Is the derivative of the unit normal vector parallel to the tangent vector? I'm reading through a proof of torsion's definition:
$\frac{dB}{ds} = \frac{d}{ds}(T \times N) = \frac{dT}{ds} \times N +  T \times\frac{dN}{ds} $
By definition, $N = \frac{dT}{\kappa ds}$ so  it is parallel to $dT$ and the cross product is zero. My confusion starts here. It seems that we could also determine $N$ and  $\frac{dT}{ds}$ are parallel by the fact that $T$ is of constant magnitude, therefore its tangent vector is always orthogonal to it, which is again a scalar multiple of $N$. My issue is that by this logic it seems to me that the  second part of the expanded equation above  can be held to similar reasoning:
$T$ is of constant unit magnitude by definition, so $\frac{dT}{ds}$ is orthogonal to $T$ for all $t$. We can unitize $\frac{dT}{ds}$ to obtain $N$, so $N$ is also of constant unit magnitude, therefore the tangent vector of $N$, $\frac{dN}{ds}$ is orthogonal to $N$ for all $t$, meaning that it must be parallel to $T$. 
This by all other indicators appears to be incorrect; it would break the proof, but I don't understand where I am going wrong. 
 A: As Spock once said, you’re exhibiting two-dimensional thinking: $N\perp T$ and $N'\perp N$ doesn’t imply that $T$ and $N'$ are parallel. Consider the standard coordinate axes: the $y$-axis is orthogonal to the $x$-axis and the $z$-axis is orthogonal to the $y$-axis, but the $z$-axis certainly isn’t parallel to the $x$-axis.
A: Let's check the definitions: let $\gamma$ be a smooth curve parametrised by arc length. Then, using $'$ to denote $d/ds$,
$$ \gamma'(s) = T(s) , $$
and by the definition of arc length, $ \lVert T \rVert = 1 $, or $ T \cdot T = 1 $, for every $s$. Differentiating this gives
$$ 0 = T' \cdot T + T \cdot T' = 2 T \cdot T' , $$
and so indeed $T' \perp T$. Therefore, if $T' \neq 0$, we can define a new vector perpendicular to $T$, the normal vector $N$. We can force this vector to be a unit vector by introducing the factor $\kappa(s) = \lVert T'(s) \rVert $, which we call the curvature; this forces $\kappa \geq 0$, and then
$$ T'(s) = \kappa(s) N(s) . $$
We now have two perpendicular vectors, and a third to complete an orthonormal basis can be defined by
$$ B(s) = T(s) \times N(s) , $$
the binormal. The derivative of $B$ is
$$ B'(s) = T'(s) \times N(s) + T(s) \times N'(s) = \kappa(s) N(s) \times N(s) + T(s) \times N'(s) = 0 + T(s) \times N'(s) , $$
since $N \times N = 0$. This is not very helpful, but we can get more information by using that $B$ is a unit vector:
$$ B \cdot B = 1 \implies B \cdot B' = 0 . $$
Since also $T \times N' \perp T $, $B$ must be a multiple of $N$. This multiple is the torsion, $\tau(s)$, and we normally choose
$$ B'(s) = -\tau(s) N(s) . $$

This leaves $N'(s)$. It's worth taking a look at all $6$ possible scalar products of the orthogonal triad, to see what happens:
$$ T \cdot T = 1 \implies T \cdot T' = 0 \implies T' \perp T \\
N \cdot N = 1 \implies N \cdot N' = 0 \implies N' \perp N \\
B \cdot B = 1 \implies B \cdot B' = 0 \implies B' \perp B  $$
The second of these implies that $ N' = a T + b B $ for some $a$ and $b$. We have already used
$$ T \cdot B = 0 \implies 0 = T' \cdot B + T \cdot B' = \kappa N \cdot B + T \cdot B' = T \cdot B' \implies B' \perp T $$
to define the torsion, which leaves
$$ T \cdot N = 0 \implies 0 = T' \cdot N + T \cdot N' = \kappa + T \cdot N' \implies T \cdot N' = -\kappa \\
B \cdot N = 0 \implies 0 = B' \cdot N + B \cdot N' = -\tau + B \cdot N' \implies B \cdot N' = \tau $$
Now we can find $N'$: since $T,N,B$ is an orthonormal basis, 
$$N' = (N' \cdot T) T + (N' \cdot N) N + (N' \cdot B) B = -\kappa T + 0 + \tau B . $$

It's worth remembering what we've used here: unit parametrisation is required for $T$ to have unit magnitude. Being in $\mathbb{R}^3$ is used when we introduce the cross product for the binormal. Then we use that $T$ and $N$ are orthonormal, which forces $B$ to have unit magnitude, and then the rest of the orthonormality is used to determine the other derivatives. In particular, the definition of the torsion, combined with $N \cdot B = 0$, forces $N'$ to have a nonzero component in the direction of $B$ if the curvature and torsion do not vanish.
