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# devexpress asp.net pdf viewer : Copy web pages to pdf control SDK utility azure wpf web page visual studio Calculus3-part761

2.1 Theslope e ofafunction
31
somealgebrawiththedierencequotientusingjustx.Theslopeofachordfrom(7;24)
toanearbypointisgivenby
p
625 (7+x)2 24
x
=
p
625 (7+x)2 24
x
p
625 (7+x)2+24
p
625 (7+x)2+24
=
625 (7+x)
2
24
2
x(
p
625 (7+x)2+24)
=
49 49 14x x2
x(
p
625 (7+x)2+24)
=
x( 14 x)
x(
p
625 (7+x)2+24)
=
14 x
p
625 (7+x)2+24
Now,canwetellbylookingatthislastformulawhathappenswhenxgetsverycloseto
zero?Thenumeratorclearlygetsverycloseto 14whilethedenominatorgetsverycloseto
p
625 72+24=48.Isthefractionthereforeverycloseto 14=48= 7=24
=
0:29167?
Itcertainlyseemsreasonable,andinfactitistrue: asxgetscloserandclosertozero,
thedierencequotientdoesinfactgetcloserandcloserto 7=24,andsotheslopeofthe
tangentlineisexactly 7=24.
itwillbeabittedious. Whatifwetrytodoallthealgebrawithoutusingaspecicvalue
forx?Let’scopyfromabove,replacing7byx.We’llhavetodoabitmorethanthat|for
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32
Chapter 2 2 InstantaneousRate e ofChange: : TheDerivative
example,the\24"inthecalculationcamefrom
p
625 72,sowe’llneedtoxthattoo.
p
625 (x+x)2
p
625 x2
x
=
=
p
625 (x+x)2
p
625 x2
x
p
625 (x+x)2+
p
625 x2
p
625 (x+x)2+
p
625 x2
=
625 (x+x)
2
625+x
2
x(
p
625 (x+x)2+
p
625 x2)
=
625 x
2
2xx x
2
625+x
2
x(
p
625 (x+x)2+
p
625 x2)
=
x( 2x x)
x(
p
625 (x+x)2+
p
625 x2)
=
2x x
p
625 (x+x)2+
p
625 x2
Now what happenswhen x x isveryclose e to o zero? Again n it seemsapparent that t the
quotientwillbeverycloseto
2x
p
625 x2+
p
625 x2
=
2x
2
p
625 x2
=
x
p
625 x2
:
Replacingxby7gives 7=24,asbefore,andnowwecaneasilydothecomputationfor12
oranyothervalueofxbetween 25and25.
Sonowwehaveasingle,simpleformula, x=
p
625 x2,thattellsustheslopeofthe
tangentlineforanyvalueofx.Thisslope,inturn,tellsushowsensitivethevalueofyis
tochangesinthevalueofx.
Whatdowecallsuchaformula? Thatis,aformulawithonevariable,sothatsubsti-
tutingan\input"valueforthevariableproducesanew\output"value?Thisisafunction.
Startingwithonefunction,
p
625 x2,wehavederived,bymeansofsomeslightlynasty
algebra,a newfunction, x=
p
625 x2,that t givesusimportant informationabout the
originalfunction. Thisnewfunctioninfactiscalledthederivativeoftheoriginalfunc-
tion. Iftheoriginalisreferredtoasforythenthederivativeisoftenwrittenforyand
pronounced\fprime"or\yprime",sointhiscasewemightwritef
0
(x)= x=
p
625 x2.
Ataparticularpoint,sayx=7,wesaythatf0(7)= 7=24or\f f primeof7is 7=24"or
\thederivativeoff at7is 7=24."
Tosummarize,wecomputethederivativeoff(x)byformingthedierencequotient
f(x+x) f(x)
x
;
(2:1:1)
whichistheslopeofaline,thenwegureoutwhathappenswhenxgetsverycloseto
0.
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2.1 Theslopeofa a function
33
Weshouldnotethatintheparticularcaseofacircle,there’sasimplewaytondthe
derivative. Sincethe e tangenttoacircleat apointisperpendiculartothe radiusdrawn
p
625 x2)hasslope
p
625 x2=x,sotheslopeofthe
tangentline is x=
p
625 x2,asbefore. It t isNOT alwaystrue that atangentline is
perpendiculartoalinefromtheorigin|don’tusethisshortcutinanyothercircumstance.
Asabove,andasyoumightexpect,fordierentvaluesofxwegenerallygetdierent
valuesofthederivativef
0
(x). Coulditbethatthederivativealwayshasthesamevalue?
Thiswouldmeanthattheslopeoff,ortheslopeofitstangentline,isthesameeverywhere.
linetoaline,butifitmakessenseatallthetangentlinemustbethelineitself. Itisnot
hardtoseethatthederivativeoff(x)=mx+bisf0(x)=m;seeexercise6.
Exercises2.1.
1. Drawthegraphofthefunctiony=f(x)=
p
169 xbetweenx=0andx=13. . Findthe
slopey=xofthechordbetweenthepointsofthecirclelyingover(a)x=12andx=13,
(b)x=12andx=12:1,(c)x=12andx=12:01,(d) x=12andx=12:001. . Now w use
thegeometryoftangentlinesonacircletond(e)theexactvalueofthederivativef
0
(12).
2. Usegeometry y tondthederivativef
0
(x)ofthefunctionf(x)=
p
625 xinthetext t for
eachofthefollowingx:(a)20,(b)24,(c) 7,(d) 15. . Drawagraphoftheuppersemicircle,
anddrawthetangentlineateachofthesefourpoints. )
3. Drawthegraphofthefunctiony=f(x)=1=xbetweenx=1=2andx=4. . Findtheslope
ofthechordbetween(a)x=3andx=3:1,(b)x=3andx=3:01,(c)x=3andx=3:001.
Now usealgebra to ndasimpleformulafor the slopeofthechordbetween(3;f(3))and
(3+x;f(3+x)). Determine e what happens when xapproaches 0. In n your r graphof
y=1=x,drawthestraightlinethroughthepoint(3;1=3)whoseslopeisthislimitingvalue
ofthedierencequotientasxapproaches0. )
4. Findanalgebraicexpressionforthedierencequotient
f(1+x) f(1)
=xwhenf(x)=
x
2
(1=x). Simplifytheexpressionasmuchaspossible. Thendeterminewhathappensas
xapproaches0. Thatvalueisf
0
(1). )
5. Drawthegraphofy=f(x)=x
3
betweenx=0andx=1:5. Findtheslopeofthechord
between (a) ) x = 1 and d x = 1:1, (b) ) x = 1 andx = 1:001, , (c) ) x = = 1and d x = 1:00001.
Thenusealgebratondasimpleformulafortheslopeofthechordbetween1and1+x.
(Usetheexpansion(A+B)
3
=A
3
+3A
2
B+3AB
2
+B
3
.) Determinewhathappensasx
approaches 0,andinyour graphofy y =x
3
drawthe straightline throughthepoint(1;1)
whoseslopeisequaltothevalueyoujustfound.)
6. Findanalgebraicexpressionforthedierencequotient(f(x+x) f(x))=xwhenf(x)=
mx+b. Simplifytheexpressionasmuchaspossible. Thendeterminewhathappensasx
approaches0. Thatvalueisf
0
(x).)
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34
Chapter 2 2 InstantaneousRate e ofChange: : TheDerivative
7. Sketchtheunitcircle.Discussthebehavioroftheslopeofthetangentlineatvariousangles
aroundthe circle. Which h trigonometric function gives s the slope ofthe tangent line at an
angle? Why?Hint: thinkintermsofratiosofsidesoftriangles.
8. Sketchtheparabolay=x
2
. Forwhatvaluesofxontheparabolaistheslopeofthetangent
oftheslopechangesfrompositivetonegativeandviceversa?
We startedthelast section by y saying, \It t isoften necessaryto knowhowsensitive the
valueofyistosmallchangesinx." We e haveseenonepurelymathematicalexampleof
this:ndingthe\steepness"ofacurveatapointispreciselythisproblem.Hereisamore
appliedexample.
Withcarefulmeasurement it might be possible to discoverthat a droppedballhas
heighth(t)=h
0
kt
2
,tsecondsafteritisreleased.(Hereh
0
istheinitialheightoftheball,
then,\Howfastistheballgoingattimet?"Wecancertainlygetaprettygoodideawitha
littlesimplearithmetic.Tomakethecalculationmoreconcrete,let’ssayh
0
=100meters
andk=4:9andsupposewe’reinterestedinthespeedatt=2. Weknowthatwhent=2
theheightis100 44:9=80:4.Asecondlater,att=3,theheightis100 94:9=55:9,
sointhat secondthe ballhastraveled80:4 55:9=24:5 5 meters. Thismeansthat t the
averagespeedduringthattimewas24.5meterspersecond. Sowemightguessthat24.5
meterspersecondisnot aterribleestimateofthespeedat t=2. Butcertainlywecan
dobetter. Att=2:5theheightis100 4:9(2:5)
2
=69:375. Duringthehalfsecondfrom
t=2tot=2:5theballdropped80:4 69:375=11:025meters,at t anaveragespeedof
11:025=(1=2)=22:05meterspersecond;thisshouldbeabetterestimateofthespeedat
t=2. Soit’sclearnowhowtogetbetterandbetterapproximations: : computeaverage
speedsovershorterandshortertimeintervals. Betweent=2andt=2:01,forexample,
theballdrops0.19649metersinonehundredthofasecond,atanaveragespeedof19.649
meterspersecond.
We can’t do thisforever, , and we stillmight t reasonablyask k what t the e actualspeed
preciselyatt=2is. Iftissometinyamountoftime,whatwewanttoknowiswhat
happenstotheaveragespeed(h(2) h(2+t))=tastgetssmallerandsmaller.Doing
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2.2 Anexample
35
abitofalgebra:
h(2) h(2+t)
t
=
80:4 (100 4:9(2+t)
2
)
t
=
80:4 100+19:6+19:6t+4:9t
2
t
=
19:6t+4:9t
2
t
=19:6+4:9t
Whentisverysmall,thisisverycloseto19.6,andindeeditseemsclearthatast
goestozero,theaveragespeedgoesto19.6,sotheexactspeedatt=2is19.6metersper
second. Thiscalculationshouldlookveryfamiliar.Inthelanguageoftheprevioussection,
wemighthavestartedwithf(x)=100 4:9x
2
f(2+x) f(2)
x
=
19:6x 4:9x2
x
= 19:6 4:9x
The algebra a is the e same, , except t that t following the e pattern n of the e previous section n the
subtraction would be e reversed, , and d we would say that the e slope e of the e tangent line is
19:6. Indeed,inhindsight,perhapswe e shouldhavesubtractedthe otherwayevenfor
the droppingball. At t t=2theheight is80.4;onesecondlatertheheightis55.9. The
usualwayto compute a\distance traveled" isto subtract the earlierpositionfromthe
laterone,or55:9 80:4= 24:5. . Thistellsusthatthedistancetraveledis24.5meters,
andthenegativesigntellsusthattheheightwentdownduringthesecond. Ifwecontinue
theoriginalcalculationwethenget 19:6meterspersecondastheexactspeedatt=2.
Ifwe interpret thenegativesignasmeaningthat themotionisdownward,whichseems
the speedof theballisthe value of thederivative ofa certainfunction, , namely, ofthe
functionthatgivesthepositionoftheball. (Moreproperly,thisisthevelocityoftheball;
The upshot isthat this s problem, nding g the speed of the ball,isexactly y the e same
problemmathematically as nding g the slope e of f a curve. Thismay y already be enough
evidencetoconvinceyouthatwheneversomequantityischanging(theheightofacurve
ortheheightofa ballorthesize of theeconomyorthedistanceofa spaceprobefrom
earthorthepopulationoftheworld)the rateat whichthequantityischangingcan,in
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36
Chapter 2 2 InstantaneousRate e ofChange: : TheDerivative
Exercises 2.2.
1. Anobjectistravelinginastraightlinesothatitsposition(thatis,distancefromsomexed
point)isgivenbythistable:
time(seconds)
0
1
2
3
distance(meters)
0
10
25
60
Findtheaveragespeedoftheobjectduringthefollowingtimeintervals: [0;1],[0;2],[0;3],
[1;2],[1;3],[2;3]. Ifyouhadtoguess s the speedatt=2 just onthe basis of these,what
wouldyouguess? )
2. Let t y =f(t)=t
2
,wheretis thetimeinseconds andy y is s thedistanceinmetersthatan
object falls ona certainairless planet. Draw w agraphofthis functionbetween t= 0and
t=3. Makeatableoftheaveragespeedofthefallingobjectbetween(a) ) 2secand3sec,
(b)2secand2.1sec,(c)2secand2.01sec,(d)2secand2.001sec. Thenusealgebratond
asimpleformulafortheaveragespeedbetweentime2andtime2+t. (Ifyousubstitute
Next,inyourformulaforaveragespeed(whichshouldbeinsimpliedform)determinewhat
happens as s t t approaches zero. This s is the instantaneous speed. Finally, , in your r graph
of y= t
2
draw thestraight line throughthepoint (2;4) whose slope is the instantaneous
velocityyoujustcomputed;itshouldofcoursebethetangentline.)
3. Ifanobjectisdroppedfroman80-meterhighwindow,itsheightyabovethegroundattime
tsecondsisgivenbytheformulay=f(t)=80 4:9t
2
. (Hereweareneglectingairresistance;
thegraphofthisfunctionwasshowningure1.0.1.) Findtheaveragevelocityofthefalling
objectbetween(a)1secand1.1sec,(b)1secand1.01sec,(c)1secand1.001sec.Nowuse
algebratondasimpleformulafortheaveragevelocityofthefallingobjectbetween1sec
and1+tsec.Determinewhathappenstothisaveragevelocityastapproaches0. That
istheinstantaneousvelocityattimet=1second(itwillbenegative,becausetheobjectis
falling). )
Intheprevioustwosectionswecomputedsomequantitiesofinterest(slope,velocity)by
seeingthatsomeexpression\goesto"or\approaches"or\getsreallycloseto"aparticular
value. Intheexampleswesaw,thisideamayhavebeenclearenough,butitistoofuzzy
torelyoninmoredicultcircumstances.Inthissectionwewillseehowtomaketheidea
moreprecise.
There is an important feature of f the e examples we have seen. Consider r again the
formula
19:6x 4:9x2
x
:
Wewantedtoknowwhathappenstothisfractionas\xgoestozero."Becausewewere
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2.3 Limits
37
as\substitutingzeroforx,"asthatwouldgive
19:60 4:90
0
;
whichismeaningless. Thequantitywearereallyinterestedindoesnotmakesense\at
wasnot immediatelyobvious. In n otherwords, we aregenerallygoingtowanttogure
outwhataquantity\approaches"insituationswherewecan’tmerelypluginavalue. If
happensto(sinx)=xasxapproacheszero.
EXAMPLE2.3.1 Does
p
xapproach1.41asxapproaches2?Inthiscaseitispossible
to compute the actualvalue
p
2 to a high precision to answerthe question. But t since
ingeneralwewon’tbeable todothat,let’snot. Wemightstartbycomputing
p
xfor
valuesofxcloseto2,aswedidintheprevioussections. Herearesomevalues:
p
2:05=
1:431782106,
p
2:04=1:428285686,
p
2:03=1:424780685,
p
2:02=1:421267040,
p
2:01=
1:417744688,
p
2:005 = 1:415980226,
p
2:004 = 1:415627070,
p
2:003 = 1:415273825,
p
2:002=1:414920492,
p
2:001=1:414567072. Soitlooksatleast t possible thatindeed
p
2:001isquiteclose. Ifwecontinuethisprocess,
however,atsomepointwewillappearto\stall." Infact,
p
2=1:414213562:::,sowewill
neverevengetasfaras1.4142,nomatterhowlongwecontinuethesequence.
Soinafuzzy, everydaysort t ofsense,it istruethat
p
x \getsclose to" 1.41,but it
doesnot\approach" 1.41inthesensewewant. Tocomputeanexactslopeoranexact
velocity,whatwewanttoknowisthatagivenquantitybecomes\arbitrarilyclose"toa
value. Consideragainthequantities
19:6x 4:9x
2
x
= 19:6 4:9x:
These two quantities s are e equal l as long g asx is s not t zero; ; if x x iszero, , the e left t hand
quantity ismeaningless,while e the right hand one is  19:6. Can n we saymore than we
we reallymake it \asclose aswewant"to  19:6? Let’strya a test case. Canwe e make
19:6 4:9xwithinonemillionth(0:000001)of 19:6?Thevalueswithinamillionthof
19:6are thoseintheinterval( 19:600001; 19:599999). Asxapproacheszero,does
19:6 4:9xeventuallyresideinsidethisinterval? ? Ifxispositive,thiswouldrequire
that  19:6 4:9x x >  19:600001. This s issomething g we can manipulate with alittle
38
Chapter 2 2 InstantaneousRate e ofChange: : TheDerivative
algebra:
19:6 4:9x> 19:600001
4:9x> 0:000001
x< 0:000001= 4:9
x<0:0000002040816327:::
Thus, we e can say y with h certainty y that if x is positive and less than n 0:0000002, then
x<0:0000002040816327:::andso 19:6 4:9x> 19:600001.Wecoulddoasimilar
calculationifxisnegative.
Sonowweknowthatwecanmake 19:6 4:9xwithinonemillionthof 19:6.But
agame.IclaimthatIcanmake 19:6 4:9xascloseasyoudesireto 19:6bymaking
x\closeenough"tozero. Sothegameis: yougivemeanumber,like10
6
,andIhave
tocomeupwithanumberrepresentinghowclosexmustbetozerotoguaranteethat
19:6 4:9xisatleastascloseto 19:6asyouhaverequested.
Nowif we actuallyplaythisgame,Icouldredo thecalculationabove foreachnew
numberyouprovide. WhatI’dliketodoissomehowseethatIwillalwayssucceed,and
evenmore,I’dliketohaveasimple strategysothatIdon’t havetodoallthatalgebra
matterwhat youspecify. Sosupposethe e numberyougive me is. Howclosedoesx
havetobetozerotoguaranteethat 19:6 4:9xisin( 19:6 ; 19:6+)?Ifxis
positive,weneed:
19:6 4:9x> 19:6 
4:9x> 
x< = 4:9
x<=4:9
SoifIpickanynumberthatislessthan=4:9,thealgebratellsmethatwheneverx<
thenx<=4:9andso 19:6 4:9xiswithinof 19:6. . (ThisisexactlywhatIdid
intheexample: Ipicked=0:0000002<0:0000002040816327:::.) Asimilarcalculation
againworksfornegativex.Theimportantfactisthatthisisnowacompletelygeneral
result|itshowsthatIcanalwayswin,nomatterwhat\move"youmake.
Nowwecancodifythisbygivingaprecisedenitiontoreplacethefuzzy,\getscloser
andcloser"languagewehaveusedsofar. Henceforward,wewillsaysomethinglike\the
limitof( 19:6x 4:9x
2
)=xasxgoestozerois 19:6,"andabbreviatethismouthful
2.3 Limits
39
as
lim
x!0
19:6x 4:9x
2
x
= 19:6:
Hereistheactual,ocialdenitionof\limit".
DEFINITION2.3.2
Limit
Supposef isafunction. . Wesaythat t lim
x!a
f(x)=Lif
forevery>0thereisa>0sothatwhenever0<jx aj<,jf(x) Lj<.
The  and  here e play y exactly the e role theydid in the preceding g discussion. The
denition says, in a veryprecise e way, , that t f(x) can n be e made e asclose e as s desired d to L
(that’sthe jf(x) Lj j < part) by making g x close enough to a(the 0 < < jx aj j < 
part). Notethatwespecicallymakenomentionofwhatmusthappenifx=a,thatis,
ifjx aj=0. . Thisisbecauseinthecaseswearemostinterestedin,substitutingaforx
doesn’tevenmakesense.
Make sure youarenot confusedbythe namesof important quantities. The e generic
thediscussionabove,thefunctionweanalyzedwas
19:6x 4:9x
2
x
:
andthe variable ofthe limit wasnotx but x. Thex x wasthevariableofthe original
function;whenweweretryingtocomputeaslopeoravelocity,xwasessentiallyaxed
quantity,tellingusatwhatpointwe wantedthe slope. (Inthevelocityproblem,itwas
literallyaxedquantity,aswe focusedonthetime2.) ) The e quantityaofthedenition
inalltheexampleswaszero: wewerealwaysinterestedinwhathappenedasxbecame
veryclosetozero.
Armedwithaprecisedenition,wecannowprovethatcertainquantitiesbehaveina