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San Jose´ State University Math 113, Spring 2010 Midterm 2 Solutions

Assigned on April 28, due on May 5, 2010, by 3 PM

Name: Granwyth Hulatberi

Explain your work 1. (25 points) Consider a surface patch

σ(u, v) = (u cos v, u sin v, log cos v + u),

π π where u ∈ R and − 2 < v < 2 .

(a) Compute the ﬁrst fundamental form of σ. π π (b) Let v1, v2 be any two numbers in (− 2 , 2 ) and deﬁne curves

γi(t) = σ(t, vi), i = 1, 2, a ≤ t ≤ b,

where a < b are arbitrary. Show that γ1 and γ2 have the same arclength.

Solution: Since σu = (cos v, sin v, 1), σv = (−u sin v, u cos v, − tan v), it follows that

2 2 2 E = kσuk = 2,F = σu · σv = − tan v, G = kσvk = u + tan v.

2 2 2 (b) Recall that kγ˙ ik = Eu˙ + 2F u˙v˙ + Gv˙ . Since u = t and v = vi for the curve γi, we have Z b √ 2 `(γi) = Eu˙ dt a Z b √ = 2 dt a √ = (b − a) 2.

Therefore, `(γ1) = `(γ2). 2. (25 points) A curve on a surface is called asymptotic if its normal curvature is everywhere zero.

(a) Determine the asymptotic curves of the surface S deﬁned by z = xy. (b) Compute the Gaussian curvature of S.

Solution: We parametrize S by σ(u, v) = (u, v, uv). Thus

σu = (1, 0, v), σv = (0, 1, u),

i j k

σu × σv = 1 0 v = (−v, −u, 1),

0 1 u so the corresponding unit normal is

σ × σ (−v, −u, 1) N = u v = √ . kσu × σvk 1 + u2 + v2 Let γ(s) = (u(s), v(s), w(s)) be a unit speed curve on S. Then w(s) = u(s)v(s), so

w00(s) = u00(s)v(s) + 2u0(s)v0(s) + u(s)v00(s).

The normal curvature of γ is therefore

00 κn = N · γ (−v, −u, 1) = √ · (u00, v00, w00) 1 + u2 + v2 1 = √ (−vu00 − uv00 + w00) 1 + u2 + v2 1 = √ (2u0v0). 1 + u2 + v2

0 0 0 0 Hence κn = 0 iﬀ u (s)v (s) = 0, which is the case if either u = 0 or v = 0. Therefore, asymptotic curves of S are the coordinate curves u = constant and v = constant.

(b) First we observe that from (a) it follows that the coefficients of the fist fundamental form are E = 1 + v2,F = uv, G = 1 + u2. Next we compute the coefficients of the second fundamental form of σ:

σuu = (0, 0, 0), σuv = (0, 0, 1), σvv = (0, 0, 0). Therefore, L = N = 0 and 1 M = N · σuv = √ . 1 + u2 + v2 The Gaussian curvature of S equals

2 1 LN − M − 2 2 1 K = = 1+u +v = − . EG − F 2 (1 + v2)(1 + u2) − u2v2 (1 + u2 + v2)2 3. (25 points) Consider the parametrized surface u3 v3 σ(u, v) = u − + uv2, v − + vu2, u2 − v2 3 3 Show that: (a) The coefficients of the first fundamental form are E = G = (1 + u2 + v2)2,F = 0. (b) If L, M, N are the coefficients of the second fundamental form, then M = 0. (You don’t have to compute L and N.) (c) The lines of curvature are coordinate curves u = constant and v = constant. Recall that a line of curvature of a surface is a curve γ such thatγ ˙ (t) is a principal vector for all t. Solution: (a) We have 2 2 2 2 σu = (1 − u + v , 2uv, 2u), σv = (2uv, 1 − v + u , −2v), so 2 E = kσuk = (1 − u2 + v2)2 + 4u2v2 + 4u2 = (1 + u2 + v2)2, 2 F = σu · σv = 0,G = kσvk = E. (b) To find M, we need to compute the unit normal defined by σ. We have

i j k 2 2 σu × σv = 1 − u + v 2uv 2u

2uv 1 − v2 + u2 −2v = (−2u(1 + u2 + v2), 2v(1 + u2 + v2), 1 − (u2 + v2)2). Furthermore,

σuu = (−2u, 2v, 2), σuv = (2v, 2u, 0) σvv = (2u, −2v, −2). Since 2 2 2 2 (σu × σv) · σuv = −2uv(1 + u + v ) + 2uv(1 + u + v ) = 0, it follows that M = 0, as claimed.

(c) Let FI and FII be the matrices of the ﬁrst and second fundamental forms of σ. It follows from (a) and (b) that both of them are diagonal, as the oﬀ-diagonal entries −1 F and M are zero. Hence the Weingarten matrix W = FI FII is diagonal. This implies that (1, 0)T and (0, 1)T are eigenvectors of W. Eigenvectors of W are principal directions. Since vectors (1, 0)T and (0, 1)T are tangent to the curves u = constant and v = constant, it follows that the coordinate curves are lines of curvature. 4. (25 points) Show that if a curve γ on a surface S is both a geodesic and an asymptotic curve, then γ is a (segment of a) straight line.

Solution: We will use the classical formula

2 2 2 κ = κg + κn,

where κ, κg and κn denote the curvature, geodesic curvature and normal curvature, respectively. Suppose that γ is a (unit speed) geodesic. Then κg = 0. If γ is also an asymptotic curve, then κn = 0. The above formula implies κ = 0, henceγ ¨ = 0. Therefore, γ(s) = sa + b, for some vectors a, b, so γ is a segment of a straight line.