49

VECTOR FORMS OF GREEN’S THEOREM

The curl and divergence operators

allow us to rewrite Green’s Theorem

in versions that will be useful in our

later work.

50


We suppose that the plane region D, its

boundary curve C, and the functions P and Q

satisfy the hypotheses of Green’s Theorem.

51


Then, we consider the vector field

F = P i + Q j

Its line integral is:

C Cd Pdx QdyF r

52


Regarding F as a vector field on with

third component 0, we have:

3

curl

, , 0

x y z

P x y Q x y

Q P

x y

i j k

F

k

53


Therefore,

curl Q P

x y

Q P

x y

F k k k

54

VECTOR FORMS OF GREEN’S TH.

Hence, we can now rewrite the equation

in Green’s Theorem in the vector form

curl C

D

d dAF r F k

Equation 12

55


Equation 12 expresses the line integral of

the tangential component of F along C as

the double integral of the vertical component

of curl F over the region D enclosed by C.

We now derive a similar formula involving the normal component of F.

56


If C is given by the vector equation

r(t) = x(t) i + y(t) j a ≤ t ≤ b

then the unit tangent vector (Section 14.2)

is:

' '

' '

x t y tt

t t T i jr r

57


You can verify that the outward unit normal

vector to C is given by:

' '

' '

y t x tt

t t n i jr r

Fig. 17.5.2, p. 1103

58


Then, from Equation 3 in Section 17.2,

by Green’s Theorem, we have:

'

, ' , ''

' '

C

b

a

b

a

ds

t t dt

P x t y t y t Q x t y t x tt dt

t t

F n

F n r

rr r

59


, ' , 'b

a

C

D

P x t y t y t dt Q x t y t x t dt

P dy Qdx

P QdA

x y

60


However, the integrand in that double

integral is just the divergence of F.

So, we have a second vector form

of Green’s Theorem—as follows.

61


This version says that the line integral of the normal component of F along C is equal to the double integral of the divergence of F over the region D enclosed by C.

div ,C

D

ds x y dAF n F Equation 13

vector calculus 17. 2 vector calculus here, we define two operations that: can be performed on...

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