# Identities related normal unitary vector, normal tangent vector and curvature.

i need help with this homework, about vector calculus.

Given a soft parametrized curve $r(t)$, we got that $T(t)=\frac {r´(t)}{\|r´(t)\|}$, the curvature is $\kappa(t)=\|\frac {dT}{ds}\|$=$\frac {\|T´(t)\|}{\|r´(t)\|}$=$\frac {\|r´(t)\times r´´(t)\|}{\|r´(t)\|^3}$. If $T(t) \ne 0$, the unit normal vector is such that $N(t)=\frac {T´(t)}{\|T´(t)\|}$ and the binormal vector is $B(t)=T(t)\times N(t)$.

a) Show $\frac {dT}{ds}=\kappa(t)N$ where $s$ is the arc length function.

b) Show that $\frac {dB}{ds}$ is perpendicular to $B$ and $T$, deduce that this rate is equal to $-\tau(s)N$ where $\tau(s)$ is an scalar called curve torsion.

c) Show that for every plane curve $\tau(s) = 0$.

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I'll give some hints. a) It's in the equation that defines $N(t)$. Change $dT/dt$ to $dT/ds$ somehow. b) This is kind of weird because you are not given what $\tau(s)$ is. Maybe you only need to prove that $N$ is the only direction that is perpendicular to $B$ and $T$. So what do you know about $T \cdot N$? Use that and $B = T \times N$ (and the fact that you're in 3D). c) There are so many ways. One is to realize that $B$ is constant (almost) always using $B = T \times N$. You can also differentiate the equation $B = T \times N$. –  Tunococ Sep 7 '12 at 5:13
As I recall, the identity $A \cdot A=1$ differentiated gives $A' \cdot A+A\cdot A'=0$ or simply $A \cdot A'=0$ is important to the argument. (here the prime could be $t$ or $s$ differentiation). Also, what does "soft" mean? Does this mean the curve is non-stop and non-linear? The Frenet formulas break down in the case the curve is too pretty. –  James S. Cook Sep 7 '12 at 5:33