Solução_Calculo_Stewart_6e

F.
274 ¤ CHAPTER 17 VECTOR CALCULUS ET CHAPTER 16
29. (a) F · dr =
1
e t2−1 ,t 5 C 0
TX.10
· 2t,
3t 2 dt =
1
1
2te t2−1 +3t 7 dt = e t2−1 + 3 0
8 t8 = 11 − 1/e
8
0
(b) r(0) = 0, F(r(0)) = e −1 , 0 ;
1
r √
1
2
= , 1
2 2 √ 1
, F r √
2
2
=
r(1) = h1, 1i, F(r(1)) = h1, 1i.
e −1/2 ,
1
4 √ ;
2
In order to generate the graph with Maple, we use the PLOT command
(not to be confused with the plot command) to define each of the vectors.
For example,
v1:=PLOT(CURVES([[0,0],[evalf(1/exp(1)),0]]));
generates the vector from the vector field at the point (0, 0) (but without an arrowhead) and gives it the name v1. Toshow
everything on the same screen, we use the display command. In Mathematica, we use ListPlot (with the
PlotJoined - > True option) to generate the vectors, and then Show to show everything on the same screen.
31. x = e −t cos 4t, y = e −t sin 4t, z = e −t , 0 ≤ t ≤ 2π .
Then dx
dt = e−t (− sin 4t)(4) − e −t cos 4t = −e −t (4 sin 4t +cos4t),
dy
dt = e−t (cos 4t)(4) − e −t sin 4t = −e −t (−4cos4t +sin4t),and dz
dt = −e−t ,so
dx 2
+
dt
2 dy
+
dt
2 dz
= (−e
dt
−t ) 2 [(4 sin 4t +cos4t) 2 +(−4cos4t +sin4t) 2 +1]
= e −t 16(sin 2 4t +cos 2 4t)+sin 2 4t +cos 2 4t +1=3 √ 2 e −t
Therefore
C x3 y 2 zds= 2π
0 (e−t cos 4t) 3 (e −t sin 4t) 2 (e −t )(3 √ 2 e −t ) dt
= 2π
3 √ √
2 e −7t cos 3 4t sin 2 4tdt= 172,704
0 5,632,705 2(1− e −14π )
33. We use the parametrization x =2cost, y =2sint, − π ≤ t ≤ π .Then
2 2
ds = dx
2
dt + dy
2dt
dt = (−2sint)2 +(2cost) 2 dt =2dt,som = kds=2k π/2
dt =2k(π),
C −π/2
x = 1
1 π/2
1
π/2
xk ds = (2 cos t)2 dt =
2πk
C 2π −π/2 2π 4sint = 4 , y = 1
1 π/2
yk ds = (2 sin t)2 dt =0.
−π/2 π 2πk
C 2π −π/2
Hence (x, y) = 4
, 0 .
π
35. (a) x = 1
xρ(x, y, z) ds , y = 1
m C m
C
yρ(x, y, z) ds, z = 1 m
(b) m = kds = k
2π
C 0 4sin 2 t +4cos 2 t +9dt = k √ 13 2π
dt =2πk √ 13,
0
1 2π
x =
2πk √ 2k √
1 2π
13 sin tdt =0, y =
13
2πk √ 2k √ 13 cos tdt =0,
13
z =
1
2πk √ 13
0
2π
0
k √
13 (3t) dt = 3
2π
2 =3π. Hence (x, y, z) =(0, 0, 3π).
2π
0
C
zρ(x, y, z) ds where m = ρ(x, y, z) ds.
C