Suppose $\gamma$ is a unit speed plane curve with a constant curvature $k\neq 0$.
Show that $\beta(s)=\gamma(s)+\frac{1}{k^2}\gamma''(s)$ is a constant curve.
To prove $\beta$ is constant curve, we have to prove $\beta'(s)=0$ for all $s\in I$.
We have $\beta'(s)=\gamma'(s)+\frac{1}{k^2}\gamma'''(s)$. I do not understand how to prove that this is identically zero.
Curvature function is given by $k(s)=||\gamma''(s)||$.
Please give only hints.
If $\gamma$ is unit-speed, $\lVert \gamma'(s) \Vert = 1$, and the usual definition of curvature gives $$ \vec{T}' = \gamma'' = \kappa \vec{N}, $$ where $\vec{T},\vec{N}$ are the tangent and normal vectors. The key lies in the other Frenet–Serret equation: since $\{\vec{T},\vec{N}\}$ is an orthonormal basis of the plane, $$ \vec{T}' \cdot \vec{N} + \vec{T} \cdot \vec{N}' = 0, $$ so we must have $$ \vec{N}' = -\kappa \vec{T}. $$