= Solution
{c}
The right-handed <material frame of a rod> has $\mathbf d_2\times\mathbf d_1=-\mathbf d_3$ and $\mathbf d_2\times\mathbf d_3=\mathbf d_1$. Hence the <director kinematics of a rod> reduce to
$$
\mathbf d_1'=-\hat u\mathbf d_3,\qquad
\mathbf d_3'=\hat u\mathbf d_1,\qquad \mathbf d_2=\mathbf e_2.
$$
Solving these constant-coefficient <ordinary differential equations> with the given initial frame gives
$$
\boxed{\begin{aligned}
\mathbf d_1(S)&=\cos(\hat u S)\mathbf e_1-\sin(\hat u S)\mathbf e_3,\\
\mathbf d_3(S)&=\sin(\hat u S)\mathbf e_1+\cos(\hat u S)\mathbf e_3.
\end{aligned}}
$$
The <inextensible rod> has $\mathbf r'=\mathbf d_3$. An <integration> using $\mathbf r(0)=\mathbf0$ yields, for $\hat u\ne0$,
$$
\boxed{\mathbf r(S)=\frac{1-\cos(\hat u S)}{\hat u}\mathbf e_1
+\frac{\sin(\hat u S)}{\hat u}\mathbf e_3.}
$$
Indeed,
$$
\left|\mathbf r(S)-\frac{\mathbf e_1}{\hat u}\right|^2=\frac1{\hat u^2},
$$
so the centreline is a <circular arc> of radius $1/|\hat u|$ in the plane perpendicular to $\mathbf e_2$. The continuous $\hat u\to0$ limit is $\mathbf d_1=\mathbf e_1$, $\mathbf d_3=\mathbf e_3$, and $\mathbf r=S\mathbf e_3$.
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