Question: I need some help on the following problem: 

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> Given that a curve $\mathbf r:I\to \Bbb R ^3$ has constant curvature $k(s)=k$, for all $s$, and constant torsion $\tau(s)=\tau$, for all $s$. Find the curve $\mathbf r$. 
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I only know that, according to the fundamental theorem, this curve exists and is unique. But, how practically find the parametric equation of the curve? 

Thanks.

Answer: 
Hint: set $r=\kappa/(\kappa^2+\tau^2)$, $h=\tau/(\kappa^2+\tau^2)$. Let $s$ be the arc-length function of $\mathbf r$ with $s(t\_0)=0$ and define $\phi(t)=s(t)/\sqrt{r^2+h^2}$. We want to construct an orthonormal basis ($a\_1$, $a\_2$, $a\_3$) and a point $p\_0$ such that 
$$\mathbf r(t)=p\_0+r\bigl(\cos(t)a\_1+\sin(t)a\_2\bigr)+hta\_3.$$

First show -- per differentiating twice -- that for $\tilde c=\mathbf r+rN$ there's a point $p\_0$ and a vector $v\neq0$ such that $\tilde c(t)=p\_0+s(t)\cdot v$. Now chose a suitable orthogonal basis $(a\_1,a\_2)$ for the orthogonal complement of $\boldsymbol Rv$ such that $\langle -N(t),a\_1\rangle=\cos(\phi(t))$ and $\langle -N(t),a\_2\rangle=\sin(\phi(t))$.
To find the arc-parameterization $\mathbf{r}=\mathbf{r}(s)$ you must resolve

$$\frac{\mathbf{r}'\cdot \mathbf{r}''\times \mathbf{r}'''}{\|\mathbf{r}''^2\|}=\tau,$$
$$\|\mathbf{r}'(s)\|=1,$$
$$\|\mathbf{r}''(s)\|=k,$$

for some injective differential map $\mathbf{r}:I\hookrightarrow{\Bbb{R}}^3$ over some open interval $I$.