Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

SECTION 16.2 LINE INTEGRALS 0 305
29. f ( x , y) = x 2 + y 2 => V f( x, y) = 2x i + 2y j . Thus, each vector V f ( x, y) lias the same direction and twice the length of
the position vector of the point (x, y), so the vectors all point directly away from the origin and their lengths increase as we
move away from the origin. Hence, V f is graph III.
31 . f (x, y) = (x + y) 2 => V f (x , y) = 2(x + y) i + 2(x + y) j. The x- andy-components of each vector are equal, so all
vectors are parallel to the line y = x . The vectors are 0 along the line y = --:x and their l~ngth increases as the distance from
this line increases. Thus, V f is graph II.
33. At t = 3 the particle is at (2, 1) so its velocity is V (2, 1) = (4, 3). After 0.01 units of time, the particle's change in
location should be approximately 0.01 V(2, 1) = 0.01 (4, 3) = (0.04, 0.0~) . so the particle should be approximately at the
point (2.04, 1.03) .
35. (a) We sketch the vector field F(x; y ) = x i - y j along with
several approximate flow lines. The flow lines appear to
be hyperbolas with shape similar to the graph of
y = ±11x, so we might guess that the flow lines have
equations y = C / x.
(b) If x = x(t) andy = y(t) are parametric equations of a Aow line, then the velocity vector of the flow line at the
point (x, y) is x' (t) i + y' (t) j . Since the velocity vectors coincide with the vectors in the vector field, we have
x' ( t) i + y' ( t) j = x i - y j => dx Nt = x, dy I dt = - y. To solve these differential equations, we kno~
dxldt = x => dx/ x = dt => In lxl = t +. C => x = ± ~ t + c = Aet for some cons~nt A, and
dy I dt = - y => dy / y = -dt => In I vi = - t + I< => y = ±e- t + K = B e- t for some constant B. Therefore
xy = A et B~- t = AB = constant. If the flow line passes through (1, 1) then (1) (1) = constant= 1 => xy = 1 =>
y = 1l x, x > 0.
16.2 Line Integrals
1. x = t 3 andy = t, 0 :::; t :::; 2, so by Formula 3
.Ia y 3 ds ~ 1\3
( ~~ Y + ( ~~ Y
2
dt = 1
t 3 y'(3t 2 )2 + (1)2 dt = 1\a yf9t 4 + 1 dt
= ..l. . 1 (9t 4 + 1) 3 / 2 ) 2 = ..l. (145 3 1 2 - 1) or ..l. (145 v'I45- 1)
:ill 3 O G4 54
3. Parametric equations for C are x = 4 cos t, y = 4 sin t, - ~ :::; t :::; ~ . Then
fc xy 4 ds = J:~~2 (4 cos t)(4sin t) 4 y'( -4sin t )2 + (4cos t)2 dt = J:~~ 2 45 cost sin 4 t yf16(sin 2 t + cos2 t ) dt
= 4 5 Jrr/ 2 (sin 4 t cos t)(4) dt = (4) 6 [l sinG t) Tr/ 2 = 2 · 40 = 1638.4
- -rr/ 2 5 - Tr/ 2 5
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