Kreyszig ; 9-1

Chapter 9 Vector Differential Calculus, Grad, Div, Curl

9.5 Curves. Arc length. Curvature. Torsion 9.6 Calculus Review 9.7 Gradient of a Scalar Field. Directional Derivative 9.8 Divergence of a Vector Field 9.9 Curl of a Vector Field

Kreyszig ; 9-2 9.5 Curves. Arc length. Curvature. Torsion

A major application of vector calculus concerns curves and surfaces. A curve C can be represented by a vector function with a parameter t.

( ) ( ) ( ) ( ) ( ) ( ) ( ) ˆˆ ˆ, , r t x t y t z t x t i y t j z t k= = + +

: parametric representation

The direction of the curve is determined by increasing values of t • Another representation of a curve ( ) ( ), , x y f x z g x= =

↑ ↑ projection of C onto xy plane projection of C onto xz plane Another representation of C can be given by an intersection of two surfaces ( ) ( ), , 0, , , 0F x y z G x y z= = Example 1 Circle The circle 2 2 4x y+ = , 0z = is given In parametric representation ( ) [ ] [ ]( ), ( ), ( ) 2cos , 2sin , 0r t x t y t z t t t= ⇒

: 0 2t≤ ≤ π

Check ( ) ( )2 22 2 2cos 2sin 4x y t t+ ⇒ + = For t=0 : ( ) [ ]0 2, 0, 0r =

For / 2t = π : ( ) [ ]/ 2 0, 2, 0r π =

… As t increases, ( )r t

moves counterclockwise.

Kreyszig ; 9-3 Example 2 Ellipse An ellipse in the form of vector function ( ) [ ] [ ]( ), ( ), ( ) cos , sin , 0r t x t y t z t a t b t= =

→ cos , sinx a t y b t= =

→ ( ) ( )2 2

2 2

2 2cos sin 1x y

t ta b

+ ⇒ + = : Conventional form of an ellipse

Example 3 straight line A straight line in the direction of a unit vector b

.

It passes through a point A. a

is a position vector.

For a point on the line is given by a position vector ( ) [ ]1 1 2 2 3 3, , r t a t b a tb a tb a tb= + ⇒ + + +

• Curves Plane curve : a curve in a plane. Twisted curve : the others. Simple curve : a curve without multiple points at which the curve intersects or touches itself. Arc of a curve : a portion between two points in the curve, simply called 'curve'. Example 3 circular helix A circular helix in parametric representation ( ) [ ]cos , sin , r t a t b t ct=

c>0 : right handed screw c<0 : left handed screw c=0 : circle

Kreyszig ; 9-4 Tangent to a Curve A tangent is a straight line touching a curve. A vector passing through both P and Q is

( ) ( )1r t t r t

t+ ∆ − ∆

Take the limit

( ) ( ) ( )0

1' lim

tr t r t t r t

t∆ →= + ∆ − ∆

↑ Tangent vector of C at P The unit tangent vector

'ˆ'

ru

r=

: Direction of increasing t

• The function for the tangent of C at P ( ) 'q w r w r= +

: and 'r r

, constant vectors. w, parameter

Example 5 Tangent to an Ellipse

Find the tangent to the ellipse 2 211

4x y+ = at ( ): 2, 1 / 2P

An ellipse, 2 2

2 2 1x ya b

+ = → a=2, b=1

The parametric presentation → ( ) [ ]2cos , sin , 0r t t t=

At ( ): 2, 1 / 2P → / 4t = π

The derivative → ( ) [ ]' 2sin , cos , 0r t t t= −

At P → ( )' / 4 2, 1/ 2, 0r π = −

The tangent is

( ) ( ) ( )( )2, 1 / 2 2, 1 / 2 2 1 , 1 / 2 1q w w w w = + − ⇒ − +

Kreyszig ; 9-5 Length of a Curve The length of the curve

0

1

lim ( ) ( ( 1) )N

tn

l r a n t r a n t∆ →

=

= + ∆ − + − ∆∑

0

1

( ) ( ( 1) )lim

N

tn

r a n t r a n tt

t∆ →=

+ ∆ − + − ∆⇒ ∆

∆∑

( )b

a

r t dt′⇒ ∫

Hence

( )b b

a a

l r t dt r r dt′ ′ ′= = •∫ ∫

Arc Length s of a Curve The upper limit is now a variable

( )t

a

s t r r dt′ ′≡ •∫ : Arc length of C. dr

rdt

′ =

(11)

Linear element ds Differentiate (11)

2 2 2 2ds dr dr dx dy dz

dt dt dt dt dt dt = • ⇒ + +

, : ˆˆ ˆ( ) ( ) ( ) ( )r t x t i y t j z t k= + +

→ 2 2 2 2( ) ( ) ( ) ( )ds dx dy dz dr dr= + + ⇒ •

: ˆˆ ˆ dr dx i dy j dz k= + +

ds is called the linear element of C. Arc Length as Parameter The parameter t can be replaced by the arc length s in the equation of a curve. For example, the unit tangent vector is given as

( ) ( )ˆ 'u s r s=

=dsrd

where

ssrssrsr

s ∆−∆+

=→∆

)()(lim)('0

Note that

( )' 1ds

r sds

= =

Kreyszig ; 9-6 Example 6 Circular Helix. Circle. Arc Length as Parameter.

A helix ( ) [ cos , sin , ]r t a t a t ct=

→ ( ) [ sin , cos , ]r t a t a t c′ = −

2 2r r a c′ ′• = +

The arc length is 2 2 2 2

0

t

s a c dt t a c= + = +∫

Replace t by s in ( )r t

2 2 2 2 2 2 2 2

ˆˆ ˆcos sins s s s

r a i a j c ka c a b a c a c

= + + + + + +

↑ ≡

r s* ( ) , a new function

• For a circle, let c=0 then /t s a=

→ 1ˆ ˆ( ) cos sin

s s sr r s a i a j

a a a ≡ = +

Curves in Mechanics. Velocity. Acceleration A curve C may represent a path of a moving body. ( )r t

: t is time in this case

( ) ( )'v t r t=

: velocity vector

The magnitude

dr ds ds

v rds dt dt

′= ⇒ ⇒

))()((dtdssu

dtrdtv

==

↑ Speed of the body along the curve, v

is parallel to the tangent of C.

)'()'( srofdirectionhastr

( v

is the velocity of the moving body along C ) ( ) ( )' ( )"a t v t r t= =

: acceleration vector

Tangential and Normal Acceleration In general tan norma a a= +

: tangential and normal acceleration vectors

Proof:

( )( )

dr t dr dsv t

dt ds dt= ⇒

: s, arc length

↑ = ˆ( )u s , unit tangent vector on C Differentiate the both sides

2 2

2

ˆˆ ˆ( ) ( ) ( )

d ds du ds d sa t u s u s

dt dt ds dt dt = ⇒ +

:

ˆd ( )ˆ( ), tangential vector. , normal vectordu s

u ss

↑ ↑ Normal acceleration. Tangential acceleration.

Kreyszig ; 9-7 Example 7 Centripetal(구심) acceleration. Centrifugal(원심) Force. The revolving path of a small body ˆ ˆ( ) cos( ) sin( ) r t R t i R t j= ω + ω

: ω, angular speed

The velocity vector : ˆ ˆ( ) sin( ) cos( ) v r t R t i R t j′= ⇒ − ω ω + ω ω

The speed of the body : v R= ω

Linear speedAngular speed =

Distance to the center : ω

The acceleration vector : 2 2 2ˆ ˆcos( ) sin( ) a v R t i R t j r′= ⇒ − ω ω − ω ω ⇒ −ω

↑ Direction toward the center → 2 2a r R= ω ⇒ω

: centripetal acceleration

If the body is to be in the circular orbit during rotation, someone should supply the centripetal force to the body. Otherwise, it will deviate from the orbit due to the centrifugal force. Example 8 Superposition of Rotations

A projectile moves along a meridian of the rotating earth. Find its acceleration. The angular speed of the earth rotation is ω.

The unit vector b

also rotates, ( )ˆ ˆ ˆcos( ) sin( )b t t i t j= ω + ω The projectile on the meridian with an angular speed of γ . Its position vector, ˆ ˆ( ) cos( ) sin( )r t R t b R t k= γ + γ

: R, radius of the earth

The derivatives ˆ ˆ ˆ( ) cos( ) sin( ) cos( )v r t R t b R t b R t k′ ′= = γ − γ γ + γ γ

2 2ˆ ˆ ˆ ˆcos( ) 2 sin( ) cos( ) sin( )a v R t b R t b R t b R t k′ ′′ ′= = γ − γ γ − γ γ − γ γ

Inserting b̂′′ , )ˆˆ)sin(ˆ)cos(''ˆ( 222 bwjwtwiwtwb −=−−= ,

2 2ˆ ˆcos( ) 2 sin( )a R t b R t b r′= −ω γ − γ γ − γ

↑ ↑ ↑ Centripetal acceleration by the earth. ↑ Centripetal acceleration by the rotation of P on the meridian. Coriolis acceleration due to interaction of two rotations. Coriolis acceleration is in the direction opposite to the earth’s rotation → The projectile deviates from the meridian toward the earth’s rotation direction in the Northern Hemisphere. (??)

If the projectile is to be in the meridian during rotation, someone should supply the Coriolis acceleration to the projectile. Otherwise, it will deviate from the meridian in the direction opposite to the Coriolis acceleration. Curvature and Torsion (skip)

Kreyszig ; 9-8

9.6 Calculus Review Chain Rule

Let all the functions be continuous and have continuous first partial derivatives in their domains. Let every point (u, v) in B has the corresponding point [ x(u,v), y(u,v), z(u,v) ] in D. (u: age, v: weight, [x,y,u]: location of participant, w: prize money in climbing competition) The function defined in B, [ ]( , ), ( , ), ( , )w f x u v y u v z u v= has first partial derivatives

w w x w y w zu x u y u z u

∂ ∂ ∂ ∂ ∂ ∂ ∂= + +

∂ ∂ ∂ ∂ ∂ ∂ ∂,

w w x w y w zv x v y v z v

∂ ∂ ∂ ∂ ∂ ∂ ∂= + +

∂ ∂ ∂ ∂ ∂ ∂ ∂.

Simple examples : If ( ), , w f x y z= and ( ) ( ) ( ), , x x t y y t z z t= = = , then

w w x w y w zt x t y t z t

∂ ∂ ∂ ∂ ∂ ∂ ∂= + +

∂ ∂ ∂ ∂ ∂ ∂ ∂.

If ( )w f x= and ( )x x t= , then

w w xt x t

∂ ∂ ∂=

∂ ∂ ∂.

Mean Value Theorem

Let ( , , )f x y z be continuous and have continuous first partial derivative in a domain D. Let two points, 0 0 0: ( , , )oP x y z and 0 0 0: ( , , )P x h y k z l+ + + , be in D and the line segment between the points be in D. Then,

0 0 0 0 0 0( , , ) ( , , )f f f

f x h y k z l f x y z h k lx y z∂ ∂ ∂

+ + + − = + +∂ ∂ ∂

The partial derivatives are evaluated at a point on the line segment

A simple example: For a function of two variables

0 0 0 0( , ) ( , )f f

f x h y k f x y h kx y∂ ∂

+ + − = +∂ ∂

Kreyszig ; 9-9 9.7 Gradient of a Scalar Field. Directional Derivative Definition 1 Gradient

The gradient of a scalar function ( , , )f x y z is defined as

ˆˆ ˆgrad f f f

f i j kx y z∂ ∂ ∂

= + +∂ ∂ ∂

: a vector function

"del" operator is defined as

ˆˆ ˆi j kx y z∂ ∂ ∂

∇ = + +∂ ∂ ∂

: differential operator

→ grad f f=∇ Directional Derivative Definition 2 Directional Derivative

Directional derivative of f at P in the direction of b̂ is defined by

( )

0

( )limb s

f Q f PdfD f

ds s→

−= ≡ ,

where Q is a variable point displaced from P by ˆ( )sb

The line L in Cartesian Coordinates ˆˆ ˆ( ) ( ) ( ) ( )r s x s i y s j z s k= + +

′ = ′ + ′ + ′r s x s i y s j z s k( ) ( ) ( ) ( ) : this is a unit vector

b̂⇒ The function f on C in parametric representation ( )( ), ( ), ( )f x s y s z s The directional derivative

ˆ ˆˆ ˆ ˆ ˆdf f x f y f z f f f x y zi j k i j k

ds x s y s z s x y z s s s∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = + + ⇒ + + • + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

→ ˆ(grad )df

f bds

⇒ •

Example 1 Directional derivative Find the directional derivative of 2 2 2( , , ) 2 3f x y z x y z= + + at the point P: (2,1,3) in the direction of ˆˆ 2a i k= −

ˆˆ ˆgrad 4 6 2f xi yj zk= + + → (2,1,3)ˆˆ ˆ(grad ) 8 6 6f i j k= + +

The unit direction vector : 1 2 ˆˆˆ5 5

a i k= −

The directional derivative : ( )1 2 ˆ ˆˆ ˆ ˆ8 6 6 1.7895 5aD f i k i j k

= − • + + = −

Kreyszig ; 9-10 Gradient Is a Vector. Maximum Increase Theorem 1 Vector Character is Gradient.

For a scalar function ( ) ( , , )f p f x y z= , its gradient is a vector function and its direction corresponds to the maximum increase of f on P. Its magnitude and direction are independent of the coordinate systems.

proof: ˆ ˆgrad grad cos grad cosbD f b f b f f= • ⇒ γ ⇒ γ

→ The maximum directional derivative for ˆ grad b f , or γ=0 grad direction of max. Dgf f .

Its max. value is gradf . • The directional derivative

0

( ) ( )lim

s

f Q f Ps∆ →

−∆

: independent of coordinates

and b̂ is fixed in space. Therefore, gradf should be independent of a coordinate system Gradient as Surface Normal Vector A surface S : ( , , )f x y z c=

A curve C : ˆˆ ˆ( ) ( ) ( ) ( )r t x t i y t j z t k= + +

Its derivative : ˆˆ ˆ( ) ( ) ( ) ( )r t x t i y t j z t k′ ′ ′ ′= + +

Tangent vector of the curve.

If C is on S, the surface equation becomes [ ]( ), ( ), ( )f x t y t z t c= .

Differentiate the surface function w.r.t t

0f f f

x y zx y z∂ ∂ ∂′ ′ ′+ + =∂ ∂ ∂

: ' /x dx dt= , …

→ (grad ) 0f r ′• =

: The tangent vector of C is also tangent to S ↑ ↑ Tangent to C and S. Surface normal vector Theorem 2 Gradient as Surface Normal Vector

When a surface S is given by ( , , )f x y z c= , grad f at a point P on S represents the normal vector of S at P.

Example 2 Gradient as a surface normal vector A cone is given by 2 2 24( )z x y= + . Find a unit surface normal vector at P : (1,0,2) The surface function is ( )2 2 24 0x y z+ − =

→ At Pˆ ˆˆ ˆ ˆgrad 8 8 2 grad 8 4f x i y j z k f i k= + − → = −

The unit surface normal vector is 8 4 ˆˆˆ80 80

n i k= −

Kreyszig ; 9-11 Vector Fields that are Gradients of Scalar Fields (Potentials) A vector function can be obtained by the gradient of a scalar function. ( ) ( )gradv P f P=

( )f P is called potential.

In this case, the field by ( )v P

is a conservative field ( no loss of energy)

9.8 Divergence of a Vector Field A differentiable vector function 1 2 3

ˆˆ ˆ( , , ) ( , , ) ( , , ) ( , , )v x y z v x y z i v x y z j v x y z k= + +

The divergence of v

is defined as

31 2divvv v

vx y z

∂∂ ∂= + +∂ ∂ ∂

Using del operator

1 2 3ˆ ˆˆ ˆ ˆ ˆdiv ( )v i j k v i v j v k v

x y z∂ ∂ ∂

⇒ + + • + + ≡∇• ∂ ∂ ∂

: a scalar function

Theorem 1 Invariance of the divergence

div v

is a scalar function independent of coordinate systems

In xyz -space with 1 2 3, , v v v In * * *x y z -space with * * *1 3 3, , v v v

→ 31 2div vv v

vx y z

∂∂ ∂= + +∂ ∂ ∂

→

** *31 2

* * *div vv v

vx y z

∂∂ ∂= + +∂ ∂ ∂

It will be proved in 10.7 • If f is twice differentiable

ˆˆ ˆgradf f f

f i j kx y z∂ ∂ ∂

= + +∂ ∂ ∂

Take the divergence

2 2 2

22 2 2div(grad )f f f

f fx y z∂ ∂ ∂

= + + ⇒∇∂ ∂ ∂

: 2 2 2

22 2 2x y z

∂ ∂ ∂∇ ≡ + +

∂ ∂ ∂, Laplace operator

→ 2div(grad )f f=∇

Kreyszig ; 9-12 Example 2 Flow of a Compressible fluid. Physical Meaning of the Divergence A small box B with a volume V x y z∆ = ∆ ∆ ∆ . No source or sink in the volume. Fluid flows through the box B. The fluid velocity vector 1 2 3

ˆˆ ˆv v i v j v k= + +

Flux density is defined as 1 2 3

ˆˆ ˆu v u i u j u k= ρ = + +

↑ Density of the fluid The flux density is the transfer of certain quantity across a unit area per unit time. ↑ Mass of fluid in this case • Loss of mass by outward flow through surfaces of B (1) Loss of mass due to flow in y direction during a time interval∆t

{ } { }2 2 2 2 2( ) ( ) ( ) ( )y y y y y yV

u x z t u x z t u u x z t u ty+∆ +∆

∆∆ ∆ ∆ + − ∆ ∆ ∆ ⇒ − ∆ ∆ ∆ ⇒ ∆ ∆

∆

↑ 2u= ∆ (2) Loss due to flow in x direction during∆t

1 1 1( ) { ( ) }x x xV

u y z t u y z t u tx+∆

∆∆ ∆ ∆ + − ∆ ∆ ∆ ⇒ ∆ ∆

∆

(3) Loss due to flow in z direction during ∆t

3 3 3( ) { ( ) }z z zV

u x y t u x y t u tz+∆

∆∆ ∆ ∆ + − ∆ ∆ ∆ ⇒ ∆ ∆

∆

The total loss of mass in B during∆t

( )31 2 Vuu u

V t tx y z t

∂ ρ∆∆∆ ∆ + + ∆ ∆ = − ∆ ∆ ∆ ∆ ∂

↑ ↑ Reduced mass inside B. Loss by flow through six side surfaces of B.

→ div ut

∂ρ= −

∂

: continuity equation, conservation of mass

For a steady flow, 0t

∂ρ=

∂ → div 0u =

For a constant ρ (incompressible) → div 0v =

, condition of incompressibility

Kreyszig ; 9-13 9.9 Curl of a Vector Field The curl of a vector function 1 2 3

ˆˆ ˆ( , , )v x y z v i v j v k= + +

is defined as

1 2 3

ˆˆ

curl

i j k

v vx y z

v v v

∂ ∂ ∂=∇× =

∂ ∂ ∂

3 2 1 3 2 1ˆˆ ˆv v i v v j v v k

y z z x x y∂ ∂ ∂ ∂ ∂ ∂ ⇒ − + − + − ∂ ∂ ∂ ∂ ∂ ∂

Example 1 Curl of a Vector Function A vector function is given, ˆˆ ˆ3v yzi zxj zk= + +

, find the curl of v

ˆˆ

ˆˆ ˆ3 2

3

i j k

v xi yj zkx y z

yz zx z

∂ ∂ ∂∇× = ⇒ − + +

∂ ∂ ∂

Example 2 Rotation of a rigid body The rotation is described by the angular speed vector w

. → Its direction : right hand rule Its magnitude : angular speed, ω The linear speed of a point on the body : v w r= ×

Let the axis of rotation be z axis, ˆw k= ω

→

ˆˆˆ ˆ0 0

i j kv w r y i x j

x y z= × ⇒ ω ⇒ −ω +ω

The curl of v

ˆˆ

ˆcurl 2

0

i j k

v kx y zy x

∂ ∂ ∂= ⇒ ω

∂ ∂ ∂−ω ω

→ curl 2v w=

Theorem 2 Grad, Div, Curl

The curl of the gradient of a scalar function is always zero ( ) 0f∇× ∇ = : irrotational The divergence of the curl of a vector function is always zero ( ) 0v∇• ∇× =

Theorem 3 Invariance of the curl

curl v is a vector independent of coordinate systems