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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.

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    Is the $k$ in the question the curvature or just a constant? – 2017-02-11
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    It is curvature of $\gamma$. @AlfredYerger – 2017-02-11

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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 $$ \vec{N}' = -\kappa \vec{T}. $$

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    We have $N'(t)=-k(t)T(t)$... As $N(t)=\frac{\gamma''(t)}{||\gamma''(t)||}$ we have $\frac{\gamma'''(t)}{||\gamma''(t)||}=-k(t)\gamma'(t)$. As $k(t)=||\gamma''(t)||$ we have $\frac{\gamma'''(t)}{k(t)}=-k(t)\gamma'(t)$ i.e., $\gamma'''(t)=-k^2(t)\gamma'(t)$ equivalently $\gamma'(t)+\frac{1}{k^2(t)}\gamma'''(t)=0$ i.e., $\beta'(t)=0$ for all $t$ which then imply $\beta$ is constant.. – 2017-02-11
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    I think that's true by accident in this case, since the curvature (and hence $\lVert \vec{N} \rVert$) is constant. Better to say that $ \vec{T}'' = (\kappa \vec{N})' = \kappa' \vec{N} - \kappa^2 \vec{T}, $ and $\kappa'=0$ in this case, so $ \vec{T}''+\kappa^2\vec{T} = 0 $. – 2017-02-11
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    The cross product is zero if two vectors are parallel (or one is zero), not perpendicular, as $T$ and $N$ are. In this case, $\gamma$ is planar, so $B$ is constant, perpendicular to the plane. – 2017-02-11
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    I am still confused... I understand that cross product is zero if two vectors are parallel... But i do not understand second line... $\gamma$ is planar so $B$ is constant? – 2017-02-11
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    How do you define planar? – 2017-02-11
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    Oh No.. It is not $B$ that is zero but the torsion that is zero... Right? As it is planar curve, there is no torison... It has nothing to do with $T,N$ right? – 2017-02-11
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    Yes, exactly. Although one might equally say that a plane curve is a continuous map $(a,b) \to \mathbb{R}^2$, and a planar curve is one in $\mathbb{R}^3$ with vanishing torsion, assuming some regularity. – 2017-02-11