Curves in $\mathbb{R}^3$ with given properties Question: Find two curves $α$ and $β$ such that $α(0) = β(0) = (1, 0, 0)$, both with
the same Frenet frame at $t = 0$ given by $T_α(0) = T_β(0) = (0, 1, 0)$, $N_α(0) = N_β(0) =
(−1, 0, 0)$, $B_α(0) = B_β(0) = (0, 0, 1)$, and moreover, $κ_α(0) = κ_β(0)$, but $τ_α(0) =
−τ_β(0) \neq 0$.
Okay, so I have found two curves given by $\alpha (t)=(1-\frac{t^2}{2}, t, \frac{t^3}{3})$ and $\beta (t)=(1-\frac{t^2}{2}, t, -\frac{t^3}{3})$. This seems to work. However, I am required to furnish unit-speed curves and it is difficult to give unit-speed parametrizations for my curves. Their speed is $\sqrt{t^4+t^2+1}$, which is creating difficulty.
I was wondering if we could use helices, but I'm not exactly able to understand how (the standard helix doesn't work). Please give me hints using a helix or otherwise, to get unit-speed curves for this problem.
 A: I think I figured out a solution while having a discussion in the comments with Ted Shifrin. Thanks, Ted!
Let us consider the curves $\alpha (t)=(3\cos(\frac{t}{5}), 3\sin(\frac{t}{5}), \frac{4}{5}t)$ and $\beta (t)=(3\cos(\frac{t}{5}), 3\sin(\frac{t}{5}), -\frac{4}{5}t)$. Some data can be computed for these two curves:
$\alpha(0)=(3,0,0), T_\alpha(0)=(0,\frac{3}{5},\frac{4}{5}), N_\alpha(0)=(-1,0,0), B_\alpha(0)=(0,-\frac{4}{5},\frac{3}{5}), \kappa_\alpha=\frac{3}{25}, \tau_\alpha=\frac{4}{25}$.
Similarly,
$\beta(0)=(3,0,0), T_\beta(0)=(0,\frac{3}{5},-\frac{4}{5}), N_\beta(0)=(-1,0,0), B_\beta(0)=(0,\frac{4}{5},\frac{3}{5}), \kappa_\beta=\frac{3}{25}, \tau_\beta=-\frac{4}{25}$.
So, we have something ''like" what we want, but we have a little more work to do. We know that applying an isometry of $\mathbb{R}^3$ (with proper handedness concerns) to a regular curve gives a congruent curve, i.e, it preserves the curvature $\kappa$ and torsion $\tau$. I'll do it for $\alpha$ and it can be done similarly for $\beta$.
Let us call the isometry as $F$ such that $\overline\alpha=F\circ\alpha$ is the curve with the required properties, i.e, $\overline\alpha(0)=(1,0,0), T_\overline\alpha(0)=(0,1,0), N_\overline\alpha(0)=(-1,0,0), B_\overline\alpha(0)=(0,0,1), \kappa_\overline\alpha=\frac{3}{25}, \tau_\overline\alpha=\frac{4}{25}$. Now, as any isometry of $\mathbb{R}^3$ is composed of a translation and an orthogonal transformation, we can write $F(x)=Ax+a$ for all $x\in\mathbb{R}^3$ where $A$ is an orthogonal matrix and $a\in\mathbb{R}^3$ is a constant. It can also be shown that the transformation $A$ is the one which satisfies $A.T_\alpha(0)=T_\overline\alpha(0), A.N_\alpha(0)=N_\overline\alpha(0), A.B_\alpha(0)=B_\overline\alpha(0)$ and $a=\overline\alpha(0)-A.\alpha(0)$.
Doing these computations, we find that $A=\begin{pmatrix}1 & 0 & 0 \\ 0 & \frac{3}{5} & \frac{4}{5} \\ 0 & -\frac{4}{5} & \frac{3}{5} \end{pmatrix}$ and $a=(-2,0,0)$.
Therefore,
$\overline\alpha(t)=(3\cos(\frac{t}{5})-2, \frac{9}{5}\sin(\frac{t}{5})+\frac{16}{25}t, -\frac{12}{5}\sin(\frac{t}{5})+\frac{12}{25}t)$.
Proceeding in a similar fashion as above, we can find $\overline\beta$. It turns out that,
$\overline\beta(t)=(3\cos(\frac{t}{5})-2, \frac{9}{5}\sin(\frac{t}{5})+\frac{16}{25}t, \frac{12}{5}\sin(\frac{t}{5})-\frac{12}{25}t)$.
It can be verified that $\overline\alpha$ and $\overline\beta$ satisfy all the required properties and hence, give a solution to the problem.
