Find the arc length of the first turn of the Archimedean

Find the arc length of the first turn of the Archimedean spiral $\rho = a\varphi$.

We apply the polar arc-length formula directly.

Step 1: Set up

$\rho=a\varphi$, $\rho'=a$, and one turn is $\varphi\in[0,2\pi]$:

\[L=\int_0^{2\pi}\sqrt{a^2\varphi^2+a^2}\,d\varphi = a\int_0^{2\pi}\sqrt{\varphi^2+1}\,d\varphi\]

Step 2: Standard antiderivative

\[\int\sqrt{\varphi^2+1}\,d\varphi = \frac{\varphi\sqrt{\varphi^2+1}}2+\frac12\ln\left(\varphi+\sqrt{\varphi^2+1}\right)+C\]

Step 3: Evaluate at the limits

At $\varphi=0$: the expression is $0$.

At $\varphi=2\pi$: $\pi\sqrt{4\pi^2+1}+\frac12\ln\left(2\pi+\sqrt{4\pi^2+1}\right)$.

Step 4: Result

\[\boxed{L=a\pi\sqrt{4\pi^2+1}+\frac a2\ln\left(2\pi+\sqrt{4\pi^2+1}\right)}\]

Step 5: Numerical value

With $\sqrt{4\pi^2+1}\approx6.362$: $\pi\sqrt{4\pi^2+1}\approx19.99$, and $\frac12\ln(2\pi+6.362)\approx1.264$, so

\[L\approx21.25a\]