Curve where torsion and curvature equal arc length I study differential geometry independently in my free time as an undergraduate. I am using the book by Do Carmo.
I recently read the section and local theory of curves and learned about torsion and curvature.
My question is, does there exist a curve that has both torsion and curvature equal to arc length? I have tried deriving such a curve, but I’ve failed.
I speculate that it must be somewhat helical in nature.
Standard equation of helix is given by $\alpha(s)=(a\cos(s/c),a\sin(s/c),b)$. The curvature of such a curve is $\kappa(s)=\frac{a}{a^2+b^2}$ and torsion is $\tau(s)=\frac{b}{a^2+b^2}$.
Clearly $a=\frac{1}{2}s^{-1}=b$.
I’m not sure if this is the right approach to take. I feel as though the curve’s normal ought to trace out a curve on a sphere, but it doesn’t.
Any help is appreciated. 
 A: In fact, given any functions $\kappa, \tau : (a, b) \to \Bbb R$ satisfying $\kappa(s) > 0$ for all $s \in (a, b)$,

*

*there is a curve $\gamma(s)$ parameterized by arc length whose curvature is $\kappa(s)$ and whose torsion is $\tau(s)$, and

*any two such curves $\gamma_1, \gamma_2$ are unique up to rigit motions of $\Bbb R^3$, that is, there is a rigid motion $A$ of $\Bbb R^3$ such that $\gamma_2 = A \circ \gamma_1$.

This appears in $\S$ 1.5 of do Carmo's text, where it's called the Fundamental Theorem of the Local Theory of Curves; see also the appendix to $\S$ 4.
In Clelland's excellent From Frenet to Cartan: The Method of Moving Frames, this is Corollary 4.15, where it's presented as a motivating special case of more general result that applies far beyond Euclidean geometry.
Setting the curvature and torsion of a curve $\gamma$ to prescribed functions $\kappa, \tau$ results in a nonlinear, third-order system in three functions (the components of $\gamma$), so for general $\kappa, \tau$ one shouldn't expect to find explicit, closed-form solutions $\gamma$.
On the other hand, the conditions $\kappa(s) = \tau(s) = s$ are tractable enough to find an explicit solution. Substituting in the usual Frenet equations in matrix form gives
\begin{align*}
\pmatrix{{\bf T}'(s)&{\bf N}'(s)&{\bf B}'(s)}
&= \pmatrix{{\bf T}(s)&{\bf N}(s)&{\bf B}(s)} \pmatrix{\cdot&-\kappa(s)&\cdot\\\kappa(s)&\cdot&-\tau(s)\\\cdot&\tau(s)&\cdot} \\
&= \pmatrix{{\bf T}(s)&{\bf N}(s)&{\bf B}(s)} \cdot s\pmatrix{\cdot&-1&\cdot\\1&\cdot&-1\\\cdot&1&\cdot} .
\end{align*}
Rearranging gives
$$\pmatrix{{\bf T}(s)&{\bf N}(s)&{\bf B}(s)}^{-1} \frac{d}{ds}\pmatrix{{\bf T}(s)&{\bf N}(s)&{\bf B}(s)} = s\pmatrix{\cdot&-1&\cdot\\1&\cdot&-1\\\cdot&1&\cdot},$$
and solving formally yields
$$\pmatrix{{\bf T}(s)&{\bf N}(s)&{\bf B}(s)} = \pmatrix{{\bf T}(0)&{\bf N}(0)&{\bf B}(0)} \exp \left[\frac{1}{2} s^2\pmatrix{\cdot&-1&\cdot\\1&\cdot&-1\\\cdot&1&\cdot}\right] .$$
Since all solutions are the same up to rigid motions, we may as well take $\pmatrix{{\bf T}(0)&{\bf N}(0)&{\bf B}(0)}$ to be any (special orthogonal) matrix we like, and it turns out to be convenient to take (cf. J.M.'s comment)
$$\pmatrix{{\bf T}(0)&{\bf N}(0)&{\bf B}(0)}
= \pmatrix{
\frac{1}{\sqrt{2}}&\cdot&-\frac{1}{\sqrt{2}}\\
\cdot&1&\cdot\\
\frac{1}{\sqrt{2}}&\cdot&\frac{1}{\sqrt{2}}.
}$$
We can also compute the matrix exponential explicitly, and putting this all together gives
$$\pmatrix{{\bf T}(s)&{\bf N}(s)&{\bf B}(s)} =
\pmatrix{
\frac{1}{\sqrt{2}} \cos \frac{1}{\sqrt{2}} s^2&\ast&\ast\\
\frac{1}{\sqrt{2}} \sin \frac{1}{\sqrt{2}} s^2&\ast&\ast\\
\frac{1}{\sqrt{2}}                            &\ast&\ast
} .$$
For a curve parameterized by arc length, ${\bf T}(s) = \gamma'(s)$, so we can recover an explicit formula for a solution $\gamma(s)$ by integrating ${\bf T}(s)$. Taking the initial condition $\gamma(0) = (0, 0, 0)$ yields the solution
$$\color{#df0000}{\boxed{\gamma(s) =
\pmatrix{
\frac{1}{\sqrt{2}} \int_0^s \cos \frac{1}{\sqrt{2}} \tau^2 d\tau\\
\frac{1}{\sqrt{2}} \int_0^s \sin \frac{1}{\sqrt{2}} \tau^2 d\tau\\
\frac{1}{\sqrt{2}} s \\
}}}.$$
Optionally, we can rewrite $\gamma$ in terms of the Fresnel integrals, $C(x) := \int_0^x \cos t^2 \,dt$ and $S(x) := \int_0^x \sin t^2 \,dt$, as
$$
\gamma(t) =
\pmatrix{
    \zeta   C\left(\zeta s\right)\\
    \zeta   S\left(\zeta s\right)\\
    \zeta^2 s
} ,
$$
where $\zeta := \frac{1}{\sqrt[4]{2}}$.
Remark The trace $s \mapsto (\zeta C\left(\zeta s\right), \zeta S\left(\zeta s\right))$ of our curve in the $xy$-plane is an Euler spiral, that is, a curve whose curvature at a point is proportional to its arc length from a reference point.
A plot of our solution $\gamma(s)$, $-12 \leq s \leq 12$:

