\[\newcommand{\u}[1]{\boldsymbol{\mathsf{#1}}} \renewcommand{\b}[1]{\boldsymbol{#1}} \newcommand{\t}[1]{\textsf{#1}} \newcommand{\m}[1]{\mathbb{#1}} \def\RR{\bf R} \def\bold#1{\bf #1} \def\mbf#1{\mathbf #1} \def\uv#1{\hat{\usf {#1}}} \def\dl#1{\underline{\underline{#1}}} \newcommand{\usf}[1]{\boldsymbol{\mathsf{#1}}} \def\bs#1{\usf #1}\]

Time derivative of the Co-rotational basis

(We will be working exclusively non-dimensional quantities in this section)

Previously, in this section, we defined the co-rotational basis vectors to be

\[\begin{align} \usf{e}_i(\tau)&=\usf{R}(\tau)\usf{E}_i \end{align}\]

The time derivatives of the co-rotational basis vectors are therefore,

\[\begin{align} \usf{e}'(\tau)_i&=\usf{R}'(\tau)(\tau)\usf{E}_i \label{eq:eider} \end{align}\]

Recall that the angular velocity matrix at the time instance $\tau$, namely, $\usf{W}_i(\tau)$ is defined as

\[\begin{align} \usf{W}_i(\tau)&=\usf{R}'(\tau)\usf{R}^{\textsf{T}}(\tau) \end{align}\]

Therefore, we can write $\usf{R}’(\tau)$ in $\eqref{eq:eider}$ as equal to $\usf{W}(\tau)\usf{R}(\tau)$. Upon doing so, we get

\[\begin{align} \usf{e}'_i(\tau)&=\usf{W}(\tau)\usf{R}(\tau)\usf{E}_i\\ &=\usf{W}(\tau)\usf{e}_i(\tau)\\ &=\usf{w}(\tau)\times \usf{e}_i(\tau) \end{align}\]

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