By Andrew Pressley

**Read Online or Download Instructor's Solutions Manual to Elementary Differential Geometry PDF**

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**Extra resources for Instructor's Solutions Manual to Elementary Differential Geometry**

**Sample text**

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). Hence, the tangent plane is x + y + √ 2z = 0.

15 (i) Any linear map R2 → R2 is obviously smooth so M is a diffeomorphism if and only if M is bijective. This holds if and only if M takes a basis of R2 , such u, v} is as {ii, j}, to another basis. e. if and only if and are linearly independent. u + vvv. v)dudv+ v is u du + 2(u dv . e. if and only if u and v are perpendicular unit vectors. e. if and only if u and v are perpendicular vectors of equal length. v)2 = 1. e. 2, the left-hand side is equal to u × v 2 . So M is equiareal if and only if u × v = 1.

N N is constant, say equal to (σ N = d. N ˜ N, the sign being that of det(J). 5, N = ±N ∂u ˜ ∂v σ˜ v˜ = σ u ∂u ∂v ˜ + σ v ∂v ˜ , we get σ ˜ u˜u˜ = σ u ˜ =± L So L ∂ 2u ∂ 2v σ + v 2 + σ uu ∂u ˜2 ∂u ˜ ∂u 2 ∂u ˜ 2 ∂u ∂u ˜ σ uv + 2σ 2 ∂u ∂v + σ vv ∂u ˜ ∂u ˜ ∂v ∂u ˜ ∂u ˜ 2 . N N = 0. N ∂u ˜ ∂u ˜ ˜ and N ˜ , are equivalent to the matrix equation gether with similar formulas for M in the question. 4 Let σ be a surface patch, P a 3 × 3 orthogonal matrix, a ∈ R3 a constant vector, σ u ×σ σ v (Proposition and σ˜ = P σ +aa.