CHAPTER 14
Vectors in three space
Team 6:
Bhanu Kuncharam
Tony Rocha-Valadez
Wei Lu
14.6 Non-Cartesian Coordinates
The position vector R from the origin of Cartesian
coordinate system to the point (x(t), y(t), z(t)) is given by the
expression
iˆ
R ( t )  x ( t )î  y( t ) ĵ  z( t )k̂
The vector expression for velocity is given by
ĵ
A Cartesian coordinate system (by MIT
OCW)
v( t )  R ( t )  x ( t )î  y( t ) ĵ  z( t )k̂
The vector expression for acceleration is given by
a ( t )  R ( t )  x ( t )î  y( t ) ĵ  z( t )k̂
http://www.wepapers.com/Papers/4521/1_Newton's_Laws,_Cartesian_an
d_Polar_Coordinates,_Dynamics_of_a_Single_Particle
14.6.1 Plane polar coordinate
P(r, θ)
r
θ
Polar Axis
Definitions:
To define the Polar Coordinates of a plane we need first
to fix a point which will be called the Pole (or the origin)
and a half-line starting from the pole. This half-line is
called the Polar Axis.
Polar Angles: The Polar Angle θ of a point P, P ≠ pole, is the angle between the Polar
Axis and the line connecting the point P to the pole. Positive values of the angle indicate
angles measured in the counterclockwise direction from the Polar Axis.
The Polar Coordinates (r,θ) of the point P, P ≠ pole, consist of the distance r of the point
P from the Pole and of the Polar Angle θ of the point P. Every (0, θ) represents the
pole.
http://www.geom.uiuc.edu/docs/reference/CRC-formulas/node5.html
Plane polar coordinate
More than one coordinate pair can refer to the same point.
o
2,
30


  2, 210 o 
2
210o
  2, 150o 
30o
150o
All of the polar coordinates of this point are:
 2, 30  n  360 
 2, 150  n  360 
o
o
o
o
http://www.geom.uiuc.edu/docs/reference/CRC-formulas/node5.html
n  0,  1,  2 ...
Plane polar coordinate
Difference quotient method to get
deˆ
deˆr
ˆ
 e and
 eˆr
d
d
R  r (t )eˆ r ( (t ))
v(t )  R  reˆr  reˆr
What is
ˆr
de
d
?
eˆ  d eˆ ( (t ))  deˆr d   deˆr
r
r
dt
d dt
d
deˆr
eˆr (   )  eˆr ( )
 lim
 0
d

Greenberg, M. D. (1998). Advanced Engineering Mathematics
(2nd ed.): Prentice Hall.
v (t )  reˆr  reˆ
ˆ
ˆr
(1 )e
de
ˆ
 lim
e



0
d

a ( t )  v (t )  rê r  rê r  r ê   rê   r ê 
Plane polar coordinate
Difference quotient method to get
deˆ
deˆr
ˆ
 e and
 eˆr
d
d
What
deˆ
is d
?
deˆ d
deˆ
d
eˆ  eˆ  (t )  
  
dt
d dt
d
deˆ
eˆ (   )  eˆ ( )
 lim
d  0

deˆ
(1 )( eˆr )
 lim
 eˆr



0
d

eˆ  eˆr
Greenberg, M. D. (1998). Advanced Engineering
Mathematics (2nd ed.): Prentice Hall.
a(t )  (r  r 2 )eˆr  (r  2r)eˆ
Plane polar coordinate
Transform method to get
deˆ
deˆr
ˆ
 e and
 eˆr
d
d
ˆ r  cos iˆ  sin ˆj
e
ˆ   sin iˆ  cos ˆj
e
ˆr
de
  sin iˆ  cos ˆj
d
ˆ
de
  cos iˆ  sin ˆj
d
i  cos eˆr  sin eˆ
j  sin eˆr  cos eˆ
dê r
  sin (cos  ê r  sin  ê  )  cos (sin  ê r  cos  ê  )  (sin 2   cos 2 )ê   ê 
d
dê 
  cos (cos  ê r  sin  ê  )  sin (sin  ê r  cos  ê  )  (cos 2   sin 2 )  ê r
d
The expressions of R, v, a in polar coordinates
ê
êr
R (t )  x (t )iˆ  y (t ) ˆj  z (t ) kˆ
v(t )  R (t )  x (t )iˆ  y (t ) ˆj  z (t ) kˆ
a (t )  v (t )  R (t )  x (t )iˆ  y (t ) ˆj  z (t ) kˆ
dR dr
deˆr
ˆ
v
 er  r
 reˆr  reˆ
dt dt
dt
dv
a   reˆr  reˆ  reˆ  reˆ  r 2 eˆr
dt
x  r cos 
y  r sin 
r 
x2  y2
  tan 1
A polar coordinate system (by MIT OCW)
http://www.wepapers.com/Papers/4521/1_Newton's_Laws,
_Cartesian_and_Polar_Coordinates,_Dynamics_of_a_Sing
le_Particle
R  r ( t )ê r (( t ))
 ê
 r
ê  r
v( t )  R
r
2

 )ê  ( r
  2r
 )ê

a ( t )  (r  r

r

y
x
14.6.2 Cylindrical coordinates
Cylindrical coordinates are a generalization of two-dimensional
polar coordinates to three dimensions by superposing a height z
axis.
(r,,z)
r

r 
( 2)
A cylindrical coordinate system
http://mathworld.wolfram.com/CylindricalCoordinates.html
Cylindrical coordinates
Definitions:
The relations between cylindrical coordinates and
Cartesian coordinates.
r 2  x2  y 2
x  r cos 
y  r sin 
r 
x2  y2
  tan  1
y
x
tan( ) 
y
x
zz
The expressions of position R, velocity v, and acceleration a in
Cylindrical coordinates are given by
R  reˆr  zeˆ z
v(t )  R (t )  reˆr  reˆ  zeˆ z
a (t )  R(t )  (r  r 2 )eˆr  (r  2r)eˆ  zeˆ z
Greenberg, M. D. (1998). Advanced Engineering Mathematics (2nd ed.): Prentice Hall.
Cylindrical coordinates
Example1:
Find the cylindrical coordinates of the point whose Cartesian
coordinates are (1, 2, 3)
Answer:
r 
5
  1.1071
z  3
Example2:
Find the Cartesian coordinates of the point whose cylindrical
coordinates are (2, Pi/4, 3)
Answer:
x  2
2
y  2
2
z  3
http://mathworld.wolfram.com/CylindricalCoordinates.html
14.6.3 Spherical coordinates
(x,y,z)
r

 z
Spherical coordinates, also called spherical polar
coordinates (Walton 1967, Arfken 1985), are a
system of curvilinear coordinates that are natural
for describing positions on a sphere or spheroid.
Define to be the azimuthal angle in the -plane
from the x-axis with (denoted when referred to as
the longitude), to be the polar angle (also known
as the zenith angle and colatitude, with where is
the latitude) from the positive z-axis with , and to
be distance (radius) from a point to the origin.
Spherical coordinates
The expressions of Spherical coordinates for velocity and acceleration
eˆ 
eˆ 
eˆ 
 sin eˆ

eˆ
 0,
 eˆ  ,
 cos eˆ



eˆ
eˆ
eˆ
 0,
 0,
  sin eˆ   cos eˆ




eˆ
 0,

eˆ
 eˆ ,
R  eˆ 
eˆ   eˆ  ( (t ),  (t ))
eˆ  d eˆ  d

) a(t )  (    2   2 sin 2  )eˆ  (   2    2 sin  cos  )eˆ


 dt
 dt
 ( sin   2  sin   2  cos  )eˆ
 eˆ    (eˆ   sin eˆ )
v  R  eˆ    (
The expressions of R, v, a in Spherical coordinates
R(t )  x(t )iˆ  y (t ) ˆj  z (t )kˆ
v(t )  R (t )  x (t )iˆ  y (t ) ˆj  z (t )kˆ
a(t )  v (t )  R (t )  x (t )iˆ  y (t ) ˆj  z (t )kˆ
x   sin  cos 
y   sin  sin 
z   cos 
R  eˆ
v(t )  eˆ  eˆ   sin eˆ
a (t )  (    2   2 sin 2  )eˆ  (   2    2 sin  cos  )eˆ
 ( sin   2  sin   2  cos  )eˆ

Figure taken from reference: http://mathworld.wolfram.com/SphericalCoordinates.html
Examples: The expressions of R, v, a in Non-Cartesian
coordinates
Example 3
Calculate the three components of the position, velocity and acceleration vectors at
t=3. The position of the point R is given by R=(t, exp(t), 3t ). Do this for the in
Cartesian coordinates, Cylindrical coordinates, and Spherical coordinates
Solution:
In Cartesian Coordinates:
R(t )  x(t )iˆ  y(t ) ˆj  z(t )kˆ
v(t )  R (t )  x (t )iˆ  y (t ) ˆj  z (t )kˆ
R (t )  tiˆ  e t ˆj  3tkˆ
v (t )  iˆ  e t ˆj  3kˆ
a(t )  v (t )  R (t )  x (t )iˆ  y (t ) ˆj  z (t )kˆ
a (t )  e t ˆj
 R(t )  3iˆ  e 3 ˆj  9kˆ, v(t )  iˆ  e 3 ˆj  3kˆ, a(t )  e 3 ˆj , or
R x  3, R y  20.08, R z  9
v x  1, v y  20.08, v z  3
a x  0, a y  20.08, a z  0
The expressions of R, v, a in Non-Cartesian coordinates
Solution:
In Cylindrical Coordinates:
 R  reˆr  zeˆ z
 R  0
put
R (t )  tiˆ  3tkˆ
i  cos eˆr  sin eˆ
j  sin eˆr  cos eˆ
into
v (t )  iˆ  e t ˆ
j  3kˆ
a (t )  e t ˆ
j
R(t )  t cos eˆr  3teˆz
get
v(t )  (cos eˆr  sin eˆ )  et (sin eˆr  cos eˆ )  3eˆz
a(t )  et (sin eˆr  cos eˆ )
 sin  
y
x2  y2

e3
32  e 6
 0.989, cos   1  sin 2   0.1468
 Rr  0.4404, R  0, R z  9
v r  20.00, v  1.959, v z  3
a r  19.86, a  2.948, a z  0
The expressions of R, v, a in Non-Cartesian coordinates
In Spherical Coordinates:
Solution:
put
ˆ
 R  e
 R  0.R z  0
R (t )  tiˆ
iˆ  sin  cos eˆ  cos  cos eˆ  sin eˆ
ˆj  sin  sin eˆ  cos  sin eˆ  cos eˆ
into
kˆ  cos eˆ  sin eˆ
get
v (t )  iˆ  e t ˆj  3kˆ
a (t )  e t ˆj
R  t sin  cos eˆ 
v  (sin  cos eˆ   cos  cos eˆ  sin eˆ )
 e t (sin  sin eˆ   cos  sin eˆ  cos eˆ )
 t (cos eˆ   sin eˆ )
a  e t (sin  sin eˆ   cos  sin eˆ  cos eˆ )
 sin  
cos  
y
x y
2
2

z
x2  y2  z2
e3
3 e
2

6
 0.989, cos   1  sin 2   0.1468
9
32  e 6  9 2
 R  0.4026, R  0, R  0
v   19.49, v  3.936, v  5.361
 0.4051, sin   1  cos 2   0.9142
a   18.15, a  2.948, a  8.045
14.6.4 Omega Method
Using the omega method derive the space derivatives of base vectors
Consider a rigid body B undergoing an arbitrary
motion through 3-space. And let A be any fixed
vector with B, that is, A is a vector from one
material point in B to another so A is constant with
time, because b is rigid. Thus A=A(t)
Fixed vector in B
Greenberg, M. D. (1998). Advanced Engineering
Mathematics (2nd ed.): Prentice Hall.
 A  A  cons tan t
  A  A A
  2A A
 0
A
A
2
 0
 A A
There exists a vector
There exists a vector
1
2
such that
  A
A
1
such that
B  1  B
Omega
Omegamethod
Method
 A  B  A B cos 
  B  A  B  0
A
 (1  A)  B  A  ( 2  B )  0
1  A  B  A   2  B  0
So(1   2 )  A  B  0
Since A is arbitrary:
(1  2 )  A  0
Since B is arbitrary:
1  2  0
So we get
   A
A
Omega method
Omega Method
In cylindrical coordinates:
Let A be êr :
  ê z
deˆr
   eˆr  eˆ z  eˆr  eˆ
dt
Using chain differentiation to write:
eˆ dr eˆr d eˆr dz
d
eˆr (r (t ),  (t ), z (t ))  r


dt
r dt  dt
z dt
eˆ
eˆ
eˆ
 r r   r  z r
r

z
eˆ
eˆ
eˆr
 r0  eˆ  z 0  r r   r  z
r

z
ˆr
ˆr
ˆr
e
e
e
ˆ ,
 0,
e
0
r

z
Similarly, let A be ê:
Let A be êz :
ˆ
ˆ
ˆ
e
e
e
ˆr ,
 0,
 e
0
r

z
ˆz
ˆz
ˆz
e
e
e
 0,
 0,
0
r

z
End of Chapter 14