determine the principal curvatures of the surface defined as the tube around a space curve using the Frenet Serret frame. Consider a regular unit speed curve $\alpha: (a,b) \to \Bbb R^3$. Then define the surface $S$ via the parametrization $x:(a,b)\times (-\pi,\pi)\to \Bbb R^3$ where
$$x(u,v) = \alpha(t) + r(N(t)\cos v+B(t)\sin v)$$
where $N$ and $B$ are the unit normal and binormal to the curve alpha respectively. This defines a tube around the curve $\alpha$. We demand that $k(t) \leq \frac1r$ as is required for the surface to be regular. Now I want to find the principal curvatures. To this end I want to find the eigenvalues of the shape operator $-dN$ where $N : S \to \Bbb S^2$ is the Gauss map. Now I tried to compute the coefficients of the first and second fundamental form but this just turns into a nightmare. I found the following expressions that I think should be useful
$$x_s = T(t) (1+rk\cos v) +r\tau (B(t)\cos v-N(t) \sin v)    ,  $$
$$x_v = r(B(t)\cos v - N(t)\sin v)$$
where $T(t) = \alpha'(t)$ is the unit tangent vector to the curve $\alpha$.
I am pretty sure the principal curvatures are just the curvature of the circle at each points, which is $\frac{1}{r}$ and the curvature of $\alpha$ which is $k(t)$. My question is, how do I determine the linear map $dN$? I am pretty sure I am meant to be using the Frenet Serret frame to locally describe the tangent plane $T_pS$ to the surface. Then any vector in this space is a linear combination of $x_s$ and $x_v$. Then... yeah I am not sure. Any help would be much appreciated.
 A: Note that the normal vector is nothing but $n(u,v)=N(u)\cos{v}+B(u)\sin{v}$. To see that write $x(u,v)=\alpha(u)+r\cdot n(u,v)$ then the tangent vectors are:
    $$x_u=T+r\cdot n_u, x_v=r\cdot n_v$$
Notice that $n\cdot n=1$ and $n\cdot T=0$ thus $n\cdot n_u=n\cdot n_v=0$ and $n\cdot x_u=n\cdot x_v=0$, which tells that $n$ is the unit normal vector, giving the Gauss map.  
It is easy to compute the following using the Frenet formula:
$$n_u=\cos{v}N_u+\sin{v}B_u=-k\cos{v}T+\tau\cdot n_v$$
$$x_u=T+r\cdot n_u=(1-rk\cos{v})T+r\tau\cdot n_v$$
Also it is easy to verify that {$T, n_v$} is an orthonormal basis of the tangent plane.
Hence
$$dn(T)=n_udu(T)+n_vdv(T)$$
$$=\frac{1}{1-rk\cos{v}}n_u-\frac{\tau}{1-rk\cos{v}}n_v$$
$$dn(n_v)=n_udu(n_v)+n_vdv(n_v)=\frac{n_v}{r}$$
So $$dn(\begin{bmatrix}T \\n_v \end{bmatrix})=\begin{bmatrix}\frac{1}{1-rk\cos{v}} & -\frac{\tau}{1-rk\cos{v}} \\0 & \frac{1}{r} \end{bmatrix} \begin{bmatrix}n_u \\n_v \end{bmatrix}$$
$$=\begin{bmatrix}\frac{1}{1-rk\cos{v}} & -\frac{\tau}{1-rk\cos{v}} \\0 & \frac{1}{r} \end{bmatrix} \begin{bmatrix}-k\cos{v} & \tau \\0 & 1 \end{bmatrix} \begin{bmatrix}T \\n_v \end{bmatrix}$$
$$=\begin{bmatrix}-\frac{k\cos{v}}{1-rk\cos{v}} & 0 \\0 & \frac{1}{r} \end{bmatrix}\begin{bmatrix}T \\n_v \end{bmatrix}$$
Thus the Jacobian matrix of the Gauss map, with respect to the orthonormal basis, is given by
$$\begin{bmatrix}-\frac{k\cos{v}}{1-rk\cos{v}} & 0 \\0 & \frac{1}{r} \end{bmatrix}$$ and the two principle curvatures are the diagonal elements.
