Spiral tangents form constant angle with polar lines.

Question

Given the logarithmic spiral $$\alpha(t) = e^{-t}(\cos(t),\sin(t))$$ I take a ray from the origin given by $\lambda(\cos \theta, \sin \theta)$ and I have to prove that in $\alpha(\mathbb{R}) \cap R_{\theta}$ the tangents form a constant angle with the vector $(\cos \theta,\sin \theta)$ (constant in the sense that it does not depend on the point nor the angle $\theta$).

My approach

I compute the tangent line as having director vector $-e^{-t}(\cos(t)+\sin(t),\sin(t)-\cos(t))$ and then the intersection of the two lines is given by the equation $\lambda_1(\cos \theta,\sin \theta) = e^{-t}((1-\lambda_2)\cos(t)\sin(t),(1-\lambda_2)\sin(t)+\lambda_2\cos(t))$.

Solving this gives me $$t = -\frac{1}{2}log\left(\frac{\lambda_1^2}{(1-\lambda_2)^2+\lambda_2^2}\right)$$ But then the angle is given by $$cos \alpha = \frac{-e^{-t_0}((\cos t_0+\sin t_0) \cos \theta + (\sin t_0 - \cos t_0) \sin \theta}{\sqrt{t}e^{-t_0}}$$ which depends on $t_0$ (the point) and $\theta$.

Perhaps I should change to polar coordinates?

Answer 1

It's much simpler: You have to prove that for all $t\in{\mathbb R}$ the angle between $\alpha(t)$ and $\alpha'(t)$ is the same. This angle $\beta$ can be computed through $$\cos\beta={\alpha(t)\cdot\alpha'(t)\over |\alpha(t)|\>|\alpha'(t)|}$$ and turns out to be ${3\pi\over4}$.

Answer 2

I assume $R_\theta$ means the ray at angle $\theta$ from the origin. So you want $t=\theta + 2\pi k$ for some integer $k$. Can you find the slope of $\alpha(t)$ at such values of $t$? You should find that $$\frac{dy}{dx} = \frac{\sin t - \cos t}{\cos t + \sin t},$$ and if you put in $t=\theta+2\pi k$ you get the same value independent of the value of $k$.

But you want more. If $\tan \psi = \frac{dy}{dx} = \frac{\sin\theta - \cos\theta}{\cos\theta + \sin\theta}$, you want to see that $\psi - \theta$ is independent of $\theta$ as well. Work out $\tan(\psi-\theta)$ using the standard formula, and you should find that $\psi-\theta = \pi/4$ (independent of $\theta$).