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Let $\alpha (s)$ , $s\in [0,L]$, be a smooth positively oriented regular Jodan curve which is arc-length parametrized. The curve $\beta(s)=\alpha (s) +\lambda n(s)$, where $\lambda$ is a positive constant and $n(s)$ is the normal vector, is called a parallel curve to $\alpha$.

I have already proved that:

1)$ \text{Length}(\beta(s))=\text{Length}(\alpha) -2\pi \lambda$

2)$K_{\beta}(s)=\frac{K{\alpha}(s)}{1-\lambda K{\alpha}(s)}$

Now, I have to compute the area of $\beta$ in terms of the area of $\alpha$: $A(\Omega_{\alpha})$

$\beta(s)=\alpha (s) +\lambda n(s)= (\alpha_1 (s) +\lambda n_1(s),\alpha_2 (s) +\lambda n_2(s))$

$\beta'(s)=\alpha '(s) +\lambda n'(s)=(\alpha _1'(s) +\lambda n_1'(s),\alpha _1'(s) +\lambda n'_1(s))$

By a corollary of Green's theorem,

$A(\Omega_{\beta})=-\int_{0}^L (\alpha_2 (s) +\lambda n_2(s))(\alpha _1'(s) +\lambda n_1'(s))ds=$$-\int_{0}^L (\alpha_2\alpha_1'+\lambda \alpha_2 n_1'+\lambda n_2\alpha _1'+\lambda^2n_2n_1') $=

Using, Frenet formulas: $n_1'(s)=-k(s)\alpha_1'(s)$ we got:

=$A(\Omega_\alpha)+\int_{0}^L \lambda k\alpha_2\alpha_1'-\int_{0}^L n_2\alpha_1'+\int_{0}^L \lambda^2kn_2\alpha_1'$

I don't know how to continue or if this is the right way to do it.

***Are the previous formulas for the curvature length and area still true for a general $\alpha$?

Thank you for your help.

Trian
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  • I haven't checked your explicit calculations, but I think you should consider the graph of $\beta$ as the boundary of a submanifold in $\mathbb{R}^2$. Then you can use Stoke's Theorem. – David Hornshaw Mar 23 '14 at 12:53
  • We haven't learned that theorem yet, so I can only use Green's theorem and some corollaries. – Trian Mar 23 '14 at 12:56
  • I use the notation as in here: https://en.wikipedia.org/wiki/Green%27s_theorem

    Choose $M = 0$, $L = 1$, and you should get Stoke's Theorem. In essence, it tells you that in order to get the area enclosed by $\beta$, you can simply integrate over the boundary of that area. It happens to be the graph of $\beta$.

     – David Hornshaw Mar 23 '14 at 13:06
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    I don't think this formula is correct. Consider a circle of radius $R$. The expanded curve will have area $\pi(R+\lambda)^2$, as expected. But now replace that for a sort of pac-man shape, where the path traces out almost a complete circle, then drops to the center and right back out (with smooth turns). Then $A(\Omega_\alpha)\approx\pi R^2$, and $L\approx2\pi R+2R$, so the formula suggests $A(\Omega_\beta)\approx\pi(R+\lambda)^2+2R\lambda$, yet it has to be contained in the circle of radius $R+\lambda$. – Mario Carneiro Apr 09 '14 at 23:16
  • Have you checked whether eqn 1 gives the correct answer for parallel straight lines? – sammy gerbil Apr 21 '20 at 03:44

1 Answers1

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Let $\gamma_t(s)=\alpha(s)+t \mathbb{n}, \forall t\in [0,\lambda]$. We have $\gamma_0=\alpha$ and $\gamma_{\lambda}=\beta$, and the area between $\alpha$ and $\beta$ is parameterized by $(s,t)$. Denote $k(s)$ as the curvature of $\alpha$ at $s$. First let's show that the tangent vector using the notation from wiki and using outer normal as the positive direction: $$\frac{\partial\gamma}{\partial t}=\mathbb{n}$$ $$\frac{\partial\gamma}{\partial s}=\dot{\alpha} +t\dot {\mathbb{n}}=T+tkT=(1+tk)T $$ Since $\{T, \mathbb{n}\}$ is an orthonormal basis the area element can be written as $$d\sigma=(1+tk)dt\wedge ds$$ Hence $$A(\Omega_\beta)-A(\Omega_\alpha)=\int d\sigma$$ $$=\int(1+t\cdot k)dt\wedge ds$$ $$=\int\int dt ds + \int t\cdot ({\int kds}) dt$$ $$=\lambda L + 2\pi\int t dt$$ $$=\lambda L + \pi \lambda^2$$

Xipan Xiao
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