A question on closed convex plane curves (from Do Carmo)

Question

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:

1. $$ \text{Length}(\beta)=\text{Length}(\alpha) +2\pi r$$

2. $$ A(\beta)=A(\alpha)+rl+\pi r^2$$

3. $$k_{\beta}(s)=\frac{k_{\alpha}(s)}{1+r}$$

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.

Answer 1

(1) Since this is a plane curve, the torsion $\tau$ is zero. Without loss of generality we may assume $s$ is the arclength parameter so that $\|\alpha'(s)\|=\|T\|=1$, with $T$ the tangent vector. Compute

$$\beta'=\alpha'-r(-\kappa T)=(1+r\kappa)T.$$

Above we used the Frenet-Serret formulas to differentiate the normal vector. Hence

$$L(\beta)=\int \|\beta'\|ds=\int(1+r\kappa) ds=L(\alpha)+2\pi r.$$

Above we used the total curvature identity $\int \kappa ds = 2\pi$.

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(2) In order to compute the area of $\beta$ we will invoke Green's theorem in the $\mathbb{R}^3$ setting with cross products. Here $B=T\times N$ is the binormal vector - note that the cross product is antisymmetric, and the area $\tfrac{1}{2}\int xdy-ydx$ for the curve $\beta$ becomes

$$A(\beta)=\frac{1}{2}\int(\beta\times\beta')\cdot B ds$$

$$=\frac{1}{2}\int(\alpha-rN)\times(1+r\kappa_{\alpha})T\cdot B ds$$

Let's FOIL this into four parts:

• $\tfrac{1}{2}\int (\alpha\times T)\cdot Bds = A(\alpha)$ because $T=\alpha'$.
• $\tfrac{1}{2}\int \alpha\times (r\kappa_\alpha)T\cdot B ds=\tfrac{1}{2}r\int \alpha\times (-dN/ds)\cdot Bds$ we can rewrite using by-parts integration as $ \tfrac{1}{2}r\int T\times N\cdot Bds=\tfrac{1}{2}r L(\alpha)$.
• $\tfrac{1}{2}\int (-rN)\times T\cdot Bds=\tfrac{1}{2}r L(\alpha)$
• $\tfrac{1}{2} \int (-rN) \times (r\kappa_\alpha T)\cdot B = \pi r^2 $

Which makes a total of $A(\alpha)+rL(\alpha)+\pi r^2$.

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(3) Now we'll compute $\kappa_{\beta}$ with a local formula:

$$\kappa_{\beta}=\frac{\|\beta'\times\beta''\|}{\|\beta'\|^3}=(1+r\kappa_{\alpha})^{-3}\left\|(1+r\kappa_{\alpha})T\times\left(r\frac{d\kappa_{\alpha}}{ds}T+(1+r\kappa_{\alpha})\frac{dT}{ds}\right)\right\|.$$

Note that $T\times T=0$ and $\|T \times T_s\| = \kappa_{\alpha}$ so this simplifies drastically to

$$\kappa_{\beta}=\frac{\kappa_{\alpha}}{1+r\kappa_{\alpha}}.$$

(I assume what you have written in the question is missing a $\kappa_{\alpha}$ in the denominator.)