Torsion Intuition Here is the formula that my instructor gave me to solve for torsion in Calc3:
$\tau = \frac{-d\vec{B}}{dS} \cdot{\vec{N}}$
However, I'm having some trouble understanding the intuition behind this formula. I understand that Torsion measures the twist of a function through space, kinda like the normal measures the turn of a function. For example, if we have a spring, then the rise of the spring would represent the torsion. However, I do not understand where the $\cdot{\vec{N}}$ part comes from. Physically, a dot product represents the magnitude of the component of a vector that lies on another vector. But in this case, why do we care about the component of the derivative of the $\vec{B}$ that's on the Normal vector?
 A: In fact this is the whole numerical value of the derivative of $\vec{B}$: since $ \vec{B} \cdot \vec{B} = 1 $, $ \vec{B} \cdot \vec{T} = 0 $ and $ \vec{B} \cdot \vec{N} = 0 $ (by definition: $\vec{B}$ is a unit vector perpendicular to the tangent and normal), we can differentiate to find that
$$ 0 = \vec{B} \cdot \frac{d\vec{B}}{ds} , $$
and
$$ 0 = \vec{B} \cdot \frac{d\vec{T}}{ds} + \frac{d\vec{B}}{ds} \cdot \vec{T} = \kappa\vec{B} \cdot \vec{N} + \frac{d\vec{B}}{ds} \cdot \vec{T} = \frac{d\vec{B}}{ds} \cdot \vec{T} , $$
so we have found that $\frac{d\vec{B}}{ds} \parallel \vec{N} $. In particular 
$$ \frac{d\vec{B}}{ds}  = \bigg(\vec{N} \cdot \frac{d\vec{B}}{ds} \bigg) \vec{N} : $$
dotting with $\vec{N}$ loses no information from the whole vector. Why do this rather than simply examine $\lVert d\vec{B}/ds \rVert$? Because the sign of $\vec{B}$ tells you which way the curve is bending out of the plane (whether your spring is wound clockwise or anticlockwise, if you like). 
(The $-$ sign is purely convention, by the way, and some people use a $+$ instead.)
A: Here's the best I can do on short notice to explain what is up with
$\tau = -\dfrac{d\vec B}{ds} \cdot \vec N. \tag 1$
First off, how does the torsion $\tau$ arise in the first place?  Usually it is introduced when we look at $d \vec N/ds$, where $\vec N$ and $\kappa$ have been defied via the first Frenet-Serret equation
$\dot {\vec T} = \dfrac{d \vec T}{ds} = \kappa \vec N; \tag 2$
here we conventionally assume that
$\kappa > 0, \; \Vert \vec N \Vert = 1 \tag 3$
so that
$\kappa = \kappa \Vert \vec N \Vert = \Vert \kappa \vec N \Vert = \Vert \dot T \Vert; \tag 4$
thus $\kappa$ is the magnitude of $\dot {\vec T}$; $\vec N$ is its direction.
The next thing we take a look at is
$\dot{\vec N} = \dfrac{d \vec N}{ds}; \tag 5$
we begin by observing that since
$\vec T \cdot \vec T = 1, \tag 6$
we have upon differentiation
$2\vec T \cdot \dot{\vec T} = 0 \Longrightarrow \vec T \cdot \dot{\vec T} = 0, \tag 7$
in light of (2), (3) this yields
$\kappa \vec N \cdot \vec T = \dot{\vec T} \cdot \vec T = 0 \Longrightarrow \vec N \cdot \vec T = 0, \tag 8$
that is, $\vec N$ is normal to $\vec T$; furthermore, differentiating again we find
$\dot{\vec N} \cdot \vec T + \vec N \cdot \dot {\vec T} = \dfrac{d{(\vec T \cdot \vec N})}{ds} = 0 \Longrightarrow \dot{\vec N} \cdot \vec T = -\vec N \cdot \dot{\vec T}, \tag 9$
and using (2) and (3),
$\vec N \cdot \dot{\vec T} = \vec N \cdot \kappa \vec N = \kappa \vec N \cdot \vec N = \kappa, \tag{10}$
whence
$\dot{\vec N} \cdot \vec T = \vec N \cdot \dot{\vec T} = -\kappa, \tag{11}$
the component of $\dot{\vec N}$ along $\vec T$; and sicnce
$\vec N \cdot \vec N = 1, \tag{12}$
We thus have that 
exactly the same argument as in (6)-(7) yields
$\dot{\vec N} \cdot \vec N = 0;, \tag{13}$
i.e. the $\vec N$-component of $\dot{\vec N}$ vanishes.  
We have thus developed the components of $\dot{\vec N}$ in the two-dimensional subspace spanned by $\vec T$ and $\vec N$; but we are working in $\Bbb R^3$ and so there is one more component of $\dot{\vec N}$ to be found; this is of course conventionally done by introducing the vector
$\vec B = \vec T \times \vec N; \tag{14}$
from this definition it immediately follows that
$\vec B \cdot \vec T = \vec B \cdot \vec N = 0, \; \Vert \vec B \Vert = 1, \tag{15}$
so that $\vec T$, $\vec N$, $\vec B$ form an orthonormal frame in $\Bbb R^3$; we may thus complete the expression for $\dot{\vec N}$ by supplying its component along $\vec B$; this is what we call the torsion $\tau$:
$\tau = \dot{\vec N} \cdot \vec B; \tag{16}$
at this point we may write
$\dot{\vec N} = -\kappa \vec T + \tau \vec B, \tag{17}$
the so-called second Frenet-Serret equation.  At his point it is easy to derive (1); we simply mimic the steps taken in (9)-(11):
$\dot{\vec B} \cdot \vec N + \vec B \cdot \dot{\vec N} = \dfrac{d(\vec B \cdot \vec N)}{ds} = 0 \Longrightarrow \dot{\vec B} \cdot \vec N = -\vec B \cdot \dot{\vec N} = -\tau, \tag{18}$
and so
$\tau = -\dot{\vec B} \cdot \vec N, \tag{19}$
which is (1).
We are now in a position from which our OP Dude 165's concerns may be addressed; we have set a context, so beloved here on math.stachexchane.com.  Thus:  the intuition behind formula (1) and (19), once we have defined $\tau$ via (16)-(17), is really more about orthogonal vectors in general than it is about $\vec T$, $\vec N$, and $\vec B$; for any two orthogonal vectors $\vec X(s)$ and $\vec Y(s)$,
$\vec X \cdot \vec Y = 0, \tag{20}$
we have
$\dot{\vec X} \cdot \vec Y + \vec X \cdot \dot{\vec Y} = \dfrac{d(\vec X \cdot \vec Y)}{ds} = 0 \Longrightarrow \dot{\vec X} \cdot \vec Y = -\vec X \cdot \dot{\vec Y}, \tag{21}$
which was applied more than once in the above, cf. (7)-(11) and (16)-(19) und so weiter.
"But in this case, why do we care about the component of the derivative of the $\vec{B}$ that's on the Normal vector?"  The best response I can give to this question is that it is essentially the same (up to a sign) as the component of $\dot{\vec N}$ along the $\vec B$ vector, $\pm \tau$, which we had to introduce in order to complete the description (17) if $\dot{\vec N}$.  As we have seen, $\dot{\vec B} \cdot \vec N$ is essentially the same thing.
