How can I express R in terms of N and B I have trouble with this vector calculus question:
If a curve $R(t)$ lies on a sphere $|R(t)| = constant$, prove that
$R = -\rho N - \frac{1}{\tau} \frac{d\rho}{dx} B$
Hint: Keep differentiating $R \cdot R = constant$, using the Frenet formulas
My effort:
Since $\rho = \frac{1}{k}$
$-\rho N - \frac{1}{\tau} \frac{d\rho}{dx} B = -\frac{1}{k} N - \frac{1}{\tau} \frac{d}{dx} \frac{1}{k} B$
$= -\frac{1}{k} N - \frac{1}{\tau} (-\frac{1}{k^2})\frac{dk}{ds} B$
Since $\frac{dB}{ds} = -\tau N$, $N = -\frac{1}{\tau} \frac{dB}{ds}$
Then $-\rho N - \frac{1}{\tau} \frac{d\rho}{dx} B = -\frac{1}{k} (-\frac{1}{\tau} \frac{dB}{ds}) - \frac{1}{\tau} (-\frac{1}{k^2})\frac{dk}{ds} B$
$= \frac{1}{\tau} \frac{d}{ds}(\frac{1}{k}B)$
Then I don't know what to do. Where is the mistake I made? And how to I use the condition $|R(t)| = r$?
Thank you for your help!
 A: There is a problem with the way this question is stated, and with the (attempted) solution our OP Matt C gives in the body of the question itself.  The issue to which I refer is that the reciprocal of the torsion $\tau$, $\frac{1}{\tau}$, is invoked, although there seems to be no restriction to $\tau \ne 0$ imposed on the curve $R(t)$.  Indeed, it is easy to construct a spherical curve with zero torsion--simply take the intersection of the sphere with an appropriate plane--one whose distance from the center of the sphere is less than the sphere's radius--and the result will be a circle lying in said plane, hence its torsion vanishes.  Clearly the formulas suggested here break down for such a curve.  As a matter of interest, it is known that spherical curves of constant non-vanishing torsion do indeed exist; see Kazaras, Demetre; Sterling, Ivan, An explicit formula for spherical curves with constant torsion, Pac. J. Math. 259, No. 2, 361-372 (2012). ZBL1264.53003;  thus the formulas developed here are not in fact vacuous, there are plenty of concrete situations to which they apply.  Also significant is the fact that such a curve has nonzero curvature, as is seen below ca. (12)-(13); thus we need only assume $\tau \ne 0$ for the program suggested in the question to go through, as is shown in detail below.
If the curve $R(t)$ lies on the sphere of radius $r$ centered at $C$, we have
$(R(t) - C) \cdot (R(t) - C) = r^2, \tag 1$
which may be differentiated with respect to $t$ to yield
$R'(t) \cdot (R(t) - C) + (R(t) - C) \cdot R'(t) = 0, \tag 2$
or
$2R'(t) \cdot (R(t) - C) = 0, \tag 3$
whence
$R'(t) \cdot (R(t) - C) = 0; \tag 4$
if we further assume that $R(t)$ is parametrized by its arc-length $s$, we obtain
$\dot R(s) \cdot (R(s) - C) = 0, \tag 5$
and since
$\dot R(s) = T(s), \tag 6$
the unit tangent vector to $R(s)$, (5) becomes
$T(s) \cdot (R(s) - C) = 0. \tag 7$
The reader will observe that in (5) and (6) I have dropped the prime notation ${}'$ for derivatives in favor of the dot $\dot {}$; I use this notation only for arc-length parametrized quantities in order to distinguish their derivatives, which play such a special role in differential geometry, from derivatives with respect to other more general curve parameters.
We differentiate (7) with respect to $s$:
$\dot T(s) \cdot (R(s) - C) + T(s) \cdot \dot R(s) = 0, \tag 8$
or again using (6),
$\dot T(s) \cdot (R(s) - C) + T(s) \cdot T(s) = 0; \tag 9$
now since $T(s)$ is a unit vector,
$T(s) \cdot T(s) = 1, \tag{10}$
and in accord with the first Frenet-Serret equation,
$\dot T(s) = \kappa(s)N(s), \tag{11}$
(9) becomes
$\kappa(s)N(s) \cdot (R(s) - C) + 1 = 0. \tag{12}$
This equation occurs ubiquitously throughout the study of curves lying in spheres, so much so that it might be termed the Fundamental Equation of Spherical Curves.  (12) may be interpreted as giving the component of $R(s) - C$ along $N(s)$, for it may be written in the form
$N(s) \cdot (R(s) - C) = -\dfrac{1}{\kappa(s)}, \tag{13}$
where we observe that (12) prohibits $\kappa(s)$ from ever taking the value $0$, so the division in (13) is always well-defined.
For the sake of notational convenience, we drop the explicit functional dependence of $R(s)$, $T(s)$, $N(s)$, $\kappa(s)$ und so weiter on $s$, simply writing $R$, $T$, $N$, $\kappa$ and so forth; here is where the $\dot {}$ notation comes in handy.
Our goal is an expression for $R(s) - C$ in terms of $N(s)$ and $B(s)$; since $T$, $N$, and $B = T \times N$ form an orthornormal triad at each point of $R$, the vector $R - C$ may be orthonormally expanded in terms of this triad, thusly:
$R - C = ((R - C) \cdot T)T + ((R - C) \cdot N)N + ((R - C) \cdot B)B; \tag{14}$
in accord with (7), this reduces to
$R - C = ((R - C) \cdot N)N + ((R - C) \cdot B)B; \tag{15}$
and in accord with (13),
$R - C = -\dfrac{1}{\kappa(s)}N + ((R-C) \cdot B)B, \tag{16}$
leaving only $(R - C) \cdot B$ to be determined; to this end we differentiate (12) yet again with respect to $s$, and find
$\dot \kappa N \cdot (R - C) + \kappa \dot N \cdot (R - C) + \kappa N \cdot \dot R = 0, \tag{17}$
or, in light of (6),
$\dot \kappa N(s) \cdot (R(s) - C) + \kappa \dot N \cdot (R - C) + \kappa N \cdot T = 0; \tag{18}$
now
$N \cdot T = 0, \tag{19}$
whence
$\dot \kappa N(s) \cdot (R(s) - C) + \kappa \dot N \cdot (R - C) = 0, \tag{20}$
and now substitution of the Frenet-Serret relation
$\dot N = -\kappa T + \tau B \tag{21}$
yields
$\dot \kappa N \cdot (R - C) + \kappa (-\kappa T + \tau B) \cdot (R - C) = 0, \tag{22}$
or
$\dot \kappa N \cdot (R - C) - \kappa^2 T \cdot (R - C) + \kappa\tau B \cdot (R - C) = 0, \tag{23}$
and once again (7) gives us
$\dot \kappa N \cdot (R - C) + \kappa\tau B \cdot (R - C) = 0; \tag{24}$
by virtue of (13) this becomes
$-\dfrac{\dot \kappa}{\kappa(s)} + \kappa\tau B \cdot (R - C) = 0, \tag{24}$
whence
$B \cdot (R - C) = 0 = \dfrac{\dot \kappa}{\kappa^2 \tau} = -\dfrac{1}{\tau}\dfrac{d}{ds}(\kappa^{-1}); \tag{25}$
we apply this to (16) and find
$R - C = -\dfrac{1}{\kappa(s)}N - \dfrac{1}{\tau}\dfrac{d}{ds}(\kappa^{-1})B, \tag{26}$
the requisite result.
