A curve is called a general helix if there exist a direction unit vector $a$ such that the tangent of the curve makes a constant angle with $a$. I would like to turn a curve $\gamma = \left(x(t),y(t)\right)$, a planar curve into $\hat{\gamma}=\left(x(t),y(t),z(t)\right)$ a space curve, where $\hat{\gamma}$ is a generalized helix over the curve. For convenience's sake, let $\gamma$ be a unit speed parameterization of the curve.
What I've tried is letting $\frac{\tau}{\kappa}=c$. Then,
$$ c = \frac{<\gamma' \times \gamma'',\gamma'''>}{\| \gamma' \times \gamma'' \|^3 } $$
My end goal here is to express $z(t)$ in terms of $x(t)$ and $y(t)$, but I just can't figure out how, and am also wondering if this is the approach I should take.
Note: I have seen this question, and this but the difference is that I need to show existence, rather than proving a generalized helix will have constant curvature/torsion. Thus that is why I am trying to find an expression for $z(t)$. Any $z(t)$ that will work.
