Kinematics in 3-D
1. Position
a) Rectangular Coordinates: x(t), y(t), and z(t)
r = xx + yy + zz, where x, y and z are constant unit vectors.
b) Spherical Coordinates: r(t), (t) and (t)
Here r is the radial coordinate, a distance measured from the origin to the location of the
point. The angle  is the angle r makes with the z-axis. The angle  is the angle the
projection of r on the x-y plane makes with the x-axis.
We need to determine the unit vectors for this system, and we need to determine the
transformation equations to the rectangular system:
r = sin()cos(x + sin()sin()y + cos()z = r(,)
 = cos()cos(x + cos()sin()y - sin()z = (,)
 = -sin()x + cos()y = ()
Notice that the  direction has a direction 90o different from the r direction (replace 
with +90o). Notice that the  direction has a direction 90o different from the projection
direction (replace  with +90o in the normal projection direction of cos()x + sin()y .
In spherical coordinates, like polar, the position vector is simply: r = rr . While the r
dependence is explicit, both the  and the  dependence are "hidden" in the unit vector, r.
2. Velocity
a) Rectangular Coordinates:
v = dr/dt = d[xx + yy + zz]/dt = (dx/dt)x + (dy/dt)y + (dz/dt)z = vxx + vyy + vzz.
b) Spherical Coordinates:
v = dr/dt = d[rr]/dt = (dr/dt)r + r(dr/dt) .
We note that the unit vector, r, is not a constant. As we did with polar, we must
determine what the derivatives of the unit vectors in spherical coordinates are. Note that
dr/dt = (r/)(d/dt) + (r/)(d/dt).
r/ = [sin()cos(x + sin()sin()y + cos()z]/ =
cos()cos()x + cos()sin()y - sin()z = .
In similar fashion, we can find the derivatives of all three unit vectors with respect to
both  and . We summarize these on the next page:

r/ = 
r/ = sin() 
/ = -r
/ = cos() 
/ = 0
/ = -sin()r - cos() .
We now have for the velocity in spherical coordinates:
v = dr/dt = d[rr]/dt = (dr/dt)r + r(dr/dt) =
r'r + r[(r/)(d/dt) + (r/)(d/dt)] =
r'r + r' + r'sin() = v .
We recognize the first term as the regular radial velocity, the second term is the circular
velocity around an axis perpendicular to the z axis (r), and the third term is the circular
velocity around the z axis - but this has a radius not of r, but of [r sin()]. On the earth
these three velocities would correspond to: 1) velocity up or down (away from or toward
the earth; 2) velocity North or South; and 3) velocity East or West.
3. Acceleration
a) Rectangular Coordinates:
a = dv/dt = d[vxx + vyy+ vzz]/dt = (dvx/dt)x + (dvy/dt)y+ (dvz/dt)z = axx + ayy+ azz .
b) Spherical Coordinates:
a = dv/dt = d[r'r + r' + r'sin()]/dt
= (dr'/dt)r + r'(dr/dt) + (dr/dt)' + r(d'/dt) + r'(d/dt) + (dr/dt)'sin() +
r(d'/dt)sin() + r'(d[sin()]/dt) + r'sin()(d/dt)
= r''r + r'[' + ' sin()] + r'' + r'' + r'['(-r) + ' cos()] + r''sin() +
r''sin() + r'[cos()'] + r'sin()'[-sin()r - cos()]
= [r'' - r'2 - rsin2()'2]r
+ [2r'' + r'' - rsin()cos()'2]
+ [2r''sin() + 2r''cos() + rsin()'']
Analysis of each term:
For the r component: 1) the r'' is just the regular straight line acceleration in the radial
direction (up-down); 2) the -r'2 term is the centripetal acceleration due to rotation
around an axis perpendicular to the z-axis (due to a North-South motion on the earth); 3)
the -rsin2()'2 term is the r-component of the centripetal acceleration due to rotation
around the z-axis, with r sin() the radius of the circular motion (due to an East-West
motion on the earth).
For the  component: 1) the 2r'' is the Coriolis term due to the radius changing affecting
the  rotational motion (N-S); 2) the r'' term is the regular tangential acceleration
causing the tangential (N-S) speed to change; 3) the -rsin()cos()'2 term is the component of the centripetal acceleration due to rotation around the z-axis (E-W), with r
sin() the radius of the circular motion.
For the  component: 1) the 2r''sin() is a Coriolis term due to the radius changing
affecting the  rotational motion (E-W); 2) the 2r''cos() is a Coriolis term due to the
-radius (rsin() changing due to the r' motion; 3) the rsin()'' is the regular tangential
acceleration causing the tangential (E-W) speed to change.
Note that each term has one r (distance) and two '' (per time squared) are required by an
acceleration.
Homework Problem #16: Given that A is a vector function, find d2A/dt2 in cylindrical
coordinates (a 3-D vector problem).
4. The del operator
a) Rectangular Coordinates:
 = ( /x) x + ( /y) y + ( /z) z
dr = dx x + dy y + dz z
df = (f/x) dx + (f/y) dy + (f/z) dz
f  dr = df = dr f
b) Spherical Coordinates:
dr = dr r + rd  + rsin()d 
(since a small distance in the radial direction is simply dr, a small distance in the  direction is rd, and a
small distance in the  direction is r sin d)
df = (f/r)dr + (f/)d + (f/)d
dr  f = df
To make this last statement work, we need for the del operator:
 = ( /r)r + (1/r)( /) + (1/r sin())( /)
5. Divergence (A)
a) Rectangular Coordinates
 = ( /x) x + ( /y) y + ( /z) z
A = Axx + Ayy + Azz
A = (Ax/x) + (Ay/y) + (Az/z)
b) Spherical Coordinates
 = ( /r)r + (1/r)( /) + [1/r sin()]( /)
A = Arr + A + A
A = r  A/r + /r  A/ + /[r sin()]  A/ =
r  [(Ar/r)r + Ar(r/r) + (A/r) + A(/r) + (A/r) + A(/r)] +
 /r  [(Ar/)r + Ar(r/) + (A/) + A(/) + (A/) + A(/)] +
 /[r sin()]  [(Ar/)r + Ar(r/) + (A/) + A(/) + (A/) + A(/)]
We can eliminate all terms that have (r/r), (/r) and (/r) since all these unit
vectors do not depend on r; we can also eliminate the term that has (/) since  does
not depend on . Recall that (r/)=; (r/)=sin(); (/)=-r; (/)=cos();
and (/)= [-sin()r - cos()].
A =
r  [(Ar/r)r + 0 + (A/r) + 0 + (A/r) + 0] +
 /r  [(Ar/)r + Ar + (A/) + A(-r) + (A/) + 0] +
 /[r sin()]  [(Ar/)r + Ar sin() + (A/) + A cos() + (A/) + A(-sin()r - cos())]
Now we can take the dot products which will further reduce the number of terms:
A = (Ar/r) + Ar/r + (A/)/r + Ar/r + A cos()/r sin() + (A/)/r sin() =
(Ar/r) + 2Ar/r + ([A sin()]/)/r sin() + (A/)/r sin() = A
where we have combined the second and fourth terms into the 2Ar/r term, and we have
combined the third and fifth terms into the ([A sin()]/)/r sin() term.