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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