Calculate the metric tensor of a tube

Question

We define the metric tensor as:

\begin{bmatrix}X_t\cdot X_t & X_t\cdot X_u &\\X_t\cdot X_u & X_u\cdot X_u \\\end{bmatrix}

So given $X(t,u)=\gamma(t)+(r\cos(u))N(t)+(r\sin(u))B(t)$, where $N(t)$ and $B(t)$ are the normal and binormal vectors of $\gamma(t)$ and $r\in\mathbb{R}$, how do we compute the metric tensor above? I think I'm getting confused on the dot product part or using the Frenet equations. Whenever I try to calculate this out, it's not what the website I'm using gets...

Answer 1

If $\gamma$ has unit speed, $$ X_t=T(1-kr\cos\ u) +\tau r\sin\ u N-\tau r\cos\ uB $$

$$ X_u=-r\sin\ u N +r\cos\ u B$$ so that $$ g_{tt}=(1-kr\cos\ u)^2+ \tau^2 r^2 $$

$$ g_{uu} = r^2,\ g_{tu}=-\tau r^2 $$ where $g$ is a metric