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Chapter 13 Vector Functions Michigan State University MTH234 1 Vector Functions and Space Curves Watch the Video at httpsmediaspacemsuedumedia131VideoNotes18j5x7ptz Media alternative to textbook to ll in the rst 3 pages of the notes before class on 13015 De nitions 1 1 1 In general a function is a rule that assigns to each element in the domain an element in the range A VatAquot 39FW Cfl39l M or fl Or 39F UAC l m is simply a function Whose domain is a set of F03 numhU S and Whose range is a set of V eLI39O PS primarily in R3 Z L39w 30 lttla113 it 3 EX ltQ o QSWVI39 2 If rt is a vector function Whose vectors are in R3 then r05 0 05 905 W fti 975J39 htk Where f g and h are real valued functions called the GAILCk RAL39IH of r Example 12 What are the component functions of rt t2 1 sin 75 31 t FC D 1 l c Sia39t h lI t 1 Example 14 MTH234 Chapter 13 Vector Functions Michigan State University Or goal will now be to develop calculus for vector valued functions Before we get to derivatives or integrals let s start with the basics limits De nitions 13 1 If rt ltftgt htgt then lim rt M490 66m kmgt provided the limit of each component function exists Recall Old Calc 1 Theorem A function f t is continuous at a if and only if 13310 it 2 A vector function r is continuous at a if and only if 4 Ear rm 39 2t Compute the limit ling lt1 3752 te3t Sm gt gt 7 lt 303L 053 kquot Slam do H lt0Agt Example 15 Is rt continuous at t 0 t 1 t2sint x9 252 if t g 0 cost tet 1 275 if t gt 0 353 Lo3 D 33 rUc lt 10 Igt 2 So no Id CDVCl MKMDWS MTH234 Chapter 13 Vector Functions Michigan State University A big part of this section is parametrizing curves as vector functions Here are the appropriate de nitions De nitions 16 1 If f g and h are continuous real valued functions on an interval I E R Then the set C of all point 51331 2 in space x ml veg t z a VB and t varies throughout the interval I is called a Mc We can also de ne I CLLTWS by removing the component where 2 Space curves are extremely similar to vector functions just like points are close UCLA Dr39s 3 Parametrizing curves in space is covered in Calc 2 section 101 of Stewart This would be nice review For those of us who have sold our Calc 2 books consider watching the 4 videos at wwwkhanacademyorgmathprecalculus parametricequations Example 1 7 Find a vector function that represents the curve of intersection of the cylinder 512 32 1 and the plane 1 z 5 X Goff 1 rg mzt 7 605 391 S j SWquotk 1 9 Cast ail ltLSS J Sinf 5 Cosgt ta toxin MTH234 Chapter 13 Vector Functions Michigan State University Group Work 1 Find a vector function that represents the curve of intersection of the surfaces 2 2 y and x2 y2 4 X a COSl 85 as39m t Q 3w5 125h gt C Ei quot ltACOS1 l wv j Ycosz l39s m kgt t QEOQT1 Max assay q 2 Suppose I want to nd vector function that represents the curve y x2 2x 1 as 6 02 note we are in R2 Which parametrization is best to start with and Why gt I a rt costsintgt 1 f e 1E b rot ltwgt H ltcgtrlttgtlttgt lt Jt1T11EIgt a J1 Then nd a parametrization f G La f t X115Ll l LXL 47 xH l 4 3 X 1 rm lt xtgt 30 4 4 MTH234 Chapter 13 Vector Functions Michigan State University 3 Suppose I want to nd vector function that represents the curve 1 32 23 1 y E 0 2 note we are in R2 Which parametrization is best to start With and Why t 3 t 6 I35 a rt 2 cost sin t 1 ms ltwgt 1 X t Qx C W lt tgt 391 Then nd a parametrization T I h t X lt t Jfgt igCD11 4 Suppose I want to nd vector function that represents the curve 1 512 32 note we are in R2 Which parametrization is best to start With and Why a rt cos tsintgt P b r05 lt75gt C W 7 75gt gt 39 Fa Then nd a parametrization 3 H Flquot l F Fs Tips to parametrizing in 2D a If A that is the curve passes the I m then start With the parametrization r 7 I v b If x that is the curve passes the Morvzo nah WE Q39slthen start With the u parametrization z vi c If the curve is Cl 0 Q starts and stops at same point then start With the parametrization V ltC95 tl S 2 1 MTH234 Chapter 13 Vector Functions Michigan State University Note sometimes there are still dif culties 2 5 Skip if need time Parametrize the ellipse 3 y2 1 in R2 Ellipus owe closed Cmsirkarl r005lt 5t25lq gt 39H QSZ in39lb SW QML W U39C 39 03 1 Si i39l l Tim39s is n6 0k rua S39iwiw l39 3 all Tm Parl culow waned 825i rid O F lLsQ So cashier rm lt1ccgtslzlsiatgt l E 01161 ul gh we am HUi Fxgr works 6 Skip if need time Parametrize the circle 52 2x y2 8 in R2 7 39L X ax H L3 cl 397 X39r Kat 3 C1 E Circle 10524 Consider l lt3 cos39l 3 Atgt 3cos39t39l39 01 qsgn lc a hKK i39 l lais 25 5 Hum 0 a So we VIM 0 X avr 1 33quot ConS M rC c Schtl j35iofgt z J 7 3 uch v as a 7 Just to assure you there is harder math out there the hyperbola x2 y2 1 in R2 is best parametrized by rlttgt cos M 5 34h tgt 6 MTH234 Chapter 13 Vector Functions Michigan State University 8 Suppose I want to nd vector function that represents the curve of intersection of the surfaces 2 taj2 32 and z 1 3 Which parametrization is best to start With and Why r f cos t sin 25 gt 1 b r05 tm 2 m jars tgt 139 paint we E lfL M s li Lar ng X quotta445 i 39Ug L Ngl tl gt in 3 1 L t I 3914 z 3 39Equoti a H U Tips to parametrizing in 3D a If if 00 and Z og that is the curve passes the X K Vaniiqu Flaw 39QS39F then start with the parametrization g 39 f gt b If X 4quot FUQ and Z Q that is the curve passes the 3 5 V64 CA quot own 1125 then start with the parametrization 73 lt k 9 c If and 3 that is the curve passes the 2 K P 39qnn then start With the parametrization lthl d If the curve is d ZA starts and stops at same point then consider a parametrization with Sin L X Cost init MTH234 Chapter 13 Vector Functions Michigan State University 9 Find a vector function that represents the curve of intersection of the two surfaces 3 422 x2 and as 22 391 lsagmh m ol ikx X is a fumeHM 0 2 0 3 is a funoj m of jws l Z usin llw alcove ii 39 f lt 11tgt 39Ez t x cquot 39 quotIiir faj 3 thlr 39tq r ltgt Gristquot D J e IR 10 Find a vector function that represents the curve of intersection of the two surfaces the semi ellipsoid x2 32 422 4 y 2 0 and the cylinder 52 22 1 lt CAS EI Aquot Simegt CosttH L sML 39I 5 q cos f 4 911 Ll t3 s r 4m ref ltcas lces t 4snquot t sint gt 8 t6 chug MTH234 Chapter 13 Vector Functions Michigan State University 2 Derivatives and Integrals of Vector Functions De nitions 2 1 1 If rt ltftgt Where fg and h are differentiable functions then r39lttgt lt f m 06 31Ll c3gt is the derivative of r 2 If rt ltftgt Where fg and h are integrable functions then Abra dt lt SHEBA g 84le gilt is the de nite integral of r Technically these should be de ned as a 1 r HL c Cir hi 39 h b n a amp rt dtn1Lngoz PH AE 1 i1 However in this class we will not focus too much on these mathematically rigorous de nitions r t and Theorem 22 Suppose u and v are differentiable vector functions 6 is a scalar and f is a real valued function Then 4 I 1 utvt 139 cl f 9 2 cut c uff lr 3 ftut l3 CH 46E 139 Jr 4 110 vlttgt 1 79c 361931 5 110 gtlt vlttgt1 33 9 Kid T Xl l gtlt INCH 11009 BURCH 43 Q MTH234 Chapter 13 Vector Functions Michigan State University Now to clear my conscious I have included a proof of 3 The book also proves formula 4 Proof of 3 d d f Wlttgtulttgt1 a flttgtltu1lttgtu2lttgtu3lttgtgt1 De Mon 0 u f 73114057 f tu2 t ftu3 De nition of scalar multiplication ftu1t ftu2t ftu3tgt De nition of vector derivative ltf tU1t ftu391ta f tU2t ftu392ta f tU3t ftu tgt Scalar function product rule from MTH 132 f 0101 t f 0102 t f 0713 75 f 0714 t f u2 t f 010 75 Vector addition ftltu1ta 10215 30 Ht W3 75 71205 7150 Scalar multiplication f tut ftu t De nition of ut and u t Cale 1 Theorem 1 f a is the slope of f1 at a U mam x 3 2 L1 f a1 a fa is the linearization of f1 at a aka it is the tangent line to f1 through 1 a Theorem 23 1 a is the direction or QHW Veg OP of rt at a 3 2 Lt riCmt CR 39139 a is the linearization of rt at a aka it is the tangent line to rt through 75 a l Example 24 5 I Q J Find parametric equations for the tangent line to the curve rt he 15225 752 through the point 0 10 Q 2 O rte 3 lt011J0gt I F Lox flL LG ltl IJ L e ltOII Jo gtlt c gz fI39l 8Qt 2 I MTH234 Chapter 13 Vector Functions Michigan State University De niti0ns 25 The LV qhyvx l VacI Dr is de ned to be r m r ltil Tt Example 26 Find the unit tangent vector of rt tee ti arctan tj Zetk when t 0 r lb l 0 Ir DDI We lt15 giro Pitt J QJgt l0lt1o 1QgtltlZgt r ol 1 L J TCO ltlll zgt VE MTH234 Chapter 13 Vector Functions Michigan State University Group Work 0 ro 1 I 16 div lt4 2 rm lt0 42gt UV 1 Consider rt t 2752 1 a Sketch the curve z El ixZ ta X VQY Fi b Findr t 3 lt I r01 lt3 3 Jam mom a c Sketch the position vector rt and the tangent vector r 75 when t 1 on the same graph as 1 Find a vector equation of the tangent line of rt through 75 1 14 m ta rm r 3 3 VI lt J 2 Lea lt2 0 lt3gt t TR MTH234 Chapter 13 Vector Functions Michigan State University 2 Consider rt sint 200815 a Sketch the curve Frm39i lma quAe ifi act39sm we cow 22 iquot is cm ipseh CsnsteLnr lo do lt07 7 s z9 70713 3 1 0 ts w e rm 0 433129 VPW L 505 Swilng us 1 m W b Find r t rick 3 ltLosE 15i gt r39 a fz agt r 130 FLm D 3 Sketch the position vector rt and the tangent vector r t when t 2 7T 4 on the same graph as a d Find a vector equation of the tangent line of rt through 75 7r 4 L lt gt gt lt gt MTH234 Chapter 13 Vector Functions Michigan State University 3 Find parametric equations for the tangent line to the curve rt 75 cost t tsint through the point 7T 7T 0 a I r00 lt1YquotNJO7 VIC l o I I 6 NH IJ IITYgt FIH39 lt 1 C05quot quot39 quot Skirt1 isiq c 39 Jcos t LH lt 11 J 17 1313 X 40 an 2 39t39 W 39Irm t 3 T 4 Evaluate the integrals quot Wk 1lt1 a 2t2tt 1tsin7rtgt dt gm Pl t 2 fat Eris3 3 I s I 1 Ut l S 39 3 Au239 QUEj Q L l gt I J i c t z Sui l M iv our k u So u 1 a 5 3 O 5 3 vinylH quot u r l la N SSm ampvggb J 1 QoS39lf T gosh 0 v 3 7cosu c a 7 Q So i Lut gre J t rF Jsin gt g 393 quot54 gt I b sec2 ttt2 13t2lntgt dt 156 i cw t l 1i quot a 8 Move Lam 5 Au quot 139 t utaint aMJFkt 5h 3 hf 2 xv14 vgt33gt 117 4 a 1 1quot3quot 2 9 3th Frat 03 1fo lt vhL 1f gt c MTH234 Chapter 13 Vector Functions Michigan State University 4 Motion in Space Velocity and Acceleration Note that we will be skipping Tangential and Normal Components of Acceleration and Kepler Laws of Planetary Motion J For this section we will take 639 i to be the position vector of a particle moving through space at time t De nitions 4 1 1 The IQlGCJl39gr PLQ39l Dr vt at time t is given by vlttgt quot3 2 The speed of a particle at time t is given by m 3 The eml m of the particle is de ned as alttgt 71 H3 Newton s Second Law of Motion 1 The vector version of Newton s second law states that F0 m U 2 In this class we will do many 2 dimensional problems with the force of gravity That is with Ft lt0 M or equivalently q aw lt0 gt Note that typical value of g on earth are 98m82 10ms2 for nice calculations or 32 ft 82 MTH234 Chapter 13 Vector Functions Michigan State University 60 j x leo a H97 5 41 5 L75 Example 42 3 05 SO 53 47 A projectile is red with muzzle speed 150m s at an a gle of elevation of 45 from a position 10m above ground level 1 Find atvt and rt OJ IDgt pltoJIogtf E vc3 qd 040 49 E IO Z a ltOJIDgttlt7s J7S gt 3 DJ6 E ltOJ395 1lt7S J7SEgtk 1 lt010gt 2 What is the projectile s maximum height S 19 55 t 754 25 43 S 7 r S L39LSgt 0 ms 16 n 45 39 M 7 139 L 3 Where does the projectile hit the ground D t i Q 1 1 94 7354 HO 0 3965 4qu ELS lz l 0 Jquot T 39I k requot 39 e39 GVSLJ exI S ltx4z 10gt X ltISEISE j a nogt 55 2m 7917375 J1gt 4 What is the projectile s speed on impact VG A z Us gt sbsm 37s MTH234 Chapter 13 Vector Functions Michigan State University The book includes the following theorem Theorem 43 Assume that oz is the angle of elevation and that v0 210 is the initial speed If we consider the initial position to be r0 2 lt0 0 and assume that gravity is the only force acting on the projectile then we have that 1 Acceleration is at 0 ggt 2 Velocity is vt lt Uo lt64 I 00 sh lt 7 3 Position is rt 4 Do 05 165 quot gt 39 1155 390 5 10 4 Horizontal distance range is 0amp2 WhiCh OCCUI S at t 3 q m 1 J 39 39 410 S m at 1 5 which occurs at t 2 It may be very tempting to use these to solve WeBWorK problems However they are very dif cult to memorize and only 5 Maximum height is apply is select situations Instead you should know how these are derived much easier and logical to memorize and will help you do well on quizzes and exams For the following please use the notation rt ltat Ideas of Proofs 39 came 1 Given by statement Only force acting on projectile isactfl EtTdn 2 Integrate at Use v0 lt Uoc39osd 03quot to solve for constant of integration 3 Integrate vt Use r0 lt0 0 Z to solve for constant of integration 4 Solve 0 for t plug into U 5 Solve fk for t plug into MTH234 Chapter 13 Vector Functions Michigan State University Group Work 1 Find the velocity accelerations and speed of the particle with position function rt 75 t2 2 ValveKw 7 3 lt1 gt Ancientst Time mt lt0 19gt 392 mm l LH 2 Find the velocity and position vectors of a particle that has the given conditions a75 120 V0 001 r1 120 mg g Timers 11 0 3 me ltoo o lto oogt 8 gt 3 400 5 OH that b 30 Six mou Sltt1tiol E fi gt E Fmltuaogtlt 39s gtZgt 2 lt t gt 39r lt gtlt Igt MTH234 Chapter 13 Vector Functions Michigan State University 3 When is the speed of rt 752 575 t2 1675 a minimum V lr lta911al gt tl1 5a i t Smu E is cm ivxcms ma Fxmc awl t is Minimi Lc39x W 81 l5 minimil bt stA 20 So msdu mim wi lvxca 1amp3 qt is at IL 167 WM MaiIOU O 3tt 39 C4 GLl l ft quotl 3 39l Clucko WARE Minimum 1 V j 7 So Lap S 4 c39S Min 4 A ball is thrown at an angle of 45 to the ground If the ball lands 30m away what was the initial speed of the ball I J Vo lt vlltns 5 l5quot 5gt ltM E M E a t Orcp Va Ram 09 tr 396 W0 ltw 39fwl 398 VG o t lvld giwl rltgt vcm ltogt IltM Mgtt 5 39ow39l Shae PLO 8 r9 2 2 8 g Va lt01gt lt VI V t weka rblt30 03 f or w 90 C5 qu us 4 co 30 V391 E I m 1 60 O 33 N f g Q c639 v gt a FEIV gt 303qugt lvlJ30 a MTH234 Chapter 13 Vector Functions Michigan State University 5 A ball is thrown eastward into the air from the origin in the direction of the positive ac axis The initial velocity is 50i 80k with speed measured in feet per second The spin of the ball results in a southward acceleration of 4 ft 82 so the acceleration vector is a 4j 32k Where does the ball land and with what speed NOTE We are in 3 3 arm s s Vargas ltoH31gt 395 Ioltsoofaogt E v00 0 H32gt r ltSocgtogt 1 a rm New lt0 2 Qt Solo t 391 r C 261 l t130t 0 i0 L 5 rosy o 1Isgt as ltSOJ gt 395 15 quot5 gt a lv53lllto k31gt5lt5 gtl IltSoJ 397OJ 160 lolt62Iegt l0 3 35 ll MTH234 Chapter 13 Vector Functions Michigan State University 3 Arc Length and Curvature Note that we are skipping Curvature and The Normal and Binormal Vectors portions of this section We will focus on length A De nitions 31 k A parametrization rt is called SMDol on an interval I if r t is Cbnlquot AOVquot and f 0 on I A curve is called smooth if it has a smooth parametrization We will use smooth parametrizations later in the course Now recall Old Theorems 1 Calc 1 Theorem If we have the velocity function vt r t notice this is not a vector function then the total distance traveled is given by b 192 wand 2 Calc 2 Theorem If a curve C is parameterized by 1t then the length of C denoted LC is given by the g formula b M0 Macm2 y39lttgtgt2dt A 3 Calc 2 Theorem REVISED If a curve C is parameterized by lt I A39L gtthen the length of C denoted LC is given by the formula 3 LC Sq l r 0 I Now it s your turn to make up a Length formula for a 3 dimensional curve Theorem 32 If a curve C is given by the vector equation rt 1ti yt j ztk then the length of C denoted LC is given by M0 WEE l H s Lx39a r W39 21 the formula MTH234 Chapter 13 Vector Functions Michigan State University Example 33 Find the arc length of rt cos t sin 75 225 from t 0 to t 5 5 5 4 lt sw l391u5EJLI t Sghtt cogf Ll cos5s1n5 10 39D O Z Jzit SJE 49 SE So we want to realize that we can treat I as a variable to get an arc length function that we will call 5 L for now Theorem 34 Arc Length Function with Base Point a la 4 55 a lrrCO l Most people don t like to use b as a variable so lets rename a few things to get our variable to be t Theorem 35 Arc Length Function with Base Point a J slttgt Sq lr ul ampK Example 36 Find the arc length parametrization of the curve rt 2 cost 2 sin 75 375 with base point 7T 1 lt 1Sivm 3u50 3gtl A T t l L15n4ul39iltokcl elk 1 43 Tmmltm gt 1 WT MTH234 Chapter 13 Vector Functions Michigan State University Group Work 1 Find the length of the curve 3 a rt lt2tt2 t 6 01 ltaat t gt 4 J7 LIF 9 4 J were 4 avch an 15312 has g 0 b rt costsint1ncostgt t E Om4 1r 8 ltquotS m b COS E Co st Sam R34 5t o v tf KW l mu 4 EWJSTZ CH 0 Sm Sad 0H 1nl8u lt39tqn tq O Crtit2jt3k 032531 ln V I lquot rz0 llt l 3 c gtd SLWOH Hut w OH H I ml39 ifh 3 17 o 77 MTH234 Chapter 13 Vector Functions Michigan State University 2 Consider the vector function r t cos t sin t t 1 a Find the arc length function with the initial point 101 l 1 O is Fain 5W 3 Ks39mucos3 an JFsinWuhcosleuyrl elk a 2 J3 b Find the time in Which a particle has traveled 7 units along the curve from 1 0 1 in the positive direction 7 HE C Find the location of the particle in ltCos Siw gt lgt MTH234 Chapter 13 Vector Functions Michigan State University 22 O 3 Parametrize the curve of intersection between the cone 2392 512 32 and 512 32 9 Sketch a picture to see What the curve looks like 2 22 e2 23 X1 least 333 11 1 e offs 4 Guess the length of the curve in problem 3 Use the arc length formula to check C hr EN 8 llt 3 quot t3 0gtI At 9 4 l 153 1 0 cos f all 8 3 M 3m 3
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