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

172 D CHAPTER 13 VECTOR FUNCTIONS
v(t) = 5(cos a:) i + [5 sin a:+ 4
g 0
(5t cos a:)( 40 - 5t cos a:)] j = (5 cos a:) i + (5 sin a:+ ~t cos a: - -fue cos 2 a:) j.
Integrating, r (t) = (5t cos a:) i + (5tsin a:+ ~t 2 cos a:- -ftt 3 cos 2 a:) j (where we have again placed
the origin at A). The boat will reach the east bank ~hen 5t cos a: = 40 =? t = ~ = - 8 -.
5 cos a: cos a:
In order to land at point B ( 40, 0) we need 5t sin a: + ~ e cos a: - ft t 3 cos 2 a: = 0 =?
5(- 8 - ) sin a:+~ (- 8 2
-)
cos a: cos a: cos a:
cos a:- -ft(- 8 3
-) .cos 2 a:= 0 =?
- 1 - (40sina: + 48 - 32) = 0 =?
cos a:
40 sin a: + 16 = 0 =? sin a: = - %. Thus a: = sin - l (-%) ~ -23.6°, so the boat should head 23.6° south of
east (upstream). The path does seem realistic. The boat initially heads
If
upstream to counte~act the effect of the current. Near the center of the river,
the current is stronger and the boat is pushed downstream. When the boat
nears the eastern bank, the current is slower and the boat is able to progress
upstream to arrive at point B.
0 1--=:::==::::::::;::>o-f'...::::::::=:=::::~ 40
- 12
35. Ifr' (t) = c x r(t) then r' (t) is perpendicular to both c and r(t). Remember that r ' (t) points in the direction of motion, so if
r' (t) is always perpendicular to c, the path of the particle must lie in a plane perpendicular to c. But r ' (t) is also perpendicular
to the position vector r(t) which confines the path to a sphere centered at the origin. Considering both restrictions, the path
must be contained in a circle that lies in a plane perpendicular to c, and the circle is centered on a line through the origin in the
direction of c.
37. r(t) = (3t - t 3 ) i + 3t 2 j =? r'(t) = (3 - 3t 2 ) i + 6tj,
lr' (t)i = J(3- 3t 2 )2 + (6t)2 = v'9 + 18t 2 + 9t 4 = J(3 - 3t 2 ) 2 = 3 + 3t 2 ,
r" (t) ·= -6t i + 6j, r' (t) x r" (t) = (18 + 18t 2 ) k. Then Equation 9 gives
- r ' (t) . r" (t) - (3 - 3e)( -6t) + (6t)(6) - 18t + 18t 3 - 18t(1 + t 2 ) - 6t
aT- lr '(t)/ - 3 +3t2 - 3 + 3t2 - 3(1 + t 2 ) -
[or by Equation 8,
I d [ 2] ] dE . LO . lr'(t) X r"(t)i 18 18e - 18(1 t 2 ) - 6
aT = v = dt 3 + 3t = 6t an quat10n g1ves aN - ir'(t)i -
3 + 3 t2 - 3 ( 1 + t 2) - ·
39. r (t) = cost i +sin t j + t k =? r '(t) = - sint i + costj + k , ir'(t)i = Vsin 2 t + cos 2 t + 1 = V'2,
r" (t) = - cost i- sin tj, r' (t) x r" (t) =sin t i - cos tj + k.
_ r'(t) · r"(t) _ sintcost - sintcost _ 0
d _lr'(t) x r"(t)l _ Vsin 2 t+cos 2 t+ 1 = .J2 = l
Then D.T - lr'(t)i - . .j2 - an aN - lr'(t)i - -/2 -/2 ·
41. r(t)=et i +-/2tj +e-t k =? r'(t)=et i+-/2j-e- 1 k, ir(t)i=v'e 21· +2+ e 2 t=J(e 1· +e t)2=e'·+e- t,
e2t _ e- 2t (et + e- t)(et _ e- t) .
r"(t) = et i + e-t k. Then aT = = = et - e:-t = 2sinht
et + e t et + e t
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