0534552080_98859

Advanced Engineering Mathematics

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Unformatted text preview: Section 12.1 - 199 Chapter Twelve Veeter DifierentialCaleulue Section 12.}. Vestal Functions of One Var. 1311:1112 (L1 %1ff—:(L1F(L11 g—[L L0L( (312151 + 1212 LLL(3t1j + SLLLLL(3L11L1: : 12. 8111(31)i+ [241 (303(31) — 3612 sin( (31;) )]j + 8[cos(3t) m 31 si11( (31) )]k 1)) £51me01 : —12L1L(3L1F(,1 + fa) (511 + 211 : —12511‘1(31)i + (~3612 Sin(3t) + 241 cos(3t)1j + [8 (505(31) — 24-13111(31)]k 2 ( a) (2: [F (t) G(i (t:)] (it [1 -—~ 31,2 cos(t)) : 1 — fitcos(t 1,1- 312 3111(1) (b 1— [—'1?(11 G(t11: (i—b‘tk) 1-(iL—cos(t1k1+(ti—3121L1L(—Lin(t11k:1—61LLL(L1+312L1L(L1 1 j k 3. (:1) Since F(t) x G(t) 2 1 '4 2 (t + 4005(1))i + (4 — 12)} — (t cos(t) + 1)i{, 1 —eos(t) t germ >< (1(1)) :(1— 45in(t))i + 211 : (LLL(L1 — LL1L(L111L 1 j k i j k (b) F’(t) x (3(1) +F(t) >< G’(t) : 1 O D + t 'i 4 1 — 003(1) 1 U 8111(1) 1 —tj — cos(t)k + (1 — 4sin(t))i — tj + tsil1(t)k i j k 4. (3) Since F(t ) x 61(1): 0 sinh(t) —t = [t3 :2 si:111(t)]i — t2j - tsin11(t)k t :2 +12 §E(F(t) x (31(1)) 2 [312 21311111(t) —-~ 12 cosl1(t)]i —-- th — [5111116 ) ~1— icosh(t)1k i j 1L 1 j k (b) F’(t) >< G(t) + F(1‘) x (33(1): 0 6.0311(1) —1 0 si1111(t) —t 2 1 1:2 —12 21,- —21 1 [—1 2(1.0511(1) —1— 12]i — tj ._ t cosl1(t)k — 21311111(t)i~tj — sin11(£)k 1: {Lad—(1 MUM 6111:2164 2131— (1—21531c0511(t)j+ +( 41331114: (1 —1583d 11+1612coLh(t1— (1:21:31L1nh(t)]j+1e 41:26 2:3 .111L (1)) )f’(t)F(t) ~1— f(t)F"() =—6£2(ti — cosh(t)j + etk)—i— (1 2 213)(i — Sinh(t)j ~1— etk) 5. (L1 $11111 . (1(111 : §1LLLL()+ 41—1—15): Lin(L1+LLLL(L1+ 4 + 5L4 (111 F’(L1G(t1+F(t1 1G (11— — (i j+21k1-(Lin(t1i—41+131L1+(ti—13+121L1(C(LLL(t1i+312k1: 8111(1) + 4 +-2t’1+tces()+ 314 i j k 7. (LL1F(L1>< (3(1) : +9 12 :2 _ —12efij —tgé ‘k so LEE-{1? 1x (1(111: r. (it e U U (2112t + t26£)j — (21'1“.E —1— tgetflc : 18%2 + 1m — 1c} . i j k i j k (1')) F’(t) xG(t)+F(L) xG'(L1 : 0 21 21 + —9 1:2 t2 : 2-18th :2tL1£k~i—tge‘j—Lgetk 8100 (23110 200 Section 12,1 8. (a) éflfit) - GB); 2 %{—16sin(t) cos(t)] 2 —15cos2(t)+1fisin2(t) (b) F’(t) - GUJ) + F(t)_- G’(t) = d-si11(t)k- (—t2i+ 4sin(t)k + (—4- cos(t)k) - (—2ti + 4cos(t)k) 9. (a) A position vector is F(t) = sin[t)i + cos(t)j + 45tk, for 0 g t S 2w; a tangent vector is F’(t) = cos(t)i — sin(t)j 42 45k. (to) 5(t) = gunman 2 fotx/fofiédg z x/2U26t. (c) Since it : mm, 0 g s g 2m/2026, a position vector is s s 45.5 G ‘ : F t I : k ‘ I + 1 . ( ) I + kl (b) ( (6)) gm (M2026) ‘ COS V2026 J M2025 so GT3) : #21}? [cos (VS???) i sin (6%)5' l 45k] and [IG'(3)|[=1. 10. (a) A position vector is F(t) = t3(i+j + k), for —1 S t g 1, a tangent vector is F’(t) 2 3t2(i +j + k}, t r (b) so) 2 jg "were 2 M/ eats 2 fat to 1.33 s . . . (G) Since t z (-2—) , a posmon vcctor 13 \/§ Ge) 2 ms» i+j + k), :E( SO 1 G’s :m— i+'+k aid Gs :1. (J ”3 .1 )1 u ()i 11. (a) A position vector is F(t) 2 t2(2i + 3j + 4k), for 1 g t S 3; a tangent vector is F’(t) : 2t(21-:~3j + 4k). (b) an) 2 / EIF’(€)id€ 2 2m] w: 2 figs? 2 1). (c) Since in iifi 471,0 5 s g {Qt/27g, a position vector is (3(3) 2 new) 2 ( S +1)(2i+ 33 + 4k), m (33(5) 2 7%(2H 33 + 4k) and "oven 2 1. 12. If F X F’ = 0, then either F(t) : O, Fft] 2: O or F6) and F’(t) are parallel for all t. We can assume F(t) ;£ 0. 1f F’(t) : 0, then there is no motion and the particle is at rest. If F(t) and F’(t) are parallel} then the velocity vector is always directed aiong the path of motion and we would have straight line motion. Section 12.2 201 Section 1.2.}; Velocity, Acceleration, Curvature, and Torsion Before giving solutions of Problems 1 through 10, we make same observations about the various quantities to be computed-which will reduce the plethora of calculations to two differentiations followed by some ioutine vect01 calculations. We assume that F(t ) is position (given) and that t is time. Clearly then velocity is V(t) — F’Ofi) land accele1ation is 1a(t )2 F”(t), both easily found. Then speed is v( (t) 2 HF’Ufi) )[l and the unit tangent is T2 —F’(t) Now recall F” 2 d9 2 31m (ing—l—n (BET) N, and observe F’ X F” 2 a- < ------ > B where this step uses the fact that T X N 2 B. (is - . . . . . . . Since (we) > U and K, >- 0 we see that B 1s a unit vector With the same direction as F’ x F” and this dt is easiiy calculated. It further follows that N is a unit vector having; the direction of (F’ x F”) x F’. 1‘1fldlly, having now found T and N we easiiy find {131— (a T) and eN— 2 (a N); also since a- N) cm 2 KRJ2 we have a— 2 ( 2 )thus completing the calculations. In summary; find V— —— F’ a— — F”; 1; then T B N am unit vectors in the lespective directions of F’ F” x F” (F’ x F”) x F’; then a (1332(5). T) GIN“ — (a- N),it2 “5%: We now apply these weeks to Problem 1 and give answers only for Problems 2 through 10. 1 1. We have V 2 F’ 2 3i+2tk,a 2 F” 2 2k, speed 2 1/9 + 49; New ‘1‘ 2 —u 3i+2tk; “—1 l 9 + 4:2 i j k we compute F“ x F” 2 —65, so B 2 —j; and (F x F”) X F’ 2 U —6 U 2 —12ti+18j‘ 3 0 2t 0 N219[ 211+ 31} Fimil c (a T) —~4t ( N)— 6 a s — _ . < 2 - 2 ;a r 2 a an #5112 +9 _ y T 9+e2 A vf—9+4t2 _ (e + 41:2)312' 2. V 2 isinU )+tcos(t] ]]i+ [cos(t) — tsin(t)]j; a 2 {‘2 cos(t) — tsin(t)]i _, l2 sin(t) + tcos(t)]j; 1 1 v 2 [1 +t2]1/2;T 2 EV 2 J—lfi [[sin(t) +tcos(t)}i + {cos(t) — tsin(t)]j] ;B 2 wk; 1 [[cos(t) 1311mm [si11(t) + tcos(t)]'] c t t? + 2 I '- .' — . ‘ ; r1 : ..... H1. 2 —'; Va +12 3 1 m N m t2 + 2 K, r. w (1 +t2)3/2 . . 1 . . 1 . . . 3. V 2 21 — 2] + k;a 2 0;?) 2 3;T 2 E(21— 23 + k);N 2 :50 +J] or any unlt vector fl perpendicular to T; B 2 —6~(-i j + 4k);flq‘ 2 O;oN 2 0; it 2 U. 4. V 2 1etflsinfl) + cos(t))i + (cos(t) «w sin(t))k1; a 2 2121[cos(t)i — sin(t]k]; e(t) 2 flat; _- fiflsirflt) )+ cosftfii + [cos(t) - sin(t)]k};]3 2 j; 1 r N Z filleosfii) — sin(t)]i — [si11(t) + cos(t)]k};aqw 2 x/ietuiw 2 fiche 2 crate" 1 i—j+2k)‘1\1 = 72(1—3') 1 5. V 2 ——3€_i i+'—2k ;a 2 38”"I i+'—2k ;'u 2 3 Be“;’1‘ 2 -----— —~ ? ( J l ( J ) f V%( x 2 292 Section 12.2 1 — i+.’ + k m” 2 ~3x/Ee_”;a 2 0;K r U V/g( ] ) l N 6. V2 —c11s111(t)i —‘— fij + crcos(t)k; a 2 —o-cos(t)i — a sin(t )k; '11— “ 1/112 + 62; 1.1.2 1+mm1;§[~—crsir1( t)i +131 + c1cos(t )k];]3— — (£3 a: +8 SJ [—fisin( (1)1 — aj + ficoe(t)k}; N— w —cos(t)i — s1n(t)k, (11-1 2 U;r1N-—— ('1 K: 2—- or any unit vector perpendicular to T; B 2 7. 'V— 2 2cosh(t)j — 2si11h(t)k;a 2 2sinh(t)j — 2cosh(t)k;’o 2 21/cosh(2f.); mm [cosl'(1(6)} —s11111(£) k;] 1:23:71" m—i [ 2 26 2 1 B Z ——i; 11.3- % 8111i ( ) _ _ ./cosh(2£)i; EN W cosl1(2t)1hi fl 2[cosh(21)]3/2' 1 1 6 . '2’ 2k 8. V2 E(i—j+2k);-a2 —t—2(i~j+2k);1r2¥;'1‘2 1—2'?+ w- si1‘1i'(i (6)3 — cosh(t )k]; ;N 2 -(w—~-—) or any unit x/E x/i \/€ 1 . vector or “Jendicuisr to T; B 2 — —i + ' + k ;a. 2 —---— r;1 — {3;1t2 0. P 1 \/§( J i T 1:9 N— 9. v = 211111 «1 111+ 1k); a = 211.1 + {11+ 1112 211N112 3% 11—2 + 12'; 1 . T = —~—~—(ai+fij +qu); N 2 any unit. vector perpendicular to T, and B 2 T x N; «a? + {32 + 1 £1112 2( (Sena) )NQQ +132 + "flaw 2 Gus 2 U. 10. V2 — [3 cos ) — 'it sin(t)]j — [391-111(6) + 36 cos(t)]k; _____ a— 2 [— 6si11(f )— 3t cos(t )]j ~— [{icos(t) — 3t si11(t)]k;c 2 3\/1—+ 1—2; 2W [cos(t) — tsin(t)]j [5.111(1) + t cos(t)]k}; 2([sin(t) + tcos(t)]j + [9.113(6) — tsin(t)}k};]3 2 —i; :1/1 +1.2 { (3121— e)2 f (31:2 + 6)? a.” — ; 1 — ; t — , , . 1 ./13 +12 N 11/1 +12 9(1+12)3/2 11. Suppose that T(s) 2 constant. Then F’(s) 2 mi + ng + (:31; for some scalars (31,62,63, where F(s) is the position vector expressed as a function of arc length 3. But then :1." (5') 2 c1,y’(s) 2 eg,z’(s) 2 133, so 31(3) 2 C18 + (11,3;(3) 2 (:25 +a2,z(s) 2 (:33 + Q3 and these are parametric equations of a. line. (£3? 12 Yes since from the first I‘1‘e11et formula if 11'. 2 D then — — —U and T( (s) is a. constant (is vector. By Problem 11, the cuwe 1s a. straight line. Section 12.3 203 Section 12.3 Vector Fields and Streamlines 8G 8G 68 3, 8G 1.-—23'_4*:___H__#4- 2- : ._ z 22. 89: 1 3;}. 83; $3 5‘2: 6 1 45:33:}, By :3 J (if—3‘2] G(2. 4) GU: 3} am; 6‘0 _ - . . 3G , 6G . 6G 3'33? ‘ 21‘” "" “(m)" "537 z 2‘“ 4-3; = Zycoswwy): + 29:5: E : 2w cos(2xy)i + j G(—2.l) y (4&2) teams}. a y a G{~w/2,1) G(l,l) 204 ‘ Section 12.3 6. Fm :— "gem-”i — 43,11; Fy : (semi + sinh(z ~§~ y)k; F2 2 sinh(z + y)k 7. F... = —422 sin(a:)i 3:32ij + 3$2yk; F9 2 “$3.33 + 323k; Fz : a: cos(s:)i 3:335 1 1 8.13223 3' w'+ '13::sz 29'2i+m' zsinh. 1; a, y1+$+y+zr yzsmlw) y it $+y+zr+m (were 1 F3 j + my sinh(:1:yz)k Zm+tj+z 9. F3 : —--yz“ cos(;ty)i + 33:42.} sinlr(z — refit; Fy : *a'z‘i eos(:ty)i + 12my3zj; F; m —42;3 sin(:r:y)i + 333y‘1j + sinh(z w- 3:)k 10. F5; 2 14-i + 29.:j + Syk; F3. 2 —2i — 2yj -+~ 5:1:k; Fz = —22:j . _ _ d3: of d 11. Since F : f1i+fal+f3k : imy23+zk, the strearnhnes satisfy T = ——y—‘: : —Z. Integrate Z d ate : ~13! to get a: 2 §+CM integrate d3: = —z to get m = ln {z[+e2. In terms of :1: we can write y z 1 the streamlines as 33 z :t, y = {I} - C} 1 (st—2) 1 . . and 1 = 89“?) so e121,c:2 = 2 and we get a: : 13,3; 2 ~-~—,z : e — Cl _ 3) — 1 of d of . . 12. Streamlines satisfy Ta: 2 "3% = 72. Integration grves y 2 ~21: + c1, 2: n :5 + (:2. To pass through (8,1,1) requires cl 21,02 = 1 to give :I: : my 2 —2s: +1,z m a: +1. (1 d d 13. Streamlines satisfy (rain: 2 E3 : —:. Integrate 3:053: : ~— as memes: = (1;: to get .33 _ e: y = are? — eI + (:1, from 3:033: : mdz we get 2:2 x “22 + Cg. To pass through (2, O, 4) we need 82 + C1 = 0 and 4- = —8 + Cg so (:1 = —e"2 and (:2 z 12. Using a: as parameter we have 1 x:$,y=me$—eI—e2,z= 412—39). 2 (£3: dy ,z : elm—C”). The particular streamline through (2,1,1) requires 1 :— 14. Streamlines satisfy e— _ eos(y) s1n(:r:) — eos(:s) + :31 : sin(y),z 2 c2. To pass through (arr/2,0, —4) we need cl 2 U and :32 : --.4 Using :1: as parameter we have I = 3:, y : sin—1} cos(:c)], z 2 ~81. dy dz 15. Streamlines satisf d3: 2 U and — —_— m . .y 26,3 eos(y) oos(y)dy = —2ezdz we get sin(y) = (:2 — 2ez. To pass through (3, arr/4, 0) we need e1 : 3 and and dz = 0. Integrate sin($)d.:t : eos(y)dy to get Integration gives :1: : c1 and from 86011101} 124 205 2 2 (:2 2 lg— + 2. With 1,: as paran1ete1 we have :1:— W 3, y— y, z— m 111 [14: + 1 — ésinwl]. d d d 11,; 1 16. St1earnli11es satisfy ::2= —~5« : 2—: Integ1ation of 3—; = -— j gives E_ _ 31111111 + (31, (1'31 dz , 1 , 1 and of —— — 3' figgwes # 1111111: #2"; + Cg. '10 pass through (2, 1, 6) we need c1 = —2~ and 12' 1 With 1 a a1nete1 have 113—2 6 C 3 — as 1 We # w—w—ww—, “' ‘ fl, 2 72 J P 1+ 6111(y) 3” iii/1+ ?21n(y) 17. Any non-zero constant vector field wiil have streamiines which a1e straight lines. If F 2 (Li —1— bj ~+~ ck it is easy to produce the streamlines (r, : at + 51:0, y 2 1512 + 110,2 2 at 1 30, which are lines. 18. Circular streamlines about the origin in the mgr-plane eou1d be written 5.22 —1— 1,12 2 1'2 at (5:1: so ads —1— ydy— — 0,01 3 = ——3;~, dz —. {1A vectol fieid having these streamlines is F— — y 311 1. mj + 0k. 19. This is impossible. IfF : f1(a",y,z)i+f2(s,y,z)j+f3(93,y,z)kl1as streamlines (3:(t),y(t),z(t)) 131ng only in the m,y-p1ane then 33’(t) = af1,y’(t) : af2,z’(t) : afg. But by assumption, 2"(15) 1" 0, so f3(:1:,y,z) : O and z : constant a: Thus F = f1(:1;,y,o:)i + f2(33,y,a.)j is a function of only two variables. Section 12.4 The Gradient Field and Directional Derivatives 6 _ 8 _ . 8 _ . . . _ . . 1 1- Whey, Z) — gay/211+ 5—y(~b312)1+ awedk 1 1121+ $21 + xyk9V10(1,111) — 1+1+ k, The maximum value of Dump(1,l,1) = |1V1p(1,1,1)1| = (/3, and the minimum value of D1190(1,111): “llVSO(33,111)l| 3 “V5- 2. V1p(:c,y,z)— (2:1:—zcos(z:r))i+$2j~11:cos(:1:z)k'V<p(1—1,7r/4)— (12—1431) +j— gig 19.11211, 11 w/41m _ 1171,0(1, —1,1T/4)11—_. (176+?T-m16ffi)/32 Dutp(1,-1,1r/4)mm _ 1—[V1p(1, 11 ,1/411— 1 —(1'.7'6+1r ~16V/21r)/32 3. V<p(:1:,y, z) 2 (2y + ez)i + 2:1:j + mezk; th(—2, 1,6) 2 (2 + efifi # 43‘ ~- 263%; Dug/(""21 1-16ij Z ileo{—21116)li : V 20 + 486 + 53121DU90(“21116)111111 : —|~1V50(#2, 1! 6)li : —\/20 + 436 + 51:;12 4. V(p(:c,y, z) : —yzsin(:1:yz)i — 11:2: sin(:1;1,iz)j — sysin(;syz)k;V1p(—1,1,1r/2) 2 «E1 — g—j — k, ”K2 DU‘P(_1:1:7r/2)max :11Vgo(—111‘/2)1|: —“ +1 .2112 Du1,0(—1,].,7r/2)min : —1|\7g0(#1,1,7r/2)1| = — .2... -+~ 1 5. V1p(s,y, z) : 2ysir111(2$y)i+ 2:1: sinh(2:1:y)j — eosh(z)k;Vgo(0,1,1) : ~ cosh(1]k; 401199“): 111)1118X—— iivwm, 1,1m1)|| # cosh(1) Du90(01111)min* —|]V1p(0,1,1)|1 S — cosi1(1) 206 Section 12.4 :1: , y , , z 1 , . rev 2,2,2 :mi+ $2+y2+z fl2+y2+z2J «$2+y2+2:2 M ) V3 Dfl@(2:212)max: I]VLP( (,2 2, 2)2II— —.. Du(p(2, 212131111 2 "~2IIV{,0(2,, ZJII_1 " + ‘ k 7. Dmflmefl) = Vflfliayfl'u = {(Bsg—zli'l'lfiwyj—Ifikl' IiI=1 J3 8- Drlflwayfll = VMMW) - u = [Sinlrc — ali + sink: — all + ezk? - I ----- 52081,: —z+16:ry - :r) i—J_+2kI e6 %(—28in(m — y) + 2ez) 2x223 +3$2y22J . ' 2' k 1 9. Dutp(:t‘,y, z) .— V<p(:1:,y,z)-n 2 [2:c:::,u:‘li+3:22:3j~1—3:r:2y,32k]AI—J + I : x/E «5‘ i— 41-: .. 1 m l - m 11. With (p(9:,y,z) : 3:2 + y2 + 2:2, the surface is the level surface 90(5c,y,z) = 4 so a normal vector to the surface at (1,1,V/ii) is N : Vg0(1,1,x/§) 2 21+ 2j + Zfik. The tangent plane to the surface at (1,1,\/§) is given by 2(:.-: — 1) + 2(3; m 1) + 2\/§(z — V?) 2 U or a: + y + v52 2 4. A normal iine to the surface is given by the parametric equations 3::1+2t,y=1+2t,z:\/§+2\/§t,—oo<t<oo. 12. With the surface written 3:2 + y — z 2 U, a normal vector at (----1, 1, 2) is N : Vtrg + y -— z)(_1,112) 2 —2i +j — k. The tangent piane is ~23: + y — z 2 1, and normal line is 55:.-‘1223'9:1+t,z=2—t,—oo<t<oo. 10 BUM-T 111,2}: W( a: a2 Zl-u ‘ —[(Z+yli+(2+$lj+(y+wlkl-I (nay—’42:) 13. N 2 V($2 — yz — 22)(1,1,c) 2 2i — Zj; tangent plane is 23: — 2y 2 U or y = 3:; normal line ism:1+2t,y21—2t,z=0,—oo<t<oo. 14. N : V(a:2 — y?’ + 32)(1,1,5} : 2i .1 2j; tangent plane is 2:1: — 2y : U or y = 3:; normal line ism:1+2t,y21—2t,z:0,~—oo<t<oo 15. N = V022: — cos(:ryz))(1,,,,1) = 21; tangent plane is :1: z 1; normal line is :1: 2 1+ 2t,1 : n,221,—oo<t<oo 16. N x was“ + 33,14 + 624)(1,1,1) : 12i+12j+ 24k; tangent plane is 12:12 + 12y + 242 z 48 or x+y+2zz4g norma} line is at: 1+12t,y=1+12t,z: 1+24t,—00 <1 < oo 17'. Write the surfaces as level surfaces in the form 33:2 + 2y2 — z 2 0 and —2:r: + Ty2 _. z 2 0 respectively. We check that the point given, (1, 1, 5), lies on both surfaces since 3(1)2+2(1)2— 5 = O and —2(1) + 7(1)2 w 5 : 0. At the point (1, 1, 5) we can find normals to the respective surfaces as N; :. V1311? + 2y2 —— z)(1’1,5) : 61+ 4} — k and N2 = V(—2:L‘ + ?y2 _ zl(1,1,5) : “2i + 14j — k. The angle between the surfaces at (1,1,5) wiil be the angle between N1 and N2. We get cos(6‘)— Nl-Ng M12+56+1 45 _,( 45 ||N1||||N2]|1_ «Ta/“”201 "MW—“10,653 3° m, “WM—10,653" 18. Normals to the two surfaces at (1, fl, 1) are given by N; 2 V(:c2 + y2 + Z2){]‘\/§‘1} 2 2i + 2\/§j + 2k, and N2 : Vb? + mgjulfi 1) 2 2i + 214. The angle between the surfaces is j 2 1.11956 radians. Section 12.5 207' N - N 4 + 4 1 the angle between N1 and N; for which we have (208(9) .— ml 2 2 ..m 2 so . ' llNlllllNQll (Qt/g 1/25 9 2 cos"1 (—1“) 2 i radians. 4 1.9 Normals to the two surfaces at (2,2, x/g] are given by N1 2 V(\/'_1:§-I—_ IIIIII +23; 2:),‘(22 (/8): 1 . —i 1 _j _ k! and N2 2 1713,12 + y?) : 431+ 113. The angle between the surfaces is V;— Vf" {2,2,4/111 N1 . N2 div/2 1 the angle between N1 and N2 for which we have (205(6) 2 SO 1141111111 ' 3141/11 """ a: 1 1T _ .—1 _ __._.I 1.. 9 2cos (v2) 4 ladlans . . 1 1 2D. Normals to the two surfaces at (2,2,1) are given by N1— ._ V (251: :1:2 + i112»? ' T232) = (221) 21 + ‘2j + 2k, and N2 2 Y/(rc + y + Z)[2,2‘1) 2- i ~1~j + k. The angle between the surfaces . N1 1 N2 6 is the angle between N1 and N2 for which we have (1011(9) 2 m— — ———«— 21 so llNillliN2ll ”21/717 6 2 cos“1(1) 2 0. Note that the two surfaces are tangent at the point (2 2 1) since their respective normals are parallel. 21. Since V90 2 1+ k for all (31,1 ,2) the normal to the surface tp(:1:,y, z) 2 K is the constant vector N ...—. i—l—k thus the surface must be :1: +z— w K a plane. The strean'ilines of the vector (1 a! fie id Vip— — i + k are given by solutions of T113 2 72 and (11,120. We easily find these to be :1: 2 2: + {11,111— 2 132,01 using 1 as pararnete1a¢ 2 t +c},y 2 132,2 — t,—oo < t < 00. These streamlines are lines in three space which are orthogonal to the surface :1: + z 2 K. Section 12.5 Divergence and Curl a ‘ a a 1 v 1 z m 2 _ -4 633141 + 6ye1+ d—z( 21—— i J k VXF2 % % % =m+m+wzm .1: 1; 22: V - (V x F) 2 . ’ 2. V - F— 2 £mja + "q" (s1nh(:1yz 5: (D) _ mzcos 11(33yz); i jy k V x F 2 3% % 6—6; 2 —.1:ycosh(:1:yz)i + 112: cosh(;1:yz)k; U sinh(:tyz) 0 1‘ 8 6 '17- (V X 1+)_ — 5(— 11:ycosh(:1:yz)) + a—(yz cosh(:1:yz}] 2 z (osh(:12yz)(—--1,1 +151) +sinh(:1:yz)(— :1:sz + $3,122) 2 0. d (3 8 _ _ J 11 MM _ y .5. V F— (3.1: —:(911y)+ a—y(:1:e)+ 62(2zj— 2y+:1:e +2; 1 j k l V1111: % 6—1 13; =(ey—21114; 2.11,! we” 22: 208 Section 12.5 V (V XF): E:(Fy—21):U 4V 1" 8( l(})+ 3(aosh-(11z)) 6% + +zj— cos}1(:1:)+ l( ) 1; . .::—— 311113: —— ., :1: — 51:25 — 8:1: 811 J 831; 1111 3:112 i j k V >< F = fi 6% g 2 [-'1 — my Sil1h(3:yz)]i —j + yz si11h(:1:yz)k; ‘ 11(1) coshmyz) —(:12+y+ 62) d V - (W >< 1‘): 6—3:(_ 1 — 3:1; sinhtyyzfl gfiyzsinm $112)): (— 1,1 --+ 1})Si11h(:1;yz) +cosl1(:1:yz_)[—:rygz—1v+21:11:22): {1. _ a .2 a y2 a _ ,. . 5.v-Fm651)18y1y)+8jz)—H11y+fla '1 j k VxF: % é % =m+m+wso 3:2 ya 2:2 V1me=0 {'3 V-F-" a-“(snhtv—zD—i— 8(2‘1)+~~a—(z—12)—0031($— )+2-11““ h 3' . — 8:1: 1 (91; J 82: J , 1 2 —cos (112—12)+ , i i j k V x F = g 2% 6—62 2 —2yi COSh<E — z)j; Si1'1h(:1:— z) 21} z — 112 11-117 M) a121+ 91—1201111— ))+3(0)—0 m 9 61 “ z & *' - 1 j R 7.1711): V(:1:----y+222) =i—j+4zk;V x (W) = if; 3% 1% = 0. 1 m1 41;: 8. ng1 = V(18$yz ~f— 6"") = (18112: + ezfi + 181132;} + 18:1:yk; i j k Vx(v¢y: % % ;% =Km$mlmfl+fimywlwy+muh*l&):0 18yz + cw 183:2 18331,: 9. ng z V(—2:1;31132) : —6:1:21;221 — 23:322j w 41:3yzk; i j k WW1): 9% 3—; 56' = ~--6:1:2yz2 —2:1:322 —4:1:3’yz (—42:33 + 4-:c3z)i —}— (—12m2'1;z + 12:132yz)j + (—6352,:2 + 6x232)k = 0. .10. ng : V(sin(:1:z)) 7—" z (:08(:1:z)i + 33c05(a:z)k; 1 j k [email protected]= % g g : zcos{:cz) [l :L’COS(:1:2) 01+ [cos(a:z) — 2:2 sin(:1:z) — cos(:1:z)+ (1:2: Sin(:1:z)]j + 0k 2 0. 11. V11) : W3 cos(:c + y + z])— m {mafia win 11 + z) — 121311101: + 1; + 2)]i — msi11(:1: + ’1; + zjj ----- :1: sin(:1: + '11 —i— z)k; Section 12.5 209 i ' j k 6‘ 3 a V X (We) : a; at E cos(:1: + y + z) m msin(:c + y + z) —2:sin($ + y + z) —:i:si11(33 «Er y + z) 2 [—a: cos(:1: + y + z) + :1: costs + y + z)]i + [~sin(:1: +y+ z) —:Lcos(:c+y+z) +sin(:r +y+ z) + 3:cos(:c+y +z)]j +[—sin(:s+y+z)—$cos(:r+y+z)+sin(:c+y+z)+$cos(:r1+y+z)]k—-O 12‘ w: WW”) =e~i+v+z< +J'+k)' i j k 8 3 6 _ ‘7 X (We) 3 a: at. "a: * €rt+y+z e$--y+z eI'l-y-l-z {ex—1.3;-“ __ ex+y+zli + [ex+y——z _ ew+y+z1j + [6m+y+z _ Bx+y+z1k E O. 13. Let F 2f1(:1r,y,z)i+ fg(m,y,z)j + f3(;c,y,z)k. Then V ' ($51?) = V ' (95f1i+ 95f2j + ¢f3kl 2 $8103.) + 6%(43f2) 56? afl 6f2 afa +£(ef3): f1+“'f2+_f3]+¢(w+5§+6?) We now recognize this first term as Vqfi - F, and the second as (NV - F), thus V . ($1?) 2 vs - F+ em - F). Alson(¢F)= 3% 6y 5?; : fibfl (We M3 :Z(¢f2)]i+[%(¢f1)— goal“ [til—mere - gym) k To identif: the above expression we apply the product rule to each term above and regroup as V X ((pF) : i j k g g—nysl [(35ny — ¢zf2li + [Cbzfl ’" (lbwalj + l¢93f2 _ (Pfiyfllk wile-ti e ti 1 e :3 14. Since the expressions on each side of this identity are vectors, we establish the equality of the i,j, k components of the vectors on the left hand and right hand sides of the identity. Let F 2 hi -l- fgj + fgk and G 3 mi +91} +g3k, then F - G E f191+f292 + fgg3 and the 1 component on the left hand side is 52mg} + f292+f393]. For the right side, we first compute F x (V X G) and G x (V X F) so we can identify the i component of each. We have k}=V¢XF+q§(VxF). 1 j k i j k x{VxG)mF>< 3%: % % = a fig 6 he a f3a 91 92 93 (“335 '5?) (62 91 as}: l (5": a?) From this we easily get the i component of 392 391 waste—w) f (% _ 9a)] 83; 62 6:: ’ 210 Section 12.5 and th...
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