Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

324 0 CHAPTER 16 VECTOR CALCULUS
31. (a) Replacing cos u. by sin u. and sin u. by cos u. gives parametric equations
x = (2 + sin v ) sin u., y = (2 +sin v) cos u., z = u. +cos v. From the graph, it
appears that the direction of the spiral is reversed. We can verify this observation by
noting that the projection of the spiral grid curves onto the xy-plane, given by
x = (2 + sin v) sin u., y = (2 + sin v) cos u., z = 0, draws a circle in the clockwise
direction for each value of v. The original equations, on the other hand, give circular
projections drawn in the counterclockwise direction. The equation for z is identical in
both surfaces, so as z increases, these grid curves spiral up in opposite directions for
the two surfaces.
(b) Replacing cos u. by cos 2u. and sin u. by sin 2u. gives parametric equations
x = (2 + sin v) cos 2u., y = (2 +sin v) sin 2u., z = u. + cos v. From the graph, it
appears that the number of coils in the surface. doubles within the same parametric
domain. We can verify this observation by noting that the projection of the spiral grid
curves onto the xy-plane, given by x = (2 +sin v) cos 2u., y = (2 + sin v) sin 2u,
z = 0 (where v is constant), complete circular revolutions for 0 ~ u. ~ 1r while the
original surface requires 0 ~ u. ~ 21r for a complete revolution. Thus, the new
surface winds around twice as fast as the original surface, and since the equation for z
is identical in both surfaces, we observe twice as many circular coils in the same
z-interval.
33. r ( u., v) = ( u. + v) i + 3u. 2 j + ( u. - v) k.
r u = i + 6u.j + k and r, = i - k, so r,. x r , = - 6u. i + 2 j - 6u k. Since the point (2, 3, 0) corresponds to u. = 1, v = 1, a
normal vector to the surface at (2, 3, 0) is -6 i + 2j - 6 k , and an equation of the tangent plane is -6x + 2y- 6z = - 6 or
3x- y + 3z = 3.
35. r (u.,v)=u.cosv i +u.sinvj +v k => r(1 , ~) = (~.4. i )·
r u = cos v i + sin v j and r , = -u sin v i + u. cos v j + k , so a normal vector to the surface at the point ( ~ ,
4 , ~) is
r u ( 1, f) x r , ( 1, i') = ( ~ i + 4 j) x (-4 i + ~ j + k) = 4 i - ~ j + k. Thus an equation of the ta~gen t plane at
(t .4,f )is4(x ·- t) - ~(y-4) + 1 (z -~ )=0 or ~x - h +z =~ .
37. r (u,v)=u 2 i +2u.sinvj +ucosv k => r (1,0)=(1,0, 1).
ru = 2u. i + 2sinvj + cosv k and r , = 2u.cosvj - usinv k , 2
so a normal vector to the surface at the. point (1, 0, 1) is
r u(1, 0) X r,(1, 0) = (2 i + k ) X (2j) = - 2 i + 4 k.
Thus an equation of the tangent plane at (1, 0, 1) is
- 2(x- 1) + O(y- 0) + 4(z - 1) = 0 or -x + 2z = 1.
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