Solução_Calculo_Stewart_6e

F.
47.
49.
TX.10 SECTION 14.3 ARCLENGTHANDCURVATURE ET SECTION 13.3 ¤ 147
d
dt [r(t) × r0 (t)] = r 0 (t) × r 0 (t)+r(t) × r 00 (t) by Formula 5 of Theorem 3. But r 0 (t) × r 0 (t) =0 (by Example 2 in
Section 13.4 [ET 12.4]). Thus,
d
dt [r(t) × r0 (t)] = r(t) × r 00 (t).
d
dt |r(t)| = d dt [r(t) · r(t)]1/2 = 1 [r(t) · 2 r(t)]−1/2 [2r(t) · r 0 (t)] = 1
|r(t)| r(t) · r0 (t)
51. Since u(t) =r(t) · [r 0 (t) × r 00 (t)],
u 0 (t) = r 0 (t) · [r 0 (t) × r 00 (t)] + r(t) · d
dt [r0 (t) × r 00 (t)]
=0+r(t) · [r 00 (t) × r 00 (t)+r 0 (t) × r 000 (t)]
[since r 0 (t) ⊥ r 0 (t) × r 00 (t)]
= r(t) · [r 0 (t) × r 000 (t)] [since r 00 (t) × r 00 (t) =0]
14.3 Arc Length and Curvature ET 13.3
1. r(t) =h2sint, 5t, 2costi ⇒ r 0 (t) =h2cost, 5, −2sinti ⇒ |r 0 (t)| = (2 cos t) 2 +5 2 +(−2sint) 2 = √ 29.
Then using Formula 3, we have L = 10
−10 |r0 (t)| dt = 10
√ √ 10
−10 29 dt = 29 t =20√ 29.
−10
3. r(t) = √ 2 t i + e t j + e −t k ⇒ r 0 (t) = √ 2 i + e t j − e −t k ⇒
|r 0 (t)| =
√2 2
+(et ) 2 +(−e −t ) 2 = √ 2+e 2t + e −2t = (e t + e −t ) 2 = e t + e −t [since e t + e −t > 0].
Then L = 1
0 |r0 (t)| dt = 1
0 (et + e −t ) dt = e t − e −t 1
0 = e − e−1 .
5. r(t) =i + t 2 j + t 3 k ⇒ r 0 (t) =2t j +3t 2 k ⇒ |r 0 (t)| = √ 4t 2 +9t 4 = t √ 4+9t 2 [since t ≥ 0].
Then L = 1
0 |r0 (t)| dt = 1
t √ 1
4+9t
0 2 dt = 1 · 2 (4 + 18 3 9t2 ) 3/2 = 1 27 (133/2 − 4 3/2 )= 1
27 (133/2 − 8).
0
2
+12 +(2t) 2 1
=
4t +1+4t2 ,so
7. r(t) = √ t, t, t 2 1
⇒ r 0 (t) =
2 √ t , 1, 2t ⇒ |r 0 (t)| =
L = 4
1 |r0 (t)| dt =
4 1
1 4t +1+4t2 dt ≈ 15.3841.
1
2 √ t
9. r(t) =hsin t, cos t, tan ti ⇒ r 0 (t) = cos t, − sin t, sec 2 t ⇒
|r 0 (t)| = cos 2 t +(− sin t) 2 +(sec 2 t) 2 = √ 1+sec 4 t and L = π/4
|r 0 (t)| dt = π/4
√
1+sec4 tdt≈ 1.2780.
0 0
11. The projection of the curve C onto the xy-plane is the curve x 2 =2y or y = 1 2 x2 , z =0. Then we can choose the parameter
x = t ⇒ y = 1 2 t2 .SinceC also lies on the surface 3z = xy,wehavez = 1 3 xy = 1 3 (t)( 1 2 t2 )= 1 6 t3 . Then parametric
equations for C are x = t, y = 1 2 t2 , z = 1 6 t3 and the corresponding vector equation is r(t) = t, 1 2 t2 , 1 t3 . The origin
6
corresponds to t =0and the point (6, 18, 36) corresponds to t =6,so
L = 6
0 |r0 (t)| dt = 6
1,t, 1 t2 dt =
6
1
0
2 0
2 + t 2 + 1
2 t22 dt =
6
1+t
0
2 + 1 4 t4 dt
= 6
0
(1 + 1 2 t2 ) 2 dt = 6
0 (1 + 1 2 t2 ) dt = t + 1 6 t3 6
0 =6+36=42