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# Multivariate Calculus MA 26100

Purdue

GPA 3.97

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This 6 page Class Notes was uploaded by Dorothea Bode on Saturday September 19, 2015. The Class Notes belongs to MA 26100 at Purdue University taught by Staff in Fall. Since its upload, it has received 12 views. For similar materials see /class/208129/ma-26100-purdue-university in Mathematics (M) at Purdue University.

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Date Created: 09/19/15

Calculus Cheat sheet Definitions Precise De nition We say limfx L if Limit at In nity We say lim fx L ifwe Ha Hm for WW 5 gt 0 there is a 5 gt OSUCh that can make fx as close to L as we want by whenever 0 lt lxeal lt 5 then lfxeLl lt 5 takingx large enough and positive Working De ni on We say 1imfx L There is a similar de nition for lim fx L Ha Hm ifwe can make fx as close toL as we want excelt we requirert large a d negative by takingx suf ciently close to a on either side 7 ofa without letting x a In nite Limit We say lggf t H co ifwe I I can make fx arbitrarily large and positive Righ hand 11m 335 Ne L Th has by takingx su iciently close to a on either side the same definition as the limit except it of a without letting x a requires x gta There is a similar de nition for limfx muo Ha Lfth dl39 39t l39 LTh39 h th e an m HT jXt 5 as 6 except we make fx arbitrarily large and same de nition as the limit except it requires negativa Relationship between the limit and oneesided limits 152404 e 3quot f x i 2f L 155 felilgifwei 2152404 lim fx limfx gt limfx Does Not Exist Ha x911 Ha Properties Assume limfx and limgx both exist andc is any number then Ha Ha I l39 139 lim x x Cft C Xgr lf lim f 7 f provided limgx 0 g x limg x Ha Ha 5 1321401 122M 3 letmm trim yew 6 1321 fella 152M Basic Limit Evaluations at i so 4 Ha 2 limfxgxlinf lgm x Ha Note sgna1ifagt0 and sgi1a7lifalt0 1 lime uo amp lim e 0 5 neven lim x uo Hm X im Hem 2 limlnxuo amp lim lnx7uo 6 nodd limx uo amp lim x 7uo Hm Hue Hm Xerm 3 Lfgt0hen1imiy0 7 neven KETmax bxcsgna co me I w 4 If r gt 0 and x39is real for negativex 839 quot Odd 132M mbxc 59101 men m 7y0 9 nodd Kimmax Cxdisgnauo Himt vlslt lttg tuturlal math lamar euu for a complete set at Calculus nutes ZEIEIE Paul Damms Calculus Cheat sheet Evaluation Techniques Continuous Functions L Hospital s Rule Iffxis continuous atathenlgf t fa If limfx 9 or m fx iiuo the Hag x 0 Hagm ice Continuous Functions and Composition 3A fix f is continuous atb and liggxb then g gr is a urnberv or T lim Hm b Polynomials at In nity H fgx f H gx px and qx are polynomials To corripute Factor and Cancel X l x2 4x712 71 x72x 6 lim p factor largest power ofx out ofboth X121 x272x 7g xx72 Him q 6 8 p and qx and then corripute limit lim x H 4 2 A M 3x74 4 3 7 3 Rationalize NumeratorDenominator 1g x 2x2 lim 11 xrm H xrmx xm7 m 327 m 327 3J Piecewise Function x X 781 X 7813 l x2 5 ifx lt 72 173x ifx 272 Compute two one sided limits limrgx limrx2 5 9 H4 H4 TX 1 where gx 9 lim lim M x2 7813 H9 x931 71 H L 186 108 Combine Rational Expressions 1 1 1 1 x7xh limi 777 lim7 7 he h xh x PHD xxh liml i lim i fir Wk xxh i H xxhi lim gx lim 173x 7 H4 H4 One sided limits are di erent so limgx x2 doesn t exist Ifthe two one sided limits had been equal then lirrgt would have existed x and had the same value Some Continuous Functions Partial list of continuous functions and the values of x for which they are continuous 1 Polynomials for a 7 cosx and sinx for all x 2 Rational function except for x s that give division by zero 8 tanx and sec x provided 3 3101 oddforallx Ruufl iyifly 4 Vnevmforallx20 2 2 2 2 5 ex for anx 9 cot x and cscx provided 6 lnx forxgt0 X quot39v27r lr07r27r Intermediate Value Theorem Suppose that fx is continuous on 0 b and let Mbe any number between fa and fb Then there exists a numbers such that a lt c ltb and fc M Vlsltl tttgtuturlalmatl1lamar edu for a complete set at Calculus nutes ZEIEIE Paul Damms Calculus Cheat sheet erivatives De nition and Notation If y fx then the derivative is de ned to bef x limw ham h If y fx then all ofthe following are Ify fx all of the following are equivalent equivalent notations for the derivative notations for derivative evaluated at x a df dy d df dy f xy iiifx Dfx a i i D a dxddeU fyxsa 1amp7 117 f Interpretation of the Derivative If y fx then 2 f a is the instantaneous rate of 1 mf a is the slope ofthe tangent change of fx at x a line to y fx at x aand the 3 If fx is the position ofan object at equation ofthe tangent line at x a is time r the f a is me velocity of given by yfafya lia39 the object at xa Basic Properties and Formulas Lf fx and gx are di erentiable functions the derivative exists 5 and n are any real numbers 1 of cf x 5 c0 239 fig fyxgyx 6 x nx quot7Power Rule 3 fgy f gfg iProduct Rule d f f f 7 fgxf gxg x 4 E e Quotient Rule This is the Chain Rule Common Derivatives d d d Ex1 Ecscxicscxcotx Eaxaxlna sinxcosx cotxicsc2x exex cosxisinx sinquotx 178 lnx xgt0 d 2 d 1 d 1 itanx secx quot 7 il x 7X 0 If dxhos x HZ a ll x 1 secxsecxtanx goaan x 171 00319 xlna x gt0 vlslt mg tuturlal math lamar edu for a Complete set of Calculus nutes ZEIEIE Paul Damms Calculus Cheat sheet Chain Rule Variants The chain rule applied to some specific functions 1 gum nmdm 5 ilcoslm cf xsinfx 2 ilef f39xe l 6 gltanimdxlxlseczixd 3 hfx 7 mg flixisecmhanw 4 sinfxf39xcosfx 8 il an39lfv h Higher Order Derivatives The Second Derivative is denoted as T e n3911 Derivative is denoted as f39x fa x 2 and is de ned as f x and is de ned as f39x f xy ie the derivative of the fl xf quotxy 711 me derivative of rst derivativevflxl the nl derivative fl x Implicit Differentiation Find y39 if ezwy x3y2 siny11x Remembery y x here so productsquotients ofx and y will use the productquotient rule and derivatives ofy will use the chain rule The trick is to differentiate as normal and every time you di erentiate a yyou tack on a y39 from the chain rule After differentiating solve for y39 ezx39gy 279y393x2y22x3yy39cosyy3911 R W 2 2 1172e 39 73xy 26quot 79 ezx399y4r3x2 2 2x3 cos 11 2 y y yy yy y my 7mm 3 2x2 7 may 7 if 7 may 7 2 2 2x yige y 9e cos yy 711 2e 3x y IncreasingDecreasing 7 Concave UpConcave Down Critical Points x c is a critical point of fx provided either Concave UpConcave Down 1 fic0 or 2 fic does t exist 1 Iff x gt0 for allx ln an lntervalIthen fx is concave up on the interval 1 IncreasingDecreasing 2 If f39x lt 0 for all x in an intervalIthen 1 If f x gt 0 for allx ln an lntervalIthen f is concave down on the interval fx is increasing on the interval 2 If f39x lt 0 for allx in an intervalIthen Emmi0 POims f x c is a in ection point of fx if the 39 h t 3 If f39 x0 for allx in an intervalIthen comm y c anges a x C x is decreasing on the interval 1 fx is constant on the interval 1 Vlslt mg tlJIEIrlal math lamar edu for a Complete set of Calculus nutes ZEIEIE Paul Damms w mm ui mpvnmnmuvpuzmpq w mm n m X mm y mammmmms nusm WW mm M mm 39 tuvnsasvp mlmp xA uab zmd mm X n mmmmmmmm mmbimjmmdw SVIQWM WF m mam was WW mammals mum mp mm mm mm munmm w mumma mmam uwmmwn mm x PWmd 39xW mmm ulesn Wmwuvmxajxsms39uvm 39x 1 mm nvmum M Wazmmwnmmmm awm mmwmm WWW WWW I w muquot mapm up m w quotmm mm x Mum I m nunuquot v11 mumm U Yll mj1ill x mmmmwwmwm WWW w Hawmm gammato xuw Mm wny quotw z mmmmmnw MW 2 law mm pmadpmmm m1 m w mm m mumg mm mum mm mm mmwmm n m mm mm mu m xmmw mmquot x Fwadhmnis 11 whp ix wmmwmmu 1 Wm Wm m Wwxw Mmmmw Z m wuuuww xJamumasqlm 1x wawm Ca1cu1us Cheat Sheet De nite Integral Suppose fx is continuous AntiiDerivative An antiderivative of fx on ab Divide ab into n subintervals of is a function Fx such that F x fx width Ax and choose xquot from each interval Indefinite Integral Ifxdx Fxc Then Ifxdx limZ xAx Hm 1 where Fx is an antiderivative of fx Fundamental Theorem 01 Calculus Part I If fx is continuous on ab then Variants 0139 ParlI gx is also continuous on ab iiXftdt y lf and g39xjftdt m i n odtgmymm Part II fxis continuous onab Fx is I2ftdtu xfux7v39jfvx an antiderivative offxie Fx Ifxdx thenJ Zfxdx raging erties Ifxigxdxjfxdxijgxdx Icfxdxcjfxdxc is a constant If xdxjfxdxrjgxdx Icfxdxcjfxdxcisaconstant Wig Ifxdx0 Ifxdxfft Ifxdxij fxdx jab044 fx dx Iffx2gx onagxgbthen Ifxdxzjgxdx Iffx20 on agxgb then Ibfxdx20 IfmexSMon agxgb then mb7aSIfxdx Mb7a Common Integrals Ikdxkxc Icosudusinuc Itanuduln secu c n 7 n1 7 f a t f J I w t 1 Ix dxiwlx cn 1 J J 4 7 7 2 7 1 7 71 u Ix dx7Idxilnx c Isec udurtanuc Immigtan c fign M f J 1 74 Jaxb a 1 1 J J Wdurs EC J lnuduu1nuiuc Icscucotudu7cscuc Ie due c Icsczudu7cotuc V sn mg Mammal math 1amar edu fur a complete set of Calcu us notes ZEIEIE Pam Dewms Ca1cu1us Cheat Sheet S andard Integration Techniques Note that at many schools all but the Substitution Rule tend to be taught in a Calculus 11 class It Substitution The substitution u gxwill convert fg x g xdx LES fu du using du g xdx For inde nite integrals drop the limits of integration Z Ex I1 5x1cosxjdx ux3 2 du3x2dx 2 xzdx du x1 gt 131 x2gt 238 I125x2 cosx3dx I18 c0sudx sin sin 87sin 27 27 Integration by Parts judvuvijvdu and I udvuv fl vdu Chooseu anddv from a a integral and compute du by di erentiating u and compute v using v Idv ExIxequotdx ux dv gt dudx v J xe39x dx 7xe39x J e39x dx ixe39xie39XJrc Ex lenxdx ulnx dvdx gt du dx vx Isl drl Vijsdrmmwr 3nx ixnxz 3 ixnx x3 51n5731n372 Products and some Quotients ofTrig Functions For Isin xcosm xdx we have the following Iquot n odd Strip 1 sine out and convert rest to cosines using sin2 x licosz x then use the substitution u cosx m odd Strip 1 cosine out and convert rest N to sines using cos2 x lisin2 x then use the substitution u sinx n and In both odd Use either 1 or 2 n and In both even Use double angle andor half angle formulas to reduce the integral into a form that can be integrated 85quot ForItan x secquot x dx we have the following 1 n odd Strip 1 tangent and 1 secant out and convert the rest to secants using tan2 x sec2 x 71 then use the substitution u sec x 2 m even Strip 2 secants out and convert rest to tangents using sec2 x 1 tan2 x then use the substitution u tanx 3 n odd and m even Use either 1 or 2 4 n even and m odd Each integral will be dealt with differently WigFormulm sin2x2sin xcosx cos2x1cos 2x sin2xr17cos2x Ex Itanz x secs x dx Itanz x secs xdx Itanz x secA x tanx sec xdx J sec2 x 71sec4 x tan x secxdx J u2 71u4du usec x lrsec7 x7 sec5xc Ex Ism5x c053 J sn5xdx7j sn4xsmdx 7J ltsm2xzsmdC c053 205 205 2 1 cos Kismde wst c053 7 H22 d 77 Huhu d r u r u 7sec2x2n cosx 7cosz xc V1511 mg mamma math 1amar edu for a complete set of Ca1cu1us notes ZEIEIE Pam Damms mmmg slmnmamislilwxnmwm memmmsm mxmmunx w Iv ammpmwmmm www3an 1144 Mama 1 444 mum 111 sum117w that 39 mu m V mum3w 421 rpmmemamah w WHw 414 c 011 w m J 39quotJ39Iv c WA Au m mum m mwmwma m mgmmquot 5 Mn mxxvmiqwmvmmsadmxmquxmex vhnw x wwqunm mm WWquot uv n mmmg wwwmme m mmwmmm w mswmuz us wx 1 MW mmdwwxusam nvmmhn Hmm xym u ammad Pi qmnxvmmjm rrudn m mznxo nxwoh4wo wiv39w mm mum m1 wmamwmmwa 7m m m Wigwam a u amapaw mny m 1 5n WWW vgf w mn quotmarlawn 1M K k quotgt11 A as Edwnnnnsmmnmu 4 re n Wmpmpmqan Wmasq39 mumm W mm H me m WFP39P LIIMH mm mm mm madm39mmr m www nmp V V wf mum Caicuius Cheat Sheet Work Ifa force ofFt moves an object Average Function Value The average value 1 17 of fx on agx Sb isfwg jja fxdx 27 inagxgbthework done is WLFxdx be Arc Length Surface Area Note that this is often a Calc 11 topic The three basic formulas are 27 L I d SA I 2er d rotate about xaxis SA I 27M air rotate about yaxis a a a where d is dependent upon the form of the function being worked with as follows air 12dx ifyfx agxgb dr i7zdt ifxftygt astgb dy gray Ek y agygb dr irzzd9 ifrf9 agegb With surface area you may have to substitute in for the x ory depending on your choice ofdr to match the differential in the air with parametric and polar you will always need to substitute Improper Integral An improper integral is an integral with one or more in nite limits andor discontinuous integrands t Integral is called convergent if the limit exists and has a finite value and divergent if the limi doesn t exist or has infinite value This is typically a Calc 11 topic In nite Limit m 2 o o 1 Ifltxgtdx1g2tfltxgtdx 2 Ifltxgtdxgstfltxgtdx 3 Ifxdx Ifxdx Icmfxdx provided BOTH integrals are convergent Discontinuous Integrand 1 Discont at a Lbfxdx fl iin fxdx 2 Discont atb Lbfxdx o c o 3 Discontinuity at a lt c ltb Lfxdx L fxdx L provided both are convergent Comparison Test or Improper Integrals If fx2gx 20 on auo then 1 LfL conv thenL gxdx conv 2 IfL gxdx divg thenL divg Useful fact If a gt 0 then Lazide converges if p gt1 and diverges for p 1 x Approximating Definite Integrals For given integral Lbfxdx and a It must be even for Simpson s Rule define Ax bf and divide ab into n subintervals xmx xhxz xwghxn with x a and x b then Midpoint Rule Ibfxdx mAx fx fx fx x is midpoint xkhxt Trapezoid Rule Ibfxdx s fxn2fx2fx22fxHfxw Ax Simpson s Rule Ifxdx m Tfx 4fx 2fx2 2fxw24fxwfxw Visit mg tumna math amar edu fur a umpiete set of Caicuius notes ZEIEIE Pam Damms

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