 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2003/paper-2/10/b/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

Write

c = cos θ

s = sin θ

. Since

θ

is constant, the multivariable chain rule gives

\partial_{x} = c \partial_{u} - s \partial_{v}

and

\partial_{y} = s \partial_{u} + c \partial_{v}

. Applying these operators twice to a twice continuously differentiable function gives

g_{xx} = c^{2} g_{uu} - 2 cs g_{uv} + s^{2} g_{vv}, g_{yy} = s^{2} g_{uu} + 2 cs g_{uv} + c^{2} g_{vv} .

The mixed terms cancel and

c^{2} + s^{2} = 1

, so

g_{xx} + g_{yy} = g_{uu} + g_{vv} .

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