# Calculate the curvature of a parametrized curve

I have started to study differential geometry and have some questions about an exercise which is probably not very difficult.

Exercise: Let $\gamma: I \rightarrow\mathbb{R}^{2}$ be a regular curve, parametrized by arclength, with Frenet frame $\{ T(s), N(s) \}$. For $\lambda \in \mathbb{R}$ we define the parallel curve $\gamma_{\lambda}: I \rightarrow \mathbb{R}^{2}$ by

$\gamma_{\lambda}(t) = \gamma(t) + \lambda N(t)$.

Calculate the curvature $\kappa_{\lambda}$ of those curves $\gamma_{\lambda}$ which are regular.

Attempt to solution:

First, we need to now when $\gamma_{\lambda}$ is regular, i.e., when $\dot{\gamma}_{\lambda} \neq 0$.

$\dot{\gamma}_{\lambda} = \dot{\gamma}(t) + \lambda \dot{N}(t)$,

so if $\dot{\gamma}(t) \neq -\lambda\dot{N}(t)$,

then $\gamma_{\lambda}$ is regular.

Secondly, we need to calculate the curvature, and we have that $\kappa(t) = \langle\dot{T}(s), N(s)\rangle$, if $\{ T(s), N(s) \}$ is the Frenet frame for a curve $\gamma(s)$, if $\gamma$ is parametrized by arclength.

However, our curve $\gamma_{\lambda}$ is not parametrized by arclength so I don't think we can use the scalar product directly to calculate the curvature. So I am not sure how to continue.

I guess we shall do something like $\langle \dot{T}_{\lambda}, N_{\lambda}\rangle$.

Any help would be appreciated!

Hint: Let us only assume that $\gamma$ is immersive and $\lambda>0$. Show that $\dot\gamma_{\lambda}=(1-\lambda\kappa)\dot\gamma$, so $\|\dot\gamma_{\lambda}\|=|1-\lambda\kappa|\cdot\|\dot\gamma\|$.
It follows that $\gamma_{\lambda}$ is immersive iff $\kappa(t)\neq\frac{1}{\lambda}$. Let's assume that from now on.
Denote $\frac{d}{ds_{\lambda}}$ the derivation in respect to the arclength of $\gamma_{\lambda}$ and $\frac{d}{ds}$ the derivation in respect to the arclength of $\gamma$ and verify $$\frac{d}{ds_{\lambda}}=\frac{1}{|1-\lambda\kappa|}\frac{d}{ds}.$$
Now it's easy to compute $T_\lambda$ and $N_{\lambda}$ and to derive that $$\kappa_{\lambda}=\frac{\kappa}{|1-\lambda\kappa|}.$$
Gimmick: verify $$\dot\kappa_{\lambda}=\operatorname{sgn}(1-\lambda\kappa)\frac{\dot\kappa}{|1-\lambda\kappa|^2},$$ which reveales that $\gamma$ and $\gamma_{\lambda}$ share the same inflection points.