ma1012_hw17_201012

ma1012_hw17_201012 - (a) œ 1 2 + Tan[y/x]==2; œ (b) V œ...

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Unformatted text preview: (a) œ 1 2 + Tan[y/x]==2; œ (b) V œ 1r h Ê dt œ 21rh dt h Ê 3. œ 1r V dtqn=x1rh dt dt c" 1r dt e br h Ê D[%, x] dt c5 e ‹ (c y 19.y[x]} (1)m/min 4 (b) (c) dr ¸ y=8 œ 2881 m/min b (a)dV ‹ (1) b Š È329qn/.{ 2)#Ä Š È229 ‹241 x, # 0 c 3, x0 b 3},{y, y0 c 3, y0 b 3}] r œ È26y c y# m dt Iœdh r h Êc614 knots dhm/sec rhSdr mplicitPlot[eqn,{ 1œ dt b 21 dt x0 29 ( V # x] œr olve[%,#y'[x]] dV dr # ts h Ê dVc) D[%, 1 14. dt r (b) V œ instant dV œ r Ê dt dt œ fh Ê dV œ "1rr isdh ft Ä x0,and x y0} y œ 5sat the 1# hh Ê of time 221rhdr 13 " # t/sec, the ladder # dt Section eqn/.{x long, y Ä œ 12, (b) V lope=y'[x]/.First[%] 1rh dt œ 3 1r œ dt 3 1 # dh dt dV Solve[%, x 1rh dr y'[x]] # ryhdœ 169 Êdr dy qn/.{20.dx dVdtœ kS œ(5) 1r#cdr œ k, a constant. dt is sliding dt # the wall 2 2 œ dV ˆ " # œ 12 ft/sec, the ladder $ 3 # hÊ 2dt œ 1re dt by Ä y[x]} 12 ‰ 4k dh , " œ 1 # h 20 12, 5 he ladder 1rh sft œ " cr15.dtÊœft/sec œ r at the16. =slope/.{x Ä H(b) V œ down7. 16. dV œ Rrh dr in$ /min. dt athematics m About 0.0239 in y[x] 3 1r1 Ê dt omework 4 (a) Vdtlong, and xdt c œ r# dt b. 3 is 13dtlope=y'[x]/.First[%] y 3 1 5 dt Minstant of time I: x0, /min.Ä xy0} h dxRelated 1 ates y About 0.0239 3 dt 3 dt Êbds œ Èxx0) b z dx b Èx byy b z dy b Èx bzy b z dz dy dx D[%, x]# ˆdh ‰ (5) œ c12 ft/sec, the ladder is sliding down the wall dt V dx œ c 12 " 2 dr dt tœ " xy Ê dA œm " ‰ Šx b y b y ‹ . The area dt anline=y==y0 dtˆ (x c dy 69 y0} dt (œ c y" œ# byrtheÄ x0, y[x] Ä y0} A Ê c) m=slope/.{x b 3 1 and 1 92 Chapter 3 is f the triangle formed 3dh dt5ladder rh dt walls Differentiationdt dV# 2 dt dr dt dt " 1r# h Ê dV œdt3 1r dt y'[x]] (b) V#œ 3 r# h Ê tanline}, dt dx c 3, 3.7 Related Rates rh dy 21. (a) ¸ 401 ft /min ImplicitPlot[{eqn, part14.92{x, x0 œ 0 x0 0.1492 m/sec 3, y0 + cm/sec ds dz Solve[%, 1 ft/min,0.1119 #m/secdIœ 11.191cm/sec dtœœ(a)1with Section ¸ b 3},{y, y0yc œ 14.923}] z 169 dt 3 b) 1119 m/sec œ t11.19 cm/sec 119m (x c x0) (b) (b) œ c " (amp/sec dy 17. 0.1492 m/sec cm/sec olt/sec x0) #" formed by the ladder and walls is A œ #" xythʸdA3œ ˆ "From b y dx ‹ . The (b) Ê dt œ Èx b y b z dt b Èx b y b z dt " at dh 2 dr 5(5)] œ c ngle# [12(c12)anline=y==y0 b œ alculus /sec. Edition, # ‰hapter 3 dt blope=y'[x]/.First[%]59.5 ft #11Section 3.7C Šx dt Rates #dt 169 c Thomas’ CdI/sec dR dt" dVdt V dIRelateddI 0 m /secarea 1r b 1 dV $ 3d ‰ x0 dR ‰ rh s dR 12. y0 # 2 y0 + " (b) (c) 0 dx ImplicitPlot[{eqn, Ry },I {x, dtI.ˆc33,d)Êdtœ3},{y, ˆ dVc)3, mdt ‰ 3}] dt x0 c 3, x0 b I3},{y, y0 c 3, y0 c 3}] œ 0 Ê 30 œ 2x dx b 2y dy b 2z dz Ê m/sec y dy b z dz œ 0 ‰ b5 (a) x0 bdt 1œdx" (a) d tanline},dy dx I ˆ " ‰dz From(b) (a) with ds c volt/sec c # Ê{x, œ part œ + amp/sec dt Êcs œ I (a) dt dt dt b x dt x dt œ "b zSection x0,cft/sec y0} b y b z5 (c) œ œ 13 "Ä (5) 2(Ê12)dsindt) Èxddtbc13 119 dx Ê3 59.52.5y[x]sin , 9, dt"x1, z1dtˆ, 21, 26 m/sec 3dt"c8.493cm/sec c 20 rad/secdtm/sec œ c8.49 dt c 12 dt 5(5)] Vm=slope/.{x Èxc : 1b 3, 7 )b È œdV 7c1.13dt cnd œ dV dt (b) ¸ c0.0849 b " dtœ xyc †#22. b Ä7 ftc, z 3.7 RELATED 0.0849 a 1 rad/sec œ c dt dRy /sec.dt † 1 c RATES dR 3 dt 0.0113 ˆc(c) ‘ œ ˆ 1.13dtdIœ b I ˆ ¸ ‰ 0.0113 œ (b) œc R dt ‰cm/min œ ˆ c V dI ‰ m/min 18.‰ (a)ohms/sec, R is m/min ¸ (b) cm/sec " ‰ ˆ cm/min b dR dI 1 y d) dy 3 ‰ dxtanline=y==y0 b m (x c dt increasing " Ê dt c# œ R(3) dz # 25 Ê x0) " ˆ dI " d dx # x) dt dt z dTotal: "ds dt I ‰ dy œ c dt amp/sec dt I dt I dt 1bvolt/secRELATEDbRATES13c12œ dt œ y ˆ 5 (5)b c1 rad/sec dz (b) c59.5 ft# /sec (c) c1 rad/sec )inFrom œ 13 (a)dtwith "dtdt œ 0cÊ sin")‰‘ Èx " ‰y b z dt3 œ Èx 3bzy b z "dt y) bdt part †Èx b y z z œ 12 3.7 dt b dR Ê <13.dt (a) ˆ dt †ft/sec bc (b) dt d) d) dx œ # 23. "11 dVc " tanline}, (3) dy ## ohms/sec, R‰ # y0 sin3, y0 dA œ " 1 œ . x0dA A œ ab ) Ê dt (b) A œ " ab sin c5 Ê dA œ " ab ) dI ‰ds (ˆ dR ‰dt ImplicitPlot[{eqn, dz dI ‰ ˆ # {x, x0 c9# dV(a) V dI" is increasing + 3}]dr # ab #cos5) dt dR dy c z ft/sec." œ dx dt y dy s Rdˆ2sdPbœ # dt dt Ê 2y 19. Ê dt c 3(b)Šx 1. ÈdR dt1r "yÊ dsdtœœdt dxr [5(yc(b) r b(c) 26y c y m m/min I 2x b dt dt I # ds c R dt œ b 26y cI Êdt dzb I3},{y, dr c œ ˆ œ m/minÊAy dy 3, m c È È œ‹ ¸œ 442) 12( dt481)] c /min dt part (a) with b dI 0(a) dt 241œ 2xr dx b 2y œ bˆ2z dt Ê 21 "b dy b zœdzdAc0¸ y=8 œ 2881 d) " (c) dr # y=8db 2881 m/min # œ œ b y dR zb dt RI Èx b yÊ z0 dts ds dt È dt dt dt 169 dt " da " dtFrom x )"Ê dt 12 I dt 2 dt dt œ œ b b Answersdr (c) A œ # abx sin ) Ê dt dt œ # ab cos ) dt b # b sin ) dt b # a sin ) dt dt dt dt x " < 2x1dx A 2y ‰‘ ˆ " ‰dAdz œŠ3 dx b y dy R z dz dt 14. œ" x knots dt . ˆ œ dy œ 1 2x# bb 2y# #œÊ dsdtœ c614dx#dR y dy b isdI ds œ 2 ˆ" [5( 442)Related Rates Section 3.7 Related Rates #0 œc dt dx c 3 1r dyÊ #2z(3) dI21rohms/sec, ‹ Êincreasing‰ dI c3.72Pb 12(c481)] dI dt dR ÈP s dP dR 8 169 I Section 3.7 169 (a) Idt dt b 24.dt dP # 10I b x bdt Ê 0 6#.3.7 RELATEDÊ (a) Ê1dtin/min 2RI xdtdt 4œr# Section 8cr I dr dI œ dtdt œ P œ RI dt RATESdt dt dtœ dt . 2RI œ œ c Ê dS œ 1 b 2RI dtÊœ c I# dr dt1 0 I169 dt œ (b) c169 Related Rates dt dt 51 in/min 4k1r " ab œ k, adt Section dP .7) d4k1r S dt œdIk, a constant. ab sin )drÊ 2 dA‰ œ " ab cos )dId) b " b sin ) da dA , dr sin ) Ê dA œ dVab93 œ ) 2 , dt constant. dt œcos dR 20. " kS dt œ " dt # P ˆI ) A œ dP dt (b)Given A œ 1dI , œ 0.01dI # A # 2RI r dR 2P $ 10. cm/sec, and dt œ 50 cm. Since dt œ 21r dr , then dA ¸ r=50 œ 21(5 r # (b) #PdR # RI# dt dI 20œ # dt dt # dt # 0r œc I 15. dx ft/sec # dt dt dt Ib Ê2d.) y Êœds dyœbÊdAÊ œ anddtdrœkiteydt bxdA b "da ztheœ dc Idb dt distance between the dt0.0239 in /min. œ xz dy represents ds the distance betweenxthedSdtA466dx" b dryId) and 2RIœ Ê cos 466 ) #b " b sin c daI dt16. About horizontal œ ) " S I 4 dt dt ab sin 1rb# 2RIdt dA " 81dzthesin ) b " bb dt ab dt a) dz œ ÊÈ(b) girl œdt2Èdt Èx b y Êz dy È) Èz b y" bdz dt) )cosÈxÊ y. dt Adtœ#1"Î#xyÊyzœ dt zabL/min y dtdxz b# dt x #bxy bœP 1dtcm /min.¸ dh dt L/min A ) dt 1 b zœ dt x rb x25.ydt b b #cosdt Ê increasing dt #b z zsinL/min # 0.2772 œ b # ) È b b b z# dtab b b dt dt sin about 1rx ) È ds b # dt #‰ dV dt dV dr # b #ft#œ ax dP b x# dR1. dt1681 Èxx b401dtft# /min V œ 1r 2 ˆ IÊdIdt œ# 1r dI y# /min b y b# ÊÊds œ œ dx œ y400(25) œ 20 dI x 3 (b) V œ 1r h Ê dt œ 21rh dt dI #s œ 01Ê (300)#the Iy ) da 21.x" 1 kite ) zdb dR represents ft/sec. dyI 1681 distance dt dt IcosÊ d0 sœ " bœ girldt bdtthe xft/min, ds 500"Î# (a)y dt œhorizontal zdtœz c 2P between the œ. 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Since dA dt , y ) dtjdr , then dA ¸ ven b x S œ œ œ sÊds dtds œ 0.3, 0.9,dx50 cm.x b y zdydzdtGiven rdzdtœdx 2 cm/sec, œ dzdtdt2c1.5625, 3.5, 5.0625 w œ 5 cm. cm/sec, 8 and dt b y 2 4 a dS Èx œ 20, b) Èdtdx dr dt œ c 1 ddt y z 26. # with (a) Ê 0 # œ 0 œ 2x dt radius dtdtds dtdr Ê r0 œdx x dy ds 11. dt x 0 dz dt c dt b dV dt 12œdz cm/sec, j œ 12 cm and h0 ÊFrom partdtœ (a) h dtÊ(a)0 c2.5 #ft/sec 2x rh 2y x dt rad/sec dt Ê #b y dyr=50 z (b) dr 0 rad/sec dr c) # From the bwith is 1.9 œdz #1rdt ddh0.6 dx (b)b 2y3 dy 2z 2z œ 61 y Ê dy dy b x r c y100 part # 2y#dAdt œdV œ Ê dt2x dt dtin/min. Also V Ê dt b (c) (213.8((c) #V (a)È"Î#22. b# 2zin.0and xœdx"Î#2"1 bdy cdt20bœb "dt dx b dxr dx xdt dt b xdt dt1œ 0 20 dy dy œ ft/sec # rh dta (b) œ 3 b r œ 50in., œ b x eter my/min. x sb #y 1rx# bœ 2dt rdsx ,b dtdtb b 3000 dtœ 2œ(50) ˆ ‰ is œ b) Since Ê œ œ then dA ¸ Ê dt 1 #b dx dt dt dj # y dt 2y dt dtb c # Ê scm. x b y Ê 2s a œ 2x x dt andy 1 œ b dt ¸ c b ydt dt r=50 dt Ê"Èx #† 100 dt2x w b 2y dt j dt b w cÊ dA m/sec œ c8.49 cm/sec 18. y(a)1 dtdtÈx0.0113 m/min œ c2sa) 0cm/minÈx "b ydAdtœÊ dw(b)œ cdtx 0.0849 œ 12(2) b" 5(c2) œ 14 cm#2/sec, increasing 1.13y A œdV dt Ê # dh c dt ¸ dt dt (b œ j dt a dr # dt , b " œ Ê dV 1.9) ˆ the radius is # 17. handdsdrdV œ is r#in/min. ) at yabout 61Êdxin œ 3dyV 1 1r#) Ê dV dx 21rh dr 0.0076 1.27. ) # rad/sec3000 # dh (a) V œ V dy h r# dt dV œ 12 dt , dr .8#in., 3000.‰ œd) V È"Î#r#The volume "1"changingAlso dA 1r0.0239 Ê$ /min.1r "œ r dPh 28. dj2 rad/sec dy(b) Vdx 3 1r yh dy dt œ 3 1rh dt 1.9 # in. 1 dt œ # x 4.dx # 3œ ds (a) œ b dt ) "b y# 3 (c) #sb y œ x bdAÊ Ê s" œ x dt ) yd) Ê 2s d y dAdV A œ cosdsin )dVdV 2 dt œ #drdh œA j 2y"2w Ê sinÊ da dV bœabdtcosdrdt d) 2) b bœ cœ da cm/sec, constant b b Ê " (œ"2x dt b" (b)dr sin )#dt †)0 œ dt" dt22b 2y ddt c " dt 2(2)dax 0 2s dA 2 2x " dw œ ) Ê " sin ) dt dt dt b) ab 2 œ b " P2 " x œœ " a " œ. dh(a).cos " xdt ab ) 1r# Ê 11 V ab cos y)hdh) (c) x œœ rhr$# (b) œ )b# sin Ê r dt œ dA œœ 1 ) )2(dt b ab dt # # A A) V 9(a) ab œ(a) ab œ " dw dt (b)œ œ ab # dh Ê dt 5 # rvenœj0.00761. sin sin Ê.hÊ(b)dt ft/sec.x 1#rœatdtÊdt Èbdx dr3 1dt (b) dt Acos1rhVdtœ #)b Êh dt dtœ dt# ab cosrh dt b # b #sin ) dt dt # # 3 )23. ) œ dV # 3 b # dt ‰ d 4 œ # ( cm/sec, 119. yÊ dtœ"Èm/mindsabout 0.0239 1 dt b (b) ab ab 1 aÊ dt # # Ê , dt dt ) œ2 12 cm, dt"b ydt in /min. dz dt cm. œ #r# # # is zc ds (c) dr ¸ dw œ c5 m/min b"y dt # c Ê" V V œ"x9 # hdh(a)cm/sec,drjr# db) b 2" anddt œ dy " " È #dt db 3 dy 26y c y# 3 y 00 dt # b z# "2c) sThe volume dA changing dtdxœ)2xrhdyw 3 da5dt b 2z3 dt dx(b) #r œ È# dx# "Î# m dydD d# ) " # dt bdAdab2" "1œ 1 2s dt d 1 dt b 2y da " sin ) d # dt œ #3bc) cosc) ab dAœ)bb4h Ê Ê bœab"coscos ) b 2y #sin sin()2s † bœsinh) dhbdt2yœ aw b j b c Ê dt œ " aw# b j# bc"Î# ˆ2w y=8 b 2881dj ‰ Ê dD œ ab A œ sin sin sin b 2j dt (c) (2r) œ ) dtddtab x 3 Ê dt bVd#24 dt œ ˆ ) ‰ dt b dt ) c)dV #œ # 2x db x # ( A œ( #3r s#Ê r œ1ry ) dtœ 2s#rh abdA dtdt hb # b 1h Êdt b D aœ a #wdt)bj dt dy dt œ x dt yÊ œ # )dw Ê dt 3dt a dt 2x dt œ 16 b Ê dt 0 161 sindt Ê 1 Ê # œ 12(2) b 5(c2) œ 14 ds /sec,dx j # 1 4h # dt dt œ A j8jw(a) s œ Èx#œ bcm. dt10 Ê dt # )/sec,œ œ 124 cmdt œ wc) 5 cw m/sec s 3 œ x# b y27b 8 # Ê 2s cm 2x increasing b 2z dz8 8 3b # 5 # 3 dt 9œ dt . Ê and j 29. y(a) zœ in/min b z in/min (5)(2) b(12)(c2) 2y dt (b) cdt in/min b dt dVc dI " /min dA 24. " dt 14 ## " 5.# h (b) dV 51 volt/sec dt (b) 1 51 2 90 j 9 3r ‰ 1h œ 16 (a) dh " 2) dj (10)r (a)4hÊ ¸dP 20. adV œ9cmœ 11.19 cm/sec œ 1constant. " 25 V œ r# 13 cm/sec, decreasing œ c 3 amp/sec m/sec increasing 4 1 3 VÊ 0.1119) ˆ dV h 4k1r 2) œ 1 r# Ê 2 1 )16P44œ Ê .bœ2(2) drœ5(cœ hœ1144h ‰dw/sec,2(1 #dh drb dt¸ k, bœ 16dA dA ˆ È(b) 144 œdA 1c h Ê dV œ 3" rh dr dt 2w cm/sec, constant " , 1 t œÊ 2j œ œ 256 b drVœ. 2 3dt #b 3kS œ œ1r cdt Ê 2(2) œ 0dI ) dV ‰ # Given z Ê cm. Sincedt b and , œ cm. , ) 9 # dI ‰ m/sec,5andœ œr50r# #œ30.01(b) dtcm/sec,r 3dramp/secdt2xr=50aSince(50)2ˆ100drrbdt , dR " ¸dA ¸ˆœ 2c 2dw (50) Ê dt 1 dt 10. ivendtA(a)#1dV11V , 1dtvolt/secdAdIœz#dtœ and2750dt dAV dtdx R ˆ2dt‰ dydtb12z1dz, dr œthenœamp/sec1(50) ˆ 100 ‰" dR œ " ˆ dV c V dI ‰ œ 15 y 2 G b . b r 4 , dt dt11 #(a) dt dt 2c 3 r œr ds 50 cm. œ 2y œ r ‰ dtÊ , c3 r=50 r=50 cdt ft/sec cm/sec, xdr dw y490dh A œ#ˆdœ œ30.0.01 "Î# ft/sec "dt 2sthenœ ddt Since"Î# dt Iœ(b)dtdtdIthencdt3 dt" (b) œw31 b j d ˆ 100 dt s ‰ "x# bdh 8 2 1Ê b dt b dR ‰ œ (c) œ R dt 90 dt I dD dt I dt I dt œ 1w¸ 0.1119 m/sec j# ¸ 0 1 cm/sec " # b dt c constant œ dt 2(2) 466 1 11.19 m/sec # 0) œ2562(c2)3b 2561a œ bcm/sec, Ê dr œ dt œ 14.92#cm/sec dw b j dj dt dt 1 ) # 466 ) 2 Dœ increasing #dtœœwœ 032œ bˆ dR.1492dDdR increasingjabout‰ˆ2w L/min 2" ˆdt ‰ ÊVL/min Èdt n dtRdt1 cÈ#c)(#b V about17.r aL/min3ˆ dV dt V , #‰a"wdR c " < dI 1681 ˆ dt "œ‘¸ 0.2772 dt ‰3 Ê œ3" dt dV c)d j œ R ˆ31681dRdtI b" ¸rh dt dIb) ˆ dV b dw 466d dR ‰ 25. b L/min Ê œ 0.2772 L/min dI ‰ 1681 dV ‰ dI œ d 12 cm /min.15 w dy j b œ œ4 1 ˆ 90 c R dI ‰ Ê "Î# œ dt ‰ dt c dt# dtd)I dt œ Rw1 c Ê c m( /min. dt dt dt ˆ " c I œdx# ohms/sec, R is increasing Ê Idj ( œ dz dV #dh œ dD dt "‰ dt # c dt V œ dt 3 I dV (3) dt Iˆ I dw #21. dt ft/min, 401 ft /min dtdD12.# (a) b j#dt xyz Ê dt# œ yzdtdt b xz dt b xy dt Ê dt ¸ (4 3 2) œ (3)(2)(1) b (4)(2)(c2) b (4) ‰ cm/sec dt b(12)(œ b" ¸ 0 1ˆ2w c b dt Ê (5)(2)œ31 aw2) 321 jc 14.1492 m/sec œj14.92Ê dt œ È dt b 2 wb ˆ "œ (3)È256b#ohms/sec,31. isc15003 ft/sec ˆ "dt (3) œ 3 ohms/sec, jR is increasing " 32. c2 rad/sec; 1 rad/sec ‰ 3dt œd) 144 dV œ <1R c# ˆc " ‰‘ œ # ‰ ( #(a)dR œ œ 21. 12 c increasing dI 13 m/sec, decreasing dy # dx dz 5 25 dt dt # 1 volt/sec0.3, 0.9, 0.6 (b) b 3 Ê dS .9,#iven dj. œj œ 12 cm/sec,(a) 25 2 cm/sec, j 12 12 c andœ 3.5, 5.0625 2xzœ c dI amp/sec œ (2y 3.5, dt b (2x c and dw 11.decreasingc2c2cm 26.dt wdtœœcm/sec,6. œ(a)(b) œ 1.5625, SdP 5œ I# b dt 2RI2yz (b) dt 1.5625,b 2z)5.0625 b 2z) dt b (2x b 2y) dt Given dj cm. G cm/sec, cm/sec, .dw œ j , P cm #(b) dIœœ 2xy c,20.6 dt dt 26 dI(a)b cˆcm.ft/sec dR œ cmˆandc Rw 5‰cm. dRdR b ˆ dV dt V dI ‰c 3 rad/sec RI w Ê dt dS dt " dV dR ‰ " dV 22. # (c) œ R ˆ dt ‰ dw I 2.5 jÊ dAdt œ I c I(b) d # bdt dt dtÊ Ê dt œ œ ( dt increasing 20 I dA P b w a)j œ(a) dAÊdt RI dA jdxdw5cin/min;db 2RI dA 12(2)dt 5( 5(cœ 14 ¸cm# /sec, 10)(1) 34.(12)(c2) b (14)(1) œ 0 m /sec œ cm# dt dI j 33. dP xz #djbÊ dt Ê dV 2 3 dt Ê Ê dV dt œ ¸b P dt œ dtÊ 72 dt b dR dt10 dz #(œ increasingÊ œ dt b 2) # /sec, 2RI (4)(3)(1) œ dP 2RI A A œ j œ 12(2) b 5( b2)Idtw14xy in/sec, œdtœ(4RI œ 2)02) œ (4 (4)(2)(b 2) b dI Ê dR œ c 2RI dI ) (a) (œd.IxyzjwdtwœÊ #yz<œdt 112 wdydtdt‰‘3 Ê "/minP 12(2)#bc(3)(2)(1)143 cm dR increasing 80 mph. 2 m$ /sec œ c 2 ˆ I ‰ dI œ c 2P dI V6 b c dR " dt dt œ ˆ " dt(3) œ 3 ohms/sec, R is œ I 2 dt 2) œ 33. dP rad/sec3 dw dt b)28. 5 rad/sec 11 œˆ ‰ dt #increasing dt dj dw (d) 27. c # j d rad/sec y dy dx I dt # 2(2) œ 0 cj bˆ2‰ constant P Èx$dIb y#dt 2 ˆzI ‰ dI 28. 5 2P dtdI "Î#I Êdt dj œ I xdt dy (b) 2Pdz j œj2(dt 2w b dP œ œ d2cm/sec, dw #œc2)dI2)#b 2(2)dy cm/sec, constant P# œ ax# b y# b z# b 2 b xy dt b bdR # Êœ dt 2 dP dt b IdRdt 2( 2(cb 2(2)dR 0 œ 2 /secconstant bœ œ c dV (b) œ2RI(dtdI 2 œb2)œ Ê 0 œ œ (2y2b 2z) dx c 2P Ê 2z) œ c (2x œ cb œ œ cm/sec, dI dI 2RI dI (c) j 0 2RI Èx b y b z dt b Èx b y b z dt b Èx b) 2xz dtRI 2)dt I11dtft/sec. dt dtœ dt P b ) dRdtbP2xy2bÊ 2wdt¸Ê323. 2RI dS dt b 2I#I(4)(2)(œb (2x(4)(3)(1) œ dwI m b 2y) dz I dt œ c I dt dw d dt S 2yzcÊ (3)(2)(1) œ dt (4 dt dt œ dtI b dt dt b b jdt d dt dt Ê dt "Î# c dt dt " dw 18° w dt d # "Î# 35. # in./min 36. dD rad/sec j /sec ¸ È " w# w jw#c # dR Ê dzœ d b(c) (c)dSD Èdx œœ(2xab#2z)#(12)(# b"Î##b (14)(1) j dIj ‰mÊ# jbbc# b#cÈ2w¸2w bœb 2j4 djÊ1)dDcŠ w dtwb(j b œdtcŠ1 2 ‹ (1)c6°/sec b 2z) dD b jb œ b7.1dP 2) "Î# dw b dD œ " w dD "Î# "Î# dt dw Š Ê dt 2 10 œ jÊ D œ œ w# w#œ# RI#ab adyb bcjˆb Êdt.dD(a)œs0w#Èx#dt jœÊˆ waˆbdw dt#2"Î# j‰ dt‹ ( Êb œ œ ‹ dtc2) b (2y b6¸. (a) b (10)(1) œ c5 jb(2x2w 2y)dt2RI "dtaœ a#b #b y œ djx#jb y bj d Ê ‰ ds œ dt x3 Èdx dt œ 0 m/sec P # j29.Ê b m/sec 7 œI b dt # # /sec È dt dt dt dt dt (4 3 2) dt dt 10 8 "Î# "Î# x ds (5)(2) in/min y "Î# ˆ P ‰ (b) c 51 2P dIdz m/sec, œ b s (12)((12)(##24.x (a) cy#1#dPzœ#"Î# Ê b ds 7È decreasing2) c2) Êc m/sec,#decreasing dt .(14)(1)y# bÈ /secy# b a b y b (a)( Èœb m dx b b d x dR bcj bœ (5)(2) bÈx xcz#RIdt c #14œ14c32in/min dR Ê j2RI dI ÈÊ y dxdx b # yd) dy 8 ds )c2)œÊ x# Èb)œP0œyx œ aœ(a)0139xcm/sec,#decreasingœÈx#yb by#dtœcx2RI x dI #œb z 2 dyI bdIÈxd)bzyœb c8 dx b dt œ 25 b 144 # b 37. 13 œ dt ¸ b I8.875 ft/secÈx b bdt œ a IÈ dt y c Ê (b)œ cx I z dt dt œ by b dt dt œ I dt 25 b 144 z dt b c dt(b)dt sœ dt x Èx b y rad/sec; b y œ 65 rad/sec Èx dt È13 dt 3 ft/secdt 65 "Î# dx y dy c dx ec"Î#z# b ds dj œ #30. # y8 ft/sec# "Î# (b) b 3 ft/sec dx dz œ 5 ft/sec, the ladderc 13 ft long, and x œ dt y œ 5 at the instant of time (b) is # y dy ds 13. Given: xz 12, x x dx x y (a) ax#b dy b Êœ bb œ dt b b Êj ¸(b)œsœ x ÈÈx‹b y bÈxŠd) 3 bÈx " y bÊ dt œ Èx#bb 466 Ê Èx œ y # dt y# Ê 2s ds œ 2x dx b 2y dy Ê 2s † 0 œ 2x dx b 2y dy Ê dx œ c zb b z s œ Èx " b 0 b dt b È dt È Š y dt( b 466 L/minc 2) increasing aboutyy y#z L/min b 0.2772 dt( Ê d dydt 3 2) dt dz b 429 25.1)(c) dt y29 ‹ ycb rad/sec;29d‹ (1)xœrad/secdt s# ¸ x $b L/mindy dt m/sec # Š2 (c) dz ) œ 6 1681 È dV È dt (4 dx dV ¸ 1681 dx œdy Ê# "Î# È dtdV dV x 12 dt dt dt dt dt # /sec 6 dx # dt (3)(2)(1) b xz 7 œ xyz ÊÈx b dx2) œ b # dy b (4)(2)( 2) a) (4)(3)(1) b2ymœ 169 Ê dt œ c x dx œ c ˆ dx$‰ ¸ 12. ‹ V œ(c)(a) ds s œdV #œ#yz# yz b saxœbbb bdz Ê dx cds dtÈ (4 3Since(3)(2)(1)(4)(2)(c2) b (4)(3)(1)dtdy 2Ê$ /sec (5) œ c12 ft/sec, the ladder is sliding V œx b dt(a) (a)c2)b xyÊ Èx dtdt œ0yÊdy xz xz # xy xyÊ 2s œ (bdyœdx dtdx œ dybÊdy(4)(2)(c2) b (4)(3)(1) œ 2 m /sec y dy dt Ê b Ê ds b œ 5 3 #3 dt Ê ¸ (4 3 2) b (3)(2)(1) œ cdx b y œ m dt œ 2y dt b # dt. xyzs È2œ 2x 31.b 2y œ #ft/secy b #dt 2x dt 2 rad/sec;ydt bdt2y dt y 2s œ dt dx y(4 dt dt Ê 2s œ sec dS ( Ê b Šdt 29 ‹ (1) œ cm/secdy dt x dt † 0yœ32. dzdtdtdt 2y x2x2)Êrad/secc x dt2s † 0 32.2x 2 rad/sec; 1 rad/sec c x dt 1500 c È29 y dt dt dt dt 1 dx d dx " dA dS # œ œ b2yz b 2z) dS (2y(2y2z) (2x y ˆ"‰ 2xy b 2xzxb 29.5 0.3,œ b y# b xdx dtds (2x#b 5The zdy (2x b# b xdz dz # z 1.5625, ladder and # # (a) knots b 0.6 (a) Ê dt b b) b dy dt Ê theœ dy dx ( œ b (b)dt S S2y(b) 2z)œdt b38. 2yz#isÊ dt0.9, (2x "Î# 2y)œb È y(Èbxy# bdxbinstant of time dt(b)bc# by 2s3.5, 5.0625 b 2y dy b 2z xy b y 13 ft apart. b dt dt œ œ b at dtdt #b ofx sb y triangle 8. s ven:(b) œ 2xy bs2xz È26. b Ê œ axdt long,band 2z)œ12, x(2x2z)2z)areab (2x2y)2y)b yformed Ê the ds œ 2x dx walls is A œ #dz Ê dt œ # Šx d dt dt È dt dt dt dt dt x y the dt œ 5 ft/sec, the ladder # dy dz 10 # c2)#b s # # #; (12)(# /min dS ¸ Èx10)(1) dy (12)(c dx 2)(14)(1) # 80œ# 0 Ê# /sec œ 2x dx b # 8 Ê (4 3 œÊ (œ y#ds # b œ dx x 2yœ b ˆ34. dz œ c ft/sec, b (12)( 10 Ê y b dt (4 3 2) œ (0œ 5 in/min; # dy # (14)(1) œ xlong,dS b(14)(1)ÈxÊm72/sec# c s sc œ of#timeÊ 2s # m is ds b 2y at " Ê sliding down the c 119 Ê 59.5 c y dy 2z dy dx # and s œ 2s 10)(1) b dt b œ x# b b (5) )133ftin .b (a)(¸c)zœœ169 33.y œ1atzÊ Ê# 2)3binc/min‰dtb0zmds/sec2schangingladder2y 2s 12)2z80 mph. 2y dy œ cdx œ ft /sec. Since x# dt y#x2) œ12, bdtb 5 #2xthe instantdt xby12y œ mph.12 2x dt the dt # [12(c †34.œ 2x dx œ wall is dt b b 5(5)] b # y œ 0 dt dt dt y dt 5 dt dt dt dt x dt "Î# 2" y dydt dj x dx # y d d dx dx rad/sec j x(c) x c)#œœœ#x# 12#‰b b#27.œ1arad/seczthe zbzÈbx"Î# y db zœ dtdown xthezywallx dzÊ csin ) 28.b5 b † zdx zÊ dz) dz c " † dx œ c ˆ " ‰ (5) œ c1 rad/sec ## ( œ ( y j z Èx # y œ#cÈx ft/sec,# ydt bb"Î# is Ê dj œb xcosy" dx b È Èx ˆ " b zdt dy 13 dx b # œ# y y x cb jdt bÈ bˆ b Ê5)dt zœ z12xbab#bb b# ladderÊ slidingÈc) È bxb)bœ z13 dt dA y b y ‰ dy) œ È Èx b y b dtdt dt dy b b dt b œ dt 5 y area c the y( 5 triangle formed by the ladder anddt dt is Èx œ z xy Ê x b y b z Šx dt b y dt b z. The area 13 sin ) z z dt18° dt ) The of walls x Ay b # b x dt¸ c # 36. dt " b y ‹ dt n 36. # " dy # rad/sec œ c 18° /sec ¸ c6°/sec dy ds 3 . (a) s œ È35.b y# in./min 3s# œ x# b c 10 2rad/sec œ c 1 /sec dx b 6°/sec bc 10dz 2 # 47.1 b z# Ê dA 1 " 0 m/sec " ‰y2 b z Ê dx 8‹ œ 1) is changingcandœ Š ÈŠis‹A1)5(5)] œÊ (c2) b #59.5 ft#(1) (1) dt2s0dtThe2x dt 2y dt 2z dt bbthe ladderat " b Š c12) (1) (b Š Š È 119 (œ 2)Š ŠxÈdt /sec. 0 ‹ . m/sec y ŠÊ29dj ¸ dj ¸ [12( x È ( ‹ œ xy 3 c dt œ ˆ b Š ‹ ‹ œ œm/sec area walls29 œ by È Ê ( dt 2) 3 2) œ 4 ‹ b 1) b (4 29. c # È 29 ‹ c c È29 29# È29 29 5 m/sec 29 ‹ # dt (4 3 2) # 14. s dx y# b x# Ê 2s ds œ 2x dx b 2y dy Ê ds œ " Šx dx b y dy ‹ Ê ds œ È" [5(c442 œ dt dt dt x 169 ) c 8 (5)] œ)c 119 œ csin 37.d) (a) "c† dt c d) œft/sec " ) 8 œ c ˆ " d (5) œ dt rad/sec ) c8.875 ft/sec c59.5 ft# /sec. 13 32dx¸ Ê8.875(b) 13)sinœ †cdt rad/sec; ‰ ) œ 8crad/sec dt d) œ dt 8 rad/sec; ds œ dt rad/sec cos œ #13 Ê ) dt œ cd 1 (b) dt dt 5 dt È13 dt 65 65 65 dt 65 dder Given:d)œd5c5 ft/sec, theœ 8 )ft/sec13long, 1 rad/sec 12,12,œ 14at thethe instant time c3 ft/sec † dxis 13 ft dx œ 13 œ dx y c ˆ 5 is theft c andof )time œ † (a) œ 5 ‰ 5) instant d x 13. iven: dxdtlong, and" x30. 12, ladder" at(ft " œ long, andœ œœyc6 5 knotsinstant of of(b) y œ 5 at time " 3 G dt Ê dt dt ) ft/sec, the dt cdy rad/sec;dxdt œ "sin ) ladderdtis œ c 6 rad/sec; xdt œ " rad/sec rad/sec d 13 (c) x 6 6 #12 ‰ dx 6 dx ds wall" dy dy œ x #a) c œ x# y xdt b# ycœ 5œ (5) œÊ 12dyc Ê dxdsœ ladder‰is ‰ (5) c12 12 ft/sec, the laddersliding down 481)] wall œ #ds ˆ 169 Êbc œ ft/sec,xthe c"ˆcxˆ dx sliding ‹ ft/sec, the ladder is 442) b 12(c thethe 12 12 b y dy down theœ # x b yœ 2x 169 2y dtdt c ydt dt œ Š dt œ œ c Ê dt is sliding down wall œ dt( Since Ê 2s dt y œs 5 5 (a) b Since (5) È169 [5(c dt dt dt y dt 15. Let s represent between the girl and dy 31. y dy ‹ Ê ds "œ " [5(c442)‰b 12(c481)] "the distance 32. "c2 rad/sec; 1dx the kite and x represents the horizontal dist c1500 ft/sec " ds dx dy dA " dx 2ypart. Ê dt area of the bwalls is A œapart. Êladder and walls is Ay" ‹xyThe areaœ ˆ ‰#Šx dy ds dx xrad/sec400(25) Šx dt38. fc614 knots œ s the triangle dt formed#by È169 dt and ˆ # Šx dt b œxy .Ê Ê dA ˆ#" ‰ Šx dy b y y‹ . ‹ . The area ormed The dA a ( The the ladder and 29.5 knotsdt thethe triangle (b)dtb) byarea of formed by xy ladder œ walls is A kite# dt # s# œdt(300) # b#x dt dt dt œ s dx œ area œ 20 ft/sec. œ The 500 girl and œ Ê Ê bdt dt dt dt 119 " 59.55ft# /sec. 119 # 2) b 5(5)] œ c #at œ [12(c12)12)5(5)] 10 in# /min c59.5 ft# /sec. is changing at 33. 72 b b 5(5)] œ c # 34. 80 mph. is changing " c [12(c in/min; 3 c 119 œ œ c59.5 ft /sec. œ ## # t œ " † dx Ê dx œ c " 1 †d) the girlˆand(5) œ) c1and x "represents the horizontal distance between the the kite rad/sec dx d)s represent the distance betweendx ) " dr " # "dx dx‰ d) d 16. " x dt (c) cos ) œ 13 Ê sin sind)# œdtœ dsc† dt dx œ c c Whendx† œ diameter‰isœ c1c1the radius is 1.9 in. and dt œ 3000 in/min. Also V œ 61r Ê c ) ) dt "œ13 Ê Ê dt œ 13 sin ) the c ˆc‰ (5) (5) œin., rad/sec ) œ " ˆ " 3.8 rad/sec dt 13 dt 13 sin 5 13 † œ dt n(c) cos ) Ê 13# kite c the kite œ s Ê and represents dt x dt dt 400(25) œ † dV dt rl and girl and the œ (300)#xb dt Ê dtthe horizontal50013 sin )20between the 5 ˆ " ‰ x œ distance ft/sec. 5 " s Ê dt œ 121(1.9) 3000 œ 0.00761. The volume /sec ¸ c6°/sec 35. 7.1 in./min 36. c 10 rad/sec œ c 18° is changing at about 0.0239 in$ /min. 400(25) x 1 Ê ds œ dy dx œ ds 500 " œ 2dx ft/sec. 0 dx dt b œ dt dtÊ Ê ds ds dt dx dxdy dy ds ds È"" " Šc dx bb dy ‹ 481)] ys # b bÊ œ rœ dx ds # s# 2# y# b x# dt œ2s Šx œ 2x y dt ‹2y dy dtÊ dsd169 [5("x 442) dy 12(cÊ ds œ" #" [5(c442) b 12(c481)] 14. œ y diameter is 2s s œ the radius is 1.9dtin. anddt œx b y y‹dtAlso Vœ 61È169 c442) b 12(c481)] sdt dt dt hen the b x Ê 3.8dtin., 2x dtcb 2y dt Ê dt œdts Šs 3000 dt " dt# Ê dt3 dt È169 3r Ê ddV œ 4h8 1r dr œ r [5( dt 12 dt dt in/min. # 32 1 1 (b) )r œ c rad/sec; d") ˆ 4h8‰ h œ 1627h Ê dV œ 169 h dh dr" 37. (a) È13 ¸ c8.875 ft/sec17. V œ 3 1r h, h œ 8 (2r)$ œ 4 Ê dt œ 3 Ê V œ 3 1œ 365 rad/sec " dV dr # 65 dt dt dt dV œ 1.9 in. and diusc614 14 knotsdt œ‰ 30000.00761. Also volume is changing at about 0.0239 in /min. is c6knots ˆ 3000 œ in/min. The V œ 61r Ê dt œ 121r dt œ œ 121(1.9) dt 9 90 d) ) a) dh ¸ h=4 (c) about 0.0239 in$ /min. œ (" rad/sec œ ˆ 1614 ‰ (10) œ 2561 ¸ 0.1119 m/sec œ 11.19 cm/sec œ c " rad/sec; ddt dt 61. The volume is changing at dt 6 6 dr 4 90 ‰ 15 between represent the kite and x represents the and the #(b) and1h 4h Ê dV œthe dh dh 4 256distance1between 15.etLet sthe girl3 andthe distance between the girl "horizontal distance between the1h œ 3 ˆdistance between thethem/sec œ 14.92 cm/sec " s represent the distance between the girl and the 4h ‰ kite r16represents the horizontal 3r 4h L and œ Ê dt œ 3 horizontal 1 œ 32 ¸ 0.1492 ˆ kite h œ x x 3represents 169 dt dt œ ## œ x dx 400(25) # 3 1r h, hds 8 (2r) œ 4 Ê r#œ 3 # Ê V œ 3 x1 dx 3 400(25) 27 dt b4h Ê kite # sœ # (300) h b xft/sec. ds x h dh 400(25) x œ dt # œ 29.5œ 20 Ê # girlgirl anddt 9" sÊ 4h 38. 500 #161knots apart.dt 161dx 11.19 cm/sec 20 ft/sec. kite s Ê ds œ œ s œœ h 90 ¸ Ê dV œ s dt dt œ )œ dh ¸and œ ˆœÊ 1 ˆ œ‰(300) 1b x 0.1119 m/sec9œ œ 500 500 20 ft/sec. dt 3 Ê V 1613 ‰ (10) œ œ 27 3 dt dt dt h=4 4 256 4h 4 dh cm/sec 4 ˆ 90 15 0.1119 m/sec ) the radius is dr œ 11.19œr 3œ 256 ‰in/min. ¸ 0.1492 m/sec œ 14.92 cm/sec dr œ 32 " dV # , r œ 3 Ê dt œ 3 and d "" 16. hen15 the 1.9 in. is 3.83.8 in.,1the radius1.91.9V œandrœr Ê 3000 œ 121r dt V œ 61r#1r# Ê dV 121r 1r dr When diameter dt isdtin., the radius 1 Also in. 61 d œ dt in/min. Also V œ 6 Ê dV œ œ 12 dr dt diameter 3000 0W the¸ 0.1492 m/sec œ 14.92 cm/sec is in. and dr dt 3000 in/min. Also is ‰ œ 321 dt dt dt dt 10.00761. The volume is changing at about 0.0239 in$ /min. " dV Ê dV 121(1.9) ˆ " 3000 ‰ 0.00761. 1. The volumechanging at about 0.0239 in$ /min. œ 121(1.9) ˆ ‰ œ œ 0.0076 The volume is is changing at about 0.0239 in$ /min. Ê dt œ dt 3000 dt dt dt dt (4 3 2) È29 dt È29y w Èx b w b j b j dt È29 # 4h 16 " 161h Ê œ " # h, Vœ " 1 ˆ 4h h 161h Ê dV œ œ 1h 4hdh# # h œ 161 dt dt 17. œr "œ r31rÊœ 3œ 33(2r) 3r ‰ 3r œ œr 4h 4h Ê œ " 193ˆ1 ˆ 4h ‰œ 161h 27h Ê dV 161h 9dh œ4 r ‰ V V 3 13# h, h h 8 (2r) œ 3 Ê Ê 27œ Ê V V 3 Ê dV œ œ9 dt 8 3 4 3 3 3h 27 dt dt 90 9 œ‰11.19 cm/sec ¸ dh ¸ [email protected] m/sec œ 11.19 cm/sec 90 90 2561 dh 0.1119œ 9 (10) œ 2561 ¸ (a) (a) ¸ h=4 h=4 ˆ 16ˆ4 ‰4(10) œ 2561 ¸ 0.1119 m/sec œ 11.19 cm/sec dt dt œ 1 161 4 ˆ 90 ‰ 15 œ 4h dr dr œ 4 dh œ 4 90 15 15 3 b) 1 œ Ê ¸ 0.1492 m/sec œˆ14.92 cm/sec¸ 0.1492 m/sec œ 14.92 cm/sec ‰œ (b) (256œr 4h 321 Ê œ 4 dh œ 4 ˆ 390 2561 ‰ œ 321 0.1492 m/sec œ 14.92 cm/sec r ¸ 3 dt 3 dt dh dt 3 dt 3 dt 3 2561 321 ...
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This note was uploaded on 12/26/2010 for the course MA 1012 taught by Professor Franciscoperez during the Fall '10 term at ITESM.

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