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Concepts in Calculus III Solutions Manual

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Concepts in Calculus III Solutions Manual Concepts in Calculus III Multivariable Calculus Solutions Manual K. DeMason and S. V. Shabanov Department of Mathematics, University of Florida, Gainesville, FL 32611 USA For the Latest Edition of the Textbook: 2019 Fist Edition of the Textbook: University of Florida Press, Gainesville, 2012 ISBN 978–1–61610–162–6 c 2017 Sergei Shabanov, All Rights Reserved Contents Chapter 1. Vectors and the Space Geometry 1 1. Rectangular Coordinates in Space 3 2. Vectors in Space 2 3. The Dot Product 41 4. The Cross Product 60 5. The Triple Product 78 6. Planes in Space 96 7. Lines in Space 108 8. Euclidean spaces 124 9. Quadric Surfaces 132 Chapter 2. Vector Functions 155 10. Curves in Space and Vector Functions 155 11. Differentiation of Vector Functions 171 12. Integration of Vector Functions 183 13. Arc Length of a Curve 193 14. Curvature of a Space Curve 203 15. Applications to Mechanics and Geometry 218 Chapter 3. Differentiation of Multivariable Functions 239 16. Functions of Several Variables 239 17. Limits and Continuity 251 18. A General Strategy to Study Limits 265 19. Partial Derivatives 280 20. Higher-Order Partial Derivatives 286 21. Differentiability of Multivariable Functions 297 22. Chain Rules and Implicit Differentiation 315 23. The Differentials and Taylor Polynomials 333 24. Directional Derivative and the Gradient 358 25. Maximum and Minimum Values 375 26. Extreme Values on a Set 390 27. Lagrange Multipliers 405 Chapter 4. Multiple Integrals 429 28. Double Integrals 429 29. Properties of the Double Integral 443 1 2 CONTENTS 30. Iterated Integrals 452 31. Double Integrals over General Regions 460 32. Double Integrals in Polar Coordinates 476 33. Change of Variables in Double Integrals 490 34. Triple Integrals 509 35. Triple Integrals in Cylindrical and Spherical Coordinates 524 36. Change of Variables in Triple Integrals 539 37. Improper Multiple Integrals 552 38. Line Integrals 568 39. Surface Integrals 575 40. Moments of Inertia and Center of Mass 592 Chapter 5. Vector Calculus 607 41. Line Integral of a Vector Field 607 42. Fundamental Theorem for Line Integrals 619 43. Green’s Theorem 633 44. Flux of a Vector Field 645 45. Stokes’ Theorem 657 46. Gauss-Ostrogradsky (Divergence) Theorem 669 Acknowledgments 681 CHAPTER 1 Vectors and the Space Geometry 1. Rectangular Coordinates in Space 1–2. Find the distance between the specified points. 1. (1, 2, 3) and ( 1, 0, 2) − − − Solution: Let d denote the distance between the given points. Then, by the distance formula, d = ( 1 1)2 + (0 2)2 +( 2+3)2 = √4+4+1= √9 = 3 − − − − p 2. ( 1, 3, 2) and ( 1, 2, 1) − − − − Solution: Let d denote the distance between the given points. Then, by the distance formula, d = ( 1+1)2 + (2 3)2 +( 1+2)2 = √0+1+2= √2 − − − 3–4. Determinep whether the given points lie in a straight line. 3. (8, 3, 3), ( 1, 6, 3), and (2, 5, 1) − − Solution: Given three points A, B, and C in Rn, we can determine if they are colinear if AB + BC = AC . | | | | | | Therefore, we must compute the distances between each pair of points and see if one is the sum of the other two. Let A = (8, 3, 3), B = ( 1, 6, 3), and C = (2, 5, 1). By the distance formula,− − AB = (8+1)2 + (3 6)2 +( 3 3)2 = √81+9+36= √126 | | − − − BC = p( 1 2)2 + (6 5)2 + (3 1)2 = √9+1+4= √14 | | − − − − AC = p(8 2)2 + (3 5)2 +( 3 1)2 = √36+4+16= √56 | | − − − − p 3 4 1. VECTORS AND THE SPACE GEOMETRY Note that AB = BC + AC , so they are colinear. To see this, consider| squaring| | the| | right| hand side, (√14 + √56)2 = 14 + 2√14√56 + 56 126 = 70 + 2√7 2 7 8 · · · 126 = 70 + 2(7)(4) = 70 + 56 = 126 4. ( 1, 4, 2), (1, 2, 2), and ( 1, 2, 1) − − − − Solution: Given three points A, B, and C in Rn, we can determine if they are colinear if AB + BC = AC . | | | | | | Therefore, we must compute the distances between each pair of points and see if one is the sum of the other two. Let A = ( 1, 4, 2), B = (1, 2, 2), and C = ( 1, 2, 1). By the distance− formula,− − − AB = ( 1 1)2 + (4 2)2 +( 2 2)2 = √4+4+16= √24 | | − − − − − BC = p(1+1)2 + (2 2)2 +(2+1)2 = √4+0+9= √13 | | − AC = p( 1+1)2 + (4 2)2 +( 2+1)2 = √0+4+1= √5 | | − − − Notep that no one distance is the sum of the other two, hence the points are not colinear. 5. Determine whether the points (1, 2, 3) and (3, 2, 1) lie in the plane through O = (1, 1, 1) and perpendicular to the line through O and B = (2, 2, 2). If not, find a point in this plane. Solution: Let P1 = (1, 2, 3) and P2 = (3, 2, 1). Let be the plane that passes through a given point O and is perpendicularP to the line through O and B. An arbitrary point P R3 will be in iff OP,OB, and BP form a right triangle. Clearly, OP∈ and OB mustP be perpen- dicular, hence we must have OB 2 + OP 2 = BP 2 | | | | | | First checking P1, OB = (2 1)2 + (2 1)2 + (2 1)2 = √3, | | − − − BP = p(2 1)2 + (2 2)2 + (2 3)2 = √2, | 1| − − − 2 2 2 OP1 = p(1 1) + (1 2) + (1 3) = √5. | | − − − p 1. RECTANGULAR COORDINATES IN SPACE 5 Since 3 + 5 = 2, P1 / . A similar argument shows that P2 / . It turns out that,6 in this∈ Pproblem, OP = OP and BP = BP∈ P. | 1| | 2| | 1| | 2| To find a point in , let P =(x,y,z). Then we must have that P 3+((1 x)2 + (1 y)2 + (1 z)2) = ((2 x)2 + (2 y)2 + (2 z)2) − − − − − − (2 x)2 (1 x)2 + (2 y)2 (1 y)2 + (2 z)2 (1 z)2 3 = 0 − − − − − − − − − − (2 x (1 x))(2 x + 1 x)+(2 y (1 y))(2 y + 1 y)+ − − − − − − − − − − (2 z (1 z))(2 z + 1 z) 3 = 0 − − − − − − 3 2x + 3 2y + 3 2z 3 = 0 − − − − 2x + 2y + 2z = 6 x + y + z = 3 Thus any point P =(x,y,z) satisfying x + y + z = 3 will be in . For example, (3, 0, 0) . In the future, you will learn an easierP way to arrive at this formula.∈ P 6. Find the distance from the point (1, 2, 3) to each of the coor- dinate planes and to each of the coordinate axes.− Solution: Let P = (1, 2, 3). To find the distance between a point − and a plane, we must find a point P ∗ on the plane such that P P ∗ is perpendicular to the plane. For the coordinate planes, this is easy – simply project the point onto the plane. Let dxz represent the distance between P and the projection of P onto the xz plane, P ∗ = (1, 0, 3). Thus, by the distance formula, − d = (1 1)2 + (2 0)2 + (3 ( 3))2 = 2 xz − − − − Similarly, dyz and dpxy are as follows d = (1 0)2 + (2 2)2 + (3 ( 3))2 = 1 yz − − − − 2 2 2 dxy = p(1 1) + (2 2) + (3 0) = 3 − − − To find the distancep between P and a coordinate axis, we must find our P ∗ and project it again onto the desired axis. For example, find the distance between P and the x-axis, we must project P onto the xz plane (P ∗ = (1, 0, 3)), and then project it again onto the x axis − (P ∗∗ = (1, 0, 0)). Thus, dx, the distance between P and the x-axis is d = (1 1)2 + (2 0)2 + (3 0)2 = √13 x − − − p 6 1. VECTORS AND THE SPACE GEOMETRY Similarly, dy and dz are as follows d = (1 0)2 + (2 2)2 + (3 0)2 = √10 y − − − 2 2 2 dz = p(1 0) + (2 0) + (3 3) = √5 − − − p 7. Find the length of the medians of the triangle with vertices A = (1, 2, 3), B =( 3, 2, 1), and C =( 1, 4, 1). − − − − Solution: The coordinates of the midpoint of a straight line segment with given endpoints are the half-sums of the corresponding coordi- nates of the endpoints. Let Pa, Pb, and Pc be the midpoints of BC, AC, and AB, respectively. Then P =( 2, 1, 0) , P = (0, 1, 2) , P =( 1, 2, 1) . a − − b − c − By the distance formula, AP = ( 2 1)2 +( 1 2)2 + (3 0)2 = √9+9+9=3√3 , | a| − − − − − BP = p(0+3)2 +( 1 2)2 +(2+1)2 = √9+9+9=3√3 . | b| − − 2 2 2 2 CPc = p( 1+1) +(2+4) + (1 1) = √0 + 6 +0=6 . | | − − 8. Let thep set consist of points (t, 2t, 3t) where <t< . Find the point of thatS is the closest to the point (3, 2, 1).−∞ Sketch the∞ set . S S Solution: I will begin by answering the second part. Consider the following: t(1, 2, 3). This is a restatement of the given set of points. Interpreting (1, 2, 3) as a vector in space, we can see that t acts as a scalar. Thus the set of points is a straight line parallel to the vector 1, 2, 3 . To minimize the distance between our curve and the given pointh Pi = (3, 2, 1), we must find a point Q on our curve such that the function d = PQ is minimized.
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