Let $c : (a,b) \rightarrow \mathbb{R}^3 $ be a space curve ( parameterized by arc length ) with curvature $k(s) \neq 0 \ \forall s \in (a,b) $. Consider the tube $X : (a,b) \times \mathbb{R} \rightarrow \mathbb{R}^3 $ with: $$X(s,v)=c(s)+r(N(s)\cos(v)+B(s)\sin(v)) $$ Remark: $N(s)$, $B(s)$ the normal, the binormal vector of $c(s)$. Moreover $r>0$.
Show that : $ X $ is regular if and only if $r< \frac{1}{k(s)} \forall s \in (a,b) $.
So first of all we know that $c$ is parameterized by arc length. So we can use frenet equations: $c'(s) = T(s) = k(s)N(s)$ , $N'(s) = -k(s)T(s) + \tau(s)B(s)$ and $B'(s) = -\tau(s)N(s)$. Remark: $\tau$ is torsion.
We want that X is regular ( if and only if $k(s) \neq 0 \ \forall s \in (a,b) $). So we have to show that $D_X(s,v)$ has full rank. If I'm right the first column of $D_X(s,v)$ is $c'(s) + r(N'(s)\cos(v)+B'(s)\sin(v))$ and the second one is $r(-N(s)\sin(v)+B(s)\cos(v))$. My idea is to show that these two vectors are linearly independent ( maybe with help of frenet?), which means that $D_X(s,v)$ has full rank. I hope that you can help me with this.
We are given the so-called tubular map
$X(s, v) = c(s) + r(N(s) \cos v + B(s) \sin v), \tag 1$
where $c(s) \in \Bbb R^3$ is parametized by its arc-length; as such, it is also a unit-speed curve, as is well-known; thus we have the unit tangent vector
$T(s) = c'(s), \tag 2$
and in accord with the Frenet-Serret view of things we also have the unit normal vector $N(s)$ and curvature scalar $\kappa(s)$ defined by
$T'(s) = \kappa(s) N(s), \tag 3$
and a third vector $B(s)$, the binormal of $c(s)$,
$B(s) = T(s) \times N(s); \tag 4$
the Frenet-Serret equations affirm that
$N'(s) = -\kappa(s) T(s) + \tau(s) B(s), \tag 4$
and
$B'(s) = -\tau(s) N(s), \tag 5$
where $\tau(s)$ is known as the torsion of $c(s)$. $N(s)$ and $B(s)$ are the vector fields along $c(s)$ used in the definition (1) of $X(s, v)$.
The tubular surface $X(s, v)$ has two distinguished tangent vector fields, each corresponding to one of the parameters $s$, $v$; they are
$X_s(s, v) = \dfrac{\partial X(s, v)}{\partial s}, \tag 6$
and $X_v(s, v) = \dfrac{\partial X(s, v)}{\partial v}; \tag 7$
these vector fields $X_s(s, v)$ and $X_v(s, v)$ are tangent to the curves in the surface $X(s, v)$ on which $v$ and $s$ are respectively constant.
The surface $X(s, v)$ is regular precisely when the two vector fields (6) and (7) are linearly independent; to investigate further what regularity means in terms of the vectors $T(s)$, $N(s)$, and $B(s)$, we may compute $X_s(s, v)$ and $X_v(s, v)$ directly from (1), using (2)-(5):
$X_s(s, v) = c'(s) + r(N'(s) \cos v + B'(s) \sin v)$ $= T(s) + r((-\kappa(s) T(s) + \tau(s) B(s)) \cos v - \tau(s) N(s) \sin v)$ $= (1 - r\kappa(s) \cos v)T(s) + r\tau(s) B(s) \cos v - r\tau(s) N(s) \sin v, \tag 8$
$X_v = -rN(s) \sin v + rB(s)\cos v; \tag 9$
comparing (8) and (9) we find the curious relationship
$X_s(s, v) = (1 - r \kappa(s) \cos v) T(s) + \tau(s) X_v(s, v); \tag{10}$
furthermore, we see that
$T(s) \cdot X_v(s, v) = 0, \tag{11}$
since
$T(s) \cdot N(s) = 0 = T(s) \cdot B(s); \tag{12}$
therefore $X_s(s, v)$ and $X_v(s, v)$ are linearly independent exactly when
$(1 - r\kappa(s) \cos v)T(s) \ne 0; \tag{13}$
or, since $T(s) \ne 0$,
$1 - r \kappa(s) \cos v \ne 0; \tag{14}$
or
$r \kappa(s) \cos v \ne 1. \tag{15}$
We note that $r \kappa(s) > 0$; now if for any value of $s$,
$r \kappa(s) \ge 1, \tag{16}$
then by selecting $v_0 \in \Bbb R$ such that
$\cos v_0 = \dfrac{1}{r \kappa(s)} \le 1, \tag{17}$
which is always possible since the range of the variable $v$ is $\Bbb R$, we have
$1 - r \kappa(s) \cos v_0 = 0, \tag{18}$
at which point $(s, v_0)$ $X_s(s, v)$ and $X_v(s, v)$ become linearly dependent and the surface $X(s, v)$ ceases to be regular. Therefore we must have
$0 < r\kappa(s) < 1, \tag{19}$
or
$r < \dfrac{1}{\kappa(s)}, \; \forall \; s \in (a, b). \tag{20}$