# Can the equation for a logarithmic spiral be derived from the equiangular constraint alone?

(Ok, I'm guessing maybe we need the assumption that the function is differentiable as well somewhere too, but that doesn't count).

I've been trying to derive the equation for the logarithmic spiral $$r = a\cdot e^{k\theta}$$ from the requirement that the angle between the spiral and a circle sharing its origin (say, for simplicity, at $$\left ( 0,0 \right )$$) be constant.

With the help of some trigonometric identities, I managed to go full circle and show that if we assume that the angle between them is some constant, we get... that the angle between them is a constant.

I also looked at some proofs that the angle is constant given the formula for the spiral but couldn't find inspiration there on how to derive the formula from the assumption of a constant angle.

• Is your definition equivalent to saying that for a particle moving on the spiral, the direction of motion (i.e. tangent to thre curve) makes a constant angle with ther radius vector? If so, just use the polar velocity components and the equation is derived very easily. Sep 29, 2022 at 15:46
• @DavidQuinn it is. I barely learned anything about polar representation in high school or college. I tried using those but I guess I didn't know enough. I will read up on it and make another attempt using polar velocity. thank you! Sep 29, 2022 at 15:56
• Well if you need a hint just let me know Sep 29, 2022 at 16:21
• The missing ingredient is a differential equation (or the geometric equivalent), but your guess is correct: A continuous path making a constant angle with rays through the origin is a logarithmic spiral. Sep 29, 2022 at 19:38
• Spiral is $r(\phi)$, then tangent of angle between spiral and circle is $\frac{r'(\phi)d\phi}{r(\phi)d\phi}$. Sep 30, 2022 at 9:40

HINT…Following on from my comment, the velocity components are $$\frac{dr}{dt}$$ radially and $$r\frac{d\theta}{dt}$$ tangentially.
So if $$\alpha$$ is the fixed angle between the direction of motion and the radius vector, we have $$\tan\alpha=\frac{r\frac{d\theta}{dt}}{\frac{dr}{dt}}=k$$ $$\implies r\frac{d\theta}{dr}=k$$ You can solve this differential equation and get the result.
• Thank you! It seems knowing $\tan\alpha=\frac{r\frac{d\theta}{dt}}{\frac{dr}{dt}}=k$ is what I was missing. Is there a name for this equality, or is it supposed to be obvious? I'd like to look up the proof for that Sep 30, 2022 at 17:24