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by: Jayde Lang


Marketplace > Rice University > Mathematics (M) > MATH 212 > MULTIVARIABLE CALCULUS
Jayde Lang
Rice University
GPA 3.86


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Class Notes
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This 12 page Class Notes was uploaded by Jayde Lang on Monday October 19, 2015. The Class Notes belongs to MATH 212 at Rice University taught by Staff in Fall. Since its upload, it has received 10 views. For similar materials see /class/224935/math-212-rice-university in Mathematics (M) at Rice University.

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Date Created: 10/19/15
Vectors Matrices and You A Quick Overview of Chapter 1 Math 212 Vectors Dot Product and the Norm Just like a regular variable is an element of R a real vector is a general element of R A vector 17 is de ned as follows v1 1 e1 H or 17 117 1727 1n vn There are special notations for vectors in R2 and R3 17 Ly and 17 17y72 to represent the coordinates along the xaxisyaxis and zaxis Vectors have a componentwise addition operation 17 117 11 1171 172 1172 1n wn They also have a componentwise scalar multiplication 0117 011717 011727 N 011 There are standard basis vectors in R3 100i 010I 001 This allows us an alternative way to write a vector in R3 17 I y 2 117 yj7 2 There is a dot product operation 0 R X R A R 17011711w1v2w2vnwn with the following properties 11717207where170170ltgt177 2a170117a17ou7170au7 3170u71717u717017and17u7o1717017u7017 417117u7017 So given this clot product we de ne the norm of a vector 17 as W W a When we refer to a vector as normalized we mean that it has norm of 1 To normalize a given vector we take 17 The norm also gives us a formula for distance between vectors dist1713H177 all Also if 9 is the angle between vectors 17 and if then a w W M cosy Note that this equation holds for any nedimensz39onal vector not just for R2 or R3 We have the CauchyBunyakovskySchwarz inequality The othogonal projection of 17 onto 6 is a E 0 17a 39Ua The norm also holds to the triangle inequality H771UH H77HH13H Matrices Determinant and the Cross Product An m by n real matrix is an array of m and rows and n columns and is an element of Rm X R A matrix is written as a11 a12 quot39 047 a21 a22 39 39 39 an A 1 1 1 or A am am 1 am 2 arm The determinant of a square n X n matrix is as follows a11 a12 am a21 a22 39 39 39 a2 n v DetA 1 1 i Zewk amDewAm 1 am 1 am2 arm in 27lilailDetAlJ for any 0 S k S m0 S l S n i1 The notation AM means the matrix made by elimating both the ith row and j h column For example in the 3 X 3 case a11 a12 as a a a a a a De all a a a1 1D6tlt 22 23gtSa12Detlt 21 23gta1 3D6tlt 21 22gt as am a31 am a31 as as as a33 The determinant has the following properties 1 Swapping two rows or columns changes the sign of the determinant 2 Adding a multiple of one row or column to another doesn7t affect the sign 3 Multiplying a row or column by a scalar also multiplies the determinant The crossproduct is a special operation gtlt R3 X R3 A R3 z7 i E 5X13Det v1 v2 v3 w1 w2 w3 Unlike the dot product the cross product only applies to 3dimensional vectors The cross product has the following properties 1 17x13 sint9 where 9 is the angle between 17 and if 217X1DL17andz7X1DL1D 317X130ltgt170ru700r17auforscalara 4i 6X137ugtlt17 5i vgtlt131i17gtltu717gtlt1iand17u7Xu17gtlt1i13gtlt 6 117 Xi a17gtlt 13 vX 113 for scalara We are able to multiply certain matrices together but when two matrices are square and of the same size we can always multiply them according to the following rule n AB am big 031 0 Where Cm Suzukka k1 In general this operation is not commutative iiei AB BA for certain matrices A and B A special matrix called the identiy matrix In an n X n matrix such that 1 0 0 0 1 0 In andforanyngtltnmatrixAAInInAA 0 0 1 We usually omit the n in cases where we know the dimension Square matrix A is invertible HA 1 3 A A 1 A 1 A I We have another important property ofinvertibility given by the determinant Square matrix A is invertible ltgt DetA y 0 One nal property of interest concerning matrix multiplication is DetAB DetADetBl A Little Geometry Parametrization and Other Coordinate Systems In R2 we have the usual equations for a line ymzbandazbyc0 In any dimension we can parametrize a line going through point 1539 in direction 17 by t 15 t1 If we want to make an equation for a line passing through points p1 and p we use the pointpoint form Zt 103 t152 7 pi Since this form has the properties p1 and 1 p3 this will be a useful form in this class In R3 we de ne a plane going through a point 1539 10 yo 20 with normal vector FL abc as P I azizobyiyoczizo 0 Using FL and as above we can nd the distance from point 5 11 yl 21 7 b 7 7 Dis p la11 10 91 yo 421 20M x a2 b2 c2 The coordinate system we usually use involving the xaxis and such is called the Cartesian Coordinate System There are other systems that will be used in this course The Polar Coordinate System for R2 is de ned by the following identities zTcost9 y Tsint9 WhereTZOandO O QTr The Cylindrical Coordinate System for R3 is de ned by the following identities z T cos 9 y T sin 9 2 2 Where T 2 0 and 0 S 9 S 27L This is essentially replacing zy with polar coordinates and leaving 2 as is The last standard system we ll consider is the Spherical Coordinate Sys tem z p cos 9 sin 45 y p sin 9 sin 45 2 p cos 45 Where2200 9 27r7and0 7n E i 3 a 1mm i ume L X9 59 is s the f amp L321 2 J V M39T r39hb 7 quotf an Shiv e x0 79 Zn quot5 by TQQ 2715 L 1 T C I 21 T t 7 ft Hli n 7 x0 1 170 4quot quota T0 50 quot quotT To r1 m J SN x x P quot it 9 11 iquot 39 m m f quot twquot l n x gt m V 139 t 77 9206 122quot 5522 H Iflt39f CMSTV 1 TG L 397quot 1quot i 3 16 t quotl LIL MATH 212 I vi 39 Calculus Spring 2006 Integration theorems If G is an oriented curve and f 31 y z is a function then Vf d5 ffinal point of G fstarting point of G i lt7 Note that the two end points of a curve can be viewed the boundary of the curve This result then already encapsulates the principle of the more complicated integration theorems If X is a curve surface solid then derivative of G G X i 8X the trick is to know what G can be function or vector field and what the derivativequot is in each situation Green s theorem Let S be a region in R2 and Fy y F1yyF2ay a vector field Then i EH72 EH71 K X L4 F am is 3 9 493 39J The boundary 68 has the following orientation Walking along the orientation the surface has to be on the left Green s theorem can be used follows A W52 lt9F 1 Anintegral 51 W is F ym In that case FZ Fld41dAareaofS f 1 y 1A can be replaced by a line integral The most common application 2 A more important application is to go the other way replace a line integral by an area integral This works if the line integral is over a closed curve Green s theorem comes in handy if the closed curve is di icult to parametrize eg the boundary of a rectangle whereas the area it bounds is easy to parametrize Stokes theorem Let S be an oriented surface ie you pick a normal vector indicating outsidequot and Fy y z a vector field Then VdeS FayzdS i i i 83 The boundary 68 has the following orientation Take your thumb of your right hand and let it point into the normal direction of the oriented surface then your fingers will give the direction of the boundary H Stokesquot theorem is mostly used to replace the integral of a vector field of the form V x F by a line integral Line integrals tend to be much easier than integrals of vector fields over surfaces Note that you can not use Stokesquot theorem unless your vector field is curl of another vector field 5 Stokesquot theorem also makes it possible to replace the integral of a vector field of the form V x F over a di icult surface by the integral of V x F over a nice surface with the same boundary OJ Note that if S has no boundary if S is a sphere or the boundary of a cube then ffoFdS0 Gauss theorem Let R be a solid and F00 y z a vector field Then divFmyzdv fi m Fmyzds Here we give ERR the normal vector which points away from R Gauss theorem is mostly used to convert a flux integral FmyzdS over a closed surface which might be di icult to parametrize the boundary of a cube into a triple integral over a solid which bounds the surface the cube and which often is much easier to parametrize Summarizing we get the following integration theorems in R3 39RdivFdV mm FdS Gaussquot theorem x FdS 2 am FdS Stokesquot theorem 0 V f d5 fend point of C fstarting point of C One can easily show that V x Vf 0 divV x F 0 ie the application of two consecutive derivativesquot give zero Here consecutivequot means derivatives which appear in consecutive integration theorems For example it is not true that diva 0 for any function This observation that the application of two consecutive derivativesquot gives zero is in fact related to the fact that if one takes twice the boundary of any geometric object then one gets nothing For example the boundary of a ball is the sphere but the boundary of a sphere is empty Idiot s guide to integrals 1 Line integrals is a If the vector field F is a gradient vector field ie F gradf for some function f then F is end point of c initial point of c 7 b If t is a closed curve in the plane then you can also apply Green s theorem E E F d5 Kisecond component of F fall st component of F L4 l a r r R 3 9 where R is the region bounded by the curve 0 cf above for details concerning the orientations c If F is not a gradient vector field then you have to find a parametrization ctt 6 11 for c and compute 5b F dS Fct c tdt l t l 511 2 Flu integrals is a If F is the curl of another vector field ie if F curlG then you can apply Stokesquot theorem FdScurlG4dS G S l l l bmmdaryaf cf above for the details on orientation b If the surface is a closed surface ie has no boundary then you can apply Gaussquot theorem F dsFdv i i l l l B where B is the 3 dimensional body with boundary 8 c If F is not the curl of another vector field and if the surface has boundary then you have to the integral by handquot Pick a parametrization Mu 1 and conlpute d5 Flt1u14 Tu X Tiadudu 3 Integral of a function over a curve surface 3 dimensi0nal body In this case there are no tricks you can apply you have to use the usual formulas


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