By Mike Swarbrick Jones

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**Additional info for A Binomial Congruence**

**Example text**

At p. If the restriction of F to S has a local maximum or a local minimum at p, so does 33 d F (γγ (t)) = 0 at p, which F (γγ (t)) for all curves γ on S passing through p, hence dt implies that ∇F is perpendicular to γ˙ , and hence perpendicular to the tangent plane of S at p. This means that ∇S F = 0. 6). 5 We find that, at θ = ϕ = π/4, σ θ = 21 b(−1, −1, 2), σ ϕ = 21 (a 2 + b)(−1, 1, 0), so the unit normal of the torus at this point is parallel to √ √ √ (−1, −1, 2) × (−1, 1, 0) = − 2(1, 1, 2).

3, there is a straight line on the hyperboloid passing through γ (u), and it takes the form (x − z) cos θ = (1 − y) sin θ, (x + z) sin θ = (1 + y) cos θ. Substituting x = cos u, y = sin u, z = 0 gives θ = u2 + π4 . The vector δ (u) = (cos 2θ, sin 2θ, 1) = (− sin u, cos u, 1) is parallel to this line, so a parametrization of the hyperboloid as a ruled surface is σ (u, v) = (cos u, sin u, 0) + v(− sin u, cos u, 1) = (cos u − v sin u, sin u + v cos u, v). 4). 2. (a) Types (vii) through (xi). (b) Types (vi) and (x).

Each part of type (vi)is diffeomorphic to the unit cylinder by (x, y, z) → x p x2 p2 , 2 + y2 q y q x2 p2 2 + y2 q , ln |z|. The two connected pieces of type (viii) are given by x > 0 and x < 0; each piece is diffeomorphic to the plane by (x, y, z) → (0, y, z). 2 with u ∈ R and −π < v < π or 0 < v < 2π. σ (u, v) 2 = sech2 u(cos2 v + sin2 v) + tanh2 u = sech2 u + tanh2 u = 1, so σ parametrizes an open subset of S 2 ; σ is clearly smooth; and σ u × σ v = −sech2 u σ (u, v) is never zero, so σ is regular.