Deducing a curve from torsion and curvature I would like an example of how to get a curve, $\alpha$, from torsion, $\tau(s)$, and curvature, $\kappa(s)$.
Lets say I have $\tau(s) = \kappa(s) = s^2$.
A curve with this torsion and curvature exists, but I want to find it.
I understand we have a system of ODEs in the frenet frame,
$T'(s) = \kappa(s)N(s) = s^2N(s)$
$N'(s) = -\kappa(s)T(s) + \tau(s)B(s) = -s^2T(s) + s^2B(s)$
$B'(S) = -\tau(s)N(s) = -s^2N(s)$
We can assume $T(0)=T_0, N(0)=N_0, B(0)=B_0$
What I don't understand is how we get our curve $\alpha$ with soley this information and nothing of what $T(s), N(s), B(s)$ are.
Maybe we do this,
Since $\alpha'(s) = T(s)$
$N'(s) = s^2T(s) + s^2B(s) <=> T(s) = \frac{1}{s^2}N'(s) - B(s)$
$\int T(s)ds = \int \frac{1}{s^2}N'(s) - B(s) ds$
$\alpha(s) = \frac{1}{s^2}N_0 - \int B(s)ds$
Help is appreciated. There are similar questions on SE, but none are very useful to me and people may benefit from an example.
 A: Fairly quickly, the Frenet-Serret formulae
$$
\frac{\mathrm{d}}{\mathrm{d}s}\begin{bmatrix}T\\N\\B\end{bmatrix}
=\begin{bmatrix}0&\kappa&0\\-\kappa&0&\tau\\0&-\tau&0\end{bmatrix}\begin{bmatrix}T\\N\\B\end{bmatrix}
$$
For general $\kappa,\tau$ this differential equation doesn't have "nice" solutions.  But in your case note that $\kappa=\tau$ means $T+B$ is constant, so we reduce to solving
$$
\begin{aligned}
\frac{\mathrm{d}}{\mathrm{d}s}\begin{bmatrix}\frac1{\sqrt2}(T-B)\\N\end{bmatrix}
&=\begin{bmatrix}0&(\kappa+\tau)/\sqrt2\\-(\kappa+\tau)/\sqrt2&0\end{bmatrix}\begin{bmatrix}\frac1{\sqrt2}(T-B)\\N\end{bmatrix}\\
&=\begin{bmatrix}0&\sqrt2s^2\\-\sqrt2s^2&0\end{bmatrix}\begin{bmatrix}\frac1{\sqrt2}(T-B)\\N\end{bmatrix}
\end{aligned}
$$
So
$$
\begin{aligned}
\frac1{\sqrt2}(T-B)&=\cos(\theta(s))\frac1{\sqrt2}(T_0-B_0)+\sin(\theta(s))N_0\\
N&=\cos(\theta(s))N_0-\sin(\theta(s))\frac1{\sqrt2}(T_0-B_0)\\
\end{aligned}
$$
with $\theta'(s)=\sqrt2 s^2$ and $\theta(0)=0$, i.e., $\theta(s)=\frac{\sqrt2}3s^3$.  So
$$\begin{aligned}
T(s)&=\frac{\frac1{\sqrt2}(T+B)+\frac1{\sqrt2}(T-B)}{\sqrt2}\\
&=\frac{\frac1{\sqrt2}(T_0+B_0)+\cos(\theta(s))\frac1{\sqrt2}(T_0-B_0)+\sin(\theta(s))N_0}{\sqrt2}\\
&=\frac{1+\cos\frac{\sqrt2}3s^3}2T_0+\frac1{\sqrt2}\sin\frac{\sqrt2}3s^3 N_0+\frac{1-\cos\frac{\sqrt2}3s^3}2B_0
\end{aligned}
$$
But then $\exp(\lambda z^3)$ doesn't have elementary antiderivatives (for nonzero $\lambda$), so unless you want to have [confluent hypergeometric function $_{1}F_{1}$] or suchlike you can only write
$$
\alpha(s)=
\alpha(0)+T_0\int_0^s\frac{1+\cos\frac{\sqrt2}3\sigma^3}2\,\mathrm{d}\sigma+N_0\int_0^s\frac1{\sqrt2}\sin\frac{\sqrt2}3\sigma^3\,\mathrm{d}\sigma+B_0\int_0^s\frac{1-\cos\frac{\sqrt2}3\sigma^3}2\,\mathrm{d}\sigma.
$$
