Crushed Notes on MATH132: Calculus

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1 Mth 13, Fll 011 Siyg Yg s Outlie Crushed Notes o MATH13: Clculus The otes elow re crushed d my ot e ect This is oly my ow cocise overview of the clss mterils The otes I put elow should ot e used to justify your swers i my quiz or fil ems Chpter Limits d Cotiuity Issue 1 Rte of chge Slope =!y y( + h) " y() =! h Issue Limit L is limit of f(), the the followig sttemet is true: As gets sufficietly close to 0, f() gets sufficietly close to L Issue 3 Sdwich theorem If f() is lwys sdwiched etwee g() d h(), d if g() d h() hve sme limit L t some poit =c, the f() lso hs limit L t =c Issue 4 Sttemet: lim f () = L! 0 Mthemticl defiitio of limit Coditio:! 0 < " # f ()! L < $ Recipe for solvig prolems: Fid! for give! y derivig ck from f ()! L < " Issue 5 Oe-sided limit Right-hd limit: lim f () = L where! mes! 0 d > 0! 0 Left-hd limit: lim f () = L where! " " 0 mes! 0 d < 0! 0 Issue 6 Limits ivolvig si (! ) ( ) si! lim = 1 whose proof eeds trick to ccout for the zero deomitor i the limit!"0! Issue 7 Cotiuity Fuctio f() is cotiuous t =c if the vlue of limit equls the vlue of the fuctio Issue 8 Itermedite vlue theorem for cotiuous fuctio f () For y y 0![ f (), f ()], there eists c![,] such tht f (c) = y 0 1

2 Mth 13, Fll 011 Siyg Yg s Outlie Issue 9 Asymptotes lim f () =, the the horizotl symptote is y =!" lim f () = k +, the the olique symptote is y = k +!" lim f () = ±#, the the verticl symptote is =! + " or! Chpter 3 Differetitio: f '()! d f () Issue 31 Derivtives: f '() = lim h!0 Defiitios of derivtive f ( + h) " f () h or f '() = lim z! Oe-sided derivtives: f '() = lim h!0 + f ( + h) " f () h f (z) " f () z " (right-sided); f '() = lim h!0 " f ( + h) " f () h (left-sided) Issue 3 Issue 33 Differetiility implies cotiuity Differetitio rules Costt rule: d C = 0 Power rule: d =!1 Trigoometric fuctio: d si() = cos() d d cos() =! si() Product rule: d ( f! g ) = f '! g + f! g' Quotiet rule: d f / g Chi rule: d f (g()) ( ) = f '! g " f! g' g ( ) = df (g()) dg()! dg() Issue 34 Recipe for implicit differetitio Step 1 Apply d o oth sides of the equtio, ie differetite oth sides with respect to Step Collect ll the terms icludig dy o oe side, d solve for dy

3 Mth 13, Fll 011 Siyg Yg s Outlie Issue 35 Recipe for solvig prolems of relted rtes Step 1 Write out the geerl form of chi rule df () dt = df ()! dt df () df () Step For the specific prolem, figure out which of the three terms,, dt,, re give or dt differetile The lgericlly solve for the third ukow qutity s sked i the prolem Issue 36 Lieriztio d equtio of lie Purpose: replce complicted curve with stright lie loclly d pproimtely Tool: Give the poit ( 0, y 0 ) d slope k t this poit, the equtio of lie c e writte y y = y 0 + k(! 0 ) Recipe: First figure out the vlues of 0, the clculte y 0 = f ( 0 ) d k = f '( 0 ) usig the give curve equtio y = f () After tht, write out the equtio of lie, d plug i the vlue of to fid y Chpter 4 Issue 41 Cocept 1 Applictios of derivtives Importt poits d cocepts of the curve Criticl poit: f '() = 0 or f '() is udefied Cocept Locl mimum: etreme vlue with dditiol f "() < 0 Cocept 3 Locl miimum: etreme vlue with dditiol f "() > 0 Cocept 4 Poit of iflectio: f "() = 0 Cocept 5 Mootoic fuctio: lwys icresig or lwys decresig Cocept 6 Cocvity: cocve up if f "() > 0 ; cocve dow if f "() < 0 Issue 4 Rolle s theorem Coditio: Fuctio is cotiuous d differetile i the domi [,] Therefore: There is t lest oe poit = c with f '(c) = 0 Issue 43 Me vlue theorem (slted versio of Rolle s theorem) Coditio: Fuctio is cotiuous d differetile i the domi [,] Therefore: There is t lest oe poit = c with f '(c) = k where k = two edig poits f ()! f ()! is the slope coectig the 3

4 Mth 13, Fll 011 Siyg Yg s Outlie Issue 44 Recipe for solvig pplied optimiztio Step 1 Write out the costrit (or coditio) Step Write out the qutity f to e optimized Step 3 Usig costrit, simplify f s fuctio of sigle vrile, ie f=f() Step 4 Clculte f '() d fid vlue of t which f '() = 0 This is the poit of optimiztio Issue 45 Recipe of Newto s method for solvig roots of f () = 0 Step 1 Mke iitil guess of 0 Step Clculte y 0 = f ( 0 ) d slope f '( 0 ) Step 3 Clculte 1 = 0! f ( 0 ) f '( 0 ) Step 4 Clculte y 1 = f ( 1 ) d slope f '( 1 ) Step 5 Similr to step 3 Fid = 1! f ( 1) f '( 1 ) Step 6 Similr to step 4 Fid y d f '( ) Step 7 Repet steps 5 d 6 util covergece, ie roots re foud Issue 46 Recipe for doig tiderivtives Step 1 Give G(), mke good guess from differetitio lws for f() which stisfies f '() = G() Step Verify how good the guess is y doig differetitio for f() Step 3 Modify d improve the guess of f() Step 4 Repet step d 3 util f '() = G() is ectly correct, d the dd costt to f() 4

5 Mth 13, Fll 011 Siyg Yg s Outlie Chpter 5 Issue 51 Itegrtio Overview ) "# Defiite itegrl: S =! f () where S is vlue$ %%% Riem sum: ( f (c k )& ' + k Itegrl = * + Ati-derivtive Idefiite itegrl: G() =! f () where G() is fuctio$ %%%%% f () = dg(), + S = G() - G() Issue 5 Fiite sum Lower sum Upper sum Midpoit rule Issue 53 Distce (D) vs displcemet (S) Displcemet c e positive or egtive d is give y S = v(t 1 )! "t + v(t )! "t + v(t 3 )! "t +! Distce c oly e positive d is give y D = v(t 1 )! "t + v(t )! "t + v(t 3 )! "t +! Issue 54 Poit 541 Sigm ottio Defiitio of sigm ottio k! = m + m+1 +!+ where >m d k is some dummy vrile Note tht k is some fuctio of k, eg k = k or k = 1 k Poit 54 Epress sum i sigm ottio k! = !+10 if k = k with ide m = 1 d = 10 k! = !+ 1 9 if k = 1 with ide m = d = 9 k 5

6 Mth 13, Fll 011 Siyg Yg s Outlie Poit 543 ( )! k ± k =! k ±! k Alger rules of sums " c! k = c! m + c! m+1 +!+ c! = c! m + m+1 +!+ " $$$$$$$ terms #! c = c + c +!+ c = c " $$$$$$$$ terms # 1! = !+ 1 = " 1 = 1 Poit 544! k = !+ = Importt result ( +1) ( ) = c" k For emple, 4! k = = 10 = 4 "(4 +1) Poit 545 Importt result! k = !+ = 4 ( +1)( +1) 6 For emple,! k = = = 30 = Poit 546 Limit of ifiite sum 4 "(4 +1)"( # 4 +1) Riem sum pproches the idefiite itegrl oly i the limit of For emple, k! = ! = 1! k = 1 ( +1) " k lim!" # + = lim = 1!" = + 6

7 Mth 13, Fll 011 Siyg Yg s Outlie Issue 55 Riem sum ( ) " f ( c k ) S = # k! k!1 where the two edig poits re deoted y 0 = d = Prtitio P implies the choice of su-itervls 0, 1,,!, P = m{ 1! 0,! 1, 3!,!,!!1 } Issue 56 Defiite itegrl { } Norm of P is defied y Limit of Riem sum is defiite itegrl, which gives the geometricl re uder the curve (d with respect to the -is d two oudries = d =) # f ( c k )! " $% &&& $ ' f ( )! Poit 561!" ###! Defiitio of itegrle fuctio Coditio: Fuctio f is cotiuous or piecewise cotiuous etwee [,] Therefore: This fuctio f is itegrle etwee [,] Notice: Piecewise cotiuous mes there re oly fiite discotiuous poits 7

8 Mth 13, Fll 011 Siyg Yg s Outlie Issue 57 Rules of defiite itegrl The two limits of itegrl is importt! f () = "! f () : itegrtig from to is opposite to itegrtig from to! f () = 0 : if itervl [,] hs zero width, itegrl is lso zero! f () +! f () =! f () : itegrl from to d the from to c is equivlet to tht from to c c c! f () "! f () =! f () : this c e esily derived from rule 1 d rule 3 ove c Cosequetly, oe c derive two more rules: c " f () = " f () if f is eve fuctio, ie f (!) = f ()! 0 " f () = 0 if f is odd fuctio, ie f (!) =! f ()! Issue 58 Bridgig defiite d idefiite itegrl (CONCEPTUALLY VERY IMPORTANT) How to formlly costruct the itegrl of fuctio f? There re two equivlet pproches usig the defiite d idefiite itegrl respectively o Idefiite itegrl pproch: G() =! f () Eg f () = the G() =! = C o Defiite itegrl pproch: G() = f (t)dt Eg f () = the G() =! f (t)dt = t dt! = t 3 3 Additiol otes for uderstdig ove two pproches:! = 3 3 " 3 3 o Wht is the meig of t 3 3 = 3 3! 3 3? I geerl, f () equls the vlue of f t = mius the vlue of f t = o Two pproches yield sme fuctio G() with differet costt (C or! 3 3 ) 8

9 Mth 13, Fll 011 Siyg Yg s Outlie Issue 59 Fudmetls of Clculus: differetil clculus d itegrl clculus Forml tetook sttemet Simplified istructor s uderstdig If G() =! f (t)dt, the G '() = d f (t)dt! d G '() = f () Itegrl Differetil f ()!!!" G()!!!!" f () The first step of itegrl correspods to! f (t)dt i the forml sttemet The secod step of differetil correspods to d i the forml sttemet Differetil Itegrl If G '() = f (), the! f () = G() " G() G()!!!!" f ()!!!" G() The first step of differetil implies dg()! f () =! o the left The secod step of itegrl implies dg()! = G() Issue 510 Applictio of itegrl: Fidig re Poit 5101 Fidig re uder oe curve f() Geometriclly, the re is eclosed y four curves: curve f(), lie of -is, verticl lie = o the left, d verticl lie = o the right Mthemticlly, re is clculted vi S =! f () Poit 510 Fidig re etwee two curves f1 d f Geometriclly, the re is eclosed y four curves: the two curve f1() d f(), verticl lie = o the left, d verticl lie = o the right Mthemticlly, re is clculted vi S = " f 1 ()! f () Issue 511 From chi rule to sustitutio method I differetil clculus, you hve chi rule: df ((g()) u=g() = df (u) du! du I itegrl clculus, you hve the correspodig sustitute method: " f (g())! g'()! u=g() = " f (u)du 9

10 Mth 13, Fll 011 Siyg Yg s Outlie The commo poit is the use of dummy vrile u=g() Poit 5111!u = g()! du = g'() ie du = g'()" Forml proof of sustitutio method Replce u y g() d du y g'() # Poit 511 $ f (u)du = $ f (g())" g'()" Prcticl recipes (VERY IMPORTANT RECIPE) Step 1 Mke good guess out u Step Do derivtive d fid reltio etwee du d Step 3 Properly replce fuctio of y fuctio of u, d y du, d itegrl limits of y those of u Step 4 Evlute the ew (ut ideticl) itegrl usig vrile u isted of i the origil itegrl Poit 5113 Cse illustrtio! Apply ove recipe to clculte " si( ) : Step 1 Guess u = Step Derivtive gives: du =! du = 0 Step 3 Lower itegrl limit chges from = 0 to u = 0, d upper itegrl limit chges from =! 1 u =! Totl sustitutio gives ew ut ideticl itegrl: " si( u)du 0! to Step 4 Direct itegrl gives ( )! si u $ " du = # cos(u) ' 0 % & ( )! 0 $ = # cos(! ) ' $ % & ( ) # # cos(0) ' % & ( ) = $ 1 ' % & ( ) # $ # 1 ' % & ( ) = 1 10