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hw-13-sec-15.5-solns - fl 384 CHAPTER 15 PARTIAL...

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Unformatted text preview: fl. 384 CHAPTER 15 PARTIAL DERIVATIVES ET CHAPTER 14 2 2 _ 2 _ (b) For (93,31) ¢ (0,0), fw(x,y) = W— _ w. If we approach (0 0) along the y- -axis, then 3 1 fx(m,y) = fz(0,y) = g; = a, so f¢(:c,y) ——» :too as (any) ——> (0,0). Thus (x $130 O)fr(av,y) does not exist and f (x, y) is not continuous at (0,0). Similarly f (x y): W = M for (as y) ¢ (0 0) and g It 7 y (332 +y2)2 ($2 +y2)2 7 7 9 )(X/\ 3 if we approach (0, 0) along the 3:- —axis, then fy (33 y): f$(x, 0) = % = 313—. Thus ( l)irn(0 0) fy (93,11) does not exist and Ivy —4> , \ fy(x, y) is not continuous at (0,0). 15.5 The Chain Rule ET 14.5 dz 8zd$ 82d 2 2 - t @z—r +y +xy, m—smt, y—e :> dt _8m dt +8’ydt: (2x+y)cost+(2y+m)e 2. z 2 605(1: +4y), m = 5t4, y = 1/t :> Eli — Qz—gfi gig- ._ 3 _ - _ #2 dt f 82: dt + 8y dt — sinks +4y)(1)(20t )+[ Sln(x+4y)(4)]( t ) 4 . 4 . . : —20t3 s1n(a: + 4y) + t—2— s1n(a: + 4y) = (i5 — 20t3> s1n(m + 4y) =x/1+x2+y2, x=lnt, yzcost => dz 82 d3: 82 dy 1 2 2 —1/2 1 1 2 2 _1/2 . 1 5c _._:————— ____:_ 2 __ __ _ : _— dt 8mdt+6ydt 2mm +y> (ac) t+2(1+w +y) (2y)( Sm) 1+m2+y ( ysint) 4. z = tan’1(y/x),cc = 623; = 1 — 6—15 => dz_8zdm 8zd_y' 1 _ _2 t _t dt'ax dt+8ydt_1+(y/x)2( W ) e +1+(y/x)2(1/x) (e X 1) _ y .t __1_.. 4—H x2+y2 e +x+y2/a§ e _ 3324—112 5.w=:cey/z $2t2,y=1~t, z=1+2t => 5 dw 8w dac 8w dy 8w dz y/ 1 y 31: 2mg __:____ ____ _= z 2 y/2 _ y/z __ _ : y/z _____ dt 8x dt+8ydt+8zdt e H” z ( 1H” ( 2) 2 6 2t 6. w : lnx/av2 +y2+z2 = %1n(w2 +y2 +z2),m=sint,y =cost,z=tant => fl[email protected]_g£+ggd_y+gtfld_z:l‘ 23: ~Cost+l- 2’5] ~(—sint)+l ____2—Z———-seczt dt ‘83: dt 8ydt 82 dt 2 m2+y2+22 2 m2+y2+z2 2 iv2+yz+z2 mcost — ysint +zsec2t ___’____—————— $2+y2+22 @zzmzys, xzscost, yzssint :> 8z 82 83: + 93% 22mg3 cost-kiln:2 y2 sint 82:8E—8-s 8y85 82 82 8x 82 8y 3 . 2 2 3 . 2 2 t __ _ _____ fl _ _ 3 cos 8t 8:1: 8t +— 8y 8t :(233y )( ss1nt)+(3m y )(5 cos t) 232:3; smt+ SCI? 1! C 385 SECTION 15.5 THE CHAIN RULE ET SECTION 14.5 )m—52+t2,y=1—25t :> s'zsamSinCE" y 62 az 81: 3281/ 25+2t ,=———+—-—: 1 2+ 1.-2t= as 82368 (93433 1—(135— y)2() s x/1(:nvy)2( )( ) x/l—(mwy)2 62: 82: 83C 32 (9y 2s+2t ,=’——+——= 1) 2t+ 1 —2 = w éhat Byw 1—m—-yv( ¢1-a— yP( ) ” Ml—m—yv 5? 2 2 i 9_z=sin0cos¢, 9:315, ¢=5t => g—:=%E%:+ +%Qfl=(0056 cos¢>)(t2—)+( sin05in¢)(2st)=t2c050cos¢—23tsin6 sind) azfazgg __8_Z____ _. . 2_ _2, _ .5? _ 66 at + 59¢ at— (cos6cos¢)(2st)+( sm651n¢)(s )—23tc0s6cosd> s sm0 smqb 1o.z=e1+2y, m=8/t, y=t/8 => 62_QZ_Q£ 828y_ z+2y z+2y _ —2 _ 111211 3—9: as#a$ 83+ay 3 e )(1/t)+(2e )( t8 )._6 t 82 Qf _ 25% ‘91 ay__1+2y —2 261+2y _ z+2y 2_ _S_ @z=ercose, Tzst’ 0:1/32+t2 : 82 8261" 8280 — 6t+T—'6-l 2“24/22 2111 9-1-9. 6 cos e( s1n ) 2(s ) (s) e cos 8 sm 52 t2 EZEE+wm = J (t cos (9 — 3 sin 6) \/ 32 + t2 t e TCOSO' 5+6 ~Slnf} 1 5 +t2 1/ 2t —58 (:Ose'e Sl!19-—————— T( > 22( )— ( ) /'——'_'_52+ 2 a=5a+wa t =er<scos0— #81110) 1/82+t2 12. z : tan(u/v), u = 23 + 315, 11 2 33 — 2t 6: _ 52 a” + 9&92 = 5602(u/v)(1/v) . 2 + sec2(u/v)(—uu—2) -3 ¢ “%$ 81183 f 2 2 2 £11 2(11 _ 212—311 2 u — 1) sec (1)) 112 sec 12)— 112 sec (1)) 95 62 0U+QEQE=seC2 (u/v)(1/v) 3+sec2(u/v)(—uv 2)- (— 2) = 2))— sec2 (3) + 22 sec2 (2) = 2n + 31) sec2 (2) v2 1) 112 v 3) = 7. By the Chain Rule (2), 13. Whent:3,x:g(3) =2andy:h( a aflm+ww nanma>[email protected]+v&v®=m E_8xdt aydt 14. By the Chain Rule (3), 96% 2 373218513 + 1811/ gig W5(1,0) : Fu(u(1>0)3v(170)) ”3(170) + F‘U(u(110)1v(170)) vs(170) : : m2, 3)u5(1, 0) + m2, 3)v5(1,0) =PUP$+WM®=W SECTION 15.5 THE CHAIN RULE ET SECTION 14.5 w = 11138.0, 7" = Way). 8 = 803.21), t = My) => w 19. //\\ @[email protected]@[email protected]@ @fl[email protected],@& W&+mw 8m — 87" 83: 83 8:17 815 8313’ 8y _ 8?" 8y 8s 8y 8t 8y r 5 /\ x/y 3‘ Y x y 20 t t : f(u7v7w)’ u : 114(1), q) T, s), v I 'U(p7 q7 7‘7 s)? w : w(p? q? r, s) : /\\ 8:__8_t_81_L+_8l8_v+ 8t8w 8t__8t_81£+8382+_8_t8__w " W 8p d 8u 8p 8118p 811) 8p’ 8q 8a 8g 81) 8g 8w 8q fi_fl@+ia+fl@za_fl@+fi@+mm 8r'8u87“ 81187“ 8w8r’8s’8u8s 8v85 8w8s @zzx2+xy3 mzuv2+w , y=u+vew => Qg_ 1283: 9319-3;— 3 8n :82: an + ay au — (2TH! )(v2 )+ (3mg 2,)(1) Es: / ” Es / Es / 52 _ 82 81 + 82 821- (23,“, 3)(2uv)+(3xy )(6 w), 22: 82 8m 82 8y_ 3 2 w — 812810 + 8y w —(2ac+y )(3w 2)4—(3333; )(ve ). Whenu=2,v= 1, andw =0,wehavem = 2,y=3, +(54)(1) —— 85, i— — (31) (4) + (54)(1)— — 178, .3713: (31)(0) + (54)(1)— — 54. so —— —+(31)(1). a WNW. 0..., " W ‘“M‘VW‘Wmo-wwwmum-Mm..w. 22. u=(7"2+52)1/2, r:y+$cost, s:x+ysint => 811 8u 81" + 8U 85 -— £03 + 32)’1/2(2T)(Cost) + 12‘0“? + 82)—1/2(2$)(1) = (rcost+ s)/\/7"2 + 32, 8;- 8r8ac 8583: a“ a“ 6" + a“ 85 — go? + 52)_1/2(2r)(1) + go"? + 52)-1/2(2s)(smt) = (7" + ssint)/W, (Ty 81" 8y 8s 8y 8 8 8 8 — ' u u T u 85 = a1? + s2)"1/2(2r)(——z sint) + %(T2 + $2)'1/2(23)(ycost) : W51. Ezfifi+wm Whenm:1,y:2,andt=0wehaveT=3ands=1 50%;: %,Z 23. R=1n(u2+v2+w2), u=as+2y, v=2$——y, w=2my => 8R 8R8u+flfl+81283_ 82: 811,833 811 82: 8112830 _ u2+v2+w2 _2u+4v+4wy _ u2+v2+w2 ’ 813 _ 813% 8R8v +85% _ Zn (3+ 21) (_1)+ 211) (213) 8y—8u8y 81183; 8uI8y—u2+vZ+w2 u2+v2+w2 ”LR—#12244112 __4u—2v+4wac — u2+v2+w2 Whenmr—y:1wehaveu=3wz1,andw=2,s0(—9§—=9—andQE:g m 7 y 7 387 l l t SECTION 15.5 THE CHAIN RULE ET SECTION 14.5 [3 389 2 = 355%, so let F(m,y, z) : 0:2 + y2 + z2 — 3xyz = 0. Then by Equations 7 :3; 3+2?” ‘ F 22: .- 3yz Syz —— 2m 82 Pg 2y — 3932 303.2 — 2y 8: i='/—:/ and ——::-—-——::—- :__—————_ E = ’ F; 2z .— 3353; 22 — 3mg 8y F2 22 —— 3mg 2z — 3mg n‘I”;geosutpyl'z). LetF(a:,y,z) =myz—cos(:v+y+z) :0, so 82 E, 9:2 + sin(z + y + z) yz+sin(a:+y+z) - ~—/ and ——._————:- . ' ’1 F2 xy+81n(tc+y+z) 6y F; $y+SIn(m+y+z) 8:1: ®,zzarctan(yz),soletF(m,y,z)=cc—z—arctan(yz)=0.Then 82_,_F1N__,_,,1.._—#_li22: 5;- F2” 1 ”1+y+y2z2 .1# 1 z T 2(2) 22 §i__&___,1_:1fl,_a-,_gfli__-_,z,f 8y F2 _ _ 1 1+y222+y 1+y+y2z2 1 2(9) /—- 1+(yz) 1+sz2 34.1,; = 111(2: + z), so let F(a:,y, z) = yz — ln($ + z) = 0. Then 1 (1 gangsflzfef and gang" 2 “gem. 6m- Fz y— 1 (1) y(:c+z)—1 6y Fz y— 1 y(a:+z)—1 $+z w+z dT ade+§_T_dy After 35. Since a: and y are each functions oft, T (:5, y) is a function of t, so by the Chain Rule, —d—t— : 3:13 dt 8y dt dw 1 3seconds,m=\/1+t=\/1+ =2, =2+lt=2+l3 =3,——= = -, y 3 3” dt 2./1‘+"t 2 1+3 4 461) + 3%) z 2. Thus the temperature is rising at a rate of 2°C/s. dT da: dy Th — : ~ — = en T423) dt +T.,(2,3) dt rainfall remains constant) causes a decrease in 36. (a) Since 8W/6T is negative, a rise in average temperature (while annual wheat production at the current production levels. Since 6W/8R is positive, an increase in annual rainfall (while the average temperature remains constant) causes an increase in wheat production. 5 rising at a rate of 0.15°C/ year, we know that dT/dt = 0.15. Since rainfall is\ (b) Since the average temperature 1 r, we know dR/dt = ~—O,1. Then, by the Chain Rule, decreasing at a rate of 0.1 cm/ yea dW 8W dT 8—W— 515 : (—2) (0.15) + (SEX—0.1) = ~1.1. Thus we estimate that wheat production will decrease 7‘36! dt 6R dt at a rate of 1.1 units /year. + 0.00029T3 + 0016]), so 93% = 4.6 — 0.11T + 0.00087T2 and 3% 2 0.016. t = 20 minutes, so 37. C = 1449.2 + 4.6T — 0.055T2 According to the graph, the diver is experiencing a temperature of approximately 125°C at 6 . . —C— : 4.6 # 01102.5) + 0.00087 (12.5)2 m .36. By sketching tangent lines att = 20 to the graphs given, we estimate 6T dB 1 dT 1 . d0_60dT BCdDN 1 1 N N — and—— N —I6.Then,by the ChamRule, dt N M dt + 8D dt N (3.36)(—w) + (0016)“) N —0.33. 71? 2 dt r is decreasing at a rate of approximately 0. Thus the speed of sound experienced by the dive 33 m/ 5 per minute. 7"" 390 CHAPTER 15 PARTIAL DERIVATIVES ET CHAPTER 14 $6 C l g“ g dV (9V dr 6V dh 27rrh — 2 — : —— ———- = 38. V —7rr h/3,so dt 87" dt + 8h dt 3 2 1.8 + 1992.5) = 20,1607r — 12,0007r = 8160a in3/s. @(a) V = Ewh, so by the Chain Rule, dV anEIBdelanh_ cw dw dh_ _ 3 H?‘ 86 dt ‘ 6w dt ' 8h dt whdtHh dt +11“ dt ’2’2’2+1'2'2+1‘2'(”3)Jim/5' (b) S = 2(611) + 6h + wh), so by the Chain Rule, dS 68012 as dw as dh dfl dw dh __ : ____ ____ _____ : 2 _ _ __ dt 86 dt + 8w dt + ah dt (“’HL) dt ”(Hm dt ”(”1” dt = 2(2 + 2)2 + 2(1 + 2)2 + 2(1 + 2)(—3) = 10 mZ/s (c) L2 = 112 +213 + h2 2 2L315 : Milg +2wg—7fl +2hd—h— : 2(1)(2) +2(2)(2) + 2(2)(~3) = 0 => dt dt dt dt dL/dtzOm/s. v 40. I _ E => 1 d1 aIdV 61dR ldV VdR 1dV [dB 1 008 ~ In — EVE + $35 — an — fiat— — m ’ an — "mm-01) “ Z0_0(0‘03) — ”(J-”0003”“ dP dT T dV 8.31 dT T dP 41. E # 0.05, E — 0.15, V — 8.31? and E # —P————t- — 8'31EE' Thus whenP — 20 andT — 320, dV _ 0.15 (0.05)(320) N dt —8.31[ 20 400 N 0.27L/s. 'et 90 and y be the respective distances of car A and car B from the intersection and let 2 be the distance between the two cars. Then dz/dt = —90, dy/dt = —80 and 22 = x2 + 3,2. When x = 0.3 and y = 0.4, 2 = \/0.2 = 0.5 and 22 (dz/dt) = 2x (dm/dt) + 2y (dy/dt) or dz/dt = O.6(—90) + 0.8(—80) = —118 km/h. 43. Let x be the length of the first side of the triangle and y the length of the second side. The area A of the triangle is given by A = émy sin 0 where 6 is the angle between the two sides. Thus A is a function of w, y, and 0, and ac, y, and 0 are each in turn functions of time t. We are given that (fl—C: 2 3, % = —2, and because A is constant, til—:1 : 0. By the Chain Rule, dA 6A doc 8A dy 8A d6 (M 1 . dm 1 , dy 1 d6 .__:__ __ __ _:_ r— _. .__ _ ..__, :20, :30, dt 8x dt + 8y dt + 86 dt :’ dt 23’5“ dt + 2”” dt + Moose dt Whe” y and 9 = 7r/6 we have 1 - 7r 1 . 7r 1 71' d6 0 : 5(30)(sm g)(3) + 5(20)(sm g)(—2) + 5(20)(3O) (cos g) a _ «3 d0 d0 — . l _ . l . ._ . __ 2 E _ _45 2 20 2+300 2 dt 2+150¢§dt . d9 . d6 —25/2 1 . . . Solvm for — 1ves — = = - , so the an 1e between the Sides lS decreasm at a rate of g dig dt ism/:3 12\/§ g g 1/(12 \/§) m 0.048 rad/s. ...
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