NAME:Mathematics 138A–001, Winter 2010, Final ExaminationAnswer Key11. [25 points] Find the curvature and torsion of the helix curve with parametrizationγ(t) = (cos t, sin t, t). [Hint: A modified arc length parametrization is given by s =√2 tor equivalently t = s/√2.]SOLUTIONTake s as in the hint. Then the unit tangent vector is given byT(s) = γ0(s) =ddscos(s/√2), sin(s/√2), (s/√2)=−(1/√2) sin(s/√2), (1/√2) cos(s/√2), (1/√2)so thatT0(s) = −12·cos(s/√2), sin(s/√2), 0.The length of this vector is the curvature, and it is equal to12; the principal normal N isthe unit vector in the same direction and hence is −( cos(s/√2), sin(s/√2), 0). It followsthat the binormal B = T × N must be(1/√2) sin(s/√2), −(1/√2) cos(s/√2), (1/√2)which implies B0(s) = −12N(s). By the definition of torsion, it follows that the torsionmust be12.22. [25 points] Let F(u, v) = (u2+ v2, v2− u2). Compute the derivative matrixDF(u, v). Find a point (a, b) with integral coordinates for which the derivative matrixDF(a, b) is zero, and find a point (c, d) with integral coordinates for which the rank ofthe derivative matrix DF(c, d) is equal to 1 (in other words, it is not invertible but stillnonzero).SOLUTIONThe derivative matrix is given byDF(u, v) =2u 2u−2v 2vand this matrix is zero if and only if (u, v) = (0, 0). Likewise this matrix has rank 1 if andonly if it is nonzero but its Jacobian is nonzero; since the Jacobian is 8uv, it vanishes ifand only if u = 0 or v = 0. Thus it has rank 1 at every nonzero point of the form (a, 0) or(0, b), and it is enough to take a or b equal to some nonzero integer.33. [20 points] Let S be the ruled surface with parametrizationX(u, v) = (t, et, 0) + u(0, cos t, sin t) .Compute the First Fundamental Form of S with respect to this parametrization.SOLUTIONThe first and second partial derivatives of X are given byX1= (1, et− u sin t, u cos t) , X2= (0, cos t, sin t)so the coefficients of the First Fundamental Form are equal to E = X1· X1= 1 + u2+e2t− 2etu sin t, F = X1· X2= 0, and G = X2· X2= 1. Hence the First FundamentalForm must be(1 + u2+ e2t− 2etu sin t) dt dt + du du .44. [25 points] Let γ(t) be a regular smooth curve with parametrizationx(t), y(t),and let S be the cylindrical surface with parametrization X(t, u) =x(t), y(t), u. Showthat the Gauss map is constant on the vertical lines in S of the formx(a), y(a), u, wherea is some fixed value of the first variable and u is arbitrary.Extra credit. [20 points] Using this, show that at every pointx(a), y(a), bon Sthe Shape Operator for S has determinant equal to zero.SOLUTIONLet Ω(t, u) = X1(t, u) × X2(t, u), so that the orientation N is given by ε|Ω|−1· Ω,where ε = ±1. We need to choose an orientation, and we shall take the orientation withthe positive sign (everything works similarly for the other choice).We have X1(t, u) =x0(t), y0(t), 0and X2(t, u) = (0, 0, 1), and thereforeΩ(t, u) = X1(t, u) × X2(t, u) =y0(t), −x0(t), 0which means that Ω, and hence also N, will only depend upon t. But this means that Nis constant on the vertical lines in S of the formx(a), y(a), u.Extra credit question. The shape operator S is given in terms of the parametriza-tion by −DN(t, u), and the shape operator has a zero determinant at p = X(t0, u0) if andonly if−DNβ(v0)[β0(v0)] =ddvNβ(v0)= 0for some regular smooth curve β in the surface with β(v0) = (t0, u0). By the main part ofthe problem, the vertical line curves satisfy this condition.55. [30 points] Let S be the graph of the function f(x, y) = x2+y2, take the standardparametrization X(u, v) = (u, v, u2+ v2), and choose the orientation N which is a positivemultiple of (−2u, −2v, 1). Find the Second Fundamental Form of S with respect to thisparametrization, compute the Gaussian curvature of S as a function of u and v, and explainwhy the formula shows that the Gaussian curvature of S is always positive. You may usethe fact that the First Fundamental Form is equal to(1 + 4u2) du du + 8uv du dv + (1 + 4v2) dv dv .SOLUTIONHere is what we are given:N =1√1 + 4u2+ 4v2·(−2v, −2u, 1) , E = 1+4u2, F = 4uv , G = 1 +4v2To simplify notation let α =√1 + 4u2+ 4v2so that N = α−1(−2v, −2u, 1); note thatα ≥ 1 everywhere.To compute the coefficients of the Second Fundamental Form, we need the secondpartial derivatives of X, which are X1,1= (0, 0, 2), X2,2= (0, 0, 2), and X1,2= X2,1=(0, 0, 0). Therefore we havee = N · X1,1= α−1· 2 = g = N · X2,2, f = N · X1,2= 0so that the Second Fundamental Form is equal to2(du du + dv dv)α.The Gaussian curvature is then given by the quotienteg − f2EG − F2where the preceding computations show that the numerator is 4/α2and the denominatoris(1 + 4u2)(1 + 4v2) − (4u2v2)2= 1 + 4u2+ 4v2= α2so that the Gaussian curvature K is equal to 4/α4.Since α is positive everywhere, the same is true for K.66. [25 points] Let A be a 2 ×2 matrix, and let I and O denote the identity and zeromatrices of the same size. Prove the identity A2− trace (A) ·A + det A · I = O (theCayley-Hamilton Theorem).SOLUTIONLet A be the matrixa bc dso that the trace and determinant of A are ad − bc and a + d respectively. Then we haveA2=a2+ bc ab + bdc d, − trace (A) ·A = −a2+ ad ab + bdac + dc ad + d2det A · I =ad − bc 00 ad − bcand if we add these three matrices we obtain the zero