Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 26J / c / Solution

From part (a),As $\alpha$ is orthogonal to $t$,To compute $\dot{\beta }\cdot \dot{\alpha }=\dot{\beta }\cdot t$, only its $t$ component matters. Differentiating $-\phi t$ contributes $-\dot{\phi }t$, while differentiating $-{\kappa }^{-1}\dot{\phi }n$ contributes $+\dot{\phi }t$ through $\dot{n}=-\kappa t-\tau b$. Every other term is normal to $t$. Hence