Question: Let $\gamma:[a,b]\to \mathbb{R}^2$ be a regular plane curve. For a constant $\epsilon\in \mathbb{R}$, define $\tilde{\gamma}:s\mapsto \gamma(s)+\epsilon \pmb{n}(s)$. Show that
i) If $|\epsilon|\ll 1$, then $\tilde{\gamma}$ is regular.
ii) What is the new curve's arc length and signed curvature in terms of $\gamma$'s.
What did I do: We want to show that $|\tilde{\gamma}'(s)|\neq 0$, therefore, $$|\gamma'+\epsilon\pmb{n}'|=\left\lvert|\gamma'|\pmb{t}+\epsilon(-|\gamma'|\kappa\pmb{t}+|\gamma'|\tau\pmb{b})\right\rvert\neq 0\iff \pmb{t}+\epsilon(-\kappa\pmb{t}+\tau\pmb{b})\neq 0$$ However, I could not continue because I am not sure how to use $\kappa$ and $\tau$ after that. How should I continue or should I change my starting point?
I search some question in here and I think most of the people use the notation $T(s)$ and $N(s)$ instead of $\pmb{t}(s)$ and $\pmb{n}(s)$. Also, $\pmb{b}(s)$ is the unit binormal which is equal to $\pmb{t}(s)\times \pmb{n}(s)$.
Recall that the signed curvature of $\gamma$ at $s$ is $$ k_\gamma(s)= \langle N'_\gamma(s), T_\gamma(s)\rangle, $$ where $T_\gamma(s) = \gamma'(s)/\Vert\gamma'(s)\Vert$ and $N_\gamma(s)\perp T_\gamma(s)$. Therefore, given $\epsilon\in\mathbb R$, we have $$ (\gamma(s) + \epsilon N_\gamma(s))' = \gamma'(s) - \epsilon k_\gamma(s) T_\gamma(s) = (\Vert\gamma'(s)\Vert - \epsilon k_\gamma(s))T_\gamma(s). $$ Now, observing that $[a,b]$ is compact, and $\gamma'(s)\ne0$, we deduce that the maximum $$ c=\max\Big\{\frac{|k_\gamma(s)|}{\Vert\gamma'(s)\Vert}\mid s\in[a,b]\Big\} $$ is attained in $[a,b]$. In particular, $\bar\gamma$ is regular for $0<\epsilon < 1/(c+1)$ because in this case $\Vert\gamma'(s)\Vert>\epsilon k_\gamma(s)$.
To find the arc-length of $\bar\gamma$ observe that, by definition, it equals $$ \ell(\bar\gamma) = \int_a^b\Vert\bar\gamma'(s)\Vert\,ds = \int_a^b(\Vert\gamma'(s)\Vert-\epsilon k_\gamma(s))\,ds = \ell(\gamma)-\epsilon\int_a^bk_\gamma(s)\,ds. $$