Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 3 / 4C / Solution

In polar coordinates, the curve is the logarithmic spiral $r=a{e}^{bu}$, $\theta =u$. It starts at $(a,0)$ and winds counterclockwise outwards through one and a half turns. It crosses the horizontal axis atThe following original sketch uses representative positive values; changing $a$ rescales it and changing $b$ changes its rate of expansion.

Differentiate the coordinates:Hence $ds/du=\sqrt{(x^{\prime }{)}^2+(y^{\prime }{)}^2}=a{e}^{bu}\sqrt{1+b^2}$. Integrating the arc length element gives, for $U\ge 0$,For a negative endpoint parameter the length between $U$ and zero uses the absolute value of the final factor.

A second differentiation givesThe numerator is positive, so the absolute value in the PDF's curvature of a plane curve formula leaves it unchanged. ConsequentlyThe speed is $1/\kappa$, and therefore $dt=ds/v=\kappa ds=du$. Thus the requested elapsed time isThis 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.