Let $\alpha (s)$ , $s\in [0,l]$, be a closed convex plane curve positively oriented. The curve $\beta(s)=\alpha (s) -rn(s)$, where $r$ is a positive constant and $n(s)$ is the normal vector, is called a parallel curve to $\alpha$. Show that:
$ \text{Length}(\beta)=\text{Length}(\alpha) +2\pi r$
$ A(\beta)=A(\alpha)+rl+\pi r^2$
$k_{\beta}(s)=\frac{k_{\alpha}(s)}{1+r}$
where the $k$'s are the curvatures of the corresponding curves.
For 1, I used the equation L(\beta)=\int_{0}^{l} \|\beta'(s)\| ds = \int_{0}^{l} (\|(1+rk_{\alpha})\alpha'(s) \| +\tau r |b(s)|) ds =L(\alpha) (1+rk)+r\tau
I don't see how from this I arrive at equation 1.
For 2, in this case it's the area enclosed by the curve, not sure what formulas to use here. Any hints?
For 3, I calculated according to the definition of the curvature, where k_{\beta}(s)=\frac{\|\gamma '(s)\|}{\|\beta '(s) \|} where \gamma (s)=\frac{\beta '(s)}{\|\beta '(s)\|}. One of my assumptions is that we're in arclength representation, so k_{\alpha}(s)=\|\alpha ''(s) \|. But I didn't arrive at the same equation, on the contrary it looks like a messy calculation that leads to nowhere. I got that \gamma '(s)= \frac{r^2[r\tau ^2 k'_{\alpha}-(1+rk_{\alpha})\tau \tau ']\alpha '(s)+[(1+rk)k+r\tau ^2][(1+rk)^2+(r\tau)^2] n(s) +r\tau '(1+rk)^2 b(s)}{[(1+rk)^2+(r\tau)^2]^{\frac{3}{2}}} I don't see how anything gets simplified here.
Thanks in advance.