= Solution
In <polar coordinates>, the curve is the <logarithmic spiral> $r=ae^{bu}$, $\theta=u$. It starts at $(a,0)$ and winds counterclockwise outwards through one and a half turns. It crosses the horizontal axis at
$$
(a,0),\quad(-ae^{b\pi},0),\quad(ae^{2b\pi},0),\quad(-ae^{3b\pi},0).
$$
The following original sketch uses representative positive values; changing $a$ rescales it and changing $b$ changes its rate of expansion.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ia/paper-3-logarithmic-spiral.png]
{title=Logarithmic spiral over one and a half counterclockwise turns}
Differentiate the coordinates:
$$
x'=ae^{bu}(b\cos u-\sin u),\qquad y'=ae^{bu}(b\sin u+\cos u).
$$
Hence $ds/du=\sqrt{(x')^2+(y')^2}=ae^{bu}\sqrt{1+b^2}$. Integrating the <arc length> element gives, for $U\geq0$,
$$
\boxed{L(U)=\frac{a\sqrt{1+b^2}}{b}\bigl(e^{bU}-1\bigr).}
$$
For a negative endpoint parameter the length between $U$ and zero uses the absolute value of the final factor.
A second differentiation gives
$$
x'y''-y'x''=a^2e^{2bu}(1+b^2).
$$
The numerator is positive, so the absolute value in the PDF's <curvature of a plane curve> formula leaves it unchanged. Consequently
$$
\kappa(u)=\frac1{ae^{bu}\sqrt{1+b^2}}.
$$
The <speed> is $1/\kappa$, and therefore $dt=ds/v=\kappa\,ds=du$. Thus the requested elapsed time is
$$
\boxed{\Delta t=\int_{2n\pi}^{2(n+1)\pi}du=2\pi.}
$$
This <travel time at reciprocal-curvature speed> is independent of $n$ because the growing arc length per unit angle is exactly matched by the growing speed.
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