Exercícios Resolvidos - Cap. 10 (Pares) - Ácidos e Bases - -Princípios de Química - Atkins
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6.1SOLUTIONS 313
CHAPTERSIX
Solutions forSection6.1
Exercises
1.
2.
3.
4.
5. BytheFundamentalTheoremofCalculus,weknowthat
Usingaleft-handsum,weestimate .Usingaright-handsum,weestimate
.Averaging,wehave
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314 ChapterSix/SOLUTIONS
Weknow ,so
Similarly,weestimate
so
Similarly,
so
Thevaluesareshowninthetable.
0
2
4
6
100 128 169 217
6. Thechangein between and isequalto .Aleft-handestimateforthisintegralis anda
righthandestimateis .Ourbestestimateistheaverage, .Thechangein between and is .
Since ,wehave .Wefindtheothervaluessimilarly.TheresultsareshowninTable6.1.
Table6.1
0 2 4 6
50
82
107
119
7. (a) Thevalueoftheintegralisnegativesincetheareabelowthe -axisisgreaterthantheareaabovethe -axis.We
countboxes:Theareabelowthe -axisincludesapproximately boxesandeachboxhasarea ,so
Theareaabovethe -axisincludesapproximately2boxes,eachofarea2,so
Sowehave
(b) BytheFundamentalTheoremofCalculus,wehave
so,
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6.1SOLUTIONS 315
8. Since isnegativefor andpositivefor ,weknowthat isdecreasingfor andincreasingfor
.Betweeneachtwointegervalues,themagnitudeofthechangeisequaltotheareabetweenthegraph and
the -axis.Forexample,between and ,weseethatthechangein is .Since at ,wemust
have at .Theothervaluesarefoundsimilarly,andareshowninTable6.2.
Table6.2
1 2 3 4 5
Problems
9. (a) Criticalpointsof arethezerosof : and .
(b) hasalocalminimumat andalocalmaximumat .
(c)
Noticethatthegraphcouldalsobeaboveorbelowthe -axisat .
10. (a) Criticalpointsof are , and .
(b) hasalocalminimumat ,alocalmaximumat ,andalocalminimumat .
(c)
11.
Notethatsince and , isalocalmaximum;since and , is
alocalminimum.Also,since and
changesfromdecreasingtoincreasingabout ,
hasaninflection
pointat .
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316 ChapterSix/SOLUTIONS
12.
Notethatsince , ,so isalocalminimum.Since and changesfrom
decreasingtoincreasingat , hasaninflectionpointat .
13.
Notethatsince , iseitheralocalminimumorapointofinflection;itisimpossibletotellwhichfrom
thegraph.Since ,and changessignaround isaninflectionpoint.Also,since
and changesfromincreasingtodecreasingabout , hasanotherinflectionpointat .
14. Between and ,theparticlemovesat10km/hrfor1hour.Sinceitstartsat ,theparticleisat
when .SeeFigure6.1.Thegraphofdistanceisastraightlinebetween and becausethevelocityis
constantthen.
Between and ,theparticlemoves10kmtotheleft,endingat .Between and ,itmoves
10kmtotherightagain.SeeFigure6.1.
(km)
(hr)
Figure6.1
Asanaside,notethattheoriginalvelocitygraphisnotentirelyrealisticasitsuggeststheparticlereversesdirection
instantaneouslyattheendofeachhour.Inpracticethismeansthereversalofdirectionoccursoveratimeintervalthatis
shortincomparisontoanhour.
15. (a) Weknowthat fromtheFundamentalTheoremofCalculus.Fromthegraphof we
canseethat bysubtractingareasbetween andthe -axis.Since ,wefindthat
.Similarreasoninggives .
(b) Wehave , , , , ,and .Sothegraph,beginningat ,
startsatzero,increasesto
at ,decreasesto
at ,increasesto
at ,thenpassesthroughazeroas
itdecreasesto at ,andfinallyincreasesto at .Thus,therearethreezeroes: , ,and .
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6.1SOLUTIONS 317
(c)
16. Wecanstartbyfindingfourpointsonthegraphof .Thefirstoneisgiven: .BytheFundamentalTheorem
ofCalculus, .Thevalueofthisintegralis (theareais7,butthegraphliesbelowthe
-axis),so .Similarly, ,and .Wesketchagraphof
byconnectingthesepoints,asshowninFigure6.2.
Figure6.2
17. Thecriticalpointsareat , , ,and .Agraphisgivenbelow.
18. Lookingatthegraphof below,weseethatthecriticalpointsof occurwhen and ,since at
thesevalues.Inflectionpointsof occurwhen and ,because hasalocalmaximumorminimumat
thesevalues.Knowingthesefourkeypoints,wesketchthegraphof asfollows.
Westartat ,where .Since isnegativeontheinterval ,thevalueof isdecreasingthere.
At wehave
areaofshadedtrapezoid
Similarly,
areaoftriangle
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318 ChapterSix/SOLUTIONS
Continuing,
and
We
now
find
concavity
of
in
the
intervals
,
,
,
by
checking
whether
increases
ordecreasesinthesesameintervals.If increases,then isconcaveup;if decreases,then isconcave
down.Thuswefinallyhaveourgraphof :
19. Betweentime andtime ,thevelocityofthecorkisalwayspositive,whichmeansthecorkismovingupwards.
At
time
,
the
velocity
is
zero,
and
so
the
cork
has
stopped
moving
altogether.
Since
shortly
thereafter
the
velocityofthecorkbecomesnegative,thecorkwillnextbegintomovedownwards.Thuswhen thecorkhasrisenasfaras
iteverwill,andisridingontopofthecrestofthewave.
Fromtime totime ,thevelocityofthecorkisnegative,whichmeansitisfalling.When ,the
velocityisagainzero,andthecorkhasceasedtofall.Thuswhen thecorkisridingonthebottomofthetroughof
thewave.
Sincethecorkisonthecrestattime andinthetroughattime ,itisprobablymidwaybetweencrestandtrough
whenthetimeismidwaybetween and .Thusattime thecorkismovingthroughtheequilibriumpositionon
itswaydown.(Theequilibriumpositioniswherethecorkwouldbeifthewaterwereabsolutelycalm.)Bysymmetry,
isthetimewhenthecorkismovingthroughtheequilibriumpositiononthewayup.
Sinceaccelerationisthederivativeofvelocity,pointswheretheaccelerationiszerowouldbecriticalpointsofthe
velocityfunction.Sincepoint (amaximum)andpoint (aminimum)arecriticalpoints,theaccelerationiszerothere.
Apossiblegraphoftheheightofthecorkisshownbelow.Thehorizontalaxisrepresentsaheightequaltotheaverage
depthoftheoceanatthatpoint(theequilibriumpositionofthecork).
height
time
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6.1SOLUTIONS 319
20. Therateofchangeisnegativefor andpositivefor ,sotheconcentrationofadrenalinedecreasesuntil
andthenincreases.Sincetheareaunderthe -axisisgreaterthantheareaoverthe -axis,theconcentrationofadrenaline
goesdownmorethanitgoesup.Thus,theconcentrationat islessthantheconcentrationat .SeeFigure6.3.
adrenalineconcentration( g/ml)
(minutes)
Figure6.3
21. (a) Thetotalvolumeemptiedmustincreasewithtimeandcannotdecrease.Thesmoothgraph(I)thatisalwaysincreasing
isthereforethevolumeemptiedfromthebladder.Thejaggedgraph(II)thatincreasesthendecreasestozeroisthe
flow
rate.(b) Thetotalchangeinvolumeistheintegraloftheflowrate.Thus,thegraphgivingtotalchange(I)showsanantideriva-
tiveoftherateofchangeingraph(II).
22. Thegraphof isshowninFigure6.4.Weseethattherearerootsat and .Theseare
thecriticalpointsof .Lookingatthegraph,itappearsthatofthethreeareasmarked, isthelargest, isnext,
and issmallest.Thus,as increasesfrom0to3,thefunction increases(by ),decreases(by ),andthen
increasesagain(by ).Therefore,themaximumisattainedatthecriticalpoint .
Whatisthevalueofthefunctionatthismaximum?Weknowthat ,soweneedtofindthechangein
between and .Wehave
Changein
Weseethat ,sothemaximumvalueof onthisintervalis
Figure6.4
23.
(a) isgreatestat
(b) isleastat
(c) isgreatestat .
(d) isleastat
(e) isgreatestat
(f) isleastat
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320 ChapterSix/SOLUTIONS
24. Both and haverootsat and .Bothhaveacriticalpoint(whichisalocalmaximum)at .
However,sincetheareaunder between and islargerthantheareaunder between and
,the -coordinateof at willbelargerthanthe -coordinateof at .Seebelow.
25. (a) Suppose istheamountofwaterinthereservoirattime .Then
Rateatwhichwater
inreservoirischanging
Inflow
rate
Outflow
rate
Thustheamountofwaterinthereservoirisincreasingwhentheinflowcurveisabovetheoutflow,anddecreasing
whenitisbelow.Thismeansthat isamaximumwherethecurvescrossinJuly1993(asshowninFigure6.5),
and
is
decreasing
fastest
when
the
outflow
is
farthest
above
the
inflow
curve,
which
occurs
about
October
1993(seeFigure6.5).
Toestimatevaluesof ,weusetheFundamentalTheoremwhichsaysthatthechangeinthetotalquantity
ofwaterinthereservoirisgivenby
Jan93 inflowrate outflowrate
or Jan93 inflowrate outflowrate
rateofflow
(millionsofgallons/day)isdecreasingmostrapidly
isincreasing
mostrapidlyismax
ismin
Outflow
Inflow
Jan(93) April July Oct Jan(94)
millions
of
gallons
Jan(93) April July Oct Jan(94)
Figure
6.5
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6.2SOLUTIONS 321
(b) SeeFigure6.5.MaximuminJuly1993.MinimuminJan1994.
(c) SeeFigure6.5.IncreasingfastestinMay1993.DecreasingfastestinOct1993.
(d) InorderforthewatertobethesameasJan93thetotalamountofwaterwhichhasflowedintothereservoirmustbe
0.ReferringtoFigure6.6,wehave
inflow
outflow
giving
rateofflow
(millionsofgallons/day)
Inflow
Outflow
Jan(93) April July Oct Jan(94) April July
Figure6.6
Solutions forSection6.2
Exercises
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16. ,whichhasantiderivative
17.
18.
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322 ChapterSix/SOLUTIONS
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34. ,so . impliesthat ,so .Thus istheonlypossibility.
35. ,so . impliesthat ,so .Thus istheonlypossibility.
36. ,so . impliesthat ,so .Thus isthe
onlypossibility.
37. ,so . impliesthat ,so .Thus istheonly
possibility.
38. ,so . impliesthat ,so .Thus istheonlypossibility.
39. ,so . impliesthat ,so .Thus isthe
only
possibility.
40. ,so . impliesthat .Thus
istheonlypossibility.
41. ,so . impliesthat ,so .Thus
istheonlypossibility.
42. .
43.
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6.2SOLUTIONS 323
44.
45.
46.
47.
48.
49.
50.
51.
52.
53.
54.
55.
56.
57.
58.
59.
60.
61.
62. .
63.
64. .
65. .
66. .
67. Since ,
.
68. .
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324 ChapterSix/SOLUTIONS
69. .
70. .
71. .
72. Since ,
73. ,since ,so
.
Problems
74.
We
have
Area
75. Thegraphcrossesthe -axiswhere
so and .SeeFigure6.7.Theparabolaopensupwardandtheregionisbelowthe -axis,so
Area
Figure6.7
76. Thegraphisshowninthefigurebelow.Since for ,wehave
Area
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6.2SOLUTIONS 325
77. Sincethegraphof isabovethegraphof (seethefigurebelow),wehave
Area
78. Theareaunder between and isgivenby .UsingtheFundamentalTheoremtoevaluate
theintegral:
Area
Sincetheareais ,wehave
Since islargerthan1,wehave .
79. Thegraphof has -interceptsof .Seethefigurebelow.Theshadedareaisgivenby
Area
Wewant tosatisfy ,so .
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326 ChapterSix/SOLUTIONS
80. Wehave
Averagevalue
We
see
in
Figure
6.8
that
the
average
value
of
for
looks
right.
Figure6.8
81. Theaveragevalueof ontheinterval is
Since ,wehave ,so .
82. (a) Theaveragevalueof over isgivenbytheformula
Average
Wecancheckthisanswerbylookingatthegraphof below.Theareabelowthecurveandabovethe -axis
overtheinterval isthesameastheareaabovethecurvebutbelowthe -axisovertheinterval
.Whenwetaketheintegralof overtheentireinterval ,weget .
(b) Since
theaveragevalueof on isgivenby
Averagevalue
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6.3SOLUTIONS 327
83. Since wewanttoevaluatetheindefiniteintegral
where isaconstant.Thus ,andthefixedcostof1,000,000riyalmeansthat
.
Therefore,
the
total
cost
is
Since dependson ,thesquareofthedepthdrilled,costswillincreasedramaticallywhen growslarge.
84. (a) CCl dumped
(b) 7years,because indicatesthattherateofflowwaszeroafter7years.
(c)
Areaunderthecurve
cubicyards.
Solutions forSection6.3
Exercises
1.
2.
3.
4.
5. Since ,wedifferentiate toseethat ,so satisfiesthedifferentialequation.Toshow
thatitalsosatisfiestheinitialcondition,wecheckthat :
6. .If ,then and .
Thus, .
7. .If ,then so .Thus, .
8. .If when ,then ,so .
Thus .
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328 ChapterSix/SOLUTIONS
9. Integratinggives
If when ,then so .Thus .
10.
We
differentiate
using
the
product
rule
to
obtain
andso satisfiesthedifferentialequation.Wenowcheckthat :
Problems
11. (a) Acceleration= m/sec
Velocity= m/sec
Height= m
.
(b) Atthehighestpoint,
so
seconds.
Atthattime, m.Weseethatthetomatoreachesaheightof m,at secondsafteritis
thrown.
(c) Thetomatolandswhen ,so
The
solutions
are
and
seconds.
We
see
that
it
lands
seconds
after
it
is
thrown.
12. (a) ,sothesolutionis .
(b)
(c) At
,wehave
andso
.Thuswehavethesolution
.
13.
Since when ,wehave ,so
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6.3SOLUTIONS 329
14. (a) Tofindtheheightoftheballoon,weintegrateitsvelocitywithrespecttotime:
Sinceat ,wehave ,wecansolvefor toget ,givingusaheightof
(b) Tofindtheaveragevelocitybetween and ,wefindthetotaldisplacementanddividebytime.
Averagevelocity ft sec
Theballoonsaveragevelocityis32ft/secdownward.
(c) First,wemustfindthetimewhen .Solvingtheequation ,weget
Thus, or .Since makesnophysicalsense,weuse tocalculatetheballoons
velocity.At ,wehaveavelocityof ft/sec.Sotheballoonsvelocityis56ft/sec
downwardatthetimeofimpact.
15. Sincethecarsaccelerationisconstant,agraphofitsvelocityagainsttime islinear,asshownbelow.
(mph)
(seconds)
Theaccelerationisjusttheslopeofthisline:
Toconvertourunitsintoft/sec ,
mph
sec
mph
sec
mph
sec
5280
ft
1
hour
1mile 3600sec
ft
sec
16. Sincetheacceleration ,where isthevelocityofthecar,wehave
Integratinggives
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330 ChapterSix/SOLUTIONS
Thecarstartsfromrest,so when ,andtherefore .If isthedistancefromthestartingpoint,
and
so
Since when ,wehave ,so
Wewanttosolvefor when :
Thisequationcanberewrittenas
Theequationcanbesolvednumerically,orbytracingalongagraph,orbyfactoring
Thesolutionsare and .Sincewearetold ,thesolutionwewantis
sec.
17. (a)
80ft/sec
5sec
(b) Thetotaldistanceisrepresentedbytheshadedregion ,theareaunderthegraphof .
(c)The
area
,
a
triangle,
is
given
by
(d) UsingintegrationandtheFundamentalTheoremofCalculus,wehave or where
isanantiderivativeof .
Wehavethat ,theacceleration,isconstant: forsomeconstant .Therefore forsome
constant .Wehave ,sothat .Puttingin , ,
or .
Thus ,andanantiderivativefor is .Sincethetotaldistance
traveledat is ,wehave which means .Finally,
,whichagreeswiththepreviouspart.
18. Sincetheaccelerationisconstant,agraphofthevelocityversustimelookslikethis:
(mph)
200
mph
A
(sec)
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6.3SOLUTIONS 331
Thedistancetraveledin30seconds,whichishowlongtherunwaymustbe,isequaltothearearepresentedby .
Wehave .Firstweconverttherequiredvelocityintomilespersecond.
200mph
Therefore .
19. (a) Sincethevelocityisconstantlydecreasing,and ,thecarstopsafter6seconds.
(sec)
(ft/sec)
(b) Overtheinterval ,theleft-handvelocityis ,andtheright-handvelocityis .Sincewe
areconsideringhalf-secondintervals, ,and .Theleftsumis ft.,andtherightsumis ft.
(c) Area inthefigurebelowrepresentsdistancetraveled.
velocity(ft/sec)
Deceleration
=5ft/sec
(seconds)
(d)
The
velocity
is
constantly
decreasing
at
a
rate
of
5
ft/sec
per
second,
i.e.
after
each
second
the
velocity
has
dropped
by5units.Therefore .
Anantiderivativefor is ,where .ThusbytheFundamentalTheoremofCalculus,
thedistancetraveled .Since isdecreasing,the
left-handsuminpart(b)overestimatesthedistancetraveled,whiletheright-handsumunderestimatesit.
Thearea isequaltotheaverageoftheleft-handandright-handsums: .The
left-hand
sum
is
an
overestimate
of ;
the
right-hand
sum
is
an
underestimate.
20. (a)
highestpoint ground(sec)
(b) Thehighestpointisat seconds.Theobjecthitsthegroundat seconds,sincebysymmetryiftheobject
takes5secondstogoup,ittakes5secondstocomebackdown.
(c) Themaximumheightisthedistancetraveledwhengoingup,whichisrepresentedbythearea ofthetriangleabove
thetimeaxis.
ft/sec sec feet
(d) Theslopeofthelineis ,so Antidifferentiating,weget . ,
so At , ft.
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332 ChapterSix/SOLUTIONS
21. Theequationofmotionis .Takingthefirstderivative,weget .
Thesecondderivativegivesus .
(a) Atitshighestpoint,thestonesvelocityiszero:
,so .
(b)
At
,
the
height
is
ft
(c) Whenthestonehitsthebeach,
So seconds.
(d) Impactisat .Thevelocity, ,atthistimeis ft/sec.Uponimpact,thestones
velocityis192ft/secdownward.
22. (a) so ,sincetheinitialvelocityis0.
(b) ,
where is
the
rocks
initial
height.
23. (a) ,where initialvelocity,and .Atthemaximumheight, ,so max.
Plugging
into
the
distance
equation
yields max max max,so max seconds,fromwhichwe
get ft/sec.
(b)
This
time
ft/sec
,
so
,
and
.
At
the
highest
point,
,
somax seconds.Pluggingintothedistanceequationyields ft.
24. Theheightofanobjectabovethegroundwhichbeginsatrestandfallsfor secondsis
where istheinitialheight.Heretheflowerpotfallsfrom200ft,so .Toseewhenthepothitstheground,
solve .Thesolutionis
seconds
Now,velocityisgivenby .So,thevelocitywhenthepothitsthegroundis
ft/sec
whichisapproximately mphdownwards.
25. Thefirstthingweshoulddoisconvertourunits.Wellbringeverythingintofeetandseconds.Thus,theinitialspeedof
the
car
is
miles
hour
hour
sec
feet
mile
ft/sec
Weassumethattheaccelerationisconstantasthecarcomestoastop.Agraphofitsvelocityversustimeisgivenin
Figure6.9.Weknowthattheareaunderthecurverepresentsthedistancethatthecartravelsbeforeitcomestoastop,
157feet.Butthisareaisatriangle,soitiseasytofind ,thetimethecarcomestorest.Wesolve
whichgives
sec
Sinceaccelerationistherateofchangeofvelocity,thecarsaccelerationisgivenbytheslopeofthelineinFigure6.9.
Thus,theacceleration, ,isgivenby
ft/sec
Notice
that is
negative
because
the
car
is
slowing
down.
ft/sec
Figure6.9:Graphofvelocityversustime
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6.4SOLUTIONS 333
Solutions forSection6.4
Exercises
1.
BytheFundamentalTheorem, .Since ispositiveandincreasing, isincreasingandconcaveup.
Since ,thegraphof muststartfromtheorigin.
2.
Since
is
always
positive,
is
always
increasing.
has
an
inflection
point
where
.
Since
, goesthroughtheorigin.
3.
Since isalwaysnon-negative, isincreasing. isconcaveupwhere isincreasingandconcavedownwhere is
decreasing; hasinflectionpointsatthecriticalpointsof .Since ,thegraphof goesthrough
theorigin.
4. Table
6.3
0 0.5 1 1.5 2
0
0.50
1.09
2.03
3.65
5. UsingtheFundamentalTheorem,weknowthatthechangein between and isgivenby
Since ,wehave .Theothervaluesarefoundsimilarly,andaregiveninTable6.4.
Table6.4
6.
(a)
Again
using
0.00001
as
the
lower
limit,
because
the
integral
is
improper,
gives
,
(b) decreaseswhentheintegrandisnegative,whichoccurswhen
7. If ,then isoftheform
Since ,wetake and ,giving
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334 ChapterSix/SOLUTIONS
8. If ,then isoftheform
Since
,
we
take
and
,
giving
9. If ,then isoftheform
Since ,wetake and ,giving
Problems
10.
11.
Weknowthat increasesfor becausethederivativeof ispositivethere.Seefigureabove.Similarly,
decreasesfor .Therefore,thegraphof risesuntil ,andthenitbeginstofall.Thus,themaximum
valueattainedby is .Toevaluate ,weusetheFundamentalTheorem:
whichgives
The
definite
integral
equals
the
area
of
the
shaded
region
under
the
graph
of
,
which
is
roughly
350.
Therefore,
the
greatestvalueattainedby is .
12. Since and ,wehave
Substituting andevaluatingtheintegralnumericallygives
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6.4SOLUTIONS 335
13. Since and ,wehave
Substituting andevaluatingtheintegralnumericallygives
14.
15.
16.
17.
18.
19. Considering asthecompositionof and ,wemayapplythechainruletoobtain
20. (a) Thedefinitionof gives .
(b) TheFundamentalTheoremgives .
(c) Thefunction isconcaveupwardwhere ispositive.Since ,weseethat isconcaveupwhere is
increasing.Thisoccursontheinterval .
(d) Thefunction decreasesfrom to andincreasesfor ,andthemagnitudeoftheincreaseis
morethanthemagnitudeofthedecrease.Thus takesitsmaximumvalueat .
21. (a)Since
,
we
have
.
(b) (Areaabove -axis) (Areabelow -axis) .(Thetwoareasareequal.)
1
-1
(c) everywhere. onlyatintegermultiplesof Thiscanbeseenfor bynoting
Area
above
-axis
Area
below
-axis
,
which
is
always
non-negative
and
only
equals
zero
when
is
an
integer
multipleof .For
sincetheareafrom to0isthenegativeoftheareafrom0to Sowehave forall
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336 ChapterSix/SOLUTIONS
22. (a) bytheConstructionTheorem.
(b) For , ,so isincreasing.Since for ,thegraphof is
concavedown.
(c)
23.
24. Ifwelet and ,thenwearelookingfor .Bythechainrule,thisisthesameas
.Since
and ,wehave
and
so
25. Ifwelet and ,thenweusethechainrulebecausewearelookingfor
.Since ,wehave
26. Wesplittheintegral intotwopieces,sayat (thoughitcouldbeatanyotherpoint):
Wehaveusedthefactthat Differentiatinggives
Forthefirstintegral,weusethechainrulewith astheinsidefunction,sothefinalansweris
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6.5SOLUTIONS 337
Solutions forSection6.5
Exercises
1. (a) Theobjectisthrownfromaninitialheightof meters.
(b) Thevelocityisobtainedbydifferentiating,whichgives m/sec.Theinitialvelocityis m/sec
upward.
(c) Theaccelerationduetogravityisobtainedbydifferentiatingagain,giving m/sec,or m/sec down-
ward.
2. Sinceheightismeasuredupward,theinitialpositionofthestoneis metersandtheinitialvelocityis
m/sec.Theaccelerationduetogravityis m/sec .Thus,theheightattime isgivenby
meters.
Problems
3. Thevelocityasafunctionoftimeisgivenby: .Sincetheobjectstartsfromrest, ,andthevelocity
isjusttheaccelerationtimestime: .Integratingthis,wegetpositionasafunctionoftime: ,
where
the
last
term,
,
is
the
initial
position
at
the
top
of
the
tower,
so
feet.
Thus
we
have
a
function
giving
positionasafunctionoftime: .
Tofindatwhattimetheobjecthitstheground,wefind when .Wesolve for ,getting
,so .Thereforetheobjecthitsthegroundafter5seconds.Atthistimeitismovingwitha
velocity feet/second.
4. InProblem3weusedtheequation tolearnthattheobjecthitsthegroundafter5seconds.Inamore
generalformthisistheequation ,andweknowthat , ft.Sothemomenttheobject
hitsthegroundisgivenby .InProblem3weused ft/sec ,butinthiscasewewanttofinda
thatresultsintheobjecthittingthegroundafteronly5/2seconds.Weputin5/2for andsolvefor :
so ft/sec
5.
.
Since
is
the
antiderivative
of
,
But
,
so
.
Since
istheantiderivativeof , ,where istheheightofthebuilding.Sincetheballhitsthegroundin5
seconds, Hence feet,sothewindowis400feethigh.
6. Lettime bethemomentwhentheastronautjumpsup.Ifaccelerationduetogravityis5ft/sec andinitialvelocity
is10ft/sec,thenthevelocityoftheastronautisdescribedby
Suppose describeshisdistancefromthesurfaceofthemoon.BytheFundamentalTheorem,
since
(assuming
the
astronaut
jumps
off
the
surface
of
the
moon).
Theastronautreachesthemaximumheightwhenhisvelocityis ,i.e.when
Solvingfor ,weget secasthetimeatwhichhereachesthemaximumheightfromthesurfaceofthemoon.Atthis
timehisheightis
ft.
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338 ChapterSix/SOLUTIONS
Whentheastronautisatheight ,heeitherjustlandedorisabouttojump.Tofindhowlongitisbeforehecomes
backdown,wefindwhenheisatheight .Set toget
Sowehave sec(whenhejumpsoff)and sec(whenhelands,whichgivesthetimehespentintheair).
7. Lettheaccelerationduetogravityequal meters/sec ,forsomepositiveconstant ,andsupposetheobjectfallsfrom
aninitialheightof meters.Wehave ,sothat
Sincetheinitialvelocityiszero,wehave
whichmeans .Ourformulabecomes
Thismeans
Since
wehave ,andourformulabecomes
Supposethattheobjectfallsfor seconds.Assumingithasnthittheground,itsheightis
sothatthedistancetraveledis
meters
whichisproportionalto
.
8. (a) ,where isthetimeittakesforan objecttotravelthedistance ,startingfrom restwithuniform
acceleration . isthehighestvelocitytheobjectreaches.Sinceitsinitialvelocityis0,themeanofitshighest
velocityandinitialvelocityis
(b) ByProblem7, ,where istheaccelerationduetogravity,soittakes secondsforthebody
tohittheground.Since ft/sec.Galileosstatementpredicts ft ft/sec
seconds,andsoGalileosresultisverified.
(c) Iftheaccelerationisaconstant then ,and .Thus
9. (a) Since ,thedistanceabodyfallsinthefirstsecondis
Inthesecondsecond,thebodytravels
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SOLUTIONStoReviewProblemsforChapterSix 339
Inthethirdsecond,thebodytravels
andinthefourthsecond,thebodytravels
(b) Galileoseemstohavebeencorrect.Hisobservationfollowsfromthefactthatthedifferencesbetweenconsecutive
squaresareconsecutiveoddnumbers.For,if isanynumber,then ,whichisthe odd
number(where1isthefirst).
10. If isthedistancefromthecenteroftheearth,
soat2meters
At100metersabovetheground,
so
Thus,tothefirstdecimalplace,theaccelerationduetogravityisstill9.8m/sec at100mabovetheground.
At100,000metersabovetheground,
SolutionsforChapter6Review
Exercises
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
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340 ChapterSix/SOLUTIONS
12. ,since
13. .
Anotherwaytoworktheproblemistoexpand to asfollows:
Thesetwoanswersarethesame,since ,whichis ,
plusaconstant.
14. .
Anotherwaytoworktheproblemistoexpand to :
Itcanbeshownthattheseanswersarethesamebyexpanding .
15.
16. Since ,theindefiniteintegralis
17. Since ,theindefiniteintegralis
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30. .If ,then andthus .So
31. Wehave .Since ,wehave ,so .So .
32. .If ,then andthus .So .
33. .If ,then andthus .So .
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SOLUTIONStoReviewProblemsforChapterSix 341
34. .If ,then andthus .So
35. .If ,then andthus .So .
36.
We
have
Problems
37. .
38. Since ,thegraphcrossestheaxisatthethreepointsshowninFigure6.10.Thetworegions
havethesamearea(bysymmetry).Sincethegraphisbelowtheaxisfor ,wehave
Area
Figure6.10
39. Theareawewant(theshadedareainFigure6.11)issymmetricaboutthe -axisandsoisgivenby
Area
Figure6.11
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342 ChapterSix/SOLUTIONS
40. Since from to and from to ,wehave
Area
41. (a) SeeFigure6.12.Since for and for ,wehave
Area
Figure6.12:Graphof
(b)
Calculating
gives
Thisintegralmeasuresthedifferencebetweentheareaabovethe -axisandtheareabelowthe -axis.Sincethe
definiteintegralisnegative,thegraphof liesmorebelowthe -axisthanaboveit.Sincethefunctioncrosses
theaxisat ,
whereas
Area
42. Sincetheareaunderthecurveis6,wehave
Thus and .
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SOLUTIONStoReviewProblemsforChapterSix 343
43. Thegraphof has -interceptsof .SeeFigure6.13.Sinceitissymmetricaboutthe -axis,wehave
Area
Wewanttheareatobe1,so
giving
Figure6.13
44. Thecurvesintersectat and .Atany -coordinatetheheightbetweenthetwocurvesis .
height
Thusthetotalareais
Another
approach
is
to
notice
that
the
area
between
the
two
curves
is
(area
A)
(area
B).
AreaB sincethefunctionisnegativeon
AreaA
Thustheareais .
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344 ChapterSix/SOLUTIONS
45. SeeFigure6.14.Theaveragevalueof isgivenby
Average
Averagevalue
Figure6.14
46. Thetotalamountofdischargeistheintegralofthedischargeratefrom to :
Totaldischarge
cubicmeters.
47. (a) Since ispositiveontheinterval andnegativeontheinterval ,thefunction is
increasingon anddecreasingon .Thus attainsitsmaximumat .Sincethearea
underthe -axisisgreaterthantheareaabovethe -axis,thefunction decreasesmorethanitincreases.Thus,the
minimumisat .
(b) Toestimatethevalueof at ,weseethattheareaunder between and isabout1box,which
hasarea5.Thus,
Themaximumvalueattainedbythefunctionis .
Theareabetween andthe -axisbetween and isabout3boxes,eachofwhichhasanareaof
5.Thus
Theminimumvalueattainedbythefunctionis .
(c) Usingpart(b),wehave .Alternately,wecanusetheFundamentalTheorem:
48. (a) Startingat ,wearegiventhat .Movingtotheleftontheinterval ,wehave ,
so
.Ontheinterval
,wehave
,so
Movingtotherightfrom ,weknowthat on .So .Ontheinterval
, so
Ontheinterval ,wehave ,so
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SOLUTIONStoReviewProblemsforChapterSix 345
(b) Inpart(a)wefoundthat and .
(c) Theintegral isgivenbythesum
Alternatively,knowing and andusingtheFundamentalTheoremofCalculus,wehave
49.
Point
ofinflection
50. InflectionpointLocalmax
LocalminInflection
point
51. representsthenetareabetween andthe -axisfrom to ,withareacountedasnegativefor
belowthe -axis.Aslongastheintegrandispositive isincreasing.Therefore,theglobalmaximumof
occursat andisgivenbythearea
At , .Figure6.15showsthatthearea islargerthanthearea .Thus for .
Thereforetheglobalminimumis .
Figure
6.15
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346 ChapterSix/SOLUTIONS
52. Since isthegraphofadecreasingfunction,thegraphofitsderivativeshouldfallbelowthe -axis.Thus, couldbe
and
couldbe
.Sincethegraphof isabovethe -axisandrepresentsadecreasingfunction,thefunction
shouldbeincreasingandconcavedown.Thus, couldbethegraphof .
53. Afunctionwhosederivativeis isoftheform
forsomevalueof
.
(a) Toensurethatthefunctiongoesthroughthepoint ,wetake and :
(b) Toensurethatthefunctiongoesthrough ,wetake and :
54. Weknowtheheightisgivenby
sothevelocityisgivenby
andtheaccelerationisgivenby
Theaccelerationduetogravityis ft/sec downward.Since ,theobjectwasthrownat ft/sec.Since
,theobjectwasthrownfromaheightof ft.
55. Thegraphof mustslopedownwardsmoststeeplywhen hasitsminimum.Thegraphof shouldhaveits
minimumabouttwo-thirdsofthewaythroughthetimeinterval(whenthegraphof intersectsthe -axis),andhave
itsfinalvalueabouthalf-waybetweenitsmaximumandminimumvalues.Apossiblegraphof isgiveninFigure6.16.
Theplacementofthehorizontalaxisbelowthegraphisarbitrary.
Figure6.16
56. Let bethevelocityand bethepositionoftheparticleattime .Weknowthat ,soaccelerationistheslope
ofthevelocitygraph.Similarly, velocityistheslopeofthepositiongraph.Graphsof
and
areshowninFigures6.17
and6.18,respectively.
Figure6.17:Velocityagainsttime Figure6.18:Positionagainsttime
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SOLUTIONStoReviewProblemsforChapterSix 347
57. (a) Since sec= min,
Angularacceleration revs/min
(b) Weknowangularaccelerationisthederivativeofangularvelocity.Since
Angularacceleration
wehave
Angularvelocity
Measuringtimefromthemomentatwhichtheangularvelocityis revs/min,wehave .Thus,
Angularvelocity
Thusthetotalnumberofrevolutionsperformedduringtheperiodfrom to minisgivenby
Numberof
revolutions revolutions
58. (a) Sincetherotorisslowingdownataconstantrate,
Angularacceleration revs/min
Unitsarerevolutionsperminuteperminute,orrevs/min .
(b) Todecreasefrom to revs/minatadecelerationof revs/min ,
Timeneeded min
(c) Weknowangularaccelerationisthederivativeofangularvelocity.Since
Angularacceleration revs/min
wehave
Angularvelocity
Measuringtimefromthemomentwhenangularvelocityis revs/min,weget .Thus
Angularvelocity
So,thetotalnumberofrevolutionsmadebetweenthetimetheangularspeedis revs/minandstoppingisgiven
by:
Numberofrevolutions (Angularvelocity)
revolutions
59. (a) Using ft/sec wehave
(sec)
0
1
2
3
4
5
(ft/sec) 80 48 16
(b) Theobjectreachesitshighestpointwhen whichappearstobeat seconds.Bysymmetry,theobject
shouldhitthegroundagainat seconds.
(c) Leftsum ft,whichisanoverestimate.
Rightsum ft,whichisanunderestimate.
Notethatweusedasmallerthirdrectangleofwidth toendoursumat .
(d) Wehave soantidifferentiationyields
But so
At , ft.,so100ft.isthehighestpoint.
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348 ChapterSix/SOLUTIONS
60. Thevelocityofthecardecreasesataconstantrate,sowecanwrite: .Integratingthisgives .
Theconstantofintegration isthevelocitywhen ,so mph ft/sec,and .Fromthis
equationwecanseethecarcomestorestattime .
Integratingtheexpressionforvelocityweget ,where istheinitialposition,so .We
canusefactthatthecarcomestorestattime aftertraveling200feet.Startwith
andsubstitute and :
ft/sec
61. (a) Inthebeginning,bothbirthanddeathratesaresmall;thisisconsistentwithaverysmallpopulation.Bothratesbegin
climbing,thebirthratefasterthanthedeathrate,whichisconsistentwithagrowingpopulation.Thebirthrateisthen
high,butitbeginstodecreaseasthepopulationincreases.
(b)
bacteria/hour
bacteria/hour
time(hours) time(hours)
Figure6.19:Differencebetween and isgreatestat
Thebacteriapopulationisgrowingmostquicklywhen ,therateofchangeofpopulation,ismaximal;
thathappenswhen isfarthestabove ,whichisatapointwheretheslopesofbothgraphsareequal.Thatpointis
hours.
(c) Totalnumberbornbytime istheareaunderthe graphfrom uptotime .SeeFigure6.20.
Totalnumberaliveattime
isthenumberbornminusthenumberthathavedied,whichistheareaunderthe
graphminustheareaunderthe graph,uptotime .SeeFigure6.21.
bacteria
bacteria
time(hours)
Figure
6.20:Numberbornbytime
is
time(hours)
Figure
6.21:
Number
alive
at
time
is
FromFigure6.21,weseethatthepopulationisatamaximumwhen ,thatis,afterabout hours.This
standstoreason,because istherateofchangeofpopulation,sopopulationismaximizedwhen ,
thatis,when .
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SOLUTIONStoReviewProblemsforChapterSix 349
62. (height)
(time)
Suppose isthetimetofilltheleftsidetothetopofthemiddleridge.Sincethecontainergetswiderasyougoup,
therate decreaseswithtime.Therefore,for ,graphisconcavedown.
At ,waterstartstospillovertorightsideandsodepthofleftsidedoesntchange.Ittakesaslongfortheright
sidetofilltotheridgeastheleftside,namely .Thusthegraphishorizontalfor .
For ,waterlevelisabovethecentralridge.Thegraphisclimbingbecausethedepthisincreasing,butata
slower
rate
than
for
because
the
container
is
wider.
The
graph
is
concave
down
because
width
is
increasing
with
depth.Time representsthetimewhencontainerisfull.
63. For[ ],theaccelerationisconstantandpositiveandthevelocityispositivesothedisplacementispositive.Thus,
theworkdoneispositive.
For[ ],theacceleration,andthereforetheforce,iszero.Therefore,theworkdoneiszero.
For[ ],theaccelerationisnegativeandthustheforceisnegative.Thevelocity,andthusthedisplacement,is
positive;thereforetheworkdoneisnegative.
For[ ],theacceleration(andthustheforce)andthevelocity(andthusthedisplacement)arenegative.Thus,the
workdoneispositive.
For[ ],theaccelerationandthustheforceisconstantandnegative.Velocitybothpositiveandnegative;total
displacement is .Sinceforceisconstant,workis .
CAS
Challenge
Problems
64. (a) Wehave and ,so,since ,
Riemannsum
(b) ACASgives
Takingthelimitas gives
(c) Theanswertopart(b)simplifiesto .Since ,theFundamentalTheoremofCalculussays
that
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350 ChapterSix/SOLUTIONS
65. (a) ACASgives
(b) Thethreeintegralsinpart(a)obeytherule
(c) Checkingtheformulabycalculatingthederivative
bytheconstantmultiplerule
bythechainrule
66. (a) ACASgives
(b) Thethreeintegralsinpart(a)obeytherule
(c) Checkingtheformulabycalculatingthederivative
bytheconstantmultiplerule
bythechainrule
67. (a) ACASgives
Althoughtheabsolutevaluesareneededintheanswer,someCASsmaynotincludethem.
(b) Thethreeintegralsinpart(a)obeytherule
(c) Checkingtheformulabycalculatingthederivative
bythesumandconstantmultiplerules
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CHECKYOURUNDERSTANDING 351
68. (a) ACASgives
Althoughtheabsolutevaluesareneededintheanswer,someCASsmaynotincludethem.
(b) Thethreeintegralsinpart(a)obeytherule
(c) Checkingtheformulabycalculatingthederivative
CHECKYOURUNDERSTANDING
1. True.Afunctioncanhaveonlyonederivative.
2. True.Checkbydifferentiating .
3. True.Anyantiderivativeof isobtainedbyaddingaconstantto .
4. True.Anyantiderivativeof isobtainedbyaddingaconstantto .
5. False.Differentiatingusingtheproductandchainrulesgives
6. False.Itisnottrueingeneralthat ,sothisstatementisfalseformanyfunctions .For
example,if ,then ,but .
7. True.Addingaconstanttoanantiderivativegivesanotherantiderivative.
8. True.If isanantiderivativeof ,then ,so .Therefore, isasolutionto
thisdifferential equation.
9. True.If isasolutiontothedifferentialequation ,then ,so isanantideriva-
tiveof .
10.
True.
If
acceleration
is
for
some
constant
,
,
then
we
have
Velocity
forsomeconstant .Weintegrateagaintofindpositionasafunctionoftime:
Position
forsomeconstant .Since ,thisisaquadraticpolynomial.
11. True,bytheSecondFundamentalTheoremofCalculus.
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352 ChapterSix/SOLUTIONS
12. True.Weseethat
13. False.If ispositivethen isincreasing,butif isnegativethen isdecreasing.
14. True.Since and arebothantiderivativesof ,theymustdifferbyaconstant.Infact,wecanseethattheconstant
isequalto since
15. False.Since and arebothantiderivativesof ,weknowthattheydifferbyaconstant,buttheyarenotnecessarily
equal.Forexample,if then but .
16. True,since .
PROJECTSFORCHAPTERSIX
1. (a) Ifthepoorest ofthepopulationhasexactly ofthegoods,then
(b) Anysuch isincreasing.Forexample,thepoorest50%ofthepopulationincludesthepoorest40%,and
sothepoorest50%mustownmorethanthepoorest40%.Thus ,andso,ingeneral, is
increasing.Inaddition,itisclearthat and .
Thegraphof isconcaveupbythefollowingargument.Consider .Thisisthe
fractionofresourcesthefifthpoorestpercentofthepopulationhas.Similarly, isthe
fractionofresourcesthatthetwentiethpoorestpercentofthepopulationhas.Sincethetwentiethpoorest
percentownsmorethanthefifthpoorestpercent,wehave
Moregenerally,wecanseethat
forany smallerthan andforanyincrement .Dividingthisinequalityby andtakingthelimit
as ,weget
So,thederivativeof isanincreasingfunction,i.e. isconcaveup.
(c) istwicetheshadedareabelowinthefollowingfigure. Iftheresourceisdistributedevenly,then is
zero.Thelarger is,themoreunevenlytheresourceisdistributed.Themaximumpossiblevalueof is
1.
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PROJECTSFORCHAPTERSIX 353
2. (a) InFigure6.22,theareaoftheshadedregionis .Thus, and,bytheFundamental
Theorem, .
(annual
yield)
(time
in
years)
Figure6.22
(b) Figure6.23isagraphof .Notethatthegraphof lookslikethegraphofaquadraticfunction.
Thus,
the
graph
of
looks
like
a
cubic.
(total
yield)
(time
in
years)
Figure6.23
(c)
We
have
(d) Ifthefunction takesonitsmaximumatsomepoint ,then .Since
differentiatingusingthequotientrulegives
so
.
Since
,
the
condition
for
a
maximum
may
be
written
as
oras
Toestimatethevalueof whichsatisfies , usethegraphof .Noticethat
istheareaunderthecurvefrom0to ,andthat istheareaofarectangleofbase and
height Thus,wewanttheareaunderthecurvetobeequaltotheareaoftherectangle,or
inFigure6.24.Thishappenswhen years.Inotherwords,theorchardshouldbecutdownafter
about50years.
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354 ChapterSix/SOLUTIONS
(annual
yield) Area
B
AreaA
(time
in
years)
Figure6.24