Proving $\dfrac{dN}{ds}=-\kappa T+\tau B$ Currently revising for a differential geometry exam. The question I am working on is one of those types where the next part of the question follows from the last. I've gotten to the point where I have proven $T\cdot \dfrac{dN}{ds}=-\kappa$, and the next part is where I got stuck, which is to prove $\dfrac{dN}{ds}=-\kappa T+\tau B$. I looked at the mark scheme and it said "Follows from previous item, and $B=T\times N$". I simply don't see how it follows, though.  
 A: Since $B = T \times N$, differentiate with respect to $s$, obtaining $$\frac{dB}{ds} = -\tau N = \frac{dT}{ds} \times N + T \times \frac{dN}{ds} = T \times \frac{dN}{ds}$$ since $dT/ds= \kappa N$ and $N \times N = 0$. There are some conclusions to be drawn from this:


*

*$dN/ds$ is not in the direction of $T$, else this cross product would be null. 

*$dN/ds$ is not in the direction of $B$ either, else you'd have $- \tau N = \kappa B$.

*It is, however, in a linear combination of $T$ and $B$, or else its cross product couldn't be $N$.


It follows that we can state that $$\frac{dN}{ds} = a_1 T + a_2 B.$$ We need to determine $a_1$ and $a_2$. Since $$T \cdot \frac{dN}{ds} = -\kappa = T \cdot (a_1 T + a_2 B) = a_1$$ we have $a_1 = - \kappa$. I've used a little trick to find $a_2$. Notice that $$\frac{d}{ds}(B \cdot N) = \frac{dB}{ds} \cdot N + B \cdot \frac{dN}{ds},$$ therefore $$B \cdot \frac{dN}{ds} = \frac{d}{ds} (B \cdot N) - \frac{dB}{ds} \cdot N,$$ but $$B \cdot \frac{dN}{ds} = a_2$$ and $$B \cdot N = 0, \quad \frac{dB}{ds} \cdot N = (- \tau N) \cdot N = - \tau.$$ Therefore $a_2 = - (-\tau) = \tau$ and $$\frac{dN}{ds} = -\kappa T + \tau B.$$ 
A: We are given a curve in ${\mathbb R}^3$:
$$\gamma: \quad s\mapsto{\bf x}(s)\ ,$$
parametrized with respect to arc length $s$. Assume that $\ddot {\bf x}(s)\ne{\bf 0}$.
Along this curve the so-called Frenet frame, a moving orthonormal frame, is defined as follows: Begin with
$$T=T(s):=\dot{\bf x}(s)\ .$$
This is a unit vector for all $s$. Therefore $\dot T\cdot T=0$, which means that $\dot T=\ddot{\bf x}\ne{\bf 0} $ is orthogonal to $T$. It follows that there is a uniquely defined unit vector $N$, depending on $s$, and a positive function $s\mapsto\kappa(s)$, the curvature of $\gamma$, such that
$$\dot T=\kappa\>N\ .\tag{1}$$
This normal vector $N$ shall be the second vector of our frame. The third vector is the binormal $B:=T\times N$, which is automatically a unit vector. We now have our orthonormal frame $(T,N, B)$.
The so-called Frenet formulas are a system of coupled differential equations for $T$, $N$, and $B$. The first of these equations is $(1)$. The second is about $\dot N$. As $N$ is a unit vector for all $s$ its derivative is orthogonal to $N$ and therefore is a linear combination of $T$ and $B$:
$$\dot N=\alpha T +\tau B\ .$$
According to the rules of vector algebra one has
$$\alpha=T\cdot\dot N={d\over dt}(T\cdot N)-\dot T\cdot N=-\kappa\ ,$$
so that we definitively can write
$$\dot N=-\kappa T+\tau B\ ,\tag{2}$$
where the function $s\mapsto\tau(s)$ is called the torsion of the curve $\gamma$.
The third of the Frenet equations is obtained by differentiating the defining equation of $B$:
$$\dot B=\dot T\times N+T\times \dot N={\bf 0}+T\times(-\kappa T+\tau B)=\tau\ T\times B\ .$$
According to the rules of vector algebra we therefore obtain
$$\dot B=-\tau\> N\ .\tag{3}$$
Equations $(1)-(3)$ can be condensed to the matrix equation
$$\left[\matrix{\dot T\cr\dot N\cr\dot B\cr}\right]=\left[\matrix{0&\kappa&0\cr-\kappa&0&\tau\cr0&-\tau&0\cr}\right]\>\left[\matrix{T\cr N\cr B\cr}\right]\ .$$
A: $\frac{dN}{ds}$ is orthogonal to $N$ by differentiating the identity $N\cdot N=1$. So, $\frac{dN}{ds}$ is a combination of $T$ and $B$, i.e. $\frac{dN}{ds}=x\,T+y\,B$. Comparing the "known identity" $T\cdot \frac{dN}{ds}=-\kappa$ with
$$
T\cdot \frac{dN}{ds}=x\,(T\cdot T)+y\,(T\cdot B)=x
$$
($T$ is perpendicular to $B$ by the definition of $B$) you get $x=-\kappa$. 
I suppose that $\tau$ hasn't been defined yet and $\tau:=y$ is then defined by this equation, altghough it was not completely clear from your question. (If $\tau$ has already been defined by the equation $\frac{dB}{ds}=-\tau N$, then you derive $y=\tau$ by differentiating $B=T\times N$.)
A: You know that dNds must be orthogonal to N. Hence dN/ds=CT+DB for some C(s) and D(s). You have shown that C=−κ. How do you think you should go about finding D?
