9.4 Vector and Scalar Functions and Fields. Derivatives

중앙대학교 건설환경플랜트공학과 교수

김 진 홍

- 7주차 강의 내용 -

9.4 Vector and Scalar Functions and Fields. Derivatives

 There are two kinds of functions,

• vector functions whose values are vectors

• scalar functions whose values are scalars )]

( ), ( ), ( [ )

(P v₁ P v₂ P v₃ P v

v 

f = f (P) * P is a point in the domain of definition.

 Vector function defines a vector field, and a scalar function defines a scalar field.

Notation. If we introduce Cartesian coordinates x, y, z, then )]

, , ( ), , , ( ), , , ( [ ) , ,

(x y z v₁ x y z v₂ x y z v₃ x y z

v 

Vector Calculus

• Convergence. An infinite sequence of vectors a_(n), n = 1, 2, …, is said to converge if there is vector a such that

(4)

lim

a is called the limit vector of that sequence, and

(5) ^{a n a}

n





 ( )

lim

• Cartesian coordinates being given, this sequence of vectors converges to a if and only if the three sequences of components of the vectors converges to the corresponding components of a.

Chap. 9 Vector Differential Calculus. Grad, Div, Curl

 Continuity. A vector function v(t) is said to be continuous at if it is defined in some neighborhood of and t t⁰

t0

(8)

lim

0

t v t v

t t





 If we introduce a Cartesian coordinate system,

k t v j t v i t v t v t v t v t

v( )[ ₁( ), ₂( ), ₃( )] ₁( )  ₂( )  ₃( )

Then v(t) is continuous at if and only if its three components are continuous

at . t⁰

t0

DEFINITION

Derivative of a Vector Function

A vector function v(t) is said to be differentiable at a point t if the following limit exists

(9)

This vector v'(t) is called the derivative of v(t).^Δt

t v t Δ t t v

v

t Δ

) ( ) ) (

(

'

lim

0



 



 In components with respect to a given Cartesian coordinate system, (10) v'  [v₁'(t),v₂'(t),v₃'(t)]

Hence the derivative v'(t) is obtained by differentiating each component separately. For instance, if v [t,t²,0], then v'[1,2t,0]

Derivative of a vector function

Chap. 9 Vector Differential Calculus. Grad, Div, Curl

 Familiar differentiation rules continue to hold for differentiating vector functions,

' )' (cv



' ' )'

(







' '

)'

(











' '

)'

(











) ' ( ) ' ( ) ' ( )'

(u v w  u vw  uv w  uvw

(11) (12) (13)

Partial derivatives of a Vector Function

 The components of a vector function

are differentiable functions of n variables . Then the partial derivative of v with respect to is denoted by and is defined as the vector function

k v j v i v v v v

v

 [

,

,

] 





t₁

,   ,

tm  /v t_m

Similarly, second partial derivatives are

t k t j v t t i v t t

v t

t v

m m

m

m  

 





 





 







1 3 2

1 2 2

1 1 2

1 2

and so on.

t k j v t i v t v t

v

m m

m

m 

 











 _{1 2 3}

Chap. 9 Vector Differential Calculus. Grad, Div, Curl

Ex. 1) Find the derivative of r(t)=(1+t³)i+te ^t j+sin2tk tk

j e t i

t t

r( )3 ² (1 ) ^-t 2cos2

And find the unit tangent vector at a point t=0.

k j r(0) 2 Unit tangent vector at t=0 ,

k k j

j r

u r

5 2 5

1 4 1

2 )

0 (

) 0 ) (

0

(  



 



 

Ex. 2) Find the first and second derivative of r

(

)  [ 2 cos

, sin

,

] ]

1 , cos , sin 2 [ )

(

r

  

 (

)  [  2 cos

,  sin

]

(

)  cos 2

 sin 2



3 3

] 4 3

2 cos 2 2 sin 2

[

dt

dr _{ s}









t

s 

k e t j t t

i t

t³

sin( 2

) 8

cos( 2

) 12

8  





Chap. 9 Vector Differential Calculus. Grad, Div, Curl

9.5 Curves. Arc Length

 Curves C in space may occur as paths of moving bodies.

Parametric representations with parameter t will be, (1) r(t) = [x(t), y(t), z(t)] = x(t)i+ y(t)j + z(t)k

Here x, y, z are Cartesian coordinate (the usual rectangular coordinates). To each value t= t₀ there corresponds a point of Cwith positive vector r(t₀), that is, with coordinates x(t₀), y(t₀), z(t₀).

 The sense of increasing t, called the positive sense on C, induces a direction of travel along C. The sense of decreasing t, called the negative sense on C.

Ex. 1) The circle x²+y²=4, z=0in the xy-plane with center 0 and radius 2 is,

r(t) = [2cost, 2sint, 0] or r(t) = [2cost, 2sint] 0 t 2

← x²+y²= (2cost) ²+(2sint) ²=4

For t=0, r(0)=[2, 0], for t= , r( )=[0, 2] and so on. ← positive sense

2

/ /2

Chap. 9 Vector Differential Calculus. Grad, Div, Curl

Parametric representation

of a curve

Circle in Example 1

Ex. 2) The vector function

(3) r(t) = [acost, bsint, 0] = acost i+ bsint j

represents an ellipse in the xy-plane with center at the origin and principal axes in the direction of the x and y axes. From (3),

0 ,

2 1

2 2

2   z

b y a x

If b = a (3) represents a circle of radius a.

Ex. 3) A straight line Lthrough a pointAwith position vector a in the direction of a constant vector b can be,

(4) r(t) = a + tb = [a₁+ tb₁, a₂ + tb₂, a₃ + tb₃]

 A plane curve is a curve that lies in a plane in space. A curve that is not plane is called a twisted curve. A standard example of a twisted curve is the following.

Chap. 9 Vector Differential Calculus. Grad, Div, Curl

Ellipse in Example 2

Parametric representation of a straight line

Ex. 4) The twisted curve C represented by the vector function

(5) r(t) = [acost, asint, ct] = acost i+ asint j + ct k (c ≠ 0) is called a circular helix. It lies on the x²+y²=a². If c>0, the helix is shaped like a right-handed screw. If c<0, it looks like a left-handed screw. If c=0, then (5) is a circle.

 A simple curve is a curve without multiple points, that is, without points at which the curve intersects or touches itself. Circle and helix are simple. An example which is not simple is [sin2t, cost, 0].

 An arc of a curve is the portion between any two points of the curve.

Left-handed circular helix

Chap. 9 Vector Differential Calculus. Grad, Div, Curl

Right-handed circular helix

 If C is given by r(t), and P and Q correspond to t and t+Δt, then a vector in the direction of L is

(6) ^{1 [r(t t) r(t)]}

t  



In the limit, this vector becomes the derivative

)]

( ) ( 1 [ )

(

'

lim

0

t r t t t r t

r

t





 







(7) provided r(t) is differentiable.

If , we call a tangent vector of C at P. The corresponding unit vector is the unit tangent vector

0 ) ( ' t 

r r' t()

(8) ^'

1 r' u  r

 The tangent to C at P is given by (9) q(w) = r + wr'

Tangent to a Curve

 Tangents are straight lines touching a curve. The tangent to a simple curve C at a point P is the limiting position of a straight line L through P and a point Q as Q approaches P along C.

Tangent to a curve

Chap. 9 Vector Differential Calculus. Grad, Div, Curl

Formula (9) for the tangent to a curve

Ex. 5) Find the tangent to the ellipse at1 4

1x2 y2  P:( 2,1/ 2)

P corresponds to since

] cos , sin 2 [ ) ( ' ] sin , cos 2 [ )

(t t t r t t t

r   

4



 t

] 2 / 1 , 2 [ )]

4 / sin(

), 4 / cos(

2 [ ) 4 /

(    

r

Hence r'( /4)[ 2,1/ 2] From (9), q(w) = r(π/4)wr(π/4)

Chap. 9 Vector Differential Calculus. Grad, Div, Curl

)]

1 )(

2 / 1 ( ), 1 ( 2 [ ] 2 / 1 , 2 [ ] 2 / 1 , 2

[ w   w w



Ex. 6) Find the tangent to the helix with parameter equations )

2 / , 1 , 0 (

: π

P

P corresponds to since

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

(t t t t r t t t

r   

2 π/ t 

], 2 / , 1 , 0 [ ] 2 / ), 2 / sin(

), 2 / cos(

2 [ ) 2 /

(π π π π π

r   r'(π/2)[2,0,1]

q(w) = r(π/2)wr(π/2)

)]

2 / , 1 , 2 [ ] 1 , 0 , 2 [ ] 2 / , 1 , 0

[ π w    w π w



t z t y

t

x 2cos , sin ,  at

Length of a Curve

 will be the limit of the lengths of broken lines of n chords with larger and larger n. This limit is given by the integral

(10)

' ' ( ' )

dt r dr dt

r r

l ^b

a

 

 

is called the length of C, and C is called rectifiable.

l

l

Chap. 9 Vector Differential Calculus. Grad, Div, Curl

Cf) (3) in 9.2

Arc Length s of a Curve

 The length (10) of a curve C is a constant and positive number. The arc length function or simply the arc length of C will be

(11)

~ ( ' ~ )

' ' )

(

r dr t

d r r t

s ^t

a

 

 

Ex. 7) Find the length of arc of the circular helix with vector equation ]

1 , cos , sin [ ) (

' t t t

r  

] , sin , [cos )

(t t t t

r  from the point (1,0,0) to the point (1,0,2π).

2 1 ) (cos )

sin ( ) (

' t   t ² t ²  r

The arc from (1,0,0) to (1,0,2π) is described by

0  t  2 π .

Curves in Mechanics. Velocity. Acceleration

 Curves C in mechanics may serve paths of moving bodies and they are represented by a parametric representations r(t) with time t as parameter.

 The tangent vector (7) of C is called the velocity vector v, which means the instantaneous direction of motion and its length gives the speed v r' r'r'ds/dt

 The second derivative of r(t) is called the acceleration vector and is denoted by a. Its length lal is called the acceleration of the motion. Thus,

(16) v(t) = r'(t), a(t) = v'(t) = r''(t)

Chap. 9 Vector Differential Calculus. Grad, Div, Curl

Thus, l ^π r'(t)dt ^2π 2dt 2 2π

0 2

0  



 

Ex. 8) Find the arc length of the circular helix in the direction of increasing t.

] , sin , [cos )

(t t t t

r  from (1,0,0)

t dt

dt t r

s ^t '( ) ^t 2 2

0

0  



 

Tangential and Normal Acceleration

 Whereas the velocity vector is always tangent to the path of motion, the acceleration vector will generally have another direction, so that it will be

(17) a = a_tan + a_norm

where the tangential acceleration vector a_tan is tangent to the path (or, sometimes, 0) and the normal acceleration vector a_norm is normal (perpendicular) to the path (or, sometimes, 0). From (16)

dt s ds dt u

ds ds dr dt t dr

v( )   ( )

(18) ( ) ( ( ) ) ( )² ( ) ²₂

dt s s d dt u

ds ds du dt s ds dt u

d dt t dv

a    

 Since the tangent vector u(s) has constant length (one), its derivative du/ds is perpendicular to u(s). Hence the first term on the right of (18) is the normal acceleration vector, and the second term is the tangential acceleration vector, so that (18) is of the form (17).

u(s) ; unit tangent vector

Chap. 9 Vector Differential Calculus. Grad, Div, Curl

Chap. 9 Vector Differential Calculus. Grad, Div, Curl

Ex. 9) The position vector is given by r(t)t³it²j

Find its velocity, speed and acceleration when t1 tj

i t t r t

v( ) ( )3 ² 2 a(t)r(t)6ti2j The speed is, v(t)  (3t²)² (2t)²  9t⁴ 4t² When t1, v(1)3i2j, a(1)6i2j, v(1)  13

Find its velocity, speed and acceleration ]

) 1 ( , , 2 [ ) ( )

(t r t t e^t t e^t

v     a(t)v(t)[2,e^t,(2t)e^t]

t t

t

t t e t e t e

e t

t

v( )  (2 )² ( )² (1 )²( )²  4 ² ² (1 )^{2 2} Ex. 10) The position vector r(t)[t²,e^t,te^t]

Ex. 11) Centripetal Acceleration. Centrifugal Acceleration

 The vector function r(t) =[Rcoswt, Rsinwt]RcoswtiRsinwtj

represents a circle C of radius R with center at the origin of the xy-plane and describes the motion of a small bodyB counterclockwise around the circle.

v = r' =^{[Rwsinwt,Rwcoswt]RwsinwtiRwcoswtj} v is tangent to C. Its magnitude, the speed, is

Rw r

r r

v  '  ' ' _{; constant}

where, w is the angular speed, the speed divided by the radiusRfrom the center.

(19) a=v'= ^{[Rw2coswt, Rw2sinwt] Rw2coswti Rw2sinwtjw2r} which means that there is an acceleration toward the center, called the centripetal acceleration of the motion. It occurs because the velocity vector is changing direction at a constant rate. Its magnitude is constant,

R w r w

a  ²  ²

 Multiplying a by the mass m of B, we get the centripetal force ma. The opposite vector -ma is called the centrifugal force. At each instant these two forces are in equilibrium. In this motion the acceleration vector is normal to C, hence there is no tangential acceleration.

 Differentiation gives the velocity vector

 Differentiating the velocity vector, we obtain the acceleration vector

Chap. 9 Vector Differential Calculus. Grad, Div, Curl

Ex. 12) Superposition of Rotations. Coriolis Acceleration

A projectile is moving with constant speed along a meridian of the rotating earth. Find its acceleration.

Sol.) Let the earth, together with a unit vector b, be rotating about z-axis with angular speed w>0. b is

wtj wti

t

b ( )  cos  sin

Let the projectile be moving on the meridian whose plane is spanned by b and k (Fig. 209) with constant angular speed >0.

Then its position vector is,

tk R

t tb R

t

r( )  cos ( ) sin

(R = Radius of the earth) (20)

) ( sin

cos )

( ' '

cos sin

) ( '

2 2

2 wti w wtj w b t

w t

b

wtj w

wti w

t b



















Chap. 9 Vector Differential Calculus. Grad, Div, Curl

Superposition of two rotations

(21)

r tb

R tb

R

tk R

tb R

tb R

tb R

v a

tk R

tb R

tb R

t r v

2

2 2

' sin 2

' ' cos

sin cos

' sin 2

' ' cos '

cos sin

' cos )

( '

























































• The first term in a is the centripetal acceleration due to the rotation of the earth. The third term is the centripetal acceleration due to the motion of the projectile. The second term is called the Coriolis acceleration and is due to the interaction of the two rotations.

• On the Northern Hemisphere, (for, by assumption), so that a_cor has the direction of -b', that is, opposite to the rotation of the earth. la_corl is maximum at the north pole and zero at the equator.

• The projectile B of mass m₀ experiences -m₀a_cor opposite to m₀ a_cor , which tends to let B deviates to the right (and in the Southern Hemisphere, where , to the left). This deviation has been observed for missiles, rockets, and atmospheric air flow.

0

sint t0, 0

0 sint

Chap. 9 Vector Differential Calculus. Grad, Div, Curl

감사합니다