Definition. A regular curve $\alpha: I \to \mathbb R^3$ is a helix if there is a unit vector $v$ that forms a constant angle with $\alpha'(t), \forall t \in I$.
We can assume $\alpha$ parameterized by arc length. If $\alpha$ is a helix, then there is a unit vector $v$ such that $\langle \alpha'(s), v \rangle$ is constant. So $\langle \alpha''(s),v \rangle = 0$, that is $k(s)\langle n(s), v
\rangle = 0$. As $k(s) \ne 0$, it follows that $v$ belongs to the plan determined by $t(s)$ and $b(s)$, for each $s \in I $. So be
$$ v= \cos\theta(s)\,t(s) + \sin\theta(s)\,b(s)$$
Differentiating and using Frenet's formulas, we get
$$0 = -\sin\theta(s)\theta'(s)t(s) + (k(s)\cos\theta(s)+\tau(s)\sin\theta(s))n(s) + \cos\theta(s)\theta'(s)b(s)$$
Therefore,$\forall s \in I$,
$$\sin\theta(s)\theta'(s) = 0,$$
$$\cos\theta(s)\theta'(s) = 0,$$
$$k(s)\cos\theta(s) + \tau(s) \sin\theta(s) = 0$$
The first two equations determine $\theta'(s) = 0, \forall \in I$. Therefore, $\theta(s)$ is constant. Also, the constant $\cos\theta$ is non-zero, otherwise we would have $\tau(s)=0$, which contradicts the hypothesis. It follows from the third equality that $\frac{k}{\tau}$ is constant. Conversely, if $\frac{k}{\tau}$ is constant, we set $\theta$ such that $tg\theta = -\frac{k}{\tau}$. So $$v=cos\theta t(s) + sin \theta b(s)$$ is a constant unit vector and $\forall s \in, \langle t(s), v \rangle = cos\theta $ is constant. So $\alpha$ is helix.
Apparently the answer is correct but incomplete, I'm not able to complete it.
Thanks for any help.
Best Answer
The curve $\alpha: \vec{r}(s)$ is parametrized by its arclenght. The associated vector functions $\vec{T}$, $\vec{N}$, $\vec{B}$ and Frenet-Serret formulas are explained in here: https://en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas. In the solution, we will show these functions without the parameter $s$.
Assume first that ${\kappa}{\tau}=c$ is a constant. Then the first and third Frenet-Serret formulas $\frac{\vec{T}}{ds}=\kappa\vec{N}$ and $\frac{\vec{B}}{ds}=-\tau\vec{N}$ give $\frac{\vec{T}+c\vec{B}}{ds}=\vec{0}$ and hence $\vec{T}+c\vec{B}=\vec{v_0}$ is a constant vector. But then, since $\vec{T}\bot\vec{B}$ and $\vec{T}\cdot\vec{T}=1$ ve have $\vec{T}\cdot\vec{v_0}=1$. That is, the tangent vector makes a constant angle with $\vec{v_0}$. We can normalize $\vec{v_0}$ to obtain a unit vector with the same property.
Conversely, assume that the tangent vector makes a constant angle with a constant vector $\vec{v_0}$. Then $\vec{T}\cdot\vec{v_0}=k_0$ for some constant $k_0$. Taking derivative of both sides, we get $\frac{d\vec{T}}{ds}\cdot\vec{v_0}=0$. Then, if $\kappa$ is not zero, from the first Frenet-Serret formula, we deduce that $\vec{N}\cdot\vec{v_0}=0$ and hence $a\vec{T}+b\vec{B}=\vec{v_0}$ for some constants $a,b$. If $\kappa$ is zero, alternatively, we can use the third formula since $\vec{B}$ makes a constant angle with $\vec{v_0}$ too. Taking the derivative of the last equation we obtained, we get $(a\kappa -b\tau)\vec{N}=\vec{0}$. Hence, $a\kappa -b\tau=0$ and ${\kappa}{\tau}=\frac{a}{b}$ is a constant. By the way, $b$ must be non-zero, by the assumption of the problem that $\tau$ is non-zero.