gabarito_lista4

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UNIVERSIDADE FEDERAL DE UBERLÂNDIA
FACULDADE DE MATEMÁTICA
Gabarito da Lista 4 - Func¸o˜es vetoriais de uma
varia´vel real
Disciplina: Ca´lculo Diferencial e Integral 2
Curso: Engenharias Aerona´utica e Mecatroˆnica
Professora: Ana Paula Tremura Galves
Monitor: Giovanni Borges
1. (a) (−1, 2].
(b) (−3,−2) ∪ (−2, 3).
2. (a)
−→
i +
−→
j +
−→
k .
(b)
−→
i + 3
−→
j − pi−→k .
(c)
(
−1, pi
2
, 0
)
.
(d)
(
0,
1
2
, 1
)
.
3. (a)
−→
r′ (t) = (t cos t+ sen t, 2t, cos 2t− 2t sen 2t).
(b)
−→
r′ (t) =
(
sec2t, sec t .tg t,− 2
t3
)
.
(c)
−→
r′ (t) = 4e4t
−→
k .
(d)
−→
r′ (t) = − 1
(1 + t)2
−→
i +
1
(1 + t)2
−→
j +
t2 + 2t
(1 + t)2
−→
k .
(e)
−→
r′ (t) = 2tet
2−→
i +
3
(1 + 3t)
−→
k .
(f)
−→
r′ (t) = (a cos3t− 3a sen3t)−→i + 3b sen2t cost−→j − 3c cos2t sent−→k .
(g)
−→
r′ (t) =
−→
b + 2t−→c .
(h) Escrever −→r (t) = t−→a ∧ (−→b + t−→c ) = t(−→a ∧ −→b ) + t2(−→a ∧ −→c ) e da´ı,−→
r′ (t) = −→a ∧ −→b + 2t(−→a ∧ −→c ).
4. (a)
−→
T (1) =
(
15√
262
,
6√
262
,
1√
262
)
.
(b)
−→
T (0) =
3
5
−→
j +
4
5
−→
k .
5.
−→
r′ (t) = (1, 2t, 3t2),
−→
T (1) =
(
1√
14
,
2√
14
,
3√
14
)
,
−→
r′′(t) = (0, 2, 6t) e
−→
r′ (t) ∧ −→r′′(t) = (6t2,−6t, 2).
6.
−→
T (0) =
(
2
3
,−2
3
,
1
3
)
,
−→
r′′(0) = (4, 4, 4) e
−→
r′ (t).
−→
r′′(t) = (8t2+12t+12)e4t−8e−4t.
7. (a) 2
−→
i − 4−→j + 32−→k .
(b) pi
−→
j + ln 2
−→
k .
(c)
−→
i +
−→
j +
−→
k .
(d)
7
3
−→
i +
16
15
−→
j − 3
pi
−→
k .
(e) et
−→
i + t2
−→
j + (t ln t− t)−→k + c, onde c e´ um vetor constante.
(f)
1
pi
senpit
−→
i − 1
pi
cospit
−→
j +
1
2
t2
−→
k + c.
8. −→r (t) = t2−→i + t3−→j +
(
2
3
t3/2 − 2
3
)−→
k .
9. −→r (t) =
(
1
2
t2 + 1
)−→
i + et
−→
j + (tet − et + 2)−→k .
10. (a) −→v (t) = (−2 sent, 3, 2 cost), −→a (t) = (−2 cost, 0,−2 sent) e |−→v (t)| = √13.
(b) −→v (t) = √2−→i + et−→j − e−t−→k , −→a (t) = et−→j + e−t−→k e |−→v (t)| = et + e−t.
(c) −→v (t) = 2t−→i + 2−→j + 1
t
−→
k , −→a (t) = 2−→i − 1
t2
−→
k e |−→v (t)| = ∣∣2t+ 1
t
∣∣.
(d) −→v (t) = et (cost− sent, sent + cost, t+ 1), −→a (t) = et (−2 sent, 2 cost, t+ 2)
e |−→v (t)| = et√t2 + 2t+ 3.
11. −→r (t) =
(
1
3
t3 + t
)−→
i + (t− sent+ 1)−→j +
(
1
4
− 1
4
cos2t
)−→
k .
12. No instante t = 4.