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LOGO
Chapter 2
Coordinate System
Transformation
Part 2
iugaza2010.blogspot.com
Express B in Cartesian
(b)B  2r sin cos ar  r cos cos a  r sin a
Ax  (sin cos ) * 2r sin cos  (cos cos ) * r cos cos
 (sin ) * rsin
Ax  (2r sin  cos  )  (r cos  cos  )  (rsin  )
2
2
2
2
2
2
2
r2
r


Ax  2 (2r sin 2 cos2 )  ( rcos 2 2 cos2 )  (r 2 sin 2 )
r
r


2
1
r
2
2
2
2
2
2
( r 2 sin 2 cos2 )  (
r
c
os


cos

)

(

sin
)
2
2
r
r

2
1
r
2
2
2

(x2 )  (
z
x
)

(
y
)
2
2
r
r

Ax 

2x2
x  y z
2
2
2

zx
(x  y )
2
2
x  y z
2
2
2

x2  y2  z 2
y2
2
2
x  y
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(a) Prove that: ax.aρ=cosφ
A  1ax
B  1a
x   cos
convert B to cartesian :
Bx  cos - sin
By   sin  cos
  
Bz  0
0
Bx  cos
0  1 
0 0
1  0
y   sin 
z  r cos
  r sin 

x2  y2
r
x2  y2  z 2
B  Bx ax  cos ax
ax.a  ax. cos ax  cos
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A   cos a  z sin  az
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(a) Transform A to rectangular
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
1
x
Ax   cos2  
 cos2  
 2 cos2  




A y  sin   cos  

1
 sin  cos    sin 


A
x2  y2
ax 
xy
x2  y2
x2  y2
xy
 cos  
x
2
y
2
x   cos
Az   z 2 sin   z 2y
x2
x2
ay  z 2yaz
A  (3,4,0)  1.8ax  2.4ay
| A | 3
y   sin 
z  r cos
  r sin 

x2  y2
r
x 2  y 2 5 z 2
A   cos a  z sin  az
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(b) Transform A to spherical
 Ar  sin 
  
 A   cos
 A  0
 
0
0
1
x   cos 
y   sin 
z  r cos 
  r sin 
cos    cos 



- sin   0

0   z 2 sin  
Ar   cos sin   z 2 sin cos
 ( r sin  ) cos sin   ( r sin  )(r cos ) 2 sin cos
 r sin 2  cos  r 3 sin  cos3  sin 
A  r sincos cos  r 3 sin2 cos2 sin 
A  0
 A  Ar ar  A a  A a
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Find the value of A at point(3,-4,0)
from part (b)
Ar  r sin  cos  r sin  cos  sin 
2
3
3
A  r sincos cos  r 3 sin2 cos2 sin 
A  0
 A  Ar ar  A a  A a
( x, y, z )  (3,4,0)  (r ,  ,  )  (5,  / 2,53.13o )
A ( 5, / 2, 53.13o )  3ar

| A | 3 as in part(a)
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Calculate the distance between the points:
(a) P1=(2,1,5) and P2=(6,-1,2)
P1P 2  P 2  P1  4ax  2ay  3az
| P1P 2 | 16  4  9  5.38
(b) P1=(3,pi/2,-1) and P2=(5,3pi/2,5)
d
  1   2 2  1  2 cos( 2- 1)  (Z2 -Z1 )
2
2
cyl .
2
2
d 2 cyl .  52  32  2(5)(3) cos( )  (6)2  100
d  10
OR
Or convert all points to Cartesian coordinates :
(3, pi/2,-1)  (0,3,-1)
(5,3pi/2,5)  (0,-5,5)
d
0  64  36  10
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(c) P1=(10,pi/4,3pi/4) and P2=(5, pi/6,7pi/4)
2
2
2
d sph. r r 2r r 2 cos 1 cos 2 - 2r r 2 sin 1 sin 2 cos( 2- 1)
2
1
1
1
d 2 sph.  99.11
Other Sol.
Convert all points to Cartesian
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H  ρz cos aρ  sin

2
At point (1,/3,0) find :
a   ρ 2 az
H  (1, pi / 3,0)  0.5 a  az
(a) H.ax
first we must convert  H to cartesian  or
 A to cyl.
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 
Hx  ρz cos  cos   sin  sin  0  0 
 0  0.433
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 2
note : We dont need to find Hy & Hz only we need Hx
H.ax  0
( Hx ax  Hyay  Hzaz).ax  Hx  0.433
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H  ρz cos aρ  sin

2
At point (1,/3,0) find :
a   ρ 2 az
(b) Hxaθ
 A  sin 
  
 A    0
 A  cos
 z
cos
0
- sin 
first we must convert H to sph  or A to cyl.
0  0 
1  1 
0  0
A  cos
A  0
Az   sin 

1
)  tan 1 ( )   / 2
z
0
A  cos 90  0
A  0
  tan 1 (
Az   sin 90  1
a
 0
H  (1, pi / 3,0)  0.5 a  az
0
H x A  H x az  (0.5 a  az) x ( az)
a
az
0.5
0
1  0.5 a 
-1
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H  ρz cos aρ  sin

2
At point (1,/3,0) find :
a   ρ 2 az
H  (1, pi / 3,0)  0.5 a  az
(c) Find the vector component of H normal to surface ρ=1
( H .a )a  0.5a  az.a a  0
aρ
ρ=1 (cyl.)
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H  ρz cos aρ  sin

2
At point (1,/3,0) find :
a   ρ 2 az
H  (1, pi / 3,0)  0.5 a  az
(d) Find the scalar component of H tangential to plane Z=0
az normal to plane Z  0
D  Dt  Dn
Dn  az
Dt  D  Dn  0.5a
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x=const.
y=const.
z=const.
  y  
  x  
  x  
  z  
  z  
  y  
Represent Infinite Plane
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ρ=const.
0  φ  2π
  z  
Circular Cylindrical
φ=const.
0ρ
  z  
semi infinite plane
z=const.
0 
0    2
infinite plane
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r=const.
0   π
0  φ  2π
Sphere
(center at origion)
θ=const.
φ=const.
0r
0r 
0   π
0    2
circualr cone infinite plane
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θ=pi/4
φ=2pi/3
x= 10
r=1 , θ=pi/3 , φ=2pi/3
ρ=3 , φ=2pi/3
ρ=3 , z=1
r=4 , φ =pi/6
r=4 , θ =pi/6
φ =pi/6 , z=10
y=1, z=0
 Cone
 semi-infinite plane
 infinite plane
 Point
 Straight line
 circle
 semi circle
 circle
 radial line
 line
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θ=pi/4
 Cone
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φ=2pi/3
 semi-infinite plane
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x= 10
 infinite plane
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ρ=3 , φ=2pi/3
 Straight line
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ρ=3 , z=1
 circle
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(1) Unit vector normal to the surface θ=30 
(2) Unit vector normal to the surface ρ=2 
aρ
ρ=2 (cyl.)
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(3) Unit vector normal to the surface φ=3pi/2 
semi-infinite plane
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J  r sin cos a r - cos2 sin a   tan

ln r a 
2
at T(2, /2, 3/2) determine the vector component of J that is :
(a) Parallel to az
( J .az) az
convert az to spherical :
Ar  cos * (1)  cos(pi/2)  0
A  -sin * (1)  -sin(pi/2)  -1
A  0
A  -a
( J .(a ))(a )  cos(2 ) sin   cos sin(3 / 2)  1
( J .(a ))(a )  (1)(a )  a
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(b) Normal to surface φ=3pi/2
( J .a )a  (tan

2
ln r )a  tan( pi / 4) ln 2a  ln 2 a
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(c) Tangential to the spherical surface r=2
D  Dt  Dn
Jt  J  Jn
Jn  r sin cos a r
Jt  J  Jn  cos2 sin a   tan

2
ln r a 
 a   ln 2 a 
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(d) Parallel to the line y=-2,z=0
( J .ax) ax
ax  a
( J .a ) a  ln 2a
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LOGO
iugaza2010.blogspot.com
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