 Solution

ID: past-exam-of-the-geometry-course-of-the-university-of-oxford/2008/a/2/iii/c/solution

r_{u} = (- v sin u, v cos u, 1), r_{v} = (cos u, sin u, 0), n = \frac{( - s i n u , c o s u , - v )}{1 + v ^{2}} .

Hence

E = 1 + v^{2}

F = 0

G = 1

. Since

r_{uu} = (- v cos u, - v sin u, 0)

and

r_{vv} = 0

, both

L

and

N

vanish. The mean-curvature formula therefore gives

H = 0.

For comparison,

M = 1/ 1 + v^{2}

, so

K = - 1/ (1 + v^{2})^{2}

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