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# Vector Calculus I MTH 254

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This 22 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 26 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

MTH 254 Simonds Chain RuleDirectional DerivativesGradients Introductory Example 1 Suppose ThaT a pill bug is rolling along a line doWn The plane 2107 x7 y Suppose specifically ThaT The bug rolls aT consTanT speed and moves from The poinT 0010 To The poinT 280 in exachy 5 seconds Suppose ThaT The linear uniTs in The model are fT Q1 WhaT including uniT are The consTanT values of and LP dt dt dt 4 A 2 4 1 h 39 I 2 39 i J n H Y 24 ill 3 I H I L Hls in Ar F U 39 it E L 3 0 quotl 5 if s s 0 0 Q2WhaT are The values including uniT of 0 2 and a2 aT every poinT on The plane 210 7 x 7 y x k M 9 R an 7 4 5 I The chain rule for one paramefer sTaTes ThaT if z fxy x gX and y gy Then dziazdx dt 0xdt 6ydt Q3Verify The chain rule for The model in Example 1 D 97 D A 5 H g H 5333 3 5 Eml uskl S J W5 Page 1 of 8 MTH 254 Simonds Chain RuleDirectional DerivativesGradients Introductom Example 2 The facT ThaT The bug was on The plane Z 10 x y was To a large exTenT a conTr ivance The paTh followed by The bug is on an endless number of planes any plane The passes Through The sTar Ting and ending poinTs of The roll would fiT The bill For example The bug39s paTh also lies along The plane 2 10 gy Ex Q4 Ver ify The chain rule for39 The model in Example 2 la J C fk if E Ll d Jl i 51 lt14 J7 E E l 3 4 l39 93 1 5 0H 55 LP 5 i Introductom Example 3 Suppose ThaT an anT is meander ing along The par aboloid z x2 y2 and iTs paTh reafl39ve 7 0 fhe xypane is given by The funcTion 17 t 2tcos t tsin t 0 lt ltgtLVltgtgt m X f quot44 g ml dz 7 Q5 Find The formula for E usIng The chain rule y U JJ Dj nu Q 96 scum airs3a 4 1 mm Realm Jk2 i b if cl 2 M w Le a W m a k1i l4 2l7sfllrlszaLkVie 14 h 511quot Page 2 of 8 MTH 254 Simonds Chain RuleDirectional DerivativesGradients Q6 Find The formula 2 and Then differenTiaTe ThaT formula Compare The resulT wiTh whaT you found in Q5 Recall ThaT zx2 y2 and 7tlt2tcosttsint 2 5 XIJAj L 1 a l wLM C 4694 a Q 3wwlhium nmti 1L6 LQgtLSCU l chlU ll 1 9 Di all 011 l i 37 Jr a m Practical Example 1 The volume m3 of a righT circular cylinder is given by The formula V rzh where ris The cylinder39s radius and h is The cylinder39s heighTm Suppose ThaT The cylinder is capable of mainTaining is righT circular shape while iTs heighT is squished aT a consTanT raTe of k Q7 Use The chain rule To deTermine a formula for The raTe of change in r Does This raTe increase decrease or remain consTanT as Time goes by lull ails Eva 3quot Dc dj39 Jr 0 Lg rl pnvLCK Jr Trquot L 9L llt U Jth Jlu39tjrph 39iwIopsbs Y all 9 I lakl Page 3 of 8 MTH 254 Simonds Chain RuleDirectional DerivativesGradients Introductom Examgle 4 Suppose ThaT an objecT is experiencing circular moTion relaTive To The xyplane governed by The funcTion Ht cosb z Esi11t2 Suppose Though ThaT The moTion is in facT Taking place along The surface z xyz 7 x2 2 In This example we are going To deTermine The change in elevaTion wiTh respecT To quotTimequot aT Two differenT Times The objecT moves Through The poinT 711 The Chain Rule is used To calculaTe This raTe of change NoTice ThaT The Chain Ru39e Bz Bz dx dy 8x 7 8y dt 7 dt The Chain rule could be sTaTed as g f dz3z dz 8x dz ay dz LeT39s find formulas for a z E and F39f IX 8x By E 3 oi LeT39s use The formulas To compleTe Table 1 f 394 3 Table 1 f xv y W 7t i Q dt39 dz 4 0 km a Q1 0 llll r alum Q8 How can The surface have 2 differenT slopes in The same direcTion aT The same poinT 39 on WWW 412 44 51quot mer 6 H lt35 i dz Nl l 5 N l l 17 ll I l W of Mr 5mm H H ue39 F U 1quot hp 039 I LL M IHMVW HM ClL Geld Gr Page 4 of8 Hal 5quot3J I39M 91 ulna 41 45 MTH 254 Simonds Chain RuleDirectional DerivativesGra ients lul l Q9 Suppose ThaT The paTh along The surface relaTive To The xy plane was differenT WhaT Two properTies could we give This paTh ThaT would ensure ThaT The slope of The surface in ThaT dz direcTion and The value of 7 would be numerically equivalenT t O lLL rold vv l o Mj Plpw Nui h lac llhw i w ull loo 3 Q10 Suppose ThaT uvuz is a uniT vecTor parallel To a line Through The poinT xmyo WhaT is a very special properTy of The paTh described by The parameTric equaTions x x0 Hit and y yo it L9 i1 alburr aqu will 04wf Spaac pawl ltuu0ll Q11 Suppose ThaT an objecT moves on The surface 2 fxy along The paTh reafVe fa re xy pare generaTed by x x0 Hit and y y0 uzt WhaT is The formula for i aT The t poinT x0y0 WhaT else musT This formula cacuaTe M k 4 Lt a 9i ii Jquot 96 elf 93 a M 341 31 3 E I l39 5 my I 5 31 4t 5 lt 3 33 gt uuu3 l v3914 lLJ and Hullltd Hquot I39M quotF F 43 Sl lpt T in La Aliroa W N 39 39 ML 5 w TM Page 5 of 8 MTH 254 S monds 7 Chem Ru eD recuona DerwatweSGradwems DefinitionfTheorem Assuming that the function fxs differentiable m all duecuons at the pomt Wu the slope 0mg surface 2 Aw atthe pomt mu m the duecuon 0mg unit vector 2 s gwen by the directional derivative value Daxovynvxnvyn a Where XovynAXnvynv6xnvyn 5 E ledmemof at Xnvyn Practical Example 2 In Me gure be uw M2 func un z fxy where xy x2 y 2 5 Shawn Mung wwh M2 pmne xry o Lev s defermme M2 s upe uf z fxy m We pmm 114 fwe muve 2H m mgm mung M2 curve uf Wersec un bevween z fxy and M2 p ane x 7 y 0 Lev s wusvme M2 resuquot an me gure A L I 7 J J U ltle J 539 339 xi 3 1x 1 DRK50 GK quot739 a anat 45 Q 9 H33 Pageema MTH 254 Q39 quot quotquot 39 Practical Example 3 The 3dimensionul figure below shows the function fxy xy 7 2x2 5 Q12 What is the slope of this function at the point 114 in the direction of the vector v 747 2 What is the angle of ciscent cit that point in that direction I J uel P 3quot i x Table 2 coordinates E is 3 4 b quot 3 4343 962x1gt H c WW DAgCLI 520 39139 quot5 u i L CJJ39D C in gt J L 35 I 4 39 m 93918 14 v IVaLuI Ld39 Toiw d 39tn 4 J rwuh bl nan Illfinch Hu Xyv lug 4 4 L i cquot n U a 3 J lt e A lt393 4 2 3114 Tit13 39M IH39UJ Nut391 3ti14iPage7ois Kw 39 XyJXTY x l 1 dxjs PALE4r gm rm gig DLI lt C7ltr 5 in 4 r I MOE Q I Mu LWWC V39 quotffj 74 39 1 t51 35 1wg v j m 6 Jr x ax crk 0331 MTH 254 Simonds Chain RuleDirectional DerivativesGradients Introductom Examg OK assuming ThaT f is differentiable in all direcTions aT ynThe slope of f aT The poinT xnynzn in The direcTion 1 is D fxnyn vfyn2 Q13 In whaT direcTion does The funcTion f have maximum slope aT The poinT xnyn WhaT is The slope in ThaT direcTion A LA 3 m 09333 Verve F frugal IN cub 3amp6an co9 r4 is MLViwikoJ uLarc coJail ukuL ecu3 4 9 o it aha gtgy 39ra a H Jirwh M 4 Mum u q 395 Xv 3 m4quot39J o D JLC 6439 J Lo Jhpbia L kr39l39 Air o mLS IV XV H Q14 Find The mamal raTe of change along The surface zx2 y 2 at The poinT 1 2 IndicaTe The direcTion in which This raTe occurs on The figure below megm J x1jA 3 q z L52 11 t hx39vapl 5quot 9 lokjb P blag nuqu 39n Lt Air ah 3 3911 11 in H 5 fei 41 Anniu iJ la ut Z Page 8 of 8 Osculating Circles and three planes assomated with motion m Just as we use lines as reference when discussing the slope of a function we use circles as a reference when discussing the curvature of a function With slope a large absolute value indicates a steep incline whereas a small absolute value indicates a shallow incline The sign on the slope indicates whether the line is increasing or decreasing Since a circle neither increases nor decreases there is no need for negative curvatures It would be nice if we had a definition for curvature that leads to large curvature values for really curvy circles and small curvature values for circles that are not so curvy De nition The curvature K of a circle of radius p is Really curvy circles have 5Q IV by Not so curvy circles have I I 4 I The ascua ng circle to a function at a given point is the circle in the osculating plane that at that point has the same tangent line as the function and that lies on the same side of the tangent line as 7t The osculating circle to function shown in Figure 1 has been drawn at the point 11 What is the curvature of the function at this point r k j At the point 4 1 the curvature of the ellipse in Figure 2 is Let39s find the center of the osculating circle and draw the circle onto the picture P 4 Page 1 of 6 OSCulatllm circles alllJ three planes associated With motion Let39s find the curvature of the ellipse in Figure 2 at the point 31 Let39s then determine the center of the escalating circle at that point and draw the circle onto the gure Finally let39s state avector function that graphs to the osculating circle One formula for curvature is Ccv3J u xrb m quot is l 3 HM 395 a 1 TL Lll w afwh n U wj 39 l J Mquot b C LxJlquot 31 J 14 J3 43941 7quot 1x J J 174quot O 4m 9 i 2KJ quotquot a Uquot aquot J Jr w 4 l i VnrL 393 9C175L an HUI rl v 1 2 M flu J Juquot j rl 05IILPT FKI39 LIEUL Aghit l 4 J J J 00 is quotat lt3 culbi 2594 Page 2 of6 Osculaling circles and three planes associated with motion The function y fx 73x18 is shown in Figure 3 Let39s find the center of the escalating A circle at the point 7271 let39s draw the circle and finally let39s state an implicit function that graphs to the escalating circle One formula for curvature is Uro rJJHL RM 1 7 XiVF l J39 quotW 39 I 39 1Jlln1aL b 39yx 21h J 71 7 Ag P n t quot 2 ML L L f39lL 39b MGquot quotquot 5 I D MM z 7 cosh OC PawL 41K Ill139 4547 5 lt3D a L j Aplnzi Ion x 3443 quot5 z 5 39I39 JLLI39II FUM 1 7 5 uniL w H39l 5394 o C kt m Page 3 of 6 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 Page 4 of 6 Figure 1a Figure 2a Figure 3 NOTE G g m m y Tne osculating plane is parallel tu tne The oscula ng plane Bantams the unit tangent vector and tne unit norrna vector Figure 1b tangent line and tne nonnal line We osculating the binonnalline binonnal vector in 5 E in 5 f 1 VA vv Vi 2pxlnaip 395Xy Tne rectifying plane euntainstne tang Figure Tne rectifying plane is parallel tu tne unit em line and WE b39quot quot39quot39 quotquot9 2b tangent vector and the binorrnal vector Tne rectifying l m the quot rquot al quot9 unit normal vector 395 y i in Figure 39 line and the binomial Ime 3b nonnal vector and the binorrnal vector ornial plane is perpendicular tne Tne norrnal plane is perpendicularthe The n 9 unit tangent vector tangent lin The engn o eau uni euul Osculating circles and three planes associated with motion Figure 9 shows the vector function 71 4si11tcos21 7 35in21 cos The normal line slope Let39s intuit the equations of the osculating normal and rectifying planes Let39s verify the equation of the binormal plane using traditional techniques Page 5 of6 MTH 25th Prtuecme Mutmn Introductory Problem Mtssy has tssues thh her thtra grade teaeher Mrs tltatz s baT Mtssy Tuaaea her shuuter stuck tt trt her muuth t art a Te recmvem th uu tank breath art Tet the b H y The erta at shuuter was three feet abuve the gruund at the ttme the baH exttea the shuuter The baH htt Mrs Katz trt the back at the eaa we feet ueth aartalzf tf th tt htehthe tgm began Ftha the tmttaT speed uf the baT Background If We Tet g represent the eurtstartt aeeeTeratturt due ta earth39s gravtty tgrture mc nvemem truths Mk2 atr reststahee a nunrspher ma gTabe mauhtaths aha vaHeys ete and de ne 7t t h uf the spttbah r seeahas atter tt takes tght theh the aeeeTerattah tuhettah ts eahstahtTy Hr o 7g assummg that We use a eaarathate system th sueh a Way that the eampahehts therease as the spttbah muves upward The teTaetty tuhettah theh ts 7rJ 7 rdr J07gdt0gt Ifwe assume that the tmttaT teTaettyts 7 VV2Men 700rg 03Vn 39 a 7 6OrgtVttvzVtrgtvz Nah rteeImmjtegttzdreltttegtzetzta Ifwe assume that the tmttaT pastttah u he spttbah ts n 05n theh Thtsgttesus V trgt2 V2tgt0xnltvxtrgt2 winsquot P3921 UT 6 570 r i N r393 1 Load x at La v N MTHZSL PvujecNeMmmn de cswnfreefhghvsubJeuun ym ap anemrygmwm una cunsmm n JECVWGS aunched fmm Cm mmd pusmun squot 1 m on Ms clf e b WHEN v2 umy an vv2 Men m pdsmdn uf we deem 2 Mme ver mm s gwen byme mnendn s 1 2 39z e trig vzlxn Furmenndre w we WHEN speed s vquot speed ms and me WHEN N QJECVDN ang e s 9 Men 41 vncns9 n and v2vnsm9 Onedrm ev suse g321i ur s g33 n3 39 965 7 s 2 vquot sme unknuwn mm speed uHhe spude Ms 2 sme unknuwn vdvdx fhgmnme uf M2 bdu s PM lt1quot mm 39r rs m 37 m u p 0er 1r mum R 11 1 van mxpllei39 yi 5 5quot f a 1 5 w Jim er Lr rh1v39quotU 397l1Tu Jr 1gtII2 a when J wry mmm 39 Ar 1 Mar fL 7 L L L39J xpml or AM 2115441 swmmtaneous equatwon Smeu 55s t 2 a sohmon on Tcacuator 2usa59742 V eltveuslts u Page 2 u e MTH 254 Projectile Motion An Olympian Tosses a shoT puT aT a heighT of 6 TT wiTh an iniTial speed of 43 fTsec aT an angle of 45 wiTh respecT To The horizonTal Find The speed aT which The shoT puT lands 6100 s UAWouts Ws a gal 4 i1 Una90quot 53m l l t Cs le5 V 30 w a 1 45quot in C llcu39I l LOS E1 4 1 51LV Lgt lt 2Lsz 1 LthLll lr i 6 lamb L A 46 a Mic4 6 1L lGS Hgii QISG 439 3 CD Lrnl quotl 7th 5 I V39Lkg l us 1321 al5figtl 3923 4727 Defs Page 3 of 6 MTH 254 Projectile Motion BaskeTbaII player Shaquille O39Neal aTTempTed a shoT while sTanding 20 fT from The baskeT IT llThe Shaqquot shoT from a heighT of 7 TT and The ball reached a maximum heighT of 12 fT before swishing Through The 10 fT high baskeT whaT was The iniTiaI speed of The ball s07f 5tm12f Shaq sassumptions 42 S Analysis 7tf 2010 v0 is The unknown iniTiaI speed of The ball fTs i9 is The unknown iniTiaI TrajecTory angle degrees tf is The unknown ToTaI flighT Time of The ball s tn is The unknown flighT Time unTiI maximum heighT s Sh aq S a riablegd fihifioffs Since There are four unknowns V0 9 tm and tf we will uTimaTey need To come up wiTh four equaTions 7101 lt11 Jami Le v Shown Ffu 5 7 lt U a V aA 1f L01L0339 zo NFL 4 Uzsma c 31110 I 39 15 1 Fl UJ5J39T3 E 3 a 7 SiaLe39 7 391 Fj TltLDI 107 6 1 711 Bf3 1 1115v0o C53 Lo 1 one 11D LO 1lv Page 4 of 6 F 3910 l 7 1d 39439 quot vzulm quot Lu J39quot 15 h a 5 4Lvw m wh 439mv 0fh Lvn warm mum Ce S a a L a Marv h E I 1r r Ic U z 9 H 5 C57 Innn 7 17 73 an l 1 2 Fl 5quot quotr quot b iii 5523 aquot T1711 W 5quot 15 H E1 amp1L Sj wh W I am 39 5 Jul 39 1 39 24531 430 3 if I J toJ39w run 9 31 ru d39 670 9 71 9 31139 m 9 Im39 11 4 a Fn E quotJ3l FM39III 1 21 51 z 39w n39 1 5 um J 3 quotI 1 gm H m biwdwnl 94391 LIL Lquot new JM 4 Mr TL MM 1J an Ant 31x4 lx MTH 254 Projectile Motion Projectile Motion Practice Problems 1 I39 A 3 01 A ball rolls off of a level TableTop 4 feeT above The ground aT The speed of 5 fTsec WiTh whaT speed does The ball hiT The ground Nancy Lopez hiTs a golf ball from The Tee wiTh an iniTial speed of 125 fTsec aT an angle of 45 wiTh r especT To The hor izonTal How far from The Tee does The ball land Sammy Sosa hiTs a ball in The AsTr odome Thr ee feeT above The plaTe and aT an angle of 24 wiTh r especT To The hor izonTal The ball jusT clear s The 9fooT fence locaTed 400 feeT from home plaTe Find The speed aT which The ball began iTs flighT and The speed aT which The ball cleared The wall Fir e breaks ouT on The roof of a 20fooT building ThaT is 25fT by 25fT aT boTh The base and Top of The building A fir eper son is sTanding aT The midpoinT r elaTive To one side of The base aiming a hose aT The roof of The building The Top of The hose is 4 feeT above ground level and 15 feeT from The base of The building WaTer is shooTing ouT of The end of The hose aT a speed of 40 fTsec AT whaT angle is The hose being held if The waTer is landing smack dab in The middle of The roof STeve Young Thr ew a pass aT a 45 angle from a heighT of 65 fT wiTh an iniTial speed of 50 fTsec AT The snap Jerry Rice began running dir echy downfield and caughT The pass aT The 26 yar dline 6 feeT above The ground Mr Young was aT The 50 yardline when he received The snap How far back from The line of scrimmage was Mr Young when he Thr ew The ball Figure 2 Football Field T T line of scrimmage ball caught at 26 yard line Page 5 of 6 MTH 254 Projectile Motion Answers to Practice Problems H I39 A 39gt 01 The ball hiT The ground wiTh a speed of abouT 1676 fTsec The ball hiT The ground abouT 4883 feeT from The Tee The iniTiaI speed of The ball was abouT 1335 fTsec and The ball cleared The wall aT abouT 13206 fTsec The hose was IT an angle of abouT 6607o wiTh r especT To The horizonTaI Mr Young Thr ew The ball abouT 662 feeT behind The 50 yard line Page 6 of 6

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