Chapter 9: Vector Differential Calculus

1

9.1. Vector Functions of One Variable

-- a vector, each component of which is a

function of the same variable

i.e., F(t) = x(t) i + y(t) j + z(t) k,

where x(t), y(t), z(t): component functions

t : a variable2( ) cos 2 3t t t j tk Fe.g.,

◎ Definition 9.1: Vector function of one variable

2

。 F(t) is continuous at some t0 if x(t), y(t), z(t) are all continuous at t0

。 F(t) is differentiable if x(t), y(t), z(t) are all differentiable

○ Derivative of F(t): ( ) ( ) ( ) ( )t x t y t z t F i j k

e.g., 2( ) cos 2 3t ti t j tk F

( ) sin 4 3t ti tj k F

e.g., 1ln , 0, 1

1G t i tk t t

t

3

○ Curve: C(x(t), y(t), z(t)), in which

x(t), y(t), z(t): coordinate functions

x = x(t), y = y(t), z = z(t): parametric equations

F(t)= x(t)i + y(t)j + z(t)k: position vector

pivoting at the origin

Tangent vector to C: ( )tF

Length of C: 2 2 2( ) ( ) ( ) ( )b a

a bt dt x t y t z t dt F

4

○ Example 9.2: cos ,sin , , 4 43tC t t t

Position vector: ( ) cos sin / 3t ti tj t k F

Tangent vector: ( ) sin cos 1/ 3t ti tj k F

Length of C:4 4 2 2

4 4( ) sin cos 1/ 9t dt t t dt

F

4

410 / 3 8 10 / 3dt

5

○ Distance function: ( ) ( )t

as t d F

2 2 2( ) ( ) ( ) ( )ds

t x t y t z tdt

F

t(s): inverse function of s(t)

○ Let ( ) ( ( )) ( ( )) ( ( )) ( ( ))s t s x t s i y t s j z t s k G F

Unit tangent vector: ( ) ( )s sT G

1 1( ) ( ( )) ( ) ( ) (t)

( )

d d dts t s t t

dsds dt ds tdt

G F F F FF

6

。 Example 9.3:

Position function: ( ) cos sin / 3 , 4 4t ti tj t k t F

( ) sin cos 1/ 3 , F ( ) 10 / 3t ti tj k t F

4( ) ( ) 10 / 3 10( 4 ) / 3

t t

as t F d d t

Inverse function: 34

10t s s

7

3( ) ( ( )) ( 4 )

10s t s s G F F

3 3 1 3cos( 4 ) sin( 4 ) ( 4 )

310 10 10s s s k i j

3 3 1 4cos( ) sin( ) ( )

310 10 10s s s

i j

Unit tangent vector:

3 3 3 3 1( ) sin( ) cos( )

10 10 10 10 10s s s G i j k

8

○ Assuming that the derivatives exist, then

(1)

(2)

(3)

(4)

(5)

( ) ( ) ( ) ( )t t t t F G F G

( ) ( ) ( ) ( ) ( ) ( )f t t f t t f t t F F F

( ) ( ) ( ) ( ) ( ) ( )t t t t t t F G F G F G

( ) ( ) ( ) ( ) ( ) ( )t t t t t t F G F G F G

( ( )) ( ) ( )f t f t f t F F

9

9.2. Velocity, Acceleration, Curvature, Torsion

A particle moving along a path has position vector

( ) ( ) ( ) ( )t x t y t z t a t b F i j k,

Distance function: ( ) ( )t

as t d F

◎ Definition 9.2:

Velocity: ( ) ( )t tv F (a vector) tangent to the curve of motion of

the particle

Speed : ( ) ( ) ( )ds

v t t tdt

v F

(a scalar) the rate of change

of distance w.r.t. time

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Acceleration: ( ) ( )t ta v or ( ) ( )t ta F

(a vector) the rate of change of velocity

w.r.t. time

○ Example 9.4: 2( ) sin 2 tt t e t F i j k

The path of the particle is the curve whose parametric equations are

sin , 2 ,tx t y e z t

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Velocity: ( ) ( ) cos 2 2tt t t e t v F i j k

Speed: 2 2 2( ) ( ) cos 4 4 .tv t t t e t v

Acceleration:

( ) ( ) ( ) sin 2 2tt t t t e a v F i j k

Unit tangent vector:

1 1( ) ( ) ( )

( ) ( )t t t

t v t

T F v

F

12

○ Definition 9.4: Curvature

( )( ) ( )

d sk s s

ds

TT

(a magnitude):

the rate of change of the unit tangentvector w.r.t. arc length s

For variable t, ( ) ( )

( ) ,d t d t dt

k tds dt ds

T T

1 1 1 , ( ) ( )

( ) ( )

dtk t t

ds ds dt t t

T

F F

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○ Example 9.7: Curve C:

2cos sin , sin cos , ,x t t t y t t t z t t > 0

Position vector:

2( ) [cos sin ] [sin cos ]t t t t t t t t F i j k

14

Tangent vector: ( ) cos sin 2t t t t t t F i j k

Unit tangent vector:

( ) 1( ) [cos sin 2 ]

( ) 5

tt t t

t

F

T i j kF

1 ( ) [ sin cos ]

5t t t T i j

Curvature:

2 2( ) 1 1 1( ) [sin cos ]

( ) 5 55

tk t t t

t tt

T

F

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◎ Definition 9.5: Unit Normal Vector

1( ) ( ), ( ) 0

( )s s k s

k s N T

i)( ) ( )

( ) ( ) , ( ) 1( ) ( )

s sk s s s

k s s

T TT N

T

ii) 2

( ) 1, ( ) ( ) ( ) 1s s s s T T T T

Differentiation

( ) ( ) ( ) ( ) 2 ( ) ( ) 0s s s s s s T T T T T T

( ), ( ) : orthogonals sT T

1 ( ) ( ), , :

( )s s N s T s orthogonal

k s N T

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○ Example 9.8: Position vector: t > 0

2( ) [cos sin ] [sin cos ]t t t t t t t t F i j k

Write ( )tF as a function of arc length s

( ) 5t t F (Example 9.7)

2

0 0

5( ) 5

2

t ts d d t F

Solve for t,

1 41 4

2, 2 5

5t s s

Position vector:

2

( ) ( ( ))

[cos( ) sin( )]

[sin( ) cos( )]

s t s

s s s

s s s s

G F

i

j k

17

Unit tangent vector:

( ) ( )s sT G

2 2 21 1cos( ) sin( )

2 2s s i j k

4

2 4 41 4

1 5 2( ) 1

4 4 5s

T

Curvature:

3 3

( ) sin( ) cos( )4 4

s s ss s

T i j

18

1 26 3

31 2

1 4 3 4

( ) ( )16 4

1 2 1 1 1 , for 0

5 54 2

k s ss s

ss s

T

21 4

2( , 5 2 )

5t s s t

3 4 3 4 1 4

1 1 1 1 1 2 1( )

55 5 52 2k t

tts

Unit normal vector: 1( ) ( )

( )s s

k sN T

3 3

3

4[ sin( ) cos( )

4 4

sin( ) cos( )

ss s

s s

s s

i j

i j

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9.2.1 Tangential and Normal Components of

Acceleration T Na a a T N

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◎ Theorem 9.1: T Na a a T N,

where 2 and T Nv ka a v

Proof:

( ) 1 ( ) ( )

tt vvt

FT v, v T

F

2

2

( )

d v v tdt

d dsv vds dt

v v s

v v k

a v T T

TT

T T

T N

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○ Example 9.9: Compute Ta and Na

for curve C with position vector

2( ) cos( ) sin( ) sin( ) cos( )t t t t t t t t

F i j k

Velocity: ( ) ( ) cos sin 2t t t t t t t v F i j k

Speed: ( ) ( ) 5v t t t F

Tangential component: 5Tdvadt

Acceleration vector:

cos sin sin cos 2t t t t t t

a v i j k

2 2 2 2 2 25 5 5, N Tt a a t t a a

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Normal component: 0tNa t

Acceleration vector: 5 t a T N

Since 2 25Na t kv t k , curvature: 15

kt

Unit tangent vector:

1 1( ) cos sin 25

t t tv

T v i j k

Unit normal vector:

1 1 1( ) ( ) ( )

5 1 sin cos sin cos5 5

dt dt s tk k ds dt kv

t t t t tt

TN T T

i j i j

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◎ Theorem 9.2: Curvature3

k / F F F

Proof: 2Ta kv a T N

2 2Ta kv kv

T a T T T N T N

2 2

2sinkv kv T a T N T N

112

, sin T N

22kv , k / v T a T a

, v T F F a F , F

2 3

1k

F FF F

FF F

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○ Example 9.10:

Position function: 2 3( )t t t t F i j k

2( ) 2 3 ( ) 2 6t t t t t F i j k, F i j

2 22 3 1 6 2 62 6 0t t t t

t

F F

i j ki j k

22 4

3 322 2 4

6 2 636 4 36( )

2 3 4 9 1

t tt tk t

t t t t

i j k

i j k

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9.2.3 Frenet Formulas

Let Binormal vector: B T N

T, N, B form a right-handed rectangular

coordinate system

This system twists and changes orientation along curve

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○ Frenet formulas:

' k k , T N, N T B B NThe derivatives are all with respect

to s.

(i) From Def. 9.5, 1

'k

N T ' k T N

(ii) 0 0, B T B T B T

( )( , : orthogonal) 0

k k

B T B T B N B NB N

0 B B

B is inversely parallel to N

Let B N, : Torsion

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(iii) N N TT N NN N BB

(a) 0 0 , N T N T N T

k k N T N T N N

(b) 0 0 , N B N B N B

( ) N B N B N N

(c) 0 N N

k N T B

＊ Torsion measures how (T, N, B) twists along the curve

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12.3 Vector Fields and Streamlines

○ Definition 9.6: Vector Field

-- (3-D) A vector whose components are functions of three variables

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

( , , )

x y z f x y z g x y z

h x y z

G i j

k

-- (2-D) A vector whose components are functions of two variables

( , ) ( , ) ( , )x y f x y g x y K i j

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。 A vector filed is continuous if each of its component functions is continuous.

。 A partial derivative of a vector field -- the vector fields obtained by taking the partial derivative of each component function

e.g., ( , , ) cos( ) ( )x y z x y x x z F i j k

sin( )x x yx

F

F i j k

sin( )y zx yy z

F F

F i, F k

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◎ Definition 9.7: Streamlines

F: vector field defined in some 3-D region Ω: a set of curves with the property that through each point P of Ω, there passes exactly one curve from The curves in are streamlines of F if at each point ( , , )x y z in Ω, F ( , , )x y z

curve in ( , , )x y z passing through

is tangent to the

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○ Vector filed: f g h F i j k

( ( ), ( ), ( ))C x y z : Streamline of F

Parametric equations -- ( ), ( ), ( )x x y y z z

Position vector -- ( ) ( ) ( ) ( )x y z P i j k

Tangent vector at ( ( ), ( ), ( ))x y z

( ) ( ) ( ) ( )x y z P i j k

--

32

( ( ), ( ), ( ))x y z F is also tangent to C at

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

( ( ), ( ), ( ))x y z F // ( )P

( ) ( ( ), ( ), ( ))t x y z P F

dx dy dztf tg th

d d d i j k i j k

, ,dx dy dz

tf tg thd d d

dx dy dz dx dy dzt

fd gd hd f g h

33

○ Example 9.11: Find streamlines

Vector field:2 2x y F i j k

From 2 2 1

dx dy dz dx dy dz

f g h x y

Integrate 2

1dxdz z c

xx

Integrate1

ln2 2

dydz y z k

y

Solve for x and y21

, zx y aez c

Parametric equations of the streamlines

21, ,zx y ae z z

z c

34

Find the streamline through (-1, 6, 2).

4 411 , 6 3, 6

2ae c a e

c

4 21

, 6 ,3

zx y e z zz

35

9.4. Gradient Field and Directional Derivatives

◎ Definition 9.8:

Scalar field: a real-valued function ( , , )x y ze.g. temperature, moisture, pressure,

hightGradient of : a vector field

x y z

i + j k

36

e.g., 2( , , ) cos( )x y z x y yz

2

2 2 2

2 cos( ) [ cos( )

sin( )] sin( )

xy yz x yz

x yz yz x y yz

i

j k

。 Properties: ( ) , ( )c c

○ Definition 9.9: Directional derivative of in the direction

of unit vector a b c u i j k

( , , )d

D x at y bt z ctdt

u

37

◎ Theorem 9.3: 0 0( ) ( )D P P u u

Proof: By the chain rule

( , , )

( ) ( ) ( )

dD x at y bt z ct

dtd x at d y bt d z ct

x dt y dt z dt

a b cx y z

u

u

0 0 0 0

0

( ) ( , , )

( )

dD P x at y bt z ct

dtP

u

u

38

○ Example 9.13:2( , , ) zx y z x y xe

0

1(2, 1, ), u (i 2 j k)

6P

2 (2 )

x y z

z zxy e x xe

i + j k

= i + j k

(2, 1, ) ( 4 ) 4 2e e = i j k

(2, 1, ) (2, 1, )D u u

1[( 4 ) 4 2 ] ( 2 )

61 3

( 4 8 2 ) (4 )6 6

e e

e e e

i j k i j k

39

◎ Theorem 9.4: ( , , )x y z has its

1. Maximum rate of change,0

( )P , in the direction of0( )P

2. Minimum rate of change,0( )P , in the direction of 0( )P

Proof: 0 0 ( ) ( )D P P u u

0 0( ) cos ( ) cosP P 　 　u

cos 1

cos 1

Max.:

Min.:

40

○ Example 9.4:

2

0( , , ) 2 , 2, 1, 1yx y z xz e z P

2( , , ) 2 (2 2 )y yx y z z e z x ze i j k

0( ) (2,1,1) 2 (4 2 )P e e i j k

The maximum rate of change at0 :P

2 20

2

( ) 4 (4 2 )

5 16 20

P e e

e e

The minimum rate of change at 0 :P

20

( ) 5 16 20P e e

41

9.4.1. Level Surfaces, Tangent Planes, and Normal Lines

○ Level surface of : a locus of points

( , , ) , :x y z k k constant

e.g., 2 2 2( , , )x y z x y z k

Sphere (k > 0) of radius k

Point (k = 0),

Empty (k < 0)

42

○ Tangent Plane at point 0P to ( , , )x y z k

Normal vector: the vector perpendicular to the tangent plane

43

○ Theorem 9.5: Gradient ( )P normal to ( , , )x y z k

at point on the level surface

Proof: Let ( ), ( ), ( )C x t y t z t : a curve passing point P

( , , )x y z k on surface

C lies on ( , , )x y z k

( ( ), ( ), ( )) ,x t y t z t k t

44

( ( ), ( ), ( )) 0d

x t y t z tdt

( ) ( ) ( ) ( ) ( ) ( )x y z

P x t P y t P z t

( ) [ ( ) ( ) ( ) ]

( ) ( ) 0

P x t y t z t

P t

i j k

T

( )P normal to ( )tT

This is true for any curve passing P on the

surface. Therefore, normal to the surface( )P

45

○ Find the tangent plane to 0( , , )x y z k at P

Let (x, y, z): any point on the tangent plane

0 0 0Vector ( ) ( ) ( )x x y y z z i j k

orthogonal to the normal vector0( )P

0 0 0 0( ) [( ) ( ) ( ) ] 0P x x y y z z i j k

The equation of the tangent plane:

0 0 00 0 0

( ) ( ) ( )( ) ( ) ( ) 0

P P Px x y y z z

x y z

46

○ Example 9.16: Consider surface sin( )z xy

Let ( , , ) sin( )x y z xy z

The surface is the level surface ( , , ) 0x y z

Gradient vector:

x y z

i j k

cos( ) cos( )y xy x xy i j k

Tangent plane at 0 0 0 0( , , )P x y z

0 0 00 0 0

( ) ( ) ( )( ) ( ) ( ) 0

P P Px x y y z z

x y z

47

0 0 0 0 0 0 0 0

0

cos( )( ) cos( )( )

( ) 0

y x y x x x x y y y

z z

0 0 0( , , ) (2,1,sin(2))Let x y z

cos2( 2) 2cos2( 1) sin 2 0x y z

48

9.5. Divergence and Curl

( ) ( , , ) ( , , ) ( , , )x, y,z f x y z g x y z h x y z F i j k

○ Definition 9.10: Divergence (scalar field)

f g h

divx y z

F

e.g., 22 ( sin( )) x yxy xyz yz ze F i j k

22 cos( ) x ydiv y xz z yz e F

49

○ Definition 9.11: Curl (vector field)

h g f h g fcurl

y z z x x y

F i j k

e.g., 2 xy xz ze F i j k

curl 2 (2 1)xx ze z F i j k

50

○ Del operator:x y z

i j k

。 Gradient:

x y zx y z

i j k i j k

。 Divergence:

f g hx y z

f g h div

x y z

F i j k i j k

F

。Curl:

x y z

f g h

h g f h g f

y z z x x y

curl

i j k

F

i j k

F

51

○ Theorem 9.6: 0

Proof: x y z

i j k

x y z

x y z

i j k

2 2 2 2

2 2

- - y z z y z x x z

-x y y x

i j

k 0

52

◎ Theorem 9.7: 0 F

Proof:h g f h g f

- - -y z z x x y

F = i j k

x y z

F i j k

h g f h g f- - -

y z z x x y

i j k

h g f h g f- - -

x y z y z x z x y

02 2 2 2 2 2h g f h g f

- - -x y x z y z y x z x z y

FORMULA

○ Position vector of curve

F(t)= x(t)i + y(t)j + z(t)k

。 Distance function: ( ) ( )t

as t d F

。 Unite tangent vector: ( )

( ) ,( )

tt

t

F

TF

( ) ( )s sT G

( ) ( ( ))s t sG Fwhere

C(x(t), y(t), z(t))

( ) ( ) ( ) ( )t t t t F G F G

( ) ( ) ( ) ( ) ( ) ( )f t t f t t f t t F F F

( ) ( ) ( ) ( ) ( ) ( )t t t t t t F G F G F G

( ) ( ) ( ) ( ) ( ) ( )t t t t t t F G F G F G

( ( )) ( ) ( )f t f t f t F F

○ Velocity:

Speed : ( ) ( ) ( )ds

v t t tdt

v F

Acceleration: ( ) ( )t ta v ( ) ( )t ta F

T Na a a T N Ta dvdt Na 2kv

or

, where = , =○ Curvature:

( )( ) ( )

d sk s s

ds

TT

1( ) ( )

( )k t t

t

T

F

3k / F F F

( ) ( )t tv F

1( ) ( )

( )s s

k sN T

B T N: B N

k' T N k N T B B N

( ) ( ) ( )f g h F i j k

( ( ), ( ), ( ))C x y z dx dy dz

f g h

○ Unit Normal Vector:

○ Binormal vector:

Torsion ○ Frenet

formulas:

○ Vector filed:

Streamline:

○ Scalar field:

Gradient:

○ Directional derivative:

○ Divergence:

○ Curl:

0 0 F

( , , )x y z

x y z

i + j k

0 0( ) ( )D P P u u

f g h

divx y z

F

h g f h g fcurl

y z z x x y

F i j k

○