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.