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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-1/section-b/6/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / Section B / 6 / Solution by

Codex  0 2026-10-20

Put

S = θ_{1} + θ_{2}

and

D = θ_{1} - θ_{2}

. The gradient has components

3 sin S + sin D

and

3 sin S - sin D

, so a stationary point requires

sin S = sin D = 0

. In the closed square,

S \in [0, 2 π]

and

D \in [- π, π]

, and the compatible possibilities are precisely the four corners and

(π /2, π /2)

.

The Hessian matrix is

(3 cos S + cos D 3 cos S - cos D 3 cos S - cos D 3 cos S + cos D) .

Its eigenvectors

(1, 1)

and

(1, - 1)

have eigenvalues

6 cos S

and

2 cos D

. The second-derivative test gives the following classification:

Point \|\| $U$ \|\| Hessian eigenvalues \|\| Classification	$(0, 0), (π, π)$ \| $- 4$ \| $6, 2$ \| Minima	$(π, 0), (0, π)$ \| $4$ \| $- 6, - 2$ \| Maxima	$(π /2, π /2)$ \| $2$ \| $- 6, 2$ \| Saddle point of a scalar function

The corner classifications also hold for the smooth extension beyond the square, and hence for the restricted region. The global bounds

- 4 \leq U \leq 4

show that the stated minima and maxima are global within it. The central point has opposite curvature along the two diagonal directions.

Figure 1.
Energy contours show two minima, two maxima and a central saddle
. The $U = 2$ contour crosses at the saddle. Filled markers locate all five points where the gradient vanishes; color and labeled level curves show how the energy changes between them. Details.

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