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.

  • $\begingroup$ Where did $\tau$ come from? If it's supposed to be the torsion, you're aware that $\tau=0$ if $\alpha$ is flat? (Frenet-Serret simplifies vastly in the planar case, and you no longer need to consider the binormal.) $\endgroup$ – J. M. is a poor mathematician Aug 19 '11 at 9:33

(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 $\int \kappa ds = 2\pi$.

(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.

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

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

$$ = \left(\frac{1}{2}\int \alpha\times T\cdot Bds\right)+ r\int B\cdot Bds+\frac{1}{2}\left(\int r^2\kappa_{\alpha}ds\right)$$

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

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


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


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

  • $\begingroup$ Thanks, anon, though in question 2) you could have just tell me the first equation to use here, nontheless it was helpful. With regard to the last equation, I didn't miss it, this is how it's written in my edition of the book. I guess there's a misprint there. $\endgroup$ – MathematicalPhysicist Aug 19 '11 at 10:39
  • $\begingroup$ in part(2) how can you get $rL(\alpha)$ $\endgroup$ – python3 Jan 29 '15 at 5:10
  • $\begingroup$ @tiandiao123 $\int ds$ is length. $\endgroup$ – anon Jan 29 '15 at 9:35
  • $\begingroup$ I mean $r\int BBds$, I am not sure how to get this, I understand the other two $\endgroup$ – python3 Jan 29 '15 at 18:50
  • 2
    $\begingroup$ I am very confused as to how you derived that cross-product formula for Green's Theorem. Could you elaborate on that? $\endgroup$ – Felicio Grande Feb 19 '17 at 2:04

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