Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 17.2 LINE INTEGRALS ET SECTION 16.2 ¤ 271
29. f(x, y) =x 2 + y 2 ⇒ ∇f(x, y) =2x i +2y j. Thus, each vector ∇f(x, y) has 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, ∇f is graph II.
31. f(x, y) =(x + y) 2 ⇒ ∇f(x, y) =2(x + y) i +2(x + y) j. Thex-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 length increases as the distance from
this line increases. Thus, ∇f is graph II.
33. At t =3the particle is at (2, 1) so its velocity is V(2, 1) = h4, 3i. After 0.01 units of time, the particle’s change in
location should be approximately 0.01 V(2, 1) = 0.01 h4, 3i = h0.04, 0.03i, 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 = ±1/x,
so we might guess that the flow lines have equations
y = C/x.
(b) If x = x(t) and y = y(t) are parametric equations of a flow line, then the velocity vector of the flow line at the
point (x, y) is x 0 (t) i + y 0 (t) j. Since the velocity vectors coincide with the vectors in the vector field, we have
x 0 (t) i + y 0 (t) j = x i − y j ⇒ dx/dt = x, dy/dt = −y. To solve these differential equations, we know
dx/dt = x ⇒ dx/x = dt ⇒ ln |x| = t + C ⇒ x = ±e t + C = Ae t for some constant A,and
dy/dt = −y ⇒ dy/y = −dt ⇒ ln |y| = −t + K ⇒ y = ±e −t + K = Be −t for some constant B. Therefore
xy = Ae t Be −t = AB = constant. If the flow line passes through (1, 1) then (1) (1) = constant =1 ⇒ xy =1 ⇒
y =1/x, x>0.
17.2 Line Integrals ET 16.2
1. x = t 3 and y = t, 0 ≤ t ≤ 2,sobyFormula3
C y3 ds =
2 dx
2
0 t3 dt + dy
2 2
dt dt =
0 t3 (3t 2 ) 2 +(1) 2 dt = 2
√
0 t3 9t 4 +1dt
= 1 · 2
36 3 9t 4 +1 3/2
2
√
= 1
54 (1453/2 − 1) or 1
54 145 145 − 1
0
3. Parametric equations for C are x =4cost, y =4sint, − π ≤ t ≤ π .Then
2 2
C xy4 ds = π/2
(4 cos t)(4 sin t)4 (−4sint)
−π/2 2 +(4cost) 2 dt = π/2
−π/2 45 cos t sin 4 t 16(sin 2 t +cos 2 t) dt
=4 5 π/2
−π/2 (sin4 t cos t)(4) dt =(4) 6 1
5 sin5 t π/2
= 2 · 46 =1638.4
−π/2 5