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

310 0 CHAPTER 16 VECTOR CALCULUS
51. The work done in moving the object is fc F · dr = fc F · T ds. We can approximate this integral by dividing C. into
7 segments of equal length t:.s = 2 and approximating F · T , that is, the tangential component of force, at a point (xi , yi) on
each segment. Since C is composed of straight line segments, F · T is the scalar projection of each force vector onto C.
If we choose (xi, yi) to be the point on the segment closest to the origin, then the work done is .
7
fc F · T d.s ~ E (F(xi, yi) · T(xi, yi)] t:.s = (2 + 2 + 2 + 2+ 1 + 1 + 1)(2) = 22. Thus, we estimate the work done to
i=l .
be approximately 22 J.
16.3 The Fundamental Theorem for Line Integrals
1. C appears to be a smooth curve, and since \J f is continuous, we know f is differentiable. Then Theorem 2 says that the value
of fc \J f · dr is simply the difference of the values off at the terminal and initial points of C. From the graph, this is
50- 10 = 40.
3. 8(2x- 3y)j8y = - 3 = 8( -3x + 4y- 8)j8x and the domain ofF is IR 2 which is open and simply-connected, so by
Theorem 6 F is conservative. Thus, there exists a function f such that \J f = F , that is, f:r:(x, y) = 2x -
3y and
f 11
(x, y) = -3x + 4y - 8. But f,.(x, y) = 2x - 3y implies f(x, y) = x 2 -
3xy + g(y) and differentiating both sides of this
equation with respect to y gives f 11
( x, y) = - 3x + g' (y). Thus - 3x + 4y - 8 = -3:~.: + g' (y) so g' (y ) = 4y - 8 and
g(y) = 2y 2 - By+ [(where I< is a constant. Hence f(x, y) = x 2 - 3xy + 2y 2 - 8y +I< is a potential function for F .
5. 8(e"' cosy)j8y := - e"' sin y, 8(e"' siny)f8x = e" sin y. Since these are not equal, F is not conservative.
7. o(ye"' + sin y)j8y =ex+ cosy= o(e"' + X cosy)f8x and the domain ofF is IR 2 • Hence F is conservative so there
exists a function f such that \J f =F. Then fx (x, y) = ye" +sin y implies f(x, y) = ye" + x sin y + g(y) and
fv(x, y) = e"' + xcos y + g'(y). But f 11 (x, y) = e"' + x cosy so g(y) = K and f(x, y) = ye" + xsin y + K is a potential
function for F.
9. 8(lny + 2xy 3 )j8y = 1/y + 6xy 2 = 8(3x 2 y 2 + xfy)j8x and the domain ofF is {(x, y) I y> 0} which is open and simply
connected. Hence F is conservative so there exists a function f such that \J f = F . Then f:z:(x, y) =In y + 2xy 3 implies
f(x,y) = xlny + x 2 y 3 + g(y) and ! 11
(x,y) = xfy + 3x 2 y 2 + g'(y). But f 11 (x,y) = 3x 2 y 2 + xjy so g'(y) = 0 =}
g(y) = [(and f(x, y) = xlny + x 2 y 3 + [(is a potential function for F .
11. (a) F has continuous first-order partial derivatives and 8
8 2xy = 2x = 8
8 (x
2 ) on JR 2 , which is open and simply-connected.
y X .
Thus, F is conservative by Theorem 6. Then we know that the line integral ofF is independent of path; in particular, the
value of fc F · dr depends only on the endpoints of C. Sin.ce all three curves have the same initial and terminal points,
J~ F · dr will have the same value for each curve.
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