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

13 D VECTOR FUNCTIONS
13.1 Vector Functions and Space Curves
1. The component functions v'4- t 2 , e- 3 t, and ln(t + 1) are a ll defined wher4 - t 2 2: 0 => - 2 ~ t ~ 2 and
t + 1 > 0 => t > - 1, so the domain ofr is ( -1, 2].
1 1
- =
t-+O t--+o sin 2 t . t.-o sin 2 t lim sin 2 t
~ t--+0 t2
3. lim e- 31 = e 0 = 1, lim~= lim -
and lim cos 2t = cos 0 = 1. Thus
t-+0
=
1 _ I_ _ 1
. t)2 - !2 - '
(
lim~
t --+0 t
lim (e-st i + 4 j +cos 2t k) = [lim e- 3 t] i + [urn 4] j + [rim cos 2t] k = i + j + k.
t-+O sm t t-+O t-+O sm t t -+O
] . 1 + t 2 1 . (1/ t2 ) + 1 0 + 1
1 1 . _ 1 t ,.
1 . 1 - e-2 t
S. ~~ 1- t2 1 . 1 1 = t~ (1/t2)- 1 = 0 - 1 = - 't 2.~ tan = 2 • t~ t = t2.~ t - -te_2_t = 0 - 0 = 0· Thus
(
1 + e 1 - e- 2 t) .
lim - 1
2 ,tan-1 t, t =(-l,f,O).
t--+oo - t
7. The corresponding parametric equations for this curve are x = sin t, y = t.
We can make a table of values, or we can eliminate the parameter: t = y =>
x = sin y, with y E JR. By comparing different values oft, we find tl)e direction in
which t increases as indicated in the graph.
X
9. The corresponding parametric equations are x = t, y = 2 - t, z = 2t, which are
parametric equations of a line through the point (0, 2, 0) and with direction vector
(1, -1, 2).
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