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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ib/paper-3/14g/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 3 / 14G / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-05

At a regular point, choose the unit normal

n = (σ_{x} \times σ_{y}) /∣ σ_{x} \times σ_{y} ∣

. The first fundamental form is the induced squared length

I = E d x^{2} + 2 F d x d y + G d y^{2}, E = σ_{x} \cdot σ_{x}, F = σ_{x} \cdot σ_{y}, G = σ_{y} \cdot σ_{y} .

The second fundamental form is

II = e d x^{2} + 2 f d x d y + g d y^{2}, e = n \cdot σ_{xx}, f = n \cdot σ_{x y}, g = n \cdot σ_{yy} .

The Gaussian curvature is the determinant of the shape operator,

K = (e g - f^{2}) / (EG - F^{2})

, independent of the normal orientation.

For the given graph,

σ_{x} = (1, 0, y)

and

σ_{y} = (0, 1, x)

, so with

S = 1 + x^{2} + y^{2}

,

I = (1 + y^{2}) d x^{2} + 2 x y d x d y + (1 + x^{2}) d y^{2}, n = \frac{( - y , - x , 1 )}{S}, II = \frac{2 d x d y}{S} .

Therefore

K (x, y) = - \frac{1}{( 1 + x ^{2} + y ^{2} ) ^{2}} .

The surface is regular everywhere. The printed description “hyperboloid” is a naming error: this graph is a hyperbolic paraboloid, and the computation uses the actual parametrization.

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