Solution
= Solution
For the parametrized curve $\mathbf x(u)=(u,f(u))$, the <curvature of a plane curve> is
$$
\kappa=\frac{|x'y''-y'x''|}{(x'^2+y'^2)^{3/2}}
=\boxed{\frac{|f''(u)|}{(1+f'(u)^2)^{3/2}}}.
$$
For the upper semicircle $f(u)=\sqrt{a^2-u^2}$,
$$
f'=-\frac{u}{f},\qquad f''=-\frac{a^2}{f^3},\qquad
1+f'^2=\frac{a^2}{f^2},
$$
and therefore \b[$\kappa=1/a$], as required for a circle of radius $a$.