 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-2/13c/b/iv/solution

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

Substitute

y = y_{0} + y_{F}

into the energy functional and integrate the resulting polynomial:

E = \int_{0}^{L} [\frac{A}{2} (y^{''})^{2} + ρ g y] d x = - \frac{ρ ^{2} g ^{2} L ^{5}}{40 A} + \frac{F ^{2} L ^{3}}{6 A} .

The term linear in

F

cancels. Hence

E_{0} = - \frac{ρ ^{2} g ^{2} L ^{5}}{40 A}, E = \frac{F ^{2} L ^{3}}{6 A}, E = E_{0} + E .

As required,

E_{0}

is independent of the force and

E

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