Let $\gamma$ be a plane curve parametrized by the arc length, having the property that its tangent vector $T(t)$ forms a fixed angle $\theta$ with $\gamma(t)$.
Before explaining where I am, let define the notations I use:
- $T(t) := \dot{\gamma}(t)$;
- $N(T)$ is the vector that is normal to $T(t)$;
- $\kappa(t)$ is the curvature of $\gamma(t)$;
- Frenet differential equation is $\ddot{\gamma}(t) = \kappa(t) N(t)$ and $N(t) = i \dot{\gamma}(t)$.
We have that $\gamma$ is parallel to $\dot{\gamma} \exp(i \theta)$, so we can write $\gamma(t) = \alpha(t) \dot{\gamma}(t) \exp(i \theta)$ where $\alpha \in C^\infty$. Taking the derivative in both sides we obtain: $$\begin{align*} \dot{\gamma}(t) &= \dot{\alpha}(t) \dot{\gamma}(t) \exp(i \theta) + \alpha(t) \ddot{\gamma}(t) \exp(i \theta)\\ &= (\dot{\alpha}(t) \dot{\gamma}(t) + \alpha(t) \ddot{\gamma}(t)) \exp(i \theta)\\ &= (\dot{\alpha}(t) \dot{\gamma}(t) + \alpha(t) i \kappa(t) \dot{\gamma}(t)) \exp(i \theta)\\ \Rightarrow 1 &= \exp(i \theta)(\dot{\alpha}(t) + \alpha(t) i \kappa(t))\\ \Rightarrow \exp(-i \theta) &= \dot{\alpha}(t) + \alpha(t) i \kappa(t). \end{align*}$$ By identifying real and imaginary parts, we obtain: $$\begin{cases} \cos(\theta) &= \dot{\alpha}(t)\\ \sin(\theta) &= -\alpha(t) \kappa(t). \end{cases}$$ Integrating on both side of the first equation we obtain $$\begin{cases} \cos(\theta)t &= \alpha(t)\\ \sin(\theta) &= -\alpha(t) \kappa(t). \end{cases}$$ Therefore $$\sin(\theta) = -\cos(\theta)t \kappa(t) \quad \Rightarrow \quad \kappa(t) = -\frac{1}{t} \tan(\theta).$$
Now my hint says that I should obtain $\kappa(t) = -\frac{1}{t} \cot(\theta)$, which I obviously don't although I'm not far. Then I'm a bit lost on what to do next. Could someone help on these two points?