Calculating the normal vector of a surface. 
Let $\alpha: I\rightarrow \mathbb{R}^3$ be a parametrized curve with non-zero curvature every where and parametrized by arc length. Let 
  $$x(s,v)=\alpha(s)+ r(n(s)\cos v + b(s)\sin v), r\not =0, s\in I$$ 
  a parametrized surface, where $n$ is the normal vector of $\alpha$ and $b$ is the vector binormal of $\alpha$. Show that when $x$ is regular then the normal vector of the surface is 
  $$N(s,v)=-(n(s) \cos v+b(s)\sin v) $$

I arrive anywhere trying to calculate
$$x_s\times x_v$$
I think this is the correct way, but I don't know how to get the expression above.
I have seen a "prove" that check the properties of the normal vector on $N(s,v)$, but I'm not sure that it is well done, because of the direction. 
Thanks!
 A: Starting from your work; let's write $t(s)$ for $\alpha'(s)$, the unit tangent vector to $\alpha$ at parameter $s$. And I'm going to use bold capitals for the three Frenet vectors:
$$
\newcommand{\TT} {{\mathbf{T}}}
\newcommand{\NN} {{\mathbf{N}}}
\newcommand{\BB} {{\mathbf{B}}}
\newcommand{\cv} {\cos(v)}
\newcommand{\sv} {\sin(v)}
\newcommand{\al} {\alpha}
\newcommand{\ka} {\kappa}
\newcommand{\ta} {\tau}
\TT, \NN, \BB
$$
and leave off the $(s)$ in general, writing $\alpha$ instead of $\alpha(s)$, etc. That'll make everything more compact. Finally, I'm going to set $r = 1$ so that it disappears from the equations; you can re-insert it yourself once you understand the general computation. 
Thus your starting point, 
\begin{align}
x_s&=\al'(s)+r(n'(s)\cos v+b'(s)\sin v) \\
x_v &=r(-n(s)\sin v+b(s)cos v)
\end{align}
becomes
\begin{align}
x_s&=\al'+\cv\NN'+\sv\BB' \\
x_v &=-\sv\NN+\cv \BB
\end{align}
And since $\al' = \TT$, we have
\begin{align}
x_s\times x_v&=(\TT+\cv\NN'+\sv\BB')\times (-\sv\NN+\cv \BB)
\end{align}
At this point in your work, $\alpha'(s) = \TT(s)$ seems to have gotten lost; I'll correct your work continuing from here:
\begin{align}
x_s\times x_v
&=(\TT+\cv\NN'+\sv\BB')\times (-\sv\NN+\cv \BB)\\
&=\left(\TT \times (-\sv\NN+\cv \BB)\right)+\cv\left(\NN'\times (-\sv\NN+\cv \BB)\right)+\sv\left(\BB'\times (-\sv\NN+\cv \BB)\right)\\
&=\left(-\sv\BB-\cv \NN)\right)+\cv\left(\NN'\times (-\sv\NN+\cv \BB)\right)+\sv\left(\BB'\times (-\sv\NN+\cv \BB)\right),
\end{align}
where I've used the fact that $\TT \times \NN = \BB$ to simplify the first term. Now I'll simplify by using the Frent-Serret formulas to write $\NN' = -\ka \TT + \ta \BB$ and $\BB' = -\ta \NN$:
\begin{align}
x_s\times x_v
&=\left(-\sv\BB-\cv \NN)\right)+\cv\left(\NN'\times (-\sv\NN+\cv \BB)\right)+\sv\left(\BB'\times (-\sv\NN+\cv \BB)\right)\\
&=\left(-\sv\BB-\cv \NN)\right)+\cv\left( (-\ka \TT + \ta \BB) \times (-\sv\NN+\cv \BB)\right)+\sv\left((-\ta \NN)\times (-\sv\NN+\cv \BB)\right)
\end{align}
The first term is almost exactly the answer we want. Let's simplify the 2nd and 3rd terms one at a time: 
\begin{align}
\text{2nd term}
&=\cv\left( (-\ka \TT + \ta \BB) \times (-\sv\NN+\cv \BB)\right)\\
&=\cv\left( (-\ka \TT \times (-\sv\NN+\cv \BB)+ \ta \BB \times (-\sv\NN+\cv \BB)\right)\\
&=\cv\left( (\ka \sv\BB + \ka\cv \NN)+ \ta \sv\TT\right)\\
\end{align}
and
\begin{align}
\text{3rd term}
&=\sv\left((-\ta \NN)\times (-\sv\NN+\cv \BB)\right)\\
&=-\ta\sv\cv \TT\\
\end{align}
The third term exactly cancels the last bit of the second term, so 
\begin{align}
x_s\times x_v
&=\left(-\sv\BB-\cv \NN)\right)+\cv( (\ka \sv\BB + \ka\cv \NN))+ \ta \sv\TT)  
-\ta\sv\cv \TT \\
&=(-\sv\BB-\cv \NN))+\cv( \ka \sv\BB + \ka\cv \NN) \\
&=(-\sv\BB-\cv \NN))+\ka\cv( \sv\BB + \cv \NN)  \\
&=(1-\ka) (-\sv\BB-\cv \NN))
\end{align}
This final vector is just $(1-\ka)$ times a unit vector, so normalized, we get that the surface normal is 
\begin{align}
-\sv\BB-\cv \NN.
\end{align}
When you redo this computation including $r$, that final factor will be $(1 - r\ka)$, I believe, showing that if $r$ is less than the radius of curvature, everything works fine. If not, then there's a sign change ... but in that case ($r$ greater than radius of curvature), the offset surface is singular, so we're not in the "regular surface" case any more. 
