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gaiaecologicaq2 2022-07-11 Answered
Assume we have an arc-length parameterized curve β : I E 2 with I a random interval. I want to show that if β ( s ) β ( s ) = 0 for arc-length parameter s, then β is (part of) a circle.

I was thinking about using the Frenet-formulas for proving this. If we want to show something is a circle, one needs to show that the curvature is constant. In the Frenet-frame ( T , N ), the curvature is: κ = T N. So to be constant, it needs to be:
κ = 0 = T N + T N
I was thinking about using the formulas T = β , T = κ N and N = κ T. But together with the given condition: β ( s ) β ( s ) = 0, I got stuck.
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Answers (2)

Mekjulleymg
Answered 2022-07-12 Author has 14 answers
Note that a curve lies on a circle if β ( s ) β ( s ) = c = r 2 for some (necessarily positive) constant c. Since I is connected, it suffices to check that the derivative of β ( s ) β ( s ) is 0. However the derivative of β ( s ) β ( s ) is 2 β ( s ) β ( s ), and we already know this is 0 by the given information.

In general, if the curve lies in R n , this shows that the curve lies on a sphere.

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Yesenia Obrien
Answered 2022-07-13 Author has 5 answers
Since | | β ( s ) | | 2 = β ( s ) β ( s ) you can use the product rule to show s | | β ( s ) | | 2 = 0.

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