Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 70 / 1 / c / ii / Solution

The smallest admissible axial stretch is ${\lambda }_{\min }=\sin 2\theta$, where the two branches merge and ${\alpha }^2=\tan \theta =s/c$. The current unit fibre vectors are $m_{\pm }=U{l}_{\pm }$. Their dot product isAt the minimum, both terms in parentheses equal $cs$, soHence the two families are orthogonal at the maximum possible contraction. There is also a geometric proof of the bound: their reference parallelogram area is $\sin 2\theta$, and its current area is $\sin 2\theta /\lambda$, because the in-plane area stretch is $1/\lambda$. Two unit vectors span area at most one, with equality precisely when perpendicular. This gives $\lambda \ge \sin 2\theta$ and the same equality condition without solving the quadratic. At $\theta =\pi /4$ the minimum is one, so no axial contraction is possible and the fibres are already orthogonal.