Here is the reasoning I've attemped in order to espablish a formula for the counterclockwise rotation of a curve.
Let $M=(x,y)$ be any point on the plane, associated with a cartesian coordinate system $<0, \vec { i}, \vec{j}>$.
The position vector of ths point is $\vec {OM} = \vec A+ \vec B= x\vec { i} + y \vec{j} $
Rotating point $M$ counterclockwise by a positve angle $\alpha$ amounts to rotate counterclockwise the components of its postion vector by the same angle.
So , with $M'$ the image of $M$ under ths rotation, we will have :
$\vec {OM'} = \vec {A'}+\vec {B'}$ with $\vec {A'}$, the image of $\vec {A}$ and $\vec {B'}$ , the image of vector $\vec {B}$.
Now ,
$\vec {A'}= x cos (\alpha) \vec i + x sin (\alpha) \vec j$.
$\vec {B'}= y cos (\alpha + \pi /2) \vec i + x sin (\alpha+ \pi/2) \vec j $
$= y ( – sin (\alpha) )\vec i + x cos(\alpha) \vec j$
$= -y sin (\alpha) )\vec i + x cos(\alpha)\vec j$.
Therefore ,
$\vec{OM'} = \vec {A'} + \vec {B'} = \Large{[} x cos (\alpha) – y sin (\alpha)\Large{]} \vec i + \Large{[}x sin( \alpha) + y cos(\alpha)\Large{]} \vec j$,
implying that
$M' = (x cos (\alpha) – y sin (\alpha) , x sin( \alpha) + y cos(\alpha))$.
My problem is that :
(1) when I apply this reasoning to a parametric curve, it seems to work , that is, if I have a curve $C= < f(t), g(t)> $ and ask for
$$C' = <f(t) cos (\alpha) – g(t) sin (\alpha) , f(t) sin( \alpha) + g(t) cos(\alpha)>$$
what I get is that $C'$ rotates counterclockwise, as desired.
Example with $\alpha = 2 $ rd: https://www.desmos.com/calculator/daaouk87yw
(2) but if I substitute the coordinates of a generic $M'$ point in a cartesian " $x-y$" equation, the curve rotates clockwise.
Example : https://www.desmos.com/calculator/vm5pbtxzge
I'm sure some confusion prevents me from understanding what is happening, but I can't manage to see which one.
Best Answer
We'll compare the two operations you describe. Instead of reasoning with coordinates, let's use directly the point $M$, and the operation $R$ applied to $M$. $R$ can be a rotation, or any other bijective function.
$R$ being bijective, $E(R(M''))=0$ is equivalent to $M=R(M'')$.
While in the first case $M'=R(M)$.
We can see the role of $R$ is reversed; hence what you remarked: if you express $M''=S(M)$, we see that $S=R^{-1}$.
This is very intuitive, in fact. Let's say it with a scaling operation; compare:
Another similar situation happens when transforming a vector, versus transforming the vector base: that's covariance and contravariance.