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How can we get $ \large w= \frac{1}{k(a)} + \frac{1}{k(a+pi)}$ by using those $4$ facts I got?

Let $y (a)$ be a simple closed planar curve with curvature $k > 0$ parametrized by $a$, where $a$ is defined as the arclength parameter of the unit tangent field $e_1$. Further assume that the width $w = <e_2 (a) , (y (a + pi) − y (a))>$ is constant. Show that:

$$ \large w= \frac{1}{k(a)} + \frac{1}{k(a+pi)}$$

Start by establishing the facts: $$\frac{dy}{da}=\frac{e_1}{k}$$ $$\frac{de_1}{da}=e_2$$ $$\frac{de_2}{da}=-e_1$$ $$e_1(a+pi)=-e_1(a)$$

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Hint: $w$ is a constant. Use this fact to get an expression for $y(a+\pi)-y(a)$. – achille hui Jan 30 '13 at 8:18
I'm confused about the"<,>",it's dot product? – user59889 Jan 30 '13 at 16:57
Yes, that "<,>" is an inner product. BTW, you should edit the pi in your post. – achille hui Jan 30 '13 at 17:53

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