To make a long story short: I have a surface $X: I \to \mathbb{R}^3$ that sends $(u ,v) \in I \subset \mathbb{R}^2$ to $X(u,v) \in \mathbb{R}^3$. Consider these two family of surfaces:
$\hat{X}(u,v,t) = \sqrt{2bt + 1}\exp\left( \frac{\log(2bt + 1)}{2b} A\right) X(u,v)$
$F(u,v,t) = \sqrt{2bt + 1}\exp\left( t A\right) X(u,v)$
where $b \in \mathbb{R}$ is a constant and $A$ is an orthogonal matrix with determinant $1$. In the thesis I'm reading (page 45, the dilation and rotation section) the author writes (not exactly as I put it here, but I just made things simpler so I could understand, there's no loss of generality) that, under $\hat{X}$, $X$ rotates with decreasing speed if $b > 0$ and with increasing speed if $b < 0$, and that under $F$, $X$ rotates with constant speed. This is supposed to be obvious but I'm not quite seeing it and I'd appreciate some help.
EDIT: Turns out it was just a matter of analysing the signs of $\dfrac{\partial \hat{X}}{\partial t}$
I couldn't find where it says that $A$ orthogonal matrix with determinant 1 is a rotation.
As food for thought, if $Q(t)$ are rotations, which are of the form
$$Q(t)=\exp(t\cdot S)$$
for skew symmetric matrices (see Wikipedia and the links there) then $Q'(t)=Q(t)\,S$ and so
$$Q^T(t)\,Q'(t)=Q^{-1}(t)\,Q(t)\,S=S.$$
So
$$\exp(h(t)\cdot Q^T(t)\,Q'(t))$$
is really again just a rotation and $\phi=h(t)$ is a parametrization of angles. Not sure if those bullet points all apply here, but if they do, then you got your explanation.