Solution Manual for Vector Calculus, 5th Edition
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CONTENTS How to Use This Book IV Acknowledgments I V CHAPTER 1 The Geometry of Euclidean Space 1 CHAPTER 2 Di ff erentiation 21 CHAPTER 3 Higher-Order Derivati ves; 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 App li cations of Integration 97 CHAPTER 7 Integrals over Paths and Surfaces 17 CHAPTER 8 The Integr al Theorems of Vector Analysis 143 CHAPTER 9 Sample Exams 161 APPENDIX Answers to Chapter Te ts and Sample Exams 167Page 2
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1 THE GEOMET RY OF EU CL ID EA N SP ACE 1.1 : VECTORS IN TW O- A ND T H RE E- D IM ENSIO N A L SP ACE GOALS 1. Be able to perform the following operations on vector s: addition, subt raction , scalar multipli cation. 2. Given a ve ctor and a point, be able to write th e 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 HINT S 1. Sp ac e notatio n. T he sy mb ol ]R or ]R I refers to all points on the real n um ber line or a one dimensional space. ]R 2 refers to all ordered pairs (X , y) which lie in the plane, a two-dimensional space. ]R 3 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. Ve ctors and scalars. A ve ctor has both length (magnitude) an d direction. Scalars are just ~u m bers . S c~l a rs do not have direction. Two vectors are equal If and only If they both ha ve the same length and the same ~~ direction . Pict orially, th ey do not need to originate from the same s ta rting point. The vectors shown here are equal. 3. V ec tor notation. Ve c to rs are often denoted by bol df ace let ters , underlined letters, arrows over letters, or by an n- tu ple (Xl, X2 , ... , x 71 ) . Each Xi of th e n-tuple is called the lth component. BEWARE that th e n- tu ple may represent either a point or a vector. T he vector (0,0, ... , 0) is denoted O. Your instructor or oth er textbooks may use other notations such as a squiggly line underneath a le tt er. A circumflex over a letter is sometimes used to represent a unit vector. 4. Vector addition. Vec tors may be added componentwise , e.g., in ]R2 (XI , yr) + (X2, Y2) = (Xl +YI,X2+Y2). Pi ctorially, two vectors may be thought of as the sides of a parallelog ra m. St ar ting from the vertex formed by the two vectors, we fo rm a n ew vector wh ich ends at the opposite corner ;;;?J v of the parallelogram. This new vector is the sum of the other two. Al ternat iv el y, one could simply translate v so that the ta il of v meets the head of u . The vector joining the tail of u U I to th e h ea d of v is u + v . £?:j U + V I ______ __ 1Page 4
CHAPTER 1 2 5 . Vector subtmction . Just as with addition, vectors may be su b tra cted component wi se. Think of this as adding a negative vector . Pi ctorially, th e ve ctors a , b a b and a - b form a triangle. To determine the correct direction , hb ~ you should I be able to add a b and b to get a. a- - Thus a b ~ go," f<om the tip of b to th e tip of 8 . 6 . Scalar multiplication. Here, each component of a vector is multiplied by the sam e scalar, e.g., in JR. 2 , r(x , y) = (rx, ry) for any real number r . The effe ct of multiplication by a positive scalar is t.o c ha nge th e 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 exce pt for a single 1. In JR.3 , i, j and k denote the vectors which lie on the x, y and z axes. T hey 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. Thi s is called the point-direction form of t he 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 p oi nt-point fo rm 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 A I V I + A2 V 2 + . .. + An V n , 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 simpli fy a proof. Try to compare vector meth ods and non-vector methods by doing example 10 witho ut vectors. SOL UTIONS TO SELECTED E X ERCISE S 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 ).Page 5
3 THE GE OM ETRY OF EU CLI D EAN SPACE 7. To sketch v, tart at the origin and mo ve 2 units al ong the x ax is , then move 3 units par al le l to th e y axis, and then move -6 units parallel to the z axi s. T he vector w is sketched anal ogously. The vector - v has th e same length as v, but it points in the opposite direction . To sketch v + w , t ranslate the tail of w to the head of v and dr aw the vector fr om the origin to the head of the translated w . The vec tor v - w go es from the y head of w to the head of v. 9. On t he y axis, points have the coordin at s (0 , y, 0) , so we must restri ct x and z to be 0. On the z axis, points have z (~~~' 51 _ . ( O,y ,z) th e coordinates (0 ,0, z) , so we must re st rict x and y to be ( x,O,z )/ / y 0. In t he xz plane, points have the coordi na tes ( :I: ,O , z). so ( ( o, y, o) we must restrict y to be 0. In the yz pla ne , points have the coordinates (0 , y , z ), so we must restrict x to be O. x 12. Every point on t he plane spanned by the giv en ve tors ca n be writ t en as aV I + bV2, wh ere a and b are real numbers; therefore, th e plane is described by a(3 , -1,1) + b(O , 3,4 ). 15 . Giv en t wo points a and b , a line thr ough 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, ya nd 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. Si nc e 7 f. 0, there are no points ( :I: , y , z) satisfying the equation and lying on v . 23. J us t as th e p ara ll el og ram of examp le 17 was described by v = sa + tb for sa nd t in [0, 1 ], the para li elpiped can be d es cribed by w = sa + t b + rc, fo r s, t and r in [0 , 1] . Let a , band c be the sides of the triangle as shown , and let Vi j denote th e vector from point i to poin j . We assume tha t each median is divided int o a ratio of 2 : 1 by the p oi nt of intersection. Then we have V I2 -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 V I S should be the sum V1 2 + 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 t he med ia n of the vector that ends on c. The other two median ' are analyzed the same way.Page 6
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 ba lance the equation componen twise: 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 solu t ion 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 ,1 2) and S is (0,8,4). Both sides of the equation add up to th e vector (6,8,16). R p y x 1.2: THE INNER PRODUCT, LENGTH AN D DI S TAN CE GO A LS l. Be able to compute a dot product. 2. Be able to explain the geometric significance of th e d ot product. 3. Be able to normalize a vector. 4. Be abl e to compute the projection of one vector onto another. STUDY HINTS l. Inne1' product. This is also commonly called the dot pro duct, and it is denoted by a . b or (a, b). The dot product is the sum 2.:::7=1 ajbj, where aj and bj a re the ~th components of a and b , respectively. For example, in lR 2, a . b = al b 1 + a2b2. N ote that the dot product is a scalar. 2. Length of a vector. The length or the norm of a ve ct or x = (x, y, z) is J x 2 + 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 v ec tor, divide the vector by its length, i.e., compute a/liali . 4. Cauchy-Schwarz inequality. Knowing that la· hi ~ Ilallll bll is most im portant for doing proofs in the optional sections of this text and in more advanced cou rses. . 5. Important geome tri c properties. Know that a· b = Ilallllbll cos e, where B is the angle between the two v ec tors. As a consequence, a . b = 0 imp lies th at a an d b are orthogonal. The zero vector is orthogonal to all vectors. 6. Proje ctions. The orthogonal projection of b onto a is th e "sh ad ow" of b falling onto a . The projection of b onto a is a vector of length (a· b)/llall , in the dir ection of a/liali. Thus, the projection of b onto a is (a ' b)) a (a· b) ( -W W=~ a . .. .Page 7
THE GEOMETRY OF EUCLIDEAN SPACE 5 7. P ro bl em solving. Since vec tors have magnitude and dir ection, they can be represented picto ri a ll y. It is often us ef ul to sketch a di agram to help you visualize a vector word problem . SOLUTIONS TO SELECTED EXERCISES 3. From t he de fin it ion of t he dot product, we get u·v (0 7 19) · (- 2 - 1 0) -7 cos (j = = ' , " = ~ -0 15 46 Il ullllv ll V7 2 + 19 2 V 22 + 12 V 41 0 V5 . . From a hand calculator, we fin d th at (j ~ 99° . 7. If w = ai + bj + ck, then I lwll = va 2 + b 2 + c 2 , and so Il u ll = vT+4 = ..)5; Il vll = v1+l = h . Using t he formula for the dot product, we g et u· v = (-1 )(1) + (2 )( - 1) = - 3. 10. Using t he sam e fo rmulas as in exercise 7, we get Il ull = VI + 0 + 9 = JiO; Il vl l = v O+ 16 + 0 = 4. Since u does n ot ha ve a j component and v does not have any i or k co mponent, the vectors are perpendic ul ar; therefore n . v = O. 12. A vector w is nor ma li zed by constructing th e ve ctor w /ll wll. For th e ve ct ors in exercise 7: u 1 ( . 2 ') v 1 (' ') W = V5 -) + J; IRI = V2 ]- J . 15. T he projection of v onto u is n·v _(-1)( 2 )+( 1 )( 1) + (1 )( - 3 )(_, , k )-_i(_, , k) Il u l1 2u - (V 1 + 1 + 1 )2 l + J + - 3 I+ J + . 16. For o rt ho gonal it y, we wa nt the dot pro du ct to be O. (a) Th e dot pro duct is (2i + bj ) . (- 3i + 2j + k ) = -6 + 2b , so b must be 3. (b) The dot pro duct is (2i + bj ) . k = 0, so b can be any real n um ber. 21. (a) Looking at the x componen s, the pilot needs to g et from 3 to 23 . His ve locity in the x direction is the i compone nt , 40 0 km/ hr . T hus, Llt = ~d = 23 - 3 = ~. v 40 0 20 Th e pilot flies over the airport (1/ 20) hour or 3 minutes later. Th e sam e answer could ha ve been obtained by an alyzing the y components. (b) We loo k at th e z components and use the formula ~d = vLlt , i.e., h - 5 = (- 1)( 1/ 2 0) , so h = 9 9/20 . Th us, the pilot is 4 .95 km above the airp or t when he passes over. 24. (a) It is con ve ni ent to draw the diagram with A on the .r axis.Page 8
CHAPTER 1 6 y A x (b) From the dia gram in part (a), we get A = 15 0i and B = (llO cos 60 0 )i + (llO sin 60 0 )j. A + B = (15 0 + llO cos 60 0 )i + ( ll Osin 60 0 )j = 20 5i + 55V3j. The angle that A + B makes with A is () = tan- 1 ( ; ) (5~ : ) ~ t an - 1 (0.46 47) ~ 25 0. = ta n- 1 Alt.ernat.ively, the definition of the dot produ ct gives us () = cos- 1 ( A · (A + B ) ) = cos- 1 (1 50)( 205) + (0)( 55 V3) IIAII IIA + BII (15 0)(J 5 1l 00) ~ 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 here y F fore, F = IIFII cos ()i + IIFII sin () j, where () is the angle from the x axis. Since t he angle from t he y axis is 7r/4, () is also 7r/4, so F = 3 J2 (i + j ). (b) We compute D = 4i+ 2 j, so F·D = ( 3)2 )(4) +( 3)2 )(2) = 18"\/2. Also, II F II = 6 and IIDII = J20 . From the de fi nition of the dot product, F · D 18 )2 3 cos() = II F II IIDII = 6J20 = vT5 ~ 0.9487, i. e., () ~ 18°, or equi valently, () ~ 0.3 22 radians. (c) From part (b), we had computed F· D = 18)2. Knowing t hat cos () = 3 /.JfO, we calculate IIFII liD II cos() = (6J20)(3/JIO) = 18)2, also . 1.3: M A TRICES , DETERMINANT S AND THE CRO SS PROD UC T 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 t he plane or normals to the plane. xPage 9
7 TH E GEOMETRY OF EUCLIDEAN SPA CE STUDY HIN T S 1. Ma tri ce s a nd determi1 wn t s. A matrix is just a recta ngula r array of numbers. The array is written between a set of br ackets. Th e determ inan t of a matrix is a. number; a matrix has no numerical value. The determ in an t is de fi ned only f or square matrices and it is denot ed by vertical bars. 2. Comput ing dete rmi nants. K now th at I ~ ! I = ad - be. Also know that abc d e f 9 h Note the min us sign in front of th e second ter m on the right-hand side. The general met hod for comput ing dete rm ina nts is described next . 3. Computing n x n det er minants. Use the checkerboard pattern shown here which begins wi th a plus sign in th e up per left corner. Choose any column or row - usua lly picking th e one wit,h the most zeroes saves work. + + Draw vertical and horizontal lines through the first number of + t he row or column. The nu mbers remaining form an (n -1) x + + (n -1) determ inant, which sh ould be multiplied by the number (with the sign det erm ined by the checkerboard) through which both lines are dra wn . Repeat for the remaining numbers of the row or column . Finally, sum the resul ts . This process , called exp ans ion by minors , works for any row or column. The best way to r em em ber the pro cess is by pr acticing. Be sure to use the c orr ect signs. 4. Simplifyi ng dete rmi nants . Deter m inant s are easie st to compute when zeroes are present. Addin g a non-zero mul tipl e of one row or one c olum n to an o ther row or column does not change the value of the deter min ant, and th is c an often simplif y the computations. See exa.m pIe 3. 5. Computing a cr oss product. If a = (aI, a 2 , a3) and b = (b 1 , b 2 , b 3 ), then J k a x b = a1 a 2 a3 b 1 b 2 b 3 The order m at ters: a x b = -( b x a). Th e cross pro du ct is n ot commutative. Also, note that a x b is a vector , not a scal ar . 6. Properties of the cross pro du c t. T he ve c tor s a , b an d a X b form a right-handed system (see figure 1.3.2 of th e text) . Th e cross product a x b is orthog o na l to both a and b . The length of a x b is II allllb ill sinOI, where 0 is th e angle between a and b. Note that th e formula for the cross pro d u ct involves sin 0, whereas t he dot product involves cos O. 7. Mor e properties. If th e cross produ ct is zero , then either : (i) t he length of one of the vectors must be zero , or (ii) sinO = 0, i. e., 0 = 0, so the vectors must be parallel. 8. Geometry. Th e absolute value of th e de ter mi nant I ~ ! I is the area of the parallelogram spanned by the vectors (a, b) and ( c, d) o riginat ing from the same point. The absolute value of a bc the determinant d e f is the vol um e of th e parallelp i ped spa nned by the vectors (a, b, c), g h i (d, e, J) and (g, h, i) originating fr om the sam e point. The length of the cross product 11a x hll is th e area of the parallelogram sp a nn ed by the vec tors a a nd b. The vector a x b gives a vector normal to th e plane s panned by a a nd b .Page 10
CHAPTER 1 8 9. Equation of a plan e. Recall t hat the equation of the plane is ax + by + cz + d = O. The vector (a , b, c) is orthogona] to the plane . Knowing two ve ctors in the plane , we can deter mine an orthogonal vector by using the cross product. Compare methods 1 and 2 of example 1 1. 10. Distance from point to plan e. You should unde rsta nd the derivation of the equation in the box preceding example 12. If necessary, review the ge om e tr ic properties of the dot product in section 1.2. SOLUTIONS TO SELECTED EXERCISES 2. (b) We subtract 12 times the third row from the first row and subtract 15 tim es the third row from the second row. 'f hen we expand along the first c ol umn: 36 18 17 o -42 41 -42 41 1 . - . 45 24 20 o -51 50 = 3 -51 50 = 3(-2100 + 20 91) = -27. 1 3 5 -2 3 5 -2 5. The area of the parallelogram is /l a x hll . We compute j k a x b = 1 -2 1 = -3 i + j + 5k 2 1 1 and so the area of the parallelogram is lin x hll = v'9 + 1 + 25 = 55 . 8. The volume of the parallelpiped is the absolute value of the 3 x 3 determina nt made up of t he vectors ' components. Expand along the first row: 1 0 0 _1 3 -1 0 3 -1 - 2 'I =- l. -1 4 2 -1 Thus , the volume is 1. 11. We want to find the cross product and then normalize it . We compute So J k v = -5 9 -4 I = 113i + 17j 103k. 789 There are two orthogonal vectors in opposing directions; they ar e given by ±v/llvll =: ±(113i + 17j - 1 03k )/ v' 236 67. Since all vectors are orthogonal to 0, the inclusion of that ve ctor into the problem does not affect the answer. 15. (a) The equation of a plane with normal vector (A ,B , C) and passing through the point (xo, Yo, zo) is A(x - xo) + B(y - Yo) + C (z - zo) = O. In this case, the equation is l(x - 1) + l(y - 0) + l(z - 0) = 0 or x + y + z = 1. (d) Here, the normal vector is parallel to the line, so it is ( -1 , - 2,3 ). Hence, t he equation of the desired plane is -1(x - 2) - 2(y - 4) + 3(z + 1) = 0 or - x - 2y + 3z + 13 = O.Page 11
9 TH E GEO METRY OF EU CLI DE AN SP AC E 16. (b) Two vectors in the desired plane are v = (0 - 1, 1 - 2, - 2 - 0) = (-1,-1 , - 2) and w = (4 - 0, 0 - 1, 1 + 2) = (4, - 1 ,3 ). The cross pr duct v x w is ort h og ona l to both vectors, and hence normal to the desired plan e. We co mpute v x w = - 5i - 5j + 5k, so the desired equation is -5( x - 1) - 5(y - 2) + 5(z - 0) = 0 or - x - y + z + 3 = O. 22. (a) Let D be the m atr ix with ro ws u, v, w . T hen Ul U2 U3 det D = u· ( vxw )= VI V2 V3 W I Wz W3 Use the fo llowing property of determinants: v· (w x u ) corresponds to two row exchanges of th e m atrix D , 80 we ha v v . (w x u ) = (-1)( -1) det D = det D and w · (u x v ) = (-1)( -1) det D = det D. To prove the other th r ee , rec all that u x v = -( v xu ). 26. The line perpendicular to the pla ne is parallel to th e normal of th e plane, so t he equ at ion of the line is l (t) = (1, - 2, - 3) + t ( 3, -1, -2 ). 2 9. Let u be the ve ctor normal to th e plane. Then u is perpendicular to 3i + 2j + 4k since v is on the pl an e. Alsu u is perpendic ul ar to 2i + j - 3k, because the vector Ai + Bj + Ck is e rp endicular to all vectors in t he plane Ax + By + Cz = D . To find u, we take t he cross produ t of 3i + 2j + 4k and 2i + j - 3k: J k u = 32 4 =- 10i + 17 j - k. 2 1 - 3 Wh en t = 0, we fi nd that a point on the plane is (- 1,1 , 2), so the equation of the plane is - 10 (x+ l) +1 7(y -l)- (z-2}=0 or - 10x + 17y-z - 25 =0 . 30 . First, fi nd the normal o ft he plane. The norm al o fthe plane is perpendicular 0 th e line passing thr ough (3, 2, - 1) and (1, - 1, 2). T he equation of the line is l(t) = ( 3,2 , - 1) + t(2, 3, - 3 ). Th e normal of t he plane is al so perpendicular to v = (1 , - 1, 0) + t( 3, 2, -2 ). T herefore, tw o ve ctors on t he desired plane are 2i + 3j - 3k a nd 3i + 2j - 2k , an the normal is (2i + 3j - 3k ) x (3i + 2j - 2k ) = - 5j + 5k. No w, we need a point on the plane, say, (3 ,2 , -1 ). Thus, the equation of t he plane is O( z - 3) - 5 (y - 2) + 5(z + 1) = 0 or - y + z = l. 34. T he plane passi ng through the origin and perpendicular to i - 2j + k is x - 2y + z = O. By th e dist ance formu la with (A, B , C , D ) = (1, - 2, 1, 0) and (X l, Y1, zt) = (6 ,1, 0), d = I Ax l + BY I + C Z l + DI = 6 - 2 _ --±- _ 2 V6 V A2 + B2 + C2 VI + 4 + 1 V6 3 ' 37. Sin ce all vectors in this exercise are unit vectors, li N x all = sin 0 1 and II N x h ll = sin O 2 . From S ne ll 's la w, n l sin 0 1 = n2 sin O 2 . Hence, nlll N x a ll = n2 11N x h ll. To e st ab li sh that N x a and N x h have the same direction, we assume t h at N, a and hall lie in th e same plane, and a and h are on the same side of N. Hence N x a and N x h both are perpendicular to this plane and parallel to ach oth er . Thus nl ll N x al l and n 211N X hll are equa l.Page 12
10 CHAPTER 1 38. First, 4 times the first row is subtracted from the second row. Next, 7 times the first row is subtracted from the third row . The next step is expansion by minors along the first column. Finally, the 2 x 2 determinant is computed. 1.4: CYLINDRICAL AND SPHERICAL COORDINA TE S GOALS l. Be able to convert back and forth between the cylindrical, spherical and cartesian coordin ate systems. 2. Be able to describe geometric objects with cylindrical and spherical coordinates. 3. Be able to describe the geometric effects of changing a coordinate. STUDY HINTS l. Review . You should review polar coordinates in your one-variable calculus text . . . ~ 2. Cylindrical coordinates. Denoted (r, {}, z), this is just like polar coordin ates except that a z coordinate has been added. Know the formulas x = rcos{}, y = rsin{}, r = Jx 2 + y2 and {} ::::; tan-1(y/x). 3. Spherical cooridinates. Denoted by (p, {}, ¢», p is the distance from th e origin, ¢> is the angle from the positive z axis and {} is the same as in cylindrical coordi nat es. Know the formulas x=pcos{}sin¢>, y=psin{}sin¢> and z=pcos¢>. Also know t. hat p = Jx2 + y2 + Z2, {} = tan-1(y/x) and ¢> = cos-1(z/ Jx 2 + y2 + z2) = cos- 1 (z/ p). 4. Graphs of r, p = constant. Note that r = constant in cylindrical coordinates describes a cylinder and that p = constant in spherical coordinates describes a sphere. You may have suspected this from the name of the coordinate system. 5. Computing {} and ¢>. Remember that, in this text, ¢> takes values from 0 to rr and {} ranges from 0 to 2iT . In some instances, it is more convenient to define {} in the range -rr to iT . You should be very careful about computing {}. If x = y = -1, t.hen tan-1(y/x) = iT/4, but plotting the point (-1,-1) in the xy plane shows that, in reality, {} = 5rr/4. This is why t he authors fuss with tan- 1 (y/x) in the definition. Plotting the x and y coordinates is very helpful for determi ni ng {}. 6. Negative r, p. Note that we have defined r an d p to be non-negative. If the distance is given as a negative number, we need to reflect the given point across the origin . 7. Unit vector's in spherical and cylindrical coordinates. The unit vectors in cylindrical coordi nates are e r , eo and e z . The vector e r points along the direction of r, while eo goes in the direction in which {} is measured , and e z = k. As one might expect., those three unit vectors form an orthogonal basis, and e r x eo = e z . Those vectors, however, lire not fixed as is the case with i, j and k , that is, if you change the point (7', {}, z), the set of unit vectors rotates. For spherical coordinates, there is also a set of unit vecto rse p , eo and e",. Those vectors , in terms of i, j, k and the cartesian coordinates of the point are worked out in exercise 7 (see below) of this section.Page 13
11 T HE GE OM ET RY OF E UCLID E AN SPAC E SOL UT IONS TO SELECTED EXER C ISES 1. (a) To conver t to re cta ngu lar coordinates, use x = r cos & and y = 1'S in 11: x = 1 cos 45° = v2/ 2 and y = 1 sin 45 ° = V2 /2. Next , use p = J x 2 + y2 + z 2 and rP = cos - 1 (z / p) to get the spherical coordin ate s: Ir. - 1 ( 1l' P :::;; V I 2" + 2" 1 + 1 = V 2 and ¢ = cos )2 1) = "4 ' Hence, for th e cylindr ic al co ordinates ( 1,45 ° ,1 ), the rect angul ar coordinates are given by ( /2 / 2 ,)2 / 2,1 ) and hes ph rical coor dinates are (V2 ,rr/ 4, 1l' / 4). (b) To co nvert to cylindrical coordin at es, we use l' = ..jX 2 + y2 and (J = tan -1(y/ x ): r = . )4+1 = v'5 and B = tan - 1 (1/ 2). Ne xt , use the same formulas as in par t (a) to get t h sp herical coordina te s: p=V4+ 4+ 1 = 3 and ¢ = cos -1(- 2/ 3). Hence, the re ctangular co or din ates (2, 1, - 2) convert to the corresponding cylindrical coord i nat s ( /5 , t a n- 1 (1/ 2 ), - 2) and t o the spherical coordi nates (3, ta n - 1 (1/ 2) ,c o s-1(- 2/ 3)). 2. (b) T his mappin g takes a point a nd ro tates it by 1l' radians about the z axis . This is follo we d by a re fl ection across the xy plan e. The net effect is that the point is re fle cted thr ough the origin. 3. (b) Reca ll that the angle ¢ is measured from the "Nor th pole." If ¢ is 7i ra dia ns , t hen t he loc at ion is at th e "S outl pole." The e ff ect. of c hang ing ¢ to 1[' - ¢ is ta king a point and r efl ecting it across the x y plane. 5. Let p ~ 0, then (p, 0, 0) is the positi ve z axis. Now, let ¢ vary from 0 to 1l'. Then (p, 0, ¢) is the ha lf plan e in the xz plane wi t h z ~ 0. By allowing & to vary from 0 to 2 1l', we rot ate t he half plan e described above. Th ere fo re, p ~ 0, ° ~ (J < 21l' and 0 ~ ¢ ~ 1l' describes all points in JR 3. If p < 0 also , the co ordinates are not unique . For exam pl e, (x, y, z) = (1, 1,0) has spherical co ordinates (V2 , 1l'/4,1l'/2) and (- V2 , 51l' / 4,1l' /2). 7. (a) Fir st , e p is th e unit vector along the ve ctor ( x, y, z) ; t herefore, the fo rm ula is xi + yj + zk e p = . J x 2 + y2 + z2 z y -----k--- --- ~ r y x x ( x, y ,O ) Next, e o is paral lel to th e xy plane and denotes t he direction in which t he angle B is measured . It is perpendicu lar to r = (x , y, 0), so ( ai + bj) . (xi + yj ) = O. Since we want B measured counterclockwise, a = -y (instead of y) and b = x. Therefore, - yi + zj eo = J x 2 + y2'Page 14
12 CHAPTER 1 To fi nd e ,p , we no te that e p , eo and e,p is a set of orthogonal vectors; they form a right-handed coor din ate sy ste m with e p x eo = - e,p. So . (x, y, z ) x (- y, x, 0) x zi + yzj - (x 2 + y2 )k ~ =- = . rp J x 2 + y2Jx2 + y2 + Z2 9. (a) The length of xi + yj + zk is J x 2 + y2 + z2 , which is the definition of p. (b) Note that Il v ll = Jx 2 + y2 + z2 = P and v· k = z; therefore, <p = cos~l(z/p) = cos-1(v. k/llvll)· (c) Note that Ilull = J x 2 + y2 , which is the cylindrical coordinate rand u · i = x; therefore, () = cos- 1 (x/r) = cos- 1 (u . i /llull). 13. Note th at <p will be between 7r / 2 and 7r because the region lies in th e lower hemisphere. From the triangle, we see th at cos a = (d/6) -;- (d/ 2) = 1/ 3; ther efore, we ha ve 7r - a ::; <p ::; 7r or 7r - cos-1(1/3) ::; ¢ ::; 7r . Now, p can be as la rge as d/ 2; ho we ver, as p gets smaller, its lower limit depends on <p . P ic k any ¢ , then ¢+ (3 = 7r and according to the diagram cos (3 = (dj6) -;- p. Rearrangement gives d/(6cos(3) = p = d/(6cos(7r - ¢)) = -d/(6 cos¢). Therefore, -d / (6 cos ¢) ::; p ::; d/2. So far , we have described the cross-section in one quadrant. The entire volum e requires a revolution aroun d the z axis, so it s description is 1 __ d_ < p < ~ 0 < () < 27r and 7r - cos- 1 ( -3 ) <_ A, < _ 7r. 6 cos ¢ - - 2 ' - - 'i' 1.5: n-DIMENSION A L EUC LI DEAN SPACE GOALS 1. Be able to ext end th e ideas of the previous sections to jR n. 2. Be able to multiply ma trices. STUDY HINT S 1. The space ]Rn . Most of this textb ook deals with the Euclidean spaces that we can visualize, jR2 and jR3. Many of the same properties hold in jRn. Ve ctor addition, scalar mu lt iplication, vector lengths, th e dot product and the triangle inequal ity are defined similarly. 2. No cross product analog. The cross pro duct in ]R3 does not have an easy analog in jRn, n ~ 4. 3. Standard basis v ec tors. Th e analogs of i, j, k are defined e j. The ve ctor ej is (0,0, . .. ,1, ... 0) with 1 in the zth position. Th e vectors ej a nd ej are orth ogonal if i =f j.Page 15
13 THE GEOME TRY OF EUCLIDE AN S PA CE 4. Matrices. A matr ix is a rect angul ar array of numbers . Un like a determinant, a matrix has no numerical value. You should remember th at an n x m m atr ix has n rows and m columns. The (i, j) entry is the number loc at ed in row i, column j. 5. Mat rix m ultiplication. You should practice until m at rix multiplication becomes second nature to you. Let th e components of A be a i j and let those of B be bkl, where A is an m x p matrix and B is a p x n matr i x. Then the components of A B are p (A B )mn = L amjbjn . j =l We can only multiply an m x p ma t r ix with a p x n matrix, i.e., [m x p ][P x n]. Note that the number of columns of A and the number of rows of B must be equal (p in this case). The result is an m x n mat rix ("cancelling" th e p). 6. Non-commutativity of m a tr ix multiplication. In general, AB # B A . In fact, AB may be defined when BA is undefined . However , m atr ix multiplication is asso ci at ive; i.e., (AB)C = A(BC) if the product AB C is defined . 7. Matrices and mappings. An m x n matr ix can represent a map ping from ]Rn to ]Rm. To see this, l et A be the mat rix and let x be a vector in ]R n, r epr esented as an n x 1 matrix, and y be a vector in ]Rm, an m x 1 m at rix. Then th e matrix A tak es a point in ]R n to a point in ]Rm by the equation Ax = y. SOLUTIONS TO SELE C TED EXERC I S ES 2. (a) Use the properties of leng hs and dot products: (x + y) . (x + y) + (x - y) . (x - y) x . x + 2x . y + y . y + x . x - 2x . y + y . y 2 x· x + 2y· y = 2 11xl12 + 211Y11 2 . The figure at the left depicts the equation geometrically. By the l aw of cosines, we have 12<tl y and We also note that a + j3 = 1r, so j3 = 1r - a . T herefore, cos j3 = cos( 1r - a) = - cos a. Adding th e two equations from th e law of cosines yields 2 11xl1 2 + 211Yl12 ;: Ilx - Yl12 + I' lx + Y112. 4. To verify the Cau chy-Schwarz inequality, we compute Ix· yl 1(1)( 3) + (0)(8) + (2 )( 4) + (6)(1)1 = 17. Ilxll '1'1 + 0 + 4 + 36 = J41. Il , yll J9 + 64 + 16 + 1 = V90. Thus , we indeed have Ix . yl = v'I7v'I7 < v4f J90 = II 'xlillyll. For the triangle inequality, we compute x + y = (4,8,6,7) and II x + yll = J16 + 64 + 36 + 49 = v'l65 < 13 . Indeed, we have Ilx + yll < 13 < 15 = 6 + 9 < AT + v'9O = Ilxllllyll.Page 16
14 CHAPTER 1 8. We compute 1 n AB~ r: -1 ~31 and A+B~[: 0 2 1 1 -1 1 Expanding by minors across the first row gives det A 3: 1 2 -1 I 11 ~ II = 6 - 2= 4, 0 1 + 1 1 detE 11 0 1 ~ 1- 11 ~ ~ 1 = -3, det(AE) 1-1 1 1_ 115 and 3 1 -1 1 ~1 1- 31 5 1 -1 1 1=-12 det(A + B) - 41 ~ ~ I = 8. 11. (a) For n = 2, I 'xall det('xA) = 'xa21 For n = 3, 'xall Aal2 'xa13 det( AA) Aa21 'xa 22 'xa23 Aa31 'x a 32 'xa33 'xall det('xAd - 'xa12 det('xA 2 ) + 'xa13 det(AA3) ,X . ,X2(a11 det Al - al2 det A2 + a1 3 det A 3) ,X3 det A, where AI , A2 and A3 are 2 x 2 matrices obtained by expanding across the first row . Assume that for n = k, det('xA) = Ak det A. The for n = k + 1, det(AA) ca n be found by a process analogous to the 3 x 3 case: det(AA) 'xall det('xAd - 'xa12 det('xA 2 ) + ... + (_l)k 'xal,k+l det('xAk+d ,Xk+l det A, where AI, A 2 , ... , Ak+l are k x k matrices obtained by expanding across the first row . By induction, det('xA) = An det A for an n x n matrix A. 14 . Assume, as in the book, that det(AE) = (det A)(det E). Then det(AEC) = det[(AB)C] = det(AE) det C = (det A)(det E)(det C). 17 . Multiply the two matrices to get the identity matrix: [ a b] 1 [d -b] e d ad - be -e a Similarly, we can show thatCONTENTS How to Use This Book IV Acknowledgments I V CHAPTER 1 The Geometry of Euclidean Space 1 CHAPTER 2 Di ff erentiation 21 CHAPTER 3 Higher-Order Derivati ves; 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 App li cations of Integration 97 CHAPTER 7 Integrals over Paths and Surfaces 17 CHAPTER 8 The Integr al Theorems of Vector Analysis 143 CHAPTER 9 Sample Exams 161 APPENDIX Answers to Chapter Te ts and Sample Exams 167 1 THE GEOMET RY OF EU CL ID EA N SP ACE 1.1 : VECTORS IN TW O- A ND T H RE E- D IM ENSIO N A L SP ACE GOALS 1. Be able to perform the following operations on vector s: addition, subt raction , scalar multipli cation. 2. Given a ve ctor and a point, be able to write th e 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 HINT S 1. Sp ac e notatio n. T he sy mb ol ]R or ]R I refers to all points on the real n um ber line or a one dimensional space. ]R 2 refers to all ordered pairs (X , y) which lie in the plane, a two-dimensional space. ]R 3 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. Ve ctors and scalars. A ve ctor has both length (magnitude) an d direction. Scalars are just ~u m bers . S c~l a rs do not have direction. Two vectors are equal If and only If they both ha ve the same length and the same ~~ direction . Pict orially, th ey do not need to originate from the same s ta rting point. The vectors shown here are equal. 3. V ec tor notation. Ve c to rs are often denoted by bol df ace let ters , underlined letters, arrows over letters, or by an n- tu ple (Xl, X2 , ... , x 71 ) . Each Xi of th e n-tuple is called the lth component. BEWARE that th e n- tu ple may represent either a point or a vector. T he vector (0,0, ... , 0) is denoted O. Your instructor or oth er textbooks may use other notations such as a squiggly line underneath a le tt er. A circumflex over a letter is sometimes used to represent a unit vector. 4. Vector addition. Vec tors may be added componentwise , e.g., in ]R2 (XI , yr) + (X2, Y2) = (Xl +YI,X2+Y2). Pi ctorially, two vectors may be thought of as the sides of a parallelog ra m. St ar ting from the vertex formed by the two vectors, we fo rm a n ew vector wh ich ends at the opposite corner ;;;?J v of the parallelogram. This new vector is the sum of the other two. Al ternat iv el y, one could simply translate v so that the ta il of v meets the head of u . The vector joining the tail of u U I to th e h ea d of v is u + v . £?:j U + V I ______ __ 1 CHAPTER 1 2 5 . Vector subtmction . Just as with addition, vectors may be su b tra cted component wi se. Think of this as adding a negative vector . Pi ctorially, th e ve ctors a , b a b and a - b form a triangle. To determine the correct direction , hb ~ you should I be able to add a b and b to get a. a- - Thus a b ~ go," f<om the tip of b to th e tip of 8 . 6 . Scalar multiplication. Here, each component of a vector is multiplied by the sam e scalar, e.g., in JR. 2 , r(x , y) = (rx, ry) for any real number r . The effe ct of multiplication by a positive scalar is t.o c ha nge th e 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 exce pt for a single 1. In JR.3 , i, j and k denote the vectors which lie on the x, y and z axes. T hey 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. Thi s is called the point-direction form of t he 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 p oi nt-point fo rm 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 A I V I + A2 V 2 + . .. + An V n , 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 simpli fy a proof. Try to compare vector meth ods and non-vector methods by doing example 10 witho ut vectors. SOL UTIONS TO SELECTED E X ERCISE S 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 GE OM ETRY OF EU CLI D EAN SPACE 7. To sketch v, tart at the origin and mo ve 2 units al ong the x ax is , then move 3 units par al le l to th e y axis, and then move -6 units parallel to the z axi s. T he vector w is sketched anal ogously. The vector - v has th e same length as v, but it points in the opposite direction . To sketch v + w , t ranslate the tail of w to the head of v and dr aw the vector fr om the origin to the head of the translated w . The vec tor v - w go es from the y head of w to the head of v. 9. On t he y axis, points have the coordin at s (0 , y, 0) , so we must restri ct x and z to be 0. On the z axis, points have z (~~~' 51 _ . ( O,y ,z) th e coordinates (0 ,0, z) , so we must re st rict x and y to be ( x,O,z )/ / y 0. In t he xz plane, points have the coordi na tes ( :I: ,O , z). so ( ( o, y, o) we must restrict y to be 0. In the yz pla ne , points have the coordinates (0 , y , z ), so we must restrict x to be O. x 12. Every point on t he plane spanned by the giv en ve tors ca n be writ t en as aV I + bV2, wh ere a and b are real numbers; therefore, th e plane is described by a(3 , -1,1) + b(O , 3,4 ). 15 . Giv en t wo points a and b , a line thr ough 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, ya nd 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. Si nc e 7 f. 0, there are no points ( :I: , y , z) satisfying the equation and lying on v . 23. J us t as th e p ara ll el og ram of examp le 17 was described by v = sa + tb for sa nd t in [0, 1 ], the para li elpiped can be d es cribed by w = sa + t b + rc, fo r s, t and r in [0 , 1] . Let a , band c be the sides of the triangle as shown , and let Vi j denote th e vector from point i to poin j . We assume tha t each median is divided int o a ratio of 2 : 1 by the p oi nt of intersection. Then we have V I2 -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 V I S should be the sum V1 2 + 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 t he med ia n 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 ba lance the equation componen twise: 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 solu t ion 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 ,1 2) and S is (0,8,4). Both sides of the equation add up to th e vector (6,8,16). R p y x 1.2: THE INNER PRODUCT, LENGTH AN D DI S TAN CE GO A LS l. Be able to compute a dot product. 2. Be able to explain the geometric significance of th e d ot product. 3. Be able to normalize a vector. 4. Be abl e to compute the projection of one vector onto another. STUDY HINTS l. Inne1' product. This is also commonly called the dot pro duct, and it is denoted by a . b or (a, b). The dot product is the sum 2.:::7=1 ajbj, where aj and bj a re the ~th components of a and b , respectively. For example, in lR 2, a . b = al b 1 + a2b2. N ote that the dot product is a scalar. 2. Length of a vector. The length or the norm of a ve ct or x = (x, y, z) is J x 2 + 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 v ec tor, divide the vector by its length, i.e., compute a/liali . 4. Cauchy-Schwarz inequality. Knowing that la· hi ~ Ilallll bll is most im portant for doing proofs in the optional sections of this text and in more advanced cou rses. . 5. Important geome tri c properties. Know that a· b = Ilallllbll cos e, where B is the angle between the two v ec tors. As a consequence, a . b = 0 imp lies th at a an d b are orthogonal. The zero vector is orthogonal to all vectors. 6. Proje ctions. The orthogonal projection of b onto a is th e "sh ad ow" of b falling onto a . The projection of b onto a is a vector of length (a· b)/llall , in the dir ection 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. P ro bl em solving. Since vec tors have magnitude and dir ection, they can be represented picto ri a ll y. It is often us ef ul to sketch a di agram to help you visualize a vector word problem . SOLUTIONS TO SELECTED EXERCISES 3. From t he de fin it ion of t he dot product, we get u·v (0 7 19) · (- 2 - 1 0) -7 cos (j = = ' , " = ~ -0 15 46 Il ullllv ll V7 2 + 19 2 V 22 + 12 V 41 0 V5 . . From a hand calculator, we fin d th at (j ~ 99° . 7. If w = ai + bj + ck, then I lwll = va 2 + b 2 + c 2 , and so Il u ll = vT+4 = ..)5; Il vll = v1+l = h . Using t he formula for the dot product, we g et u· v = (-1 )(1) + (2 )( - 1) = - 3. 10. Using t he sam e fo rmulas as in exercise 7, we get Il ull = VI + 0 + 9 = JiO; Il vl l = v O+ 16 + 0 = 4. Since u does n ot ha ve a j component and v does not have any i or k co mponent, the vectors are perpendic ul ar; therefore n . v = O. 12. A vector w is nor ma li zed by constructing th e ve ctor w /ll wll. For th e ve ct ors in exercise 7: u 1 ( . 2 ') v 1 (' ') W = V5 -) + J; IRI = V2 ]- J . 15. T he projection of v onto u is n·v _(-1)( 2 )+( 1 )( 1) + (1 )( - 3 )(_, , k )-_i(_, , k) Il u l1 2u - (V 1 + 1 + 1 )2 l + J + - 3 I+ J + . 16. For o rt ho gonal it y, we wa nt the dot pro du ct to be O. (a) Th e dot pro duct is (2i + bj ) . (- 3i + 2j + k ) = -6 + 2b , so b must be 3. (b) The dot pro duct is (2i + bj ) . k = 0, so b can be any real n um ber. 21. (a) Looking at the x componen s, the pilot needs to g et from 3 to 23 . His ve locity in the x direction is the i compone nt , 40 0 km/ hr . T hus, Llt = ~d = 23 - 3 = ~. v 40 0 20 Th e pilot flies over the airport (1/ 20) hour or 3 minutes later. Th e sam e answer could ha ve been obtained by an alyzing the y components. (b) We loo k at th e z components and use the formula ~d = vLlt , i.e., h - 5 = (- 1)( 1/ 2 0) , so h = 9 9/20 . Th us, the pilot is 4 .95 km above the airp or t when he passes over. 24. (a) It is con ve ni ent to draw the diagram with A on the .r axis. CHAPTER 1 6 y A x (b) From the dia gram in part (a), we get A = 15 0i and B = (llO cos 60 0 )i + (llO sin 60 0 )j. A + B = (15 0 + llO cos 60 0 )i + ( ll Osin 60 0 )j = 20 5i + 55V3j. The angle that A + B makes with A is () = tan- 1 ( ; ) (5~ : ) ~ t an - 1 (0.46 47) ~ 25 0. = ta n- 1 Alt.ernat.ively, the definition of the dot produ ct gives us () = cos- 1 ( A · (A + B ) ) = cos- 1 (1 50)( 205) + (0)( 55 V3) IIAII IIA + BII (15 0)(J 5 1l 00) ~ 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 here y F fore, F = IIFII cos ()i + IIFII sin () j, where () is the angle from the x axis. Since t he angle from t he y axis is 7r/4, () is also 7r/4, so F = 3 J2 (i + j ). (b) We compute D = 4i+ 2 j, so F·D = ( 3)2 )(4) +( 3)2 )(2) = 18"\/2. Also, II F II = 6 and IIDII = J20 . From the de fi nition of the dot product, F · D 18 )2 3 cos() = II F II IIDII = 6J20 = vT5 ~ 0.9487, i. e., () ~ 18°, or equi valently, () ~ 0.3 22 radians. (c) From part (b), we had computed F· D = 18)2. Knowing t hat cos () = 3 /.JfO, we calculate IIFII liD II cos() = (6J20)(3/JIO) = 18)2, also . 1.3: M A TRICES , DETERMINANT S AND THE CRO SS PROD UC T 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 t he plane or normals to the plane. x 7 TH E GEOMETRY OF EUCLIDEAN SPA CE STUDY HIN T S 1. Ma tri ce s a nd determi1 wn t s. A matrix is just a recta ngula r array of numbers. The array is written between a set of br ackets. Th e determ inan t of a matrix is a. number; a matrix has no numerical value. The determ in an t is de fi ned only f or square matrices and it is denot ed by vertical bars. 2. Comput ing dete rmi nants. K now th at I ~ ! I = ad - be. Also know that abc d e f 9 h Note the min us sign in front of th e second ter m on the right-hand side. The general met hod for comput ing dete rm ina nts is described next . 3. Computing n x n det er minants. Use the checkerboard pattern shown here which begins wi th a plus sign in th e up per left corner. Choose any column or row - usua lly picking th e one wit,h the most zeroes saves work. + + Draw vertical and horizontal lines through the first number of + t he row or column. The nu mbers remaining form an (n -1) x + + (n -1) determ inant, which sh ould be multiplied by the number (with the sign det erm ined by the checkerboard) through which both lines are dra wn . Repeat for the remaining numbers of the row or column . Finally, sum the resul ts . This process , called exp ans ion by minors , works for any row or column. The best way to r em em ber the pro cess is by pr acticing. Be sure to use the c orr ect signs. 4. Simplifyi ng dete rmi nants . Deter m inant s are easie st to compute when zeroes are present. Addin g a non-zero mul tipl e of one row or one c olum n to an o ther row or column does not change the value of the deter min ant, and th is c an often simplif y the computations. See exa.m pIe 3. 5. Computing a cr oss product. If a = (aI, a 2 , a3) and b = (b 1 , b 2 , b 3 ), then J k a x b = a1 a 2 a3 b 1 b 2 b 3 The order m at ters: a x b = -( b x a). Th e cross pro du ct is n ot commutative. Also, note that a x b is a vector , not a scal ar . 6. Properties of the cross pro du c t. T he ve c tor s a , b an d a X b form a right-handed system (see figure 1.3.2 of th e text) . Th e cross product a x b is orthog o na l to both a and b . The length of a x b is II allllb ill sinOI, where 0 is th e angle between a and b. Note that th e formula for the cross pro d u ct involves sin 0, whereas t he dot product involves cos O. 7. Mor e properties. If th e cross produ ct is zero , then either : (i) t he length of one of the vectors must be zero , or (ii) sinO = 0, i. e., 0 = 0, so the vectors must be parallel. 8. Geometry. Th e absolute value of th e de ter mi nant I ~ ! I is the area of the parallelogram spanned by the vectors (a, b) and ( c, d) o riginat ing from the same point. The absolute value of a bc the determinant d e f is the vol um e of th e parallelp i ped spa nned by the vectors (a, b, c), g h i (d, e, J) and (g, h, i) originating fr om the sam e point. The length of the cross product 11a x hll is th e area of the parallelogram sp a nn ed by the vec tors a a nd b. The vector a x b gives a vector normal to th e plane s panned by a a nd b . CHAPTER 1 8 9. Equation of a plan e. Recall t hat the equation of the plane is ax + by + cz + d = O. The vector (a , b, c) is orthogona] to the plane . Knowing two ve ctors in the plane , we can deter mine an orthogonal vector by using the cross product. Compare methods 1 and 2 of example 1 1. 10. Distance from point to plan e. You should unde rsta nd the derivation of the equation in the box preceding example 12. If necessary, review the ge om e tr ic properties of the dot product in section 1.2. SOLUTIONS TO SELECTED EXERCISES 2. (b) We subtract 12 times the third row from the first row and subtract 15 tim es the third row from the second row. 'f hen we expand along the first c ol umn: 36 18 17 o -42 41 -42 41 1 . - . 45 24 20 o -51 50 = 3 -51 50 = 3(-2100 + 20 91) = -27. 1 3 5 -2 3 5 -2 5. The area of the parallelogram is /l a x hll . We compute j k a x b = 1 -2 1 = -3 i + j + 5k 2 1 1 and so the area of the parallelogram is lin x hll = v'9 + 1 + 25 = 55 . 8. The volume of the parallelpiped is the absolute value of the 3 x 3 determina nt made up of t he vectors ' components. Expand along the first row: 1 0 0 _1 3 -1 0 3 -1 - 2 'I =- l. -1 4 2 -1 Thus , the volume is 1. 11. We want to find the cross product and then normalize it . We compute So J k v = -5 9 -4 I = 113i + 17j 103k. 789 There are two orthogonal vectors in opposing directions; they ar e given by ±v/llvll =: ±(113i + 17j - 1 03k )/ v' 236 67. Since all vectors are orthogonal to 0, the inclusion of that ve ctor into the problem does not affect the answer. 15. (a) The equation of a plane with normal vector (A ,B , C) and passing through the point (xo, Yo, zo) is A(x - xo) + B(y - Yo) + C (z - zo) = O. In this case, the equation is l(x - 1) + l(y - 0) + l(z - 0) = 0 or x + y + z = 1. (d) Here, the normal vector is parallel to the line, so it is ( -1 , - 2,3 ). Hence, t he equation of the desired plane is -1(x - 2) - 2(y - 4) + 3(z + 1) = 0 or - x - 2y + 3z + 13 = O. 9 TH E GEO METRY OF EU CLI DE AN SP AC E 16. (b) Two vectors in the desired plane are v = (0 - 1, 1 - 2, - 2 - 0) = (-1,-1 , - 2) and w = (4 - 0, 0 - 1, 1 + 2) = (4, - 1 ,3 ). The cross pr duct v x w is ort h og ona l to both vectors, and hence normal to the desired plan e. We co mpute v x w = - 5i - 5j + 5k, so the desired equation is -5( x - 1) - 5(y - 2) + 5(z - 0) = 0 or - x - y + z + 3 = O. 22. (a) Let D be the m atr ix with ro ws u, v, w . T hen Ul U2 U3 det D = u· ( vxw )= VI V2 V3 W I Wz W3 Use the fo llowing property of determinants: v· (w x u ) corresponds to two row exchanges of th e m atrix D , 80 we ha v v . (w x u ) = (-1)( -1) det D = det D and w · (u x v ) = (-1)( -1) det D = det D. To prove the other th r ee , rec all that u x v = -( v xu ). 26. The line perpendicular to the pla ne is parallel to th e normal of th e plane, so t he equ at ion of the line is l (t) = (1, - 2, - 3) + t ( 3, -1, -2 ). 2 9. Let u be the ve ctor normal to th e plane. Then u is perpendicular to 3i + 2j + 4k since v is on the pl an e. Alsu u is perpendic ul ar to 2i + j - 3k, because the vector Ai + Bj + Ck is e rp endicular to all vectors in t he plane Ax + By + Cz = D . To find u, we take t he cross produ t of 3i + 2j + 4k and 2i + j - 3k: J k u = 32 4 =- 10i + 17 j - k. 2 1 - 3 Wh en t = 0, we fi nd that a point on the plane is (- 1,1 , 2), so the equation of the plane is - 10 (x+ l) +1 7(y -l)- (z-2}=0 or - 10x + 17y-z - 25 =0 . 30 . First, fi nd the normal o ft he plane. The norm al o fthe plane is perpendicular 0 th e line passing thr ough (3, 2, - 1) and (1, - 1, 2). T he equation of the line is l(t) = ( 3,2 , - 1) + t(2, 3, - 3 ). Th e normal of t he plane is al so perpendicular to v = (1 , - 1, 0) + t( 3, 2, -2 ). T herefore, tw o ve ctors on t he desired plane are 2i + 3j - 3k a nd 3i + 2j - 2k , an the normal is (2i + 3j - 3k ) x (3i + 2j - 2k ) = - 5j + 5k. No w, we need a point on the plane, say, (3 ,2 , -1 ). Thus, the equation of t he plane is O( z - 3) - 5 (y - 2) + 5(z + 1) = 0 or - y + z = l. 34. T he plane passi ng through the origin and perpendicular to i - 2j + k is x - 2y + z = O. By th e dist ance formu la with (A, B , C , D ) = (1, - 2, 1, 0) and (X l, Y1, zt) = (6 ,1, 0), d = I Ax l + BY I + C Z l + DI = 6 - 2 _ --±- _ 2 V6 V A2 + B2 + C2 VI + 4 + 1 V6 3 ' 37. Sin ce all vectors in this exercise are unit vectors, li N x all = sin 0 1 and II N x h ll = sin O 2 . From S ne ll 's la w, n l sin 0 1 = n2 sin O 2 . Hence, nlll N x a ll = n2 11N x h ll. To e st ab li sh that N x a and N x h have the same direction, we assume t h at N, a and hall lie in th e same plane, and a and h are on the same side of N. Hence N x a and N x h both are perpendicular to this plane and parallel to ach oth er . Thus nl ll N x al l and n 211N X hll are equa l. 10 CHAPTER 1 38. First, 4 times the first row is subtracted from the second row. Next, 7 times the first row is subtracted from the third row . The next step is expansion by minors along the first column. Finally, the 2 x 2 determinant is computed. 1.4: CYLINDRICAL AND SPHERICAL COORDINA TE S GOALS l. Be able to convert back and forth between the cylindrical, spherical and cartesian coordin ate systems. 2. Be able to describe geometric objects with cylindrical and spherical coordinates. 3. Be able to describe the geometric effects of changing a coordinate. STUDY HINTS l. Review . You should review polar coordinates in your one-variable calculus text . . . ~ 2. Cylindrical coordinates. Denoted (r, {}, z), this is just like polar coordin ates except that a z coordinate has been added. Know the formulas x = rcos{}, y = rsin{}, r = Jx 2 + y2 and {} ::::; tan-1(y/x). 3. Spherical cooridinates. Denoted by (p, {}, ¢», p is the distance from th e origin, ¢> is the angle from the positive z axis and {} is the same as in cylindrical coordi nat es. Know the formulas x=pcos{}sin¢>, y=psin{}sin¢> and z=pcos¢>. Also know t. hat p = Jx2 + y2 + Z2, {} = tan-1(y/x) and ¢> = cos-1(z/ Jx 2 + y2 + z2) = cos- 1 (z/ p). 4. Graphs of r, p = constant. Note that r = constant in cylindrical coordinates describes a cylinder and that p = constant in spherical coordinates describes a sphere. You may have suspected this from the name of the coordinate system. 5. Computing {} and ¢>. Remember that, in this text, ¢> takes values from 0 to rr and {} ranges from 0 to 2iT . In some instances, it is more convenient to define {} in the range -rr to iT . You should be very careful about computing {}. If x = y = -1, t.hen tan-1(y/x) = iT/4, but plotting the point (-1,-1) in the xy plane shows that, in reality, {} = 5rr/4. This is why t he authors fuss with tan- 1 (y/x) in the definition. Plotting the x and y coordinates is very helpful for determi ni ng {}. 6. Negative r, p. Note that we have defined r an d p to be non-negative. If the distance is given as a negative number, we need to reflect the given point across the origin . 7. Unit vector's in spherical and cylindrical coordinates. The unit vectors in cylindrical coordi nates are e r , eo and e z . The vector e r points along the direction of r, while eo goes in the direction in which {} is measured , and e z = k. As one might expect., those three unit vectors form an orthogonal basis, and e r x eo = e z . Those vectors, however, lire not fixed as is the case with i, j and k , that is, if you change the point (7', {}, z), the set of unit vectors rotates. For spherical coordinates, there is also a set of unit vecto rse p , eo and e",. Those vectors , in terms of i, j, k and the cartesian coordinates of the point are worked out in exercise 7 (see below) of this section. 11 T HE GE OM ET RY OF E UCLID E AN SPAC E SOL UT IONS TO SELECTED EXER C ISES 1. (a) To conver t to re cta ngu lar coordinates, use x = r cos & and y = 1'S in 11: x = 1 cos 45° = v2/ 2 and y = 1 sin 45 ° = V2 /2. Next , use p = J x 2 + y2 + z 2 and rP = cos - 1 (z / p) to get the spherical coordin ate s: Ir. - 1 ( 1l' P :::;; V I 2" + 2" 1 + 1 = V 2 and ¢ = cos )2 1) = "4 ' Hence, for th e cylindr ic al co ordinates ( 1,45 ° ,1 ), the rect angul ar coordinates are given by ( /2 / 2 ,)2 / 2,1 ) and hes ph rical coor dinates are (V2 ,rr/ 4, 1l' / 4). (b) To co nvert to cylindrical coordin at es, we use l' = ..jX 2 + y2 and (J = tan -1(y/ x ): r = . )4+1 = v'5 and B = tan - 1 (1/ 2). Ne xt , use the same formulas as in par t (a) to get t h sp herical coordina te s: p=V4+ 4+ 1 = 3 and ¢ = cos -1(- 2/ 3). Hence, the re ctangular co or din ates (2, 1, - 2) convert to the corresponding cylindrical coord i nat s ( /5 , t a n- 1 (1/ 2 ), - 2) and t o the spherical coordi nates (3, ta n - 1 (1/ 2) ,c o s-1(- 2/ 3)). 2. (b) T his mappin g takes a point a nd ro tates it by 1l' radians about the z axis . This is follo we d by a re fl ection across the xy plan e. The net effect is that the point is re fle cted thr ough the origin. 3. (b) Reca ll that the angle ¢ is measured from the "Nor th pole." If ¢ is 7i ra dia ns , t hen t he loc at ion is at th e "S outl pole." The e ff ect. of c hang ing ¢ to 1[' - ¢ is ta king a point and r efl ecting it across the x y plane. 5. Let p ~ 0, then (p, 0, 0) is the positi ve z axis. Now, let ¢ vary from 0 to 1l'. Then (p, 0, ¢) is the ha lf plan e in the xz plane wi t h z ~ 0. By allowing & to vary from 0 to 2 1l', we rot ate t he half plan e described above. Th ere fo re, p ~ 0, ° ~ (J < 21l' and 0 ~ ¢ ~ 1l' describes all points in JR 3. If p < 0 also , the co ordinates are not unique . For exam pl e, (x, y, z) = (1, 1,0) has spherical co ordinates (V2 , 1l'/4,1l'/2) and (- V2 , 51l' / 4,1l' /2). 7. (a) Fir st , e p is th e unit vector along the ve ctor ( x, y, z) ; t herefore, the fo rm ula is xi + yj + zk e p = . J x 2 + y2 + z2 z y -----k--- --- ~ r y x x ( x, y ,O ) Next, e o is paral lel to th e xy plane and denotes t he direction in which t he angle B is measured . It is perpendicu lar to r = (x , y, 0), so ( ai + bj) . (xi + yj ) = O. Since we want B measured counterclockwise, a = -y (instead of y) and b = x. Therefore, - yi + zj eo = J x 2 + y2' 12 CHAPTER 1 To fi nd e ,p , we no te that e p , eo and e,p is a set of orthogonal vectors; they form a right-handed coor din ate sy ste m with e p x eo = - e,p. So . (x, y, z ) x (- y, x, 0) x zi + yzj - (x 2 + y2 )k ~ =- = . rp J x 2 + y2Jx2 + y2 + Z2 9. (a) The length of xi + yj + zk is J x 2 + y2 + z2 , which is the definition of p. (b) Note that Il v ll = Jx 2 + y2 + z2 = P and v· k = z; therefore, <p = cos~l(z/p) = cos-1(v. k/llvll)· (c) Note that Ilull = J x 2 + y2 , which is the cylindrical coordinate rand u · i = x; therefore, () = cos- 1 (x/r) = cos- 1 (u . i /llull). 13. Note th at <p will be between 7r / 2 and 7r because the region lies in th e lower hemisphere. From the triangle, we see th at cos a = (d/6) -;- (d/ 2) = 1/ 3; ther efore, we ha ve 7r - a ::; <p ::; 7r or 7r - cos-1(1/3) ::; ¢ ::; 7r . Now, p can be as la rge as d/ 2; ho we ver, as p gets smaller, its lower limit depends on <p . P ic k any ¢ , then ¢+ (3 = 7r and according to the diagram cos (3 = (dj6) -;- p. Rearrangement gives d/(6cos(3) = p = d/(6cos(7r - ¢)) = -d/(6 cos¢). Therefore, -d / (6 cos ¢) ::; p ::; d/2. So far , we have described the cross-section in one quadrant. The entire volum e requires a revolution aroun d the z axis, so it s description is 1 __ d_ < p < ~ 0 < () < 27r and 7r - cos- 1 ( -3 ) <_ A, < _ 7r. 6 cos ¢ - - 2 ' - - 'i' 1.5: n-DIMENSION A L EUC LI DEAN SPACE GOALS 1. Be able to ext end th e ideas of the previous sections to jR n. 2. Be able to multiply ma trices. STUDY HINT S 1. The space ]Rn . Most of this textb ook deals with the Euclidean spaces that we can visualize, jR2 and jR3. Many of the same properties hold in jRn. Ve ctor addition, scalar mu lt iplication, vector lengths, th e dot product and the triangle inequal ity are defined similarly. 2. No cross product analog. The cross pro duct in ]R3 does not have an easy analog in jRn, n ~ 4. 3. Standard basis v ec tors. Th e analogs of i, j, k are defined e j. The ve ctor ej is (0,0, . .. ,1, ... 0) with 1 in the zth position. Th e vectors ej a nd ej are orth ogonal if i =f j. 13 THE GEOME TRY OF EUCLIDE AN S PA CE 4. Matrices. A matr ix is a rect angul ar array of numbers . Un like a determinant, a matrix has no numerical value. You should remember th at an n x m m atr ix has n rows and m columns. The (i, j) entry is the number loc at ed in row i, column j. 5. Mat rix m ultiplication. You should practice until m at rix multiplication becomes second nature to you. Let th e components of A be a i j and let those of B be bkl, where A is an m x p matrix and B is a p x n matr i x. Then the components of A B are p (A B )mn = L amjbjn . j =l We can only multiply an m x p ma t r ix with a p x n matrix, i.e., [m x p ][P x n]. Note that the number of columns of A and the number of rows of B must be equal (p in this case). The result is an m x n mat rix ("cancelling" th e p). 6. Non-commutativity of m a tr ix multiplication. In general, AB # B A . In fact, AB may be defined when BA is undefined . However , m atr ix multiplication is asso ci at ive; i.e., (AB)C = A(BC) if the product AB C is defined . 7. Matrices and mappings. An m x n matr ix can represent a map ping from ]Rn to ]Rm. To see this, l et A be the mat rix and let x be a vector in ]R n, r epr esented as an n x 1 matrix, and y be a vector in ]Rm, an m x 1 m at rix. Then th e matrix A tak es a point in ]R n to a point in ]Rm by the equation Ax = y. SOLUTIONS TO SELE C TED EXERC I S ES 2. (a) Use the properties of leng hs and dot products: (x + y) . (x + y) + (x - y) . (x - y) x . x + 2x . y + y . y + x . x - 2x . y + y . y 2 x· x + 2y· y = 2 11xl12 + 211Y11 2 . The figure at the left depicts the equation geometrically. By the l aw of cosines, we have 12<tl y and We also note that a + j3 = 1r, so j3 = 1r - a . T herefore, cos j3 = cos( 1r - a) = - cos a. Adding th e two equations from th e law of cosines yields 2 11xl1 2 + 211Yl12 ;: Ilx - Yl12 + I' lx + Y112. 4. To verify the Cau chy-Schwarz inequality, we compute Ix· yl 1(1)( 3) + (0)(8) + (2 )( 4) + (6)(1)1 = 17. Ilxll '1'1 + 0 + 4 + 36 = J41. Il , yll J9 + 64 + 16 + 1 = V90. Thus , we indeed have Ix . yl = v'I7v'I7 < v4f J90 = II 'xlillyll. For the triangle inequality, we compute x + y = (4,8,6,7) and II x + yll = J16 + 64 + 36 + 49 = v'l65 < 13 . Indeed, we have Ilx + yll < 13 < 15 = 6 + 9 < AT + v'9O = Ilxllllyll. 14 CHAPTER 1 8. We compute 1 n AB~ r: -1 ~31 and A+B~[: 0 2 1 1 -1 1 Expanding by minors across the first row gives det A 3: 1 2 -1 I 11 ~ II = 6 - 2= 4, 0 1 + 1 1 detE 11 0 1 ~ 1- 11 ~ ~ 1 = -3, det(AE) 1-1 1 1_ 115 and 3 1 -1 1 ~1 1- 31 5 1 -1 1 1=-12 det(A + B) - 41 ~ ~ I = 8. 11. (a) For n = 2, I 'xall det('xA) = 'xa21 For n = 3, 'xall Aal2 'xa13 det( AA) Aa21 'xa 22 'xa23 Aa31 'x a 32 'xa33 'xall det('xAd - 'xa12 det('xA 2 ) + 'xa13 det(AA3) ,X . ,X2(a11 det Al - al2 det A2 + a1 3 det A 3) ,X3 det A, where AI , A2 and A3 are 2 x 2 matrices obtained by expanding across the first row . Assume that for n = k, det('xA) = Ak det A. The for n = k + 1, det(AA) ca n be found by a process analogous to the 3 x 3 case: det(AA) 'xall det('xAd - 'xa12 det('xA 2 ) + ... + (_l)k 'xal,k+l det('xAk+d ,Xk+l det A, where AI, A 2 , ... , Ak+l are k x k matrices obtained by expanding across the first row . By induction, det('xA) = An det A for an n x n matrix A. 14 . Assume, as in the book, that det(AE) = (det A)(det E). Then det(AEC) = det[(AB)C] = det(AE) det C = (det A)(det E)(det C). 17 . Multiply the two matrices to get the identity matrix: [ a b] 1 [d -b] e d ad - be -e a Similarly, we can show that