= Solution
{c}
For the <cardioid>, $r'=-a\sin\theta$. The expression under the <square root> simplifies to
$$
r^2+(r')^2=a^2(2+2\cos\theta)=4a^2\cos^2(\theta/2).
$$
Since $0\leq\theta\leq\theta_2\leq\pi$, the <cosine> is nonnegative. Therefore
$$
\boxed{s(\theta_2)=\int_0^{\theta_2}2a\cos(\theta/2)\,d\theta=4a\sin(\theta_2/2).}
$$
The upper half of the <cardioid> starts at $(2a,0)$, passes through $(0,a)$, bends to negative $x$, and ends at the cusp $(0,0)$ when $\theta_2=\pi$. Its full upper-half <arc length> is $4a$; reflecting gives $8a$ for the full <cardioid>.
The <fundamental theorem of calculus> gives
$$
\frac{ds}{d\theta_2}=2a\cos(\theta_2/2).
$$
This decreases from $2a$ to zero, so \b[the <arc length> increases fastest at $\theta_2=0$], interpreted as the right-hand rate at the initial endpoint. If only positive endpoints are allowed, the rate has supremum $2a$ as $\theta_2\downarrow0$, rather than an attained interior maximum.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-iv-q3-cardioid.png]
{title=A partial <cardioid> with $\theta_2=2\pi/3$, and the continuation to its cusp}
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