= Solution
{c}
Use the <polar radius> $r=at\in[0,1]$ as the parameter of the <Archimedean spiral>. Its <angle> in <polar coordinates> is $r/a$. The radial and angular contributions to its <arc length> are <orthogonal>, so
$$
aL_a=\int_0^1\sqrt{a^2+r^2}\,dr.
$$
There is no need to evaluate this <integral>. The simple bounds $r\leq\sqrt{a^2+r^2}\leq r+a$ imply
$$
\frac12\leq aL_a\leq\frac12+a.
$$
The <squeeze theorem> therefore gives the <scaled length of an Archimedean spiral>:
$$
\boxed{\ell=\lim_{a\to0^+}aL_a=\frac12.}
$$
Geometrically, the radial contribution becomes negligible after multiplying the <arc length> by $a$; the remaining contribution is the area under the <function graph> $r\mapsto r$ on $[0,1]$.
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