Reconstruct a Euclidean motion from its action on a curve I have an issue with an exercise regarding the reconstruction of a euclidean motion.
Let $\gamma: [0,L] \to \mathbb{R}^2$ be an arclength-parametrized closed curve and $A: \mathbb{R}^2 \to \mathbb{R}^2$ a Euclidean motion. Define $\tilde{\gamma}:=A \gamma$ and assume that $\tilde{\gamma}(0)=0$ and $\tilde{\gamma}'(0)=(1,0)$. Determine $A$ from $\gamma(0)$ and $\gamma'(0)$. Find a regular homotopy between $\gamma$ and $\tilde{\gamma}$.
This is my attempt at a solution. Since $A$ is a Euclidean motion it takes the form
$$ A: \mathbb{R}^2 \to \mathbb{R}^2, A(v)=Sv+b
$$
where $S \in O(2)$ (that is, $S$ is an orthogonal transformation) and $b \in \mathbb{R}^2$. We have
$$
(0,0)=\tilde{\gamma}(0)=S \gamma(0)+b \\ 
(1,0)=\tilde{\gamma}'(0)=S \gamma'(0).
$$
We can derive the following equations
$$
\tilde{S} b=-\gamma(0) \tag{1} 
$$
$$
\tilde{S} (1,0)=\gamma'(0) \tag{2}
$$
where $\tilde{S}$ denotes the inverse of $S$. (2) implies that the first row of $\tilde{S}$ is equal to $\gamma'(0)$. Since $|\gamma'(0)|=1$ we have $\gamma'(0)=(\cos(\alpha),\sin(\alpha))$. This means that $\tilde{S}$ is a rotation where
$$
\tilde{S}=
\begin{pmatrix}
\cos(\alpha) & -\sin(\alpha) \\
\sin(\alpha) & \cos(\alpha)
\end{pmatrix}
$$
So
$$
S=
\begin{pmatrix}
\cos(\alpha) & \sin(\alpha) \\
-\sin(\alpha) & \cos(\alpha)
\end{pmatrix}.
$$
But now I am not sure how to use (1) to determine $b$ as I do not see where the rotation maps the vectors. I know that $S$ causes a clockwise rotation by $\alpha$ and $\tilde{S}$ causes a counterclockwise roatation by $\alpha$. But I do not see how to derive a formula for $b$ from this.
\Edit: Regarding the formula for b it seems that indeed (1) yields the solution $b=-S \gamma(0)$. Then $A$ takes the form $A(v)=Sv-S\gamma(0)=S(v-\gamma(0))$. So $\tilde{gamma}$ is a rotated version of $\gamma$ that has been translated into the origin. This seems to be a valid answer as the exercise asks for $A$ to be determined from $\gamma(0)$ and $\gamma'(0)$.
