3
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.