# Show that a curve has constant curvature and constant torsion.

I'm learning differential geometry, specifically the theory of curves, and need help with the following exercise:

Consider the curve $\boldsymbol\alpha(s)$ of $E^3$ with (unit) normal vector given by

$$\mathbf N(s) = (\sin(s), 0, \cos(s)).$$

$(a)$ Show that $\boldsymbol\alpha(s)$ has constant curvature and constant torsion.

$(b)$ Identify the type of the curve.

If we assume that $(a)$ is true, i.e. if both curvature and torsion are constant, then $\boldsymbol\alpha(s)$ is a circular helix. This result is mentioned in my textbook by means of an example (without rigorous explanation).

I still need to prove $(a)$ but I don't know how to do so. Since we are not given the parametrization of $\boldsymbol\alpha(s)$ explicitly, we can't compute the well-known formulas for curvature and torsion. I thought I could use the Frenet formulas but I could not come up with something useful.

One has $$\frac{d\mathbf{n}}{ds}=-\kappa \mathbf{t}+\tau \mathbf{b}$$ differentiating this and using $\frac{d^2\mathbf{n}}{ds^2}=-\mathbf{n}$ we have
$$-\mathbf{n}=-\dot{\kappa}\mathbf{t}-\kappa^2\mathbf{n}+\dot{\tau}\mathbf{b}-\tau^2\mathbf{n}$$
Now because $\mathbf{t},\mathbf{n},\mathbf{b}$ are independent, $\dot{\kappa}=\dot{\tau}=0$.