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Naval Architecture & Ocean Enginee ring

Engineering Mathematics 2

Prof. Kyu-Yeul Lee

Department of Naval Architecture and Ocean Engineering, Seoul National University of College of Engineering

[2008][06-1]

October, 2008

(2)

Naval Architecture & Ocean Enginee ring

Vector Calculus (2)

: Vector Functions, Vector Fields

Vector Functions Partial Derivatives

Directional Derivatives

Tangent Planes and Normal Lines

Vector Fields and Fluid Mechanics

(3)

Vector Functions

 Parametric Curve

x

y

(x(t0),y(t0))

c

( ) ( ), ( )

0 0

0 x t y t

tr

x

y

)) ( ), ( ), (

(x t0 y t0 z t0

c

) ( ), ( ), ( )

(t0x t0 y t0 z t0

z

r

( ) ( ), ( ) ( ) ( ) t x t y t

f t g t

 

r

i j

 Vector Function

( ) ( ), ( ), ( )

( ) ( ) ( )

t x t y t z t

f t g t h t

  

r

i j k

2-D

3-D

For a given parameter , is a position vector t r ( t ) of a point on a curve P C ( ) ( )

x tf t ( ) ( ) y tg t

( ) ( ) x tf t

( ) ( ) y tg t

( ) ( ) z th t

P

0

P

0

o

o

(4)

Vector Functions

Limit of a Vector Function

If , , and exist, then Definition 9.1

) ( lim f t

a

t

lim g ( t )

a

t

lim h ( t )

a t

) ( lim ), ( lim ),

( lim )

(

lim t f t g t h t

a t a

t a

t a

t

r

  

Properties of Limits

If and , then Theorem 9.1

1 1

( )

lim rL

t

a

t

lim r

2

( )  L

2

t

a t

scalar a

is ,

) ( lim

)

( i c

1

t c

1

c

a

t

rL

1

( )

2

( )

1 1

lim )

( rrLL

t t

ii

t a

) ( )

( lim )

( iii r tr tLL

(5)

Vector Functions

Continuity

A vector function is said to be continuous at if is defined

exists, and Definition 9.2

) ( )

( i r a

r

) ( )

( lim )

( iii t a

a

t

rr

Derivatives of Vector Function The derivative of a vector is

for all for which the limit exists Definition 9.3

r t

( ) ( )

lim 1 )

(

0

t t t

t t

t

r r

r   

 

 

a t  )

( ) ( ii r t

dt

t d r

r  ( ) 

(6)

Vector Functions

Differentiation of Components

If where and are differentiable, then Theorem 9.2

) ( ), ( ), ( )

( tf x g x h x

r f , g h

) ( ), ( ), ( )

( tfx gx hx r

 Proof

 

t

t h t

t h f

t

t g t

t g t

t f t

t f

t h t g t f t

t h t t

g t t

t f

t t

t t t

t t t

 

 

 

 

) ( )

, ( ) ( )

, ( ) ( )

lim (

) ( ), ( ), ( )

( ), (

), 1 (

lim

) ( )

1 ( lim )

(

0 0

0

r r

r

(7)

Vector Functions

 Smooth curve

When the component function of a vector function have continuous first derivatives and for all in the open interval , then

is said to be a smooth function and the curve traced by is called a smooth curve

r 0

) ( 

t

r t ( a , b )

r C r

(8)

Vector Functions

 Not Smooth curve ex.)

3

3

2

1

( ) 2 (4 ) 4 2 cos 2 (sin )

2 2

F t t  

t

t



i

  t

j

t

k

(9)

Vector Functions

Differentiation of Components

If where and are differentiable, then Theorem 9.2

) ( ), ( ), ( )

( tf x g x h x

r f , g h

) ( ), ( ), ( )

( tfx gx hx r

 High-Order Derivatives

) ( ), ( ),

( )

( tf  x g  x h  x r 

Also obtained by differentiating its components.

Ex) Second-order

(10)

Vector Functions

 Example 4

Tangent Vectors

Graph the curve C that is traced by a point P whose position is given by r(t)=cos2ti+sintj, 0≤t≤2π.

Graph r′(0) and r′(π/6).

t

t t

t t

t x

2

2 2

2 2

sin 2 1

sin )

sin 1

(

sin cos

2 cos

x

y r  ( 0 )



 

 2 ,1 2 1

) 0 , 1 (

 

 

 

6 r

t y  sin

) 2 0

(  t  

1 1

, 2

1 

2

  

y x

x

j i

r  ( t )   2 sin 2 t  cos t j

r  ( 0 ) 

j i

r 2

3 3 6    

 

  

Solution)

(11)

Vector Functions

 Example 5 Tangent Line

Find parametric equations of the tangent line to the graph of the

curve C whose parametric equations are x=t

2

, y=t

2

-t, z=-7t at t=3.

k j

i

r ( t )  t

2

 ( t

2

t )  7 t

k j

i

r  ( t )  2 t  ( 2 t  1 )  7 k j

i

r  ( 3 )  6  5  7 k j

i

r ( 3 )  9  6  21

   

i   j i k   j i j kk

r r

t t

t

t t

7 21 5

6 6

9

7 5 6 21

6 9 )

3 ( )

3 (

 

t z

t y

t

x  9  6 ,  6  5 ,   21  7

Solution)

(12)

Vector Functions

 Example 6

Derivative of a Vector Function

If r(t)=(t 3 -2t 2 )i+4tj+e -t k,

then r′ (t)=(3t 2 -4t)i+4j-e -t k

and r″(t)=(6t-4)i+e -t k.

(13)

Vector Functions

Chain Rule

If is a differentiable vector function and is a differentiable scalar function, then, the derivatives of with respect to is

Theorem 9.3

r s u (t )

) (s r

) ( ) ( s u t dt

ds ds d dt

d rrr  

t

(14)

Vector Functions

 Example 7 Chain Rule

If r(s)=cos2si+sin2sj+e

-3s

k, where s=t

4

, then

 

. 12

) 2 cos(

8 ) 2 sin(

8

4 3

2 cos 2 2

sin 2

3 4

3 4

3 4

3

3 3

k i

k j

r i

t s

e t j

t t

t t

t e

s dt s

d

(15)

Vector Functions

Rules of Differentiation

Let and be differentiable vector functions a differentiable scalar function.

Theorem 9.4

r

1

u (t )

) ( )

( )]

( )

( [ )

(

1

t

2

t

1

t

2

t

dt

i d rrr   rr

2

) ( ) ( )

( ) ( )]

( ) ( [ )

( u t

1

t u t

1

t u t

1

t dt

ii d rr    r

) ( )

( )

( )

( )]

( )

( [ )

(

1

t

2

t

1

t

2

t

1

t

2

t

dt

iii d rrr   rrr

) ( )

( )

( )

( )]

( )

( [ )

(

1

t

2

t

1

t

2

t

1

t

2

t

dt

iv d rrr   rrr

Differentiation of Cross Product

(16)

Vector Functions

 Integral of Vector Functions

k j

i

r ( t )  f ( t )  g ( t )  h ( t )

If and are integrable, then the indefinite and definite integrals of a vector function are defined, respectively, by

g

f , h

   i   jk

r   

( t ) dt f ( t ) dt g ( t ) dt h ( t ) dt

k j

i

r

 

 

 

 

 

 

 

 

 

    

ab

b a b

a b

a

( t ) dt f ( t ) dt g ( t ) dt h ( t ) dt

) ( )

( ,

)

( t dt R c where R t r t

r    

(17)

Vector Functions

 Example 8

Integral of a Vector Function

If r(t)=6t

2

i+4e

-2t

j+8cos4tk then

     

     

, 4

sin 2 2

2

4 sin 2 2

2

4 cos 8 4

6 )

(

2 3

3 2

2 1

3

2 2

c k j

i

k j

i

k j

i r

t e

t

c t c

e c

t

dt t dt

e dt

t dt

t

t

t t

where c=c

1

i+c

2

j+c

3

k.

(18)

Vector Functions

 Example 9

Consider the helix of Example 1.

Find the length of the curve from r(0) to an arbitrary point r(t).

Find the parametric equations of the helix.

9.1 Example 1.

. 0 ,

sin 2 cos

2 )

( tt it jt k tr

k j

i

r ( t )  2 cos t  2 sin tt k j i

r  ( t )   2 sin t  2 cos t

  

2

2 2

2 2

( ) 2sin 2 cos 1

4sin 4 cos 1 5

t t t

t t

    

   

r

5 )

( 

t r

t du

du u

s

t

( )

t

5 5

0

0

  

  r

k j

i

r 2 sin 5 5

cos 5 2

)

( s s s

s   

sin 5 2 ) (

cos 5 2 ) (

s s g

s s f

Solution)

(19)

Partial Derivatives

Function of two variables

zf ( x , y )

x

y

z

) ,

( x y f

z

Domain of

) ,

( x y

) ,

( x y f

z

) ,

( )

, ,

( x y z where zf x y

y x,

z

: independent variables

: dependent variable

(20)

Partial Derivatives

) (x f y

x

x f x

x f dx

dy

x

 

) ( )

lim (

0

) , ( x y f

z

Partial Derivatives

0

( , ) ( , )

lim

x

x x

z f x x y f x y

x x

f z f

x

 

    

 

   

0

( , ) ( , )

lim

x

y y

z f x y y f x y

y y

f z f

y

 

   

  

   

partial derivative with respect to

x

partial derivative with respect to

y

treating as a constant

x

Ordinary Derivatives

(21)

Partial Derivatives

 Example 3

Partial Derivatives

If z=4x

3

y

2

-4x

2

+y

6

+1, find ∂z/∂x and ∂z/∂y.

x y

x x

z  12

2 2

 8

5 2

3

6

8 x y y y

z  

(22)

Partial Derivatives

 Example 4

Partial Derivatives

If F(x,y,t)=e

-3πt

cos4x sin6y, then the

partial derivatives with respect to x, y, and t are, in turn,

. 6 sin 4

cos 3

) , , (

6 cos 4

cos 6

) , , (

6 sin 4

sin 4

) , , (

3 3

3

y x

e t

y x F

y x

e t

y x F

y x

e t

y x F

t t

t y

t x

(23)

Partial Derivatives

Chain Rule

yf (u )

dx du u y dx

dy

 

) ( , ug x

Chain Rule

If is differentiable and and have continuous first partial derivatives, then

Theorem 9.5

) , ( x y g

u  )

, ( v u f

zvh ( x , y )

x v v z x

u u

z x

z

 

 

y v v z y

u u

z y

z

 

 

 ( ( )) y f g x

 

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

zf u vf g x y h x y

(24)

Partial Derivatives

 Example 5 Chain Rule

If z=u

2

-v

3

and u=e

2x-3y

, v=sin(x

2

-y

2

), find ∂z/∂x and ∂z/∂y.

Solution)

   

) cos(

6 4

) cos(

2 3 2

2

2 2

2 3

2

2 2

2 3

2

y x

xv ue

y x

x v e

u

x v v z x

u u

z x

z

y x

y x

 

 

( 2 ) cos( )

3 ) 3

(

2 u e

2 3

v

2

y x

2

y

2

y v v z y

u u

z y

z

y

x

  

 

 

(25)

Partial Derivatives

special case

zf ( v u , )

dt dv v z dt

du u z dt

dz

 

 

) ( ,

) (

, ug t vh t x

v v z x

u u

z x

z

 

 

y v v z y

u u

z y

z

 

 

Chain Rule

zf ( v u , ) , ug ( x , y ) , vh ( x , y ) ( , ) ( ( , ), ( , ))

zf u vf g x y h x y

( , ) ( ( ), ( ))

zf u vf g t h t

(26)

Partial Derivatives

If r(x,y,z) = x 2 +y 5 z 3

and x(u,v)=uve 2s , y(u,v)=u 2 -v 2 s, z(u,v)=sin(uvs 2 ), find ∂r/∂s.

 

2 4 3 2 5 2 2

2 (2

s

) 5

z

( ) 3 2 cos( )

r r x r y r z

s x s y s z s

x uve y v y z uvs uvs

      

  

      

   

Chain Rule example)

(27)

Partial Derivatives

generalization

zf ( u

1

, u

2

,..., u

n

)

dt du u

z dt

du u

z dt

du u

z dt

dz

n

n

 

 

 

 

2

2 1

1

) (

) (

) (

2 2

1 1

t g u

t g u

t g u

n n

special case

zf ( v u , )

dt dv v z dt

du u z dt

dz

 

 

) ( ,

) (

, ug t vh t

( , ) ( ( ), ( ))

zf u vf g t h t

(28)

Partial Derivatives

x v v z x

u u

z x

z

 

 

y v v z y

u u

z y

z

 

 

Chain Rule

zf ( v u , ) , ug ( x , y ) , vh ( x , y )

generalization

zf ( u

1

, u

2

,..., u

n

)

n

x u u

z x

u u

z x

u u

z dx

dz

 

 

 

 

1 2

) ,..., ,

(

) ,..., ,

(

) ,..., ,

(

2 1

2 1 2 2

2 1 1 1

k n

n

k k

x x

x g u

x x

x g u

x x

x g u

( , ) ( ( ), ( ))

zf u vf g t h t

(29)

Gradient of a Function

Apply Vector differential operator

z y

x

y x

 

 

 

 

 

k j

i

j i

k j

i j i

z F y

F x

z F y x F

y f x

y f x f

 

 

 

 

 

) , , (

) , ( )

, , (

) , (

z y x F w

y x f z

Differentiable function

Gradients of function Introduce a new vector based on partial differentiation

: “del” or “nabla”

f

: “grad “

f

(30)

Directional Derivatives

x

y

z

i j

) , ( x y f

z

:Rate of change of in the -direction

f i f

x

i

j

(31)

Directional Derivatives

x

y

z

i j

) , ( x y f

z

:Rate of change of in the -direction

f j f

y

i

j

(32)

u

j

Directional Derivatives

x

y

z

i j

) , ( x y f

z

u

f y

The rate of change of

f

in the direction of :

y

f cos

x

f

x

The component ofin the direction of :

u

cos( ) 2 sin f

y f y

 

 

 

f

y

The component ofin the direction of :

u

f x

The rate of change of

f

in the direction of :

x

= f cos

x

sin

f

y

+

The rate of change of

f

in the direction given by the vector :

u D f

u

f

f y

j i y

f x

y f x

f

 

 

 ( , )

i

j

+

The rate of change of

f

in the direction given by the vector :

u D f

u

=

(33)

Directional Derivatives

x

y

z

i

j

) , ( x y f

z

u

P

secant tangent

y x f h

y h

x f y

x f y

y x x

f ( , ) ( , ) (  cos ,  sin )  ( , )

 

  

h f ( x   x , y   y )  f ( x , y )

x

y

) 0 , , ( x y

) 0 , ,

( x   x y   y

u vh

Slope of indicated secant line is

(34)

Directional Derivatives

x

y

z

i

j

) , ( x y f

z

u

P

secant tangent

h f ( x   x , y   y )  f ( x , y )

x

y

) 0 , , ( x y

) 0 , ,

( x   x y   y

u vh

Slope of indicated secant line is

(35)

Directional Derivatives

x

y

z

i

j

) , ( x y f

z

u

P

secant tangent

y x f h

y h

x f y

x f y

y x x

f ( , ) ( , ) (  cos ,  sin )  ( , )

 

  

h f ( x   x , y   y )  f ( x , y )

x

y

) 0 , , ( x y

) 0 , ,

( x   x y   y

As we expect the slope of secant line would be tangent line u

vh

 0

h

(36)

Directional Derivatives

h

y x f h

y h

x f h

y x f y y

x x

f

( , ) ( , ) (

cos ,

sin )

( , )

 

 

Directional Derivative

The directional derivative of in the direction of a unit vector is

provided the limit exists.

Definition 9.5

) , ( x y f

z

0

( cos , sin ) ( , ) ( , ) lim

h

f x h y h f x y

D f x y

h

 

  

u

j i

u  cos   sin 

Slope of secant line

x

y

z

i

j

) , (x y f z

u

P secant

tangent

h f(xx,yy) f(x,y)

x

y ) 0 , , (xy

) 0 , , (xx yy

u vh

x z h

y x f y h x y f

x f D

h

 

 

) , ( )

, lim (

) , ( implies

0

i 0

( , ) ( , )

f x y h f x y z

   

(37)

Directional Derivatives

) , ) (

, ( )

sin ,

cos lim (

) 0 ( )

0 lim ( )

0

(

0 0

D f x y

h

y x f h

y h

x f h

g h

g g

h

h

   

u

 

 

Computing a Directional Derivative If is differentiable function of x and y and then,

Theorem 9.6

) , ( x y f

z

u

f ( x , y )   f ( x , y )  u D

j i

u  cos   sin 

Proof)

) sin ,

cos (

)

( t f x ty t

g   

by definition of a derivative

by the Chain Rule,

1 2

1 2

( ) ( cos , sin ) ( cos ) ( cos , sin ) ( sin )

( cos , sin cos ( cos , sin ) sin

d d

g t f x t y t x t f x t y t y t

dt dt

f x t y t f x t y t

     

     

        

      

(0) ( , cos ( , )sin

g f x yf x y

   

(38)

Directional Derivatives

Computing a Directional Derivative If is differentiable function of x and y and then,

Theorem 9.6

) , ( x y f

z

u

f ( x , y )   f ( x , y )  u D

j i

u  cos   sin 

Proof)

) sin ,

cos (

)

( t f x ty t

g   

( , ) (0) ( , cos ( , ) sin

[ ( , ) ( , ) ] (cos sin ) ( , )

x y

x y

D f x y g f x y f x y

f x y f x y f x y

 

 

    

   

  

u

i j i j

u

j i y

f x

y f x

f

 

 

 ( , )

(39)

Directional Derivatives

 Example 2

Gradient at a Point

If F(x,y,z)=xy

2

+3x

2

-z

3

, find ∇F(x,y,z) at (2,-1,4).

Solution)

k j

i

k j i

2 2

3 2

2

3 2

2

3 2

2

3 2

) 6 (

) 3

(

) 3

(

) 3

( )

, , (

z xy

x y

z x

dz xy

z x

dy xy

z x

x xy z

y x F

 

 

 

 

k j

i 4 48 13

) 4 , 1 , 2

(    

F

(40)

Directional Derivatives

Computing a Directional Derivative If is differentiable function of x and y and then,

Theorem 9.6

) , ( x y f

z

u

f ( x , y )   f ( x , y )  u D

j i

u  cos   sin 

( , ) [ ( , ) ( , ) ] (cos sin ) ( , )

x y

D f x y f x y f x y f x y

 

   

  

u

i j i j

u

j i y

f x

y f x

f

 

 

 ( , )

y

z

i j

) , ( x y f

z

j

(41)

Directional Derivatives

 Example 3

Directional Derivative

Find the directional derivative of

f(x,y)=2x

2

y

3

+6xy at (1,1) in the direction of a unit vector whose angle with the positive x-axis is π/6.

Solution)

j i

j i

) 6 6

( ) 6 4

( ) , (

2 2

3

y x y x

xy y f x

y f x f

 

 

j i 12 10

) 1 , 1

(  

f

j i

j i

u sin 6

cos 6 sin

cos       

(1,1) (1,1)

3 1

(10 12 )

2 2

5 3 6 D f   f

 

         

 

u

u

i j i j

(42)

Directional Derivatives

 Example 4

Directional Derivative

Consider the plane that is

perpendicular to the xy-plane and passes through the points P(2,1) and Q(3,2). What is the slope of the tangent line to the curve on intersection of this plane with the surface f(x,y)=4x

2

+y

2

at (2,1,17) in the direction of Q?

Solution)

j i

u 2

1 2

1 

PQ

2

4

2

) ,

( x y x y

f  

j i y x

y x

f ( , )  8  2

j i 2 16 )

1 , 2

(  

f

(2,1) (2,1)

1 1

(16 2 )

2 2

9 2

D f   f

 

       

u

u

i j i j

(43)

Directional Derivatives

 Example 5

Directional Derivative

Find the directional derivative of F(x,y,z)=xy

2

-4x

2

y+z

2

at (1,-1,2) in the direction of 6i+2j+3k.

Solution)

2 2

2

4

) , ,

( x y z xy x y z

f   

k j

i

k j i

z x

xy xy

y

z y x z xy

z y x y xy

z y x x xy

z y x f

2 ) 4 2

( ) 8 (

) 4

(

) 4

(

) 4

( )

, , (

2 2

2 2

2 2

2 2

 

 

 

 

 

 

k j

i 6 4 9

) 2 , 1 , 1

(    

f

7 3

2

6 ijk

k j

i k

j i

u 7

3 7

2 7

) 6 3 2 6 7 (

1      

6 2 3

(1, 1, 2) (9 6 4 )

7 7 7

54 7

D F           

 

u

i j k i j k

(44)

Directional Derivatives

Points in the direction of maximum increase of at P

f f

The rate of change of f in the direction given by the vector : u

( , ) [

x

( , )

y

( , ) ] (cos sin ) ( , ) D f x y

u

f x y if x y j   i   j   f x yu

cos cos , :angle between and

1 cos 1

D f ff   f

    

  

u

u u

The maximum value of

D f

u

D f

u

  f

, When

cos   1,   0

has the same direction of

f u

is the direction of maximum increase of at P

f f

is the direction of maximum decrease of at P

 f f

(45)

Directional Derivatives

 Example 6

Max/Min of Directional Derivative

In Example 5 the maximum value of the directional derivative at F at (1,-1,2) is .

133 )

2 , 1 , 1

(  

F

The minimum value of D

u

F(1,-1,2) is then  133 .

(46)

Directional Derivatives

 Example 7

Direction of Steepest Ascent

Each year in Los Angeles there is a

bicycle race up to the top of a hill by a road known to be the steepest in the city. To understand why a bicyclist with a modicum of sanity will zigzag up the road, let us suppose the graph of

shown in Figure (a) is a mathematical model of the hill. The gradient of f is

where r=-xi-yj is a vector pointing to

Thus the steepest ascent up the hill is a straight road whose projection in the xy-plane is a radius of the circular base.

Since D

u

f=comp

u

∇f, a bicyclist will

zigzag, or seek a direction u other than

∇f, in order to reduce this component.

r j

i 2 2 2 2

2 2

3 / 2 3

) 2 , (

y x y

x y y

x y x

x

f  





 

 

, 4 0

3 , 4 2 ) ,

(x y   x2y2zf

(47)

Directional Derivatives

 Example 8

Direction to Cool Off Fastest

Solution)

The temperature in a rectangular box is approximated by

If a mosquito is located at (½,1,1), in which direction should it fly to cool off as rapidly as possible?

. 3 0

, 2 0

, 1 0

) 3 )(

2 )(

1 ( )

, , (

z y

x

z y

x xyz

z y x T

j i

k j

i

) 2 2 )(

3 )(

1 (

) 2 1 )(

3 )(

2 (

) , , ( )

, , ( )

, , ) (

, , (

y z

x xz

x z

y yz

x z y x T y

z y x T x

z y x z T

y x T

 

 

 

4 k 1 1 , 1 2 ,

1  

 

T 

To cool off most rapidly, the mosquito should fly in the direction of -1/4k;

that is, it should dive for the floor of

the box, where the temperature is

T(x,y,0)=0.

(48)

Directional Derivatives

x

y

z

) , ( x y f

z

surface

c z

plane

c y

x

f ( , ) 

Surface

x y

Level Curves

increasing value of

f

Level Curves

c  10

11 c

12

c

(49)

Directional Derivatives

x

y

z

) , ( x y f

z

surface

c z

plane

c y

x

f ( , ) 

Surface

x

y

Level Curves

increasing value of

f

Level Curves

c  10

11 c

12 c

P

0

u

1

P

1

u

2

P

2

u

3

P

3

(50)

Level Curves

x y

Directional Derivatives

) , ( x y f

z

x

y

z

surface

c z

plane

c y x f

( , )

Surface

increasing value of

f

Level Curves

c  10

11 c

12 c

( , ) D f x y

u

The rate of change of in the direction given by the vector : f u

1

( , )

D f x y

u

 0

1 0

( ) ( ) 10 10 f Pf P    0

P

0

u

3

u

1

u

2

P

1

P

2

P

3

(51)

Directional Derivatives

) , ( x y f

z

x

y

z

surface

c z

plane

c y x f

( , )

Surface

x y

Level Curves

increasing value of

f

Level Curves

c  10

11 c

12 c

P

0

u

3

u

1

u

2

( , ) D f x y

u

The rate of change of in the direction given by the vector : f u

2

( , ) Vs.

3

( , ) D f x y

u

D f x y

u

2 0 3 0

2 0 3 0

( ) ( ) ( ) ( )

f P f P f P f P

P P P P

  

P

1

P

2

P

3

P P

is the shortest path between c=10 and c=11

참조

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