Vector Calculuspart I

By Dr. Michael Medvinsky, NCSU

online lectures 03/2020

Vector Field (definition)

• Definition: Vector Field is a function F that for each (x,y)\(x,y,z) assign a 2\3-dimensional vector, respectively:

• Examples of VF: gradient, direction field of differential equation.

• Vector field vs other functions we learned:

2 3

: function of 1,2,3variables

: vector (of size 1,2,3) valued function, e.g. parametric curve

: parametric surface

: vector field ( 2,3)

n

n

n n

f n

r n

r

n

→ =

→ =

→

→ =F

( ) ( ) ( )

( ) ( ) ( )

, , , ,

( , , ) , , , , , , , ,

x y P x y Q x y

x y z P x y z Q x y z R x y z

=

=

F

F

Vector Field (how to sketch it)

• We draw VF as vectors

starting at points

Examples:

( ) ( ) ( ) ( ) ( ), , , \ , , , , , , , ,P x y Q x y P x y z Q x y z R x y z

( ) ( ), \ , ,x y x y z

x

y

Q

P

General case

Uniform\Constant VF:F=<1,2>Uniform\Constant VF: F=<2,1>

Vector Field (how to sketch it)

• More examples:

( ), , ,f x y xy f y x= = =F ( ), , , ,0y x

x y zz z

= −F( ), ,x y x y=F

Line Integral (the idea)

• Consider smooth curve C be given by:recall: smooth means is continuous and

• Divide [a,b] into subintervals

• Denote a piece of C corresponding to and displacement as .

• Denote a sample point on .

• Consider that function f(x,y) is defined along C.

• What do you think about the followingRieman-like sum

( ) ( ) ( ) , , ,r t x t y t t a b=

tt0 tn

,i

b at a i t t

n

−= + =

1,i it t−is

* *,i ix yis

is

( )* *

1

,n

i i i

i

f x y s=

( )'r t ( )' 0r t

sub−arc 𝑠𝑖

Line Integral (definition)

• Let f be a function defined along a curve C given by

then the line\contour\path\curve integral is defined by

• Recall: Arclength formula:

( ) ( ) ( ) ( ) ( ) ( ) ( ) , , , , ,r t x t y t or r t x t y t z t t a b= =

( ) ( ) ( ) ( )* * * * *

1 1

, lim , , , lim , ,n n

i i i i i i in n

i iC C

f x y ds f x y s or f x y z ds f x y z s→ →

= =

= =

( )2 2 2 2 2 '

b b b

t t t t t

a a a

L x y dt or L x y z dt or L r t dt= + = + + =

Line Integral (2 theorems)

• 1) Let f be continuous function along curve C (defined as before):

regardless of the parameterization as long as the curve traversed exactly once between a and b.

• 2) Let be piecewise smooth curve, then

( ) ( ) ( )( ) ( ) ( )( )

( ) ( ) ( ) ( )( ) ( ) ( )( )

( ) ( )( ) ( ) ( ) ( )

2 2

2 2 2

, , , or

, , , , , or

' , for either , or , ,

b b

t t

C a a

b b

t t t

C a a

b

C a

dsf x y ds f x t y t dt f x t y t x y dt

dt

dsf x y z ds f x t y t z t dt f x t y t x y z dt

dt

f x ds f r t r t dt x x y x x y z

= = +

= = + +

= = =

1 2 ...C C C=

1 2

...C C C

f ds f ds f ds= + +

Line Integral(examples)

• 1) Let (half circle, r=2), evaluate

Solution: We have the arclength is thus

( ) 2cos ,2sin ,0r t t t t = 2

C

x yds

( ) ( )( ) ( )2 2 2, 2cos ,2sin 4cos 2sin 8cos sinf x y x y f r t f t t t t t t= = = =

( )' 2 sin ,cos 2r t t t= − =

( )( ) ( )

1

2 2 2 2

cos'0 0 1

328cos sin 2 16 cos sin 16

3u tr tC f r t

x yds t t dt t tdt u du

=−

= = = =

Line Integral(examples)

• 2) Letevaluate

Solution:

( ) ( )1 1 2 28 ,4 ,5 ,0 1 and 1,2, 5 , 1 0C r t t t t t C r t t t= = = = − −

1 2C C

x y zds

+ +

( )( )( )

( )

( ) ( )( )

( )( )( ) ( )

1

2

1 2

11 1 2

0 0 0'

00 0 2

1 1 1'

7 9 88 4 5 8 16 25 7 9 8 7 9 8

2 2

1 25 551 2 5 0 0 25 5 3 5 5 3 5 5 3 5 15

2 2 2 2

C r tf r t

C r tf r t

C C

tx y zds t t t dt tdt

tx y zds t dt tdt t

x y zds

− − −

++ + = + + + + = + = + =

+ + = + − + + = − = − = − − − = + =

+ +

( )1 2

7 9 8 55 7 859

2 2 2C C

x y zds x y zds+

= + + + + + = + = +

Line Integral(examples)

• 3) Evaluate where C is a line between (1,2) and (3,-1)

Solution: parameterize

to get

Alternative Solution: parameterize

to get

13C

xyds

( ) ( )1 1,2 3, 1 1 2 ,2 3r t t t t t= − + − = + −

( )( )1 2 2 3

13 13C

t txyds

+ −=

( )( )

4 9

f r t

+

( )

1 1

2

0 0'

12 6

2r t

dt t t dt= + − =

( ) ( )1 3, 1 1,2 3 2 , 1 3r t t t t t= − − + = − − +

( )( )3 2 1 3

13 13C

t txyds

− − +=

( )( )

4 9

f r t

+

( )

1 1

2

0 0'

13 11 6

2r t

dt t t dt= − + − =

Line Integral(theorem + definition)

• In previous example we traversed a curve (line) in two opposite direction and got the same result – it didn’t happen by an accident.

• Theorem: Denote by -C the same curve as C, but with different direction:

• Definition: Denote a line integral with respect to arclength, a line integral with respect to x: ,with respect to y:and analogically in 3D .They often occur together, e.g.

C Cf ds f ds

−=

Cf ds

( ) ( ) ( )( ) ( ), , 'b

C af x y dx f x t y t x t dt=

( ) ( ) ( )( ) ( ), , 'b

C af x y dy f x t y t y t dt=

( ) ( ) ( ), , , , , , , ,C C C

f x y z dx f x y z dy f x y z dz

( ) ( ) ( ) ( ), , , ,C C C

f x y dx g x y dy f x y dx g x y dy+ = +

Line Integral(example)

• Evaluate

Solution:

2 ,4 ,5 ,0 1

t t tt

ydx zdy xdz+

+ +

( )

( )

( )

( )

( ) ( )

( )

( )

1 1 1

2 ,4 ,5 , 0 0 00 1 ' ' '

1 1

0 0

4 2 5 4 2 5

4 20 5 2 29 10 24.5

y zt t tx

t x t y t z t

d d dydx zdy xdz t t dt t t dt t t dt

dt dt dt

t t t dt t dt

+

+ + = + + + +

= + + + = + =

Line Integral of Vector Field

• Reminder:• A work done by variable force f(x) in moving a particle from a to b along the x-

axis is given by .

• A work done by a constant force F in moving object from point P to point Q in space is .

• Unit tangent vector:

• Consider now a variable force F(x,y,z) along a smooth curve C. • Divide C into number of a small enough sub-arcs so that the force is roughly

constant on each sub-arc.

• The displacement vector becomes unit tangent (T) times displacement ( ):

( )b

aW f x dx=

W PQ= F𝐅 is ~c𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑜𝑛 𝑠𝑖

( ) ( ) ( )( ) * * * *

1, , , ,j i i i i i iPQ s x t y t z t t t t−= T

is

( )( )

( )

'

'

r tt

r t=T

Line Integral of Vector Field(cont)

• Finally the work of F(x,y,z) along C is given by

• Denote

( ) ( ) ( )( ) ( ) ( ) ( )( )

( ) ( )

* * * * * *

1

lim , , , ,

, , , ,

n

i i i i i i jn

i

C C

W x t y t z t x t y t z t s

x y z x y z ds ds

→=

=

= =

F T

F T F T

( ) ( )' or 'd r t dr r t= =r

( )( )( )

( )

'

'C

r tds r t

r t = F T F ( )'r t

( )( ) ( )( )( ) '

'

b b b

drr ta a a

dt r t r t dt d= F

F F r

Line Integral of Vector(example)

• Let VF be given by and

• the curve C byEvaluate .

Solution:

, ,x x y x y z= + + +F

Cd F r

( ) sin ,cos ,sin cos ,0 2r t t t t t t = +

( )

( )( ) ( )

( ) ( )

( )

22

'

2 2

2cos2sin

2

0

sin ,cos sin ,2 sin cos cos , sin ,cos sin

32cos 3sin 1 cos 2 1 cos 2

2

3 sin 2 31 cos 2 1 cos 2

2 2 2

r t dtr t

tt

C

d t t t t t t t t t dt

t t dt t t dt

td t t dt t

=

==

= + + − −

= − = + − −

= + − − = + −

F

F r

F r

2

0

sin 2

2

tt

− = −

Line integral of Vector vs Scalar fields

• Letandthen

( ) ( ) ( )( , , ) , , , , , , , ,x y z P x y z Q x y z R x y z=F

( )( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

'

, , , , , , , , ' , ' , '

, , ' , , ' , , '

, , , , , ,

b

C a

b

a

b

a

b

a

d r t r t dt

P x y z Q x y z R x y z x t y t z t dt

P x y z x t Q x y z y t R x y z z t dt

P x y z dx Q x y z dy R x y z dz

=

=

= + +

= + +

F r F

( ) ( ) ( ) ( ), , ,r t x t y t z t a t b=

Fundamental Theorem for Line Integrals

• Recall: Fundamental Theorem of Calculus (FTC)

• Definition: A vector field F is called a conservative vector field if there exist a potential, a function f, such that .

• Theorem: Let C be a smooth curve given by . Let F be a continuous conservative vector field, and f is a differentiable function satisfying . Then, .

f= F

( ) ,r t a t b

f= F

( )( ) ( )( )C C

d f d f r b f r a = = − F r r

( ) ( ) ( )'b

aF x dx F b F a= −

Fundamental Theorem for Line Integrals(cont)

• Proof:

( )( ) ( )

( )( )

( )( ) ( )( )

' , , , ,

b b

x y z

C a a

b b

x y z

a a

dx dy dzf d f r t r t dt f f f dt

dt dt dt

dx dy dz df f f dt f r t dt

dt dt dt dt

f r b f r a

= =

= + + =

= −

r

Fundamental Theorem for Line Integrals(cont)

• Definition: Let F be continuous on domain D. The integralis called independent of path in D if for any two curves C1, C2 with the same initial and end points, we have

• Corollary: A line integral of a conservative vector field is independent of path.

• Definition: A curve C is called closed if its terminal points coincides.

2 1C Cd d = F r F r

𝐶1

𝐶2

𝐴

𝐵

Cd F r

Fundamental Theorem for Line Integrals(cont)

• Theorem: The integral is independent of path in D if and only if on any closed curve C.

Proof: (➔) Let is independent of path in D. Let C be arbitraryclosed curve. Choose any two points on C, A and B. Let C1 be the curve from A to B, and C2 from B to A, so that , then

() Let on any closed curve C in D. Choose A,B∈D and let C1, C2 be arbitrary paths from A to B. is closed curve, thus

Cd F r

1 2C C C=

1 2 1 2

0C C C C C

d d d d d−

= + = − = F r F r F r F r F r

0C

d = F r

Cd F r

0C

d = F r

1 2C C C= −

1 2 1 2 1 2

0C C C C C C C

d d d d d d d−

= = + = − + = F r F r F r F r F r F r F r

Fundamental Theorem for Line Integrals(cont)

• Theorem: Suppose F=<P,Q> is continuous vector field on an open connected region D. If is independent of path in D, then F is conservative vector field in D, that is there is f such that .

Proof: Let (a,b)∈D be arbitrary fixed point. Define .

Due to independency of path we can choose path C from (a,b) to (x,y) that crosses (x1,y)∈D, x1 is const.

Similarly,

Cd F r

f= F

( )( )

( ),

,

,

x y

a b

f x y d= F r

( )( )

( )

( )

( )1, ,

, ,

,

x y x y

x

a b a b

d df x y d d

dx dx= = F r F r

( )

( )

1

0, no ,

,

xx y

x y

d dd Pdx Qdy

dx dx

=

+ = + F r

( )

( )

1 1

, 0

,

x y xdy

FTCx y x

dPdx P

dx

=

= =

( )( )

( )

( )

( )

( )

( )1

1

, , ,

, , ,

,

x y x y x y

y

a b a b x y

d d df x y d d d

dy dy dy= = + F r F r F r

0, no y

dPdx

dy

=

=( )

( )

1 1

, 0

,

x y ydx

FTCx y y

dQdy Qdy Q

dy

=

+ = =

(x,y1)

(x1,y) (x,y)

(a,b)