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

118 0 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
Substituting gives {13 cos (} - 3.5) 13
:in(} = 2 => 39 cos 8- 10.5 = 26 sin(} (1). Squaring both sides, we have
1521 cos 2 8 - 819 cos 8 + 110.25 = 676 sin 2 () = 676 (1 - cos 2 9)
2197 cos 2 (} - 819 cos(} - 565.75 = 0
The quadratic formula gives
_ 819 ± v'C--819)2- 4{2197)( - 565.75)
cos
8
. - ·2{2197)
· = 819 ± ~ 5 g~ 42 ' 572 :::::: 0. 72699 or - 0.35421
The acute value for(} is approximately cos-- 1 (0.72699) :::::: 43.4°. Thus the boatman should steer in the direction that is
43.4° from the bank, toward upstream.
Alternate solution: We could solve ( 1) graphically by plotting y = 39 cos 8- 10.5 andy = 26 sin 8 on a graphing device
and finding the appoximate intersection point (0.757, 17.85). Thus 0:::::: 0.757 radians or equivalently 43.4°.
(b) From part (a) we know the trip is completed when t = ·- 3
~
(} .· But 8 :::::: 43.4°, so the time required is approximately
1 sm
. 3 13sm 43 . 4
0
:::::: 0.336 hours or 20.2 minutes.
41 . The slope of the tangent line to the graph ofy = x 2 at,the point {2, 4) is
dy I = 2xl = 4
· dx x=2 x=2
and a parallel vector is i + 4j which has length Ji + 4j J = .)1 2 + 42 = ..JPi, so unit vectors parallel to the tangent line
are ±~ (i + 4 j ).
~ _.. --t ,_.. -+ ------+- ------+- ----+- ._. ---+ ---+ ( --+)
43. By the Triangle Law, AB + BC = AC. Then AB + BC + CA = AC + CA, but AC + CA = AC + -AC = 0.
--+ --+ --+
SoAB+BC+CA = 0.
45. (a), (b) (c) From the sketch, we estimate that s :::::: 1.3 and t:::::: 1.6.
(d) c = sa + t b # 7 = 3s + 2t and 1 = 2s - t.
Solving these equations gives s = t and t = V.
47. Jr- roJ is the distance between the points (x, y , z) and (xo, y 0 , z 0 ), so the set of points is a sphere with radius 1 and
center (xo, Yo, zo).
Alternate method: Jr - ro J = 1 # J(x- xo) 2 + (y- vo)2 + (z- zo) 2 = 1 #
(x - xo? + (y - yo) 2 + (z - zo? = 1, which is the equation of a sphere with radius 1 and center (xo, yo, zo).
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