The Kinematic transport theorem is a very basic theorem relating time derivatives of vectors between a non rotating frame and another one that's rotating with respect to it with a uniform angular velocity.
I was trying to prove it for the special case of $3$ dimensions, and everything seems straightforward apart from the fact that I'm getting into a difficulty with obtaining the exact form of the final expression.
Following is my attempted proof:
Let's assume without loss of generality that we have a frame $\widetilde{O}$ rotating with angular velocity $\mathbf{\Omega} = (0,0,\omega)$ about the $z$ axis of another frame $O$. Let $\bf{f}$ be a vector seen by frame $O$ and $\widetilde{\mathbf{f}}$ the same vector as seen by an observer in the $\widetilde{O}$ frame, and further assume that this observer is located at position $\widetilde{\bf{r}}$ with respect to this rotating frame, at a distance $R$ away from the origin. Note that this distance $R$ is correct for both frames since their origins coincide by assumption. It is clear that this observer, as seen from the $O$ frame is given by the position vector $\bf{r}$ as follows:
\begin{equation} \textbf{r} = R(\cos{\omega t}, \sin{\omega t}, 0) \end{equation}
Hence the vector $\widetilde{\bf{f}}$ is given by:
\begin{equation} \widetilde{\textbf{f}} = \textbf{f} - \textbf{r} = (f_x - R\cos(\omega t), f_y - R\sin(\omega t), f_z) \end{equation}
Differentiating the above with respect to time we get:
\begin{align} \dot{\widetilde{\textbf{f}}} &= (\dot{f_x} + \omega R\sin(\omega t), \dot{f_y} - \omega R\cos(\omega t), \dot{f_z}) \\ & = \dot{\textbf{f}} + \omega R (\sin(\omega t), -\cos(\omega t), 0) \\ & = \dot{\textbf{f}} + \omega (r_y, -r_x, 0) \end{align}
Isolating $\dot{\mathbf{f}}$ we get:
\begin{equation} \dot{\textbf{f}} = \dot{\widetilde{\textbf{f}}} + \omega (-r_y, r_x, 0) \end{equation}
Now as you can see, what I'm getting is slightly different than what I'm supposed to get according to the theorem, the above in fact reads:
$$\left(\frac{d\mathbf{f}}{dt}\right)_O = \left( \frac{d\mathbf{f}}{dt}\right )_\widetilde{O} + \mathbf{\Omega} \times \mathbf{r} $$
Where I've replaced $\dot{\mathbf{f}}$ and $\dot{\widetilde{\mathbf{f}}}$ with the "abstract" notation that denotes differentiating with respect to each frame $O$ and $\widetilde{O}$, just to make it look more similar to how the theorem is usually stated.
However, what I'm supposed to get according to the theorem is in fact:
$$\left(\frac{d\mathbf{f}}{dt}\right)_O = \left( \frac{d\mathbf{f}}{dt}\right )_\widetilde{O} + \mathbf{\Omega} \times \mathbf{f} $$
Where is my error? I suspect that it may have to do with how I am interpreting the operation of differentiating the vector "in the rotating frame", for example I'm not totally sure it's correct to say that: $\dot{\widetilde{\textbf{f}}} = \left( \frac{d\mathbf{f}}{dt}\right )_\widetilde{O}$
Also it's very weird that the final expression I got depends on the position of the observer in the rotating frame, but I can't find what causes this error either.