If $0 < \theta < \frac{\pi}{2}$, then $\gamma$ is a logarithmic spiral

Question

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?

Answer 1

Okay I might have found something.

Let $0 < \theta < \frac{\pi}{2}$ and let's use the same notations that I defined in the question.

1. Let $\gamma$ be parametrized by the arc length. As done in the question, we find that $$\begin{equation} \kappa(t) = \pm\frac{1}{t} \tan(\theta). \end{equation}$$ depending in which "side" turns $\gamma$. If it turns in the trigonometric "side", it is "+", otherwise it is "-".
2. Let $\gamma_2(t) = (ae^{bt}\cos(t), ae^{bt}\sin(t))$. We have $$\kappa_{\gamma_2}(t) = \frac{\langle N(t), \ddot{\gamma_2}(t) \rangle}{\|\dot{\gamma_2}(t)\|^3}= \cdots = \frac{1}{ae^{bt}\sqrt{b^2+1}}.$$ What we want to do next is show that $\kappa(t) = \kappa_{\gamma_2}(t)$ for all $t$. For that we must look at the curvature at the same point. The problem is that $\gamma(t) \neq \gamma_2(t)$ for all $t$. Therefore we want to find a bijection $\phi(t)$ such that $\gamma(\phi(t)) = \gamma_2(t)$ for all $t$. We take $$\phi(t) = \int_{-\infty}^t\|\dot{\gamma_2}(s)\|ds = \cdots = \frac{1}{b} a \sqrt{1+b^2}e^{bt}.$$ Why this function? Because $\gamma(t)$ starts at $(0,0)$ and we want to know the length of $\gamma_2$ between $(0, 0)$ and $\gamma_2(t)$. If we know that length, as $\gamma$ is parametrized by the arc length, we know the $\tilde{t}$ for which $\gamma$ will be at the same point as $\gamma_2$, that is $\gamma(\tilde{t}) = \gamma_2(t)$.
   Let's remember that $\tan^2(\theta) +1 = \frac{1}{\cos^2(\theta)}$, thus $\tan(\theta) = \pm\frac{1}{b}$. Now we can compare the curvatures: $$\pm\kappa(\phi(t)) = \frac{1}{\phi(t)}\cdot \frac{\pm1}{b} = \frac{\pm b}{a\sqrt{1+b^2}e^{bt}b} = \frac{\pm 1}{a\sqrt{1+b^2}e^{bt}} = \pm \kappa_{\gamma_2}(t).$$ As the curvatures are the same, they define the same curve, therefore $\gamma$ is a logarithmic spiral.

Answer 2

Your result is correct. Note that $\alpha(t)$ denotes the distance of the moving point from the origin. When $0<\theta\ll1$ then the tangent vector $T(t)$ is almost parallel to $\gamma(t)$, which means that almost all of the arc length goes into increasing $|\gamma(t)|=\alpha(t)$. Therefore $\dot\alpha(t)=\cos\theta\doteq1$ makes sense, etcetera.