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# CALCULUSIIPROB&MATRIC MATH115

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This 26 page Class Notes was uploaded by Claudine Friesen on Monday September 28, 2015. The Class Notes belongs to MATH115 at University of Pennsylvania taught by Staff in Fall. Since its upload, it has received 61 views. For similar materials see /class/215398/math115-university-of-pennsylvania in Mathematics (M) at University of Pennsylvania.

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Date Created: 09/28/15

The Inverse of a Matrix A square an matrix A is invertible if there exists an nxn matrix B such that AB 2 BA 2 Im IN is the ngtltn identity matrix A 1 is unique B is called the multiplicative inverse of A Not all matrices are invertible The Symbol used for B is Air A matrix that is not invertible is called singular example 2 5 3 5 You only need to check one 2 5 AA 1 direction since AZL 3 1 3 1 2 AIM M71 2 235 1 2 552 AA mull on n byA A 13 5 133 1 ABAA multisassoc 1 2 AAA 1 0 AA71I gtBAI sinceAIA 0 1 7lt2 Finding the Inverse of a 2 X 2 Matrix a 7 Let A c d 0 Switch a and d A 1 6 Negate b and c 9 Calculate D 2 ad be C 6 0 Divide every entry by D 1 2 4 7 2 7 2 1 A J A 1 2 2 3 4 372 172 2 D 2 ad bc D1423 AA11 D 2 3 4 72 0 1 Finding the Inverse of a 3 X 3 or larger Matrix Method 1 Let A be nxn Adjoin attach the ngtltn Identity matrix 11 012 39 39 39 In 1 O 39 39 39 0 Perform row operations onA untilIis obtained all ail aZYL 0 1 E E V 39 0 7 l aYLl aYLZ r r 0 1 v E A I By simultaneously applying thesarne row operations to 1we get A4 The 3 Elementary Row Operations a Multiply a row by a number nonzero b Switch rows c Add a multiple of one row to another row EH Row you want not changing to replace 2 0 1 LetA 72 3 4 FindAquot 75 5 6 2 0110 0 2R 4 0 22 0 0 R1R NEWR gt 715 82 0113 7234010 7234010 7234010 7556001 7556001 7556001 1 am 1 75 7872 0 71 2 3 4 0 1 0 2RR2NewR2gt 5 5 6 0 0 1 5RampNewamp3 2R 2 410 415 4 o 42 5R 5 42s 40 410 o is gtR27234010R37556001 NewRZIO 47 412 4 1 42 NewR40 42o 434 I 410 o 4 1 75 78 72 0 71 1 75 78 72 0 7 0 7 12 4 1 2 4R2 o 21 36 12 73 6 ampR2Nmamp 0 720 734710 0 4 0 720 734710 0 4 l 75 78 72 0 71 0 l 2 2 73 2 0 720 734 710 0 74 0 LetA 72 3 FindAquot continued 5 75 mpH 1 5 8 2 0 1 5R2 R1 NewRl gt 0 1 2 2 3 2 0 720 734 710 0 74 20R2 Hg NewKl 5R2 0 5 10 I 10 15 10 ZORZ 0 20 40 I 40 60 40 2 R1 1 5 8 I 2 0 1 2 R3 0 20 34 I 10 0 4 NewR1 0 2 I 8 15 9 NewR40 0 6 I 30 60 36 1 0 28 715 9 1 0 28 715 9 72RARINele 0 1 22 73 2 0 1 22 73 22R Rz vewR2 0 0 630 60 36 0 0 15 10 6 ZR3 0 0 2 I 10 20 12 ZR3 0 0 2 I 10 20 12 2 R1 1 0 2 I 8 15 9 2 R1 0 1 2 I 2 3 2 NewR41 0 0 I 2 5 73 NewR40 1 0 I 8 17 10 1 0 02 5 3 2 5 3 2 0 1 72 5 73 1 0 0 0 1 078 17 710 Aquot 78 17 710 2 3 4 8 17 710 0 1 0 0 0 15 10 6 5 10 6 75 5 6 5 10 6 0 0 1 AZ A AA 1 1 1 2 LetA 2 4 5 FindAquot 6 0 3 1 71721 0 0 2 4 50 1 072RR2NewR2gt 6 0 730 0 176RR1NewR1gt 1 1 721122417200 7500 gtRZ 2 4 5 I010 o 01 NewR20 5 9 72 1 o 76 o 1 17172100 06 97210 0 6 976 01ampampNewamp Further reduction will always yield another matrix with a row of zeros This matrix can never be turned into L 1 71 72 A 2 4 5 is notinvertible Ais singular 6 0 73 Properties of the Inverse o A4 quot A k 0 A quotAquot kgt0 O cAquot 0 A3quot BquotAquot 7 0 471211441 HHTRIXKH 3 x3 x 2 v a 1 z z a 1 2r 5 s Use the Inverse to Solve a System of Equations 2x xi 1 2 o 1 x 1 722 3x2 413 72 2 2 3 4 t2 2 4 5 53 3 is 5 5 i Ax b andAis invertible l l Earlier we found A A Ax A b I e x 72 5 73 x A lb zz is 17 z 5 710 To solvethe systemfoxx T T 11151 fmd the Inverse and muluply 11 by z m the 011112114427 R 1 P E D g 39U E 3 60 r 61 55 62 69 Frostbite Times 30 minutes 10 milnmos El 5 mlnutu Wind chill on 3574 0621 51r 3575M 51 04275Tw9 61 WhereT AirTemperature PF V Wind Speed mph Effective l norm T 2 temperature in OF V 2 Wind speed in miles per hour Wind Chill 3574 06215T 35 75V016 04275TV03916 Wind Chill 2 quotfeels likequottemperature in OF httpwwwweathergovoswindchiindexshtm Ma h115 Rimmer fag x w 151 Functions of Several Variables Section 151 9 Figure 10ab Computer generated graphs of functions 3 fx x2 3y2ex2y2 fxy x2 3y2ex2y2 Math 115 Rimmer 151 Functions of Several Variables Section 151 9 Figure lOcd Computergenerated graphs of functions C fx y sinx sin y sin x sin y d f x y mi Math 115 Rimmer 151 Functions of Several Variables Maple Commands gt withplots gt plot3dsqrtxy x11 Y11 Math 115 Rimmer age 151 Functions of Several Variables Section 151 9 Figure 11 Level curves being lifted up to graphs of functions w Math 115 Rimmer 151 Functions of Several Variables Section 151 0 Figure 13 World mean sealevel temperatures in January in degrees celsius Math 115 Rimmer 151 Functions of Several Variables Section 151 0 Figure 17 The graph of hx y 4 x2 y2 is formed by lifting the level curves a Contour map b Horizontal traces are raised level curves Math 115 Rimmer 151 Functions of Several Van39ables Section 151 9 Figure 19ab Level curves and two views of fx y xyex2 y2 Math 115 Rimmer 151 Functions of Several Van39ables Section 151 9 Figure 19cd Math 115 Rimmer 151 Functions of Several Variables Section 151 O Exercise 30 y X y afxyxy b fxyxy 1 2 c foe y m a fee y x2 yz e my x y2 f my sinxy FEE Math 115 Rimmer 151 Functions of Several Variables Section 151 9 Exercises 53 58 C 53 F 54 55 III 56 57 VI 58 z sinqx2 y2 22 szZyZe x y 1 z x24y2 Z x3 3xy2 Z sinxsin y 2 1 2 Z sm x1y Math 115 Rimmer 151 Functions of Several Variables PARTIAL DERIVATIVES 154 Tangent Planes and Linear Approximations In this section we will learn howto Approximate functions using tangent planes and linear functions TANGENT PLANES Suppose a surface S has equation 2 fx y where fhas continuous first partial derivatives Let Px0 yo 20 be a point on S TANGENT PLANES Equation 2 Suppose fhas continuous partial derivatives An equation of the tangent plane to the surface 2 fx y at the point Px0 yo 20 is Z 20 fXX0 YoXXo fyXOa Yoy YO TANGENT PLANES Example 1 Find the tangent plane to the elliptic paraboloid 2 2X2 y2 at the point 1 1 3 Let fX y 2X2 y2 Then fXX y 4X fyX y 2y TANGENT PLANES Example 1 80 Equation 2 gives the equation of the tangent plane at 1 1 3 as 2 3 4X 12y 1 or z4X2y 3 TANGENT PLANES The figure shows the elliptic paraboloid and its tangent plane at 1 1 3 that we found in Example 1 LINEAR APPROXIMATIONs In Example 1 we found that an equation of the tangent plane to the graph of the function fX y 2X2 y2 at the point 1 1 3 is z4x2y 3 LINEAR APPROXIMATIONs Thus in view of the visual evidence in the previous two figures the linear function of two variables Lx y 4x 2y 3 is a good approximation to fX y when X y is near 1 1 LINEARIZATION amp LINEAR APPROXIMATION The function L is called the linearization off at 1 1 The approximation fX y 4x 2y 3 is called the linear approximation ortangent plane approximation of fat 1 1 LINEAR APPROXIMATIONs For instance at the point 11 095 the linear approximation gives f11 095 411 2095 3 33 I This is quite close to the true value of f11 095 2112 0952 33225 LINEAR APPROXIMATIONs However if we take a point farther away from 1 1 such as 2 3 we no longer get a good approximation In fact L2 3 11whereas f2 3 17 LINEAR APPROXIMATIONs In general we know from Equation 2 that an equation of the tangent plane to the graph of a function fof two variables at the point a b fa b is z fa b fxa bx a 13a by b LINEARIZATION Equation 3 The linear function whose graph is this tangent plane namely LX y fa b fXa bX a fya by b is called the linearization of fat a b LINEAR APPROXIMATION Equation 4 The approximation waz aMMamv m uamg m is called the linear approximation or the tangent plane approximation of fat a b LINEAR APPROXIMATIONS Theorem 8 If the partial derivatives fX and fy exist near a b and are continuous at a b then fis differentiable at a b LINEAR APPROXIMATIONs Example 2 Show that fx y xeXY is differentiable at 1 O and find its linearization there Then use it to approximate f11 O1 LINEAR APPROXIMATIONS Example 2 The partial derivatives are MMWGWWWV fy X y XQeXy Mnm1 ML m1 Both fX and fy are continuous functions So fis differentiable by Theorem 8 LINEAR APPROXIMATIONs Example 2 The linearization is LX y f1OfX1OX 1fy10y O 11X 11y xy LINEAR APPROXIMATIONS Example 2 The corresponding linear approximation is xeXYz X y So f11 O111 O1 1 Compare this with the actual value of f11 O1 11e4l11 z 098542 DIFFERENTIALS For a differentiable function of one variable y fx we define the differential dx to be an independent variable That is dX can be given the value of any real number DIFFERENTIALS Equation 9 Then the differential of y is defined as dy f X dX See Section 310 DIFFERENTIALS The figure shows the relationship between the increment Ay and the differential dy yt 51V 0 a aAx tangent line f39ax a DIFFERENTIALS Ay represents the change in height of the curve y fX dy represents the change in height of the tangent line yt when xchanges by an amount dX AX d dxAx y I l l l 0 a aAx tangent line y fa f39lallx a nnnnnnnnnnnnn nun XV DIFFERENTIALS For a differentiable function of two variables 2 fx y we define the differentials dx and dy to be independent variables That is they can be given any values DIFFERENTIALS If we take dx AX X aand dy Ay y b in Equation 10 then the differential of 2 TOTAL DIFFERENTIAL Equation 10 Then the differential dz also called the total differential is defined by is dz dz dz fxx y dx 106 My gd gdy dz ua bx a fya by b l Compare Equation 39 SO in the notation Of differentials the linear approximation in Equation 4 can be written as Sometimes the notation dfis used in place of dz 77X Y z Ra b d2 DIFFERENTIALS It shows the geometric interpretation of the differential dz and the increment Az DIFFERENTIALS The figure is the threedimensional counterpart of the previous figure aAxbAyfa Axb Ay a AXJ Ayfa Ax b Ay stuface z fx y surface 2 fX y tangent plane z nmm m w awdnmmu m tangent plane z mm mmu mnmww m DIFFERENTIALS Az represents the change in height of plane the surface 2 fx y when x y changes from a b to a AX b Ay DIFFERENTIALS dz is the change in height of the tangent aAxbAyfa Axb Ay a AXJ Ayfa Ax b Ay stuface z fx y surface 2 fxy Z F 39 39 WV gt a AX b Ay 0 tangent plane ab0 Aydy Z fig b fra3 biix a fya biiy b tangent plane Z fl b fxlfl b I lm My 11 DIFFERENTIALS Example 4 a lf 2 fX y X2 3xy y2 find the differential dz b If xchanges from 2 to 205 and ychanges from 3 to 296 compare A2 and dz DIFFERENTIALS Example 4 a Definition 10 gives Bz BZ d d d Z Bx x By y 2x3ydx3x 2ydy DIFFERENTIALS Example 4 b Putting X 2 dX AX 005 y 3 dy Ay 004 we get dz 22 33005 32 23 oo4 065 DIFFERENTIALS Example 4 b The increment of zis Az f205 296 f2 3 2052 3205296 2962 l22 323 32 06449 Notice that A2 dz but dz is easier to compute DIFFERENTIALS In Example 4 dz is close to A2 because the tangent plane is a good approximation to the surface 2 X2 3xy y2 near 2 3 13 DIFFERENTIALS Example 5 The base radius and height of a right circular cone are measured as 10 cm and 25 cm respectively with a possible error in measurement of as much as 01 cm in each Use differentials to estimate the maximum error in the calculated volume of the cone DIFFERENTIALS Example 5 The volume Vof a cone with base radius r and height h is V l39rrZhS So the differential of Vis Z dV aldraldh dr ldh ar 3h 3 3 DIFFERENTIALS Example 5 To find the largest error in the volume we take the largest error in the measurement of rand of h I Therefore we take dr 01 and dh 01 along with r 10 h 25 DIFFERENTIALS Example 5 Each error is at most 01 cm So we have Al1 s 01 Ah s 01 DIFFERENTIALS Example 5 That gives dV 50070 11007170 1 3 3 207139 I So the maximum error in the calculated volume is about 2077 cm3 6 cm3 FUNCTIONS OF THREE on MORE VARIABLES The differential dwis defined in terms of the differentials dX dy and dz of the independent variables by aw aw aw d dx d d w ax ay y az z MULTIPLE VARIABLE FUNCTIONS Example 6 The dimensions of a rectangular box are measured to be 75 cm 60 cm and 40 cm and each measurement is correct to within 02 cm I Use differentials to estimate the largest possible error when the volume of the box is calculated from these measurements MULTIPLE VARIABLE FUNCTIONS Example 6 It the dimensions of the box are X y and 2 its volume is V xyz Thus dVa dea dea de ax By az yzdxxzdyxydz MULTIPLE VARIABLE FUNCTIONS Example 6 We are given that IAXI s 02 Ay s 02 Az s 02 I To find the largest error in the volume we use dx 02 dy 02 dz 02 together with X75y60 z40 MULTIPLE VARIABLE FUNCTIONS Example 6 Thus AVdV 6040O2 754002 7560O2 1980 MULTIPLE VARIABLE FUNCTIONS Example 6 So an error of only 02 cm in measuring each dimension could lead to an error of as much as 1980 cm3 in the calculated volume I This may seem like a large error I However it s only about 1 of the volume oft e 0x zfh x g0 y W c g dt dx dt dy dt Math 115 7 Rimmer 3 1155 The Chain Rule zf d xgst yhst Q as 8x as 8y as xZ lt Ht S PEI Math 115 7 Rimmer 33 a9155 The Chain Rule i a ax a By a X y F St S t mm a W amm m 5 m 8F zf u f a ydefined implicitly asafunction ofx dx 8i Fy y 300 By zfh h Set the function equal to 0 w F x y Math 115 7 Kim 171551155 The chm Rule 8F Wflxayazl Ling z implicitly as a function of x and y ax F z a z ghwl Z Set the function equal to 0 8i az 8y F y pFx y Z a F 8y 7 FZ az w Math 115 7 R39mmer 2 a155 The Chain Rule 16391 Double Integrals Single variable integral area under the graph of the function and above Double variable integral volume under the graph of the function surface and above the xyplane found by using the volume of infinitely many rectangular prismsr V HfxydgtcdyHfxydydx When the region R is rectangular the integration is simplified 1 2 1 2 1 x2 J4ixiydxjdy 4x777xy r l672ydy6yiy2 5 2 dyj87272y70dy This is called an iterated integral Double Integrals over general regions J39 7v 39 gtth x y l x x I a a dydx HfxydA igXfxydydx R N H 4xy y3dA R the region between y J and y x3 R H 4xy y3dydx R DZ Dr El U2 DA DE DE 1 1J2 1 3 2 y 4xy ydydx ny ix 03 0 4 x3 1 Z 12 1 Z 12 3 8 13 1 j 2x27 7 2x77 dxj 7i72x7i dx hii i 0 4 4 0 4 4 12 4 52 0 f h 7 1 1 1 1 55 4 77777777 12 4 52 3 52 156 163 pm 163 M 3 He ydA Rthereg10n1 y 2y SxSy R 25 2 g 2 g y 15 y 15 1 i 1 i as n5 2 A J J exydxdy 3 R 2y cy 2 cy ya 2 y2 1 J Je dxdy ye L dyJ1 ye yedy Area of a 2 dimensi0nal region using double integration Use f1mgt1 j dA R 82X i 01de i82Xg1xdx dh2y 1 1 dxdy fh2ltygt mygtdy 163 163 163 Order of Integration dictated by a the integrand 2 2 2 le y dydx 0 x b the region dA Rx2 y 4 x2 H R C Both the integrand and the region R triangle bounded R x y byyxy2xx3

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