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Chapter 15 - SECTION 15.1(PAGE 811 R A ADAMS CALCULUS...

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SECTION 15.1 (PAGE 811) R. A. ADAMS: CALCULUS CHAPTER 15. VECTOR FIELDS Section 15.1 Vector and Scalar Fields (page 811) 1. F = x i + x j . The field lines satisfy dx x = dy x , i.e., dy = dx . The field lines are y = x + C , straight lines parallel to y = x . y x Fig. 15.1.1 2. F = x i + y j . The field lines satisfy dx x = dy y . Thus ln y = ln x + ln C , or y = Cx . The field lines are straight half-lines emanating from the origin. y x Fig. 15.1.2 3. F = y i + x j . The field lines satisfy dx y = dy x . Thus x dx = y dy . The field lines are the rectangular hyperbolas (and their asymptotes) given by x 2 y 2 = C . y x Fig. 15.1.3 4. F = i + sin x j . The field lines satisfy dx = dy sin x . Thus dy dx = sin x . The field lines are the curves y = − cos x + C . y x Fig. 15.1.4 5. F = e x i + e x j . The field lines satisfy dx e x = dy e x . Thus dy dx = e 2 x . The field lines are the curves y = − 1 2 e 2 x + C . y x Fig. 15.1.5 6. F = ∇ ( x 2 y ) = 2 x i j . The field lines satisfy dx 2 x = dy 1 . They are the curves y = − 1 2 ln x + C . y x Fig. 15.1.6 570
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INSTRUCTOR’S SOLUTIONS MANUAL SECTION 15.1 (PAGE 811) 7. F = ∇ ln ( x 2 + y 2 ) = 2 x i + 2 y j x 2 + y 2 . The field lines satisfy dx x = dy y . Thus they are radial lines y = Cx (and x = 0) y x Fig. 15.1.7 8. F = cos y i cos x j . The field lines satisfy dx cos y = − dy cos x , that is, cos x dx + cos y dy = 0. Thus they are the curves sin x + sin y = C . y x Fig. 15.1.8 9. v ( x , y , z ) = y i y j y k . The streamlines satisfy dx = − dy = − dz . Thus y + x = C 1 , z + x = C 2 . The streamlines are straight lines parallel to i j k . 10. v ( x , y , z ) = x i + y j x k . The streamlines satisfy dx x = dy y = − dz x . Thus z + x = C 1 , y = C 2 x . The streamlines are straight half- lines emanating from the z -axis and perpendicular to the vector i + k . 11. v ( x , y , z ) = y i x j + k . The streamlines satisfy dx y = − dy x = dz . Thus x dx + y dy = 0, so x 2 + y 2 = C 2 1 . Therefore, dz dx = 1 y = 1 C 2 1 x 2 . This implies that z = sin 1 x C 1 + C 2 . The streamlines are the spirals in which the surfaces x = C 1 sin ( z C 2 ) intersect the cylinders x 2 + y 2 = C 2 1 . 12. v = x i + y j ( 1 + z 2 )( x 2 + y 2 ) . The streamlines satisfy dz = 0 and dx x = dy y . Thus z = C 1 and y = C 2 x . The streamlines are horizontal half-lines emanating from the z -axis. 13. v = xz i + yz j + x k . The field lines satisfy dx xz = dy yz = dz x , or, equivalently, dx / x = dy / y and dx = z dz . Thus the field lines have equations y = C 1 x , 2 x = z 2 + C 2 , and are therefore parabolas. 14. v = e xyz ( x i + y 2 j + z k ) . The field lines satisfy dx x = dy y 2 = dz z , so they are given by z = C 1 x , ln | x | = ln | C 2 | − ( 1 / y ) (or, equivalently, x = C 2 e 1 / y ). 15. v ( x , y ) = x 2 i y j . The field lines sat- isfy dx / x 2 = dy / y , so they are given by ln | y | = ( 1 / x ) + ln | C | , or y = Ce 1 / x . 16. v ( x , y ) = x i + ( x + y ) j . The field lines satisfy dx x = dy x + y dy dx = x + y x Let y = x v( x ) dy dx = v + x d v dx v + x d v dx = x ( 1 + v) x = 1 + v. Thus d v/ dx = 1 / x , and so v( x ) = ln | x | + C . The field lines have equations y = x ln | x | + Cx . 17. F = ˆ r + r ˆ θ . The field lines satisfy dr = d θ , so they are the spirals r = θ + C .
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