where φ is the latitude angle.

An arbitrary vector A has representations in both coordinate systems; i.e.,

A = A_x I + A_yJ + A_zK
A = A'_x i + A'_yj + A'_zk
 

When considering the change in the vector A with time the crucial difference between the two coordinate systems is that in the inertial system the unit vectors are constant whereas in the rotating system they are not. Thus,

dA/dt = (dA_x/dt)I + (dA_y/dt)J + (dA_z/dt)K
but
dA/dt = (dA'_x/dt)i + (dA'_y/dt)j + (dA'_z/dt)k
+ A'_xdi/dt + A'_ydj/dt + A'_zdk/dt
 

The terms

(dA'_x/dt)i + (dA'_y/dt)j + (dA'_z/dt)k
 

represent the apparent time rate of change of A in the rotating coordinate system, which can be denoted as d_rA/dt. Thus

dA/dt = d_rAdt) + A'_xdi/dt + A'_ydj/dt + A'_zdk/dt
 

The Time Derivatives of the Tangent Plane Unit Vectors of a Rotating Coordinate System

The point of origin of the tangent plane coordinate system can be expressed in terms of the inertial frame either in rectangular coordinates (x, y, z) or spherical coordinates (r, θ, φ), where r is the radius of the sphere, θ is the longitude and φ is the latitude. The relationship between the two representations is:

x = rcos(φ)cos(θ)
y = rcos(φ)sin(θ)
z = rcos(φ)
 

The local vertical unit vector k at (x, y, z) in terms of the inertial frame is:

k = (x/r)I + (y/r)J + (z/r)K
so
dk/dt = (1/r)((dx/dt)I + (dy/dt)J + (dz/dt)K)
 

Since r and φ are constant and θ = Ωt

dx/dt = -rcos(φ)sin(θ)(dθ/dt) = -rcos(φ)sin(θ)Ω
dy/dt = rcos(φ)cos(θ)(dθ/dt) = rcos(φ)cos(θ)Ω
dz/dt = 0.
 

Therefore

dk/dt = Ω[-cos(φ)sin(θ)I + cos(φ)cos(θ)J]
 

On the other hand, Ωxk is equal to the determinant:

|	I	J	K	|
|	0	0	Ω	|
|	(x/r)	(y/r)	(z/r)	|

 

which reduces to

I((-y/r)Ω) - J((-x/r)Ω) + K(0)
= Ω((-y/r)I + (x/r)J)
but
y/r = cos(φ)sin(θ) and x/r = cos(φ)cos(θ)
so
Ωxk = dk/dt
 

The local east-pointing unit vector i expressed in the inertial frame is

i = -sin*θ)I + cos(θ)J
and therefore
di/dt = -cos(θ)ΩI - sin(θ)ΩJ
= -Ω(cos(θ)I + sin(θ)J<
 

But Ωxi, given by the determinant

|	I	J	K	|
|	0	0	Ω	|
|	-sin(θ)	cos(θ)	0	|

 

reduces to I(-Ωcos(θ) - J(sin(θ) + K(0) which is the same as -Ω(cos(θ)I + sin(θ)J which is Ωxi. Thus,

di/dt = Ωxi
 

And finally there is the local north-pointing unit vector j, which expressed in the inertial frame is

j = -sin(φ)cos(θ)I - sin(φ)sin(θ)J + cos(φ)K
and therefore
dj/dt = sin(φ)sin(θ)ΩI - sin(φ)cos(θ)ΩJ
Ω(sin(φ)sin(θ)I - sin(φ)cos(θ)J).
 

But Ωxj in determinant form is

|	I	J	K	|
|	0	0	Ω	|
|	-sin(φ)cos(θ)	-sin(φ)cos(θ)	cos(φ)	|

 

which reduces to

I(sin(φ)sin(θ)Ω - J(sin(φ)cos(θΩ) + K(0)
= Ω(sin(φ)sin(θ)I - sin(φ)cos(θ)J).
 

But this the same as dj/dt so

dj/dt = Ωxj.
 

The General Form of the Time Derivative of a Vector

It was found in the previous section that for a general vector

dA/dt = d_rA/dt + A'_xdi/dt + A'_ydj/dt + A'_zdk/dt
 

If the time derivatives of the local unit vectors, di/dt, dj/dt and dk/dt, are replaced by their values as Ωxi, Ωxj and Ωxk, and the rotation vector Ω factored from the cross products the result is:

dA/dt = d_rA/dt + Ωx(A'_xi + A'_yj + A'_zk)
 

which is none other than

dA/dt = d_rA/dt + ΩxA

This says that the time derivative of a vector can be constructed from its apparent time derivative in the rotating frame plus the vector which is the vector cross product of the rotation vector for the frame and the vector itself. There are number of places in the literature where the time derivatives of the unit basis vectors are derived from the above formula on the basis of the argument that such unit vectors are just special cases of position vectors to which the formula applies. This is in valid because the formula has to be derived from the determination of the time derivatives of those basis vectors. The formula does of course apply to the basis vectors but it is logically invalid to derive its application to the basis vectors from the formula itself.