Vtyjkyuk

To make a long story short: I have a surface $X: I to mathbbR^3$ that sends $(u ,v) in I subset mathbbR^2$ to $X(u,v) in mathbbR^3$. Consider these two family of surfaces:

$hatX(u,v,t) = sqrt2bt + 1expleft( fraclog(2bt + 1)2b Aright) X(u,v)$

$F(u,v,t) = sqrt2bt + 1expleft( t Aright) X(u,v)$

where $b in mathbbR$ 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 $hatX$, $X$ 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 $dfracpartial hatXpartial t$

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

$$Q(t)=exp(tcdot S)$$

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.