Finding the curvature and torsion of a curve 

For this question:
I'm not able to find what the torsion is for the curve gamma. 
In my notes I'm given that first derivative of b, (b dot) =-tau multiplied by n.
so in the solution I don't understand why they are working out n dot when we should be working out b dot and so torsion tau=-b dot/n. So I'm not able to understand what they have done to work out torsion. Why are they working out tau=n dot multiplied by b when according to my notes tau should be equal to -b dot/n?
Any help would be much appreciated.
 A: If $f$ is any reasonable function, then
$$
\frac{d}{dt}\left( \int_a^t f(u)du   \right) = f(t)
$$
This is the "Fundamental Theorem of Calculus".
So, in your case
$$
\vec{\mathbf{t}} = \gamma'(t) = \frac{1}{\sqrt2}
\left(  \sin \pi t^2, 1, \cos \pi t^2, \right)
$$
I expect you can take it from there.
As the comment said, the angle bracket in the last line is a dot product of two vectors: $\langle \mathbf u, \mathbf v \rangle = u_xv_x + u_yv_y + u_zv_x$. 
There are several different formulas for torsion. One of them (the one you know, apparently) is 
$$
\tau = - \mathbf{N} \cdot \dot{\mathbf{B}}
$$
However, you also probably know one of the Serret-Frenet formulae, which says that
$$
\dot{\mathbf{N}} = \tau \mathbf{B} - \kappa \mathbf{T}
$$
If you take the dot product of this formula with $\mathbf{B}$, you get
$$
\dot{\mathbf{N}} \cdot \mathbf{B} = 
\tau (\mathbf{B} \cdot \mathbf{B}) - \kappa (\mathbf{T} \cdot \mathbf{B}) =
\tau (1) - \kappa (0) = \tau
$$
so it's also true that $\tau = \dot{\mathbf{N}} \cdot \mathbf{B}$, and that's the formula they used in the solution you showed.
