Proving Constant Curvature of r(s) is a Circle

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The discussion confirms that if a vector-valued function r(s) has constant curvature (K ≠ 0), then r(s) represents a circle. The proof involves demonstrating that the derivative of the unit tangent vector |T'(s)| equals the curvature K, leading to the expression <-Kcos(s), -Ksin(s)>. The integration of this expression component-wise is necessary, particularly when the function is defined in R², as higher dimensions complicate the proof.

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hholzer
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If we parameterize the arc length of a vector valued
function, say, r(s) and r(s) has constant curvature
(not equal to zero), then r(s) is a circle.

Thus, |T'(s)| = K but to prove it we would need
to show |T'(s)| = K => <-Kcos(s), -Ksin(s)>
and integrate component-wise two times,
right?
 
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If your function is into R^2...else it could be way more complicated. I'd try to find a point x such that |r(s)-x| is constant.
 

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