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by: Reyes Glover


Marketplace > University of Texas at Austin > Mathematics (M) > M 408D > SEQ SERIES AND MULTIVAR CALC
Reyes Glover
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This 75 page Class Notes was uploaded by Reyes Glover on Sunday September 6, 2015. The Class Notes belongs to M 408D at University of Texas at Austin taught by Staff in Fall. Since its upload, it has received 17 views. For similar materials see /class/181460/m-408d-university-of-texas-at-austin in Mathematics (M) at University of Texas at Austin.

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Date Created: 09/06/15
CALCULUS 11 Three Dimensional Space Paul Dawkins Calculus H Table of Contents Profarp ii Three 1quot 39 anr9 3 T 1 3 The 3D Coordinate Sv tem 5 Equations ofT ine 11 Equations of Planes 17 Quadric Surfaces 20 Functions of Several Variable 26 Vector Functions 33 Calculus with Vector Functions 42 Tangent Normal and Binormal Vectors 45 Arc Length with Vector Function 49 Curvature 52 Velocity and t39 54 Cylindrical Coordinate 57 Spherical Coordinates 59 2007 Paul Dawkins i httptutorialmathlamaredutermsaspx Calculus II I Preface Here are my online notes for my Calculus II course that I teach here at Lamar University Despite the fact that these are my class notes they should be accessible to anyone wanting to learn Calculus II or needing a refresher in some of the topics from the class These notes do assume that the reader has a good working knowledge of Calculus I topics including limits derivatives and basic integration and integration by substitution Calculus II tends to be a very dif cult course for many students There are many reasons for this The rst reason is that this course does require that you have a very good working knowledge of Calculus I The Calculus I portion of many of the problems tends to be skipped and left to the student to verify or ll in the details If you don t have good Calculus I skills and you are constantly getting stuck on the Calculus I portion of the problem you will nd this course very difficult to complete The second and probably larger reason many students have difficulty with Calculus II is that you will be asked to truly think in this class That is not meant to insult anyone it is simply an acknowledgment that you can t just memorize a bunch of formulas and expect to pass the course as you can do in many math classes There are formulas in this class that you will need to know but they tend to be fairly general You will need to understand them how they work and more importantly whether they can be used or not As an example the rst topic we will look at is Integration by Parts The integration by parts formula is very easy to remember However just because you ve got it memorized doesn t mean that you can use it You ll need to be able to look at an integral and realize that integration by parts can be used which isn t always obvious and then decide which portions of the integral correspond to the parts in the formula again not always obvious Finally many of the problems in this course will have multiple solution techniques and so you ll need to be able to identify all the possible techniques and then decide which will be the easiest technique to use So with all that out of the way let me also get a couple of warnings out of the way to my students who may be here to get a copy of what happened on a day that you missed Because I wanted to make this a fairly complete set of notes for anyone wanting to learn calculus I have included some material thatI do not usually have time to cover in class and because this changes from semester to semester it is not noted here You will need to nd one of your fellow class mates to see if there is something in these notes that wasn t covered in class N In general I try to work problems in class that are different 39om my notes However with Calculus II many of the problems are dif cult to make up on the spur of the moment and so in this class my class work will follow these notes fairly close as far as worked problems go With that being said I will on occasion work problems off the top of my head when I can to provide more examples than just those in my notes Also I often 2007 Paul Dawkins ii htggtutorialmathlamarcdutermsaspx Calculus II don t have time in class to work all of the problems in the notes and so you will nd that some sections contain problems that weren t worked in class due to time restrictions LN Sometimes questions in class will lead down paths that are not covered here I try to anticipate as many of the questions as possible in writing these up but the reality is thatI can t anticipate all the questions Sometimes a very good question gets asked in class that leads to insights that I ve not included here You should always talk to someone who was in class on the day you missed and compare these notes to their notes and see what the differences are 5 This is somewhat related to the previous three items but is important enough to merit its own item THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS Using these notes as a substitute for class is liable to get you in trouble As already noted not everything in these notes is covered in class and often material or insights not in these notes is covered in class I Three Dimensional Space Introduction In this chapter we will start taking a more detailed look at three dimensional space 3D space or R3 This is a very important topic for Calculus III since a good portion of Calculus III is done in three or higher dimensional space We will be looking at the equations of graphs in 3D space as well as vector valued functions and how we do calculus with them We will also be taking a look at a couple of new coordinate systems for 3D space This is the only chapter that exists in two places in my notes WhenI originally wrote these notes all of these topics were covered in Calculus II however we have since moved several of them into Calculus III So rather than split the chapter up Ihave kept it in the Calculus H notes and also put a copy in the Calculus III notes lVIany of the sections not covered in Calculus III will be used on occasion there anyway and so they serve as a quick reference for when we need them Here is a list of topics in this chapter The 3 D Coordinate SVstem 7 We will introduce the concepts and notation for the three dimensional coordinate system in this section Eguations of Line 7 In this section we will develop the various forms for the equation of lines in three dimensional space Eguations of Planes 7 Here we will develop the equation of a plane Quadric Surface 7 In this section we will be looking at some examples of quadric surfaces 2007 Paul Dawkins iii htgptutorialmathlamarcdutermsaspx Calculus H Functions of Several Variables 7 A quick review of some important topics about functions of several variables Vector Functions 7 We introduce the concept of vector functions in this section We concentrate primarily on curves in three dimensional space We will however touch brie y on surfaces as well Calculus with Vector Functions 7 Here we will take a quick look at limits derivatives and integrals with vector functions Tangent Normal and Binorm al Vectors 7 We will de ne the tangent normal and binormal vectors in this section Arc Length with Vector Functions 7In this section we will nd the arc length of a vector function Curvature 7 We will determine the curvature of a function in this section Velocity and Acceleration 7In this section we will revisit a standard application of derivatives We will look at the velocity and acceleration of an object whose position function is given by a vector function Cylindrical Coordinates 7 We will de ne the cylindrical coordinate system in this section The cylindrical coordinate system is an alternate coordinate system for the three dimensional coordinate system Spherical Coordinates 7 In this section we will define the spherical coordinate system The spherical coordinate system is yet another alternate coordinate system for the three dimensional coordinate system 2007 Paul Dawkins 4 htggtutorialmathlamarcdutermsaspx Calcu us We39ll cuurddnate sysLems can ebange the gmph ufan equatmn m3 m2 A L y n A s den ted by m Alsu as yuu mgqt have guessed men a general n dmmsdnd cuurdmate sysLEm ds u en deduced by mquot Nan let39s take a dudek luuk at the has cuurdmate sysLEm z as needed space Lheputh WeszythathxtsmLhexgrplane Thexyrplanecun39ewundstu allthepumtswhmh xrcuurdmate ufzem cbueedyely Lhexy xz andyzrplznes are sbmedmes called the cuurdmateplznes lnthe make sure Lhatyuu can Alsud1eputhxs u enrefenedtu asLhe pmje lun umethexyrplane Likewxsa Rxs the prujechun ufP m theyzrplzne and sds the pm eean ume the xzrplane m m3 FunnsLance the distance between tvm pmnts m m2 s gven by mm mm DavInns 5 buy mutanal mudme edsmmsdw Calcu us ma xrxa wmf thethe m ancebetwem any twu pmnts m m3 15 gvm by 2 2 2 M332 xrxx 022 zrzx Mk w hJc xihf ysk and the general equanunfuraspha39ethh center Mm andradxusns gwen by xrhzykzzrlzr 2 mm m extent Cunslder the full uwmg Example Examplel Graph p3 m m m2 and m3 Salu an In R 7 A In W the equauun x 3 tells us m gzph all the pmntsthat are m the farm 3y Ths s a vemcal lme m a 271 cuurdmate system In In the equanun x 3 tellsustu gaph all Lhepumts that are m the farm 3yz Ifyuu gm excepthsnmewehzve 3 x s ad 5 m an cuurdmate sysLEm thsxs aplane that Wm be parallel m theyzsplane and pass mmugh the was at x 3 Here sthegaphufx3 mm 71 n 1 2 3 Haexsthegaphufx3 m m2 mm mm DavIons 5 my mum mathlamar enhuxmsayx Calcu us many here 15 me gaph uf x 3 m m3 Nuts that we39ve presemed Lhs gaph m twu mffaent s 1es 39 and st we39ll u m anusst m gaphs and sketches 7mg 2 2 E z 71 2 es 5 A x wen usmgths sales 2 0 Weplane y 0 neplme x 0 yzeplane Let39s take smek at a shgmy mure general example znm m DavInns my mutanal mathlamar eewmms aw Calcu us Exme Graph y2x3 m m2 and m3 Solution we can39tbem 2171 Space In W thsxs alme wnh slupe z and aymtacept em Huwever m mzthsxsnutnecessanly alme Because wehzvenut speci ed avalue ufzwe are ths 1an Sn the graph 15 men avemcal plane Lhathes uverthe lme gwen by y 2x7 3 m the Xyrplane Harms me gaph m m2 here is the gaph m m3 422 1H4 m m2 mm mm DavIons 2 m muzaml mathlamar eonum x Calcu us uurdmate systems Example3 Graph y24 mm2 and m3 Solution In m2 LhIs Is a nrcle cantered at the ungm wnhradms 2 In In huwever as wmn the prevmus example LhIs may Dr may nut be a nrcle sInse we have nut museum any Way we must assume thatz can take un anyvalue In Dunes vmrds at any Here are the graphs In LhIs example N h m I I mmle 39nm m2 mm m Dawhns 9 hug mutanal munmnn sununnst Calculus H We need to be careful with the last two examples It would be tempting to take the results of these and say that we can t graph lines or circles in R3 and yet that doesn t really make sense There is no reason for there to not be graphs lines or circles in R3 Let s think about the example of the circle To graph a circle in R3 we would need to do something like x2 y2 4 at Z 5 This would be a circle of radius 2 centered on the Zaxis at the level of Z 5 So as long as we specify a Z we will get a circle and not a cylinder We will see an easier way to specify circles in a later section We could do the same thing with the line from the second example However we will be looking at line in more generality in the next section and so we ll see a better way to deal with lines in R3 there The point of the examples in this section is to make sure that we are being careful with graphing equations and making sure that we always remember which coordinate system that we are in Another quickpoint to make here is that as we ve seen in the above examples many graphs of equations in R3 are surfaces That doesn t mean that we can t graph curves in R3 We can and will graph curves in R3 as well as we ll see later in this chapter 2007 Paul Dawkins lO htggtutorialmathlamarcdutermsaspx Calculus II Equations ofLines In this section we need to take a look at the equation of a line in R3 As we saw in the previous section the equation y mx b does not describe a line in R3 instead it describes a plane This doesn t mean however that we can t write down an equation for a line in 3D space We re just going to need a new way of writing down the equation of a curve So before we get into the equations of lines we rst need to brie y look at vector functions We re going to take a more in depth look at vector functions later At this point all that we need to worry about is notational issues and how they can be used to give the equation of a curve The best way to get an idea of what a vector function is and what its graph looks like is to look at an example So consider the following vector function t t 1 Avector function is a function that takes one or more variables one in this case and returns a vector Note as well that a vector function can be a function of two or more variables However in those cases the graph may no longer be a curve in space The vector that the function gives can be a vector in whatever dimension we need it to be In the example above it returns a vector in R2 When we get to the real subject of this section equations of lines we ll be using a vector function that returns a vector in R3 Now we want to determine the graph of the vector function above In order to nd the graph of our function we ll think of the vector that the vector function returns as a position vector for points on the graph Recall that a position vector say 7 a b is a vector that starts at the origin and ends at the point a b So to get the graph of a vector function all we need to do is plug in some values of the variable and then plot the point that corresponds to each position vector we get out of the function and play connect the dots Here are some evaluations for our example 331gt 1L1 2lt21gt 551gt So each of these are position vectors representing points on the graph of our vector function 61mm 34 lt ugt 21 51 are all points that lie on the graph of our vector function If we do some more evaluations and plot all the points we get the following sketch 2007 Paul Dawkins l l htggtutorialmathlamarcdutermsaspx Calcu us well as the shave each pm we u sad fur each evaluanun meckshke m Lhs casethe gaph ufthevecturequanunxsmfad Lhelme y 1 Ham39s mutherqulck Example Harms the graph uf 71 6cosz3smz y hunh rank ufvecmfuncuuh dc nuthzvetu he lmes as the example abuve shuws vecm fuhchch Imagne that apm lpm 15 attached in the md ufthe pushun vecm and as we sketches nut the curve funhe vecm funcuuh equahun ufalmem m3 and as suggested bythewurk abuvewewxll heed ave ths Tu see huw 1 39 my funchuntu he he gumgtu ac Lhs at s Lhmk abuutwhat Weneedtu wme dawn the mm m thhms 12 buy mutanal manhme ecummshw Calcu us equatlun ufalmem m2 Intvm dxmehsmhs we need the 51an m and apumtthatvas unthe hhe m urdertu Wnte duwnthe equanun h m3 slupe dmmduhd slupe sh we need sume umg that Wm alluqu m desmbe a du39e lun Lhath putanhally m three dxmehsmhs We already have a quantity that wdl de he furus Veems ave remuns and caubethree dmehsmhd abjects E Knynznzndthat7 155xssumevedurthatxspzrallelEthelan Nuts1nal hkehhuud a mllhmheuhmehhenself Weunlyneed v tubeparalleltuthelme FmallyJet P x W he anypumtunthelme Nuw smee uur slupequot s avedurlet39s 315 represent the tvm pmnts eh Lhelme asveems wen dd 5 wnh pushed vecturs sh let 39n and 7 he the padded vecturs fur p and p mwecnvely Alsu furnu apparehneasuh let39s dedhe Em bathe vecturthhrepresemauun ETP We nuwhave the fulluwmg sketch wnh all these pmhcs and vecturs eh w z veemr w anh mm n WW n mthehhe Nan hunee that we can wme 7 as fuuuws anthmencsecnun Nuw nuncethatthevecturs a and 3 arepzra el Tha39efuretha39exsa number I sueh that mm mm DavInns u buy mutanal mudme edmemsew Calcu us a 15 We huw have Knynzngttltabcgt khuwmsthet Nuheethat 1 7 WAllbeavemurthathesalungthelmeandxttellsushuwfzr hum Ifhsp the mheehuh hf t3 hght m an sketeh aha m is hegahve we m u e uhghal pmht t th uppusttemeetauhufs le muursketch Ashantesuverallpussxblevalues 11 wnh the vectur mm and an a shght rewnte 7ltxnynzntltabc xyzgt xn tayn 2zn 2 The my way furtwu veeth tn he equal ls fur the eumpuhehtstu he equal 1h uthehwuhaa xxnta yyntb zzntc Ths set hf equanuns is called the pmmmic fm39m nthe equau39m nra line Nance as well that e 14 httE mutanal mathxamae enhuxmsayx 2nn7 Pam Dawhms Calculus II To get a point on the line all we do is pick a t and plug into either form of the line In the vector form of the line we get a position vector for the point and in the parametric form we get the actual coordinates of the point There is one more form of the line that we want to look at Ifwe assume that a b and c are all nonzero numbers we can solve each of the equations in the parametric form of the line for t We can then set all of them equal to each other since 1 will be the same number in each Doing this gives the following This is called the symmetric equations of the line If one of a b or c does happen to be zero we can still write down the symmetric equations To see this let s suppose that b 0 In this case 1 will not exist in the parametric equation for y and so we will only solve the parametric equations for x and Z for t We then set those equal and acknowledge the parametric equation for y as follows x x0 Z 20 y 2 yo a 6 Let s take a look at an example Example 1 Write down the equation of the line that passes through the points 2 l 3 and l 4 3 Write down all three forms of the equation of the line Solution To do this we need the vector 7 that will be parallel to the line This can be any vector as long as it s parallel to the line In general 7 won t lie on the line itself However in this case it will All we need to do is let 7 be the vector that starts at the second point and ends at the first point Since these two points are on the line the vector between them will also lie on the line and will hence be parallel to the line So a 1 56 Note that the order of the points was chosen to reduce the number of minus signs in the vector We could just have easily gone the other way Once we ve got 7 there really isn t anything else to do To use the vector form we ll need a point on the line We ve got two and so we can use either one We ll use the first point Here is the vector form of the line 2 13gttlt1 56 2t 1 5t36tgt Once we have this equation the other two forms follow Here are the parametric equations of the line 2007 Paul Dawkins 15 htggtutorialmathlamarcdutermsaspx Calculus H x2t y 1 5t z36t Here is the symmetric form x 2 y 1 z 3 1 5 6 Example 2 Determine if the line that passes through the point 0 3 8 and is parallel to the line given by x 10 3t y 121 and Z 3 t passes through the xZplane If it does give the coordinates of that point Solution To answer this we will rst need to write down the equation of the line We know a point on the line and just need a parallel vector We know that the new line must be parallel to the line given by the parametric equations in the problem statement That means that any vector that is parallel to the given line must also be parallel to the new line Now recall that in the parametric form of the line the numbers multiplied by t are the components of the vector that is parallel to the line Therefore the vector 6 312 1 is parallel to the given line and so must also be parallel to the new line The equation of new line is then 0 38t312 13t 312t8 t Ifthis line passes through the xZplane then we know that the y coordinate of that point must be zero So let s set the y component of the equation equal to zero and see if we can solve for I If we can this will give the value of tfor which the point will pass through the xZplane 312t0 3 t1 4 So the line does pass through the xZplane To get the complete coordinates of the point all we need to do is plug 1 2 into any of the equations We ll use the vector form Flt3l3128gtG l Recall that this vector is the position vector for the point on the line and so the coordinates of the 3 3 1 point where the line will pass through the xZplane are 2 0 2007 Paul Dawkins 16 htggtutorialmathlamarcdutermsaspx Calcu us h the rst semun enhs chapter we saw a cuuple uf equanuns ufplznes Huwever hehe uf luuk atm twu demehsmhs We vmuldhke emehe general eduaheh furplznes Sulet39s start by assuming Lhatwe knuw apmntthatxs eh heplahe g e szn Let39s alsu suppusethatwehzve avemurthahs unhugunal papendxmlzr utheplzne a ltabc The Enemy smee we are gumg m he whdhhg wnh deems mmally we39ll let a and 7 he the peddeh deems fur Pa and p respemvely Harms asketch ufall theseveems Nehee that we added m the vectur 77 wheh wdl he cumpletely m the plane Alsu hehee that much the plane m any Way New because a 15 unhugunal m heplehe ms 215 enhegehet m any vecturthathesm the plane 1h pamwlant39s unhugunal m 77 Recall 39nm the Bat Pruduct seem that cwh uthugunal vecturs wdl have a de phedua efzem 1h utha wehds 2 u wn drea The 15 called the mm emuuh nf39hEplanE 2nn7 Pam dehhs 17 hug mutanal mathlamar edwmmsew Calculus H A slightly more useful form of the equations is as follows Start with the rst form of the vector equation and write down a vector for the difference ltabcgtltltxyzgt ltxoyo20gtgto ltabcgtoltx x0y y0z zogt0 Now actually compute the dot product to get ax x0by y0cz zo0 This is called the scalar equation of plane Often this will be written as ax by 02 d where d ax0 by0 020 This second form is often how we are given equations of planes Notice that if we are given the equation of a plane in this form we can quickly get a normal vector for the plane A normal vector is ltabcgt Let s work a couple of examples Example 1 Determine the equation of the plane that contains the points P 1 2 0 Q 314 and R 0 12 Solution In order to write down the equation of plane we need a point we ve got three so we re cool there and a normal vector We need to nd a normal vector Recall however that we saw how to do this in the Cross Product section We can form the following two vectors from the given points 17Q39234gt 17R 112 These two vectors will lie completely in the plane since we formed them from points that were in the plane Notice as well that there are many possible vectors to use here we just chose two of the possibilities Now we know that the cross product of two vectors will be orthogonal to both of these vectors Since both of these are in the plane any vector that is orthogonal to both of these will also be orthogonal to the plane Therefore we can use the cross product as the normal vector 139 j k 139 j 17Qx1TR 2 3 4 2 32f 8j395 1 1 2 1 1 The equation of the plane is then 2007 Paul Dawkins 18 htggtutorialmathlamarcdutermsaspx Calculus H 2x 1 8y25z 0 0 2x 8y52 18 We used P for the point but could have used any of the three points Example 2 Determine if the plane given by x 22 10 and the line given by 7 lt5 2 t 10 4tgt are orthogonal parallel or neither Solution This is not as difficult a problem as it may at rst appear to be We can pick off a vector that is normal to the plane This is 1 0 2 We can also get a vector that is parallel to the line This is V 0 1 4 Now if these two vectors are parallel then the line and the plane will be orthogonal If you think about it this makes some sense If and 7 are parallel then 7 is orthogonal to the plane but 7 is also parallel to the line So if the two vectors are parallel the line and plane will be orthogonal Let s check this I 139 j ma 1 0 2 1 0 2F4jl 6 4 0 1 So the vectors aren t parallel and so the plane and the line are not orthogonal Now let s check to see if the plane and line are parallel If the line is parallel to the plane then any vector parallel to the line will be orthogonal to the normal vector of the plane In 0 er words if and 7 are orthogonal then the line and the plane will be parallel Let s check this q70088 0 The two vectors aren t orthogonal and so the line and plane aren t parallel 4 So the line and the plane are neither 39 nor parallel 2007 Paul Dawkins l9 htggtutorialmathlamarcdutermsaspx ma Harms asketch ufatypmal eene Calcu us Quadric Surface quot3 and are 215 used fauly regularly and 5 we need m take aluuk at Lbuse me gena39al mm sz ny any equauun that can beputmtu f Azz8y2 ExzFyzGzHyIzJ0 whats A fare cunstants here s 21st ufsume quhemure eummun quadm surfaces Ellipsnirl Here 15 the general equanun uf an empsma 1 Harms asketch ufatypxcal empsma sgx 5 men wemuheve asphere centered un the ewe m ne my Dr muther me Here is the generat equauun uf a eene mm mm DavInns my mutanal mathlamar eemsms aw Calcu us mama w zua V 1 EM equanun Ths Wm be the case furthe ESL quhe surfaces that we39ll be luukmg at m Lhs secuun as well w m mquot the equanun Furmust quhe fulluwmg surfaces we Wm nut ave me utherpusslble furmulas We Wm huwsver surface Cylinda39 Here 15 the general equauun uf a cylmuer If a b secuun s a cucle wen be deang wnh Lhuse kmds ufcylmders mure than the general farm 5 me equanun uf a cylmaer wnh a cucula truss secuun 15 x2 y2 392 Here s a sketch uftypxcal cylinderthh an empse cmss secuun 2uu7 Pam DavIons 21 my mutanal mathlamar ecummsaw Calcu us equanun Be careful m nut cunfuse Lhs wnh a male In twu mmmsuns n s 2 m1 but m um mmmsuns ms 2 cylinder Hyperhnlnid nf One Sheet Harms the equanun ufahyperbulmd ufune sheet 2 2 2 mm mm DavIons 22 my mumml mathlamar gunmme Calcu us Hyperhnlnid nf Twn Shea Ha els the equauun ufahyperbulmd uftvm sheets 2 t t y z 7 7 a b 5 H3315 ssketeh ufatypxcal hypetheletd uftwu sheets 2 EY YA VAW yet 37 7 aura sheets ts the stghs th unt quhe vanshtes They are exac y the uppestte stgts Elliptic 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msuahze Recall that the eguanun ufa glans xs gven by axbycz d unfwe sulvedms furzwe can wme xtmtezms uffunmun nutanun Ths gves fxy Ax8yD p1 ans gwen by AM 1273x713 mm mm DavInns 25 my muzaml mathlamar eewmmsew cumust Furpurpuses uf gaphmg Lhs n wuuld pmbably be easier m Wnte Lhs as 212i3x74y 3x4yz12 Nuw r rm n x y 0 Sn the three mta sechun pmnts are New tn extend Lhs nu gapns nffuneunns ufthe fnnn wxyz wuuldbe fuur a c We next want tn talk abqu the dumzms uf funeunns ufmure than nne vanable Recall that y f V numberlme Dr une nmensmnel spaee T4 Vzxvyv 4 my real number mm mm DavIons 27 hug mutanal nemxemn eewmnsew Calcu us W M y JJ W cfxyln9ix279y2 W lu ml that we must require xy20 Harms asketch cme graph quhxsregun RemmtaPmblems 0quot x20 and yZU funmmn Herexsthesketchufthxsregun v RemmtaPmblems mm mm DavIons 22 my mumml mathlamar gunmme Calcu us 2 97x279y2gt0 y2lt1 Herels a sketch quhls regmn Remmta Pmblems Nuts that dumzms hf luhehuhs quhree vanahles w fxyz wlll he reguns m three hmehsluhsl space Example 2 Salu an ln ths case we have m deal wnh the square mut and alwsluh byzem lssues These wlll requlrE x2y2zzels gt0 x2y7zz gt16 The nEXttuplE that we shuuld luuk at ls that uflevel curves m cuntmlr curves The level curves quhe luhehuh z xy are cwh almehsmhsl euweswe getby semhg z k Whatele any number 5 the equafmns ufthelevel curves are NW k Nulelhel sumehmeslhe fxyz 0 arexyk0 If curves fur the luhehuh that glves the elevauun quhe 12nd lh that area Of cuurse we pmbably Let39s an aqmck Example ufthls mm m thlhhs 29 buy mmml mathlamar eelsmmshw Calcu us Example3 xaenufy melevel curves uf fxy lx2y2 Sketch afewufthem Solution FxrsL erase sake ufpramce let39s menafy whatthxs surface gvmby fxy 5 Tu an Lhs let39s man n as z y z xzy2 sewehaveaeene uratleastapumunufacune Smeewe we39veunly W half ufa eune surface and Lhs may eeme m handy dawn the mad me equanun are fuund by subsumung z k In the ease ufuur example Lhs is k 2 2 xzy2 kz Sn m Lhs ease ungm 2nn7 Pam DavIons an my mutanal mathlamar eewmmsaw Calcu us z fxy and Lheplzne z k The cuntuur Wm representthemtersechun ufthe surface and Lheplane Furfunmmns quhe farm fxyz wewm uncasmnallyluuk atlzval mrfzczs The equanuns uflevel surfaces are gven by New k Whereka any numba39 As nuted abuve we can Lhmk ufcuntuurs as the mta39secuun quhe surface gven by z f w and Lheplane z k Traces ufsurfaces are curvesLhatrepresemtbemtersemun quhe surface and Lheplanegvmby Fa Dr yb Let39s takeaqumkluuk at an Example uftxaces Exka Sketchthetxacesufxy 10sz2 7y furtheplzne x1 and y 2 Solution wen 5m vmb xl We can get an equanun furthetxaceby pluggmg x1 mm the equanun Dumg nus gves 1y1oe41t 2 z and Lhs Wm be gaphed m the plane gven by x 1 p1 mthxspart mm mm DavInns 31 buy mutanal mathlamar eewmmsew Calcu us 212 12 3 u 3 Kw Fm y 2 wewxll du pretty much the szmethmgthat we ma wnh the rstpart Herexsthe equahunufthetxace zx2 1074x222 2674x and here are the sketches fur Lhs case 212 212 3 In 3 4y mm mm DavIons 32 my mumml mathlamar eonterms aw Calculus II Vector Functions We rst saw vector functions back when we were looking at the Equation of Lines In that section we talked about them because we wrote down the equation of a line in R3 in terms of a vector function sometimes called a vector valued function In this section we want to look a little closer at them and we also want to look at some vector functions in R3 other than lines Avector function is a function that takes one or more variables and returns a vector We ll spend most of this section looking at vector functions of a single variable as most of the places where vector functions show up here will be vector functions of single variables We will however brie y look at vector functions of two variables at the end of this section Avector functions of a single variable in R2 and R3 have the form 7tltftagtgt itltftagtahtgt respectively where g I and 111 are called the component functions The main idea that we want to discuss in this section is that of graphing and identifying the graph given by a vector function Before we do that however we should talk brie y about the domain of a vector function The domain of a vector function is the set of all t s for which all the component functions are defined Example 1 Determine the domain of the following function t ltcostm4 tJt1gt Solution The rst component is defined for all t s The second component is only defined for t lt 4 The third component is only de ned for t Z 1 Putting all of these together gives the following omain 1 4 This is the largest possible interval for which all three 1 are de ned Let s now move into looking at the graph of vector functions In order to graph a vector function all we do is think of the vector returned by the vector function as a position vector for points on the graph Recall that a position vector say 7 a b c is a vector that starts at the origin and ends at the point a b 6 So in order to sketch the graph of a vector function all we need to do is plug in some values of t and then plot points that correspond to the resulting position vector we get out of the vector function Because it is a little easier to visualize things we ll start off by looking at graphs of vector functions in R2 2007 Paul Dawkins 33 htggtutorialmathlamarcdutermsaspx cumust m Saluuan h 71 123 710 7 Enlmmn Solution 2 71 11 0k Hare are a few 773e31 771711 7221 7551 721 in 21 51 Hare is a sketch quhs Vectur funeunn In Lhs sketch we39ve maluded many mure evaluauuns that Justthuse abuve Alsu nuts that we39ve Nuts In Lhs case n luuks hke we39ve gutthe mph ufthe lme y 1 RemmtaPmblems h7tltt1371017 Hare are a cuuple uf evaluanuns fur Lhs vectur funcuun 773lte310 7 16 gt 1 71 4004 73 W su m quotn w amk mm 2nn7 Pam DavInns 24 hug mutanal nemxenn enhuxmsayx Calcu us Bath quhe vectur funmmns m the abuve Example were m the farm 71 v E 1 y gm Let39s mph 2 uuple uf uLhEr vectur funmmns that an nut m1 mm um panem a we Salmm wnh thatsam here maths sketches ufeach ufthese mm mm DavIons 35 my mutanal mathlamar maxaxme Calcu us se m Lhs case xtluuks hke we39ve gut an empse Rem mm b 7zzezsmzzz Hare39s the sketch fianth Vedur funmzun Wins 3 veem funeuen we sketched m the prevmus example 7 mm mm DavInns 35 z6eosz3smz The reamazwe my mutanal mathlamar eewmms aw Calculus II got an ellipse here should not come as a surprise to you We know that the first component function gives the x coordinate and the second component function gives the y coordinates of the point that we graph If we strip these out to make this clear we get x6cost y3sint This should look familiar to you Back when we were looking at Parametric Equations we saw that this was nothing more than one of the sets of parametric equations that gave an ellipse This is an important idea in the study of vector functions Any vector function can be broken down into a set of parametric equations that represent the same graph In general the two dimensional vector function 71 lt f I g can be broken down into the parametric equations xft ygt Likewise a three dimensional vector function 71 lt f I g 1 111 can be broken down into the parametric equations xft ygt zht Do not get too excited about the fact that we re now looking at parametric equations in R3 They work in exactly the same manner as parametric equations in R2 which we re used to dealing with already The only difference is that we now have a third component Let s take a look at a couple of graphs of vector functions Example 4 Sketch the graph of the following vector function W 2 4t 15t3t Solution Notice that this is nothing more than a line It might help if we rewrite it a little m lt2 13gttlt 4 51 In this form we can see that this is the equation of a line that goes through the point 2 1 3 and is parallel to the vector 7 4 51 To graph this line all that we need to do is plot the point and then sketch in the parallel vector In order to get the sketch will assume that the vector is on the line and will start at the point in the line To sketch in the line all we do this is extend the parallel vector into a line Here is a sketch 2007 Paul Dawkins 37 htggtutorialmathlamarcdutermsaspx Calcu us am aswell FunnstauEE 71lt105mt 310costgt at willbeamrcleuframusl cmtereduntheyrzxxsand yr Inumavmrawlmgasm mm mm DavIons 32 my mumml mathlamar gunmme Calcu us Let39s take aluuk at amm canun ufdms FunnsLanEE 7 tltt6ost65mt s ahehx Lhatmtates aruund the was be mffaent we Wm get ellipses Furexzmple 71 905Lt25mt mm mm DavIons 39 my mumml mathlamar gunmme Calculus II will be a helix that rotates about the yaXis and is in the shape of an ellipse There is a nice formula that we should derive before moving onto vector functions of two variables Example 7 Determine the vector equation for the line segment starting at the point P xly1zl and ending atthe point Q x2y2zz Solution It is important to note here that we only want the equation of the line segment that starts at P and ends at Q We don t want any other portion of the line and we do want the direction of the line segment preserved as we increase I With all that said let s not worry about that and just find the vector equation of the line that passes through the two points Once we have this we will be able to get what we re after So we need a point on the line We ve got two and we will use P We need a vector that is parallel to the line and since we ve got two points we can nd the vector between them This vector will lie on the line and hence be parallel to the line Also let s remember that we want to preserve the starting and ending point of the line segment so let s construct the vector using the same orientation 7 ltx2 x1y2 y1z2 zlgt Using this vector and the pointP we get the following vector equation of the 1ine t x1y1zlgttltx2 x1y2 y1z2 zlgt While this is the vector equation of the line let s rewrite the equation slightly 17tx1y1zlgttltxzy2zzgt tltx1y1zlgt 1 tltx1y1zlgttltx2y2zzgt This is the equation of the line that contains the points P and Q We of course just want the line segment that starts at P and ends at Q We can get this by simply restricting the values of t Noticethat 0ltx1 y1 21gt 1ltx2 yzizzgt So if we restrict t to be between zero and one we will cover the line segment and we will start and end at the con ect point So the vector equation of the line segment that starts at P x1 y1 Z1 and ends at Q x2y2zz is tl tltxly1zlgttltx2y2zzgt OStSI 2007 Paul Dawkins 40 htggtutorialmathlamarcdutcrmsaspx Calculus H As noted brie y at the beginning of this section we can also have vector functions of two variables In these case the graphs of vector function of two variables are surfaces So to make sure that we don t forget that let s work an example with that as well Example 8 Identify the surface that is described by 17 x y x y x2 y2 Solution First notice that in this case the vector function will in fact be a function of two variables This will always be the case when we are using vector functions to represent surfaces To identify the surface let s go back to parametric equations x x y y z x2 y2 The rst two are really only acknowledging that we are picking x and y for free and then determining Z form our choices of these two The last equation is the one that we want We should recognize that function from the section on quadric surfaces The third equation is the equation of an elliptic r 39 loid and so the vector function represents an elliptic l 39 39 39 39 As a nal topic for this section let s generalize the idea from the previous example and note that given any function of one variable y f x or x hy or any function of two variables 2 g xyx g yz or y g xz we can always write down avector form ofthe equation For a function of one variable this will be xxifxj yhyy and for a function of two variables the vector form will be xyxiygxyk yzgyzyzk xz xgxzfzl depending upon the original form of the function For example the hyperbolic paraboloid y 2x2 522 can be written as the following vector function 7xzxz2x2 522jzlg This is a fairly important idea and we will be doing quite a bit of this kind of thing in Calculus 111 2007 Paul Dawkins 41 htggtutorialmathlamarcdutermsaspx Calculus H Calculus with Vector Functions In this section we need to talk brie y about limits derivatives and integrals of vector functions As you will see these behave in a fairly predictable manner We will be doing all of the work in R3 but we can naturally extend the formulaswork in this section to Rquot ie ndimensional space Let s start with limits Here is the limit of a vector function 131170lggltftagtahtgt 1112ftgtgt1i2gtgtali3hlttgtgt i31ftig1gtfii hrl 39 So all that we do is take the limit of each of the components functions and leave it as a vector Example 1 Compute F0 where 171 ltf w e2tgt Solution There really isn t all that much to do here 11mm ltlimt3limwlimeugt tgt1 tgt1 tgt1 1 tgt1 3 cos 3t 3 ltlim t3 lim lim ezlgt a1 tgt1 1 2 1 3 e2 Notice that we had to use L Hospital s Rule on the y l Now let s take care of derivatives and after seeing how limits work it shouldn t be too surprising that we have the following for derivatives 39t f tg39th39tgt f rg39tfh39tl Example 2 Compute 39t for 71 t6 lsinlt2tgt lnt 1I Solution There really isn t too much to this problem other than taking the derivatives 739t6t5200s2t Most of the basic facts that we know about derivatives still hold however just to make it clear here are some facts about derivatives of vector functions 2007 Paul Dawkins 42 htggtutorialmathlamarcdutermsaspx Calculus H Facts 7 397 oily cil fltrgtaltrgtgtfewerow 7 itquot x7il 739 gtlt7 39xv gtlt7 ltfltrgtgtr39ltrwltrltrgtgt There is also one quick de nition that we should get out of the way so that we can use it when we need to A smooth curve is any curve for which 39I is continuous and 71 72 0 for any 1 except possibly at the endpoints A helix is a smooth curve for example Finally we need to discuss integrals of vector functions Using both limits and derivatives as a guide it shouldn t be too surprising that we also have the following for integration for inde nite integrals It ltIftdtIgtdtIhtdtgtE jtjftdrjgtdr jjhtdr 13 and the following for definite integrals rawltJfltrgtdrrgltrgtdnlfhltrgtdrgt IffltrgtdrJffltrgtdrflfglttgtdtNW With the indefinite integrals we put in a constant of integration to make sure that it was clear that the constant in this case needs to be a vector instead of a regular constant Also for the de nite integrals we will sometimes write it as follows 17 I may 1 mm j gown Ihtdtgta Lbtdtjftdt 39gtdt J39htdt 1 In other words we will do the inde nite integral and then do the evaluation of the vector as a whole instead of on a component by component basis 17 a 2007 Paul Dawkins 43 htggtutorialmathlamarcdutermsaspx Calculus H Example 3 Compute 171 dt for 7t ltsint 6 4t Solution All we need to do is integrate each of the components and be done with it J39Ftdt lt cost6t2t2gt5 Example 4 Compute 7t dt for 7t ltsint 6 4t Solution In this case all that we need to do is reuse the result from the previous example and then do the evaluation 0170M lt cos 6t 2 cos 1 6 2 1 0 0 1 cos162 2007 Paul Dawkins 44 htgptutorialmathlamarcdutermsaspx Calculus H Tangent Normal and Binormal Vectors In this section we want to look at an application of derivatives for vector functions Actually there are a couple of applications but they all come back to needing the first one In the past we ve used the fact that the derivative of a function was the slope of the tangent line With vector functions we get exactly the same result with one exception Given the vector function I7 I we call 39I the tan gent vector provided it exists and provided 739 I 72 6 The tangent line to I7 I at P is then the line that passes through the point P and is parallel to the tangent vector 71 Note that we really do need to require 739 I 72 6 in order to have a tangent vector If we had 70 awe would have a vector that had no magnitude and so couldn t give us the direction of the tangent Also provided 71 72 6 the unit tangent vector to the curve is given by t t 39t T um While the components of the unit tangent vector can be somewhat messy on occasion there are times when we will need to use the unit tangent vector instead of the tangent vector Example 1 Find the general formula for the tangent vector and unit tangent vector to the curve given by 7t t2 2sint2costk Solution First by general formula we mean that we won t be plugging in a specific I and so we will be nding a formula that we can use at a later date if we d like to nd the tangent at any point on the curve With that said there really isn t all that much to do at this point other than to do the work Here is the tangent vector to the curve 39t 2tf2costf 2sintg To get the unit tangent vector we need the length of the tangent vector H70 l4t2 4cos2 t 4sin2t 4t2 4 The unit tangent vector is then 1 Tt J4t24 2 2t i 2cost 2sint I J4t24 V4t24 J4t24 2tf2costf 2sintlg 2007 Paul Dawkins 45 htggtutorialmathlamarcdutermsaspx Calculus H Example 2 Find the vector equation of the tangent line to the curve given by F0 2 f2sintf2005tl at t Solution First we need the tangent vector and since this is the function we were working with in the previous example we can just reuse the tangent vector from that example and plug in t 7 E 2 FZCOS E j Zsin E IgZ Tf J3 3 3 3 3 3 We ll also need the point on the line at t so 2 71 71 a F iJ3 39k 3 9 J The vector equation of the line is then tltZA lgttlt2 1 J gt Before moving on let s note a couple of things about the previous example First we could have used the unit tangent vector had we wanted to for the parallel vector However that would have made for a more complicated equation for the tangent line Second notice that we used 70 to represent the tangent line despite the fact that we used that as well for the function Do not get excited about that The 71 here is much like y is with normal functions With normal functions y is the generic letter that we used to represent functions and I7 1 tends to be used in the same way with vector functions Next we need to talk about the unit normal and the binorm al vectors The unit normal vector is de ned to be The unit normal is orthogonal or normal or perpendicular to the unit tangent vector and hence to the curve as well We ve already seen normal vectors when we were dealing with Equations of Planes They will show up with some regularity in several Calculus IH topics The de nition of the unit normal vector always seems a little mysterious when you rst see it It follows directly from the following fact Fact Suppose that 71 is a vector such that c for all t Then 71 is orthogonal to 71 2007 Paul Dawkins 46 htggtutorialmathlamarcdutermsaspx Calculus H To prove this fact is pretty simple From the fact statement and the relationship between the magnitude of a vector and the dot product we have the following 7ttHt 2 02 for allt Now because this is true for all t we can see that d a a 61 E0 tr E62 0 Also recalling the fact from the previous section about differentiating a dot product we see that 7t Vt 7t VtVt 739t WWW Or upon putting all this together we get 2 rr0 2 tFt0 Therefore 39t is orthogonal to 7l The de nition of the unit normal then falls directly from this Because T t is a unit vector we know that l for all t and hence by the Fact T39t is orthogonal to However because TO is tangent to the curve T39t must be orthogonal or normal to the curve as well and so be a normal vector for the curve All we need to do then is divide by to arrive at a unit normal vector Next is the binormal vector The binormal vector is de ned to be EtTrxNI Because the binormal vector is de ned to be the cross product of the unit tangent and unit normal vector we then know that the binormal vector is orthogonal to both the tangent vector and the normal vector Example 3 Find the normal and binormal vectors for 17 I t 3sin t 3 cos I Solution We rst need the unit tangent vector so first get the tangent vector and its magnitude 7390 1 3cos t 3 sin I 739t m The unit tangent vector is then Tt icost isint J J J5 1 The unit normal vector will now require the derivative of the unit tangent and its 2007 Paul Dawkins 47 htggtutorialmathlamarcdutermsaspx Calculus H f39t lt0 sint costgt 10 f39ltrgtJWg The unit normal vector is then NO lt0 sint costgt 0 sint cost Finally the binormal vector is Etftx t l 37W 7 39 f j 1 3 3 1 3 m Ecost smt J1 Ecost 0 sint cost 0 sint icosz tf Lsintlgcost j isin2 I J5 J5 J10 J5 3 a 1 a 1 i cost 39 sintk m m J6 2007 Paul Dawkins 48 htggtutorialmathlamaredutermsaspx Calculus H Arc Length with Vector Functions In this section we ll recast an old formula into terms of vector functions We want to determine the length of a vector function on the interval a S t S b We actually already know how to do this Recall that we can write the vector function into the parametric form xft ygt F110 Also recall that with two dimensional parametric curves the arc length is given by La trmfwgmzw There is a natural extension of this to three dimensions So the length of the curve I7 I on the L VUw nw nwvw There is a nice simpli cation that we can make for this Notice that the integrand the function we re integrating is nothing more than the magnitude of the tangent vector quot70 llf39tl2 g39tl2 h39tl2 Therefore the arc length can be written as interval a S t S b is F39tHdt L Let s work a quick example of this Example 1 Determine the length of the curve 71 lt2t 3sin2t 3 cos 2tgt on the interval 0 S t S 271 Solution We will first need the tangent vector and its magnitude 7t lt26cos2t 6sin2tgt H70 4 36cos2 2t36sin2 2 W 245 The length is then 2007 Paul Dawkins 49 htggtutorialmathlamarcdutermsaspx Calculus H L l I02 2J10 dt 471 F39tdt We need to take a quick look at another concept here We define the arc length function as slttgtIJHVltugtlldu Before we look at why this might be important let s work a quick example Example 2 Determine the arc length function for F0 Zr 3 sin 2t 3 cos 2t Solution From the previous example we know that m 2J5 The arc length function is then stJ2J du2 u 2l 0t l 0 Okay just why would we want to do this Well let s take the result of the example above and solve it for t s 2J5 Now taking this and plugging it into the original vector function and we can reparameterize the function into the form I7 For our function this is Fltrltsgtgtl sml l ml l So why would we want to do this Well with the reparameterization we can now tell where we are on the curve after we ve traveled a distance of 3 along the curve Note as well that we will start the measurement of distance from where we are at t 0 Example 3 Where on the curve I70 21 3 sin2t 3 cos 2t are we after traveling for a distance of nil 0 Solution To determine this we need the reparameterization which we have from above 2007 Paul Dawkins 50 htggtutorialmathlamarcdutermsaspx Calculus H Fltrltsgtgtlt 3sml l ml l 71 into this and we ll Then to determine where we are all that we need to do is plug in S wltaasm l3cos agtWad 3 get our location 10 So after traveling a distance of n 3 along the curve we are at the pomt 51 htggtutorialmathlamarcdutermsaspx 2007 Paul Dawkins Calculus H Curvature In this section we want to brie y discuss the curvature of a smooth curve recall that for a smooth curve we require 39t is continuous and 71 72 0 The curvature measures how fast a curve is changing direction at a given point There are several formulas for determining the curvature for a curve The formal definition of curvature is L ds where T is the unit tangent and s is the arc length Recall that we saw in a grevious section how to reparameterize a curve to get it into terms of the arc length In general the formal definition of the curvature is not easy to use so there are two alternate formulas that we can use Here they are Will 2 Hf39rgtxquotltrgtll Hml quot nmlr These may not be particularly easy to deal with either but at least we don t need to reparameterize the unit tangent Example 1 Determine the curvature for 17 I t 3 sin I 3 cos I Solution Back in the section when we introduced the tangent vector we computed the tangent and unit tangent vectors for this function These were 739 t 1 3 cost 3 sin I T0 L icost isint J10 quot10 J10 The derivative of the unit tangent is T39t lt0 isint icostgt mm The magnitudes of the two vectors are l9cos2 t9sin2t m The curvature is then 2007 Paul Dawkins 52 ht tutorialmathlamarcdutermsas x Calculus H In this case the curvature is constant This means that the curve is changing direction at the same rate at every point along it Recalling that this curve is a helix this result makes sense 2zquottk Example 2 Determine the curvature of I7 I 1 Solution In this case the second form of the curvature would probably be easiest Here are the first couple of derivatives F39t2tl t2 Next we need the cross product F f I F f 39txquott2t 0 1 2t 0 2 0 0 2 0 2 The magnitudes are 39IXquottll 2 Hm m The curvature at any value of t is then 2 K 3 2 42 1 There is a special case that we can look at here as well Suppose that we have a curve given by y f x and we want to find its curvature As we saw when we rst looked at vector functions we can write this as follows x szfxT39 If we then use the second formula for the curvature we will arrive at the following formula for the curvature Ifquot MI 3 1f39ltxgt2E K 2007 Paul Dawkins 53 httgtutorialmathlamarcdutermsaspx Calculus II Velocity and Acceleration In this section we need to take a look at the velocity and acceleration of a moving object From Calculus I we know that given the position function of an object that the velocity of the object is the first derivative of the position function and the acceleration of the object is the second derivative of the position function So given this it shouldn t be too surprising that if the position function of an object is given by the vector function I7 I then the velocity and acceleration of the object is given by WWW 5tquott Notice that the velocity and acceleration are also going to be vectors as well In the study of the motion of objects the acceleration is often broken up into a tangential component 17 and a normal component IN The tangential component is the part of the acceleration that is tangential to the curve and the normal component is the part of the acceleration that is normal or orthogonal to the curve If we do this we can write the acceleration as 5 aTT aNN where f and IV are the unit tangent and unit normal for the position function If we de ne V 7 then the tangential and normal components of the acceleration are given by 39rgtxquotrll KV HWOH llr39rll N where K is the curvature for the position function There are two formulas to use here for each component of the acceleration and while the second formula may seem overly complicated it is often the easier of the two In the tangential component v may be messy and computing the derivative may be unpleasant In the normal component we will already be computing both of these quantities in order to get the curvature and so the second formula in this case is de nitely the easier of the two Let s take a quick look at a couple of examples find the objects IE and the initial Example 1 Ifthe acceleration of an object is given by 5 2 I 6t velocity and position functions given that the initial velocity is 7 0 position is N0 i 2 31 Solution We ll first get the velocity To do this all well almost all we need to do is integrate the 2007 Paul Dawkins 54 htggtutorialmathlamarcdutermsaspx Calculus H 17t await J72f6tl dt t2t393t2 IEE To completely get the velocity we will need to determine the constant of integration We can use the initial velocity to get this j IE 60 a The velocity of the object is then 7z t2t3tZ BEE IE t2t13t2 1 We will find the position function by integrating the velocity function t await It2t13t2 1Idt 1 7 e a Et2 t2 t t3 tk c Using the initial position gives us f 2j393 0 6 So the position function is ttz 1t2 t 2t3 t3k Example 2 For the object in the previous example determine the tangential and normal components of the acceleration Solution There really isn t much to do here other than plug into the formulas To do this we ll need to notice that 7t t72t13t2 1I 7390 F2f6tl Let s first compute the dot product and cross product that we ll need for the formulas 39toquott t22t16t3t2 118t3 t2 2007 Paul Dawkins 55 htggtutorialmathlamarcdutermsaspx Calculus H F f I f 39 739tx tt 2t1 3t2 1 t 2t1 1 2 6t 1 2 6r2t13t2 1j392t 6 j 23t2 1 2r1 6t26r2 3 1 13 Next we also need a couple of magnitudes ll7ltrgtll r2lt2r1gt23r2 12 39ltrgtxquotltrgtll 6r26r223t21y1m The tangential component of the acceleration is then 18t3 t 2 a7 4 2 V9l t 4t2 The normal component of the acceleration is a 4455 72 66t2 24t 6 455 72 66t2 24t 6 N J9t4 t24t2 9t4t24t2 2007 Paul Dawkins 56 htggtutorialmathlamarcdutermsaspx Calcu us yvz uurdmate system In the last m semuns quhls chapter we39ll be luukmg at sums alternate suumnazes systems furthree mmmsmnsl space n w mm and Sara the same as wnh pulzr cuurdmates Harms asketch ufapumt m m3 atpulzr cuurdmates Sn xfwe have apmnt m sylmanssl cuurdmates the cmesm uurdmates can be nund by usmg the fulluwmg sunvasmns pulzr cuurdmates s the same V k w usmg the full uwmg cunversmns mm mm DavIons 57 my mum mathlamar enhuxmsayx Calculus H r x2y2 OR r2x2y2 9tan l1 x zz Let s take a quick look at some surfaces in cylindrical coordinates Example 1 Identify the surface for each of the following equations a r 5 b r2 22 100 c z r Solution 21 In two dimensions we know that this is a circle of radius 5 Since we are now in three dimensions and there is no Z in equation this means it is allowed to vary freely So for any given Z we will have a circle of radius 5 centered on the ZaXis In other words we will have a cylinder of radius 5 centered on the ZaXis b This equation will be easy to identify once we convert back to Cartesian coordinates r2 z2 100 x2 y2 z2 100 So this is a sphere centered at the origin with radius 10 c Again this one won t be too bad if we convert back to Cartesian For reasons that will be apparent eventually we ll first square both sides then convert Z Z Z r zzx2y2 From the section on quadric surfaces we know that this is the equation of a cone 2007 Paul Dawkins 58 htggtutorialmathlamarcdutermsaspx Calcu us gemngused m It39s pmbably meg m 5m Lhmgs uffwnh a sketch Spherical cuumnazes cunslst cme fulluwmg um quantities Fusnherexs p Thsxsthemstance 39umtheungntuLhepmntzndwewxllreqmre p20 NmLherexs e Thssthesameanglethatwesawmpulzrcylmdncalcuurdmates msme palercylindrical cuurdmates There 2mm resmmuns Bu 9 Fmallytherexs p Thxsxstheangtebetwemthepusmvezraxls andthelme 39umtheungntu Lhepumt Wevnllreqmre eggagn In summary p sthe mstznce 39nmthe ungm quhepumL p s the angledan Weneedtu mtate was m get m the gum cuurmnatesandaskwhatthecylindncalcuurdmates quhepumtaxe su weknuw we and whattu and 092 Ofcuursewe really unlyneedtu ena39 andz me 9 mm samexnbuth uurdmate sysLems mm mm DavIons my mumml mathlamar gunmme Calculus H z p cos p r p sin p and these are exactly the formulas that we were looking for So given a point in spherical coordinates the cylindrical coordinates of the point will be rpsinltp 99 zpcosltp Note as well that r2 z2 p2 cos2 p p2 sinZ p p2 cos2 psin2 p p2 Or p2 r2 22 Next let s nd the Cartesian coordinates of the same point To do this we ll start with the cylindrical conversion formulas from the previous section xrcos9 yrsin9 zz Now all that we need to do is use the formulas from above for r and Z to get x psinltpcos9 y psinltpsin zpcosltp Also note that since we know that r2 x2 y2 we get p2x2y2zl Converting points from Cartesian or cylindrical coordinates into spherical coordinates is usually done with the same conversion formulas To see how this is done let s work an example of each Example 1 Perform each of the following conversions 21 Convert the point g J5 from cylindrical to spherical coordinates Solution b Convert the point 1 1 E from Cartesian to spherical coordinates S olution 2007 Paul Dawkins 60 htggtutorialmathlamarcdutermsaspx Calculus H Solution a Convert the point from cylindrical to spherical coordinates We ll start by acknowledging that 9 is the same in both coordinate systems and so we don t need to do anything with that Next let s nd p pr2z2 62 22 Finally let s get p To do this we can use either the conversion for r or Z We ll use the conversion for Z z pcosltp 3 cosltp 3 p cos 1l p 2J5 2 3 Notice that there are many possible values of p that will give cos p however we have restricted p to the range 0 S p S 71 and so this is the only possible value in that range So the spherical coordinates of this point will are 2 Return to Problems b Convert the point 1 1 2 from Cartesian to spherical coordinates The rst thing that we ll do here is nd p p ix2 y2 z2 x112 2 Now we ll need to find p We can do this using the conversion for Z z x2 1 2 371 zpcosltp 3 cosltp 3 pcos p 2 2 4 As with the last parts this will be the only possible p in the range allowed Finally let s nd 9 To do this we can use the conversion for x or y We will use the conversion for y in this case sink 3 91093 psinltp2 J5 2 4 T 2 Now we actually have more possible choices for 9 but all of them will reduce down to one of the two angles above since they will just be one of these two angles with one or more complete rotations around the unit circle added on We will however need to decide which one is the correct angle since only one will be To do 2007 Paul Dawkins 61 htggtutorialmathlamarcdutermsaspx Calculus H this let s notice that in two dimensions the point with coordinates x 1 and y 1 lies in the second quadrant This means that 9 must be angle that will put the point into the second quadrant Therefore the second angle 9 must be the correct one 3 3 The spherical coordinates of this point are then 2 T Return to Problems Now let s take a look at some equations and identify the surfaces that they represent Example 2 Identify the surface for each of the following equations a p 5 Solution b p Solution 6 9 2 Solution d p sin p 2 Solution Solution a p 5 There are a couple of ways to think about this one First think about what this equation is saying This equation says that no matter what 9 and p are the distance from the origin must be 5 So we can rotate as much as we want away from the Zaxis and around the Zaxis but we must always remain at a xed distance from the origin This is exactly what a sphere is So this is a sphere of radius 5 centered at the origin The other way to think about it is to just convert to Cartesian coordinates p 5 p2 25 x2 y2 z2 25 Sure enough a sphere of radius 5 centered at the origin Return to Problems 77 b ltP 3 In this case there isn t an easy way to convert to Cartesian coordinates so we ll just need to think about this one a little This equation says that no matter how far away from the origin that we move and no matter how much we rotate around the Zaxis the point must always be at an angle of from the Zaxis This is exactly what happens in a cone All of the points on a cone are a fixed angle from the Z 2007 Paul Dawkins 62 htggtutorialmathlamaredutermsaspx Calculus II axis So we have a cone whose points are all at an angle of from the ZaXis Return to Problems 271 9 C 3 As with the last part we won t be able to easily convert to Cartesian coordinates here In this case no matter how far from the origin we get or how much we rotate down from the positive ZaXis the points must always forrrr an angle of with the xaxis Points in a vertical plane will do this So we have a vertical plane that forms an angle of with the positive x axis Return to Problems d p sin p 2 In this case we can convert to Cartesian coordinates so let s do that There are actually two ways to do this conversion We will look at both since both will be used on occasion Solution 1 In this solution method we will convert directly to Cartesian coordinates To do this we will rst need to square both sides of the equation p2 sin2 p 4 Now for no apparent reason add p2 cos2 p to both sides p2 sin2 p p2 cos2 p 4 p2 cos2 p p2 sin2 pcos2 p 4 pZ cos2 p p2 4p cosltp2 Now we can convert to Cartesian coordinates x2 y2 z2 4z2 x2 y2 4 So we have a cylinder of radius 2 centered on the Zaxis This solution method wasn t too bad but it did require some not so obvious steps to complete Solution 2 This method is much shorter but also involves something that you may not see the first time around In this case instead of going straight to Cartesian coordinates we ll rst convert to cylindrical coordinates This won t always work but in this case all we need to do is recognize that r p sinltp and we u39 will get we can recognize Using this we get 2007 Paul Dawkins 63 htggtutorialmathlamaredutermsaspx Calculus H p sin p 2 r 2 At this point we know this is a cylinder remember that we re in three dimensions and so this isn t a circle However let s go ahead and finish the conversion process out r2 4 x2 y2 4 Return to Problems So as we saw in the last part of the previous example it will sometimes be easier to convert equations in spherical coordinates into cylindrical coordinates before converting into Cartesian coordinates This won t always be easier but it can make some of the conversions quicker and easier The last thing that we want to do in this section is generalize the first three parts of the previous example WQQ sphere of radius a centered at the origin cone that makes an angle of a with the positive 2 axis CD S D vertical plane that makes an angle of f with the positive x axis 2007 Paul Dawkins 64 htggtutorialmathlamaredutermsaspx FHaSr NOTES 0amp1 POQER SEEMES 4 De m39c v quot A War series is a J m e Lorrix variab z at D39F Hq Form 30 M 2 Chx quot Co C XC2xH 50 whenquot Cal 6quot I Ca are cm flan Caled quotHa coePQ39cim39 s 6 1 pawer series 4450 Sf ar hhg ainrO I39S OP39hMO L VJ KO Z 00 3 EK MPl E Z 11quot 1 x Lkz xu quotquotquotquotquotquotquotquotquotquot n02 1 Here cnvzm The Power 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