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# Vector Calculus I MTH 254

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This 74 page Class Notes was uploaded by August Feeney on Monday October 19, 2015. The Class Notes belongs to MTH 254 at Portland Community College taught by Staff in Fall. Since its upload, it has received 31 views. For similar materials see /class/224646/mth-254-portland-community-college in Math at Portland Community College.

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Date Created: 10/19/15

28 May 2009 gm 5 Applicaunns Dame Integals Densny Funeuons Lmea Densny Fune1mnlte g A quantitynun lenth Surface Area Densny Fune1mnlte g n quantitynun area Vulume Densny Funeaen 2 g e quantitynun vulume m Imus m e IIuxydA m e IIIpxyzdV gm 5 Applicaunns menuble Inlegals lecm Charge Densny Am hm 7 Surface Density no y AA m AA Hm ma Mass m 322 whimsy asas z J39J39uxydA 1 Example Let my be the surface charge dansxty ufamsk ufradwsR sunhthat Dxyuz x N2 N am ndthetutal ehage Qm Cuulumbs QTIkequotwrdlde Tdefequot rdz an n n r my l R 4 gm 5 Applicaunns meme Integals Elecmc Potential 7 Example Appneauon Elecm putsnhal Vat a dmance R 39nm a ehage elemem 19 15 peepemena m dQR The elem putmnal ufacharge dmnbunums the sum quheputennals 39nm eaeh ehage element Cunsxdera Circular 645k ufradxus a wnh umfurm charge dmnbunun Um charge3628231 Cuulumbs per squaremetxe and Vat aputh un the axis the eemxe the charged disk dA rdldeanddQUdA R P dVEmvde 211 L R E 1 Kidoguchi Kenneth 12 5 Applications ofDouble Integrals Fond Moments from MTH7252 65 e point P on which athin plate lamina of arbitme shape will balance is called the centre ofmass Consider amassless rod supportin two masses m andng Ifthe fulcrum is located at midi d2 the rod will be in balancequot X2 Ford emee anile2 e em XI dl d2 m m a 2 pm l l 2 2 2 m m m m l 2 l 2 2m fulcrum 2 kl kaxk is called the moment about the origin a L331 ufthe Lever hy Archimedes k4 2 5 Applications ofDouble Integrals Fond Moments from MTH7252 65 For a system of discrete mass elements the coordinates ofthe centre of Emit mt kl Where m is the total mass ofthe lamina and quot J distance from axis quot 1 M i 2 mm y or MS My m J distance from xaxis are the moments about thexaxis and yaxis respectively we 28 M ay 2009 12 5 Applications ofDouble Integrals Fond Moments from MTH7252 For athin lamina ofm fs mass density 6 the coordinates ofthe centre ofmass are M y yam m Kidoguchi Kenneth 12 5 Applications ofDouble Integrals First Moments for MTH7254 For athin lamina ofmass density ow the coordinates ofthe centre of mass are m Ham UuxydA o o M Um UNWW mm o D My mm mot ydA o o where D is the bounded region ofthe lamina 2 5 Applications ofDouble Integrals First Moments and Centre of Mass 7 General Case For athin lamina of surface mass density 0xy the coordinates ofthe centre ofmass are M IIde IIxcxydA x 13 D m I dm IIcxydA D D Jdistance yam IchxydA fromyaxis y 7 1 dm Timmy D D J distance dm GXydxdy y where M is the rst moment abnutthexaxis Mx y crydA My is the rst moment abuuttheyaxis My ammol D 2 5 Applications ofDouble Integrals First Moments and Centre 0 Mass 7 Example Computation Consider athin lamina that can be described as the region bounded by the equations y o and y re x r 4 Ifthe mass density ofthe lamina is uniform c 15 gmcm2 and amp y are signed distances in centimetres present the analysis to nd a m the total mass ofthe lamina b M the moment about the yaxis e NIX the moment about the eaxis and d em yam the coordinates ofthe centre ofmass Reanalyse the problem above for anonuniform mass density of 0xy x y gmcm2 28 M ay 2009 Kidoguchi Kenneth o va Moment oanemaeT The secund mnment uf ut an ems rs de ned tn be m2 where M5 the perpmd cular arstanee quhe parade tn the ems 2m Mm ygtdA o rim X1UxydA secund mnmentahnutyr 2x15 12 5 Apphcauans ufDauhle Integrsz he SeeondMoment a arseretepsmele ufmassm aha ecund mnment ahnutxe sxrs I I ular mnment ennaua n n xed pmnt at enn krnene energy uf energynfthe 19sz v 12 Seeond Moment amp Rotatrona Energy for Drserete Masses censraertw b d sufm s where 1 ma 3 papendxculzr arstanee rntannn Tharefnre the tnte1 kmenc systems 5 Apphcauans ufDauhle Integrsz e as m and m urbmng a stmt angular meea m The Lhamass m xs and ms the hem the ems uf ZI m1lm1 12 s Seeond Moment amp Rotational Energy for a Thm Rod 5 Apphcauans ufDauhle Integal m 1 Izld m m z 392 e MLZ m K12L1m2LM Mtnta1 mass L ntenength Axis uf 4 2 rntannn m M Lu2 thogucm 28 May 2009 Kenneth l2 5 Applications ofDouble Integrals Parallel Axis Theorem Let m be the moment of inertia about an axis through the centre of mass The moment of inertia about a parallel axis is 11cm mhZ where m is the total mass of the body and h is the perpendicular distance of the axis of rotation to the axis which goes through the centre of mass grzdmgxra2yrb2dm xiyzdmmazbzdm72a xdmrlb ydm D D D D Was yzdm Hal bzdm72amxm72bmym D D Hxzyzdm az bzdm D D 1 th mm 13 mm 28 May 2009 l2 5 Applications ofDouble Integrals Rotational Energy If the system is conservative the total energy Pm in the system remains constant Therefore AE U7K0 Vycm where AE is the total change in energy U is the potential energy K is the kinetic energy and U mgh K v1710f Where g is the acceleration due to gravity and I is the second moment of the rotating body mm lt m w r l2 5 Applications ofDouble Integrals Rotational Energy 7 Example Analysis Consider a thin circular lamina of radius r 2 1d moment total mass m that rolls down an inclined plane as shown Assume that the system is conservative and show that 1 z Zgh I v where 1 11 m2 2 if yt is the vertical distance travelled at time t then dyi 2gh a lmsml l 2mm 15 Kmnmmmtt Kidoguchi Kenneth Miscellaneous stufffrom sections 114 and 116 The lmglicit Function Theorem The slope at any point glory the level curve fxy k is given by 3 7 fxExJ Note The Implicit Function Theorem has lathg to do with the slope along a surface 2 fxy The Implicit Function Theorem gives you the slope of a curve in the xy pzme Example 1 Let39s illustrate the Implicit Function Theorem using the curve 2xZ y 7 4x y2 7 4 at the point 714 213 ltjquot t 770 quotf6 J 7 a 9 3 0mg Ax j qx j39 a Q be 3 M T 43053 9 3wquot QJ 41539 3 u a l 3 l mJ 4 a m a3 IS Lawn quotIi 2x3 he 3 l Figure 1 Page 1 of 5 Miscellaneous stqu from sections 114 and 116 Fact Jack Assuming everyThing relevanT is differenTiable fx0y0 is perpendicular To The TangenT line To The level curve fxy fx0y0 aT The poinT x0y0 Please noTe ThaT This acTion is Taking place I39ll re xy pare DespiTe ThaT This facT does have repercussions vis a vis The surface 2 fxy The maximal slope along The surface aT The poinT x0y0zo always occurs perpendicularly To The level curve aT ThaT poinT relaTive To The xyplane Example 2 LeT39s illusTraTe Jack39s FacT for The funcTion z 3 7 2x 7 2y 7 4y2 aT The poinT 071 Figure 3 shows The level curves 271 21and 23 44 L fiiw l Co I 3 uD 439 Y I Sh J L Ionl can e gun39hn39a OJ l ml LA bzup J l 3 hxllijljIJI 5 Lu 095 IL I T 399 L EHC bLmaSng 9 4393 A 3Lgt3 3 3 J Fquot S1quot D ltvb Law Aquot is J Page 2 of5 Figure 3 Miscellaneous stuff 39om sections 114 and 116 Examp le LeT39s prove Jack39s FacT SW Md 4 Ht 5 7 Ithl Luru Lquot39SL Lu 3 ix lely g YIj llt 6 IF LL s v n4 39 l 29 5 VRU quot 437ml V C X0 w 393 lt x Mn31 xui 7 whick is 39Pyallbl 41 quotr 0quot0 ILtul curIL r LX731 35 4 Q In um Hoquot klxnya 39 Q6 Example4 The level curves in Quads I and II in Figure 4 are from boTTom To Top z11 z8 25 3 z 2 and z 71 The slope of The TangenT line To The level curve Through The poinT 23 is Z LeT39s esTimaTe vf2 3 it1mm I 39P LF bQ ALb In M r39m M39 3 341 quot 4051 TT Ur WHEN quot2 3 7 lt 347 5me 39739 arm lt 3 I 6 PageS m5 Miscellaneous stuff from sections 114 and 116 Definition The gradienT of The four dimensional funcTion wfxyz aT The poinT xmymza is The Three dimensional vecfor Vfx0y020 ltLltxgyygyzg7 fyx07y07207fzx07y0720 AT any poinT The gradienT exisTs The gradienT has boTh of The following properTies Vfx0y0za poinTs in The direcTion The funcTion wfxyz increases mosT rapidly Through The poinT xmya 20 fx0y0 20 The insTanTaneous raTe of change in w fxyz aT The poinT x0y020fx0y020 in The direcTion of Vfx0y0za is Vfltx0y0za Consequenle Vfltx0y020 is The maximum raTe of change along w fxy z aT The poinT xmymza Example 4 Missylaneous is one saIT seeking microbe A sample of The microbe is sTuck in a Tank whose saIT 5 concenTraTion ppm aT The poinT xyz is given by The funcTion fxyz xsiny x2 x z where an axis sysTem has been esTainshed so ThaT a Three independenT variables always have values beTween O and 3 Assuming ThaT The microbe always moves in The direcTion ThaT The saIT concenTraTion increases mosT rapidly eT39s deTermine The direcTion aT which The crobe moves when iT is aT The poinT 1 L SJ embarkth Lyn nnH39 f b39i l lj 6 I I 9 39 v i a an L IN 4 10 0 lt1 Minn ltSic13fgt32 xwyb 2wgt 9 3 UNI3 liHbquot vwad 5 JLJ quotJILL V th 3439 Jung hmu39l e13quot UkM 5 m n I jrrI39J 2A T fW C l Page 4 of5 Miscellaneous stuff from sections 114 and 116 Word Jill Assuming everyThing relevanT is differenTiable Vfx0y020 is perpendicular To The TangenT plane To The level surface fxy z fx0y0 20 aT The poinT xmymzo Examgle 6 LeT39s use Jill39s Word To help us find an equaTion for The TangenT plane To The surface z x2 y aT The poinT 7171 0 x i j 3 lled 1 bciiw mm s MLT j quotf 71 LCu uo Iquot gt S I H4141quot 19 1510 Joulw 0 wl iw PMDWT rim 1Ca i L love feralch Jo Ly ulekj P fig C ylj 39l 7 lt 2095 47 1 C 4H c mI f 71ijle M J h 05 quot1 MJ 39 39 vrl un lt s w Hi Sh Lb a uglIon HI5 2 0quot 39o 1 Wquot J ZzleW39ltxw3Jilogt zu 3 1 2 1 isu 139 I 72 1 u U D 71 72 391 X Home is DUPMA Wquot l 36 I r w J kquot 399 gr Iquot l Page 5 of5 I Extreme Va ues on 3d men5 0na Surfaces 2 Fwd M2 absu uve exvreme va ues uf M2 func un fxy3xy76xi3y27 aver M2 cmst mangumr regmn Wm ver ces av M2 paws 0030and 05 F idol ulh zu 39139 ukuv JM 9 51533 E33 L a 3x 1 0 11 j 341 1 rah 01 Gm 1 ij cI 1 0quot 3 LEFT 0 a wa1 quot Gll l thl P75 L y But S nlxj 33gt 1 m nan 1w Dr C j gx 501xi S J 3xf 39 3 15 u x J 35pm H7 534 qu nu ma ms j I j gx S I L x quotSr39x4s 04 147 39 7 r SO x 1quot X chf f 395 nMLta uh ILri BOP xj j 12 D 13 I M5914 I 239 39 I 1 1 Lu 0K 7 391 73 lt3 J 94 L nag LAHLIJ P w 5 1 0U VL JiKw V r 3 quotI LL 0839 39c ag cl he ah ah Unlda Iquot L5 r39mx u lot Extreme Va ues on 3d men5 0na Surfaces 3 Fwd M2 absumve ememe venues fur M2 func un fxyx2y22 1 Ever M2 c used regmn de ned by M2 2 2 empse LL1 1 4 n E E A 2 n P l ft s 13031 3 51quot a ongxv39 r 13 Ian on 7 393 MU T 1 IJ 3Lr9W 3 H A 0 Wu hump L Jr Alma 39lefo o e xo 5r L kJ JF 73 4 L m 4 Sgt 91quotquot Yzo 5 Bw J Ed si 61 rd 05quot Pages ms C ltgtltjgt WWquot X 093 44c 51 f m Outr r io Uw dt D4 g c o J Lu 2 0 I39D L L3 Ht 0x 00 no 03 Lo O LJ 4 l up b J Double Integrals using Cylindrical Coordinates PIoT The olar funcTion r115sin l9 P r Table lr1 l9 IN IN o Q S g Page 1 of 8 Double Integrals using Cylindrical Coordinates Fact J ldA finds the numerical area ofthe region R R Double Integrals in Cylindrical Coordinates r Z A double integral in cylindrical coordinates always has the form 52m roux Jam In frt9rdrdt9 where rm fm 9 is the boundary curve closest to the pole rm f0ul 9 is the boundary curve farthest from the pole the region of integration is swept out exactly once in the counterclockwise direction from 9 9 to 9 952 and 11 rdrd Start Note Ifthe region of integration includes the pole then r 0 m Examgle 1 Find the area of the region inside the circle r3sint9 and outside the convex Iimacon r2 sint9 3 339 V Jr e 3 5249 Page 2 of 8 Double Integrals using Cylindrical Coordinates Examgle 2 Examgle 3 Find The area beTween The loops of The Iimacon r 1 Zsint9 em L C r a f ro 95 C n o 3quot I hung quotav 3 llshm e 3 9rprrb ITL 1 9H0 i5 HM Firsl vulva 954 i klt 7quot f L 9 av H39U 0 11va r2 0 J OV luL uHv ITK 95quot 2 7L Club 1 tww r 0 We ham 9 asiqkob i A 94 S r449 1 463 L quot rr L gtr QEQQEngB IL 95 77 T 1 57 11 I rJrJa k 70 Double Integrals using Cylindrical Coordinates Examgle 4 Find The area of The region common To The inTer39ior39s of The dimpled Iimacons r 15 sint9 and r 15 cost9 95439s NJ ecu DcLur IALG39C LI JLLM lS LML z Shh Cul e 9 S39SIA0 IScam rrq lI1o g r D IiL Examgle 5 Find The volume of The solid bounded by The paraboloids Z 4 x2 y2 and Z 3x2 3y2 JJL klEJ39 R Ic jgn VF 5 L iLt ULH 393 W 5 l y Ly J q l r at l l Lo 3 Sur hsJ 139 34quot quotk39hd39 7 l lxquotY39L 3xz4 2y1 I 13 H1 2 2 labJt K j 4 V Tur a D u139v G39qu K Uf 0 5 1 5 I Li quotL 3 erJD nal 0 A 1 l I O 6 Quill in r31 392 SolTr gvl qr Lili afe34jES Ja Lf A gt 39 I l fx 0 3911 u 1 ou x wxvy quot loc l uc 0 1 Double Integrals using Cylindrical Coordinates ML quot Polar To Rectangular Rectangular To Polar o J x os9 ygin9 P r2 x2 y2 A 2 2 2 h y x y r tan94 Conver r each funcTion To polar form x7 y3 rcu gl 7 dude 3 v 3LCe 392 7L J9l 9 A gt l on 61 0 9 ln x2y29 yx 72 T WKquot A ll arm 1 3 039 Rc qdk1 j d wi la fLoslla 3 g r x Sula LuL6 dour Me pl 1 SDII K vq u lo x2y24x0 a xzxzyzy2 139quot 4 LfLoUo rLL 5L9r1 15nalta r r Jquotquotquot39 439l39f quot rL LLoJIe 3951n161 1 q 5 u 439 La Yquot quot r La 9 M11 L fl 392 02 03439 L e F u dkr 014 PLJ MJM r L4 thrx r 0L1 r20 50 r33 1 rgJJ L KM 9 TILJ o Page50f8 r quotl U56 39 E Dunih39l A L x i LfVL 3 3 quotIL kt Double Integrals using Cylindrical Coordinates EvaluaTe each double in regral offer firs r conver ring To a cylindrical double inTegral J2 J Izt yz 6amp2 Hzde EL P l 39IIA39 0 oLqo L 139 L 5 Q JL A 75 H ecx 7 chLJ mm H mm u H K LIT L L In t cquot 39 V J39 Jquot 0 t 3 ht 11T C l i J0 5117 49 u 1 39 1 a o f r CV L all v 3 L JI 117 v J I C L6 u Page60f8 I e I 7T3 Double Integrals using Cylindrical Coordinates 0 J17 LL cos x2 y2 dx 93 9H 3W1 7 Ul f S LoSKxW 3x4 1 10 I e V 7 R u fm SJCJLJrJe Page 7 of 8 quot all 5 4171 L7 1 Double Integrals using Cylindrical Coordinates l0 IF m 9 7 XLHF Us A 9x x x 1 4 z s x 1quot r 5 39 gt r m 3 1 1249 3 r if chch 86L M L J IJquot 1 Swle K TI 5 LS A 46 I39M340 D 3 u l 9 Page 8 of 8 10 3 Are Length 52 Curvature Arc Length 7 MTH7252 63 Revisited Suppose a curve c is described by the parametric equations r y7gt velogtgo i a g z b C is said to be smooth The curve C can be divided into n line segments Let 91 be the linear distance from PD to P Then i 2 Ain Continuing this process we Imate L t e arc length as P bgb a M P2 05 April 2009 10 3 Are Length 52 Curvature Arc Length 7 MTH7252 63 Revisited Ax Ay 2 J J J J ij ijefm anol ijerm w w 2 2 o 2 2 ea AL W M377 iii mt P P3 Kb 85 9v 10 3 Are Length 52 Curvature Arc Length 7 MTH7254 Extension ofMTHjSZ arc length definition to 3Space Lfa smooth curve with ararnetn39c equations r 7m y 7gt an 2 7 MI 1 g I b is traversed exactly once as t increases from a to b then its length is i 2 2 2 LI 11 d y 1 2 d a d2 d2 d2 MTHJS4 arc length de nition IfCis asrnooth curve given by 72 7 x0 ya 22 agtgb and C 39 4 n as39 t thenitlenthi 39 m254 arcLength mw Kidoguchi Kenneth 10 3 Are Length 52 Curvature Arc Length 7 Computation Example A particle travels along the space curve 72 R coscu2 Rsincu21 where R32 are RR gt 0 Find L the distance traveled by the particle over one cycle Solution T 2n L 7 2012 where T amp 72 uRJsinzmt cos2 cu2 In I w y r I uRdt E39uRT RE 03 27 05 April 2009 10 3 Are Length 52 Curvature The Arc Length Function IfCis apiecewise smooth curve given by 72 x2 y2 22 1 r th en thearc length function 5 is r r2 7a ldu 2 To nd St 1 Select an arbitrary point on c as a reference point 2 choose one direction as the positive 39reetr39on c 3 For any point P on c St represents quotv 39 ned arelength on c J By the Fundamental Theorem of Calculus d 4 r m r 10 3 Are Length 52 Curvature The Arc Length Function amp Reparametrisation Example A particle travels along the space curve 72 Rcosru2 R sinru21 are R R gt 0 Find 52 the arc length function for the particle motion and 75 the position vector ofthe particle as afunction of Sol ution Let s00 ands22o WhenOStSTTZnw r st I7tdt where 7t wR lsin2wr cos2wr an r 11R 7s2 ltR cosEb RsinEb 1gt 75 ltR cos R sin 1gt e Repararnetn39sed position vector Kidoguehi Kenneth 10 3 Are Length 52 Curvature Curvature c mmre of space curve c is the rate ofcharlge ofthe unit tangent vector Ta with respect to 5 distance traveled on C dT K d Alternate forms of curvature K d Tquot dTdt 72 01 ldsdtl lf t 7tgtlt739t l t 3 Special case curvature computation for plane curvey X f quotgt0 m I 1 f a 05 April 2009 10 3 Are Length 52 Curvature Unit Normal amp Binorrnal Vectors From 10 2 the unit tangent vector is rquot f 0 The unit normal vector Nt for a smooth curve c is 7390 T39 By de nition the unit tangent vector M has unit magnitude therefore Tquottl 2 77 0 Tquot and Nate orthogonal To The unit binormal vector Em for asmooth curvec at C 32 Ttgtlt Nt A Nt 10 3 Are Length 52 Curvature Normal amp Osculating Plane The 0sculating Plane ofthe curve c at a point P on c is determined by the 39 T andN 39 r 39 c lines that are perpendicular to the unit binormal vector 2quot TheNonnal Plane ofthe curve c at apoint P on c is determined by the r IQ and g ml I u c 11 mo r lines that are perpendicular to the tangent vector T Kidoguchi Kenneth 10 3 Are Length amp Curvature Curvature Radius of Curvature and the Osculating Circle For a smooth curve c the curvature tr is the rate of change ofthe unit tangent vector at apoint P on c The radms nfemvotme p 1m is the radius ofthe usculatzrtg mete at P The usculatzrtg mete aim the El 7512 cf Curvature is the circle in the osculating plane that touches the curve at P an whose tangent vector is parallel to the tangent vector ofC at P 05 April 2009 10 3 Are Length amp Curvature TNB 7 Samgle Comgutation Given a particle that travels on the curve c de ne by 72 ttytzt 2 221 w a Find irr the curvature of c and evaluate um Plot irr on the interval 7 1 gr Find pr the radius of cumture and evaluate p0 Overlay a plot of pr onto the plot omr E 0amp88 Find the unit tangent and unit normal vectors at t 0 rr u u I i i i forCatXZO Kidoguchi Kenneth Finding Local Extrema for the Function Z fx y We are assuming ThaT The funcTion is differenTiable in all direcTions aT all poinTs in The domain N Find The criTical poinTs The poinTs where Vfxy AT each criTical poinT ab evaluaTe D f a bfyya b ya b2 and sTaTe The appropriaTe conclusion 0 fab is a local minimum if boTh D gt 0 and fuabgt 0 fab is a local maximum if D gt 0 and fuablt0 z fxy has asaddle poinT aT ab if D lt 0 No conclusion can be drawn if D 0 Finding Absolute Extrema over a closed region for the function z fxy We are assuming ThaT The funcTion is differenTiable in all direcTions aT al poinTs in The domain 1 Find The global criTical poinTs ThaT occur in The inTerior of The region 2 Find The local criTical poinTs ThaT occur along The inTeriors of The boundary curves Find The criTical poinTs ThaT arise from being inTersecTion poinTs of boundary curves Make a Table ThaT lisTs all of The criTical poinTs and The value of z aT each of The criTical poinTs The greaTesT value of z in The Table is The absoluTe maximum value of The funcTion over The region and The leasT value of z in The Table is The absoluTe minimum value of The funcTion over The region A F Page 1 of 5 Extreme Values on 3dimensional surfaces 1 CaTegorize The behavior of each funcTion a r each of iTs criTical poin rs a fxyy3 73x2y73y273x2 1 39Ll v kcbl BI41 PL J Lara A 3 Q s 063 o i ijLx0 Efl 77y 33quot 339xquot Cj 0 62quot Finn ij39lL GZD 3 743 0 C 3630 g 5 7c 0 3 33quot Gj 7 77 bLS LO 5 3039eamp31 ll 3 1 Lbl fv q QM 013 545 614 I 1 1 Swag 1785 x 9 quotW 9 63 6 C QQl r DC W 2c M4 cmcgnhaao D 3 39DL rij l 71ar3quot97 C 3923 aquot vdaerqgaj A 16 4J1 039quot J J y 9 l a 04 upk3 z 3 3v 0 v613 x 39 05 93 5 quot quotPJ Extreme Values on 3dimensional surfaces b fxyxzyZ 7 Zny7 2xyZ 4xy La 7441 b Paful org 1 ch 73 4593 73 3927 E 1x3 3 gt 51 0 gym1 ay wlx 3 C twl 96 5 11 103 m 0 J39oJ 0 1 1 Algebraic solution and quotDquot analysis begins on next page Ils 13 4 3 12 I l JScJJ 1 Suing WRAC39fNi Hamel DKK Page 3 of 5 139 L 7 Extreme Values on 3dimensional surfaces N Cafegor39ize The behavior of The function fxyxzyZ 72xzy7 2xyZ 4xy 39 u 11139 each of ifs cr ifical points 2 391 y y 25 3 1 Figure1 fxyxzyz72x2y72xyz4xy I Lg Cf Pr Hg I Iu KIM h V CX 7 o lyy39l qyj j qu o 31jyj1x j 1 0 ij 1y Wj WM 0 gt 2 3 X 3977 9 2 0 Tlmug wc q SjS I NI 5 czu on 1 i g 7quot xjaxjna j39x 1j quot x IDIIL Ill Lion Shula J E M L39H IJ A gxy 3918 er a 57 JJU39lw39k I 1 Fr 6 Itquot VJ gt lj1 L 139 J P 7 Jy 1 yzyi EUL J 13 N SFh L E LAMP394 35 9g 4 wvj 439 0 jay dyad E a 35 AF 61quot ubjo l39 V do on Matt 7 X1J 1 o Pageeof 5quot 1 11 010 Extreme Values on 3dimensional surfaces E Q A r ltJLULB EXTRVH 51 61quot X J39L j qo J quotG 5 W 5 x U39L ULLIT fIw39J 91 3 xsl z j Lj4 Lo 1er 97L 3 1 v VL gtlt l IV L 0 l s I raw1 y 2quot Jr OAT o TIAO 14quot CP 134 3 W39L Iyxl7 Ca puJWw alo 0 0210 Sf w o quotC m0 5f 03quot O a quotN0 ff on k q W Wan 17 0 ll 1110 f funny ijj 459 L2 qx l v I 41quotq Law2 3 4 Page 7 of 10 Extreme Va ues on 3d men5 0na Surfaces Fwd M2 absumve ememe venues uf M2 func un fxy3xy7 6x7 3y 27 Ever M2 c used mangumr regmn WM 2 pmm ver ces av m s 0030and 05 PageA ms Extreme Va ues on 3d men5 0na surfaces Fwd me absumve exvreme venues fur M2 mum xyx2y22 aver M2 c used regmn de ned by M2 2 2 y 1 4 x em 527 P 1 Pages ms Um TangenL Nuvmakand a y w c uvs Oscma mg P ane Ame evauun cumpunems ngure 3 shuws a graph mm vecmr mmquot 712cosl42ost325mt m quots uscmmmg pmne F Euve 3 WW 5 M2 eqummn u he uscmmmg p an27 zaq de 5 nail xL 3 4er gm 3 gt wLi L ME 7341 1 U 5417 fquot R 393 5 lt 1J loy Paw 1 44 4 4c t 11 A Lquot 39lr39tquot 11 739 quot cargal quot lt 3 J51quot 05mmth 39 L s a 0 PageBDMD Fm Rbu Hm m5 EL E PM 7 W 0439 OJLulab flyM 0 4Lquot OS J54 5 39l ovg 62 01quot mw 1 2 7 5 Mquot 3 L lt75 392 Unit Tangent Normal and Binormal Vectors Osculating Plane Acceleration components Let39s use our calculator to verify I 9 7f As we illustrated on Figure 3 the osculating plane to r t at t3 Is perpendicular to A 7139 Let39s use classic techniques to confirm the equation of the osculating plane TV I lt 7 WJoIltLJJIJO LOIP a 14 5 e of Iquot 1ug hli bF Of39 lt 1 quot39J 0 39V39b 3 4J433957 0 QLq Jl qogt giggt6 UHLA Page 10of10 Osculating circles and three planes associated with motion Figure 4 shows a graph of The vecTor funcTion 7t 15 cos t3 2 cos I 4 2 sin in iTs oscuIaTing plane y 2x LeT39s find The cenTer39 of The oscuIaTing cir39cle To The curve when If t LeT39s also find a vecTor39 funcTion ThaT describes This oscuIaTing cir39cle One formula for curvature is Unify a5wlhl391 rj PIA c 7 l5HuLH 2 udfl X l wield39 3 2 Wm gt 1L V 39L 7133 W310 quot39 Hquot a 5 gt 7 J a F l 75 dw 450m 1 7 mnMWLil 2 E 5 1 397TrLHJ ijltv1 7 i lLL L A PJ u 03quot L 9 A 1139 lP13uLlt3955 1 7 3 s ulHn vJiu 5 Lu u U 0J cl quot SJ 3 5 39tr1p J 139 o 0 bquotquot f 390 J Lo HealthJ JWIquot IA Page 4 of 6 F Mx P M 3 4 SL lh l39b LC x LilL 3 3 L 3 1 11 UK 625 LX L51 i lx 3139 61 as 455quot 615 ML 3x4 JJS Mixquot oI39LXq 44 r 1quot SUE 3xlrlt3 2 mg 5 7cS1 4 C 5 6J5 a J 91 61 A 9 UV 4 39 4 Dr 2015 usL E 001A 3 44 x M 75cojLLwg 3 gpc 7 5 swel r a J 1 3E WH o Licekw fdao39kn arm Male 4 Limit 15 PM lt Lg J 3 Lona 3543554 V J E39 J 39 Osculating circles and three planes associated with motion Figure 9 shows the vector function 714si11tcos21 7 35in21 cos MIj 7 The normal line slope 0110 Let39s intuit the equations of the osjlj ating normal and rectifying planes 39 r i a la 4 l D JXkltu v Lf L In ELLwad PUBC SI nbc D39Mk fluac is J 439 niwld j flan1 13 7 lko nannyl fl la laquot Sane 4uu l AS AC 39Hr r pl line 33 l l 39 Jlx Let39s verify the equation of the binormal plane using traditional techniques Page 5 of6 ln39rndudinn 39n Multivarinble Fundinns Domains of Multivariable Functions Conswdermefuncuon fgtcy J9 e x2 e y2 Whatarethevames of f721 and f6727 JJI qq A M4 m 4 3 Fwd me domam of fgtcy and sketch the domam omo ngureW ILL imm u Lv so A ms 0295va 4w XLLZ gt 12 X1yu Lt LNAJU A 44 iAuquotq 3 Xi39j1 7 Fwd me domam of t mncuon gxy sinquotx e y 2 and sketch the domam omo Home The Amt1 Cos 1xa 39 w elgttJ 17 Swot 4quot l ex r 1 4 AL Liv inU 439 44 Juu JrL x j H 3 x j 1 13 X sj Page 1 of 4 ln39rndudinn in Multivarinble Fundinns I K A Fwd the domain of hrylnx274 cosquotryrtanquotx2y274 and sketch the domain onto Figure 3 L A is H M39 PM Lyn sua i L5 4 gt XL 9 lt2 lt H c 396 J6 39 Jlm law404quot 0439 Hquot tiny prc xgl Xqu 3 5 Kim 4 quot 55 1 111 4quot x I Whatdoesme domain ofme mncuon f xyz ook We 922 I J A x H e gl Ji Pb 35 fr 4 U ny g 5 MA 39Hd q x yL 49 gt0 gt q gt x l quot quot 391quot A quotquotquot 5 4quot 14 i Pal439 7549 4quot d39IpIui Practice on nding domains of multivariable functions y yy k 7 Find and skefch The domain of each of The following funcfions a fxy sinquot 31 lumpy b fxy lx2 71 7 cosquotr2 y2 71 x c fxyquotrxy Jr2 7y Page 2 of 4 lmrodudion Io Multivarinhle Funnions Level curves of Multivariable Functions Three dimensional funcTions z fxy always have an aTTendanT seT of level curves of form zk where k is a real number The level curve 2 k is The seT of all poinTs in The xyplane where The oquuT from The funcTion is fx y k This illusTraTed in Figure 1 idea is FIGKELLEVELClRVESFOR z x2 yz 10 For each of The funcTions in Table 1 four level curves are shown in one of The aTTached figures 1 6 and a ploT of The acTual surface is shown in idenTify The figure numbe one of The aTTached figures 712 Fo r each funcTion r ThaT corresponds To iTs level curve poT The figure number ThaT corresponds To iTs surface poT and The plane equaTion for each of The four level curves shown in iTs level curve poT Table 1 Problem 1 Worksheet Function AQUA JV ts t quot12 E k k 1153 ks Page 3 of4 D Introduction to Multivariuble Functions Level surfaces of Multivariable Functions The Temperature degrees Calvin aT a poinT inside a vaT of grease is given by The formula 002 Txyz The axis sysTem for The vaT has been seT up so ThaT one corner of The vaT x y corresponds To The poinT 222 and all poinTs in The vaT have coordinaTes ThaT are each greaTer Than 2 A level surface wk for This funcTion is The seT of all poinTs in The vaT ThaT saTisfy Txyz k ConTeXTual These level surfaces are called isoThermal surface WhaT is The shape of The isoThermal surfaces in This vaT Page 4 of 4 Finding Domains Level Curves Surface Plot worksheet kmgj gmquot Q ch I L 11 if Z EMQ39LL S J l x n r L I W t X 5 sum LWA39uI lws LawLi hm 21 C T g 139 7 gtjcsw Lino 4 anl wig p quot 3 0 F53 L zlny7x z cos391xy Page 1 of 3 Finding Domains Level Curves Surface Plot worksheet Page 2 of 3 Figure 7 Figure 8 Figure 9 Figure 10 Figure 11 Figure 12 Hams Lem Curves z mum Lem Curves z3210 name Lem Curves z3210 4 s Run4 1 A 3 wksJ L s ng ywl Finding Domains Level Curves Surface Plot worksheet Answers to finding domains and Level Curves of multivariable functions x3 x3 Figure 1a Solution to Problem 1a y yx2 Figure 1c Solution to Problem 1c Table 1 Problem 2 Worksheet F4 x1 Figure 1b Solution to Problem 1b Function Level Curves Surface Plot Red Curve Blue Curve Green Curve Violet Curve 1 2 sm Figure5 Figure9 2 1 27 271 22 x 2 2 2 2 z 5521 Figure1 Figure8 20 22 21 23 2 X Figure6 Figure7 22 23 21 20 x 21nyx Figure3 Figure12 21 271 20 22 2cosquotxy Figure4 Figure11 23 20 21 22 x2 21nE 2 Figure2 Figure10 22 21 271 20 y Page 1 of 1 28 May 2009 12 7 Tnplelntegrals Areas and Volumes 7 Geornetne Interpretataon Area efreetangnlar regen ln the Xyrplane Anselm Area ufbuunded reglun ln the Xyrplzne A militide Vulume unda a surface 2 zy mlaquot areetangnlar regen m the xyrplane ELnrlrndzdxdy Vulume unda a surfacez 1m and mlaquot a buunded regen ln the xy plan mm e VZEEnm rnrlmdzdydx 12 7 Tnplelntegrals Tnple Integral Over Box 3 and Fublnl s Theorem r If zyz ls a eentanueus luneaen that leeates a pelnt m abux 5 tlnen IIIWM 4 aw Fubl m s Theeren Iffls eentanueus un tlne reetangular bux 5 115 gtlt X r x than Illz yzdV Iiimm xdymmz xmydxdp 12 7 Tnple Imegals Tnple Integral 7 Type 1 Solld Reglon Type IPlane Reglon AsellaregenEls calledtype l lf E KW my 6 D Wkly 52511181 z Z MW Snll Regn E MOW r ryn n we n gnr xb km W t IImodwl I ltxyzgtdzay 12 e M n Kidoguchi Kenneth 12 7 Tnp1e1mega1 Tnp1e Integral 7 Type 1 501101 Regmn Type 11 Plans Regmn AsuhdregunEu calledtype 1 1f E Rh 1 my 6 D 1013052 lt 114w Z W San Raga E i y Mvw x W llx NW 2a yzgtdzdxdy 7m 28 May 2009 12 7 Tnplelmegals Tnp1e Integral eType 2 SohdRegmn AsuhdregunEu calledtype 2 1f E KW W12 6 Dv 1vzixi 2042 12 7 Tnp1e1mega1 Tnp1e Integral eType 3 SohdRegmn AsuhdregunEu calledtype 31f E WI 1 M E D 1812 S V9181 z s and EHwowJE wJ m Kidoguchi Kenneth 12 7 39Il39iple Integrals Triple Integral 7 Sample Computation Find the Volume ofthe solid bounded by z x 2y and z 0 With region D bounded byy 2x2 and y l x2 ulution Evaluating this integral With D as a Type I region We have x17 ygzxgt zx 1y Vj j jdzdydx gzxlxz xa ygx zEI 2 Find the area of the bounded region recall MTH 252 xl7 yl x2 V JX2ydydx xa y2x2 Vy2yl x2gb gw xz ED y2x2 39 2x mvmnv mm quotmm 28 May 2009 12 7 39Il39iple Integrals Triple Integral 7 Sample Computation V lww yz dx Elmx21xz27x2xzWW Find the Volume under the bounded region V x39 ifax lix3 2x2xllix xrl s 4 3 2 1 7317x 2x x x 5 4 3 2 1 2 15 2x mvmnv 12 7 39Il39iple Integrals Volume Computation Example Find the Volume of the solid bounded by the elliptic cylinder 42 22 4 and the planesy0 andyz 2 Bounding surfaces Bounding curves 2 I y0 272417 I yz2 I z2llix2 xl z 2 17x2 yz 2 J I I Fquot 272 17x2 y 22 pk yzz 2 V j j j 1th dz 24 pipJ y V47r 2x mvmnv mm quotmm Kidoguchi Kenneth 28 May 2009 12 7 Triple Integrals Another Volume Computation Example Find the volume of the solid bounded by the elliptic cylinder 4y2 z2 4 and the V quot planesx0andxz2 39 V T iffy Zacle 9y x0 1 272 17y 47I zaMavzuua m we mm K math 12 7 Triple Integrals Yet Another Volume Computation Example Find the volume of the solid bounded by the elliptic cylinder 4y2 x2 4 and the f planesz0andzx2 39 y1 x z 17y V j j 132 W 2 ZEMay2EIEI 11 mm c x2 21 12 7 Triple Integrals Volume Mass amp First Moments Let pxyz mass density function Volume V V E Total mass Within V m ygt ZdV E Moment about the yzplane Myz IIprx y zdV E Moment about the xzplane Mu y zdV E Moment about the xyplane Mxy IIIZPQJ zdV E Centre of mass coordinates xcm ycm zcm Mm ZEMay2EIEI 2 Km mm K math 4 Kidoguchi Kenneth 12 7 Triple Integrals Centre of Mass and Second Moments Moment ofInenia about hexaxis Ix y zZJpryzdV c Moment ofInertia about theyaxis I 21 pxyzdV r Moment ofInenia about the zaxis I yz pxyzdV 5 Polar Moment oflnertia In ripxyzW r where r is the J distance from the axis ofrotation 28 M ay 2009 12 7 Triple Integrals Calculus Moments Givenasolidboundedbyzrl z 1x 1 1and 7x with mass density pryz lgram per cubic centimetre gmch and all distances are in centimetres present the analysis to nd a the volume ofthe solid body b the total mass ofthe solid body c rm yam 2m the coordinates for the centre ofmass d the rotational kinetic energy ofthe body ifit rotates with angular frequency w 1 radsec about an axis perpendicular to the ryplane and through the point 000 and the rotational kinetic energy ofthe body ifit rotates with angular frequency w 1 radsec about an axis perpendicular to the ryplane and through the centre ofmass Kidoguchi Kenneth Unit Tangent Normal and Binormal Vectors Osculating Plane Acceleration components If we think of the function 71 as describing a particle moving along a curve then the unit tangent vector to the function 71 when 11U is the unit vector that points in the precise direction the particle is moving at I In this vector is denoted as TOD This idea is most easily understood when the motion is confined to a plane 8si112 t A 65in t A A graph of the function r7 I i j is shown in Figure 1 The J3 J5 7F tangent line to the curve at the point where I determine is also shown Let39s use the figure to The first thing we must determine is any Vector Ihaf pairI in he drecIo of moion a It 3 T39 t g ltv 5J3 Figure 1 Tangent line and unit normal vector to 71 at the point where 1J5 9 t Now we need to normalize our directon vector 9 rru Xxquot 4 5 39 l sm s Page 1 of 10 Unit Tangent Normal and Binormal Vectors Osculating Plane Acceleration components 85in 2J t A 65in Er A A Let39s now use re formula 7t j to determine 3 9 Let39s be brave and do so by hand The first thing wemust determine is any yecfor fraf pail7 s I39ll fire aVrecfM of 117 m5 1 J I J 111k 5 9 T LJ39I 39 VCR lt C s 1 3 w GQ f L065C2 f160 1gt n5 39 tugr713 L 065 NJ 0095 ltCv 2JJ 4351 1 lt 3 37 VF k5 Ar aweDual TIMT F 39r Cf v mac44 71 Now we need to normalize our aVrecfM yecfor I r39 f quot91 11 r IrILTI 73 a V Page 2 of 10 Unit Tangent Normal and Binormal Vectors Osculating Plane Acceleration components The osculating plane to 71 at lt is the plane through F00 that is parallel to both 7 and t0 assuming neither derivative vector is If all of the motion defined by 170 is confined to a plane the osculating plane is simply the plane of motion At any given point on t one of the 2 unit vectors that is parallel to the osculating plane and perpendicular to the unit tangent vector is called the unit normal vector More information 39 regarding which of the 2 is the unit normal vector 1700 is forthcoming M Again referring to Figure 1 what are the only Iwo A 3 passIb Ie for N 5 2 Vebi L 4 ltinl 339 5 W0 l5 cin 4447 393 M l39h 539 J gt Mw Figure 1 d Pr 7 Km Tangentline and unit normai vector to 71 at the point where I f Assuming none of the vectors are one of the following relationships must be true Which one How do we know Options r f tu Lftu vs w 7 00 L 710 Si 390 I T Uctl l3 constwt LIL rvuSi39 199 i ho OU U l A A Tag T39H l16 C 055043 gmeg VJ JILL9 D Page 3 of 10 Um anuem mm and Bmuvma vmms escmaung P anE Acce evalmn rumpunems N7 m mguseamawcmmmw N 9 J wasquot Npquot WI Stave We Vuvmu a 1m 7 r s y r E3 5 mu 4 u s was 5 u mquot 2 m 1mg unit Fwd memma my i r r and 51mg 5 as 2 mm m m same ammn as y rquot vemv the Vuvmumuv f r WWW umwym zm 2N n ixvhe unit 9 7774 vuvr m w some mmquot as i39 n me An m liege 15 Mumemmmv Tami mif m befa f w 39139 JDT39 ALDW g F 394 Unit Tangent Normal and Binormal Vectors w w M Lquot 4 Osculating Plane Acceleration components By definition the uni binomial veclor BUD to 71 at t In is fln gtlt NU Let39s draw onto m 39 and N J and use the righthand Figure 1 vectors indicating the directions of 9 9 x 6 747 I A IRE c m AI39 rvakhquot rule to help discern B 9 J H m quotT 39 A A 39 4 In In II39IV n ix N rm i I u l 0 a o iJ Fi gure 1 Tangent line and unit normal Vector to 7r 3 r I atthe point where IT 32Jtlt001 3 L T u I I quot1 7quot 5 Let39s calculate 7 Ir and indicate its e s ca cu ate 9 general direction on Figure 1 mamm K E l crossP8 3I 39 2 d dt2 l lt 9 J What does the position of Fquot 9 relative to T and N tell us about the K rquot ul39n motion along 71 at t 9 m 00 H lw35 C9 5 ctAU lnbrpps W5 j A 9 1 3 Page 5 of 10 Unit Tangent Normal and Binormal Vectors Osculating Plane Acceleration components 1 Assuming none of the vectors are 6 if fltt0 NM and the acceleration vector r to are drawn tail to tail then Auto and 7t0 always lie on the same side of the line drawn through ft0 Let39s illustrate below the implication the relative location of 7quot1 0 has on the motion described by 71 at t to Please note that I have drawn fltt0 pointing right and Auto pointing up so we can simplify the language For a given function at a given point the implications described need to be adapted to the relative directions of 71 and ZWtO Keep in mind that 4 assuming no zero vectors HID 204 and r to are always parallel to a common plane fire oscua ng plane fa 71 07 t to If 7 t0 points in the direction of If 7 t0 points in the direction of this quadrant what is happening to the speed this quadrant what is happening to the speed tail5 m Jr11 what sort of turn Is occurring what sort of turn is occurring up 5 A Vf u n 4 AM If 7quot1 0 points in the direction of 772t0 what is happening to the speed fir i If 739t0 points in the direction of NM what is happening to the speed V If 7quot1 0 points in the direction of fltt0 what is happening to the speed LU O J U Daubquot poopn quotr quot quot339 pr quot what sort of turn Is hat sort 0 turn Is what sort of turn Is occurring occurring occurring NMQJ Uyqu but Please note that if 7quot1 0 points below the line I39ve drawn through 100 then the universe has imploded and this entire conversation is not really taking place Page 6 of 10 Unit Tangent Normal and Binormal Vectors Osculating Plane Acceleration components Numerically we quantify the position and effect of the acceleration vector by the tangential and normal components of acceleration 171 and aNt If 7 1 0 points in the direction of If 7 t0 points in the direction of this quadrant this quadrant 0 H gt O 0 a D 0 TM If 7quot1 0 points in the If 739t0 points in the direction of 77111 If 7quot1 0 points in the direction of Nit direction of fltt0 5x1 lto1 I aTzo 0H 7quot a 39rl A m lll T a r 39 0J 0 MA 395 b We leave it to the reader to verify that a formula for the tangential component of acceleration 7tol7 to a7 to is law and that a formula for the normal component of acceleration is r 0 V Z all t awe M l739to Let39s show that IN I 0 V tfor the linear function 71 x0 at y0 171 391 Sg Lsny IEs ru b 1quot gt07 ol 39JltxF39Jc V AML lo 5H vu 4 owl a Le zgt chgtA 9L 31ew Page 7 of 10 A 5 k Seal T growl 390 W a af Unit Tangent Normal and Binormal Vectors sculating Plane Acceleration components Figure 2 shows the function Ft4sin Cosf2 along with its unit normal vector unit 44 i5 gnu rt 2 Let39s use our calculator 39 acceleration vector o m re 394 sirluzj 39cost2quot90 Done HJ r lehnul ia 1 Plane x 4 I Figure 2 d2 m2 ut3t 2 Let39s use our calculator to find the components of acceleration and illustrate thh componentson Figure 7 IL 39 i quot b 0T 395 5 see next page for v ll calculator screens A lLf Rll 7 If MW Let39s find the instantaneous rat e of change in the speed at f What do we observe a J raw ol UMJI 964 Ill 1 5 35 93 2 Mil 41w 52 u 5 2 cw l J M L H J 9 tJLU W s9 U39 l n sfsd l K Hquot Lul kt 690 3 H l a a r Page 8 of10 ummm Navma m BmavmaHedms swam mm Ancemahan campmm 45m rcu rgm2wamqm wamnvp 5m m Vavmu a av 0 Fmamemmmam andsmve s x m m Vavmu a av 2 r a n 2 m t z n mm 5 z H 2 a 57quot 43 7quot ngt2 z an n mum Um Tangent Nuvmak and Bmuvma Vectuvs Qscu atmg P ane Annemath cumpunems ngure 3 shuws a graph u he vecmr func un 7z 2cost42cost325mt m w uscmmmg pmne F Buve 3 WW 5 M2 equahun umg usau mmg p an27 PageSDMD Unit Tangent Normal and Binormal Vectors Osculating Plane Acceleration components Let39s use our calculator to verify I 9 7 As we illustrated on Figure 3 the osculating plane to r t at t3 is perpendicular to A 7139 Let39s use classic techniques to confirm the equation of the osculating plane Page 10of10 x 003 3 55 U 39L 1 X J1 J VecTor39 Valued FuncTion Curves Consider sin2tcos2tcos2t WhaT is The projecTion of The curve onTo The xyplane SM x sziLLe VJ uJU V39L39J 39 w LL 4 CaJL JAN 439 V 1 Lo P1160049 J y 4 quot39 Sh ll c5104 2 44 uxJ drvquot WhaT is The projecTion of The curve onTo The xzplane xLyj We x 53LLH 6w 004040 lLt mum s x 151 0 39 5 But WhaT is The projecTion of The curve onTo The yzplane 11quot 39 NJ 3 4 i5 9 jswcw g1 371077 9 cry l 53 quot I On whaT surfaces does The curve lie kw p Mquot 4quot39 4 ltsmgt u RCC x UM Page 1 of 10 I ur x X5f W VecTor Valued FuncTion Curves f t lt cost sin I t WhaT is The projecTion of The curve onTo The xyplane go Oh L 1 WhaT is The projecTion of The curve onTo The xzplane X i D 5 k k E gtgtlta31 gt WhaT is The projecTion of The curve onTo The yz plane j 3 S I n X j 3 S quot 23 3c On whaT surfaces does The curve lie FLC xth R L S39stoiiwi X JG9 jollk 3 sum UM C39 liaJvcrs Page 2 of 10 VecTor Valued FuncTion Curves 50 lttsin tttcos t j WhaT is The projecTion of The curve onTo The Joyplane u D 25 OI o X gt x 1 45 E 397 haT is The projecTion of The curve onTo The xzplane L 1 x mm Cw Q 3 a 1 m 4 new v 1 Ls k UJILir Q 395 39U J U3 1 1 X 4 i t WhaT is The projecTion of The curve onTo The yz plane 139 g j 391 j 3 tfitms 1 3V qr jw j On whaT surfaces does The curve lie Rbhi Mfiii iquot shuanl 5ich X 2 j J39th faremil 17 3 3 oJLq C bid 3 CDnO IHJ39 ZjL Page 3 of 10 VecTor Valued FuncTion Curves 761 ltt2tsin tgt WhaT is The projecTion of The curve onTo The xyplane 3 s l 1c 9 P WhaT is The projecTion of The curve onTo The xzplane X 3 Jc 9 5 i 3 2 524 gt WhaT is The projecTion of The curve onTo The yz plane e 3 24quot Z Q myL On whaT surfaces does The curve lie Pih f l 352 Ljoliow3 age 9c 55M 3 Q33 scishLVJU39W Page 4 of 10 3 L V 3 Lji U 1 Vesiur Vaiued Funstmn Curves X 77ttr4t22t22gt 391 what is the projection of the curve onto thexyplune x i 3 M i 4 1 3 mt 7 3 Lll39 1 what is the projection of the curve onto themplane 1 D1H o 97 LX144 939 what is the projection of the curve onto theyzplune a 4 3 J 5 A On what surfaces does the curve lie Lu it ru b Wet 7 J I 4 a PM K lt 1 7 3quot 9 L 3 L Llsci a quot 0quotquot 1 3 l39I J 7 Page5uno x1 W slb x H ass399 WWW 11 as e VecTor Valued FuncTion Curves 72t551nt713 cost12 51110 WhaT is The projecTion of The curve onTo The xyplane 1 s 39 e Sn L V 4w HA ngsm l 441 S o V e H 1 31JI3LJLk 55104 x f 1 13 g 439 Lquot L WhaT is The projecTion ofThe curve onTo Thexz plane 1 1 Y IL1 xrscau E JuL k I j D 2 2391112Lk llr x 5 70 WhaT is The projecTion of The curve onTo The yz plane l3 3 same 7 mm L 1 1k L4 392 3 1 IlsUlt 5 su c 71 I m On whaT surfaces does The curve lie L 11 s I IJr If 3 J39w39 quotJ quot 39 J of m 39L 11 Plum 35 quot5 ll an VL s n2 q bh u Fab J 1 1 quot JSszu Mi H IMLV H L k I 6 TM 5 quotc L a m Lsmu w WNW IL9 MO Page 6 of 10 7 ILu em Ii J o 4 Surgwi 25 2 j VecTor Valued Fundion Curves 781sintcostsin21 WhaT is The projecTion of The curve onTo The xyplane x ss Uc 59 Hams Mt Xb gt L 1 6K j quot L0 id 39L uu WhaT is The projecTion of The curve onTo The xzplane Yquot S39 L k srn LL ki 3 95 H MM 1 5 gnuv tgsnLEm Iquot kJ tQ yVS L WhaT is The projecTion of The curve onTo The yzplane S 39 n i Mquot 4 a m 3 139 39 f 3 On whaT surfaces does The curve lie Page 7 of 10 VecTor Valued FuncTion Curves Find a vecTor valued funcTion ThaT models moTion abouT The curve of inTersecTion beTween The solids described in The capTions for Figure 1 4 Use TrigonomeTric funcTions where appropriaTe Figure 1 x2y2 2210 andxy4 Page 8 of 10 Vchor39 Valued Funcfion Curves Figure 2 22 y2 andz4 Figure 3 xy4clndx2y2zzl6 Page 9 of 10 Vector Valued Function Curves Figure 4 4x2 y2 4z2 16 undxy2 Page 10 of 10

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