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Write a proper Euclidean motion of Euclidean three-space as $E(x)=Ax+b$, where $A\in SO(3)$. ThenBecause $A$ preserves norms, dot products, and cross products, the formulasshow directly that curvature and torsion are unchanged by $E$.

For the shifted curve ${\gamma }_{s_0}(s)=\gamma (s-s_0)$, every derivative at $s$ equals the corresponding derivative of $\gamma$ at $s-s_0$. ConsequentlyBoth are therefore pointwise Euclidean invariants in the stated sense.

No. For sufficiently smooth curves,is unchanged by proper Euclidean motions and transforms correctly under translations of the arc-length parameter, so it is a pointwise Euclidean invariant of a curve. It is not determined by the two numbers $\kappa (s)$ and $\tau (s)$.

For a concrete comparison near $s=0$, use the plane curve reconstructed from curvature construction withThe resulting unit-speed planar curves both have $\kappa (0)=1$ and $\tau (0)=0$, but their curvature derivatives at zero are respectively $0$ and $1$. Thus no single function $f$ can satisfy $Q(\gamma ,s)=f(\kappa (s),\tau (s))$ for all curves.

Indeed, the covariance conditions alone even permit nonlocal examples such as $Q(\gamma ,s)={\kappa }_{\gamma }(s+1)$. This is the distinction recorded by pointwise Euclidean invariants need not depend only on current curvature and torsion.