= Solution
{c}
In <Cartesian coordinates>, a <parametrization> of the <curve> is
$$
x(\theta)=f(\theta)\cos\theta,\qquad y(\theta)=f(\theta)\sin\theta.
$$
Differentiate using the <product rule>:
$$
x'=f'\cos\theta-f\sin\theta,\qquad y'=f'\sin\theta+f\cos\theta.
$$
Squaring and adding cancels the mixed terms, giving $(x')^2+(y')^2=(f')^2+f^2$. The <arc length> is the integral of the speed, so
$$
\boxed{L=\int_0^{2\pi}\sqrt{f(\theta)^2+f'(\theta)^2}\,d\theta.}
$$
This assumes the <parametrization> is piecewise continuously differentiable and counts length with multiplicity if the <curve> is retraced.
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