Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 1 / 4 / a / Solution

The distance from the origin is $r=4\pi -t$. The direction angle, measured counterclockwise from the positive horizontal axis, is $\theta =\pi /2-t$ modulo $2\pi$. Thus the path is an Archimedean spiral, turning clockwise through two complete revolutions as its radius decreases linearly from $4\pi$ to zero.

It starts at $(0,4\pi )$, reaches $(7\pi /2,0)$ at $t=\pi /2$, crosses $(0,-3\pi )$ at $t=\pi$, returns to the positive vertical axis at $(0,2\pi )$ when $t=2\pi$, and ends at the origin when $t=4\pi$. These crossings fix both the scale and direction of the sketch.