Timeline for radius of tubular neighborhood
Current License: CC BY-SA 3.0
8 events
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| Mar 10, 2017 at 9:42 | history | edited | CommunityBot |
replaced http://www.nd.edu/ with https://www.nd.edu/
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| Oct 26, 2012 at 17:39 | comment | added | Liviu Nicolaescu | Let me write it in a different way $x(t)=e^{-t}\cos t$ $y(t)=e^{-t}\sin t$, $t\in [0,2\pi N]$. As $t$ runs in the interval $[0,2\pi N]$ the point $(x(t),y(t))$ keeps on orbiting around the origin, while the distance to the origin decays exponentially to $0$. But I know believe it may not be a counterexample to your conjecture. | |
| Oct 26, 2012 at 16:28 | comment | added | Jay | so.. that is an extra restriction on $\theta \in [0, 2\pi]$? :) | |
| Oct 26, 2012 at 15:33 | comment | added | Liviu Nicolaescu | If it decreases going one way along the curve, it increases going the other way. The example of a spiral given in polar coordinates by the equation $r=e^{-\theta}$, $0\leq \theta \leq 2\pi N$, may provide a counterexample to your conjectured formula. It seems plausible that if the curve is the $C^2$-boundary of a convex domain then the largest tubular radius is the inverse of the maximal curvature. | |
| Oct 26, 2012 at 15:13 | comment | added | Jay | Thanks a lot! Maybe that is what I want: if the curvature of the curve is always decreasing, then the radius of the tubular neighborhood is given by the injective radius of (left) end point, it seems true, right? | |
| Oct 26, 2012 at 12:33 | comment | added | Pietro Majer | actually this is exactly what the OP is saying... | |
| Oct 26, 2012 at 12:17 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
added 3 characters in body
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| Oct 26, 2012 at 12:00 | history | answered | Liviu Nicolaescu | CC BY-SA 3.0 |