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I can't prove formally this idea which I'm almost 100% sure:

If a curve is inside a circle than it can't have any point with a Smaller curvature than the circle itself.

How can I prove this simple and intuitive idea formally?

user42912
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  • I assume you're talking about a curve that goes on forever? – Landuros Dec 28 '17 at 11:48
  • if you consider a square? – user Dec 28 '17 at 11:49
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    What you are trying to prove is false. Indeed, every curve contained inside a circle has a point with curvature greater than that of a circle. Is that perhaps what you want to prove? – Wojowu Dec 28 '17 at 11:50
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    Not following the question. Take a smaller circle inside the big one as an obvious counterexample. – lulu Dec 28 '17 at 11:51
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    Are you talking about curvature or the radius of curvature? IIRC those are reciprocals of each other. A smaller circle bends faster than a bigger one, and thus has higher curvature. But, you can easily have a curve inside a circle such that it goes more or less straight at some point (infinite radius of curvature) and turns very fast elsewhere (low radius of curvature). In other words, what you are trying to prove is false. – Jyrki Lahtonen Dec 28 '17 at 11:57
  • @Landuros I'm sorry I meant "smaller" see my edit please – user42912 Dec 28 '17 at 13:48
  • @lulu I'm sorry I meant "smaller" see my edit please – user42912 Dec 28 '17 at 13:49
  • @JyrkiLahtonen see my edit please. I meant "smaller" – user42912 Dec 28 '17 at 13:50
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    But that is also false, as a curve can contain an arbitrarily small straight segment (hence points of curvature $0$). – lulu Dec 28 '17 at 13:59
  • Look at the suggestion of @Wojowu . If you have a simple closed curve inside a circle of radius $r$ then it has a point of curvature greater than $\frac 1r$ ("proof": pick a point $P$ in the interior of your curve and consider the family of circles centered at $P$. The first point of contact with the curve gives you a point on the curve with curvature greater than $\frac 1r$.) – lulu Dec 28 '17 at 14:02

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Within the unit circle, consider the curve defined by $$ (4x)^2 + (2y)^2 = 1 $$ or parametrized by $$ t \mapsto (\frac{1}{8} \cos t , \frac{1}{2} \sin t) $$ This is an ellipse with major axis $1$ and minor axis $\frac{1}{4}$.

At $t = 0$, or $(x, y) = (\frac{1}{8}, 0)$, the radius of curvature is $R = (1/8)^2/(1/2) = 2/64 = 1/32$; hence the curvature is $32$.

At $t = \pi/2$, or $(x, y) = (0, \frac{1}{2})$, the radius of curvature is $(1/2)^2/(1/8) = 8/4 = 2$; hence the curvature is $1/2$.

So the curvature can be both smaller and larger than the curvature of the unit circle (which is $1$).

John Hughes
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