Suppose f is a unit speed curve. Assuming a and b are constants, what can be said about the Frenet Serret apparatus for f? 
Suppose $f(s)$ is a unit speed curve.  We know that $({\bf T}, {\bf N}, {\bf B})$ is an orthonormal basis for $R^3$.  As such, we can write $f(s)$ in the form:
  $$f(s) = a(s) {\bf T} + b(s) {\bf N} + c(s) {\bf B}$$
  Assume $a(s)$ and $b(s)$ are nonzero constants. What can be said about $a$, $b$, $c$, $\kappa$, and $\tau$?

I've made the observation that $a(s)$ is the inner product of $f(s)$ with $\bf T$, 
which I will write $\langle f(s), {\bf T} \rangle$.
Similarly, $b(s) = \langle f(s), {\bf N} \rangle$ and $c(s) = \langle f(s), {\bf B} \rangle$.
I think that $b(s) = -1 / \kappa$. Am I on the right track?
 A: With
$f(s) = a T + b N + c(s) B, \tag 1$
and
$\Vert f(s) \Vert = 1, \tag 2$
it follows that $s$ is the arc-length along $f(s)$; thus
$T = \dot f(s) = a \dot T + b \dot N + \dot c(s) B + c(s) \dot B; \tag 3$
now introducing the Frenet-Serret equations
$\dot T = \kappa N, \tag 4$
$\dot N = -\kappa T + \tau B, \tag 5$
$\dot B = -\tau N, \tag 6$
and inserting them into (3) we find
$T = a(\kappa N) + b(-\kappa T + \tau B) + \dot c(s) B -c(s) \tau N$
$= -b\kappa T + (a\kappa - c(s) \tau)N + (b\tau + \dot c(s))B; \tag 7$
comparing the coefficients of the three orthonormal vectors $T$, $N$, and $B$ yields
$-b\kappa = 1, \tag 8$
$a\kappa - c(s) \tau = 0, \tag 9$
$b\tau + \dot c(s) = 0; \tag{10}$
we thus see, in accord with our OP Gary Zoldiek, that
$\kappa = -\dfrac{1}{b}, \; \text{a constant}, \tag{11}$
which, since we conventionally assume $\kappa >0$, implies that
$b < 0; \tag{12}$
from (9), (10) (11) we also have
$c(s) \tau = a\kappa = -\dfrac{a}{b}, \tag{13}$
and
$\dot c(s) = -b\tau; \tag{14}$
now since
$a \ne 0 \ne b, \tag{15}$
(13) implies that
$c(s) \ne 0 \ne \tau, \tag{16}$
whence we may write
$\tau = -\dfrac{a}{bc(s)}, \tag{17}$
hence from (14),
$\dot c(s) = -b\tau = \dfrac{a}{c(s)}; \tag{18}$
thus,
$\dfrac{d}{ds} (\dfrac{1}{2} c^2(s)) = c(s) \dot c(s) = a, \tag{19}$
or
$\dfrac{d}{ds} (c^2(s)) = c(s) \dot c(s) = 2a; \tag{20}$
we integrate 'twixt $s_0$ and $s$:
$c^2(s) - c^2(s_0) = 2a(s - s_0); \tag{21}$
$c^2(s) =  c^2(s_0) - 2a(s - s_0), \tag{22}$
$c(s) = \pm \sqrt{c^2(s_0) - 2a(s - s_0)}, \tag{23}$
as long as the signs allow; it then follows from (17) that 
$\tau(s) = \mp \dfrac{a}{b \sqrt{c^2(s_0) - 2a(s - s_0)}}. \tag{24}$
So, what have we learned about $a$, $b$, $c$, $\kappa$ and $\tau$?
Well, $b < 0$, $c(s)$ is given by (23), $\kappa = -1/b$ is constant, and $\tau$ is as in (24); apparently $a$ and $c(s_0)$ are the two free parameters upon which all else swings.  We should also observe that $c(s)$ and $\tau(s)$ will become singular when $s$ passes through
$s = s_0 + \dfrac{c^2(s_0)}{2a} \tag{25}$
where $c(s) = 0$; depending on the sign of $a$, the curve $f(s)$ may be extended arbitrarily far in the direction of either increasing or decreasing $s$. 
I think there are many more intriguing facts about $f(s)$ which will be revealed by further scrutiny of these results, but for now I haven't the time to say more.
A: Hint Differentiating the relation
$$f = a {\bf T} + b {\bf N} + c {\bf B}$$
and using the Frenet-Serret Identities gives
\begin{align}
{\bf T}
&= a {\bf T}' + b {\bf N}' + c' {\bf B} + c {\bf B}' \\
&= a (\kappa {\bf N}) + b (-\kappa {\bf T} + \tau {\bf B}) + c' {\bf B} + c (-\tau {\bf N}) \\
&= -b \kappa {\bf T} + (a \kappa - c \tau) {\bf N} + (b \tau + c') {\bf B} .
\end{align}
Comparing like coefficients gives the system
\begin{align}
1 &= - b \kappa \\
0 &= a \kappa - c \tau \\
0 &= b \tau + c' .
\end{align}
The first equation gives the suspected (constant) value $\kappa = -1 / b$. After substituting we can solve the remaining two equations for $\tau$ and (by integrating), $c$.
