= Solution
This vector field is the negative of the one in part (a). For the same positive-definite <quadratic form>,
$$
\dot V=2r^2(1-V),\qquad\dot\theta=2\cos^2\theta+\sin^2\theta>0.
$$
The ellipse $V=1$ is invariant and has no stationary point. The angular variable increases around it, with a finite return time $\int_0^{2\pi}d\theta/(2\cos^2\theta+\sin^2\theta)=\sqrt2\pi$. It therefore explicitly supplies a \b[periodic solution]. Alternatively, the <Poincaré-Bendixson theorem> applies to any compact invariant annulus $c\leq V\leq C$ with $0<c<1<C$, because its boundaries point inward and there is no equilibrium in the annulus.
If a periodic orbit had a point with $V<1$, then $V$ would strictly increase along it and could not return; a point with $V>1$ would make it strictly decrease. A periodic orbit meeting $V=1$ remains on that ellipse by uniqueness of solutions. Consequently \b[the ellipse is the unique periodic orbit], up to time translation. It is attracting, and all nonzero trajectories approach it.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ii/paper-1-quadratic-basin.png]
{title=The invariant quadratic ellipse separates attraction from finite-time escape in the first system and is the unique attracting periodic orbit in the reversed system; inner and outer circles show the optimal attraction and escape radii}
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