MATH 104/184 F E R SESSION
Transcript of MATH 104/184 F E R SESSION
MATH104/184FINALEXAMREVIEWSESSION
BYRAYMONDSITU
TABLEOFCONTENT
I. Relatedratesin3dimensionsII. OptimizationIII. LocallinearapproximationIV. TaylorpolynomialsV. CurvesketchingVI. Somefunlimits
RELATEDRATESAplaneistakingofftherunwayataspeedof300km/hdueNorth.Theangleofelevationis30degrees.Acaristravellingdueeastat150km/honastraight,level,road.Howfastisthedistancebetweentheplaneandthecarincreasingwhentheplanehasreachedanaltitudeof3km,assumingtheybothstartedfromthesamepointandreachedtheirrespectivevelocitiesinstantly.(redisplane,greeniscar,blueisdistances).IheardtherewasafunrelatedratesonthemidterminvolvingtrianglessoImadethisquestionswithlotsoftriangles.
Fig1
Fig2 Fig3
Wewanttofindtherateofchangeofdistance.Aswecanseeinfig2andfig3,distanceisthehypotenuseofgrounddistanceandheightdistance.Togetanequationfordistanceweneedanequationforheightdistanceandgrounddistance.Lookingatfig1,grounddistanceisthehypotenuseofthehorizontaldistancetravelledbythecarandtheplane.Tofindthehorizontaldistancetravelledbytheplanewecanusetrig.Tofindhorizontaldistancetravelledbythecaritisthevelocityofthecarmultipliedbytime(thecaronlyhasvelocityin1direction).Whentheplaneisatanaltitudeof3kmhowmanysecondshaspassed?Tofindthatoutweneedtofindouthowlongittakestheplanetoreach3kmaltitude.
1. sin '(= *++*,-./
01+*./23,/= 4
+562/.768/59-,.62:/
;<= 4
+562/.768/59-,.62:/
๐๐๐๐๐๐ก๐๐๐ฃ๐๐๐๐๐ ๐ก๐๐๐๐ = 6๐๐
2.๐ก๐๐๐ = +562/.768/59-,.62:/8/5*:-.1
= (4LL
= 0.02h
3. cos '(= 69R6:/2.
01+*./23,/= +562/0*7-S*2.659-,.62:/
(
โ4<= +562/0*7-S*2.659-,.62:/
(
๐๐๐๐๐โ๐๐๐๐ง๐ก๐๐๐๐๐๐๐ ๐ก๐๐๐๐ = 3โ3๐๐
๐๐๐๐๐โ๐๐๐๐ง๐๐๐ก๐๐๐ฃ๐๐๐๐๐ก๐ฆ = +562/0*7-S*2.659-,.62:/.-Y/
= 4โ4L.L<
= 150โ3๐๐/โ
๐๐๐๐๐๐ฃ๐๐๐ก๐๐๐๐๐ฃ๐๐๐๐๐ก๐ฆ = +562/0/-\0.9-,.62:/.-Y/
= 4L.L<
= 150๐๐/โ
5.Carhorizontaldistance=time*carvelocity=0.02*150=3km
6.๐๐๐๐ข๐๐๐๐๐ ๐ก๐๐๐๐< = ๐๐๐๐๐โ๐๐๐๐ง๐๐๐ก๐๐๐๐๐ ๐ก๐๐๐๐< + ๐๐๐โ๐๐๐๐ง๐๐๐ก๐๐๐๐๐ ๐ก๐๐๐๐<
๐๐๐๐ข๐๐๐๐๐ ๐ก๐๐๐๐ = `(3โ3)< + 3<
=c9(3) + 9=โ36 = 6๐๐
7.findingrateofchangeforgrounddistanceForsakeofsimplicity(justforthispart)Ipluginnumbersfromstepsabovesoifyouarewonderwhereanumbercamefrom.Lookup!
grounddistance=cplanehorizontaldistance=acarhorizontaldistance=b
๐< = ๐< +๐<nowwederive
2๐(9:9.) = 2๐(96
9.) + 2๐(9f
9.)
2(6)(9:9.) = 2(3โ3)(150โ3) + 2(3)(150)
12(9:9.) = 2700 + 900
h9:9.i = 4(LL
;<= 300๐๐/โrateofchangeofgrounddistancewhenplanehaselevationof3km
8.Heightdistance=3km(given)
9.๐๐๐ ๐ก๐๐๐๐< = โ๐๐๐โ๐ก๐๐๐ ๐ก๐๐๐๐< + ๐๐๐๐ข๐๐๐๐๐ ๐ก๐๐๐๐<
๐๐๐ ๐ก๐๐๐๐ = โ3< + 6< = โ45
10.Forthissectiononlywewillusethefollowingvariablenames,nottobeconfusedwithpart7.Ipluginnumbersfromstepsabovesoifyouarewonderwhereanumbercamefrom.Lookup!
Distance=cheightdistance=agrounddistance=b๐< = ๐< +๐<nowwederive
2๐(9:9.) = 2๐(96
9.) + 2๐(9f
9.)
2โ45(9:9.) = 2(3)(150) + 2(6)(300)
2โ45(9:9.) = 900 + 3600
2โ45(9:9.) = 4500
9:9.= klLL
<โkl
9:9.= <<lL
โkl๐๐/โfinalanswer
OPTIMIZATIONThefirstyearstudentsatUBC(UniversityofBurnabyCoquitlam)havecriedariverafterfailingtheirMATH140/148midterm.Theriveris5kmwideandthestudentsneed togetacross to theotherside toapoint that is10kmdownstream fromthepointdirectlyacrossfromtheircurrentposition.Thestudentscanwalkataspeedof5km/h.Theriverisalsososaltythatthecurrentisnolongerflowingwhichallowsthe students to swimat a speedof3km/h.What is themost efficientway for thestudentstoโgetoveritโ?(bothphysicallyandpsychologically)
Wewanttominimizetime.0โคxโค10Thereare2separatesections(swimmingandwalking)Thetimeofswimmingsectionisdefinedby:
๐ก = 9-,.62:/8/5*:-.1
= โlmnom
4
Thetimeofthewalkingsectionisdefinedby:๐ก = 9-,.62:/
8/5*:-.1= ;Lpo
l
ThetotaltimetogettopointBwouldbe:๐ก = โlmnom
4+ ;Lpo
l
๐กq = ;4;<
<oโlmnom
โ ;l
๐กq = <o(โlmnom
โ ;l
0 = o4โlmnom
โ ;l
;l= o
4โlmnom
3โ5< + ๐ฅ< = 5๐ฅsquarebothsides9(5< + ๐ฅ<) = 25๐ฅ<9 โ 5< + 9๐ฅ< = 25๐ฅ<9 โ 5< = 16๐ฅ<
๐ฅ = `uโlm
;(
๐ฅ = โuโlm
โ;(
๐ฅ = 4โlk
๐ฅ = ;lk
Testboundarycasesaswell(x=0andx=10)
๐ก = โlmnLm
4+ ;LpL
l
= โlm
4+ ;L
l
= l4+ 2
= ;;4
๐ก = `lmn(vw
x)m
4+
;Lpvwx
l
๐ก = `mwvnmmwvy
4+
mwx
l
๐ก = `ymwvy
4+ l
k
๐ก = mwx
4+ l
k
๐ก = ;L4
๐ฅ = ;l
kgivesusthesmallestt.
๐ก = โlmn;Lm
4+ ;Lp;L
l
๐ก = โ<ln;LL4
๐ก = โ;<l4
Wecancomparethiswith10/3bysquaringbothofthem.125/9>100/9
LOCALLINEARAPPROXIMATIONGiven: 65z
a) Findthelinearapproximationb) Usethelinearapproximationtoapproximatethedesiredvaluec) Determineifitisanoverestimateoranunderestimated) Whatistheboundontheerror?
Originally,IhadaveryfunquestionplannedforyouguysbutProfessorDesaulnierssaidโIthinkthisproblemmayconfusethemโsoIhadtochangeittoasimplerone:(
a) ๐ฟ(๐ฅ) = ๐(๐) + ๐q(๐)(๐ฅ โ ๐)Wecanchoosea=64.๐(๐) = 4
๐q(๐ฅ) = ;4(๐ฅ
}mz )
๐q(64) = ;4h ;;(i = ;
k~
๐ฟ(๐ฅ) = 4 + ;k~(๐ฅ โ 64)
b) ๐ฟ(65) = 4 + ;
k~(65 โ 64)
๐ฟ(65) = 4 + ;k~= k
;hk~k~i + ;
k~= ;u<
k~+ ;
k~= ;u4
k~
c) ๐q(๐ฅ) = ;4(๐ฅ
}mz )
๐qq(๐ฅ) = ;4hโ <
4i (๐ฅ
}wz ) = โ <
u(๐ฅ
}wz )
๐โฒโฒ(64) =โ29(64
โ53)๐qq(64) = ๐๐๐๐๐ก๐๐ฃ๐Thesecondderivativeisnegativemeaningitisconcavedown.Thismeansitisanoverestimate.
d) Youneedtoknowthat|๐qq(๐)| โค ๐inotherwordsweneedtofindthemaximumvalueof|๐qq(๐)|๐๐๐กโ๐๐๐๐ก๐๐๐ฃ๐๐[64,65]
๐qqq(๐ฅ) = โ <uhโ l
4i h๐ฅ
}๏ฟฝz i = 0
h๐ฅ}๏ฟฝz i = 0
;
๏ฟฝo๏ฟฝz๏ฟฝ= 0๐๐๐ ๐๐๐ข๐ก๐๐๐, ๐๐๐กโ๐๐ข๐โ๐คโ๐๐๐ฅ = 0๐qqq(๐ฅ)๐๐ ๐ข๐๐๐๐๐๐๐๐
Therearenocriticalpointsinourintervalsowecanjusttesttheboundaries
Noticethat:๐qq(๐ฅ) = โ <uh๐ฅ
}wz i = โ <
u;
(owz)
xisinthedenominatormeaningthatsmallerx=biggerfโโ(x).Therefore,x=64willgiveusthelargervalue
|Error|โคh;<i hโ <
uih(64)
}wz i (|65 โ 64|)<
TAYLORPOLYNOMIALSWhatisthe2nddegreeTaylorpolynomialof:
๐ ๐ฅ = sin(๐ฅ<) + log( on;)z( ๐ฅ + 1)(Lata= ๐
๐(๐ฅ) = sin(๐ฅ<) + 20simplifytheweirdlogattheendto20.
๐ถL = ๏ฟฝ(โ')L!
= ๐๏ฟฝโ๐๏ฟฝ = 0 + 20 = 20
๐ถ; = ๏ฟฝ๏ฟฝ(โ');!
= ๐q๏ฟฝโ'๏ฟฝ = cos hโ๐<i๏ฟฝ2โ๐๏ฟฝ = โ2โ๐
๐ถ< = ๏ฟฝ๏ฟฝ๏ฟฝ(โ')
<!= ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ'๏ฟฝ
<= ;
<๏ฟฝโ sin hโ๐
<i ๏ฟฝ2โ๐๏ฟฝ๏ฟฝ2โ๐๏ฟฝ + cos hโ๐
<i (2)๏ฟฝ = p<
<= โ1
The2nddegreeTaylorpolynomialis:
๐๏ฟฝ(๐ฅ) = ๐ถL + ๐ถ;(๐ฅ โ ๐) + ๐ถ<(๐ฅ โ ๐)<
๐๏ฟฝ(๐ฅ) = 20 โ 2โ๐๏ฟฝ๐ฅ โ โ๐๏ฟฝโ(๐ฅ โ โ๐)<
CURVESKETCHING
Sketchthecurve๐ ๐ฅ = 3๐ฅl โ 5๐ฅ4
1.Domain:allrealnumbers2.Asymptotes:
Vertical:noneHorizontal:takelimitasx->infinityandnegativeinfinityandyouwillnotgetafinitenumberasaresult.Therefore,nohorizontalasymptotes.
3.Intercepts:xintercept:
0 = 3๐ฅl โ 5๐ฅ40 = ๐ฅ4(3๐ฅ< โ 5)
x=0,x=ยฑ`l4
(0,0),(โ`l4, 0), (`l
4, 0)
yintercept:๐ฆ = 3(0l) โ 5(04)y=0(0,0)
4.Intervalsofincreaseanddecrease: ๐q(๐ฅ) = 15๐ฅk โ 15๐ฅ<
0 = 15๐ฅ<(๐ฅ< โ 1)x=00= ๐ฅ< โ 1 ๐ฅ< = 1 ๐ฅ = ยฑ1
๐โฒ(โ2)>0 ๐โฒ(โ0.5)<0 ๐โฒ(0.5)<0 ๐โฒ(2)>0
Increasingon(โโ,โ1)๐๐๐(1,โ)Decreasingon(โ1,0)๐๐๐(0,1)
5.Localmax/minAtx=-1.๐(โ1) = 3(โ1)4 โ 5(โ1)4=โ3 โ (โ5) = 2Localmax(-1,2)Atx=1.๐(1) = 3(1)4 โ 5(1)4=3 โ 5 = โ2Localmin(1,-2)
6.Concavity ๐q(๐ฅ) = 15๐ฅk โ 15๐ฅ< ๐qq(๐ฅ) = 60๐ฅ4 โ 30๐ฅ 0 = 30๐ฅ(2๐ฅ2 โ 1) ๐ฅ = 00 = 2๐ฅ< โ 1 ๐ฅ< = ;
<
๐ฅ = ยฑ`;<
๐โฒโฒ(โ1) < 0 ๐โฒโฒ(โ0.1) > 0 ๐โฒโฒ(0.1) < 0 ๐โฒโฒ(1) > 0
Concavedownon๏ฟฝโโ,โ`12๏ฟฝ ๐๐๐ ๏ฟฝ0,`12๏ฟฝ
Concaveupon๏ฟฝโ`12, 0๏ฟฝ ๐๐๐ ๏ฟฝ`12, โ๏ฟฝ
7.sketch
Sketch๐ ๐ฅ = oomn;
,given๐q ๐ฅ = ;(omn;)z
and๐qq ๐ฅ = โ 4o(omn;)w
1.Domain:Allrealnumbers2.AsymptotesVertical:noneHorizontal-โ limoโp๏ฟฝ
oโomn;
= ๏ฟฝ๏ฟฝ ยกโ}ยข
o
๏ฟฝ๏ฟฝ ยกโ}ยข
โomn;
=
๏ฟฝ๏ฟฝ ยกโ}ยข
o
๏ฟฝ๏ฟฝ ยกโ}ยข
โomn;=
๏ฟฝ๏ฟฝ ยกโ}ยข
o
๏ฟฝ๏ฟฝ ยกโ}ยข
|o|= โ1
Horizontalasymptoteof-1as๐ฅ โ โโ
Horizontalโ limoโ๏ฟฝ
oโomn;
=๏ฟฝ๏ฟฝ ยกโยข
o
๏ฟฝ๏ฟฝ ยกโ}ยข
โomn;
=
๏ฟฝ๏ฟฝ ยกโยข
o
๏ฟฝ๏ฟฝ ยกโยข
โomn;=
๏ฟฝ๏ฟฝ ยกโยข
o
๏ฟฝ๏ฟฝ ยกโยข
|o|= 1
Horizontalasymptoteof1as๐ฅ โ โ
3.Interceptsxโintercept0= o
โomn;
x=0intercept(0,0)
yโintercepty= L
โLmn;
y=0intercept(0,0)
4.Intervalsofincreaseanddecrease๐q(๐ฅ) = ;
c(omn;)z
0 = c(๐ฅ< + 1)4and0= ;c(omn;)z
havenosolutions.
Wecancheckthederivativeatx=0togetapositivenumbermeaningthefunctionisincreasingfrom(-โ,โ)withnocriticalpoints.5.Localmax/minNone6.Concavity๐qq(๐ฅ) = โ 4o
c(omn;)w
0=-3x0=c(๐ฅ< + 1)lx=0nosolution๐qq(โ1) = ๐๐๐ ๐๐ก๐๐ฃ๐๐qq(1) = ๐๐๐๐๐ก๐๐ฃ๐Concaveupon(-โ,0),concavedownon(0,โ),inflectionpointat(0,0)
SOMEFUNLIMITS
Evaluatethelimit: limoโ;}
๐๐ ๐ฅk โ 1 โ ๐๐ โ1
Letf(x)= sin ๐z ๐๐๐ฅ = ๐'๐'๐๐๐ฅ โ ๐'
,evaluatethelimit limoโ/ยค
๐(๐ ๐ฅ ).
Giveyouranswerinacalculatorreadyform
Supposethatlimoโ:
๐ ๐ฅ = ๐ง,wherezisapositiveinteger,whichofthefollowingstatementsareguaranteedtobetrue?(Youmaycirclemultipleoptions)
a)๐q ๐ ๐๐ฅ๐๐ ๐ก๐ b)๐ ๐ฅ ๐๐ ๐๐๐๐ก๐๐๐ข๐๐ข๐ ๐๐ก๐ฅ = ๐c)๐ ๐ฅ ๐๐ ๐๐๐๐๐๐๐๐๐ก๐ฅ = ๐d)๐ ๐ = ๐งe)๐)๐๐๐๐)f)๐)๐๐๐๐)g)๐๐๐๐๐๐๐กโ๐๐๐๐๐ฃ๐
Startwiththeinsidef(x)firstthenworkoutwards limoโ/ยค
๐(๐(๐)).
AsxAPPROACHES๐' ,whichalsomeansxisNOT๐' ,thenvalueoff(x)willbe๐' (bottomcase)
Nowweareleftwithf(๐')whichwillgiveusavalueofsinโ๐z
Therefore,thefinalansweris๐ฌ๐ข๐งโ๐๐
limoโ;}
๐๐|๐ฅk โ 1| โ ๐๐ |โ1|
= ๐๐|(1p)k โ 1| โ ๐๐ 1(ln1=0)= ๐๐|(1p) โ 1|(somethingslightlysmallerthan1tothepowerof4isstillsomethingslightlysmallerthan1)= ๐๐|0p|=๐๐0n=-โ(graphofln,youshouldknowhappensatln0)
g)itispossiblefor๐(๐ฅ)tohaveaholeatx=c.Thismeansb),c),andd)arefalse.Thereisnot
enoughinformationtosaya)isalwaystrue.Example:๐(๐ฅ) = ยญ(op<)(op<)(op<)
ยญwithc=2.Checkitout
ingraphingcalculator.