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# Short Calculus MAT 016C

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This 69 page Class Notes was uploaded by Otilia Murray I on Tuesday September 8, 2015. The Class Notes belongs to MAT 016C at University of California - Davis taught by Peter Malkin in Fall. Since its upload, it has received 85 views. For similar materials see /class/187338/mat-016c-university-of-california-davis in Mathematics (M) at University of California - Davis.

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Date Created: 09/08/15

LECTURE NOTES Math 16C Short Calculus Fall 2008 Section 1 Dr Peter Malkin October 97 2008 Section C1 Solutions of differential equations Points to cover Section Cl 0 What is a differential equation 0 What is a general solution of a DE 0 What is a particular solution of a DiEi De nition 1 A differential equation is an equation involving a di erentiable function eg d2 y and one or more of its derivatives eg y i and y 5 Eg y 7 4y By 0 or equivalently 227 7 437 By 0 De nition 2 A solution of a DE is afunction y fx that satis es it Ego y 4e Sega is a solution of Question Verify that y 4e Se is a solution of Answer dy d d d 7 x 3m x 3m x 3 yidI dx4e 5e 4dze5dxe 4e 15ezi d2y d dy d 7 x 3m x 3m y 7 dz2 dz dz dI 4e 15e 4e 45e i Then7 putting y y and y into 967 we have y 7 4y By 46 45e31 7 4 lief 15e31 3 46 Sew 4e 459 716e 7 60e3 136 159 471612e 4576015e3 i ls this the only solution of NO The function y 72e llegx is also a solution of Check this yourselves ln fact7 there are in nitely many different solutions of Consider y Ae Beg Where A and B are constants Then7 Z7 Ae SBeSE and g9 Ae 9Be39 So7 y 7 4y By Ae 9Be31 7 4 Ae SBeSE 3 lief Sew Ae 9Be31 7 4Aex 712Be3 3Aex SBeSE A74A3Ae 9B712B3Be3 i There is one solution of for every choice of A and Bi The function y Aer Beg is called a general solution of the DE A particular solution is a solution Where A and B are xed 0 A general solution of a DiEi has unspeci ed constants c A particular solution of a DiEi has no unspeci ed constants Ego y 4e Sega and y 72ex llesx are particular solutions of The peiticulei solutions of a D E eie obtained fzom the genezal solution and some initial condi tions on the genezal solution and its deiivetives We use the initial conditions to nd the Values of the constants in the genezal solution Question Flnd the peiticulei solution ofy r 4y 3y 0 satlsfylng y 3 and y 8 When Answer We know the genezal solution is y Asz 8532 We must nd A and B such that y3andg 8whenz0 yAe Beszgt3AB A37B 11Asz 3853z 5 A 38 Putting A 378 into 8 A3B gives 8 378 313 5 ZB B g and thus A So the peiticulei solution saLleylng the initial conditions is y y 532 D E s eie Vexy useful Many dieleient dynamlc piocesses can be modeled using dieleientiel equar tions Maths is the language o natures ton s law of cooling The iete of change 0V6 time in the tempeietuie T of an object is piopoitionel to To 7T wheie To is the tempeietuie olsuiiounding envuonment So lcT0 e T alt wheie k is some positive constant Question Let T0 50 Venfy that T 50 e 05 is a genezal solution of k50 7T Flnd the peiticulei solution satisfying T 30 when t 0 and T 40 w en t 1 Also gieph the peiticulei solutions coiiesponding to k 1 end 0 730 720 7100 1020313 Answer We have 50 7 057m ICC and k507T 1450450705716 kCE a k 0 7 5 cc lT30whent 0 then 30 507050 0 20 soT5072057 IfT40When t 1 then 40 50 7 20H x 710 720540 x 0 5 s40 ln0 5 4c k m2 So T 50 7 205480 The peiticulei solutions also called solution cuives loi k 1 end 0 730 720 7100 1020313 die as o ows L Question Venfy that 4 7 e e1 0 is a solution of the D E 4y L z 0 wheie 0 is a cons a and gieph the solution cuives coiiesponding to 0 0i1i4 Hint use implicit dieleientietion Answer Implicit differentiation gives y 4 assuming y f 0 Then putting this solution for y into 4yy 7 z 0 gives 4y 7 z z 7 z 0 Thus 4y2 7 12 C is a solution of 4yy 7 z 0i Section C2 Separation of Variables Points to cover Section C2 0 Solving DiEis using the technique separation of variables Solving DEs There isnlt one general solution technique for DiEis In general it is very hard to guess solutions of DiEis We will see ho to solve DiEis in some special cases Case 1 Consider when y jig simplest form of a The general solution is y f1d Eigi y 21 The general solution is y fZIdz 12 C where C is a constant DO NOT FORGET THE CONSTANT OF INTEGRATION Case 2 Consider when y 2 where f and g are continuous functions The general solution is given by l 7d 1 dz yy y f assume gy 0 It s easy to remember d i fltzgtgltygt i 1 dz y dy cheatmg 7dy g 9 Question Find the general solution of y 21y Answer Here fx 2x and gy yi Then idy ldy mm 0 99 y 21dz 12 C where C and C are constants So equating the left and right side we have ln y 12 C where C C 7 C is some constant We then solve for y lnlyl I2 C emzc e0 752 y Ag 2 where A is a constant This is a general solution of the DE Check it A more direct way is as follows try to get all y variables on one side of the equation and the 1 variables on the other side 21y ldy21dz lnlylI2C yAexz I y where C and A are constants Question Find the particular solution of 3y2y z 0 where y 5 when I 0i Answer Separate the variables dy dy 2i 7 2i SyderziO SydIiiz 3y2dy izdz cheating 3y2dy 7zdz 3 1 2 y fir C C 1s a constant 1 y 7EIQ C Note that 2 where 71 and gy Then since y 5 when I 0 y 712 C 5 6 C 125 Thus the particular solution is y 712 125 Check it Question Can these DiEis be solved using separation of variables dy x OgimiNOl o 2 lzzyyi YESl This follows because 1zzyy1zy1z 1z1y So where l I and gy 1 Examples of using separation of variables First we will do a couple of examples of the separation of variables techniquei Application in Psychology One model of stimulusresponse asserts that the rate of change of the reaction R with respect to the stimulus is inversely proportional to the intensity of the stimu us dB 7 k g 7 E where k is some constant In other words you would notice changing a light bulb from 20W to 60W more than changing a light bulb from 100W to 140W Question What is the response function dR k 1 1 3 dRkgdS dRk dS Rkln5c where C is a constant Note that S 2 0 Question Geometric problem Find an equation for the function of a graph that passes through the point 82 and has slope 2 if assuming I f 0r Answer w dy dmlty0gt a 31H m 21nM c c is a constant gt1ny lnas20 a m AW 555W Am A is a constant y3Am2 or 7y AaQ ya A332 a y Ba B is a constant Now 5i nco y 2 when a 8 We have 2 B 8 a B 5 Thus the equation ofthe graph is a y Section C3 Linear FirstOrder differential equations Points to cover Section CS 0 What is a linear rst order DE 0 How can we solve LlNFODE The order of a DiEi is the highest order of the derivatives that appear in it Big y If 12 is a rst order DE y By 4y cosz is a 3rd order DE A rst order linear DE is a rst order DiEi in which y and y appear to the rst power Eigi y If 12 is NOT linear but it is rst order The standard form of a LlNFODE is y PIy QW where Pz and are functions of 1 Every LlNFODE can be written in this formi A LlNFODE can be solved using an integrating factori Let af x integrating factor then the general solution is y Qzurdm Note that you must always start from the standard formi Question Find the general solution of 2 7 2y ext Answer Here Pz 72 and Qz ext Then MI efPxdx 67211 n0 EC 6721 A6721 where A and C are constants So 1 l ex Ae 21dz e l j e e QEdz we can ignore constants e l 772167de e 6121 54 C C is a constant 21 7671 C fem C eh The general solution is thus y e C ehi Note how important the constant of integration is Also note that the constant in the integrating factor will always cancel and can be ignored Question Which method can we use to solve the following DiEis ll 2 y3 7 4y cos 1 Separation of variables This is not a LlNFODEi 2i 2 gy 17 sin 1 This is a LlNFODEi We cannot separate the variables Note that sometimes we can use either method eg 3 zyi Examples of solving a linear rst order DE Question Find a particular solution for z2y 7 41y 10 where y 10 when I 1 Answer Step 1 We put the DE into standard form 4 10 y 7 41y 10 42gt y 7 7y 72 assuming I f 0 i z I Step 2 Compute the integrating factor MI efPxdx ef T ldac e74lnlxl elmr4 174 Step 3 Find the general solution y 14731174dz So y 7g C14 is the general solution Step 4 Find the particular solution Then from the initial condition we have 2 107T014 1072C C12l So the particular solution is y 7g 1214 Question Find the general solution of zy 3y15 cos I where z 2 0 Also nd the particular solution satisfying y 0 W en 1 27f Answer Step 1 Put into standard form y 7 3y I5 cosz I f 0 So 131 7g and 15 cos I Step 2 Compute integrating factor MI efPxdx ef dx e731nx 173 Step 3 Find the general solution 7 1 7 3 5 73 y 7 7 z cosz 1 dz 1312 cos zdz 13 I2 sinz 21 cosz 7 2 sinz C integration by parts I5 sinz 214 cosz 7 213 sinz C13 So the general solution is y I5 sinz 214 cos I 7 213 sinz C13 Step 4 Find the particular solution Since y 0 when I 2 we have 0 2705 sin27r 227r4 cos27r 7 227r3 sin27r C27r3 gt 0 20704 C27r3 ltgt C 74W So7 the particular solution is y I5 sinz 214 cosz 7 213 sinz 7 47mg Integration by parts revision fudv uv 7 fvdu i 12 cos zdz 12 sinz 7 sinz 21dzC C is a constant VWH VV V u 11 u U U du Then we must apply integration by parts again 21 sin zdz 21 7 cos I 7 7 cos I 2dr C 721cosz2 sinzC C is a constant VW V 2 u 011 u v v u So 7 12 coszdz IQsinz 7 721cosz2sinz C C IQsinz21cosz 72sinz C where C is a constant Question Solve Iy y x2 lnx where z gt 0 Answer Step 1 Put it in standard form y i z lnzi Step 2 Compute integrating factor af x 6f elm 1 Step 3 Find the general solution 7iQltgtltgtd 73lt1gtd y7u1 rum 17 xnzxz z 112 lnzdz z l 113 lnz 7 113 C integration by parts I 3 9 121nz 7 312 iC So7 the general solution is y 12 lnz 7 12 iCi Integration by parts revision again 1 l l l l lnz z2dz lnz gzs 7 zs 7dr lnz z3 7 zg C Y 011 TV Vid Section C4 Applications of DEs Differential equations are very usefuli Many different dynamic processes can be modeled using differential equations Math is the language of nature Exponential growth and Decay The rate of change of a function yt is directly proportional to We can model radioactive decay7 population growth7 and compound interest in this way dy E ky 1dy kdt ldy kd lny ktC C is aconstant y 9 y ek ec y Aekt A is a constant A is the initial value of y ie when t 07 y A k is the constant of proportionality k gt 0 implies exponential growth7 and k lt 0 implies exponential decay Since there are two constants7 we need two initial conditions to nd a particular solution Modeling population growth The increase in a fruit y population is directly proportional to the size of the population expo nential growth model Q 7 kP a P 7 C 1 dt 7 7 e where t is the time in days After 2 days7 there are 100 ies7 and after 4 days7 there are 300 les So7 P 100 when t 2 and P 300 when t 4 How many ies are there in 5 days Thus 100 0er a C 33272 and 300 064k a 300 1 a 3 W a k 1116 Inn23y Then 100 052k a 100 Gem a 100 03 a C So7 the exponential model is p elnb l 3 Thus after 5 days it 57 the population is P elquot5 m 514 Newton7s law of cooling The temperature T of a cooling object drops at a rate proportional to the difference To 7 T where To is the constant temperature of the environment of the object dT 7 k T 7 T t dt 0 Solve via separation of variables g kT0 7 T 1 dT kdt 7 lnT0 7 T kt L L is a constant dt To 7 T To 7 T e kHL Ae k A is a constant T Ae k To 11 Forensics A dead body was found at midnight The temperature of the body was 946013 Assume that the temperature at the time of death was 986013 The temperature of the room when the body was found was 70 F One hour after the body was found its temperature was 934013 Let T be the temperature of the body so T Ae k 70 where t is the time in hours after the person died When did the person die Now T 986 when t 0 so 986 A60 70 7 A 286 Thus T 2866 70 Let h be the number of hours from the time of death until midnight So T 946 when t h Then 246 7kh 7kh 946 7 28 70 7 286 6 1 Also T 934 when t h 1 Then 234 934 286e kh1 70 7 7 6W 2 286 Now dividing equation 1 by equation 2 gives 246 671 k 246 7234 7676h4r1 e 7kln723 4 005 Thus T 2866 03905 70 So we can now solve for t at midnight and then we know when the person died 246 1 246 946 286 005 70 246 286 005 7 005 t 771 7 m 3 e 1 2 e 2 286 e 2 005 n286 Therefore the time of death was around 9pm Modeling chemical mixture A solution containing lb of salt per gallon ows into a tank at the rate of 2 gallons per minute The wellstirred mixture ows out of the tank at the same rate The tank initially holds 100 gallons of solution containing 51bs of salt Question How much salt is in the tank after 30 minutes And how much salt is in the tank as 7700 Answer 0 Salt owing in 2 gallonsmin gtlt lbsgallon 1lbsmin 0 Salt owing out 2 gallonsmin gtlt lbsgallon lbsmin Step 1 nd the differential equation g in 7 out 1 7 5570 Step 2 nd the general solution 1 7 0 is a linear rst order differential equation In standard form it is g 1 So Pt and 1 Then the integrating factor is um emedr El n 6amp3 Then 1 S utQtdt e tdt 671 50651701 C C is a constant 50 Ce Therefore7 the general solution is S 50 06W Step 3 nd the particular solution The initial condition is that S 5 When t 0i So7 5 50Ce 6245 Therefore7 the particular solution is S 50 7 4565173 Step 4 solve for given values When t 307 we have 5 50 7 45e 30 m 25 301bsi 30 Also7 as t A 007 we have S A 50 since 456 5170 A 0 as t A 00 Section 71 The 3dimensional coordinate system Points to cover 71 0 Sr memjonal coordinate systern 0 distance between points in 3d space 0 midrpojnts of line segments 0 equations of spheres p1anes 3dimensional coordinate system The points 134 is located at the corner of a 1 X 3 X 4 box Distance between two points Suppose thet we went to know the distance fzom 1 3 4 to the ougm 000 thet 1s we went to know thet velue of d m the diagzam We oen me Pythagozas Lheozem twme d 42 w end w 12 32 so d 12324i Tn nywmdt r 9 yew 222 Now suppose thet we went to know the distance tween two pomts eyz end 2932 2 i 0 ya 2 be Thend z27z12w2 endw e 2yr 005 be 39h 1339 1 Thme 3 402 i 601 32 31 M i a Fox exemple nd the distance between 7325 end 471 2 7125722 f499 x Midpoint between two points Whet s the midrpomt between two points eyz end muggy Let aomymzm be the e n b midrpomt T mtst be helfrwey bewteen at end e2 so em Aso em mtst e helfrwey between 3 end 32 so gm my Slmllazly zm mtst be halfway between z end zp so zm E 7 e i Thts 7 n0 yay2zdz2 newm 2 2 2 Eg Whet es the midpoint bewteen 7325 end 4127 734 2152 7131 2 2 2 2 2 2 0m 9m ze Equation of a sphere Qhemh u we x e mm Wm Wm 15 39he set m en 12mm 3 eamehee 7 from We med Ansng A shhee eeneeea 01112 The the eqmuunnf esphee eeneeea at We 5 eeayegebwaeeyyweeawmweu Eg The equAlmanLhAsphere wuheeme 2733 em mdns 5152722y321782 zs L L Inch eveee amm mvmthe summon2 e nee Ayzz eeee 15 o wheh s eh equsuunm39 e sphee m we 2mm em mane Duhe ephee mam ere ieneew eeeyee sue sum wee ezeeeeweeeemen14 e eneezeeyexeeeeesee1 yemeeeaeeyenees eemegezyeaewea o umueewemmuespheeemmmmme eewqyezyeueefeae seme memmespheeexs1ezeehduemusm 3 a Planes The gmeml gunman m e pm In Srduxmmnm 1 A2 By c 17 whee ABC17 are mums nut is 39he set men hum eemmhg 39he gunman e e pxehe v 1 Wu pbm Eg Drawz2y31a acrmtezceptlsy0z0 a9 5 ymtezceptlsap0z0y3 armtexceptlsy0w oe e2 7 But Lhexe Menu always mans 01 grams 01 moms mtexcep39s E3 13an the plehee e e a b y 4 end c e 5 e e a The plehe he paxallel Lo Lheyeplehe Thexe axe he yams oz meme mtexcep39s 39 1 e y 4 The plehe he pamllel to the eeplehe Theze axe he meme oz meme mtexcep39s Section 72 Surfaces in 3 dimensional space Points to cover o Traces of surfaces o Quadric surfaces Every quadric surface has an equation of the form A12ByZszDzEyFzG0 where A BCDEFG are constants Traces of surfaces Tip To graph a surface first nd and graph the intersection of the surface with different planes such as the planes z a y z a where a is some constant The intersection of a surface with a plane z a y or a or c is called a trace Paraboloids Elliptic Paraboloid We graph the quadric surface 2 12 y2 0 We can graph the intersection of the surface with the plane z 0 is the parabola z yz lying on the yzeplane This trace is called the yzrtrace 0 We can graph the intersection of the surface with the plane 3 0 is the parabola z 12 lying on the zzrplane This trace is called the zzrtrace I We can graph the intersection of the surface with the plane 2 0 is the point acyz 000 since 2 0 implies 0 x2 y2 This trace is called the acyitrace I We can graph the intersection of the surface with the plane 2 1 is the circle or radius one centered at 001 which lies on the plane 2 1 I Similarly we can graph the intersection of the surface with the planes 2 2 3 4 each of which giVes a circle of radius J5 J1 with center 00 2 003 0 04 respeci tiVely The surface 2 x2 y2 has a bowl shape It is called an elliptic paraboloid The general equation of the surface is i 7 20 L 12 L 77290 20 where a and 17 are nonzero constants called scaling factors that stretch or contract the graph in the a and y directions respectively and where 33030Zo is the point at the base of the bowl Note that the coefficient of 27 270 may be negative in which case the bowl is upsiderdown For example the surface 72 r2 y is an upsiderdown bowl shape It is possible to change the roles of the as and 2 variables giving the same type of surface an elliptic paraboloid which is orientated differently in sedimensional space We distinguish among the different orientations by noting which axis meaxis yraxis or zraxis goes through the center of the paraboloid The axis of the elliptic paraboloid z r2 32 is the zraxis The surface 2 z is an elliptic cone with axis the yraxis see below Also a 32 22 is an elliptic cone with axis the xeaxis Hyperbolic Paraboloid L L gzeyzw 39 l ddl ltiscalledahyperholic paraboloid The following table lists the traces given by intersecting the hyperbolic paraboloid z 32 7 32 with a plane a is acons an The general equation of the surface is 7 x T r02 3 T 302 iltl 7 2 7 T 7 T where a and 17 are scaling factors and where am0 20 is the point at the center of the saddle Ellipsoid The basic forrn of an ellipsoid is m2 32 22 l which is a sphere So a sphere is an ellipsoid T he following table lists the traces given by intersecting an ellipsoid m2 y2 22 l with a plane a is a constant Also the surface 3 y2 r l is also an ellipsoid which looks like a sphere that has been a stretched in the aeaxis direction and squished in the loads direction The general equation of an ellipsoid is mire2 11on lilo2 TTT 0 r o gt 1 then the ellipsoid looks like sphere that has been stretched direction respectively Similarly if a lt 1 17 lt 1 or o lt l t like a sphere that has been squished i where a 17 and o are scaling factors and where am0 z is the point at the center of the ellipsoid Note that ifa gt1 gt 39 39 39 in the re is or zraxl s the ellip axis peak hen oid looks n the aeaxis yams or zraxl s direction respectlvelyt Hyp erb oloids A hyperboloid has the basic forrn x2 32 7 12 where o is sorne constan hyperboloids depending on the value of o o 0 Elliptic and Hyperboloid of 2 sheets a e V There are three cone o Hyperboloid of one sheet Elliptic Cone The basic form of an elliptic cone is r2 32 7 12 0V The following table lists the traces given by intersecting the elliptic cone r2 32 7 22 0 with a plane c is a constant The axis of the elliptic cone r2 32 7 22 0 is the 278x13 which goes through the center of the cone Note that for r2 22 7 y2 0 is an elliptic cone with axis the y7axls and 32 12 7 32 0 is an elliptic cone with axis the aeaxis The general form of an elliptic cone is 7 2 x 0 7 a2 17 Hyperboloid of one sheet The basic form of a hyperboloid of one sheet is r2 32 7 22 1V The following table lists the traces given by intersecting the hyperboloid 32 y2 7 22 1 with a plane a is a constant V The axis of the hyperboloid r2 32 7 22 7 1 is the a7axis which goes through the cehter of the cOhe Note that for 32 22 7y 7 1 is hyperboloid with axis the y7axis arid 32 22 7 r2 7 1 is a hyperboloid with axis the reaxis The general form of a hyperboloid of me sheet is mire2 11on size2 TT7T1 Hyperboloid of two sheets The basic form of a hyperboloid of two sheets is r2 32 7 22 7 71 The following table lists the traces given by intersecting the hyperboloid 32 32 7 22 7 71 with some piahe a is a constant The axis of the hyperboloid r2 y2 7 22 7 71 is the a7axis which goes through the cehter of the cOhe Note that for r2 22 7y 71 is a hyperboloid with axis the y7axis arid 32 22 7 is a hyperboloid with axis the 3373315 The general form of a hyperboloid of me sheet is m i me y 302 7 7 202 a2 72 Classifying quadric surfaces Here is summary of types of quadric surfaces one sheet hen classifying a quardic surface the rst step is to decide Whether it is an ellipsoid parpaboloid or hyperboloid and then What type it is and nally We must decide What the axis of the surfac is ie ziaxis yiaxis or ziaxis note that an ellipsoid has no axis 2 1V Classify the surface 7 7 2 2 2 2 z y z z 7 7 7 7 0 5 7 2 4 32 This surface is an elliptic cone with axis the yiaxis Recall that the axis of the elliptic cone is the axis that goes through its middle and is the Variable with a negative sign 2 2 LLU 2 32 4 2V Classify the surface 7 if 7 139 This surface is an ellipsoid 3 Classify the surface 422 7 f 22 7 4 0 PI E EEr39AEEE I 4z27y22274z0 4227y22722740 2 2 2 1 1 2 7 ltgtz 727T71 2 1 32 112 ltgtz T72771 This surface is a hyperboloid of one sheet with axis the y axis The axis of a hyperboloid is the axis that goes through its center and is the Variable with a negative sign Section 73 Rinctions or more than one variable 234 12 uy Wm B 2 mang pug day s tunth mung 3 g E my Htsac mamm at Dbeset 01mm punts a mnwjm mgb hm ymmmnmmma Mmm maeuxmevamblsmmsmlv m 59 D 5 431153 the datum a 1 and the 491mm 59 Ma Fugue whiz a my raa ypmm a m Dismerangeolf xmxmhpmgy mDJJtaexs y m thesauh pumsfmwmh my Mme mmnngemmmtm 123 4 mm amen z Wmizwmgmwmzuynzo mmn mesaDz AM 3243oywmmnmmwmggmmwgnzo mmeuutpm gunman wnmmaamwammmw A r mtpzaw zuynmmmexmtmm mm in mm Wmneatmwdmm 1w ZN 2 26 Answer Ncle lhel lhe lcgeruhrh ls de ned chh lcr posluve numbers Sc lhe dcrhem ls D ey 97 2 gt 0 The rehge ls idem 9 Contour maps and Level curves For topographch rheps cchlcur lures also celled level curves ere hhes cl equel eleveuch For Weelher rheps rscbers also celled level curves ere hhes cl equel pressure Cchlcur rheps ere Zrdlmenslonal deplcuons ofsulfues m Srdlmens Eg Cchsrder lhe sulfue e KM Sraazryz ll ls err elllpuc perebclcrd upslde down The rhlersecuch cl lhe perebclcrd wuh lhe plehe e A celled e Hue ls lhe crrcle A s e e2 e y a e2 32 e level curve A collectlon ol level cums lOl evenhl spaced planes 1 a ls called a contour map De nin39on 3 A Mal curve 0 a sumMa z zy LS the curt2 a fzy m the zwzme allme a 15 some cmcmc De nin39on A A m thuV map 0 a Suianal e fzy LS a WWW olwel curt25 a zy mm mm memo may syacad mmmvm a g nsldel the uncLlon z 2 e 92 llac ale the level curves lol 1 72 r 01 29 None that lOl z Lhe level culve ls a y e 12 whlch ls a hypelbola lOl a e o and IL ls a closs lOl a o Note that level Gunes fox dlffezenc constants c nevez moss smce that would mean that the funcclon fzy has two dlffezenc Values at the same polnL zy whlch 15 not pOSSlble Also nole lhel me close me adjacent level cumes ale to each olhel me steepex me sulface Section 74 Partial derivatives Potnts to covet Sectton t A o Palttal denvattves ot tuncttons ot mote than one vattable Let a mt be a tunctton de ned at euh potnt tn some tegton D The pal al deliva ve ot wtth tespect to e wtttten as g and also wtttten as ts the as tt tt wete a constant stmtlalh the pmtal deivative ot wtth tespect to y wtttten as ts the otdtnaly dettvattve ot wtth tespect to y as tn the one vanable tt wete a consta 3 and also wtttten as fy case whete vanable as tt nt we tteat the 9 Eg Constolet the tunctton ot two vanables m y e2 32 933 Then the palttal denvattve of f t t t t t f 8 2 A 5 3 E A 1a93a9 y andaiy fy 1Aat y A palttal denvattve ot a tunctton ot two vanables ot a ts agaln a tunctton ot two vanables so we can evaluate the palttal dettvattve at 30ml wt ya The nolaugn tot evaluatton a palttal dettvattve at a potnt ob ts the tollowtng 62M Mwnt ya and ailkm Mwntyn tolet agatn the tunctton ot two vanables mtg Eg cons at 32 933 We can evaluate the palttal denvattves ot wtth tespect to 9 at the potnt 12 as tollows 1 2 21 a122 so and Ml 2 1 A1323 a What are pamal delivativas T e palttal dettvattve ot wtth tespect to e evaluated at the potnt awn wntten ggtm fnyn ts the slope ot the suttace e m y at the potnt wt ya tn the acrdlrecuon stmtlalh the palttal denvattve ot wtth tespect to 3 evaluated at the potnt my wntten lm Mewquot ts the slope ot the surfue a y at the potnt when m the y ohtectton We can easily extend this concept of partial derivatives of functions of two variables to functions of three or more variab esi Eigi Consider the function of three variables fzyz zemy22i It has three rst order deriva tives7 one for each variable exy2z Iyexy22 81 z2exy2z 19y E 6i Zrem z Higher Order Partial Derivatives As with ordinary derivatives7 we can nd partial derivatives of higher order The partial derivative of fz7 y with respect to either I or y is again a function ofz and y so we can again take the partial derivative of the partial derivative Consider the function fz7 The second order derivatives of f are as follows 82167 8 8f 7 7 w 7 a 7 f 7 my 62f 6 8f ayg Big fyy fyy 62 6 6 f 7 fxyfxy 399sz 8y I 82f 7 6 8f 7 7 My 7 a 7 fyx 7 mm The last two second order derivatives are called mixed partial derivatives Eigi Consider the function of two variables 31f 7 2y 512g The rst order partial derivatives are as follows 8 8 if 3y2 101y2 and if 61y 7 2 1012yi 81 By The second order partial derivatives are as follows 62f a Bf EL Sf m 10y2 E E a 29 03y 2 g 619 2 1012y 61 1032 59quot 6 g My 2 10129 6y20acy 52 g g 3y2 10 6y201y Note that 32f 7 3331 This is true in general for a function if it has continuous secOnd 313 order partial derivatives Section 75 Extrema of functions of two variables Points to cove 7 5 Finding eiitieel pomts of functions of two veiiebles ClaSifying eiitieel pomts of functions of two veiiebles Extrema of functions of two Variables De nition 5 Let fa y he af39zmctwn da nad et acmyu W2 say fwuyu 15 e helm mmm 0T ell he m 0f f n thaw mete e cz rcula r when R wtth cmta r emn such that AM 2 emn f pamts my Szmzlmly en winsu Wk th zf at May S mtu fw all pamts my m R Note thet if e pomt is e ieletive minimum then it is not neeesseiily the global minimum of the function It is insteed the minimum Within some loeel eiee Suiiounding the pOint 1r hum is e ieletive minimum in ieletive maximum and if Memw end fmmyu eie nite then fewuyu 0 and femiyu 0 membei that Memen is the slope of the suireee z AM at the pomt acmyu in the rdiiection and Memen is theslope in the ydiiection De nition 6 A Wet new 15 e mtml Wet 0f the functwn HM zf Mewu 0 and Memyu 0 0E tf 2mm falta uyu M fywuyu ts mda nai e g dzmswn by t7 All ieletive maxima end ieletive minime eie eiitieel pomts but not ell eiitieel pomts eie ieletive maxima oz ielatlve mlnlma F0 exemple considei HM e2 7 32 seddle shepe The pOint 00 is called e seddle pomt IL is a animal point but it is not a zelauve minimum 0 zelauve maximum Finding ielative minima and ielative maxim has many applications Theorem 2nd paitial deimtives test Let e Nagy have continuous paitial deimtives and y and also continuous 2nd oidei aitial deimtives M y and xy in a iegion R containing e b which is a oiitioaj point oi Let deb e b fyyab 7 gym 172 then 1 ii deb gt 0 and e b gt 0 then eb is a ielative minimum ii deb gt 0 and e b lt 0 then eb is a ielative maximum ii deb lt 0 then e b is a saddle point it deb 0 then Inconcluswe tiy othei methods Quastion Fmd and classny the oiitioaj poins oi the unotion Me 3 e210eayi 713 Answer Fiist we nd the oiitioal points ey2e1ooie and MM or 13 0 y 3 So theie is one and onh one oiitioal point is 3 Next we class the oiitioal point We must compute the 2nd oidei paitial deimtives in oidei to do this femiy 2 we 3 Jam 3 0 we 3 Then MPH MPH fyy5 3 1y5 32 2X5 0Z 12gt 0 since 75 a 2 gt 0 the ciitical point is a ielative minimum Quastion Fmd and class the oiitical points ofthe function Nagy yLaWL ayieaez Answer Fiist we nd the oiitioal points f r wyr 5w eweJr 1 and y 33 7 anti 7 5y 33 7 7 23 The hist paitial deimtives aie de ned oi all points M so the onh oiitioal points aie when 0 and y 0 Now ese 1 0 s e 0 oi y 71 Next we considei the cases hen e 0 and when 3 1 sepaiateh First ifz0 then fy3y271272y0 3y2727y0 y00ry2 Sowehave two critical points 0 0 and 02 Second ify71 thenfy 3y271272y0 3171220 13730 z V3 or z 7i So we have another two critical points 7 1 and 7 71 Next we have to classify the critical points fawn 69 67fyy 6y 7 67 and facy fyac 61 So dzy 76y 716y 7 1 7 7612 36 7 36y2 7 3612 3617 y2 7 12 Critical Point d 361 7 y2 7 12 f 76y 7 6 0 quot3 quot 00 361707036gt0 7607676lt0 reli max 0 2 3617 22 7 0 7 7108 lt 0 saddle 71 3617 71 7 3 7 7108 lt 0 saddle 7 71 3617712 7 7 7 7108 lt 0 saddle Question Find and classify the critical points of the function my 7 I4 7 iye Answer First we nd the critical points fxr7y y 7 13 and fyzy I ya The critical points occur when f7 zy y 7 13 0 and fyz y z 7 y3 0 Then y7130ltgtyzgi Putting this into I 7 y3 0 we have 33 z7y30ltgtz7z 0ltgtz7zg0ltgtz17180 so I 0 or z 1 or z 71 Then using the fact that y 13 we have that if z 0 then y 0 and if z 1 then y 1 and if z 71 then y 71 So there are three critical points 0 0 11 and 7171 Next we must compute the second order derivatives to classify the critical points f11lt17 y 73127fyyr7y 73y27 and f1ylt17y fyxlt17y1 Thus we classify a critical point according to the value of d fmmy fyylt17y 7 fxy17y2 P3100392 1 Critical Point d 7 Ma y w y 7 fxyr7 M Ma y 039 392 r 32 7302 71 71 lt 0 saddle point 11 d 73127312 7 8 gt 0 7312 lt 0 reli maxi 11 d 737127371271 8 gt 0 73712 lt 0 reli maxi Question Find and classify the critical points of 12 f Answer First we nd the critical points 1 2 1 2 fag gQIz2y2 3 andfy 2yz2y2 3 Note that the partial derivatives are unde ned at the point 00i So 00 is a critical point We cannot use the 2nd partial derivative test since f fyy g and fygc are discontinuous at the point 0 0 We must therefore use other techniques to decide such as graphing the function The graph of the surface is as follows 13 Aquot 1 n m h m 39 MM 2 o and oo o m 00 musch axelauve rmmmum Section 76 The Lagrange Multiplier Method Points to cover 0 Finding relative extrema of functions subject to constraints The Lagrange Multiplier Method With One Constraint Problem We want to nd the optimal solution of a maximization or minimization problem with constraints Eigi Find the box with maximum volume such that one corner of the box is the origin and the diagonally opposite corner lies on the plane 61 4y 2 24 ie maximize f1y2 1y2 subject to the constraint 61 4y 32 24 Important Result Any maximum or minimum of a function subject to the constraint that g1y 0 will occur at a critical point of the function F1 y A f1 y 7 Ag1yi Similarly for functions of 3 variables the maximum or minimum of a function f1y2 subject to the constraint that g1 y 2 0 will occur at a critical point of FI7y727 fI7y72 7 Aywywi The variable A is called a Lagrange multiplieri Note that the method nds critical points but does not say how to classify them eg whether the critical point is a maximum or minimum or neither To classify the critical points we must examine the function Note that at a critical point of F all partial derivatives are 0 we assume that all partial derivatives are de ned so we have FA 791 y 2 0 so a critical point of F must satisfy the constraint g1y2 0 as one would expect Eigi Find the box with maximum volume such that one corner of the box is the origin and the diagonally opposite corner lies on the plane 61 4y 32 24 ie maximize f1y2 1y2 subject to the constraint 61 4y 32 24 Let g1 y 2 61 4y 32 7 24 and let 1797230 fr7y72 7 Ay ww 1y2 7 A61 4y 32 7 24 We must nd the critical points of F To nd the critical points of F we set the rst partial derivatives to zero F y2760 l Fy12740 2 Fz1y730 3 FA 7614y32724 0i 4 Note that the equation F 761 4y 32 7 24 0 means that any critical point of F must satisfy the constraint 61 4y 32 0 Equation 1 implies A y2 and substituting this into equation 2 gives 2 2 3 127 y20 217gy0 200ryE1 36 Now if 2 0 then fzy0 0 which cannot be a maximum of f so we can ignore this case Substituting A yz into equation 3 gives 1 zy7 yz0 y00rz2zi Again we can ignore the case when y 0 Now putting 2 2x and y 3 into 4 gives 4 761616z7240 z Then y 2 and 2 So the point at which f is maximum subject to 61 4y 32 0 is 2 g and the maximum value is f2 2 g i Note that the most dif cult part of the Lagrange multiplier method is to solve the system of equations to nd the critical points of Ft Unfortunately there is no one general way to proceed Eigi Find the minimum of fzyz 12 y2 22 subject to the constraint 31 2y 2 2 Let Fz y z A 12 y2 22 7 A31 2y 2 7 2 Next we must nd the critical points of Ft FE2173A0 1 Fy2y72A0 2 Fz227A0 3 FA7312y2720 4 Equation 1 implies A Putting A gr into equation 2 gives y 7 gr 0 y Putting A gr into equation 3 gives 227 gr 0 2 Then putting y 1 and 2 z into 4 gives 4 1 l4 3 731 z z760 Ez2 z The Lagrange Multiplier Method With Two Constraints The Lagrange Multiplier method works for more than one constraint The maximum or minimum of a function fzy2 subject to the constraints that gzyz 0 and hzyz 0 occurs at a critical point of the function FI7y7277 7 f y7 2 7 AWWJ 7 WW y7 2 Note that at a critical point of F we have FA 7gzy2 0 and FM 7hzyz 0 so a critical point of F must satisfy the constraints as one would expect The maximum or minimum of a function fz y 2 subject to the constraints that gzy2 0 and hzyz 0 occurs at a critical point of the function FI7y7277 7 f y7 2 7 AWWJ 7 WW y7 2 Note that at a critical point of F we have FA 7gzy2 0 and FM 7hzyz 0 so a critical point of F must satisfy the constraints as one would expect Eigi Find the minimum of fz y 2 12 y2 22 subject to the constraints 1 2y 7 2 2 and 2x 7 y 7 2 i Let Fzyz7 AM 12 y222 7 Az2y7 27 2 7 M21 7 y7 2 7 2 We must nd the critical points 0 i F 21772M0 5 Fy2y72u0 6 FZ2ZM0 7 F 7z2y7z720 8 FH77217y72720 9 13 21227M0 10 22gtlt 3 2y423u0 11 73gtlt6 6z2y1020 31y5y0 12 7472gtlt5 51732760 13 758 5142720 14 109 7240 27 15 Substituting 2 7 into 9 gives 51 1772 7 6 0 51 2 1 Then substituting 27 andzinto 4 gives2y720 21 y So7 the only critical point of F is 7 7 Which must be the minimum of the function F since F must have a minimumi Thus the minimum value of F is F57 7 2 2 35415 7 9 i 4 Section 78 Double integrals and area in the plane Points to cover 0 multivariate integration 0 double integrals 7 area of regions We know that the inverse operation to differentiation is integration so it is natural to ask what is the inverse operation to partial differentiation Suppose that there is a function fz y such that f7 zy 612y3i What could be We can integrate with respect to I as in the one variable case by treating y as a constant my fxltz7ygtdz 6z2y3dz 3mg Cltygt where Cy is some function of y Here Cy is called the constant of integration Remember that when taking partial derivatives not only constants become zero but also functions involving only y also become zeroi So fxy 313 This is called partial integration with respect to 1 Similarly we can perform partial integration with respect to y We want to nd fx y given that fyzy emyi Then 1 m y fyltz7 my e ydy gem om where Cz is a function of 1 Again Cz is called the constant of integration We can also evaluate de nite partial integrals with respect to I as in the one variable case you can omit the constant of integration y 2 yeryd1 Few 16112 7 160 E 0 y 0 y y 9 Note that the limits of integration here can be functions of y since y is treated as a constant when integrating with respect to 1 Similarly with respect to y we have Eexydy leg 1 leg 7 le0 ex 71 0 z 0 z z I So there are two types of de nite integrals of functions of two variables 921 9211 fr7ydy and fr7yd1 911 9111 Double integrals Note that a de nite partial integral with respect to y 921 f 067 ydy7 911 39 gives a function of I which itself can be integrated Similarly a de nite partial integral with respec o 2 929 fltIgtydIgt my gives a function of 3 which also can be integrated An mtagml 0f an mtagml is called a double integral Evg 4 w 4 w 4 4 A 22 2ydydz 2zyy20 dz 2x2 22dz 3z2dz u 64 0 0 0 0 0 In the above example We integrated with respect to y then I it is possible to do the reverse 4 4 4 A 4 A 2x2ydzdy 22 2zylydy 16 81 7 Swizz 163 4112 Wm 54 o y o 0 So for a function of 2 Variables there are two types of de nite integrals m 419 fz39ydydz 9100 fl ydy dz and Abjgyfzydydz fzydz W 1a Note that if We integrate rst with respect to y then the limits of integration are functions of 1 since I is consi ered as a constant when integrating with respect to 3 Similarly if We integrate rst with respec to I then the limits of integration are functions of y si ce 3 is considere as a constant when integrating with respect to z The outer limits of integration must be constants ou V Areas of regions Recall that in the one Variable case the area of the region R under the curve 3 92 and over the interval 21 ive R zy a S I S 50 S y S 9Tl15 re the area of the region R hetW en the curves y 911 and y 5121 and over the Furtherrn interval 11 ivev R zy a g I S 591I y 921l15 17 E W www also express the area of the region R zy a g I g bg1z lt y g 921 as the e c double integral 5 was l7 l7 dydz 1 9200111 5121 7 911dz a g a a 1 a W For example nd the area of R zy 0 g y g 120 7 z lt 2 2 x2 dydz o o NOW We can reverse the roles of the and y Variables The area of the region R between the curves 1 919 and y 929 and OVEI 012 19 121 i a S y S MM S I S gzy is 19 Note that R m 0 g y g aw g a g 2 w y g a g 20 g y g 4 le the two leglohs m the two examples wele ldehtloal so we would expect to get the same ahswel lo the azea Thus lt ls posslole to desouoe leglohs m two oleleleht ways leadmg to two dl eleht appzoaches fol oomputmg the azea of the legloh As we saw above some leglohs can be desouoed m 2 dl eleht Waysglvmg 2 oleleleht methods fol oaloulatmg axea Fol example console the followmg legloh Descube R m 2 dl ezent ways 1 hwy osyslrgwseszh Ray03322ya03y31 2 17 1 27211 awe dandy n n n n When we want to be able to specify the mtegzal without specifying the ozdex of mtegzauon we wute dA R whexe dA daody oz dA dydao It IS NOT always posethle to descube a legion In two way Fox example considez the followxrg eglon syswsagzhbu g b smce the function 92y does not ex We can wute R ay m 71 twe cannot wute R m the fozm 92 1539 R May My S at S Section 79 Applications of double integrals Points to cover Volumes of solids Average value of a function over a region Volumes of Solids Recall that ih ohe variable f mm is the area uholer the curve y 933 above the ihterval e 17 m the two variable case R f frydA where dA drdy or dA dyda is the volume uholer the surface 2 fay ahd above the region R Note that we assume that fay is cohtihuous ahd nonrnegative over R Note that ih ohe variable we integrate over ihtervals ahol ih two variables we integrate over regiohs Also the doublelntegral R f f2ay7 f1aydA is the volumebetweeh the surfacesz f2ay and 2 f1ry and above the region R For example calculate the volume of the following pyramid So we mmt calculate the volume of the sohd undez the stufaoe z 7 177 7y and above the legion Rzy Ogyglemgzg So the volume IS 172d2 Anothez example nd the volume of the sohd S that hes below the suzface z 4 r y and above the IeglonR zy 0 so 3403 3 32gt So the volume of s is 2 y 2 4 7 y2dady 43 7 33 dy 0 0 0 Sometimes the order of integration matters because sometimes one order of integration is much easier to solve then the other order of integration For example nd the volume under the surface 2 a and above the region RmyySmSLUSzS1myUSy m0 mslv 2 Note that the intersection of the surface 2 5 with the lasrplane is as follows The volume 5 fu z 22dzdy Cf 552de The second muegal ls easlel we evaluate the f mmyal 52212 is not an elementary lunctlon Average value or a lunctlon over a region Rmzem E g Fmd the average value ol the Mellon KW 2N ovel the zeglo u u n Ovel R s Rlltwgt Uszswswll Nomthatthe areaolRlsag The average value ol Ovel R s 3 zydA2 yzzwdzdy a e u u 2 Mlle Section 101 Sequences Points to cover 0 Sequences o The limit of a sequence 0 Pattern recognition Sequences A sequence an is a function Whose domain is the set of positive integers The function values a1 a2 a3 M an are called the terms of the sequencer 72ni1 72i17 74i1777 For example cons1der the sequence de ned by an 7 3n41 Then a1 7 34 7 7 a2 7 54 7 10 7 1 7 7 7 9 7 11 57 as 7 E7 a4 7 E7 a5 7 37 em For example consider the sequence an 71 1371 Then a1 a 77 a3 7 a4 77 a5 m etc The limit of a sequence What happens to the value of an as n get larger and larger as n goes to in nity Consider the sequence an 1 a i Then limnn00 an limnnmi 01 Consider the sequence Where an 7 Then lim an lim lim n7oltgt n7oltgt 3n 4 n7oltgt l 71 71 SW 2 l 2 0 2 nm 7777 7003 30 3 3l 31 A sequence an converges if limnn00 an exists sequences that have no limit are said to diverge Consider the sequence Where an 71 11 Then an 0 if n is odd and an 2 if n is even So an 0 202 0 Then limnn00 an does not exist so the sequence an diverges 2quot 24816326412826 Cons1der the sequencenwhere an 771 Then an 1Z EEEEmi So limnn00 an limnn00 2 001 Thus the sequence an diverges Below are some more examples 7 72n25 11 an 7 7n8 1 Then 72 2 5 72 i lim an lim A lim L8 lim 72n7oo n7gtoo n7gtoo n 8 n7gtoo 1 00 so the sequence an diverges 21 an 3332 Then 139 7139 739 n 7139 74 7 4391 nir oan n ogn2n n o 7g 7 n7m1 so the sequence an converges 3 an 813 Then n 10quot 10 n 10 e hm an hm hm hm 74 naoo name namicpej namlJreT 00 so the sequence an divergesi 4 an Note that nl 1 2 34 n71 n The expression nl is pronounced n factoriali 24816 32 64128 an T E 67 n7 m7 m 7040 So nl grows much faster than 2 as n A 00 Thus limnn00 an limnn00 1 0i 1 n71 2773 L2 17 2 72 1 5i an Ti wheren22i Soann ng nin i Then 7271 1 i i 00 hmanhm7hm 32711111770 ngtoo ngtoo n2 7 n ngtoo 1 7 E ngtoo 1 0 There are two approaches to computing limits we write down the sequence explicitly and determine whether it converges or ii when an is given as a fraction we divide the numerator and denominator by the fastest growing term in the denominator Pattern recognition for sequences You are given a sequence explicitly and you must gure out the formula for the nth termi Eigi an 4 g i wi Consider the numerator and denominator separatelyi So 2n71 an 3quot Note that usually sequences are constructed from the following set of standard sequences nth term Sequence 471 41741741w AW 1771717171771 n 12345678 anb ab2ab3ab4abm n 1P2P3p4p 517W a aa2a3a4a5w nl 126241207205040m Other sequences are constructed by adding multiplying and dividing the above sequences together term by termi E g an 7 777w 07 an 3311 Big Consider the function Let an f l ie the nth derivative of f evaluated at 1 Generating sequences by evaluating derivatives of some function is a common way and o generating sequences We must write down the formula for ant First we write down explicit the rst few terms of the sequence and try to see a pattern o f1z 7E1 a1 71 o f2z 539 a2 2 o f3z 72 a3 76 0 NM a4 24 so an mm Section 102 Series Points to cover 0 Convergence and divergence 0 Properties of in nite series 0 nthterm test for divergence o geometric series Series Sigma notation 221 n l2345l In general we write ai a1a2a3waNl We call 239 the index of summation and l is the lower limit of summation and N is the upper limit of summationl Note that other limits of summation are possible eg 0 is also a common lower limit of summationl Big 74 441 7 4 4 21471i2il The in nite summation the upper limit of summation is in nity is called an in nite series ie 1 an a1 a2 a3 a4 Note that we use sometimes just say series when we mean in nite series The sequence of partial sums of an in nite series ai is denoted Sn where Sn 21 ail ie 51 a1 52 a1 a2 53 a1 a2 a3 and so on We say an in nite series ai convergesdiverges if the sequence of partial sums con vergesdiverges 00 2a 7131103 i1 Please note the difference between an in nite series and a sequence A sequence is an in nite list 00 of terms an a1a2a3a4 whereas an 1n n1te ser1es 1s an 1n mte sum of terms 211 ai a1a2a3a4w Elgi Consider the in nite series i44114 n12n 2 4 8 16 This in nite series has partial sums 5391 5392 E 53 g and S4 gr There is a pattern arising here In fact Sn 21l So to determine whether 221 2 we need to determine whether limnn00 Sn divergesl Then 2n 71 17 2i 17 0 BM 23520 W 71320 T 1 So 221 2 converges and 221 2 1 Eg Consider the in nite series l l l l l l l So 51 l 52 2 5393 3 etc and so Sn n Thus limnn00 Sn limnn00 n diverges and thus the in nite series 211 divergesl Properties of in nite series We can use the following properties of in nite series to help When determining Whether an in nite series converges or diverges and if it converges What it converges to Consider two convergent in nite series 221 an A and 221 12 B1 Then 11 221 can CA1 M8 00 can ca1ca2ca3m ca1a2a3w cZan cAi 711 S 1 21 2101 bn A B1 01 bn a1 b1 a2 b2 as b3 711 a1a2agmb1b2bgm 2 an 2 bn 711 711 AB Note that both of the in nite series must converge in order to use the above propertiesi Elgi Find the sum of the in nite series 221 oo oo 5 1 Z 27 5 Z 27 5 1 5 711 711 Elgi Find the sum of the in nite series 221 i 1 1 1 1 1 Ellt27i2n1gt 27 i2n1 1 711 00 m 1 1 1 227 227 711 711 1 1 71 2 5 2 nth term test for divergence Consider the in nite series 221 ani 1f limnn00 an 0 then the in nite series divergesi Roughly speaking the reason for this is that if limnn00 an c then aN m c for large enough N Which implies that 221 an m 221 5 Which diverges if c f 01 Big 221 This in nite series diverges since 71 1 1 1 lim 7 lim 7 7 01 nam2n1 ngtoo2 20 2 52 Eigi 221 Here limnn00 2 01 So the nth term test tells us nothing Note that if limnn00 an 0 then the series 221 an may or may not converge Eigi The in nite series 221 diverges but limnn00 01 Geometric Series The in nite series 220 aT a aT aT2 aT3 is called a geometric series There is a nice formula for the nth partial sum 5 20 aTi where T f 1 Sn a17 TWA 17 T We can use this formula to nd the sum of an in nite series because a1 7 TWA 1 1 00 E aTn lim Sn lim 0 n7gtoo n7gtoo 7 T n So if lTl lt 1 then the in nite series converges to lir otherwise the in nite series divergesi SnaaTaT1aT2waTn a Ta aT1 aT2 aTnil a Ta aT1 aT2 aTn 7 aTn1 aTSn 7aT 1 So SnaTSn7aT 1 Sn17T a7T 1 Sn mifimi Eigi Consider the series 221 This is not in standard form for geometric series since the lower bound of summation is 1 instead of 01 We must rewrite it into the standard form 220 aT h 1 1 1 11 127252n712 3927 oo Now the series End 1 in 1s a geometr1c series where a l and T So the series converges since lTl lt 1 and it converges to the sum 8 S H 00 1 1 1 i a 5 2 1 1 1 n0 2 2 1 7 T 1 7 5 5 Eg Consider the geometric series 2 3 7g n1 So a 3 and T 71 This geometric series 7 converges since lTl lt 1 and the sum of the geometric series is 11 3 21 I 7 Eg Consider the series 220 So a 1 and T Thus this geometric series diverges since lTl 3 211 Section 103 p Series and the Ratio Test Points to cover 0 pseries o the ratio test pseries Let p be a positive constant A pseries is an in nite series of the form 1 7 1 1 1 Z 2737 n1 In the special case Where p 17 the series 221 is called the harmonic series A p series converges if p gt 17 and it diverges if 0 lt p S 11 Big Does the harmonic series converge The harmonic series is 221 so p l7 and it divergesl The ratio test Let 221 an be an in nite series With nonzero elements 11 The series converges if limnn00 21 lt 11 21 The series diverges if limnn00 gt 11 11 an1 an 31 The test is inconclusive if limnn00 Note that7 although the ratio test as Written above is for series of the form 221 an it also applies to series in the form 220 an or any other integer lower limit of summation 1 Eg Consider 220 Let an Applying the ratio test7 we have 2n1n11 a 3TH nlggo 2nnl 2n1 nl 520 27 39 n11l lim 2 1 naoo n1 0 S07 the series convergesl Eigi Consider 221 3 Let an 7 i Applying the ratio test7 we have an1 n1 hm hm n136 ngtoo new n 36n n4 6 nigon3396n1 1 A lim L91 namlJr 6 1 7 0 6 Since the lt 17 the series convergesi Eigi Consider 22171 Ei Let an fly i Applying the ratio test7 we have aw lt71gtn13nvltltn1gt2lt2n11 320 an algal 1n3nn22n1 7 1 71n1 3n1 n2 2n1 331 fly 3n n12 2n11 1 1j1T lim 71 3 HOV H my 127 q 1 0 l 1393gt39W39ml 3 2 Since 3 gt 17 the series diverges Eigi Consider the p series 221 Let an Applying the ratio test7 we have l l 1 lim Lnll hm ngtoo an naoo 1nz7 mp 31 WW n p hm 7 naoo n 1 1 p hm 71 naoo 1 E 1i So7 for p series7 the ratio test is inconclusive Series tests review Test Series Converges Diverges nthterm test 221 an no test limnn00 an 0 Geometric 220 a39r M lt 1 M 2 1 p series 221 nip 10 gt 1 0 lt p S 1 Ratio 221 an 11mm 31 lt 1 11mm 31 gt 1 Determine the convergence or divergence of the following series 1 220 1 Diverges by the n th term test since 2n1 2 20 nl goy2mnl goimnl o n VHF 21 221 Converges by the p series test since 7r gt 11 31 221 253101 Diverges by the n th term test since limnn00 001 41 220 112 014 Diverges by the Geometric series test since 2 112n 014 212n 2014n 710 710 710 and the geometric series 220014YL converges since M 014 lt 17 but the geometric series 220011371 diverges since M 112 lt 17 and the sum of a convergent and divergent series divergesi 51 221 lizni Converges by the ratio test 2n 11 471 an1 m 731330 7300 2n1 4n 1 2n1 14n 71 Zn 3914n1 1 1 1 11111 2157471 1 naoo W1 7 20 0 7 2 01 6 221 2222 Converges by the ratio test hm naoo 2n 13n hm 2n 2 I 3n H x 201 1 3n 1 hm ZnJr 42n 3 naoo 1 an 1 hm 7 naoo an 1 393n33n23n1 hm M inaoo 3n33n23n1 Section 104 Power series and Taylor s Theorem Points to cover 0 Power series 0 Taylor and Maclaurin series Power Series We now examine in nite series where the terms of the series contain a variable For example7 12 IS I4 15 7 E 2 757 n 771z nl M8 0 S H Note that 0 1 These types of series are called power series De nition An in nite series of the form 00 Z aux a0 mm agz2 313 n0 is called a power series in I An in nite series of the form 00 anz 7 c a0 a1z 7 c agz 7 c2agz 7c3 0 n is called a power series centered at c A power series is like a very long polynomial If we set I to a particular value7 we get a usual in nite series Why Power series give an alternate way of representing or approximating hard functions by a polynomial A power series can be viewed as a function of z oo Zanz 7 5 710 The domain of f is the set of all z for which the power series converges Every power series at least converges at its center aEyAampwVZywwmw mgWm n0 n0 Note that 00 1 Convergence of a power series 00 For a power ser1es centered at 57 En true 0 anz 7 c 7 precisely one of the following statements is l The series converges only at z c 2 The series converges for all z 3 There exist a real number R gt 0 such that the series converges for 11 7 cl lt R and diverges for 11 7 cl gt R Here R is called the radius of convergence ie the series converges on the interval 0 7 Rc R In the special case of 11 7 cl R it is dif cult to determine convergence or divergence except in simple cases In the rst case we say that the radius of convergence is 0 and in the second case we say that the radius of convergence is 001 We can use the ratio test to determine the radius of convergence 0 Applying the ratio test we have zn1 n 1 znnl Eigi Consider the power series 220 anz E n1 a 11 hm 7 717 lim 071 n7gtoo lim n7gtoo nl n11 lim 71700 0 n1l Since the limit is 0 for any I the power series converges for all 1 Its radius of convergence is 00 Recall that for a power series centered at 5 220 anz 7 c precisely one of the following statements is true 1 The series converges only at z c radius of convergence is 2 The series converges for all z radius of convergence is 0 3 There exist a real number R gt 0 such that the series converges for 11 7 cl lt R and diverges for 11 7 cl gt R R is the radius of convergence We use the ratio test to determine the radius of convergence Eigi Consider the power series 220 721 1 1 It is a power series centered at 711 Applying the ratio test we have n1 hm 71n11 1n12n1 wool lt71gtnltz 1gtW 2 nlggollt4gt2n1ltz1gtl 1 an 11 hm T n7 1 an In nliggolltz1gtl 1 z1 i So the power series converges for all values of z where 1 z1 lt1 lz1llt2 72ltI1lt2 73ltzlt11 So the radius of convergence is 2 Also the power series diverges for all z where 11 1 gt 2 What happens when 11 1 2 I 1 or z 73 We must check speci cally each case 59 If I 1 then the power series is 00 oo oo 1 1 Z 2 lt1 w 7 Z 2 2n 7 2w 710 710 710 which divergesi If I 73 then the power series is E 212273 1n Z 72n72n Z 31 fly I 2n 1 which diverges So the series converges for all z where 11 1 lt 2 and diverges where 11 1 2 1 Eg Consider the power series 220 7 4 i The radius of convergence is an1z 7 4 1 7 n1l3n1 7 nlggo an I 7 4 T nlggo nlS I 4 7 n1 3 7 71320 T 39 3w 39 I 4 1 lin1 n1 z74 oo So by the ratio test this series diverges everywhere except the center I 4 The radius of convergence is 0 Eigi Consider 220 lezni The radius of convergence is n5quot n1 71 n1 1 5n1 hm hm lt gt ltltngt gt4 n7oo an I 71700 71 n5 71n1 n 5n l1n1 700 1 n1 5 1 1 1 lin1 1 1 71 we 1 5 1 7r 5 So by the ratio test the series converges if zl lt 5 75 lt z lt 5 So the radius of convergence is 5 and the interval of convergence is 75 5 1 Taylor and Maclaurin Series A Taylor series is a representation of a function as a power series Let f be a function that is in nitely differentiable around or Then the power series 00 fn C n f C f C f C Tz 7 c 7 c 7 c2 7 c3 is called the Taylor series of f centered at c where f c denotes the nth derivative of f evaluated at or The Taylor series need not in general be a convergent series but it often is The sum of a convergent Taylor series need not in general be equal to the function value but it often is 60 Aside If we perform termbyterm differentiation of Tz we nd that T c f c for all n 2 0 so we can think of Tz as an approximation of f around 5 based on its derivatives evaluated at or A Maclaurin series is a Taylor series centered at 0 00 n m 230 foolzn f0 1 1672sz ngolzg Eigi Find the Maclaurin series for e 1 6 so ll 2 f1z 6 so f10 ll 3 f2z 6 so f20 ll 4 f3z 6 so f30 ll 5 f z 6 so 16000 1 Thus the Maclaurin series for e is 721 gizn lz12zsw Next we compute the interval of convergence m L an 1 hm lim 71 n7gtoo new n l 1n1l lnl I lim 71700 Oi So the this in nite series converges for all 1 In this case we have ex 220 for all 1 Eg Find the Taylor series of centered at 1 Then use the result to evaluate ll i so ll 2 f1z 7 so f1l 71 3i f2z 2 so f2l 1 7727 4 f3z 70 so f3l ill 5 sz lt71gtmilso fquot171 n1 The Taylor series of centered at 1 is Tz 7 i wa 71 7 i71nz 71 17 z 71 z 712 7 z 713 n0 nl n0 To evaluate we set I and nd the sum of the in nite series we 710 T ZHW 71gt Emma Note that 220 is a geometric seriesi What is the interval of convergence of Tz 220 71 1 7 1 7 n1 7 n1 hm lt 1 nltz 1 00 1 I 1 By the ratio test the series converges if 11 711 lt 1 71 lt z 71 lt 1 0 lt z lt 2 The radius of convergence is 1 and the interval of convergence is 0 2 i lim 17171l 11711 Big Find the Maclaurin series for 6752 We could write down the Taylor series directly 2 by computing successive derivatives of e but there is a quicker way We know that the Maclaurin series of e is Do 1 12 13 T 7 i n i i 6 72mm 1z23w 710 So to nd the Maclaurin series for 612 we can simply substitute 12 for z in the series for ea Thus 00 4 5 x2 7 l 2n 7 2 17 I7 6 72mm 71z 2 3 11 710 So using Taylor series of elementary functions we can construct Taylor series for other more complicated functions 1 E i F1rst Eigi Find the Maclaurin series for 82 2 1 2 21 22 2 23 3 17 n7 n e in0nl2x in0nlx 7111I211 TSlIHH Then 00 2 21 22 23 21 7 n7 2 3 e 717Hz 7iz z n d an e214 2 1 21 221 232 7 in if i i z 21an 1112111311 n Here is a list of basic Taylor series 11e1722 701 m 2i l722 071nz71n17z7iz7127z713w ac CA3 00 7 quot1 x7 2 x7 3 x7 4 11711 En7lz71n7z7177W 1 116 1 k1 kltk271gt12 kltk713gtltk72gtxs kltk71gtltk472gtltk7sgtx4 i sinz 220 EBSMQWAi cosz 220 1271 F o1 53gt From this basic list of Taylor series we can determine the Taylor series for other functions by the operations of addition t quot 39 quot 39 division quotm term by term integration term by term and composition Eigi We can differentiate term by term the series for 7 to obtain a series for 1 1 oo oo 77 7 7 7 7 7 Zlt7lgtnltz 71gt 7 Zlt7lgtn1ltz 71gt 0 0 I I n 62 Then Section 105 Taylor Polynomials Points to cover Taylor polynomials Approximating a definite integral Taylor Polynomials We b v s n that it is sometimes possible to obtain a power series representation of a function fx called a Taylor series Tm Using this power series we can evaluate the function at a point a a ut it is usually not feasible to evaluate the infinite series lnstead we use partial sums of the infinite series to approximate the value of the function at the point a a The nth degree Taylor polynomial of a function f that has derivatives tnrougb order n at A aw 3 Mo 72 f0 2 rec2 rec3 rem 7 7600 p Z kf NHquot flt gtf1 ria 100 V So a Taylor polynomial is a partial sum of a Taylor series 11 e that we can derive the Eg Consider x a m 7 3 e V N by substituting er for a in the Taylor series of 1 ao ai aylor series of w from the Taylor series of cw Eat The Utn degree Taylor polynomial for ca is Sm 1 The 1st degree Taylor polynomial for ca is 530 17 a This is the equation of the tangent ine 0V to the graph cw at a The 2nd degree Taylor polynomial for e1 is Sm 1 e a a 2 The 3rd degree Taylor polynomial for e is 53 1 7 ac 372 7 N The 4th degree Taylor polynomial for e is 5455 1 7 ac 0372 7 T There are two important observations to make here 1 A Taylor polynomial becomes a better and better approximation of the function as the degree of the Taylor polynomial increases7 2 The Taylor polynomial becomes a better and better approximation of the function as we get closer and closer to the center of the Taylor series For example7 in the following table we evaluate the Taylor polynomials for degrees 1 to 4 at the points ac 0705710715720 ac e Eigi Find sin0i2i The Maclaurin series of sinz is M I3 5 7 l sinz i zwwl 7I7iiii 7 We 27171 7 3 5 7 Using a calculator we can compute that sin0i2 m 0198669330795 Using the Taylor polynomial of degree 7 we can compute that 02 3 0 2 5 0 2 7 sin0i2 m 02 7 7 7 0198669330794 Eigi Find sin100i Using a calculator we can compute that sin100 m 7050636564111 Using the Taylor polynomial of degree 7 we can compute that 1005 x 1007 7 5 7 7 100 3 sin100 m 100 7 719 7581030746 Approximating a de nite integral Use the 8th degree Taylor polynomial for 677 2 to approximate the de nite integral fol e712dzi The Taylor polynomial for 677 2 is 00 4 3 712 7 Dn 2n 7 2 I I e 7E 7171 7 im We can obtain this by substituting 712 for z in the Taylor series for 6 So the 8th degree Taylor polynomial for 677 2 is 16 8 I4 z 531liz2 7 1i Using this polynomial we can approximate the de nite integral as follows This value has an error less than 0001 Section 106 Newtol s Method Polnts to covel o Newton s Method Newton s Method F ndlng the zelos of a functlon l e ndlrg 2 such that m 0 ol ndlrg the iraxls lntelcept of the functlon ls a fundamental ploblem ln algebla and calculus Sometlmes we can nd the zelos of a functlon algeblalcally e g quadlatlc equatl P ons but often ln leal llfe the functlons ale deallng wlth ale too c m llcatedt be solved algeblalcally ln whlch case we can e an appxoxlmatlon method such as Newton s method Newton s method Consldez a functlon f such as the one ln the dlagzam below Wm WMquot 19M K 9 A 1 h L g OWM I0CW The lst step ls to guess a value fo the zelo c m closel to c by ndlng the lntelsectlon of the tangent 0 say zl Then we tly to nd a polnt ls The equatlon of the tangent culve at 21 ls culve at 2 wlth the zrax y f0 WWW T 21 So the lntelsectlon of the targent cuxve Wlth the zeaxls occuzs when 0 f 1 WWW 11 zl 2 2 7 2 1 M1 1 ml gt2 2 7 M21 So oul next guess ls 22 Note that 512 m1 Mame 2 ls the hlst deglee Taylol polynomlal appxoxlma lon t of the functlon f centeled at the polntac1 So the tangent culve a pIOleaLlon of the functlon fz at the polnt 21 We can lepeat thls plocess and constluct anothel guess by ndlrg the lntelsectlon of the tangent culve of the functlon at 22 wlth the zraxls o a we MM In neral7 we can construct the n 1th approximation from the nth approximation using the following formula A M 7 f f n Wh n e stop constructing approximations We stop when lxn 7 xWHl is less than some desired accuracy ie consecutive approximations are Very close together So7 now when can write down Newton7s method Newton s Method 1 Make an initial approximation 1 of the zero of a function 2 Determine a new approximation using the formula aw 1 M Hm 3 If lxn 7 xWHl is less than some desired accuracy7 then STOP7 otherwise redo step 2 ach successive approximation is called an iteration Note that we only need the rst derivative of to be de ned around the zero of f Eg Find a zero offac x272 ie ac Use an 1 as an initial guess Here7 fac 2x 07 x L72 xWH M 7 7 2 Note that e 1 414214 so aftei only 4 itezationsofNewton sMethod we have an appioxihnation of J azezo of f to within 0000002 Convergence of Newton s Method When consecutive appioxihnations of Newton s Method appxoach a zeio of the function f then Newton s Method is said to converge But Newton s Method does not always conveige it may actually convezge fox some initial appioxihnations and divezge fox otheis Theze aie two cases to considez fox when Newton s Method does not conveige 1 If faa 0 then a 1 an e is unde ned ie the tangent cuzve to f at the point an is pazallel to the aeaxis so the tangent cuxve nevex inteicepts the aeaxis e m N m 2 maeaa ea does not exist This may happen when the appioxihnations cyc1e between di eie ent values on it can happen when the appioxihnations become fuzthex and fuzthez away fzom a zelo of f 69

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