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I have a smooth curve of length $L$ with a single cusp point $P$ occuring at length $s = L_P$. Let the curve in arc length parametrization be $\alpha_t(s) \equiv (X_t(s),Y_t(s)) $. They are actually a family of curves as parameter $t$ varies, but the length of the curve remains same with changes in $t$, except the tangents of the curve rotate with different angular speeds, with variations in $t$. The cusp point $P$ at $s= L_P$ remains as cusp for all $t$.

I want to measure how straight the curve at every instant $t$, by its moment of inertia about its center of mass, given as, $$I_{0,L}(t) = \int_{0}^{L}(t) ((X_t(s)-X_{cm})^2 + (Y_t(s)-Y_{cm})^2) ds, $$ where $X_{cm} = \frac{1}{L}\int_{0}^{L}X_t(s)ds$ and $Y_{cm} = \frac{1}{L}\int_{0}^{L}Y_t(s)ds$.

Question : Its a bit subjective. Does it make more sense mathematically, to break the curve into two, the segments between $(0,L_P)$ and $(L_P,L)$, and then measure the straightness of the two curves separately as $I_{0,L_P}(t)$ and $I_{L_P,L}(t)$ using the above formula. Of course we do not get the straightness of entire curve this way, but I just want a reason to abandon the idea of using straightness of entire curve for any mathematical purpose and using straightness of each segment, but for a slightly different purpose, to derive more meaningful mathematical conclusions, which would make more mathematical sense.

The curvature as a function of a parameter, of the specific curve I am interested in is at this link

PS : I heard arc length parametrization is not possible for curve with a cusp, but I guess we are not using an local properties of the curve, so assume it is not a hindrance.

PS : Moment of inertia about center of mass is maximum when the curve is a straight line, as per this question

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  • $\begingroup$ A smooth arc-length parametrization is not possible for a curve with a cusp. However it is possible if you remove either "smooth" or "arc-length" from the requirements. $\endgroup$ – robjohn Jan 2 '15 at 4:01
  • $\begingroup$ @robjohn : Can you expand a bit with some reasons, guess that would potentially be an answer. $\endgroup$ – Rajesh D Jan 2 '15 at 15:21
  • $\begingroup$ @TedShifrin : Could you expand, i don't understand what you mean by inverse, partly due to my ignorance. $\endgroup$ – Rajesh D Jan 2 '15 at 23:55
  • $\begingroup$ Deleted my answer because of the uneasy feeling that it's a misfit to the question. $\endgroup$ – Han de Bruijn Jan 7 '15 at 17:40

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