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000 chapter 14.pdf - 14.1 Vector pitterentiation and...

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Unformatted text preview: 14.1 Vector pitterentiation and Differential Operators 913 Mfitfishm HISTORICAL NOTE I The Jacobian matrix is named :n honor of the German mathematiCian Car' GuaavJJacobMiBOtnBSU.whonmdeanumberohmponwflconnmuhons to nineteenth—century mathematics. He was one of the first to give a systematic theory of determinants and dealt with detJ,. which is called the Jacob/an determinant, rather than With the matrix J,- per se. He also worked with mflerennalequanOns. f Pierre~Simon Laplace did extenswe work in physrcs, astronomy, and chemistry. as well as mathematics. He used the equation Vli = 0 in his work . on the gravitational attraction between planets in the solar system. This work helped him explain the regularity and stability of planetary motion. Exercises 14.1 In Exercises 1—4. find limxfl F(.t) for the given function. Is In Exercises 13—16. compute the curl of the given field F at the function continuous at the given point x = (1? Why or why x and at a. not? l3. Fix): 3x3i + xyj + :k121 = (—3, 4. 2) x4 — i“ , . .1 7 . 3 . I. fix. )1.) = , 7 ‘ ‘ i + (x- + .VJlJ. + Q‘s-i“; a = (0‘0, 14. The I‘ and a of EXLI’CISC 9 X‘ + _l" W . . ,. . . . 15. Fix): 9‘ stnycoszi + e costi'coszj — e* smysm : k; . x3+l ._,1 7'; 2. Fix. y) : tun nxyi <— e“'3"j + .2 7] k; a = i2. 3) J -‘ ~11..th 6’ ‘ 7 _ 3’ + 10. Fix) = iyej"z + Znyw‘”, xe‘“, Eye”? 3 = (I. 2. 3) x3 — \‘l . 3. 17x, v = ~ 'A i + it:e""' + ixl * i'zz‘ + r“: k: . , . ( ' ) x -i- y ’ ‘I ‘ ( ) lii Lxermses 17—24, Show that the given functions are harmonic. a = i0, 0, 2) l7. fix, y) = In \;3+ 3'2: ix, y) ¢ ii), 0) 4. Fix, y, :) of Exercise 3; n = i1. — l. 0) 7 j 18. fix. y) z e“ cosy + 3x- — 3)" ln Exercises 5—8. find JF(a) for the given f and :i. '9. fix. )‘I = sin x sinh y 20- fix. H = COSh 2X COS 3y 5. Fix. y. :) = (x2 + 6‘" -‘.- zili + isinxy -i )1erle 2" fix. ‘y‘ :l = x1 “L y: — 2:: u = (I, n. — l) 22. fix. ,i‘. :) : 3x1) 2): — 5:3 6. Pix. y, 2 = ln(.\"’ + y: + ::“")i + Semj; a = (l. 0. l) 23. fix, y. z) = ”Viv: gt y: + :7; (x. y. :l s5 (0. 0.0) x cos 3.): — y: sin 7.x 24. fit, y. :) = sin .\' sinhy + cos xcosh : 7- Fl~Y,,1'.L')= Vii: + 4)‘2 + (‘2 ; :i = (l, 1,0) 25. Laplaciiin in Polar Coordinates. Suppose thzit : = fix. y). . TC. . _ 2‘ .-2 _ , tan ’0 ' 4“ and we introduce polar coordinates x = roost) and A") — : _i' = rsin 0. Then : = firms 0. rsin 0) = gir. 0). Show that 8. Fi.\‘.y. :) = x2 — :3 :11 = i1. — l. li‘Z) V’ [39 1 (lg 15g ‘ ‘ ‘*‘*' ’5=:» 773*“: tomxi-‘ (r' T r’ (0' r tr _ _ ' (Hint: See Example 4, Section 12.6.) In Exermses 9—12. compute the divergence of the given vector _ . ‘ field F at x and the given a. 26. Laplucmn in Cylindrical Coordinates. Suppose that W = fix, y. :J. iiiid we introduce cylindrtcal coordinates 9. Fix) = xi + (x: + 3)j + (y: + x)k: n = (2. —2. 3) v = roost), y : I‘Sil’l 0. and : = :. Then w = fir cos 0. rsin 0. r) = gir. 0. :l. Show that , 10. Fix) = ix,2 + x1" + x31 -:- ,\‘f.x,.\'3.\'3.\',., XIX: — xij, x,.\'4 — xle); a = (l. — l, 2. 3) ‘ If?! 520 153,0 5.4 7 7 V3ftX.y.:)=r r'+'--1+' .efirq,» ll. l“(x)=xyl:2i+:‘sinyj+x‘e’k;a=(l.x. —'_’) rr'r “'- r‘ (’0' ‘3' , x + : r _ x _- _ r (Hint: Sec Example 4, Section 12.6.) 12-F‘(Xi=-3—7‘§i+j" 1 a.i+—u' a} 1k; . . v. . . , x + y' :‘ X + i" + :“ ." + y' + :‘ 27. Lupluciun in Spherical Coordinates. Suppose that a : (1,0. — 2) W = fix. ‘i'. :l. and we introduce spherical coordinates 914 Chapter 14 Vector Calculus 28. x = psingbcos 9. y = psin (psinB. and : = pcostp. Thcn w = f(p sin (1: cos 0, psin (,bsin t), p cos 4)) = g(p. (p, 0). Show that N l (b) Use (a) to find curl (-—— x) without further calculation. IXI In Exercises 29—33. prove the indicated result. 29. 31. 33. 34. Theorem l.3(b) Theorem |.5la) Theorem l.9(b) 30. Theorem 1.3tc) 32. Theorem 1.9la) Prove that if all second partial derivatives of the coor- dinate functions fhfz, and f3 of F are continuous, then curllcurl F) = thiv F) — Vzli‘. where VZF = (szlfi + (Vlfzh' + (Vlfl) k. . Prove Theorem 1.9lc). Use Exercise 34 to compute curl(cur| I") if le,y, 2) = (x: + ylli + (xy + :3)j + lxz + y:)lt. . Verify the formula for divlF x G) in Exercise 35. if F(x, y, z) = xii + ylj + :3k and Glx,» 2) = yzi + :“‘j + xlk. 38. 39. 40. The trace of a 3-by-3 matrix it” “t: “is A: an an (123) , 53g 2 5g 1 (=19 c054) 5g 1 Pg “3% a3: "33. VL‘v:—‘-fii'l'——V +i'l 7 + 7 ~7—TVVV+_T"- 177 “7-1" ' ‘ ‘ ' 8p" p 5p p“ (tidy p“ sm 4) (345 p'sin' ¢ (34,» IS defined to be all + an T1133, the sum of the mam l diagonal entries. If F: R3 —’ R3, then what matrix discussed (a) Show that V(]x|) = — x if x = xi + yj + z k ;é 0. in this section has trace div F? Refer to Exercise 38. Iff(x. y) has continuous second par- tial derivatives, then express V'Vf as the trace of a matrix we have met. A fluid with velocity field F is called incompressible if div F(.\) = O. A fluid flow is called solenoidal if l"(x) = curl Gtxl for some G. What relationship can you conclude exists between incompressible and solenoidal llows? (Assume appropriate differentiability hypotheses.) ln Exercises 41—44, sketch the given vector field as in Figure 1.1. 41. F(x,y) = —é_\'i + lxj l 42 F., ‘ = ":7 . (r J) (YAFINLNTU) —l 43. FLY, ). I) = 7 14.; (xi + yj + :1“ \,/.—\'Z + y' :- 44. l<‘(x. y. :l = iyi + .rj + k 14.2 Line Integrals The double and triple integrals of scalar functions f: R" —> R, where n = 2 or 3. were developed in the last chapter as analogues of the ordinary definite intc‘ gral [2 f (x) dx of a function f over an interval [0, b]. They are defined over regions D having the same dimension—n—as the domain of f: R" —» R. This corresponds to the fact that a closed interval [(1, b] on the real line has the same dimension—l— as the domain of the function f in [g ftx)dx. In this section we define a different kind of generalization of I: f(x)dx, this time for vector functions F: R" -—> R” over a one—dimensional subset ofR". namely, a parametrized curve C in R”, for n = 2 or 3. Such a curve is a generalization of a one-dimensional closed interval [(4. b] in R‘. We define the new integral in such a way that when n = l and C = [a. b]. we get the ordinary definite integral [2 f(x)dx. Before constructing this new integral, we briefly review the main facts about parametrized curves that are needed for its construction. (You may also wish to review Sections 8.5 and 11.1.) A parametrized curve C in R2 or R3 is a continuous vector function g: R a R” that is defined on some closed interval [(1, b] of the real line. We restrict attention to curves that are piecewise smooth. This means that [a, b] can be partitioned into a finite number of subintervals [ab b] on each of which g is smooth. that is, g’ exists and is continuous on Exercises 14.2 In Exercises 1—14. evaluate the given line integral. 1. ic F ‘ dx, where F(x. y) = xzyi + (x3 — y)j. and C is the curve .\'(I) = (t, l — t). 0 s t S l. 2. ic F ~ Tds. where F(x, y) = (x + 2yli + (x: — y1)j, and C is the triangular path from (O. 0) to (1,0) to (1. It to (0, OJ. 3. 1"- F ' dx, where F(x, y) : vxz + y: i + l — xzj, and C is the upper half of the unit circle .\'3 + y2 = l. traversed from (1,0) to (— LO). 4. ic F - dx, where F(x, y) = J1 — yli -:~ Vii — xzj. and C is the curve of Exercise 3. 3. (C xydx + (y: + lldi'. where C is the curve y: = x from (0, 0)IO(1.1). 6. L-(xl — y‘l)dx + 2nd}; where C is the curve y = x2 front (I, —l) to (2, 4). - v x . . 7. g: ' . dx + , , 11y. where C is the Circle ‘ .\" + y' X‘ + y' x3 + y: = 9 traversed counterclockwise from (3, 0). (This exercise is referred to in Section 4.) . + = x — ~ . . 8. f j ), dx — 7, 'i, dv. where C is the path in ‘ .\" + y" x‘ +1" ' Exercise 7. 9. (C F ~ dx, where l<'(x. y. z) ; yi + :j + x k, and C is the line segment joining (O. 0, 0) to (2. 4, — 1). 10. ic F ' Tds, where F(x. y. :) = xzi + ylj + :I k. and C is the curve x(t) = (t. {1. t3). 0 s I s 1. 11. If x dx — ydy + :d; where C is the helix x(t) = (cos i, sin i, Uri), 0 s t s 27:. 12. if F - tlx where Fixia x2, x3, 35.x) = (x1 _ x;. x1 _ x3, 324 " x], xi-‘zxs-H), and C is the curve xltl = (I, :1. r", :4), 0 s z s 1. l3. [C(x: + y3)dx + (y: — ledy + (.\'y - :3)t1:. where C is the line segment from (1, 2. 3) to (4, 3. 5) followed by the segment from (-1. 3. 5) to (l, l, I). 14. L» ydx + : dy + x d:. where C is the curve of intersection 'of the plane x + y = 2 and the sphere (x — l)2 + (y — 1)] + :3 = 2, traversed in the counterclockwise sense viewed from the point (2. 2, 0). In Exercises 15—20. verify Theorem 2.4 by calculating i-f F - dx for the vector function F and curve C of the indicated exercise. 15. Exercise 1 16. Exercise 3 17. Exercise 5 18. Exercise 7 19. Exercise 9 20. Exercise 11 21. 24. 26. 27. 28. 30. ”:2 Eve 'ntsére‘s 923 Find the work done by a force 1"(x, y) = (2 — y)i + xj in moving a particle along one arch of the cycloid given by .\‘(r) = (r —sint, 1 — costl.0 31$ 2:. . Find the work done by a force 1-‘(x. y) = (l — y)i + xj in moving a particle along one arch of the cycloid x(t) = 2(t # sin I, I — cost). 0 s t s 2n. . The repelling force between a charged particle P at the origin and an oppositely charged particle Q at (x, y) is I F. =.——w.— .'+ ‘. (w) (Xz+yz),,2[“ .vi] Find the work done by F as it moves Q along the line segment from (1,0) to (—1, 2). Refer to Exercise 23. Suppose that F moves Q along the polygonal path from (1.0) to (l. l) to (— l. 21. What is the work done? The line integral ofa scalar function f: R" —> R over a curve C is defined by (Clix-ids = fl J‘ixiniix'tnl dr. (Recall that ds = |x'(z)|dt.) lff(x(t)) gives the density of a wire in the shape of C at the point x(t). then the mass of the wire is defined to be (cflxids. Formulate a definition of the center of mass of the wire. Formulate a definition of the moment of inertia of a wire in the shape of x = x0) relative to an axis [at distance dlx. y. :1 from (x, y, :) on C. (Assume that the wire has density fix, 3'. 3).) Use Exercise 25 to compute the mass ofa helical wire x(1) = (cos I, sini, t), 0 st 5 27:, if its density is given by 1 (5(_\‘, y, z) = X2 + y" + :'. Find the center of mass of the wire in Exercise 27. Find the moment of inertia of the wire in Exercise 27 about the :-axis. A force 1“ acts on a particle to move it from x(tol to still along the path x = .\'(1). Show that the work done is the change in kinetic energy. i.e.. w’ : leth .)|2 — imivirolil' (This is one form of the law of conservation of energy.) [Hintx Compute (d/dt)(|v(t)|3) and use F = mx“(t). Newton's second law] . If C is a smooth. but not necessarily simple. closed curve that does not pass through the origin, then the winding number of C is defined as (l,/21r)j'(~ F - dx. where i r x Fix. .v) = , 7,7» i + . . x- + i" x‘ + _r' j. It is shown in advanced calculus that the winding number is always an integer. which gives the number of times C winds around the origin. If it is positive. then 924 Chapter 14 Vector Calculus C encircles the origin in a counterclockwise sense. It. it is 32. If F(.\'. y) = —y' mi + x’ “lj then compute [CF - dx negative, then C encircles the origin in a clockwise sense. over the hypocycloid of four cusps x3” + y" 3 z 1 Compute the winding numbers of parametrized in the counterclockwise sense (see Cii XI!) = (3 cos I. 23m“ 0 :1 t S 67:. Exercrse 41. Section 8.5). and ('3: x0) = (cost, —sint), 0 s I 5 2r 14.3 Green's Theorem This section takes its name from a result that extends the fundamental theorem of calculus to two-dimensional vector tields. To describe the extension precisely, it is helpful to state Theorem 4.2 of Chapter 4 as follows: lff is a smooth function (that is. f’ is continuous) on [(2, b], then b , . (ll Lftxth=ftb)-J(a). Qualitatively. this says that the definite integral of the derivative of f over FIGURE 3.1 ." 3_.1 DEFINITION [11, b] is completely determined by the mines of f on the boundary of [0, b], that is, on the two-point set {(1, b}. The various forms of Fubini’s theorem in the last chapter (starting with Theorem 1.9) constitute an extension of this result to scalar functions of two or three variables. In particular, partially antidiffer- entiated functions, such as jf(x. y)d_v, are evaluated just on the boundary of the region D ofintegration. such as the curves y = gl(x) and y = gz(x) in Figure 3.1. Green’s theorem is the first of several analogues of(l) to be presented in this Chapter. It says roughly that the double integral of a certain kind of derivative of F over a standard type of region D is given by the line integral of F over the boundary of D. This lets us evaluate some complicated double integrals as line integrals. and sometimes makes it possible to compute a complicated line integral by working out a less-involved double integral. To state the theorem precisely, we must specify the “standard type” of region and the “certain kind" of derivative mentioned above. We begin with the type of region. A plane region I) is simple if it is the part of R3 enclosed by a piecewise smooth simple closed curve that intersects any vertical or horizontal line either in at most two points .4 and B or in one line segment AB. As the name suggests, simple regions are geometrically uncomplicated. Fig- ure 3.2 shows three examples of simple regions. The region in Figure 3.2(a) is bounded by a smooth curve, the circle x3 + yl = 1. The regions in Figure 3.2(b) and (c) are bounded by piecewise smooth curves C. Points where x’ fails to be continuous are labeled. In all three cases, vertical and horizontal lines intersect C in 0, l, or 2 points. Figure 3.3 shows another type of simple region, in this case a square. Notice that some lines parallel to a coordinate axis (such as x = a and y = c) intersect C in at most two points, while others (including x = — I and y = 1) meet C in entire line segments. Figure 3.4 shows a nonsimple region. Although some lines (like x = a and y = c) intersect C in at most two points. there are others (like y = d) that intersect C in four points. We try to picture a general simple region in Figure 3.5. The regions D and R in Figure 3.6 fail to be simple. because the lines shown in Figure 3.7 intersect 14.3 Green's Theorem 933 EF by .\'(I) = (x, y) = (I), t). a S t S [3. Then dx/dt = 0, so 11x 2 0. We therefore get I” Ptx, _v)d.\' = 0. Similarly, I,” Ptxt y)dx = 0. Thus from Definition 2.3, ('l 1) JC P(x, _y)dx = In: P(.\‘, y)tlx + I” Ptx, y) dx. Since a vertical line crosses each of AE and BF exactly once. these arcs arc the graphs of functions y = gl(x) and y = gltx) for x e [(1, b]. So they can be parametrized as curves C. for i = 1, 2, by x = I, y = gitt). It is natural then to use x as a parameter instead of I. Since C2 is traced out as x goes from a to b, C; joins B to F. Since C traverses the boundary of D in the counterclockwise sense, C proceeds from F to 8. Thus we have from (1 1) and Theorem 2.4 J; Ptx. y) dx = Thus (9) is proved. f6] P(x. y) dx + f-c; Ptx, y)d.\' Ll P(x, y) dx — .i‘cz P(x, y) dx L" Plx. gitxndx ~ f ax, 91mm — I 1" [Pm yam — Ptx. 91mm»: b y=g2(.x) . _ b 92“} (3})(1‘. Y) _ f; P(x.y)]y=gl(xld.\ _ . [W] a), .1) dx 5P _ff 4:7 d4. by Fubini's theorem 1) (y The approach to (10) is quite similar and is left as Exer~ cise 37. Addition of (9) and (IO) gives (2). [Qtan| Exercises 14.3 In Exercises l—l-i. use Green’s theorem to evaluate the given integrals. 1. *5c 21de + 3.\‘ (1y, where C is the polygonal path from (—1, 2) to (3, 2) to (3.5) to (—l. 5) to t—l,2) h) - {tr (312 — ISM-V + (xi + yl)dy. where C is the unit circle X: + y2 = l traversed counterclockwise (a) . {3(- F . (Ix, where I“ is as in Exercise 2, but C is the circle x1 + y2 = 9 traversed clockwise 4. {T —ye‘dx + xe" dy, where C is the polygonal path from (I, 1) to tl. 1) tot—l. —l) to (t, —1) toll. 1) U! . {if yzefilx + 2yt"dy. where C is the path of Exercise 1 6. If (8" + y -— 2.\‘l)d.\' + (7x — sin y) dy. where C is the triangular path from (O, 0) to (0, l) to (l. 0) to (0.0) 7. [C(xz + y)d.\’ + (2x — y)r1y, where C = Cl u C3. Here Cl is the polygonal path of Exercise 1, and C2 is the unit circle parametrized in the clockwise sense. 8. j} (x3 + sin x + y)dx + (23‘ — sinycosyhly, where C is as in Exercise 7 9. §(sin y — x3y)d.\' + (x cos y + xy): rly. where C is the curve of Exercise 2 10. {6 tr — xy)dx + (y3 + l)tl_v, where C is the polygonal path from (I, 0) to (2.0) to (2. l) to (l, l) to (1.0) —v x . . 11. ¢ , ‘ , dx + —, 1(1y, where C 15 the ellipse f .1" + y' x‘ + y‘ 4x: + y2 = 16, traversed counterclockwise . _ V 11g). , ' ,dx-l— , , L x‘ + y' x‘ + y' 5x2 + y: = 25, traversed clockwise (1y. where C is the ellipse x t‘ .. 11f , ,dx+ ,' —;cly,where( =C1uC:.Here C x’ + y' .\" + y’ Cl is the parabolic path from (— 1,0) to (,2. 3) along y = x3 — I, and Cl is the line segment from (2. 3) to (— 1,0). 934 Chapter 14 Vector Calculus I4. x r I , , dx 4 . , ' , dy. where C = Cl u C1. Here C X‘ + y' X" + y‘ ‘ Cl is the parabolic path from (—2, 0) to (3, 5) along j.‘ = x1 — 4, and ('2 is the line segment [3, 5) to ( —- 2, 0|. In Exercises 15—20. find the area of the given region. IS. l6. 17. 18. I9. 20. 2]. l0 DJ 24. 26. 27. The region D enclosed by the ellipse 4x3 + ",2 = lb The region D enclosed by the hypocycloid x13 + y“ = I. (See Exercise 32. Section 2.) The region D enclosed by the hypocycloid x1 3 + y2 3 = 4 The triangle with vertices (l, 2). (3. 2). and (3. 5) The annular region between x: + y: (Hint: Think before resorting to (8)!) 2 l6 and x3 + y: = 9. The annular region between 3:3 + y: = 4 and x3 + y: = 16 Show that if th. y) = y‘lexi + 2ye*j is the velocity field ofa fluid flow. then there is no tendency for a vortex around the origin. . Repeat Exercise 2! for F(x. y) = (x3 + y3 —1 y sin xli + (3xyI — cosxlj. [f D is a region satisfying the hypotheses of Theorem 3.3. then find a formula for f and y". the coordinates of the centroid of D. as a line integral. Generalize Example 5. namely. let (‘P LQ F(x) = Ptxti + thlj. where e; — — for x 1 0. t‘)‘ t X Suppose that Cl and C2 are smooth simple closed curves with the same orientation, which enclose regions D; such that 0 6 D' 2 D2. Then show that (tic! F ‘ dx =g5t“ I" ~ tfx. . Generalize Exercise 11. namely. prove that if y . N ____.V._ I .3. Fix}: . , x’+y‘ «.l, x2 + y" then for any smooth simple closed curve C ifO is enclosed by C iffl is not enclosed by (I ‘3- g) F ¢1\—_{" lfC is asmooth curve parametrized by x = xi!) = mt). y(r)) for l in [(1. b] then show that 1 di dx N‘ = ., e e l!) Mill (d1 d1 1) is a unit normal vector to C at every point on the curve Use Exercise 26 to show that iii ’ de + de =(fi [7 - Nds. If F: P(x y)i + Q(x y)j is the velocity field ofa fluid flow in R' this integralis called the flux across C Explain how it measures the amount of fluid flowing across C. (This is referred to in Sections 6 and 8.) 28. 29. Use Exercise 27 and Theorem 3.3 to show that, under suitable hypotheses. if div 1? (1.4 = {D (In Section 8 we will see how Green‘s theorem in this form can be extended to three-dimensional vector fields F: R3 —v R3.) 1“ - N (Is. Prove Green' s fit st identity an analogue of integration by parts If f: R“ —> R is diflerentiable on a simple region Dand all second partial derivatives of g: R’ —» R are continuous on D, then use Exercise 27 to show ffrixiv’igod4=f jmivg Ny—ds £1fo ngA =j:.,,f(xiDth)dx—Hwygda, D where D\(g) is the directional derivative of g in the direction of N . ln Exercise 29 assume that all second partial derivatives of f and g are continuous on D. Then prove Green '5 second identity. fflfhlengl—gixwzflxl] dA=fw f(-\')1Vg ' Nids D — f,” yth . Nids {LDUDtxlyl—gDstfllds. What can you conclude if f and g are harmonic (Definition 1.7)? . Suppose that a flowing fluid in the plane has density 5(.\', y, r) at the point (x. y) at time I. Show that ‘ic 6h: y. Ill‘ ~ Nds is the rate of loss of mass per unit time ...
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