= Solution
A <cyclic coordinate> $q_j$ is absent explicitly from the <Lagrangian>: $\partial L/\partial q_j=0$, even though $\dot q_j$ may occur. The <Euler-Lagrange equation> then gives
$$
\boxed{\frac{d}{dt}\left(\frac{\partial L}{\partial\dot q_j}\right)=0,\qquad p_j=\frac{\partial L}{\partial\dot q_j}=\text{constant}.}
$$
Thus a continuous coordinate-translation symmetry gives a conserved <conjugate momentum>. The conserved quantity need not be ordinary linear <momentum>: an absent azimuthal angle can give conserved <angular momentum>. The derivative is taken holding the other coordinates and <velocities> fixed.
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