Let $\gamma$ be a regular plane curve and let $\lambda$ be a constant. The parallel curve $\gamma^\lambda$ of $\gamma$ is defined as $$\gamma^\lambda(t)=\gamma(t)+\lambda n_s(t)$$ where $n_s$ is the normal of the tangent line.
Question:
Let $\kappa_s(t)$ be the signed curvature of $\gamma$ at $t$.
Show that if $\lambda\kappa_s(t)\neq1$ then $\gamma$ is a regular curve and its signed curvature is $\frac{\kappa_s}{|1-\lambda\kappa_s(t)|}$.
My work so far:
$\gamma^\lambda(t)=\gamma(t)+\lambda n_s(t)$ $\implies$ $\frac{d\gamma^\lambda(t)}{d s^\lambda}\frac{ds^\lambda}{dt} = \frac{d\gamma(t)}{ds}\frac{ds}{dt}+\lambda\frac{dn_s(t)}{ds}\frac{ds}{dt}$.
Let $T_\gamma(t)$ denote the tangent vector of $\gamma$ at $t$. Since $n_s'=-\kappa_sT$ we have:
$T_{\gamma^\lambda}\frac{ds^\lambda}{dt} = T_\gamma\frac{ds}{dt}+\lambda(-\kappa_sT_\gamma)\frac{ds}{dt}$ $\implies$ $T_{\gamma^\lambda}\frac{ds^\lambda}{dt} = (1-\lambda\kappa_s)T_\gamma\frac{ds}{dt}$.
I think that implies that $\frac{ds^\lambda}{dt} = |(1-\lambda\kappa_s)|\frac{ds}{dt}$, since $T_{\gamma^\lambda}$ and $T_\gamma$ are co-linear, and the variance of one is described in relation to the other.
I don't know where to go from here, but this feels like the right direction. Also, I'm not really sure what the geometric sense of what I'm proving is. So any advice is much appreciated. Thanks.
I think you read Do Carmo, and this is an excercise. Let me use $\beta (t)$ for $\gamma^\lambda$, $$\beta'(t)=(1-\lambda k(s)).\frac{ds}{dt}.\frac{d\gamma}{ds}\\ \beta'(t)=(1-\lambda k(s)).\frac{1}{k(s)}.\frac{d\gamma}{ds}\\ \frac{d\beta}{ds^\lambda} \frac{ds^\lambda}{dt}=(1-\lambda k(s)).\frac{1}{k(s)}.\frac{d\gamma}{ds}\\ \frac{d\beta}{ds^\lambda} \frac{1}{k_\lambda (s^\lambda)}=(1-\lambda k(s)).\frac{1}{k(s)}.\frac{d\gamma}{ds}\\$$ Note that $|\frac{d\beta}{ds^\lambda}|=1$ and $|\frac{d\gamma}{ds}|=1$ and finally that $1-\lambda k(s)\ne0$, then the result.