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## MA 16200 Condensed Course Notes

by: Arnold Chris Toppo

285

0

13

# MA 16200 Condensed Course Notes 16200

Marketplace > Purdue University > Math > 16200 > MA 16200 Condensed Course Notes
Arnold Chris Toppo
Purdue
Pl Anly Geo Calc II
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Condensed notes for each lecture and every equation you will need for the exams and quizzes for MA 16200 (Calculus 2).
COURSE
Pl Anly Geo Calc II
PROF.
No professor available
TYPE
Bundle
PAGES
13
WORDS
KARMA
75 ?

## Popular in Math

This 13 page Bundle was uploaded by Arnold Chris Toppo on Friday January 9, 2015. The Bundle belongs to 16200 at Purdue University taught by a professor in Fall. Since its upload, it has received 285 views. For similar materials see Pl Anly Geo Calc II in Math at Purdue University.

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Date Created: 01/09/15
121 3dimensional Coordinate systems 1R2 Set of all ordered pairs ab in twodimensional space Ie P24 gt X2 y4 1R3 Set of all ordered pairs abc in threedimensional space Pabc xcoordinate ycoordinate zcoordinate What s the difference between IR2 and 1R3 1R2 Equation for a x2 y2 4 xy plane Circle of radius 2 with center at the origin Distance Formula a2 b2 D2 D2c2a2 b2c2d2 Igt a2 b2c2d2 gt Equation of a Sphere 2 2 2 2 x y z r With center at origin 00 with r as the radius of the sphere Le x 22 y 32 z 12 16 r2 Radius r 16 4 Center at 2 3 1 P1 x1Y1Z1FP2 ixzihizzi ie 1 mi I axjs 1R3 Equation for a x2 y2 4 xyz plane In this case 2 can have any along its axis and thusly becomes a cylinder I39ll 39Il39ul139II1I I 2 22 mnn nnnnnnd H r l FUEUDEUQEJEEQEEEEWEWI I Distance between two points P1 24 1P2 1 6 3 IP1P2 J2 12 4 62 1 32 IP1le J32 22 42 5 With center at abc with r as the radius of the sphere Le x2y2zz8x 4y2212O x28xy2 4yzzZz 12 x28x16y2 4y42222 121Ti16 How did we obtain these extra numbers 2 x2 8x 16 16 18 y2 4y 4 4 15 4 2 Q i 2 Z 22 1 1 E 2 x 42 y22 z 12 9 Center at 4 2 I and Radius of 3 122 Vectors E C Vector Addition B D Triangle Law Parallelogram Law A C Components of Vectors Magnitude x y Z components 0 f 51 ie a lt 123 gt a 12 22 32 14 Standard Basis Vectors Vector Addition and Multiplication a i 39 k Addition ai31 3k gt alt100gtlt010gtlt001gt B 5 j6k gt alt111gt 65i 513j j 3k6k 4i2j3k Properties of Vectors Multiplication aaa2 az3j 3k akaa E 5z j6k w W Ta2i6j 6k c 13c 13 E 15i 3j18k u 13 O if O 0r13 0 or if they are perpendicular Ta 17i 3j 12k lt 17312 gt 123 Dot Product a lt u1u2u3 gt 13 lt 171 v gt 6 2 angle between the two vectors ie Find a unit vector in the same direction as the vector V 4139 2 j 4k 422242 V3 6 13 4i 2j4k2 12klt2 1 2 13 6 3 3 3 3 3 3 Finding the angle between two vectors 1 m7 m7 COS x WWI 6 A lt denved from FM cos6 ieu lt 211 gt v lt 13 1 gt a 13 2 3 1 4 v221212x6 M J12 32 12 V11 6 cos1lt 4 cos1lt 4 xEV11 V66 Vector Projection Scalar Projection Projection of 1 onto a Component of I in the direction of a A 55 A A 55 P101561 2 WM Compga W Projection of 6 onto 1 Component of a in the direction of I 124 Cross Product vectors HAVE to be in different directions at N WL H m N Ht 44 8 6j 4 3 8f 14 lt 8 141 gt ie Find a vector of length 2 perpendicular to El and I alt21 2gtElt324gt A A f f k aXb 2 1 2 w 3 2 4 W44i 86j4 3E8i 14j1 WV8214212J2761 8 14 A 1 A A A 1 a Vector of length 1 u Tm 8116 141 m m V267 gt Vector oflength 2 v 2a lt m m V267 gt ie Find a vector perpendicular to Area of a parallelogram the plane containing the following A101 B124 C312 E B A lt 023 gt ie Area formed between 13 lt 101 gt W lt 123 gt TCC Alt211gt I j I A A 1 f f I 1 A A 11 0 2 3 23l06J04k 1 o 1 0 2L 3 12 Okvxw 2 1 1 1 2 3 6j 4E 13m 27 2j21gt I x lmm Volume formed by a solid with a base of a parallelogram h 1 m cos9 gt h a cos9 61 Area between two curves f x Top Curve a 2 First intersection point gx 2 Bottom Curve b 2 Second intersection point ie Find the area bound by ie Find the area bound by fxex a0 ysinx x gxx 91 YCosx 3 1 x2 1 Find the area bounded by the 2 functions and lines f0 ex xdx 2 ex y1cosx x2 1 x 4 9 A Y1 Yzldx 0 x0 n 22 x 7 x O 2 W y1ldx lt 1 1 3 e e 2 2 4 E E ie Find the area between the following curves 2 sinx cosxg sinx 00530 4 x1 y2 TopCurve xE 1 1 2 x y2 1 Bottom Curve gtkNeed to take integral with respect to y The other examples were fX and gX This is fy and gy 1 y2 y2 1 y 1 y 1 1 1 A39Kra y o1 owyj2 m my 1 1 2 2st 2 224 y 3y 1 3 3 3 3 62 Volumes Squares A 32 gt 8 fx A H902 Circles A nrz gt 7quot fx A 7Tfx2 aka DiskWasher Method 2 Semicircles A gt 7quot fx 2 3 32 gtSfx Equilateral Triangles A gt A7UW Volumes of solids with known crosssections DiskWasher Method VERY IMPORTANT If the question says Rotate around Xaxis gt integral with respect to X Rotate around yaxis gt integral with respect to y gt ie fx a Ax n 2 nx DiskWasher Method with rotation around an axis other than the X or y axis ie y x y 3 x 0 Revolved abouty 4 A 7I4 x2 7r4 32 V n f034 x2 12dx 18H ie Find the volume of the solid obtained by rotating the region bounded by y 2x2 x3 and y 0 about the yaXis using the Shell method 2x2 x3 Owhenx 0x 2 2 1 2 3 1 3 4 V f0 2nx2x x dx 2n f0 2x x dx 2 x4 x522 lt8 32167I 2 20 5 5 DiskWasher Method with Two Curves ie The region Renclosed by y x and y x2 is rotated about the XaXis Find the volume of the resulting solid Ax nxz 7tx22 7Tx2 354 y x is the outer radius y x2 is the inner radius V f01Axdx f017t x2 x4dx nlx3 x51n1 12n 3 50 3 5 15 63 Volume by Cylindrical Shells ie Find the volume of the solid obtained by rotating the region bounded by x equot 2 and yaXis about the XaXis from 0 to 1 using the Shell method Axis of i mmlminn 64 Work W quot11 3 Springs Hooke s Law F kx spring constantdistance stretched Most problems will have the following steps 1 Use Hooke s Law to nd k 2 Substitute into W f kxdx 3 Find integral 4 Solve 65 Average Value of Functions ie Average value of f x 1 x2 over 1 2 1 2 favg mf11 x2dx 1 ixx3i 2 71 Integration by Parts 1 How do you choose u LIATE Logarithmic Inverse Algebraic Polynomial Trigonometric Exponential 2 Find derivative of u du and antiderivative of dv v 3 Substitute into f udv uv f vdu ie fxe6xdx u x v 2 ie du dx dv e6xdx fxe6xdx 2 6 f ew dx 6x x e e6x C 6 36 Le p448 Length of CableType question How much work is required to lift a ZOOlb cable vertically to a height of 100 ft 2001b lb looft 2 ft denStty W f1 Zxdx x2100 10 000 f 5 0 0 lb 3939 mug ll U 139 t m I Mean Value Theorem MVT 66 33 If f is continuous then there exists a number 0 to satisfy the above statements Using the function on the left to illustrate MVT favgfc2 favgfc2 1x2gt1c22 c221 ci1 ie f exsinxdx excosx f excosxdx u equot v cosx du exdx dv Sinxdx fexcosxdx exsinx f exsinxdx u equot v 317100 du exdx dv cosx fexsinxdx excosx exsinx fexsinxdx 2 f exsinxdx excosx exsinx fexsinxdx exsinx cosx C 72 Tri onometric Integrals If power of sin is odd fsin3 xcosxdx fsinxcosx1 cos2xdx If power of cos is odd fcos3xsinxdx fcosxsinx1 sin2xdx 73 Trigonometric Substitutions V42 x2 x 2sin9 V92 x2 x 3tan9 162 a2 x 4sec9 2 Steps to nd the integral using trigonometric ie f1 3fode x 2sin9 dx 2C089d9 Substitution V sin9 9 z 1 7T 1 F1nd 1dent1ty E sm9 9 g 2 2 Find derivative of identity ff Mai 962 dx 4 39 26 2 056 d6 3 Replace into function in terms of 5m X C 4 45111209 7T3 4sin262Cos6d6 7T3 45in2 62cos6d6 trlgonometrlc substltutlon 7W6 J41Sin26 W6 ZJCOSZW 7T 4 Use usubstitution to evaluate the integral 2 t 4sin2 9d9 7T 5 Find new bounds for the uintegral 2 t 4 gtIlt 1 C0s29 d9 Or g 2n 2n 7T 7T Replace trlgonometrlc functlon w1th the 29 sm29 Sln E Sln 6 triangle forms and use the original bound 2 g g g g x2x 1 x2x 1 5 i L 74 Partlal Fractlons 1e f m dx gt m x x1 x1 x2x 1 1 qu1x 1 Bx 1x 1 pawl 1 x m xl x2x 1 Ax1x 1Bxx 1Cxx1 x2x 1 Ax2 1Bxx 1Cxx1 Irreducible Form xABC1 x B C 1 1 A 1 A1B lCl 2 2 1 fx2x 1dxfltl 1 2 dx xx2 1 x1 96 1 2 llnlxl ln C fxndx1C nil fidxlnxC fexdxzexC faxdxl C flnxdx xlnx x C fsinxdx cosx C fcosxdx sinx C fcsc2 xdx cotx C f sec2 xdx tanx C f cscxcotxdx cscx C fsecxtanxdx secx C fsecxdx lnlsecx tanx C fsec3 xdx secxtanx lnlsecx tanX C ftanxdx lnlsecx C fcscxdx lnlcscx cotx C fcotxdx lnlsinx C f ajfxzzsin1 c agt0 f tan4 c f lnlx xzia C f ilnl lC 77 Approximate Integration Ax n Midioint Rule Traiezoidal Rule Simison s Rule 78 Improper Integrals ie floox lzdx limtoo ffxizdx lim limtoo 1 1 1 t 1 t gtoo Convergent if limit exists Divergent if limit doesn t exist 81 ArcLength 82 Area of a Surface of Revolution Length of f x 2 Area of f x ltRotation about Xaxis Length of gy Area of gy ltRotation about yaxis 83 Application to Physics and Engineering P pgd 6d My 2 21mx midi mzdz Mx 21mi3 i b x My M xfltxgtdx 00 i i b 1 x M pf 5 fx2dx 111 Se uences If limit exists sequence is convergent and bounded If limit does not exist sequence is divergent and unbounded Squeeze Theorem in Limits limnaoo an 2 limnaoo cn L gt limnaoo bn L L Ho ital s Rule an is monotonic if increasing or decreasing Use firstderivative or secondderivative test if xfxdx RI H i jgmxnzdx 7 Table of Properties of Limits limnawmn bn limnaoo an limnaoo bn limnoo can 2 c limnoo an limnawmnbn limnaoo an limnaoo bn limnawmn limnaoo anp gt an is convergent if the sequence is bounded and monotonic nthterm test Take limit If nonzero then divergent If zero then not necessarily convergent Geometric Series Test If r is less than 1 then convergent Divergent otherwise If there is a constant move it outside sigma ie 2 1 2200 i zzw ln gtr l an Sums s2 1 1 1 1Z an 2 first term of series Ratio Test Find an by replacing n with n1 M Arrange an Take absolute value and find the limit Same method to find convergence and divergence as Root Test Direct Comparison Test If you have a series an find a similar series bn 1 If 0 lt an S bn and 22 20 bn converges an converges If 0 lt bn S an and 220 bn diverges an diverges Pseries Test If p is greater than 1 then series converges Otherwise it diverges Alternating Series Test Absolute value of series has to be decreasing Limit of absolute value of series is 0 Integral Test Must be continuous positive and decreasing If the integral converges the series converges If the integral diverges the series diverges Remainder of integral Rn 2 IS Snl S an lt Error Root Test Take absolute value of series Ianl 1 Raise to the power of Ian DE Take limit If less than 1 series is convergent If greater than 1 series is divergent Inconclusive if 1 and different test is needed Mainly used when there is something to the nth power 118 Power Series RegularStandard Form Power series centered around a Finding the Radius and Interval of Convergence x 3 quot 1e G1ven that an 2 a Use Ratlo Test quot J391 an n1 x 3 n n om an cancels n n1 ac3quot K in numerator of an1 Tl ltx sgt n1 Take limit as n approaches 00 n n gm1ltx3gtn11Iltx3gtImn1x3 Given series is convergent when x 3 lt 1 Given series is divergent when x 3 gt 1 gt Radius of convergence at 1 gt Interval of convergence 24 Limit Comparison Test Take limit of as n approaches infinity n and it must be greater than 0 If one converges the other converges If one diverges the other diverges w 3 H Find interval of convergence by finding the X value for the equation x3lt1 gt 1ltx 3lt1 2ltxlt4 x20rx4 To determine if the bound is openclosed replace xquot with the values and test for convergence 4 3 quot 1quot oo 2200 Diverges Ptest n0 n oo 2 3n oo 1n n0 n 211207 limnaoo 0 Absolute value of Converges Alternating Series Test is decreasing ie 2n 1quotx gt Ratio Test M 2271012 n0 I 1n1x2n1 I 22nn2 x2 llmnaoo 22n1n12 1nx2n 11m39n OO 4n12 O lt 1 119 Representations of Power Series by Functions Recognize Geometric Sequences 1110 Taylor Series Maclaurin Series aka Taylor Series 1n13xn n ie e5x 30 5 0quot ie fx ln1 3x 22 21 11 1quot5x2quot 2n 1n15x2n1 ie Sin1 2200 27141 ie f x cosSX 2200 1 iefx3w1x1x 1x Definition of a Planar Curve 101 Parametric Equation where t is the variable and aka The set of points Xy as t varies over the interval I Together this is the plane curve denoted by C HOWTOSOLVE iext2 4 t2y x2y2 4 x4y2 4 t 31 5 102 Finding Derivatives dy 3 dy 2 Edy3t2 3 1ey t 3t cit 3t 3 dx 2t 2 dquot x t dt 21 d 3t2 3 gt ie dt 2t 21 Area under curves Surface Area dx 2 dy 2 d3 a a 0 Arc Length 103 Polar Coordinates Tangents to Polar Curves d y sint9rcost9 rf6xrcos6yrsin6 d y d9 dx E cos0T5in9

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