Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider a $C^m$ curve $\gamma$ in $\mathbb{R}^3$ (with $m\geq 3$). Locally, at $0$ for convenience, one can express the curve as

$$ \gamma(s) = \gamma(0)+sT+(s^2/2!)kN+(s^3/3!)(\dot{k}N-k^2T+k\tau B)+O(>3) $$ ($k,\tau$ and the vectors $T,N,B$ are evaluated in $0$). Taking the first order terms we get a tangent line, taking the second order terms we get a parabola

$$\gamma(0)+sT+\frac{s^2}{2!}kN $$

that gives a local approximation of the curve near $s=0$. If we gather the terms in order to collect all the coefficients in $T$ and $N$ we have the following approximation for $\gamma$:

$$\gamma(0)+s(1-\frac{s^2}{3!}k^2)\;T+\frac{s^2}{2}(k+\frac{s}{3}\dot{k})\;N. $$

Now... the point is that the draft notes I am reading report that the above represents a parabola; this obviously does not convince me. How do you prove that the last expression defines a parabola in the $TN$ plane?

share|improve this question
1  
That's not a parabola because there are terms of degree $3$ that don't cancel. I presume either your notes mean the second-order approximation represents a parabola or the notes are wrong. –  anon Nov 23 '11 at 18:32
    
I think this might be a parabola if you observe that the same coefficient behind $s^3$ (modulo a multiplicative constant) appear in front of $T$ and $N$. –  user17090 Nov 23 '11 at 20:05
    
I think the report says that according to $\gamma(0)+sT+\frac{s^2}{2!}kN$, it represents a parabola in $TN$ plane. –  Paul Nov 23 '11 at 21:29
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.