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.

Question

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.

Mekjulleymg · Accepted Answer

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.

Yesenia Obrien · Accepted Answer

Since       |        |    β  (  s  )      |              |        2    =  β  (  s  )  ⋅  β  (  s  ) you can use the product rule to show       ∂          ∂      s            |        |    β  (  s  )      |              |        2    =  0.

Assume we have an arc-length parameterized curve β : I →<!-- →

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