Solução_Calculo_Stewart_6e

F.
TX.10
14 VECTOR FUNCTIONS ET 13
14.1 Vector Functions and Space Curves ET 13.1
1. The component functions √ 4 − t 2 , e −3t ,andln(t +1)are all defined when 4 − t 2 ≥ 0 ⇒ −2 ≤ t ≤ 2 and
t +1> 0 ⇒ t>−1, so the domain of r is (−1, 2].
ln t
3. lim cos t =cos0=1, lim sin t =sin0=0, lim t ln t = lim
t→0 + t→0 + t→0 + t→0 + 1/t = lim 1/t
t→0 + −1/t = lim −t =0
2 t→0 +
[by l’Hospital’s Rule]. Thus lim hcos t, sin t, t ln ti = lim cos t, lim sin t, lim t ln t = h1, 0, 0i.
t→0 + t→0 + t→0 + t→0 +
5. lim e −3t = e 0 t 2
=1, lim
t→0 t→0 sin 2 t =lim
t→0
1
sin 2 t
t 2 =
lim
t→0
1
sin 2 =
t
t 2
and lim
t→0
cos 2t =cos0=1. Thus the given limit equals i + j + k.
lim
t→0
1
sin t
t
7. The corresponding parametric equations for this curve are x =sint, y = t.
2
= 1 1 2 =1
We can make a table of values, or we can eliminate the parameter: t = y
⇒
x =siny,withy ∈ R. By comparing different values of t,wefind the direction in
which t increases as indicated in the graph.
9. The corresponding parametric equations are x = t, y =cos2t, z =sin2t.
Note that y 2 + z 2 =cos 2 2t +sin 2 2t =1, so the curve lies on the circular
cylinder y 2 + z 2 =1.Sincex = t, the curve is a helix.
11. The corresponding parametric equations are x =1, y =cost, z =2sint.
Eliminating the parameter in y and z gives y 2 +(z/2) 2 =cos 2 t +sin 2 t =1
or y 2 + z 2 /4=1.Sincex =1, the curve is an ellipse centered at (1, 0, 0) in
the plane x =1.
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