Solução_Calculo_Stewart_6e

F.
5.
1
1
1
0 (t2 i + t cos πt j +sinπt k) dt = dt
0 t2 i + t cos πt dt 1
j + sin πt dt k
0 0
=
1
t3 1
i + 3 0
where we integrated by parts in the y-component.
TX.10
t
sin πt 1
− 1
π 0 0
CHAPTER 14 REVIEW ET CHAPTER 13 ¤ 159
1
sin πt dt j + − 1 cos πt 1
k
π π 0
= 1 3 i + 1
π 2 cos πt 1
0 j + 2 π k = 1 3 i − 2
π 2 j + 2 π k
7. r(t) = t 2 ,t 3 ,t 4 ⇒ r 0 (t) = 2t, 3t 2 , 4t 3 ⇒ |r 0 (t)| = √ 4t 2 +9t 4 +16t 6 and
L = 3
0 |r0 (t)| dt = 3
0
√
4t2 +9t 4 +16t 6 dt. Using Simpson’s Rule with f(t) = √ 4t 2 +9t 4 +16t 6 and n =6we
have ∆t = 3−0 = 1 and
6 2
L ≈ ∆t
3 f(0) + 4f 1
2 +2f(1) + 4f 3
2 +2f(2) + 4f 5
2 + f(3)
= 1 6
√0+0+0+4· 4
1 2
+9
1 4
+16
1 6
+2· 4(1)
2 2
2
2 +9(1) 4 + 16(1) 6
≈ 86.631
+4·
+4·
4 3
2
4 5
2
2
+9 3
2
2
+9 5
2
4
+16
3 6
+2· 4(2)
2
2 +9(2) 4 +16(2) 6
4
+16
5 6
+
4(3)
2
2 +9(3) 4 +16(3) 6
9. The angle of intersection of the two curves, θ, is the angle between their respective tangents at the point of intersection.
For both curves the point (1, 0, 0) occurs when t =0.
r 0 1(t) =− sin t i +cost j + k ⇒ r 0 1(0) = j + k and r 0 2(t) = i +2t j +3t 2 k ⇒ r 0 2(0) = i.
r 0 1(0) · r 0 2(0) = (j + k) · i =0. Therefore, the curves intersect in a right angle, that is, θ = π 2 .
11. (a) T(t) = r0 (t)
|r 0 (t)| =
t 2 ,t,1
|ht 2 ,t,1i| =
t 2 ,t,1
√
t4 + t 2 +1
(b) T 0 (t) =− 1 2 (t4 + t 2 +1) −3/2 (4t 3 +2t) t 2 ,t,1 +(t 4 + t 2 +1) −1/2 h2t, 1, 0i
−2t 3 − t
=
t 2 ,t,1 1
+
h2t, 1, 0i
(t 4 + t 2 +1) 3/2 (t 4 + t 2 +1)
1/2
−2t 5 − t 3 , −2t 4 − t 2 , −2t 3 − t + 2t 5 +2t 3 +2t, t 4 + t 2 +1, 0
=
|T 0 (t)| =
(c) κ(t) = |T0 (t)|
|r 0 (t)|
(t 4 + t 2 +1) 3/2 =
√ √
4t2 + t 8 − 2t 4 +1+4t 6 +4t 4 + t 2 t8 +4t
=
6 +2t 4 +5t 2
and N(t) =
(t 4 + t 2 +1) 3/2 (t 4 + t 2 +1) 3/2
√
t8 +4t
=
6 +2t 4 +5t 2
(t 4 + t 2 +1) 2
13. y 0 =4x 3 , y 00 =12x 2 and κ(x) =
|y 00 |
[1 + (y 0 ) 2 ] = 12x 2
12
,soκ(1) = 3/2 (1 + 16x 6 )
3/2
17 . 3/2
2t, −t 4 +1, −2t 3 − t
(t 4 + t 2 +1) 3/2
2t, 1 − t 4 , −2t 3 − t
√
t8 +4t 6 +2t 4 +5t 2 .
15. r(t) =hsin 2t, t, cos 2ti ⇒ r 0 (t) =h2cos2t, 1, −2sin2ti ⇒ T(t) = 1 √
5
h2cos2t, 1, −2sin2ti
⇒
T 0 (t) = 1 √
5
h−4sin2t, 0, −4cos2ti ⇒ N(t) =h− sin 2t, 0, − cos 2ti. SoN = N(π) =h0, 0, −1i and