Solução_Calculo_Stewart_6e

F.
plane is −6(x − 0) + 1(y − π)+0(z +2)=0or y − 6x = π.
TX.10 SECTION 14.3 ARCLENGTHANDCURVATURE ET SECTION 13.3 ¤ 151
T 0 (t) = 1 √
37
h−18 sin 3t, 0, −18 cos 3ti ⇒ |T 0 (t)| =
182 sin 2 3t +18 2 cos 2 3t
√
37
= 18 √
37
⇒
N(t) = T0 (t)
= h− sin 3t, 0, − cos 3ti. SoN(π) =h0, 0, 1i and B(π) =
|T 0 √
1
(t)| 37
h−6, 1, 0i×h0, 0, 1i = √ 1
37
h1, 6, 0i.
Since B(π) is a normal to the osculating plane, so is h1, 6, 0i.
An equation for the plane is 1(x − 0) + 6(y − π)+0(z +2)=0or x +6y =6π.
47. The ellipse 9x 2 +4y 2 =36is given by the parametric equations x =2cost, y =3sint, so using the result from
Exercise 40,
|ẋÿ − ẍẏ| |(−2sint)(−3sint) − (3 cos t)(−2cost)|
6
κ(t) =
=
[ẋ 2 + ẏ 2 ]
3/2
(4 sin 2 =
t +9cos 2 t) 3/2 (4 sin 2 t +9cos 2 t) . 3/2
At (2, 0), t =0.Nowκ(0) = 6 = 2 , so the radius of the osculating circle is
27 9
1/κ(0) = 9 and its center is 2
− 5 , 0 2
. Its equation is therefore
x + 5 2
2
+ y 2 = 81 . 4
At (0, 3), t = π ,andκ
π
2 2 =
6
= 3 . So the radius of the osculating circle is 4 and
8 4 3
its center is 0, 5 3
. Hence its equation is x 2 + y − 5 3
2
= 16 9 .
49. The tangent vector is normal to the normal plane, and the vector h6, 6, −8i is normal to the given plane.
But T(t) k r 0 (t) and h6, 6, −8i kh3, 3, −4i, so we need to find t such that r 0 (t) kh3, 3, −4i.
r(t) = t 3 , 3t, t 4 ⇒ r 0 (t) = 3t 2 , 3, 4t 3 kh3, 3, −4i when t = −1. So the planes are parallel at the point (−1, −3, 1).
dT
51. κ =
dT
ds = dT/dt
ds/dt = |dT/dt| and N = dT/dt
ds/dt |dT/dt| ,soκN = dt dT
dt
dT
dt ds = dT/dt
ds/dt = dT by the Chain Rule.
ds
dt
53. (a) |B| =1 ⇒ B · B =1 ⇒ d ds
(b) B = T × N
⇒
(B · B) =0 ⇒ 2
dB
ds
· B =0 ⇒
dB
ds ⊥ B
dB
ds = d ds (T × N) = d dt (T × N) 1
ds/dt = d dt (T × N) 1
|r 0 (t)| =[(T0 × N)+(T × N 0 1
)]
|r 0 (t)|
=
T 0 × T0
+(T × N 0 1
)
|T 0 |
|r 0 (t)| = T × N0
|r 0 (t)|
⇒
dB
ds ⊥ T
(c) B = T × N ⇒ T ⊥ N, B ⊥ T and B ⊥ N. SoB, T and N form an orthogonal set of vectors in the threedimensional
space R 3 . From parts (a) and (b), dB/ds is perpendicular to both B and T, sodB/ds is parallel to N.
Therefore, dB/ds = −τ(s)N,whereτ(s) is a scalar.
(d) Since B = T × N, T ⊥ N and both T and N are unit vectors, B is a unit vector mutually perpendicular to both T and
N. Foraplanecurve,T and N always lie in the plane of the curve, so that B is a constant unit vector always
perpendicular to the plane. Thus dB/ds = 0,butdB/ds = −τ(s)N and N 6=0,soτ(s) =0.