 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-1/9d/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Differentiate

G = \sum_{i} p_{i} \cdot r_{i}

\dot{G} = \sum_{i} (F_{i} \cdot r_{i} + p_{i} \cdot \dot{r}_{i}) = \sum_{i} F_{i} \cdot r_{i} + 2 T .

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