Find the arc length of the logarithmic spiral rho =

Find the arc length of the logarithmic spiral $\rho = e^{\alpha\theta}$ (with $\alpha>0$) from the origin to the point $M(\theta,\rho)$.

The spiral winds in toward the origin as $\theta\to-\infty$, so "from the origin" means the improper integral from $-\infty$.

Step 1: Differentiate

\[\rho=e^{\alpha\theta} \implies \frac{d\rho}{d\theta}=\alpha e^{\alpha\theta}=\alpha\rho\]

Step 2: Arc-length element

\[\sqrt{\rho^2+(\rho')^2}=\sqrt{\rho^2+\alpha^2\rho^2}=\rho\sqrt{1+\alpha^2}=e^{\alpha\theta}\sqrt{1+\alpha^2}\]

Step 3: Set up and evaluate the integral

\[L=\sqrt{1+\alpha^2}\int_{-\infty}^{\theta}e^{\alpha\tau}\,d\tau = \sqrt{1+\alpha^2}\left[\frac{e^{\alpha\tau}}\alpha\right]_{-\infty}^{\theta}=\frac{\sqrt{1+\alpha^2}}{\alpha}\,e^{\alpha\theta}\]

(the lower limit vanishes since $\alpha>0$).

Step 4: Express in terms of $\rho$

Since $\rho=e^{\alpha\theta}$ at $M$:

\[\boxed{L=\frac{\sqrt{1+\alpha^2}}\alpha\,\rho}\]

Step 5: Interpretation

The logarithmic (equiangular) spiral has the striking property that its arc length from the origin to any point is simply proportional to the radius $\rho$ at that point, with constant of proportionality $\frac{\sqrt{1+\alpha^2}}\alpha$ — a self-similarity famously admired by Jacob Bernoulli.