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# Chapter 12 Secs. 1-6 MATH 114 - 002

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This 37 page Bundle was uploaded by Jasmine Jackson on Sunday October 4, 2015. The Bundle belongs to MATH 114 - 002 at University of Pennsylvania taught by Professor Ching-Li in Summer 2015. Since its upload, it has received 27 views. For similar materials see Calculus II in Mathematics (M) at University of Pennsylvania.

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Date Created: 10/04/15

Math 114 Rimmer wring 121 3D Cartesian Coord 121 Three Dimensional Coordinate Systems Cartesian le 3 Coordinate planes Thomson Higher Education b Points have coordinates Math 114 Rimmer ww ga 121 3D Cartesian Coord xayaz Example Z A456 39 B3 3 l A T456 C 2 20 quot 139 lt gt C i 2 2a 0quot 7 I f 39 I ifxi y x J 1 f I I f Lquot J B 1 x 3 3 l Math 114 Rimmer Distance between points ID1 x1 y1 Z1 121313 Cartesian coma P 2 X20535 A456 d z PIPZ 4052 x12 y2 y12 z2 zl2 B3 3 1 C 2 20 Find AB and AC Which is larger 2 iABi Hfs 326 lt 1 AB 16449 5 210677 iACiJ4 2gt2s 2gt2lt6 ogt2 AC x364936 11 Math 114 Rimmer Equation Of a SQheI e39 M33 121 3D Cartesian Coord x h2y k2z m2 r2 Center k m Radius r Find the equation of the sphere with center at 0 3 6 and radius xg x2y32z 62 3 Find the center and radius of the sphere that has the given equation 4x2 4y2 4z2 4x8y 3 0 4x2 4x4y2 8y44z2 3L4 4x2 xi4y22y14z23Li 4x 24y124z2 8 x 2 y12 z2 2 Center l0 Radius J5 Math 114 Rimmer I m3quot 121 3D Cartesian Coord yA 5 0 2 c y 5 a line in R2 myquot E uud v x 5 N J Jquot myiaMwemR3 Thurmaun ngher Educaliull x r i a V T Math 114 Rimmer ID F1 hdl the distance ham 3 Ti 5 me each 0139 he millewmg e39 f quot 121 3D Cartesian Coord 1 The xyaplane h The yrspllane e The IZ p Ci The xa xie e The yaxis f The axis a 5 b 3 c 7 d P300 d3 327 02502 d49T 7 4 e P09790 f P0 0 5 d3 027 72 5 02 3 027 02 5 5 d9 253 4 mag 2 d d G r t w a Math 114 Rimmer 11 Find an equatmn at the largest sphe e th39h center 3 4h 9 that m39ijze l 121 3D Camsian Coord is emitaimed in the First nelamL The radius of the sphere will be the shortest distance to any of the coordinate planes r 4 Center 549 x 52y 42 z 92 16 1 a m w Math 114 Rimmer bind art Equatt eat the set tit all permits equidistant ter the PEItiequot 121 3D Cartesian Coord points At t 5 3 and 3m 2 392 Describe the set Let Px 2 z be a generic point that is equidistant from the given points PAPBI x12y 52z 32 x 62y 22z22 x12 y 52 z 32 x 62 y 22 z22 2x1 10y25 6z9 12x36 4y44z4 2x1 10y25 6z9 12x36 4y44z4 14x 6y 10z3644 1 25 9 14x 6y 10Z 9 Math 114 Rimmer 122 Vectors Quantities such as area volume mass and time can be characterized by a single real variable Other quantities such as displacement velocity and force involve both magnitude and direction To represent these quantities we use a vector represented by a directed line segment arrow B The magnitude of a vector terminal point is represented by M orHvH Any other vector u that has the same magnitude and direction as v initial point 1s called an equlvalent or equal vector gt u v Math 114 Rimmer we 122 Vectors We can multiply a vector by a real number 6 This is called scalar multiplication cv has a magnitude that is cl times the magnitude of v cv has has the same direction as v if c gt 0 cv has has the opposite direction as v if c lt 0 1142013 PM Math 114 Rimmer We can add a vector v to another vector u 122 Vectors This is called vector addition v u Connect the terminal point of the first vector to the initial point of the second vector V 11 11 When connected this way the sum is the vector from the initial point of the first vector to the terminal point of the second vector V V Vector subtraction V u is just vector H V addition in disguise V u V u This can all be summed up using the parallelogram determined by V and u H V V II u 11 H V II V a M th 114 R39 So far we have studied vectors geometrically 12 Vectors lmmer We now want to look at vectors algebraically A v lta a gt 11 612 l 2 V i The magnitude of V is found by E a 2 2 2 i V ai a2 4 gt a1 V l gt the vector is called a unit vector V Standard unit V6Ct01 S i lt10gt i lt01gt Now V can be written as r1 xH va11a21 azj all ll42013 Math 114 Rimmer Now for 3 d1men310ns we have 122 Vectors ilt100gt V alia2ja3k jlt0l0gt I k lt0 Vltala2a3gt apaz and a3 are called a the components of V More specifically a1 the i component of V if 3quot a2 the j component of V x n39 I I I I I I I I I I I I I L 1 5 a3 the k component of V A Math 114 Rimmer w n 122 Vectors V Scalar Multiplication V lta1 a2 a3gt scaled by a factor 6 CV ltcal ca2 ca3gt multiply each component by 6 Vector Addition v 611612613 added to u 191192193 V 11 lt01 91 612 92 613 93gt add componentwise I 39 FIJI 1 Iquot 125 ll 39 I l 2 quot I 39 quotf ew wt LFr I y 3 r 2 39 di ISJIa ma yl 39 will I39IIII x1 X2 1142013 Math 114 Rimmer 122 Vectors Vector from a point A to another point B A 191192 V B Vltb1 alb2 a2gt In 3dimensions A a1 a2 V ltb1 alb2 a2b3 a3gt A unit vector u in the same direction as another vector V A 4 Vlt3a4gt Scale the vector v by the reciprocal of M v5 u V 1 lt 2 Cl a 61gt 21A 1 2 3 lt5 5gt I gt M M 3 M IV V C Find the component form and magnitude of the vector V With the I m Math 114 Rimmer 122 Vectors initial point 320 and terminal point 415 0 Find 3V C Find a unit vector in the direction of V vlt4 31 25 V ltl l5gt 0gt 3v 3lt1 15gt gt 3 M V 12 1252 x1125E 3v lt3 315gt HZZLLJlt1L5gt lt115gt 1142013 Math 114 Rimmer 122 Vectors A 200 lb traffic light supported by two cables hangs in equilibrium As shown in figure b let the weight of the light be represented by w and the forces in the two cables by F1 and F2 As shown in figure c the forces can be arranged to form a triangle Equilibrium implies that the sum of the forces is 0 Find F1 and F2 and Find the magnitudes of F1 and F2 F F2 F 1 F FHcos20 Fsin20 ltI 152 0 L Hmnsmeo 1lt1 1 gt F2cos15 lFllcos20 F2 lt F2HCOS15 F2Hsin15 gt w 0 200 a HFZHSin15 5 HF1cos 20 F2 cos 15 0 gt HF1sin20 F2Hsin15 200 0 F2Hcos15 cos20 F F w F2 1 2 HF2cos15 F F2 w 0 200 s1n20 HF2Hsin15 2000 COS F1 HF2cos15 tan20 sin15 200 W e HF2 200 x 32766 lbs cos150tan200sin150 z 33639811bs 1142013 Math 114 Rimmer 123 Dot Product We can add two vectors what about multiplying two vectors Since adding two vectors yields another vector where the corresponding components are added will the same work for multiplication No the product of two vectors yielding another vector where the corresponding components are multiplied is meaningless There are actually two vector products that yield meaningful results but neither of these give a new vector using component wise multiplication 123 Dot Product 124 Cross Product Math 114 Rimmer 1E In u I Zrm ui The dot product of u ltulu2u3gt and V ltv1v2v3gt is ll 39 V bill1 M2112 M3113 Note the result is a number scalar not a vector Example u lt2 34gt and v 065 uV2O 3654 uV 1820 uV2 Math 114 Rimmer Propert1es of the dot product 123 Dot Product Let 11 V and W be vectors and let 6 be a scalar 1 ll 39 V V ll Commutative Property 2 uVWuvuW DistibutivePropertY 3 cuvcuvucv 4 0VO 2 5 vv v Math 114 Rimmer Is the followmg eXpress1on meamngful 123 Dot Product scalar vector N0 scalar vector Yes c laww No scalarxscalar NO VCCtOI 39 VCCtOI 6 139 V W Yes scalar vector f lu 39 V 39 W YES scalar 39 scalar Math 114 Rimmer rim 0 lg tgn F1nd the angle between two vectors 12393D Pquot d Let u and V be nonzero vectors then Law of Cosines 2 2 2 lv u u lv 2 UHVCOS9 u why24w th A uvuv cost9 V V V ll2 V ll V ll WV 2 cost9 rv u ltv ugtv ltv ugtu qu Vu 2vvuvvuuu w v u2v2IuIZ2uv 11 V Osasz Math 114 Rimmer Find the angle between two vectors 12393D Pquot d ll V u lt3 12gt and v lt 40 2 COS 6 2 u M uov3 4 1022 124 2 8 u 32 1222 x9T 2M vy 420222 x164 m2J 8 Z 4 70 4 2 cosl9 2 mm 70 35 Math 114 Rimmer Find a vector that points in the same 12393D Pquot d 0 If t t direction as V but has a magnitude of 10 You W a V60 or in the direction of V V 2 lt5 3 4gt with magnitude k C 39ut l b lvl 52 32422591 x50 52 JS WV ylvl We can first scale V down to be a unit vector l lt 5 3 4 gt M 5J5 s 5J5 Finally we scale this unit vector up to have magnitude 10 v 50 30 40 10 6 8 10Mlt5J 5 5J gtZlt gtZlt5 3 4 gt 10 Essentially we just scaled the original vector by xE 5J5 Alternate form of dot product Math 114Rimmer 123 Dot Product uV qu COSQ For nonzero u and V ulgt0 and Vgt0 gt u V and cos 6 will always have the same sign 0lt9lt 6 acute gtCOSHgt0 gtuVgt0 glt6lt7560btusegtcos6lt0gtuVlt0 la 6 6right gtcos60gtuV0 30 The vectors u and V are orthogonal if u V 0 Math 114 Rimmer 123 Dot Product Direction angles of v a the angle between v and i 8 the angle between v and j 7 the angle between v and k Direction cosines of v 003al cos6v 2 cosy2V 3 cos49 V V V 1qu components of the unit vector in the direction of V i lt1 0 0 Math 114 Rimmer vector projection of u onto V 12393D0 Pmdw projvu has magnitude equal to ll COS H the magnitude of the projection vector is called the scalar projection of 11 onto v or the component of 11 along v pfojvu COmlel compvu u cos 9 remember u v uv cos 9 uv M the vector projection of u onto v has compvu V ll39V so lul cos 9 H M We saw earher that 1f you want a vector 1n the d1rectlon of v w1th magmtude k Just scale v by v as its magnitude and goes in the same direction as v Math 114 Rimmer 123 Dot Product When a constant force F moves an object a distance d in the same direction of the forcethe work W done is WFd If a constant force F applied to a body acts at an angle 6 to the direction of motion then the work done W is W COS 6 where d is the displacement vector l Using the dot product we have w F d A toy wagon is pulled by exerting a force of 25 pounds on a handle that makes a 200 angle with the horizontal Find the work done in Pulling the wagon 50 feet F lt25cos20 253in20 gt d lt500gt W Fd 1250cos200 zll75 ft lbs or 1593 joules Math 114 Rimmer 124 Cross Product The cross product of two vectors is a vector with the special quality of being orthogonal to both original vectors The cross product yields a vector in contrast to the dot product that yields a scalar number The cross product of u ltu1u2u3gt and V ltv1v2v3gt is 11 X V ltM2V3 M3112 M3111 bill 3 bill 2 M2111 gt The definition only applies to threedimensional vectors The cross product is not defined for twodimensional vectors Instead of memorizing what gets multiplied by what there is a convenient way to calculate ugtlt v using the determinant form with cofactor eXpansion Math 114 Rimmer Determinant 2 x 2 Math 114 Rimmer v ii an llotormlnants Reduces to finding 3 2X2 determinants using cofactor expansion on the first row Take each entry in the first row we will multiply each of these entries by a 2X 2 determinant The 2X 2 determinants are found by crossing out that entry39s column and row One last thing is to mutiply the 2nd entry by 1 1 3 2 9 4 l 4 l 9 1 9 413 1 3 1 5 12 5 3 5 3 l 19 123 1202 345 35784 2138 W sumof sumof forward backward Copy 15 2 columns diagonal diagonal 1 6 72 1 6 products products 3 L3ci 1 272 30 361584059 19 e e 4 53 o 2 o4 85 Math 114 Rimmer How to find the cross product using determinants 124 Cross product i j k u u u u u u ugtltvu1 L12 L13 2 31 11 311 2k V2 V3 V1 V3 v1 v2 H X V 2V3 3V2 i 1V3 3V1 1V2 2V1k 11 X V ltlt2V3 M3112 M3111 bill393 bill 2 M2111 gt Let u 1 21gt and v lt31 2gt Find ugtlt v i j k ugtltvl 2 1 2 1i 11 1j12 l 2 3 2 3 l 3 1 2 4 1i 1 2 3j 16k ugtlt v 3i5j7k lt357gt Alternatively a Math 114 Rimmer 124 Cross Product 1 i 3 11 X v 1 39j21f391 11 2 4i3jk2ji 6k 3 h L2 03 a 11 X v 4 1i32j 16k ugtlt V 2 lt3 5 7gt Verify that the cross product is orthogonal to both 11 and V uxvu lt357gtlt1 21gt 3 107 0 uXVV 357lt31 2 95 14 0 Math 114 Rimmer Algebralc Propertles of the cross product 124 Cross Product Let 11 V and W be vectors and let C be a scalar luxv vxu Zlvdvwuxvuxw 3c vu a xvux w 40XVVXO0 U1 vxv0 6uvx zuxv 7 7 ugtltvgtltwuWv uvw Math 114 Rimmer O m 124 Cross Product Place your 4 fingers in the direction of the first vector curl them in the direction of the second vector Your thumb will point in the direction of the cross product H X V V X H by switching the order you get a vector in the opposite direction right hand Math 114 Rimmer Geometric Properties of the cross product 124 Cross Product Let u and V be nonzero vectors and let 6 be the angle between u and V l ugtlt V is orthogonal to both u and V 2 ugtlt vi lu lV sin 6 3 ugtlt V 0 if and only if u and V are scalar multiples of each other 4 ugtlt V 2 area of the parallelogram u determined by u and V lvl 5 ux V 2 area of the triangle having u and V as adjacent sides 553 Math 114 Rimmer 124 Cross Product A nice online java applet for the cross product can be found here httpvwwvphysyreducoursesjavasuitecrossprohtml Math 114 Rimmer saw 124 Cross Product Volume of the parallelepiped determined by the vector ab and c Area of the base 2 lbxc Height 2 compbxca 2 la cos 6 Icompthal l Volume 2 lb gtlt cl a cos 6 Volume 2 a b X c e this stands for absolute value a bxc is called the scalar triple product The vectors are in the same plane coplanar if the scalar triple product is 0 a1 a2 a3 abgtltcb1 92 93 The scalar triple product can be written as a determinant Math 114 Rimmer 124 Cross Product Letult201gtvlt111gt and wlt022gt Find uVXW 1 1O1 22 2012 0 01 2 3 Q oquot I 32 a0 o o uVXW 1 402 040 26 4 9 Q a 9 o 9 o v 9 o o Math 114 Rimmer mcif an 124 Cross Product iszk 1 jgtlti k jgtltki kgtltj i J kgtltij V igtltk j Fde me we 1mm malt wi h de emm imammm but by wing pmpem ea at ma a murnd ii K i ijgtlti j igtltiigtlt jjgtltijgtlt j 0 0 igtltjjgtlti k k 2k Math 114 Rimmer 124 Cross Product In physics the cross product is used to measure torque Q s e rs 0 IF Consider a force F acting on a rigid body at a point given by a position vector r The torque 239 measures the tendency of the body to rotate about the origin point P TrXF T 2 er 2 r F sin6 9 is the angle between the force and position vectors A r L 6 Math 114 Rimmer swan 124 Cross Product In Exercises 2 5 and fin the lime ni 39 e 39 1111i exerted by F on the bull er 1 F HPQl in an HFI 3E lbs rmer in feet x 5 peimds quot o 0 Wm M th 114 Ri 125 Equatlons of L1nes and Planes 122 Equation T nes and Planes In order to find the equation of a line we need A a point on the line P0 x0 yo Z0 B a direction vector for the line V ltab c 139 2 1390 IV vector equation of line L Z ltxyzltx0y0z0tltabc equating components xx0at y y0bt zz0ct parametric equations of the line L y eliminating the parameter t solve for t r0 lt 350 yo Z0 1n each then equate the results l ltXyZgt X XOZy yOZZ ZO a b 6 P0P W tltaa 19 C symmetric equations of the line L Math 114 Rimmer swan 125 Equations of Lines and Planes Find parametric equations of the line containing 51 3 and 3 2 4 In order to find the equation of a line we need A a point on the line P0 x0 yo zO Plck 61th 190111t B a direction vector for the line v ltab c Use the vector from d one point to the other pomt 1rec 3 x0 at vlt3 5 2 14 3gt y 2 yo M vlt 2 31gt Z Z0 ct L x 5 2t y 1 3t z 3 t Math 114 Rimmer 125 Equations of Lines and Planes Two lines in 3 space can interact in 3 ways A Parallel Lines their direction vectors are scalar multiples of each other B Intersecting Lines there is a specific I and S so that the 1 4 lines share the same point C Skew Lines 44121 their direction vectors are not parallel and there is no values of t and S that make the lines share the same point Math 114 Rimmer Determine whether the l1nes LI and L2 are parallel skew 125 Equations of Lines and Planes or intersecting If they intersect find the point of intersection Li Lz x 3 I x 8 25 y 5 3t y 6 43 z 1 4t z 5 3 Set the xS 2 Check to make sure that thez values are e ual for this t and s 3 t825 1 4t q5 S t52s t 5 2 1 4 15 2 S 323 check Set the y 39 S This Sho d happen at Now find the pt of intersection the same time I using Ll 53IZ 6 4S soplugint 5 25 x 3 1 53 5 2S 6 4S y 5 1 5 15 652 6 45 z 1 41 10 6S 6 4S 2S4 t 5 2 2 S2 t 1 Math 114 Rimmer 125 Equations of Lines and Planes Determine whether the lines LI and L2 are parallel skew or intersecting If they intersect find the point of intersection L1 L2 x 4 I x 3 25 y 8 2t y 1 S z 121 Z 3 35 Set the x S 2 Check to make sure that the z values are equal for this t and S 4 2 I 3 2S z 12S 12033304 Setthey39s 36 0 This should happen at 8 2t 2 1 S the same time I This means that the x and y 8 21 2S 2 1 S SO plug int125 values are the same for t 3 and S 1 but the z values are 8 2 4S 2 1 S different I 3 The hnes are skew 5S 2 5 S 1 Math 114 Rimmer 3 S wagequot 125 Equations of Lines and Planes In order to find the equation of a plane we need A a point on the plane P0 x0 yo Z0 this vector is called B a vector that is orthogonal to the plane n lta b cgt the normal Vector to the plane Pr F 0 11 Z nr nr0 PWY Z vector equation of the plane nr r00gt 1 0 axx0by y0czzo0 y scalar equation of the plane r02ltx0aymzogt 139 x Z ax ax0by by0cZ CZO0 x lt y gt axbycz ax0by0cz00 P0Pltxx0 yy0 ZZ0gt axbyCZd0 n lta 9 Cgt linear equation of the plane Determine the equation of the plane that contains the lines L1 and L2 Math 114 Rimmer 59 125 Equations of Lines and Planes Ll L 2 In order to find the equation of a plane we need x 3 I x 8 25 39 y 5 3 y 6 4S A ap01nt on the plane z l 4t z 5 s B a vector that is orthogonal to the plane We have two points in the plane n lt61 2 c fromL1 35 1 and from L2 8 65 We have two vectors in the plane fromLl lt 13 4 and fromlZ lt2 41 Let u lt 13 4gt and v 2 4 1 Find uXV J k 3 4 1 4 1 3 ugtltv 1 3 4 1 1 J 2 4k 2 4 1 4 1 2 1 3 16i 1 18j 4 6k ugtlt v 13i 7j 2k n 13x 7y 2zd0 13x 7y 2z720 133 75 2 1d0 39 352d0gtd72 13x7y2Z 720 39 39 Math 114 Rimmer Deternnne the equation of the plane that passes through 125 Equations of Lines and Planes 123321 and 1 22 P R In order to find the equation of a plane we need A a point on the plane FE lt3L2213gt PE 2 lt1L22 23gt we have 3 to choose from B a vector that is orthogonal to the plane 11 a b c we need 2 vectors in the plane P Q 2 0 2gt lt 2 4 1gt Let u lt20 2gt and v lt 2 4 1gt Find ugtlt v i j k 0 2 4 1 2 2 2 1J 2 0 2 4 2 4 1 k 0 8i 1 2 4j 8 0k ugtlt V 8i6j 8k n orn4i 3j4k inlowest terrns 4x 3y4zd0 41 3243d0 4 612d0 gtd 10 4x 3y4z 10O Math 114 Rimme1 Two distinct planes 1n 3 space either are 125 Equations of Lmes and Planes parallel or intersect in a line Z mm Math 114 Rimmer Find the line of intersection of the two planes 125 Equations of Lines and Planes x 2 y Z 0 Use the elimination method 2x 4y 2z 0 2 2 2x3y 2Z0 x y Z O 2x3y2z0 2x 3y 2z O 4 4x y 0 y x Take the first equation x 2yz0 andpluginy4x x 24xz0gtz7x x can be anything and we have y and z as functions of x Let x t then y 4t and z 7t L x y4t Z Math 114 Rimmer 125 Equations of Lines and Planes If two planes intersect then you can determine the angle between them 4 between planes 2 4 between their normal vectors n2 6 n1 39 2 111 cos 6 1an 6 Fin the angle between the planes 6 nl 112 6 6 540 x y z cos z 2x3y 2z0 Iannz 102 gt 111 lt1 21gt and n2 lt2 3 2gt n1 6 and n2 gtn1n22 6 2 6 Math 114 Rimmer swan 125 Equations of Lines and Planes Distance between a point and a plane 1 Take any point in the plane call it P0 x0 y0z0 pm x1 y1 Z1 2 Letb ltn xoyl yozl z0gt 0ij 3 Take projnb The length of this vector 2 D H Remember projnb compnb n 4 Since D must be positive take the absolute value 11 39 bl D Inl axbvczd 0 V n39blta b cgt39ltx1xovyi YO7Z1Z0gt ax1xoby1yoCZ1Z0 ax1by1 CZ1 axo byo CZ0 Er J gtnb ax1by1czld d D axl 191 CZ1d xa2 192 62 Math 114 Rimmer 125 Equations of Lines an es 42 E2 2 25 mm m f 4 2 x 1 391 h A i 0 Xl h g z 7 j j w b c f 5 O Xl yt awl d worm 3m d4 lt5 D K 3 3 HM a and Planes P 9A w Qua nn w zb V L UI F Warmm frum 3 Paint S m a LinI Thmugh F Familial h 1 r z lmwvl Distances n Exemiae 33 33 fian 39th xdj 39tau39uct d j39mn Me mth In Me Him2 33 EL 13 IE JI 4 139 2r 34 1 Humquot 139 3 5 2 1 31 JI 3ft 2 1 H 3 1 3 4 1 4 r39 3139 3 2L 33 L 4 3 139 ID 4 4 HM g 2 Iquot 5 3L 1 y I 3955 pII 2r y5 5 3 39 3 Z 43 d Yszlt 0l jvgt 10 gt A 11gtng M r gg Si ch1 3 3 22 3 m 3 S 3 D L X0J l l 53 Fiquot 125 y 2K9 131 0 Math 114 Rimmer 125 Equations of Lines and Planes J c Q y quotgt lLVV Psxvm J MM 1172012 Ell maid mis ttztxges All traces are ellipses abcgtsphere r 1 Math 114 Rimmer d t 126 Quadric Surfaces m A m Horizontal traces are ellipses Vertical traces are hyperbolas Axis of symmetry corresponds to the variable with negative coefficient 1172012 I W a h immer Hyperholmd two sheets 325u13drissmaces Vertical traces are hyperbolas Horizontal traces are ellipses z k k gtc Axis of symmetry corresponds to the variable with positive coefficient Cone elliptic 2213 39335 7 Horizontal traces are ellipses Vertical traces are hyperbolas k i O x k gt hyperbolas y k gt hyperbolas if Llquot k O 3 lines Axis of symmetry corresponds to the 2 2 2 variable on the left side Math 114 Rimmer m a 126 Quadric Surfaces Pagraboloid ell italic Horizontal traces are ellipses Vertical traces are parabolas Axis of symmetry corresponds to the variable raised to the first power Math 114 Rimmer 15 126 Quadric Surfaces Farahinlaid hyperbolic Horizontal traces are hyperbolas Vertical traces are parabolas 1172012 1172012 I J Math 114 Rimmer is 126 Quadric Surfaces El jiptic cylind r 39 i Math 114 Rimmer air is 126 Quadric Surfaces Hy pEFh 39 it cy in 1131 Pamhml if cylind r ww Math 114 Rimmer W 126 Quadric Surfaces 1 18 Match the Equati n with its graph la39h ed I VHUI Give rcasuns Far yum clm cesm II x1 4y 9 1 VII 21 9x 43 21 I IV 23 x2 32 2392 1 24 ix 32 2 I y 2x1 22 VI 26 32 x2 232 111 222 I VIII g Math 114 Rimmer waging 126 Quadric Surfaces III W 1172012 I L I Math114 Rimmer Reduce the equation to one oi the Sl llt l l itl rrnsa clai smly 1K 1260uadricSurfaces 1 r l he surface and sketch it Ell A 2112 32 32 4x 7 y 442 U 31 43 iv3 4 4 24 36 o 34 423 3 4 20 l 2 2 2 2 31 xyTZT or iy Z 5 g 6 3 2 elliptic paraboloid vertex 000 opens in the positive x direction 2 2 32 x y ZT hyperbolic paraboloid I I Math114 Rimmer Reduce the equation to one oi the Sl llt l l tormsa sulagmly 1260uadricSurfaces the surface and sketch it Ell A 2112 32 32 4x 7 y l 33 43 1393 42 39 f4 3924 3f l 34 4 33 20 o 33 4x2y2 4yi4z2 24z 36A 4x2y2 4y4L422 6z 36ig 4x2y 224Z32 C2y 22 z 32 l l 4 l ellipsoid center 023 1172012 I Math 114 Rimmer Reduce the equation to one at the Slitllt l l d lowing classify 1260uadricSurfaces the surface and sketch itquot 3 32 32 4 if 42 U 33 42 y 42 4 3924 3n U 34 4f 20 o 34 4y2 l6y z2 4zA 20XJ6A 4y2 4y4Lz2 4zi 20x i 4y22Z 22 L i y22z22 4 l 4 elliptic paraboloid vertex 02 2 m Math 114 Rimmer I 391 1 ua ric ur aces Summary 5 2 2 2 7 7 1 Ellipsoid K x2 y2 Z2 H b 1 39d f h 1 yper 001 o ones eet a2 b2 c2 all variables present x2 y2 Z2 gt g b 2 1 Hyperboloid of two sheets all varlables squared 2 2 2 Z x 3 Cone elliptic c2 a2 b2 a z x2 y2 Z g Parabolold 6111mm all variables present 2 2 39 E x2y2 Paraboloid hyperbolic one varlable c a b not squared one variable not present gt cylinder opening in the direction of the missing variable 2 2 2 2 1 Elliptic Cylinder E 1 Hyperbolic cylinder z ax2 Parabolic cylinder 1172012

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