Solução_Calculo_Stewart_6e

F.
TX.10 SECTION 14.3 ARCLENGTHANDCURVATURE ET SECTION 13.3 ¤ 149
21. r(t) =t 2 i + t k ⇒ r 0 (t) =2t i + k, r 00 (t) =2i, |r 0 (t)| = (2t) 2 +0 2 +1 2 = √ 4t 2 +1, r 0 (t) × r 00 (t) =2j,
|r 0 (t) × r 00 (t)| =2.Thenκ(t) = |r0 (t) × r 00 (t)| 2
|r 0 (t)| 3 = √
4t2 +1 3 = 2
(4t 2 +1) . 3/2
23. r(t) =3t i +4sint j +4cost k ⇒ r 0 (t) =3i +4cost j − 4sint k, r 00 (t) =−4sint j − 4cost k,
|r 0 (t)| = 9+16cos 2 t +16sin 2 t = √ 9+16=5, r 0 (t) × r 00 (t) =−16 i +12cost j − 12 sin t k,
|r 0 (t) × r 00 (t)| = 256 + 144 cos 2 t +144sin 2 t = √ 400 = 20. Thenκ(t) = |r0 (t) × r 00 (t)|
|r 0 (t)| 3 = 20
5 3 = 4 25 .
25. r(t) = t, t 2 ,t 3 ⇒ r 0 (t) = 1, 2t, 3t 2 . The point (1, 1, 1) corresponds to t =1,andr 0 (1) = h1, 2, 3i ⇒
|r 0 (1)| = √ 1+4+9= √ 14. r 00 (t) =h0, 2, 6ti ⇒ r 00 (1) = h0, 2, 6i. r 0 (1) × r 00 (1) = h6, −6, 2i, so
|r 0 (1) × r 00 (1)| = √ 36 + 36 + 4 = √ √
76. Thenκ(1) = |r0 (1) × r 00 (1)| 76
|r 0 (1)| 3 = √ 3 = 1
19
14 7 14 .
27. f(x) =2x − x 2 , f 0 (x) =2− 2x, f 00 (x) =−2,
κ(x) =
|f 00 (x)|
[1 + (f 0 (x)) 2 ] 3/2 = |−2|
[1 + (2 − 2x) 2 ] 3/2 = 2
(4x 2 − 8x +5) 3/2
29. f(x) =4x 5/2 , f 0 (x) =10x 3/2 , f 00 (x) =15x 1/2 ,
15x
|f 00 1/2
(x)|
κ(x) =
[1 + (f 0 (x)) 2 ] =
3/2 [1 + (10x 3/2 ) 2 ] = 15 √ x
3/2 (1 + 100x 3 ) 3/2
31. Since y 0 = y 00 = e x , the curvature is κ(x) =
|y 00 (x)|
[1 + (y 0 (x)) 2 ] = e x
3/2 (1 + e 2x ) = 3/2 ex (1 + e 2x ) −3/2 .
To find the maximum curvature, we first find the critical numbers of κ(x):
κ 0 (x) =e x (1 + e 2x ) −3/2 + e x − 3 2
(1 + e 2x ) −5/2 (2e 2x )=e x 1+e 2x − 3e 2x
(1 + e 2x ) 5/2 = e x 1 − 2e 2x
(1 + e 2x ) 5/2 .
κ 0 (x) =0when 1 − 2e 2x =0,soe 2x = 1 or x = − 1 ln 2. Andsince1 − 2 2 2e2x > 0 for x− 1 ln 2, the maximum curvature is attained at the point − 1 ln 2 2 2,e(− ln 2)/2 = − 1 ln 2, √2
1
.
2
Since lim
x→∞ ex (1 + e 2x ) −3/2 =0,κ(x) approaches 0 as x →∞.
33. (a) C appears to be changing direction more quickly at P than Q, so we would expect the curvature to be greater at P .
(b) First we sketch approximate osculating circles at P and Q. Usingthe
axes scale as a guide, we measure the radius of the osculating circle
at P to be approximately 0.8 units, thus ρ = 1 κ
⇒
κ = 1 ρ ≈ 1 ≈ 1.3. Similarly, we estimate the radius of the
0.8
osculating circle at Q to be 1.4 units, so κ = 1 ρ ≈ 1
1.4 ≈ 0.7.