 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ii/paper-1/7d/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

The principle of stationary action says that the physical path makes the action stationary under smooth variations

q + η h

with fixed endpoints

h (t_{1}) = h (t_{2}) = 0

. It need not minimize the action. Differentiating and integrating by parts gives

0 = δ S = \int_{t_{1}}^{t_{2}} (L_{q} h + L_{\overset{q}{˙}} \dot{h}) d t = \int_{t_{1}}^{t_{2}} (L_{q} - \frac{d}{d t} L_{\overset{q}{˙}}) h d t .

\frac{d}{d t} \frac{\partial L}{\partial q ˙} - \frac{\partial L}{\partial q} = 0 .

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