Rotation of a Curve In an Analytical Mechanics text I have read that, given a regular plane curve $\gamma(t): ]a,b[ \to \Bbb R^2$, if $\underline{t}(s)$ is the tangent vector wrt to the natural parameter $s$, then the curve obtained by integrating $(\underline{t}(s) + \underline{c})$, where $\underline{c}$ is a constant in $\Bbb R^2$, is just a rotated $\gamma$ (by rotation I mean that there exists a matrix $R \in SO(2)$ such it transforms $\gamma$ in the integrated curve).
Does anyone know how to prove such claim?

As always, any comment or answer is much appreciated and let me know if I can explain myself clearer.
 A: In their proof of the Fundamental Theorem of Curve Theory the authors of [1] start with the following second order ODE for the tangent vector $\mathbf{t}$ w.r.t. the natural parametrization $s$:
$$\tag{1.17}
\frac{\rm{d}^2\mathbf{t}}{\rm{d}s^2}-\frac{k'(s)}{k(s)}\frac{\rm{d}\mathbf{t}}{\rm{d}s}+k^2(s)\mathbf{t}=0\,;
$$
where $k(s)>0$. They write "after integration this yields $\mathbf{t}=\rm{d}\mathbf{x}/\rm{d}s$ up to a constant vector (i.e. a rotation of the curve)."
I find that a bit odd. What I can understand is that (1.17) is a linear ODE that determines the solution $\mathbf{t}$ only up to an affine transformation. Since in the natural parametrization the tangent vector $\mathbf{t}$ always has length one the only
possible affine transformations that preserve the length of $\mathbf{t}$ are rotations or reflections, i.e., matrices in $O(2)$.
Clearly these matrices act on $\mathbf{x}$ (the curve itself) in the same way
as they act on the tangent vectors $\mathbf{t}$.
[1] A. Fasano, S. Marmi, Analytical Mechanics. Oxford 2006.
A: The claim is false. For example, let us consider $\underline{c}=(-1,0)$ and $\gamma(t)=(t,0)$ (a line): then
$$\underline{t}(s)+\underline c=0$$
and so the curve obtained by integrating is actually a point (i.e. a constant curve)!
So in general the curve is not transformed by a rotation, or by a bijective map.
