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i was reading the book of Docarmo of differential geometry and I Have a question at the end of the proof that given the curvature and the torsion of a curve, the curve it´s unique , I only omitted the part where it shows that a rigid motion does not alter the curvature and torsionenter image description here

My question is in the red rectangle, why this equality it´s true? i did not understand it, sorry for my questions...

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The author proved that the derivative of $|t-\bar{t}|^2+|n-\bar{n}|^2+|b-\bar{b}|^2$ is $0$. When the derivative of a function is $0$, it means the function is constant. But since the expression is identifiably $0$ at the initial point $s_0$, it must be identically $0$ for all $s$. The only way that's possible is if its constituent parts $|t-\bar{t}|^2,|n-\bar{n}|^2,|b-\bar{b}|^2$ are all $0$. The only way $|t-\bar{t}|^2$ is always $0$ is if $t=\bar{t}$ always holds. But $t=d\alpha/ds$ and $\bar{t}=d\bar{\alpha}/ds$, so we have the equality in red.

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If I derivate the curve with respect the parameter, why it return an scalar t? i don´t understand it –  Daniel Aug 10 '11 at 3:20
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$t$ is not a scalar, it's the tangent vector, which is by definition $d\alpha/ds$. How can you even read about Frenet math if you don't know what $t,n,$ or $b$ stand for? –  anon Aug 10 '11 at 3:24
    
@anon: I found the notation of Do Carmo slightly confusing myself, in any event; I'd have used capital letters for those three vectors, since I use $t$ as a generic non-arclength parameter when I deal with curves... –  J. M. Aug 10 '11 at 3:29
    
@J.M.: I agree that lowercase was a very poor choice on the author's part (reading it felt like scraping a chalkboard), but the author also quite explicitly and clearly defines $t,n,b$ as the Frenet frame so there shouldn't be any confusion in the final analysis. –  anon Aug 10 '11 at 3:37
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Ok, hah sorry for that, and thanks, a problem of notation –  Daniel Aug 10 '11 at 3:39
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