If we parameterize the arc length of a vector valued(adsbygoogle = window.adsbygoogle || []).push({});

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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# Constant curvature

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