H5s - 5.4 Vector Derivatives 1. These are all simple...

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Unformatted text preview: 5.4 Vector Derivatives 1. These are all simple computations. 2. These are simple computations too. 4. (Also 5, 6, 7'.) These are straightforward calculations, but the verifications of (5.25), (5.27")= and (5.33) require a little masochism. It is less fiustrating to start with the complicated expressions (on the right in (5.25) and (52?), on the left in (5.33)) and work toward the simpler expressions on the other side. 9. This follows immediately fi'om (5.29) and (53(1). 5.1 Arc Length and Line Integrals 1. (a) In“ |g'[t}| a: : f0“ 1/o2{ain2 t + c032 t) + a! at = I 0"” o’a‘i + 52a: : amt/a2 + bi. (b) If: |g'[t)|dt = If [1:2 —1)'2 +4t2di = fag“? + 1M: 2 1%. (c) ff |g'(t]| an : ff xii-2 + 4 + 4:2 at = fie—1 + 21:] [fl 2 1+ r22 — 1 : e2. (de If: |g'[t)| a: = If Jae + as: + are + see :11: = a fin + t) (a = 24. 3. The element of arc length is do = 1.! l + ainhl‘z 1' dm : cosha? (in. Thus the arc length or “mass” of the curs-e is L11 cosh a: do: 2 sinh 1. and the y—moment is 3; d3 2 L11 cosh; a: do = %If_11(1 + coahflm} do: : . [2 + sinhfi}. Thus fi : {2 + ainhBM-‘l ainh l. and E : ll b}? symme— try. 4. as = |g'[t}|dt = 1liaisingt+=1coa'*t+alt'fidt: 2M1+t2dth so [Curses = fiat/1 +£2.11: = go + 4W2J3tl2 — 1]. bah—e 5. (a) Parainetrize C by? g(t} 2 (LL. t]. f] 5 1‘. 5 1; then F(g(t]} jg. F - dx = [013:2 an 2 1. [’32. 32-. t2} and shit} [1,1,1]. so 2 + [si111r13]2m}d:r = [ — lie—12 — {Efrrjmcosa'rr + 6. (a) fame—3“ do: + aina'rrdy} : Jrflll(ri:e_I 2 {2ftr2)51111r£:|; = — E_1)+[2}l'fl'). (b) femurs +zdy +333; dz] = [51— sin:a t +tcoat+aintcoa t} dt : coa 2t — %+ i aint +coat+ %Sln2 t]? 2 —r. (c) On the segment from (0.0) to (1.0) we have 3; : fl and dy = 0; on the segment fi'om (1,0) to (1,1) we have a: : l and (iii? 2 It}. and on the segment fi'om (1.1) to (0,0) we have 3*; : a: and d3; = r137. Hence the integral is fol odi- +If01(—2)dy + [int-i=2 - 233)de = 4 + é = ‘i- 5.2 Green’s Theorem 1. (a) Let D be the unit disc; then f0 F r ab: = — fin 2 dry :1"? = —2a' (the minus sign is there because the circle has the wrong orientation). (billet-3.12 =11: - “laden = .1531 fall-2 - 2a} do d3 = lull-233 - 23"!) rim = -%- M (63' fc:[{$2+1flmix+a2) dm+tfiri+5~ryl is] = In? .f§[{101+5a}-{1fl$+2a)] d1" d3; = if is fog 3:: d3; = 2 ‘ 6 : 12. (d) fi1[233ycos 3:2] 2 firflycoa y? = flflftmg ain 152}, so the integrand of the double integral in Green=s theorem vanishes. . Directly= by using the polar angle 9 as the parameter for both circles: It?” 32 coaflflsin‘afldfl— It?” 2 cos2 Gain? ddfl : [02” sin2 26' dd 2 %a. By using Green‘s theorem: Hflyl‘! + 32] d3? d1 = 2a" 23 _ E_fi f0 f1 r drdfl—Eir- 4 — 211'. . If S is the region inside C, fchf drt' + (3.1: — .123) dy] : HS 3(1 — re” — pg) d3: dy. The integrand is positive inside the unit disc and negative outside. so the integral is maximized by taking S to be the unit disc and C to be the unit circle. . Take the arch given by [l g t g 211': the region under it is bounded on the bottom by the segment [H.I 2a] of the ra—axis {where d3; = D) and on top by the cycloid (traversed from right to left). Thus the area is fag 3: dr 2 f2“: Rub! — sint} sintdt or — fa“; 3,! da: : In” REIEI — cos t]:2 dt; both integrals are equal to 37rR2. 5.3 Surface Area and Surface Integrals 1- A : _U_.Es+y2{fl2 V 1 + y? + $2 (til: I'll? = 1 [En- ‘le' 1 + PETr‘ft‘Em = %?T[{1+ Hula” — 1]. 2. A = flinging v’l + 42:2 + 4y” dr flit = f5” j; «r + as rdr at: = %a[(1+ titties” — 1]. 4. U! . Clearly f The integral can be set up in two ways; as in Example 1 232). First way: The upper half of the surface is z : (Ma) [:2 — $2 — yL’. so the area is + [h2 — 112:”‘2 2 dd 4 a —‘14 r “f0 The substitution .s : Va? — r2 simplifies this to (41cm) [03 ago? + {a2 — :52]32 do. and the substi- tution u = sin. turns this into 41m In] :52 + [a2 — 521112 do. Second way: Use modified spherical coordinates to parametrize the ellipsoid. With a: : osingacos 6'. r 2 asin sin (3'. and a : boos go. the formula {5.20) for area yields the integral 21111. I; cringe b2 sin2 (,9 + o2 cos2 (price? and the sub— 52 2 2 1+_I.Lyi . , rdr. a as _ mg _,y.a stitution “H. : coacp turns this into 21m b2 + {a2 — till-it'd dit. This is the same as the integral obtained in the first way since L11 : 2 D1 for even functions. Finallya one uses a trig substitution (H : [b tantflv‘og — b2 ifa. :> b, it = {trainfluted):2 — o2 if?) > a) or a table of integ'als to eval- uate the integral as 21m2 + 2a‘n.t::2C',f«tfIa2 — trial. where C" : logflo + v’og — b2]be if a. > b and C : arcainhfi?2 — [12 f b} if a < 1:. (Both expressions for the area have the limiting value 41mg as h —:~ :1.) fl = U by symmetry. The z-moment, in spherical coordinates. is flszdfl = 1 '2 . . . J2 . _ 2:7 for” coagosm gadga : it amt gala“ : it. and the area 15 211'; hence a : E. 8. (a) nclA = (—yi—mj +11) dydsr, so 'fstmdfl = fl; Elli—3:23;” —:ay) dyds: : ffl—gmz—flm} do: = “ 12—3. 9 s (1)) Since lines through the center of a sphere are perpendicular to the sphere, the 1mit normal to the Lmit sphere S at a point x E S is simply the vector x : xi + yj + zk. Hence JIS F + ndA : fig .123 div-1, which 1vanishes since m3 is an odd function. Alternatively, in spherical coordinates one finds ndA : [sin [pcoa Hi + sin (,9 sinIE-‘j + cos cpk} sin go all? dip; so F - ndfil : If; Inga: sin‘1 [,9 cos3 Halt} dip : l]. (e) 3 : 51 r": 32 where 51 and 52 are the portions of the sphere z : V“ 2 — 3:2 — y“ and the pai'aholoid z = :1?” + y"! with m2 + y"3 E 1= oriented 1with the normal pointing up and flown: respectively. As in (h)= the normal at apoint x on the sphere is so F‘ r n = = Also? the element of area in spherical coordinates is. r2 singsriprifl — Eairlpdprfll. so fill-L F- min-'1 — Err jfifl’lq Easiutpriup — :rr{2'=’-"E — 4). Go the paraboloid, 1with 2;! y- as paramerers: we have u an - [237i + 2p] — In} {remember that the z-CUDJPDHE‘HT must be negans-‘ej- so F unlit — fl'rfiwsgl [El-'3 + 2H: _ [.172 + #23] fry £11: _ 2n 1":5ri'r' — 43w. Hence F - flri'J‘i — Tl'IiEF'I'lI-‘i — ...
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H5s - 5.4 Vector Derivatives 1. These are all simple...

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