Derivative of rotating unit vector in coordinate free maner

Question

How can we calculate the derivative of a rotating unit vector in a coordinate free way? That is, we want to compute the derivative -with respect to time- of a vector function $\vec{\varepsilon}: [0, +\infty) \rightarrow \mathbb{R}^{2}$, whose length is kept constant and equal to one $|\vec{\varepsilon}(t)|=1$ for any $t \in [0, +\infty)$.

(Such computations are frequently useful in physics, especially in kinematics, when calculating velocities, accelerations etc, especially in many-body problems where the use of a single coordinate system might prove disfunctional).

Below, one answer is posted but it would be nice to see other approaches as well.

Answer 1

We will use $t$ to denote the independent variable since it normally stands for time.
Thus, at any given instant $t$, the value of the function will be a unit vector $\vec{\varepsilon}(t)$. We will apply the definition of the derivative, so we need to compute the vector \begin{equation} \label{def1} \frac{d}{dt}\vec{\varepsilon}(t) = \lim_{\Delta t \rightarrow 0} \frac{\vec{\varepsilon}(t + \Delta t) - \vec{\varepsilon}(t)}{\Delta t} \end{equation} First note that since $|\vec{\varepsilon}(t)|^{2} = \vec{\varepsilon}(t) \cdot \vec{\varepsilon}(t) = 1$ by differentiating both sides we get $$ \begin{array}{c} \frac{d}{dt} \big( \vec{\varepsilon}(t) \cdot \vec{\varepsilon}(t) \big) = \frac{d \vec{\varepsilon}(t)}{dt} \cdot \vec{\varepsilon}(t) + \vec{\varepsilon}(t) \cdot \frac{d \vec{\varepsilon}(t)}{dt} = \\ \\ 2 \frac{d \vec{\varepsilon}(t)}{dt} \cdot \vec{\varepsilon}(t) = 0 \Leftrightarrow 2 \frac{d \vec{\varepsilon}(t)}{dt} \cdot \vec{\varepsilon}(t) = 0 \end{array} $$ Thus $\frac{d \vec{\varepsilon}(t)}{dt} \cdot \vec{\varepsilon}(t) = 0$ which is equivalent to the fact that $\vec{\varepsilon}(t)$ and its derivative vector $\frac{d \vec{\varepsilon}(t)}{dt}$ are perpendicular \begin{equation} \label{def2} \vec{\varepsilon}(t) \bot \frac{d \vec{\varepsilon}(t)}{dt} \end{equation} The situation can be presented in the following picture: $\Delta s$ is the length of the arc spanned by the edge of the unit vector $\vec{\varepsilon}(t)$ while it rotates by an angle of $\Delta \varphi$ during time $\Delta t$. Since $\varphi$ is measured in radians we have $\Delta s = \Delta \varphi |\vec{\varepsilon}(t)|$ which implies that $$ \Delta s = \Delta \varphi $$ Notice that (in the figure) we also have: $|\vec{\eta}(t)|=1$ and $ \ \vec{\eta}(t) \bot \vec{\varepsilon}(t)$. Now, during that time interval $\Delta t$ the vector has changed by $$ \Delta \vec{\varepsilon} = \vec{\varepsilon}(t + \Delta t) - \vec{\varepsilon}(t) $$ While $\Delta t \rightarrow 0 $ (the notation $dt \rightarrow 0$ will be used instead from now on) we can notice two things: i). the direction of the vector $d \vec{\varepsilon} = \vec{\varepsilon}(t + dt) - \vec{\varepsilon}(t)$ "tends" to become perpendicular to $\vec{\varepsilon}(t)$ and thus parallel to the direction specified by $\vec{\eta}(t)$ and ii). the length of the vector $d \vec{\varepsilon}$ "tends" to become equal to the length of the arc $ds$ thus \begin{equation} \label{def3} |d \vec{\varepsilon}| = |\vec{\varepsilon}(t + dt) - \vec{\varepsilon}(t)| = ds = d \varphi \end{equation} (where $d \varphi$ is measured in radians). Now, combining the above, and substituting in $\frac{d}{dt}\vec{\varepsilon}(t)$ we get $$ \begin{array}{c} \frac{d}{dt}\vec{\varepsilon}(t) = \lim_{dt \rightarrow 0} \frac{\vec{\varepsilon}(t + dt) - \vec{\varepsilon}(t)}{dt} = \lim_{dt \rightarrow 0} \frac{d \varepsilon}{dt} = \frac{d \varphi}{dt} \vec{\eta}(t) \end{array} $$ Finally, if we define a vector $\vec{\omega}$ with direction perpendicular to the plane defined by $(\vec{\varepsilon}(t), \vec{\eta}(t))$ and with length $|\vec{\omega}|=\frac{d \varphi}{dt}$ then we have $$ \frac{d}{dt}\vec{\varepsilon}(t) = \vec{\omega} \times \vec{\varepsilon}(t) = \frac{d \varphi}{dt} \vec{\eta}(t) $$