Question

please solve example 3

The Definition of the Gauss Map and Its Fundamental Properties 141 Example 3. Consider the cylinder {(x, y, z) € R’; x2 + y2 = 1). By an gument similar to that of the previous example, we see that N = (x, y,0) and N=(x, y,0) are unit normal vectors at (x, y, z). We fix an orienta- tion by choosing N = (-x, -y, 0) as the normal vector field. By considering a curve (x(t), y(t), z(t)) contained in the cylinder, that is, with (x(t))2 + (y(t))% = 1, we are able to see that, along this curve, N(t) = (-x(t), -y(t), 0) and therefore dN (x"(t), y'(t), z'(t)) = N'(t) = (-x'(t), –y'(t),0). We conclude the following: If v is a vector tangent to the cylinder and parallel to the z axis, then dN(v) = 0 = Ov; if w is a vector tangent to the cylinder and parallel to the xy plane, then dN(w) = -w (Fig. 3-6). It follows that the vectors v and w are eigenvectors of dN with eigenvalues 0 and -1, respectively (see the appendix to Chap. 3). Ten or de bos tortor . --- world

Answer 1

& {(2, 7, 7) ER² : 2242 - 17, es the exlinder IF(ny,7) x² + y21 normal vecten - (22, 27,0) OFIL 1402, 492 Not roll = (22, 27,0) ru (1²21) = (22, 24,0) 22 and (-2, -3,0)

Considered a curve og (204),YCH), 911) Contained in the cylinder le act)) (Y CH) )2 = 1. AN(all), Y'It), 7 (t)) = N'(t) = (-a(t), - y(t), o) r is parallel totan i- ex=0, y con then dN (4) = (0,0,0), dass die son. = 0 o v [dN(v)=0.0 2.0 is eigenvalue and vis eigen vecten. now w is a vectens panalled tony Plare ie AN(W)= (-x'(t), – Y'(t), o) - (ACH), 41(t): 0) (txo - cu i. W is eigen recten, al is eigenvalue as Thorn you or