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## Chapt 13 Secs 1-5

by: Jasmine Jackson

60

0

33

# Chapt 13 Secs 1-5 MATH 114 - 002

Jasmine Jackson
Penn
GPA 3.0
Calculus II
Professor Ching-Li

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Chapter 13 which covers everything from Vector Functions to Torsion.
COURSE
Calculus II
PROF.
Professor Ching-Li
TYPE
Bundle
PAGES
33
WORDS
KARMA
75 ?

## Popular in Mathematics (M)

This 33 page Bundle was uploaded by Jasmine Jackson on Sunday October 4, 2015. The Bundle belongs to MATH 114 - 002 at University of Pennsylvania taught by Professor Ching-Li in Summer 2015. Since its upload, it has received 60 views. For similar materials see Calculus II in Mathematics (M) at University of Pennsylvania.

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Date Created: 10/04/15
Math 114 Rimmer 131 Curves in Space and their Tangents 31 vecmrFunwonsltspace curves anmeir angents VECTORVALUED FUNCTION 0 vector function for short sometimes called space curves 0 takes one or more real variables and returns a vector 0 we ll focus on single variable t vector functions that give three dimensional vectors represents time in most applications rt ltftgthtgt ftigtjhtk real valued functions called component functions of r We want to look at the calculus of these vector functions M Math 114 Rimme 6 131 Vector Functions Space Curves and Their quotlt mute Tangents Interesting vector valued functions r t ltcos t sin t tgt called the circular heliX For a nice animation go to 1 4 httpmathbuedupeoplepaul225CircularheliXhtml rt lt2coscost2cossintsingt called the trefoil knot For a nice animation go to httpmathbuedupeoplepaul225trefoilknothtml o o 0 Math 114 Rimmer 131 Zzgor Functions Space Curves and Their W ltft gt htgt Limit of a vector function taken component wise liml t f t g t provided the component tea tea tea tea function limits exist Letrtltsmteztln1 tgt Find limrt t teO 1imrt lt1im 8mm1ime2t1im1n1 tgt teO teO t teO teO WWEK J 1 1 O 1imrt 110 ij te0 M Math 114 Rimmer leferentlatlon of Vector Functlons 193 13391Ve quot squota e Cmm m r rtltftagtahtgt Derivative of a vector function taken component wise rt ltft gthtgt providedthe component function are differentiable Letrtltsin1t 1 t2ln13tgt Find r t 1 J lij g x it a 7 r x W Math 114 Rimmer ll 7 7 r 131 Vector Functions Space Curves and Their F l L w mfy Tangents l 39 A if 77977 7 N l r t is the position vector r t is the tangent vector 39 r t will now be called the velocity vector v t r39t is called the speed of the particle r t is called the acceleration vector at rt ltsin t2costgt t r t ltcos t 2 sintgt t r t lt sint 2costgt r t xcos2 t 4sin2 I la ltMgt vale w anae 3gt V a Math 114 Rimmer O O 0 131 Vector Functions Space Curves and Their Differentiation Rules Tangents 1 i utvtu tv t dt d 2 E cutcu t 3 ftut f tut ftu t 4 z vtu tvtut v t 5 utgtltvt tgtltvtutgtltv t Show that if rt c constant then r t J rt for all t Assume rt c 2 The fact that rt rtrt gives us rtrtc 2 d Taklng the derivative of both s1des gives ED t I 0 Using the properties of differentiation we get Iquott I39 I I I Iquott 0 So r trt0and thus r tJrt forallt ath 114 Rimmer M Most vector functions do not have constant Eingsmwunwonsepace curvesmnmeir magnitude The heliX example from earlier I39 t ltCOS t sin t tgt rt 052 t sinz tt2 rt 1t2 It s magnitude isn t constant it depends on t Consider its derivative however r39t lt Sin t COS t lgt r39t sin2tcos2 n1 z r39t has constant magnitude so 1quott J r t So for the heliX the velocity vector will always be orthogonal to the acceleraton vector don39t think that this always happens W Math 114 Rimmer 131 Vector Functions Space Curves and Their and Tangents Show that if r t is a vector function such that r t exists then d I t X t t X t mm ro ro ro Using the properties of differentiation we get rtgtltr t r39tgtltr39trtgtltr t anytime you cross a vector With itself you get the zero vector So dirtgtltr39t r tgtltr t t When taking the derivative of the cross product of a vector function and its velocity you can just take the cross product of the vector function and its acceleration o W Math 114 Rimmer Geometry of The derivative it quot squota eCmesmdm vector i Z secant vector rth rt P Tangent vector rt1hi rzhZ rz Unit tangent vector Tltzgt 8 it39s the velocity vector With magnitude 1 M th 114 Rimmer 3 a 3911 Illect0rFuncti0nsSpace Curves andTheir nts rtltftgthtgtftigtjhtk attt0 passes through the point f to g t0ht0 and has the same direction as the velocity vector at to ltft0 gt0 ht0gt39 So the parametric equations of the tangent line are x ft0 fto y 8a 3 00t ht0 h t0t 7 1 miquot 7 7 e li 39 39 l 39 y 7 Math 114 Rimmer l 7 t l 7 J L 132 Projectile Motion 1 g A projectile is launched we want to find the parametric equations for the path the projectile has mass m assume that gravity is the only force acting on the projectile after it is launched neglect air resistance L A39 Force due to gravity F mg j gt mg J ma Newton39s Second Law F ma tdrj gjdt gtjC1 1 1 tdt gtjC1dt Egt21C1tC2 Egt2JV0tC2 gtjC1 1010 Vt 3 vo rt gt2j v0tr0 w Math 114 Rimmer 132 Projectile Motion Often you are given an initial height an initial speed and the angle 9 at which the projectile is launched rt E gt2j vot 1 0 initial speed v0 ltlvolcos 6 volsin 6gt V0 1 0 lt0vhgt agj atlt0aggt i vt lt0 gtgtlt volcost9 volsin6gt vt ltlvolcos 6 v0 sin 6 gtgt rt lt0 gt2gtltlvolcosH V0 sin6gttlt0hgt rt ltv0cos 6th volsin 6t gt2gt A baseball is hit 3 feet above the ground level at 100 feet per second Math 114 Rimmel Projectile Motion and at an angle of 45 with respect to the ground Find the maximum h 3 6 4 5 0 height reached by the baseball Will it clear a 10 foot fence located 300 feet from home plate V 0 i 1 g 3 2 I t ltv0cos9thv0sin 9t gtzgt lt100cos45 t3 lOOsin45 t 32t2gt rtlt50J2t 350J2t 16t2gt The maximum height occurs when the j component of the velocity vector is zero 2amp5 Vtlt50 50 32tgt gt SOJE 32t0 3tYz 221sec 2 2500 6252 1298 The maximum height 3 50xE 16 3 V 81125 ft The ball is 300 ft from home plate When the i component of the position vector is 300 50 2t 2 300 3 6 Plug this time into the j component of the t E 3 5 z 424 360 position vector to find the height of the ball The height of the ball at the fence 3 50232 163J22 3 300 288 15 ft I Yes the ball clears the 10 ft fence for a home run I w Math 114 Rimmer 132 Projectile Motion Height Flight Time and Range fer Ideal F39mjeetile Marian 0 0 Fur tideell pmejeeti le multinn wlhte39r t am abjee t i115 Immehed remn 1eer1igin ever 21 litter ilemlrtal w aee with initial 5111313111 113 and Iaunel u angle 11 Ifl L39 em 1er M rf tigtttig height LLMm 2 211g 51mm Firg rnme 1 g 11quot 1 Rang H z 5M1 are 1 quot 3 39 M10 ax wt 11W HA hull vsci k J0 ml it cormvl Math 114 Rimmer Projectile Motion 3 Firing golf walla A wring gmu m g l urn Itweal Ifn rvc golf ball mi 21m ainy u m ma E This haw lewd I LIQJI m away 2 OMS Q ID IL Fm Em 531ml initial f imi hw tl a39itring gem that a g U n1le1 azrrtr12h rang 1 m 3 39 Wharf W515 liH39Lr Ijlau l lqs initiai gt ng 03 a y f 2 VO y x gt C 6 3quot 0391 0 v0lt7 z m zei Ag m 901 we amidquot V0 r z 7 03935 3 39 a 19 3 o m 1 SWIMT spin w Math 114 Rimmer 132 Projectile Motion WM 513 0mm 3 9 cowIla W1 SWIM 30M ms Sm M t ginok 6091 3x91 gas 15 t 35 EMSM 719 Math 114 Rimmer Projectile Motion 3 g39 J Voltl b 2391 gt ro ltOJ3 gt 3 2 ln Ci tisens Bank P39arl 1 where the Philadelphia lPhilllies play besehalL the right held fence is 33 feet fresh hunte plate where the hatter stands when he hits and the fenee is about 13 feet high First baseman Illewward hits a heme run ever the right field fence that starts an eet MIN heme late with heariaent y teward the well and initial uEward veleieity 34 ft By haw many feet duties the hall elear the temp 0f the fencequot Assmne the aeeeleratien clue to gravity is 3 2 ft see2 and ignere Wind resistanee 9 i934 Miloquot Math 114 Rimmer 3gp 133134 Arc Length and Curvature In the plane 2 d yfx Arc Lengthidlf xzdx parametric equations x f I and y g I t 9 Are Length j f t2 g t 2dr Arc Length i r39t dt x rtltftagt htgt Arc Length WW2 81012 fl0126 Arc Length Function Sm r39udu 1 A a immer r 1 lt 3 c os 3 uxt Ar Length and Curvature r tJd 39le Mad Lb 4 rim Ruth Xm rw f Est 4108 15 M s quot3 m r U 1quot TV Slr ml t 3 O 6 W HEWwt 99 r g 09 Fm Math 114 Rimmer S V g gain Arc Length and Cu ature rqir 2 asquotm Er E39j m l r 2 1 l 1 M 3991 9t fit gt 66 393956 b k w x W393 Kiwi 7E 78 1 Math 114 Rimmer H9 3 m t3 gt 21 m an H 0 Z A We 39gt Mm 1 Rt NH W mica N 0N1 r M 3W Cm SRWQ 1 WM 9 I 5 V 3 Izt je qt g quot T IVA x H9quot r Mil Lila 31H HM 9 1261 l quu lr ml WW W 1 gt 1 s z Af2 7 grlh SAJ camp i 77Qv na 3t gt r lnt lmEW T ath 114 Ri er wha 1 e gth a ture 39 3 Lt f3i3ei Hmt Lijgth ti i 39 f qgttv 56 l 3WM 95quot 3 9M l 3p Math 114 Rimmer I M dbl 133134 Arc Length and Curvature s 2 distance and t time r t gives position on the curve as a function of time like driving and looking at the clock to tell where you are after you have traveled a certain time t its more natural to look at the odometer or the mile markers to tell where you are after you have traveled a certain distance 5 we would like to have t as a function of s so that we could reparametrize in terms of 5 Using the arc length function we can find t as a function of 5 giving us rts orjust rs PM Math 114 Rimmer 1 t 2 lt3 Sin t 4 3 COS tgt 133134 Arc Length and Curvature Reparametrize the curve with respect to arc length measured from the point where t 0 in the direction of increasing t r t lt3 cost 4 3 sin t Iquott S 9cos2tl69sin2t 9cos2tsin2tl6 916 2 5 OL N r u du 35st 2 5t 0 S s5tgtt w Ks lt3 r T quotgt 5 rslt3sin 43cos gt SS1f flyquot WW g r 5 is the position vector of the point 5 units of length along the curve from its starting point Math 114 Rimmer 133134 Arc Length and Curvature Curvature a measure how quickly a curve changes direction at a point another way to look at curvature is how sharply the curve bends at a point Q curvature P gt curvature Q In order to calculate curvature KimWe need Tt to use the unit tangent vector T and the arc length parameter s Curvature the magnitude of the rate of change of the unit tangent vector T With respect to the arc length parameter s Q ds TIAt K39 T S M Math 114 Rimmer 133134 Arc Length and Curvature rslt3sina4a3cosgt r s 215cos2 215sin2 215cosz sin2 2 215 T S lt sin90 COSgt K lT sl ash 6082 3 PM Math 114 Rimmer CurVature In terms Of t 133134 Arc Length and Curvature r udu stj Fund Theorem 0f Calculus rt lt3sint4t3c0stgt 3 r t lt3c0st4 3 sintgt T t g 3 K my 5 0 5 25 Iquotl T l i i i Ix lt5c0st 5 5 s1ntgt T 2 sin 230 3c0s tgt gt T39t 215sin2t cos2 2 2 scalar Fm Math 114 Rimmer curvature In terms Of t function 133134 Arc Length and Curvature r Vector Tt gtr z r t Tt gtr z Tt Take a derivative to find r 2 dzs dS Z l T l T l M m2 0 dt 0 Consider r39z xr t 2 2 zj d T T Tgtltd T Tx rj dt dt dt dt dt dt dt 2 2 d TgtltT j TxT dt dt 0 dt r39tgtltr t r39tgtltr t 02 TXT Iquotl scalar funct 2 TxT Math 114 Rimmer 133134 Arc Length and Curvature Curvature in terms of t Section 142 T X T T Show that if Mr 2 c constant then r t Jrt for all t sin 6 Tt 1 s0 T t 1 Tt for all 2 TLT39jsin621 TXT T T m 1 TXT zT r txr rr rllexT I ll 2 r t 3 T r tgtltr t r t T rt rr Mt nesM30 3 K M r tgtltr t r t K f i i g i gquot and Curvature rt lt3sint4t3costgt r t lt300st4 3sintgt r t 5 r t lt 3 sintO 3 cost i k r l gtltr l 3COSZ 4 3sinl lt 1200st 9c0s2tsin2t12sintgt 3 sint O 3 cost r tgtltr t lt 1200st912sintgt r39tgtltr t 144cos2 t sin2 t 81 xE r tgtltr t 15 r tgtltr t 15 r t3 125 i 25 Math 114 Rimmel 133134 Arc Length and Curvature ILL uuw i il f i InnHjj 3k PM sp t Hus wast HJm f H a 8 m 41 q i O lt05quot km Sw i39JCMJJC Imd P r m xosu rsnaj WT W L 393 k 9 MR Mk 0 2 0 6 Q IO t1OSt em3 I Wk 49 a Jtto 0 4 1 gt g 0 0 k M Math 114 Rimmer 133134 Arc Length and Curvature Let rt be a smooth curve in the xy plane rt ltf t g Show K Ifzg ffg K r rxr r f 2g 2m WW rtltftgt0gt r tgtltr t lt00f g ffg gt 1 tltf If g W yr39rgtxr rgtW ffg f g Ir ltrgt f ltrgt2g39ltrgt2 Irt 3 gt232 In terms of t Planar parametric curve 1 t ltftagtgt remember T t is called the principal unit tangent vector 1 gt T J T39 but T is not necessarily a unit vector B TU making T a unit vector gives a new vector t T called the principal unit normal vector N t The vector B t TtgtltN t is called the binormal vector Bt is orthogonal to both Tt and Nt Bl 1 B is useful for finding the torsion degree of twisting of the curve TNB frame Frenet frame moves along the curve has applications in differential geometry Math 114 Rimmer 133134 Arc Length and Curvature The plane determined by the vectors T and N is called the osculating plane The plane determined by the vectors N and B is called the normal plane The circle that a lies in the osculating plane of the curve C at the point P b has the same tangent as C at P c lies on the concave side of C side towards which N points d has the reciprocal of curvature as its radius is called the osculating circle M Math 114 Rimmer 133134 Arc Length and Curvature Elfiir rg39le of 12 ll 39tiiiirll l39 Came i 31 L2 quot39n i irll 39 H Halli I LS 211T murmurs J 10 Math 114 Rimmer 7 rm 3 r39 139 133134 Arc Length and Curvature ML IWn rgi gem r rm E m ninth an the paint ml I The name purqu trigeu t hignuph and 2 5m 3 in the 1quot 7 Ian 1 auliur39i Ilw chm LugIquot g l39h7 ELLF L39qu Iith guru Planar function X7 W y fx If x WW 0 W35 W quot riK K r lt 5w lgt K 397 PM Math 114 Rimmer Egg 133134 Arc Length and Curvature 1 gt 7quot339 11 Acceleration can be broken down into two components the part of acceleration the tangential component that acts in the line of motion the part of acceleration the normal component that acts perp to the line of motion compTa aT xxxIN a Ar T x We want to find aT and aN compNa aN 4 IT and IN are related by the following formula IN a2 aT 2 compTa aT V aTaTcos 1 aTac036 4 CO5 a39NaNCOS aaTTaNN vVT aaTTaNN Math 114 Rimmer quot 2251 135 Velocity amp Acceleration vgtlta VTgtltaTTaNN vgtltavTgtltaTT vgtlta 0 VgtltaVaNTgtltN vgtlta lvlaNTgtltN vgtlta VaN VTgtltaNN VaTTgtltTVaNTgtltN since TgtltN TNSin WWW 1 1 1 aKTv TUV FWN V T dt 2 v d fT T dt dt 2 adT K Ntj dt dt dt 2 2 ad TK j N dt dt 3 aTTaNN Math 114 Rimmer 39 6 3 135 Velocity amp Acceleration J t 4 E 0 FM 4 Aux rm lt amt l 6 51352 rm a 1 39m ltcnu1 emgtfm 2535 aw99 mammal 43 egt a r v m HM5 ewf Jmstt rd7ltvzgpgt fem aghast m tl lt A65 mo lt o 3 3 wkfmu V VMMo X a o 2 5 T 39 Mun gt Ionxl Summary Formulas involving T and N aTaT aNaN Formulas involving v and a quot 39 Math 114 Rimmer x 135 Velocity amp Acceleration m Tm lumi a L Hi5 gait i5 z JM339IHj r 1 The Euruanre Ell P a i5 LEFITH E39JL 1 il39liEI39E aEEa N I H EIEFIHITI H LEE T THE l T i l li l i Ufa1 Sml lh EUWE iaquot a ain Na 5quot scalar triple product 1 t jerk r tgtltr t vaxjerk 2W am0F vxa2 139 139 MIN Math 114 Rimmer m Find the torsion at any time t r t lt3 sin t4t3cos tgt r39tlt3cost4 3sintgt r39tgtltrquott 15 3 sin t O 3 cos tgt lt lt quot39t 3 cos tO3sintgt 3cost 4 3sint r39tr tgtltr t 3Sint O 3cost 3cost0 4 9sin2t 9cos2 t 3sint0 3cost O 3sint 49Sin2tcoszt r tr tgtltr t 36 36 66 22 4 T 2 139 15 1515 55 25 Find the torsion at any time t rt ltt2costtsin tsint tcostgt 1 Math 114 Rimmer 2 135 Velocity amp Acceleration r tgtltr t t2 r t lt2ttcosttsintgt r t lt2cost tsin tsinttcos tgt 1quot t lt0 sint sint tcostcostcost tsintgt lt0 25int tcost2cost tsintgt 2t tcost tsint r39tr tgtltr39 t 2 cost tsint sinttcost 0 2int tcost 2cost tsint 2tcost tsint2cost tsint sinttcost 28int tcost tcost22cost tsint O tsint2 28int tcost 2t20082 t Mt Wostt2 SinZt 28in2t M8t 2ts39 cost t2 cos2 t 4t 082 tW 4t sin2 t W 2t 20032 t sin2 t t2 sin2 t 0082 in 4t 0032 t sin2 t Math 114 Rimmer quot 2251 135 Velocity amp Acceleration 2 r39t r tgtltr39 t 2t2cos2 t sin2 t t2 sin2 t cos2 tn 4tcos2 t sin2 t 2t2t2 4t jl2t3 4 r t r tgtltr t M r tgtltr t t2 r39tgtltr t2 5t4 r rr txr t 2t3 2 r tgtltr t2 5t4 5t

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