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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:/
𝑝𝑙𝑎𝑛𝑒𝑡𝑟𝑎𝑣𝑒𝑙𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 6𝑘𝑚
2.𝑡𝑖𝑚𝑒 = +562/.768/59-,.62:/8/5*:-.1
= (4LL
= 0.02h
3. cos '(= 69R6:/2.
01+*./23,/= +562/0*7-S*2.659-,.62:/
(
√4<= +562/0*7-S*2.659-,.62:/
(
𝑝𝑙𝑎𝑛𝑒ℎ𝑜𝑟𝑖𝑧𝑡𝑜𝑛𝑎𝑙𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 3√3𝑘𝑚
𝑝𝑙𝑎𝑛𝑒ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙𝑣𝑒𝑙𝑜𝑐𝑡𝑦 = +562/0*7-S*2.659-,.62:/.-Y/
= 4√4L.L<
= 150√3𝑘𝑚/ℎ
𝑝𝑙𝑎𝑛𝑒𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙𝑣𝑒𝑙𝑜𝑐𝑡𝑦 = +562/0/-\0.9-,.62:/.-Y/
= 4L.L<
= 150𝑘𝑚/ℎ
5.Carhorizontaldistance=time*carvelocity=0.02*150=3km
6.𝑔𝑟𝑜𝑢𝑛𝑑𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒< = 𝑝𝑙𝑎𝑛𝑒ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒< + 𝑐𝑎𝑟ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒<
𝑔𝑟𝑜𝑢𝑛𝑑𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = `(3√3)< + 3<
=c9(3) + 9=√36 = 6𝑘𝑚
7.findingrateofchangeforgrounddistanceForsakeofsimplicity(justforthispart)Ipluginnumbersfromstepsabovesoifyouarewonderwhereanumbercamefrom.Lookup!
grounddistance=cplanehorizontaldistance=acarhorizontaldistance=b
𝑐< = 𝑎< +𝑏<nowwederive
2𝑐(9:9.) = 2𝑎(96
9.) + 2𝑏(9f
9.)
2(6)(9:9.) = 2(3√3)(150√3) + 2(3)(150)
12(9:9.) = 2700 + 900
h9:9.i = 4(LL
;<= 300𝑘𝑚/ℎrateofchangeofgrounddistancewhenplanehaselevationof3km
8.Heightdistance=3km(given)
9.𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒< = ℎ𝑒𝑖𝑔ℎ𝑡𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒< + 𝑔𝑟𝑜𝑢𝑛𝑑𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒<
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = √3< + 6< = √45
10.Forthissectiononlywewillusethefollowingvariablenames,nottobeconfusedwithpart7.Ipluginnumbersfromstepsabovesoifyouarewonderwhereanumbercamefrom.Lookup!
Distance=cheightdistance=agrounddistance=b𝑐< = 𝑎< +𝑏<nowwederive
2𝑐(9:9.) = 2𝑎(96
9.) + 2𝑏(9f
9.)
2√45(9:9.) = 2(3)(150) + 2(6)(300)
2√45(9:9.) = 900 + 3600
2√45(9:9.) = 4500
9:9.= klLL
<√kl
9:9.= <<lL
√kl𝑘𝑚/ℎfinalanswer
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 thestudentsto“getoverit”?(bothphysicallyandpsychologically)
Wewanttominimizetime.0≤x≤10Thereare2separatesections(swimmingandwalking)Thetimeofswimmingsectionisdefinedby:
𝑡 = 9-,.62:/8/5*:-.1
= √lmnom
4
Thetimeofthewalkingsectionisdefinedby:𝑡 = 9-,.62:/
8/5*:-.1= ;Lpo
l
ThetotaltimetogettopointBwouldbe:𝑡 = √lmnom
4+ ;Lpo
l
𝑡q = ;4;<
<o√lmnom
− ;l
𝑡q = <o(√lmnom
− ;l
0 = o4√lmnom
− ;l
;l= o
4√lmnom
3√5< + 𝑥< = 5𝑥squarebothsides9(5< + 𝑥<) = 25𝑥<9 ∗ 5< + 9𝑥< = 25𝑥<9 ∗ 5< = 16𝑥<
𝑥 = `u∗lm
;(
𝑥 = √u√lm
√;(
𝑥 = 4∗lk
𝑥 = ;lk
Testboundarycasesaswell(x=0andx=10)
𝑡 = √lmnLm
4+ ;LpL
l
= √lm
4+ ;L
l
= l4+ 2
= ;;4
𝑡 = `lmn(vw
x)m
4+
;Lpvwx
l
𝑡 = `mwvnmmwvy
4+
mwx
l
𝑡 = `ymwvy
4+ l
k
𝑡 = mwx
4+ l
k
𝑡 = ;L4
𝑥 = ;l
kgivesusthesmallestt.
𝑡 = √lmn;Lm
4+ ;Lp;L
l
𝑡 = √<ln;LL4
𝑡 = √;<l4
Wecancomparethiswith10/3bysquaringbothofthem.125/9>100/9
LOCALLINEARAPPROXIMATIONGiven: 65z
a) Findthelinearapproximationb) Usethelinearapproximationtoapproximatethedesiredvaluec) Determineifitisanoverestimateoranunderestimated) Whatistheboundontheerror?
Originally,IhadaveryfunquestionplannedforyouguysbutProfessorDesaulnierssaid“Ithinkthisproblemmayconfusethem”soIhadtochangeittoasimplerone:(
a) 𝐿(𝑥) = 𝑓(𝑎) + 𝑓q(𝑎)(𝑥 − 𝑎)Wecanchoosea=64.𝑓(𝑎) = 4
𝑓q(𝑥) = ;4(𝑥
}mz )
𝑓q(64) = ;4h ;;(i = ;
k~
𝐿(𝑥) = 4 + ;k~(𝑥 − 64)
b) 𝐿(65) = 4 + ;
k~(65 − 64)
𝐿(65) = 4 + ;k~= k
;hk~k~i + ;
k~= ;u<
k~+ ;
k~= ;u4
k~
c) 𝑓q(𝑥) = ;4(𝑥
}mz )
𝑓qq(𝑥) = ;4h− <
4i (𝑥
}wz ) = − <
u(𝑥
}wz )
𝑓′′(64) =−29(64
−53)𝑓qq(64) = 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒Thesecondderivativeisnegativemeaningitisconcavedown.Thismeansitisanoverestimate.
d) Youneedtoknowthat|𝑓qq(𝑐)| ≤ 𝑀inotherwordsweneedtofindthemaximumvalueof|𝑓qq(𝑐)|𝑖𝑛𝑡ℎ𝑒𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙[64,65]
𝑓qqq(𝑥) = − <uh− l
4i h𝑥
}�z i = 0
h𝑥}�z i = 0
;
�o�z�= 0𝑛𝑜𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛, 𝑎𝑙𝑡ℎ𝑜𝑢𝑔ℎ𝑤ℎ𝑒𝑛𝑥 = 0𝑓qqq(𝑥)𝑖𝑠𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
Therearenocriticalpointsinourintervalsowecanjusttesttheboundaries
Noticethat:𝑓qq(𝑥) = − <uh𝑥
}wz i = − <
u;
(owz)
xisinthedenominatormeaningthatsmallerx=biggerf’’(x).Therefore,x=64willgiveusthelargervalue
|Error|≤h;<i h− <
uih(64)
}wz i (|65 − 64|)<
TAYLORPOLYNOMIALSWhatisthe2nddegreeTaylorpolynomialof:
𝑓 𝑥 = sin(𝑥<) + log( on;)z( 𝑥 + 1)(Lata= 𝜋
𝑓(𝑥) = sin(𝑥<) + 20simplifytheweirdlogattheendto20.
𝐶L = �(√')L!
= 𝑓�√𝜋� = 0 + 20 = 20
𝐶; = ��(√');!
= 𝑓q�√'� = cos h√𝜋<i�2√𝜋� = −2√𝜋
𝐶< = ���(√')
<!= ����√'�
<= ;
<�− sin h√𝜋
<i �2√𝜋��2√𝜋� + cos h√𝜋
<i (2)� = p<
<= −1
The2nddegreeTaylorpolynomialis:
𝑃�(𝑥) = 𝐶L + 𝐶;(𝑥 − 𝑎) + 𝐶<(𝑥 − 𝑎)<
𝑃�(𝑥) = 20 − 2√𝜋�𝑥 − √𝜋�−(𝑥 − √𝜋)<
CURVESKETCHING
Sketchthecurve𝑓 𝑥 = 3𝑥l − 5𝑥4
1.Domain:allrealnumbers2.Asymptotes:
Vertical:noneHorizontal:takelimitasx->infinityandnegativeinfinityandyouwillnotgetafinitenumberasaresult.Therefore,nohorizontalasymptotes.
3.Intercepts:xintercept:
0 = 3𝑥l − 5𝑥40 = 𝑥4(3𝑥< − 5)
x=0,x=±`l4
(0,0),(−`l4, 0), (`l
4, 0)
yintercept:𝑦 = 3(0l) − 5(04)y=0(0,0)
4.Intervalsofincreaseanddecrease: 𝑓q(𝑥) = 15𝑥k − 15𝑥<
0 = 15𝑥<(𝑥< − 1)x=00= 𝑥< − 1 𝑥< = 1 𝑥 = ±1
𝑓′(−2)>0 𝑓′(−0.5)<0 𝑓′(0.5)<0 𝑓′(2)>0
Increasingon(−∞,−1)𝑎𝑛𝑑(1,∞)Decreasingon(−1,0)𝑎𝑛𝑑(0,1)
5.Localmax/minAtx=-1.𝑓(−1) = 3(−1)4 − 5(−1)4=−3 − (−5) = 2Localmax(-1,2)Atx=1.𝑓(1) = 3(1)4 − 5(1)4=3 − 5 = −2Localmin(1,-2)
6.Concavity 𝑓q(𝑥) = 15𝑥k − 15𝑥< 𝑓qq(𝑥) = 60𝑥4 − 30𝑥 0 = 30𝑥(2𝑥2 − 1) 𝑥 = 00 = 2𝑥< − 1 𝑥< = ;
<
𝑥 = ±`;<
𝑓′′(−1) < 0 𝑓′′(−0.1) > 0 𝑓′′(0.1) < 0 𝑓′′(1) > 0
Concavedownon�−∞,−`12� 𝑎𝑛𝑑 �0,`12�
Concaveupon�−`12, 0� 𝑎𝑛𝑑 �`12, ∞�
7.sketch
Sketch𝑓 𝑥 = oomn;
,given𝑓q 𝑥 = ;(omn;)z
and𝑓qq 𝑥 = − 4o(omn;)w
1.Domain:Allrealnumbers2.AsymptotesVertical:noneHorizontal-∞ limo→p�
o√omn;
= �� ¡→}¢
o
�� ¡→}¢
√omn;
=
�� ¡→}¢
o
�� ¡→}¢
√omn;=
�� ¡→}¢
o
�� ¡→}¢
|o|= −1
Horizontalasymptoteof-1as𝑥 → −∞
Horizontal∞ limo→�
o√omn;
=�� ¡→¢
o
�� ¡→}¢
√omn;
=
�� ¡→¢
o
�� ¡→¢
√omn;=
�� ¡→¢
o
�� ¡→¢
|o|= 1
Horizontalasymptoteof1as𝑥 → ∞
3.Interceptsx–intercept0= o
√omn;
x=0intercept(0,0)
y–intercepty= L
√Lmn;
y=0intercept(0,0)
4.Intervalsofincreaseanddecrease𝑓q(𝑥) = ;
c(omn;)z
0 = c(𝑥< + 1)4and0= ;c(omn;)z
havenosolutions.
Wecancheckthederivativeatx=0togetapositivenumbermeaningthefunctionisincreasingfrom(-∞,∞)withnocriticalpoints.5.Localmax/minNone6.Concavity𝑓qq(𝑥) = − 4o
c(omn;)w
0=-3x0=c(𝑥< + 1)lx=0nosolution𝑓qq(−1) = 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒𝑓qq(1) = 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒Concaveupon(-∞,0),concavedownon(0,∞),inflectionpointat(0,0)
SOMEFUNLIMITS
Evaluatethelimit: limo→;}
𝑙𝑛 𝑥k − 1 − 𝑙𝑛 −1
Letf(x)= sin 𝑒z 𝑖𝑓𝑥 = 𝑒'𝑒'𝑖𝑓𝑥 ≠ 𝑒'
,evaluatethelimit limo→/¤
𝑓(𝑓 𝑥 ).
Giveyouranswerinacalculatorreadyform
Supposethatlimo→:
𝑓 𝑥 = 𝑧,wherezisapositiveinteger,whichofthefollowingstatementsareguaranteedtobetrue?(Youmaycirclemultipleoptions)
a)𝑓q 𝑐 𝑒𝑥𝑖𝑠𝑡𝑠b)𝑓 𝑥 𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠𝑎𝑡𝑥 = 𝑐c)𝑓 𝑥 𝑖𝑠𝑑𝑒𝑓𝑖𝑛𝑒𝑑𝑎𝑡𝑥 = 𝑐d)𝑓 𝑐 = 𝑧e)𝑎)𝑎𝑛𝑑𝑑)f)𝑏)𝑎𝑛𝑑𝑐)g)𝑁𝑜𝑛𝑒𝑜𝑓𝑡ℎ𝑒𝑎𝑏𝑜𝑣𝑒
Startwiththeinsidef(x)firstthenworkoutwards limo→/¤
𝑓(𝒇(𝒙)).
AsxAPPROACHES𝑒' ,whichalsomeansxisNOT𝑒' ,thenvalueoff(x)willbe𝑒' (bottomcase)
Nowweareleftwithf(𝑒')whichwillgiveusavalueofsin√𝑒z
Therefore,thefinalansweris𝐬𝐢𝐧√𝒆𝟑
limo→;}
𝑙𝑛|𝑥k − 1| − 𝑙𝑛 |−1|
= 𝑙𝑛|(1p)k − 1| − 𝑙𝑛 1(ln1=0)= 𝑙𝑛|(1p) − 1|(somethingslightlysmallerthan1tothepowerof4isstillsomethingslightlysmallerthan1)= 𝑙𝑛|0p|=𝑙𝑛0n=-∞(graphofln,youshouldknowhappensatln0)
g)itispossiblefor𝑓(𝑥)tohaveaholeatx=c.Thismeansb),c),andd)arefalse.Thereisnot
enoughinformationtosaya)isalwaystrue.Example:𝑓(𝑥) = (op<)(op<)(op<)
withc=2.Checkitout
ingraphingcalculator.