Exam (elaborations) TEST BANK FOR Vector Calculus, Linear Algebra and Differential Forms 2nd Edition By Hubbard J.H (Solution manual)

Exam (elaborations) TEST BANK FOR Vector Calculus, Linear Algebra and Differential Forms 2nd Edition By Hubbard J.H (Solution manual) CONTENTS How to Use This Book IV Acknowledgments IV CHAPTER 1 The Geometry of Euclidean Space 1 CHAPTER 2 Differentiation 21 CHAPTER 3 Higher-Order Derivatives; Maxima and Minima 43 CHAPTER 4 Vector-Valued Functions 63 CHAPTER 5 Double and Triple Integrals 77 CHAPTER 6 The Change of Variables Formula and Applications of Integration 97 CHAPTER 7 Integrals over Paths and Surfaces 17 CHAPTER 8 The Integral Theorems of Vector Analysis 143 CHAPTER 9 Sample Exams 161 APPENDIX Answers to Chapter Te ts and Sample Exams 167 1 THE GEOMETRY OF EUCLIDEAN SPACE 1.1: VECTORS IN TWO-AND TH REE-DIM ENSIONAL SPACE GOALS 1. Be able to perform the following operations on vectors: addition, subtraction, scalar multiplication. 2. Given a vector and a point, be able to write the equation of the line passing through the point in the direction of the vector. 3. Given two points, be able to write the equation of the line passing through them. STUDY HINTS 1. Space notation. The symbol ]R or ]R I refers to all points on the real number line or a onedimensional space. ]R 2 refers to all ordered pairs (X, y) which lie in the plane, a two-dimensional space. ]R3 refers to all ordered triples (x, y, z) which lie in three-dimensional space. In general, the "exponent" in ]R71 tells you how many components there are in each vector. 2. Vectors and scalars. A vector has both length (magnitude) and direction. Scalars are just ~umbers. Sc~lars do not have direction. Two vectors are equal If and only If they both have the same length and the same ~~ direction. Pictorially, they do not need to originate from the same starting point. The vectors shown here are equal. 3. Vector notation. Vectors are often denoted by boldface letters, underlined letters, arrows over letters, or by an n-tuple (Xl, X2 , ... , x71) . Each Xi of the n-tuple is called the lth component. BEWARE that the n-tuple may represent either a point or a vector. The vector (0,0, ... ,0) is denoted O. Your instructor or other textbooks may use other notations such as a squiggly line underneath a letter. A circumflex over a letter is sometimes used to represent a unit vector. 4. Vector addition. Vectors may be added componentwise, e.g., in ]R2 (XI , yr) +(X2, Y2) =(Xl +YI,X2+Y2). Pictorially, two vectors may be thought of as the sides of a parallelogram. Starting from the vertex formed by the two vectors, we form a new vector which ends at the opposite corner ;;;?J v of the parallelogram. This new vector is the sum of the other two. Alternatively, one could simply translate v so that the tail of v meets the head of u. The vector joining the tail of u U I to the head of v is u + v. £?:j U + V I ______ __ 1 CHAPTER 1 2 5. Vector subtmction. Just as with addition, vectors may be subtracted componentwise. Think of this as adding a negative vector. Pictorially, the vectors a, b a b and a -b form a triangle. To determine the correct direction, hb ~you should Ibe able to add a b and b to get a. a-- Thus a b ~ go," fom the tip of b to the tip of 8. 6. Scalar multiplication. Here, each component of a vector is multiplied by the same scalar, e.g., in JR. 2, r(x , y) = (rx, ry) for any real number r . The effect of multiplication by a positive scalar is t.o change the length by a factor. If the scalar is negative, the lengthening occurs in the opposite direction. Multiplication of vectors will be discussed in the next two sections. 7. Standard basis vectors. These are vectors whose components are all 0 except for a single 1. In JR.3, i, j and k denote the vectors which lie on the x, y and z axes. They are (1,0,0), (0, 1,0) and (0,0, 1), respectively. The standard basis vectors in JR. 2 are i and j , which are vectors lying on the x and y axes, and their respective components are (1 , 0) and (0,1). Sometimes, these vectors are denoted by 8. Lines. (a) The line passing through a in the direction of v is l(t) = a +tv. This is called the point-direction form of the line because the only necessary information is the point a and the direction of v. (b) The line passing through a and b is l(t) =a+t(b-a). This is called the point-point form of the line. To see if the direction is correct, plug in t = 0 and you should get the first point. Plug in t = 1 and you should get the second point. 9. Spanning a space. If all points in a space can be written in the form AIVI + A2V2 + . .. + An Vn , where Ai are scalars, then the vectors VI, ..., Vn span that given space. For example, the vectors i and j span the xy plane. 10. Geometric proofs. The use of vectors can often simplify a proof. Try to compare vector methods and non-vector methods by doing example 10 without vectors. SOLUTIONS TO SELECTED EX ERCISES 1. We must solve the following equations: -21-x -25 23 -6 y. We get x =4 and y =17, so (-21,23) -(4,6) =(-25,17). 4. Convert -4i +3j to (-4, 3,0), so (2,3,5) -4i +3j =(2,3,5) + (-4,3,0) =(-2,6,5). 3 THE GEOMETRY OF EUCLIDEAN SPACE 7. To sketch v, tart at the origin and move 2 units along the x axis, then move 3 units parallel to the y axis, and then move -6 units parallel to the z axis. The vector w is sketched analogously. The vector -v has the same length as v, but it points in the opposite direction. To sketch v + w, translate the tail of w to the head of v and draw the vector from the origin to the head of the translated w. The vector v -w goes from the y head of w to the head of v. 9. On the y axis, points have the coordinat s (0, y, 0), so we must restrict x and z to be 0. On the z axis, points have z (~~~'51_.(O,y,z) the coordinates (0,0, z) , so we must restrict x and y to be (x,O,z)// y 0. In the xz plane, points have the coordinates (:I:,O,z). so ( (o, y, o) we must restrict y to be 0. In the yz plane, points have the coordinates (0, y , z), so we must restrict x to be O. x 12. Every point on the plane spanned by the given ve tors can be written as aVI + bV2, where a and b are real numbers; therefore, the plane is described by a(3, -1,1) + b(O , 3,4). 15. Given two points a and b, a line through them is l(t) = a + t(b -a). In this case, a = (-1 , -1, -1) and b = (1, -1,2), so we get l(t ) = (-1, -1, -1) + t(2, 0, 3) = (2t -1, -1, 3t -1). 19. Substitute v =(x, y, z) =(2 + i, -2 + t, -1 + i) into the equation for x, yand z and get 2:1: -3y + z -2 2(2 + t) -3(-2 + t) + (-1 + t) -2 4 + 2t + 6 -3t --1 + t -2 = 7. Since 7 f. 0, there are no points (:I:, y, z) satisfying the equation and lying on v. 23. Just as the parallelogram of example 17 was described by v = sa + tb for sand tin [0,1], the paralielpiped can be described by w = sa + tb + rc, for s, t and r in [0,1]. Let a, band c be the sides of the triangle as shown, and let Vij denote the vector from point i to poin j. We assume that each median is divided into a ratio of 2 : 1 by the point of intersection. Then we have VI2 -a/2 =-(c -b)/2; V23 = (1/ 3)(a/ 2 + b); V34 (-2/ 3)(a+b/ 2); 28 . 1 V45 (a + b)/ 2. The vector VIS should be the sum V12 + V23 + V34 + V45, or c -b 1 (a ) 2 ( b) a+ b c V1 5 =--2-+ 3 2 + b -3 a+ 2' + -2-= b -2' which is the median of the vector that ends on c. The other two median ' are analyzed the same way. CHAPTER 1 4 30. (a) Using x, the number of C atoms; y, the number of H atoms; and z, the number of 0 atoms, as coordinates, we get p(3, 4, 3) + q(O, 0, 2) =r(l, 0, 2) + s(O, 2,1). {b) To find the smallest integer solution for p, q, rand s, we balance the equation componentwise: 3p= r (equating x) 2s =4p i. e., s =2p (equating y) 2q + 3p = 2r + s ~ 6p+ 2p i.e., q = (5/2)p. (equating z) Let p =2, then the smallest integer solution is p =2, q = 5, r =6, s =4. (c) In the diagram, P is (6,8,6), Q is (0,0,10) , R is (6,0,12) and S is (0,8,4). Both sides of the equation add up to the vector (6,8,16). R p y x 1.2: THE INNER PRODUCT, LENGTH AND DISTANCE GOALS l. Be able to compute a dot product. 2. Be able to explain the geometric significance of the dot product. 3. Be able to normalize a vector. 4. Be able to compute the projection of one vector onto another. STUDY HINTS l. Inne1' product. This is also commonly called the dot product, and it is denoted by a . b or (a, b). The dot product is the sum 2.:::7=1 ajbj, where aj and bj are the ~th components of a and b , respectively. For example, in lR2, a . b =al b1 + a2b2. Note that the dot product is a scalar. 2. Length of a vector. The length or the norm of a vector x = (x, y, z) is Jx2 + y2 + z2. It is denoted by Ilxll and is equal to VX-:-X. This is derivable from the fact that x . y = X1YI + X2Y2 + X3Y3 with x = y. 3. Unit vector. These vectors have length l. You can make any non-zero vector a unit vector by normalizing it. To normalize a vector, divide the vector by its length, i.e., compute a/liali. 4. Cauchy-Schwarz inequality. Knowing that la· hi ~ Ilallllbll is most important for doing proofs in the optional sections of this text and in more advanced courses. . 5. Important geometric properties. Know that a· b =Ilallllbll cos e, where B is the angle between the two vectors. As a consequence, a . b =0 implies that a and b are orthogonal. The zero vector is orthogonal to all vectors. 6. Projections. The orthogonal projection of b onto a is the "shadow" of b falling onto a. The projection of b onto a is a vector of length (a· b)/llall, in the direction of a/liali. Thus, the projection of b onto a is (a 'b)) a (a·b) ( -W W=~a. .. . THE GEOMETRY OF EUCLIDEAN SPACE 5 7. Problem solving. Since vectors have magnitude and direction, they can be represented pictorially. It is often useful to sketch a diagram to help you visualize a vector word problem. SOLUTIONS TO SELECTED EXERCISES 3. From the definition of the dot product, we get u·v (0719)·(-2-10) -7 cos (j = =' , " = ~ -0 1546 Ilullllvll V72 + 192V22 + 12 V410V5 . . From a hand calculator, we find that (j ~ 99° . 7. If w =ai + bj + ck, then Ilwll = va2 + b2 + c2 , and so Ilull = vT+4 = ..)5; Ilvll = v1+l = h. Using the formula for the dot product, we get u· v = (-1)(1) + (2)( -1) =-3. 10. Using the same formulas as in exercise 7, we get Ilull =VI + 0 + 9 = JiO; Ilvll = vO + 16 + 0 =4. Since u does not have a j component and v does not have any i or k component, the vectors are perpendicular; therefore n . v =O. 12. A vector w is normalized by constructing the vector w/llwll. For the vectors in exercise 7: u 1 ( . 2 ') v 1 (' ') W = V5 -) + J; IRI = V2 ]-J . 15. The projection of v onto u is n·v _(-1)(2)+(1)(1) + (1)(-3)(_, , k)-_i(_, , k) Ilul12u -(V1+1+1)2 l+J+ -3 I+J+ . 16. For orthogonality, we want the dot product to be O. (a) The dot product is (2i + bj ) . (-3i + 2j + k) = -6 + 2b , so b must be 3. (b) The dot product is (2i + bj ) . k =0, so b can be any real number. 21. (a) Looking at the x componen s, the pilot needs to get from 3 to 23. His velocity in the x direction is the i component, 400 km/ hr. Thus, Llt = ~d = 23 -3 = ~. v 400 20 The pilot flies over the airport (1/20) hour or 3 minutes later. The same answer could have been obtained by analyzing the y components. (b) We look at the z components and use the formula ~d=vLlt, i.e., h -5 = (-1)(1/ 20) , so h =99/20. Thus, the pilot is 4.95 km above the airport when he passes over. 24. (a) It is convenient to draw the diagram with A on the .r axis. 6 CHAPTER 1 y A x (b) From the diagram in part (a), we get A = 150i and B = (llO cos 600)i + (llO sin 600)j. A + B = (150 + llO cos 600 )i + (llO sin 600)j = 205i + 55V3j. The angle that A + B makes with A is () =tan-1 ( ; ) (5~:) ~ tan -1(0.4647) ~ 250. =tan-1 A, the definition of the dot product gives us () = cos-1 ( A · (A + B) ) = cos-1 (150)(205) + (0)(55V3) IIAII IIA + BII (150)(J51l00) ~ cos-l(O.gO~g) ~ 25° . 27. (a) Geometrically, we see that the i component of F is IIFII cos (). Similarly, the j component of F is IIFII sin B. T herey F fore, F =IIFII cos ()i + IIFII sin ()j, where () is the angle from the x axis. Since the angle from the y axis is 7r/4, () is also 7r/4, so F =3J2(i + j ). (b) We compute D = 4i+2j, so F·D = (3)2)(4) +(3)2)(2) = 18"/2. Also, IIFII =6 and IIDII = J20. From the definition of the dot product, F· D 18)2 3 cos() = IIFII IIDII = 6J20 = vT5 ~ 0.9487, i. e., () ~ 18°, or equivalently, () ~ 0.322 radians. (c) From part (b), we had computed F· D =18)2. Knowing that cos () =3/.JfO, we calculate IIFII liDII cos() =(6J20)(3/JIO) =18)2, also. 1.3: MATRICES, DETERMINANTS AND THE CROSS PRODUCT GOALS 1. Be able to compute a cross product. 2. Be able to explain the geometric significance of the cross product. 3. Be able to write the equation of a plane from given information regarding points on the plane or normals to the plane. x 7 THE GEOMETRY OF EUCLIDEAN SPACE STUDY HINT S 1. Matrices and determi1wnts. A matrix is just a rectangular array of numbers. The array is written between a set of brackets. The determinant of a matrix is a. number; a matrix has no numerical value. The determinant is defined only for square matrices and it is denoted by vertical bars. 2. Computing determinants. Know that I ~ ! I=ad -be. Also know that abc d e f 9 h Note the minus sign in front of the second term on the right-hand side. The general method for computing determinants is described next. 3. Computing n x n determinants. Use the checkerboard pattern shown here which begins with a plus sign in the upper left corner. Choose any column or row -usually picking the one wit,h the most zeroes saves work. + + Draw vertical and horizontal lines through the first number of + the row or column. The numbers remaining form an (n -1) x + + (n -1) determinant, which should be multiplied by the number (with the sign determined by the checkerboard) through which both lines are drawn. Repeat for the remaining numbers of the row or column. Finally, sum the results. This process, called expansion by minors, works for any row or column. The best way to remember the process is by practicing. Be sure to use the correct signs. 4. Simplifying determinants. Determinants are easiest to compute when zeroes are present. Adding a non-zero multiple of one row or one column to another row or column does not change the value of the determinant, and this can often simplify the computations. See Ie 3. 5. Computing a cross product. If a = (aI, a2, a3) and b = (b 1, b2, b3), then J k a x b = a1 a2 a3 b1 b2 b3 The order matters: a x b =-(b x a). The cross product is not commutative.