Interpretation of the Derivative of a Quaternion

Question

Considering this definition of the derivative of a quaternion: $$dq/dt = 1/2 w q$$ If we're considering $q$ to be a unit quaternion representing an orientation in 3D with $(cos(theta/2), sin(theta/2)*axis)$, given the values in $w$ which can be arbitrarily large, $dq/dt$ will not be a unit quaternion and so could not be interpreted back the way $q$ can. So... how should one interpret $dq/dt$ in a 3D rotational type of way?

Answer 1

Lets use the definition of derivative

$$ \frac{d}{dt} q(t) = \lim_{\Delta t\rightarrow0}\frac{q(t + \Delta t) - q(t)}{\Delta t} $$

This leads to the result $$ \dot{q} = \frac{1}{2} \left[ \begin{array}{cc} 0 \\ \omega\end{array}\right]^{\oplus} q$$

Or something similar according to the adopted convention. This result can be seen as a perturbation applied to the current quaternion, in which the perturbation is dependent on the rate at which the angle changes, since

$$ \omega = \lim_{\Delta t \rightarrow 0} \frac{\Delta \theta}{\Delta t}$$

Hence the $\omega$ will not be arbitrary large. The definition you gave is missing of the infinitesimal nature of the (local) perturbation used to compute the derivative.

If you want to see the full passages ask in the comments and I will add them.