[PDF] Antiderivatives




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[PDF] Drill problems on derivatives and antiderivatives - Arizona Math

Drill problems on derivatives and antiderivatives 1 Derivatives Find the derivative of each of the following functions (wherever it is defined):

[PDF] Antiderivatives

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[PDF] Antiderivatives

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19 oct 2011 · An antiderivative of a function f on an interval I is another function F such that F/(x) = f (x) for all x ? I Examples:

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[PDF] Antiderivatives 14228_2Lecture41.pdf

Antiderivatives

40.1Introduction

Sofarmuchofthe termhas beenspentfinding derivatives orratesof change.But insomecir cumstancesw ealreadyknowthe rateofchangeandwe wishtode- terminetheoriginalfunction.For example,metersor dataloggersoften measure ratesofchange, e.g.,milesper hourorkilo wattsper hour. Ify ouknowthevelocity ofanobject,canyou determinethe positionofthe object.Thiscould happenina car,sa y,wher ethe speedometerreadings were beingrecor ded.Canthepositionofthecarbe determinedfr omthisinfor mation? Similarly,canthepositionof anairplane bedetermined fromthe blackbox which recordstheairspeed?

Moregenerally,given f

! (x)canwe findthefunctionf(x).Ify outhinkabout it,thisis thesort ofquestionI have askedyou todoon labs,tests,and homework assignmentswhere Igavey outhegraph off ! (x)andsaiddra wthegraph off(x).

Orwhere Igaveyou thenumberline informationforf

! (x)andf !! (x)andasked youtoreconstructthe graphoff(x).Canw edo thissamethingifwe startwitha formulaforf ! (x)?Canw egetan explicitformulaforf(x)?We usuallystatethe problemthiswa y. DEFINITION40.1.Letf(x)beafunction definedonan interv alI.We saythatF(x)isan antiderivativeoff(x)onIif F ! (x)=f(x)forallx"I.

EXAMPLE40.1.Iff(x)=2x,thenF(x)=x

2 isanantideriv ative offbecause F ! (x)=2x=f(x).

Butsois G(x)=x

2 +1or, moregenerally,H(x)=x 2 +c. Arethere'other' antiderivativesoff(x)=2xbesidesthoseof theform H(x)= x 2 +c?We canusetheMVTtosho wthatthe answeris 'No.'Thepr oofwillr e- quirethreesmallsteps.

THEOREM40.1(Theorem1).IfF

! (x)=0for allxinan interval I,thenF(x)=kisaconstant function. Thismakes alotof sense:Ifthe velocityof anobjectis 0,thenits positionis constant(notchanging). Here's the Proof.ToshowthatF(x)isconstant,w emustsho wthatanytwooutput valuesof

Farethesame,i.e.,F(a)=F(b)forallaandbinI.

SopickanyaandbinI(withaF(b)#F(a) b#a =F ! (c)$F(b)#F(a)=F ! (c)[b#a]=0[b#a]=0.

Thismeans

F(b)=F(a),

inotherw ords, Fisconstant. THEOREM40.2(Theorem2).SupposethatF,Garedifferentiable ontheintervalIand F ! (x)=G ! (x)forallxinI.Thenther eexistksothat

G(x)=F(x)+k.

Proof.Considerthefunction G(x)#F(x)onI.Then

d dx (G(x)#F(x))=G ! (x)#F ! (x)=0.

Therefore,byTheorem1

G(x)#F(x)=k

so

G(x)=F(x)+k.

THEOREM40.3(Theorem3:Familiesof Antiderivatives) .IfF(x)andG(x)arebothantideriva- tivesoff(x)onaninter valI,thenG(x)=F(x)+k.

Thisisthe theorem wew anttoshow.

Proof.IfF(x)andG(x)arebothantiderivativ esoff(x)onaninter valIthen G ! (x)=f(x)andF ! (x)=f(x),thatis, F ! (x)=G ! (x)onI.Thenb yTheorem

2G(x)=F(x)+k.

DEFINITION40.2.IfF(x)isanyantideriv ative off(x),we saythatF(x)+cisthegeneral antiderivativeoff(x)onI.

Notationfor Antiderivatives

Antidifferentiationisalsocalled'indefiniteintegration.' ! f(x)dx=F(x)+c. • " istheintegration symbol •f(x)iscalledthe integrand •dxindicatesthe variableof integration •F(x)isaparticular antiderivative off(x) •andcistheconstant ofintegration. •Wereferto " f(x)dxasan 'antiderivative off(x)'oran 'indefiniteintegralof f.' math130,day41in troducti ontoantiderivatives3

Herearesev eralexamples.

!

2xdx=x

2 +c ! costdt=sint+c ! e z dz=e z +c ! 1 1+x 2 dx=arctanx+c

Antidifferentiationreversesdiffer entiationso

! F ! (x)dx=F(x)+c anddiffer entiationundoesantidifferentiation d dx # ! f(x)dx $ =f(x). Differentiationandantidifferentiationar ere verseprocesses,soeachderivative rule hasacorr espondingantidiffer entiationrule.

DifferentiationAntidifferentiation

d dx (c)=0 " 0dx=c d dx (kx)=k " kdx=kx+c d dx (x n )=nx n#1 " x n dx= x n+1 n+1 +c,n%=#1 d dx (ln|x|)= 1 x " x #1 dx= " 1 x dx=ln|x|+c d dx (sinx)=cosx " cosxdx=sinx+c d dx (cosx)=#sinx " sinxdx=#cosx+c d dx (tanx)=sec 2 x " sec 2 xdx=tanx+c d dx (secx)=secxtanx " secxtanxdx=secx+c d dx (e x )=e x " e x dx=e x +c d dx (arcsinx)= 1 & 1#x 2 " 1 & 1#x 2 dx=arcsinx+c d dx (arctanx)= 1 1+x 2 " 1 1+x 2 dx=arctanx+c

VariationsandGeneralizations

Noticewhathappens whenwe useaxinsteadof xinsomeof thesefunctions. We multiplyby awhentakingthe derivative, sowe havetodividebyawhentaking theantideriv ative.

DifferentiationAntidifferentiation

d dx (e ax )=ae ax " e ax dx= 1 a e ax +c d dx (sinax)=acosax " cosaxdx= 1 a sinx+c d dx (tanax)=asec 2 ax " sec 2 axdx= 1 a tanax+c d dx (arcsin( x a )= 1 & a 2 #x 2 " 1 & a 2 #x 2 dx=arcsin( x a )+c d dx (arctan( x a ))= a a 2 +x 2 " 1 a 2 +x 2 dx= 1 a arctan( x a )+c

Tryfillingintherulesfor

" sinaxdxand " sec(ax)tan(ax)dx. 4

EXAMPLE40.2.Hereareafe wexamples.

! cos(4x)dx= 1 4 sin(4x)+c ! e z/2 dz=2e z/22 +c ! 1 16+x 2 dx= 1 4 arctan x 4 +c

40.2Problems

1.Determineantiderivativesof thefollowingfunctions.Takethederiv ativeof youransw er

toconfirm thatyouarecorr ect.Whyshould youadd+ctoanyansw er?Basics: (a)7x 6 (b)x 6 (c)2x 6 (d)e x (e)2e x (f)2e 2x (g)e 2x (h) 1 x (i)# 8 x (j)cosx(k)4cos x(l)4cos (4x)(m)cos(4x)(n)# 1 x +4cosx (o)#2secxtanx(p)sin4 x(q)4sec 2 x(r)2x(s)x (t)8x(u)6 x ln6(v)6x#6 x ln6 (w)6(x)6+sec 2 x(y) 6 1+x 2 (z) 2 & 1#x 2

2.Nowtrythese.Think itthrough.

(a)x 8 +2(b)e 8x #c(c)#2sin4x(d)cos2 !#sec 2 2! (e)x 2 +3x#1(f) cosx 2 (g)x 5/2 (h)x #3/5 (i) 1 4 & x 7 (j) 4x 3 x 4 +12 (k)14x(x 2 +5) 6 (l)e x cos(e x )(m)12xe x 2 +9 (n)

12(lnx)

5 x

Answers

1.Answers

(a)x 7 +c(b) x 7 7 +c(c) 2 7 x 7 +c(d)e x +c(e)2e x +c (f)e 2x +c(g) 1 2 e 2x +c(h)ln|x|+c(i)#8ln|x|+c (j)sinx+c(k)4sin x+c(l)sin(4x)+c(m) 1 4 sin(4x)+c(n)#ln|x|+4sinx+c (o)#2secx+c(p)# 1 4 cos4x+c(q)4tan x+c(r)x 2 +c(s) 1 2 x 2 +c (t)4x 2 +c(u)6 x +c(v)3x 2 #6 x +c (w)6x+c(x)6x+tanx+c(y)6ar ctan(x)+c(z)2ar csin(x)+c

2.Answers.

(a) 1 9 x 9 +2x+c(b) 1 8 e 8x #cx+d(c) 1 2 cos4x+c(d) 1 2 sin2!# 1 2 tan2!+c (e) 1 3 x 3 + 3 2 x 2 #x+c(f) sinx 2 +c(g) 2 7 x 7/2 +c(h) 5 2 x 2/5 +c (i)# 4 3 x #3/4 (j)ln(x 4 +12)+c(k)(x 2 +5) 7 +c(l)sin(e x )+c (m)6e x 2 +9 +c(n)2(lnx) 6 +c

GeneralAntiderivativeRules

Thekey ideaisthateachderivativ erulecan bewrittenas anantiderivativ erule. We've seenhow thisworkswithspecificfunctions likesinxande x andnow weexaminehowthe generalderivativ erulescanbe'reversed.'

FACT41.1(SumRule).Thesumrule forderivativ essays

d dx (F(x)±G(x))= d dx (F(x))± d dx (G(x)).

Thecorresponding antiderivativeruleis

! (f(x)±g(x))dx= ! f(x)dx± ! g(x)dx. FACT41.2(ConstantMultipleRule) .Theconstantmultiple ruleforderiv ativessa ys d dx (cF(x))=c d dx (F(x)).

Thecorresponding antiderivativeruleis

! cf(x)dx=c ! f(x)dx.

Examples

! 8x 3 #7 & xdx= ! 8x 3 dx# ! 7x 1/2 dx=8 ! x 3 dx#7 ! x 1/2 dx= 8x 4 4 # 7x 3/2 3/2 +c=2x 4 # 14x 3/2 3 +c !

6cos2x#

7 x +2x #1/3 dx=6 ! cos2xdx#7 ! 1 x dx+2 ! x #1/3 dx=6· 1 2

·sin2x#7ln|x|+

2x 2/3 2/3 +c=3sin2x#7ln|x|+3x 2/3 +c. ! 3e x/2 # 8 & 9#x 2 dx=3 ! e x/2 dx#8 ! 1 & 9#x 2 dx=3·2e x/2 +8arcsin x 3 +c.=6e x/2 +8arcsin x 3 +c.

Rewriting

Rewritingtheintegrandcangr eatlysimplifythe antiderivative process. ! 2 5 & t 2 #6sec 2 tdt= ! 2t 2/5 #6sec 2 tdt= 2t 7/5 7/5 #6tant+c= 10t 7/5 7 #6tant+c. ! x 4 +2 x 2 dx= ! x 2 +2x #2 dx= x 3 3 + 2x #1 #1 +c= x 3 3 #2x #1 +c. ! 6x 2 (x 4 #1)dx= ! 6x 6 #6x 2 dx= 6x 7 7 + 6x 3 3 +c= 6x 7 7 +2x 3 +c. ! 2 3 & t 5 dt= ! 2t #5/3 dt= 2t #2/3 #2/3 +c=#3t #2/3 +c. ! 8x 2 +7 & x dx= ! 8x 3/2 +7x #1/2 dx= 8x 5/3 5/3 + 7x 1/2 1/2 +c= 24x
5/3 5 +14x 1/2 +c. 6

41.3Evaluating'c '(InitialValueProblems)

Sofarweha vebeen calculatinggeneralantiderivativesof functions.Whatthismeansis thatifw ekno wthevelocityv(t)ofthecar wear edrivingin, wecandeterminethe position p(t)ofthecar ,upto aconstantcifwe canfindanantiderivative forv(t).Ifw ehav emore information,thepositionof thecarat aparticulartime say, thenwe areable todetermine theprecise antiderivative.

EXAMPLE41.3.Supposethatf

! (x)=e x +2xandf(0)=3.Findf(x).f(0)is sometimescalledthe initialvalueandsuchquestions arer eferredto asinitialvalue problems.[How couldyouinterpret thisinformation intermsofmotion?]

SOLUTION.f(x)mustbean antiderivative off

! (x)so f(x)= ! f ! (x)dx= ! e x +2xdx=e x +x 2 +c.

Nowusetheinitialv aluetosolv eforc:

f(0)=e 0 +0 2 +c=3$1+c=3$c=2.

Therefore,f(x)=e

x +x 2 +2.

EXAMPLE41.4.Supposethatf

! (x)=6x 2 #2x 3 andf(1)=4.Findf(x).

SOLUTION.Againf(x)mustbean antiderivativ eoff

! (x)so f(x)= ! f ! (x)dx= ! 6x 2 #2x 3 dx=2x 3 # x 4 2 +c.

Nowusethe'initial'v aluetosolv efor c:

f(1)=2#1/2+c=3$c=2.5.

Therefore,f(x)=2x

3 # x 4 2 +2.5.

EXAMPLE41.5.Supposethatf

!! (t)=6t #2 (thinkacceleration)with f ! (1)=8(think velocity)andf(1)=3(thinkposition). Findf(t).

SOLUTION.Firstfind f

! (t)whichmustbe theantiderivativ eoff !! (t).So f ! (t)= ! f !! (t)dt= ! 6x #2 dt=#6t #1 +c.

Nowusethe'initial'v alueforf

! (t)tosolve forc: f ! (1)=#6(1)+c=8$c=14.

Therefore,f

! (t)=#6t #1 +14.Now wearebackto theearlierproblem. f(t)= ! f ! (t)dt= ! #6t #1 +14dt=#6ln|t|+14t+c.

Nowusethe'initial' valueof ftosolve forc:

f(1)=#6ln1+14(1)+c=3$6(0)+14+c=3$c=#11.

Sof(t)=6ln|t|+14t#11.

InClassPractice

EXAMPLE41.6.Findfgiventhatf

! (x)=6 & x+5x 3

2wheref(1)=10.

SOLUTION.f(x)mustbean antiderivative off

! (x)so f(x)= ! f ! (x)dx= ! 6 & x+5x 3 2 dx=4x 3/2 +2x 5/2 +c.

Usethe'initial' valueto solve forc:

f(1)=4+2+c=10$c=4.

Therefore,f(x)=4x

3/2 +2x 5/2 +4. math130,day41in troducti ontoantiderivatives7

EXAMPLE41.7.Findfgiventhatf

!! (!)=sin!+cos!wheref ! (0)=1andf(0)=2.

SOLUTION.Firstfind f

! (!)whichmustbe theantiderivativ eof f !! (!).So f ! (!)= ! f !! (!)d!= ! sin!+cos!=#cos!+sin!+c.

Nowusetheinitialv alueforf

! (!)tosolve forc: f ! (0)=#cos0+sin0+c=#1+0+c=1$c=2.

Therefore,f

! (!)=#cos!+sin!+2. f(!)= ! f ! (!)d!= ! #cos!+sin!+2d!=#sin!#cos!+2!+c.

Nowusetheinitialv alueofftosolve forc:

f(0)=#sin0#cos0+2(0)+c=0#1+c=2$c=3.

Sof(!)=#sin!#cos!+2!+3.

MotionProblems

41.4Introduction

Earlierinthe term weinter pretedthefirstandsecond derivativesasvelocity andaccel- erationinthe contextofmotion. So let'sapplythe initialvalue problemr esultstomotion problems.Recallthat •s(t)=positionat timet. •s ! (t)=v(t)velocityattimet. •s !! (t)=v ! (t)=a(t)accelerationat timet.

Therefore

• " a(t)dt=v(t)+c 1 velocity. • " v(t)dt=s(t)+c 2 positionattime t. Wewillneedtouse additionalinfor mationtoe valuatethe constantsc 1 andc 2 . EXAMPLE41.8.Supposethatthe accelerationof anobjectis givenb ya(t)=2#cost fort'0with •v(0)=1,this isalsodenoted v 0 •s(0)=3,thisis alsodenoteds 0 .

Finds(t).

SOLUTION.Firstfind v(t)whichisthe antiderivativ eofa(t). v(t)= ! a(t)dt= !

2#costdt=2t#sint+c

1 .

Nowusetheinitial valuefor v(t)tosolve forc

1 : v(0)=0#0+c 1 =1$c 1 =1. Therefore,v(t)=2t#sint+1.Now solvefors(t)bytakingtheantiderivativ eofv(t). s(t)= ! v(t)dt= !

2t#sint+1dt=t

2 +cost+t+c 2

Nowusetheinitialv alueofstosolve forc

2 : s(0)=0+cos0+c=3$1+c=3$c=2.

Sos(t)=t

2 +cost+2t+2. EXAMPLE41.9.Ifaccelerationis given bya(t)=10+3t#3t 2 ,findthe position functionifs(0)=1ands(2)=11.

SOLUTION.First

v(t)= ! a(t)dt= !

10+3t#3t

2 dt=10t+ 3 2 t 2 #t 3 +c. Now s(t)= ! 10t+ 3 2 t 2 #t 3 +cdt=5t 2 + 1 2 t 3 # 1 4 t 4 +ct+d. Buts(0)=0+0#0+0+d=1sod=1.Thens(2)=20+4#4+2c+1=11so

2c=#10$c=#5.Thus,s(t)=5t

2 + 1 2 t 3 # 1 4 t 4 #5t+1. 10 EXAMPLE41.10.Ifaccelerationis given bya(t)=sint+cost,findthe position functionifs(0)=1ands(2")=#1.

SOLUTION.First

v(t)= ! a(t)dt= ! sint+costdt=#sint+cost+c. Now s(t)= ! #sint+cost+cdt=#cost#sint+ct+d.

Buts(0)=#1+0#0+0+d=1sod=2.Thens(")=#1+0+"c+2=#1so

2"c=#2$c=#1/".Thus,s(t)=#cost#sint#

t " +2.

41.5ConstantAcceleration:Gravity

Inmanymotion problemsthe accelerationisconstant. Thishappenswhenanobjectis thrownordroppedandtheonly accelerationisdue togravity.Insuch asituation weha ve •a(t)=a,constantacceleration •withinitial velocityv(0)=v 0 •andinitialposition s(0)=s 0 . Then v(t)= ! a(t)dt= ! adt=at+c. But v(0)=a·0+c=v 0 $c=v 0 . So v(t)=at+v 0 . Next, s(t)= ! v(t)dt= ! at+v 0 dt= 1 2 at 2 +v 0 t+c.

Attimet=0,

s(0)= 1 2 a(0) 2 +v 0 (0)+c=s 0 $c=s 0 .

Therefore

s(t)= 1 2 at 2 +v 0 t+s 0 . EXAMPLE41.11.Supposeaball isthro wnwithinitial velocity96 ft/sfromarooftop

432feethigh.The accelerationdueto gravityis constanta(t)=#32ft/s

2 .Findv(t) ands(t).Thenfind themaximumheight oftheball andthetime whentheball hits theground.

SOLUTION.Recognizingthatv

0 =96ands 0 =432andthat theaccelerationis constant,w emayusethegeneral formulaswejustde veloped. v(t)=at+v 0 =#32t+96 and s(t)= 1 2 at 2 +v 0 t+s 0 =#16t 2 +96t+432.

Themaxheight occurswhenthe velocity is0:

v(t)=#32t+96=0$t=3$s(3)=#144+288+432=576.

Theballhits theground whens(t)=0.

s(t)=#16t 2 +96t+432=#16(t
2 #6t#27)=#16(t#9)(t+3)=0.

Sot=9and(t%=#3).

math130,day41intr oducti ontoantiderivatives11 YOUTRY IT41.1.Astoneis thrown upward withaninitialvelocityof48ft/sfrom theedge ofaclif f64ftabov eariver.(Rememberaccelerationdueto gravityis #32ft/s 2 .) (o)Findthe velocityof thestonefor t'0. (p)Findthe positionofthe stonefor t'0. (q)Findthe timewhen itreaches itshighestpoint (andtheheight). (r)Findthe timewhenthe stonehitsthe ground.Answers:v(t)=#32t+48,s(t)= #16t 2 +48t+64,maxht: 100ftat
t=1.5s,hitsgr oundatt=4s.EXAMPLE41.12.Apersondr opsastone fromabridge.Whatis theheight(in feet)of thebridgeif thepersonhears thesplash5secondsafterdr oppingit?

SOLUTION.Here'swhatwekno w.v

0 =0(dropped) ands(5)=0(hitsw ater).And weknowaccelerationis constant,a=#32ft/s 2 .We wanttofindtheheight ofthe bridge,whichis justs 0 .Useour constantacceleration motionformulas tosolve fora. v(t)=at+v 0 =#32t and s(t)= 1 2 at 2 +v 0 t+s 0 =#16t 2 +s 0 .

Nowweusethe positionweknow:s(5)=0.

s(5)=#16(5) 2 +s 0 $s 0 =400. Noticethatw edidnot needtousethevelocity function. YOUTRY IT41.2(ExtraCredit) .Inthepr evious problemdidyoutakeintoaccountthat sound doesnottra velinstantaneously inyourcalculationabove?Assume thatsoundtra velsat

1120ft/s.What istheheight (infeet)of thebridgeif thepersonhears thesplash5seconds

afterdropping it? EXAMPLE41.13.Here'savariation andthistime wewillusemetricunits. Supposea ballisthr own withunknowninitialvelocityv 0 ft/sfrom arooftop49meterhighand thepositionof theballat timet=3sis s(3)=0.Theacceleration duetogra vityis constanta(t)=#9,8m/s 2 .Findv(t)ands(t).

SOLUTION.Thistimev

0 isunknown buts 0 =49ands(3)=0.Againthe acceleration isconstantso wema yuse thegeneralformulasforthis situation. v(t)=at+v 0 =#9.8t+v 0 and s(t)= 1 2 at 2 +v 0 t+s 0 =#4.9t 2 +v 0 t+49.

Butwe knowthat

s(3)=#4.9(3) 2 +v 0 (3)+49=0 whichmeans 3v 0 =4.9(9)#4.9(10)=#4.9$v 0 =#4.9/3. So v(t)=#9.8t# 49
30
and s(t)=#4.9t 2 # 49
30
t+49.

Interpretv

0 =#4.9/3. EXAMPLE41.14.MoGreen isattemptingtorunthe 100mdashin theGenev aInvi- tationalTrack Meetin9.8seconds.Hew antstorun inaway thathisaccelerationis constant,a,ov ertheentirerace.Determinehisv elocityfunction?( awillstill appear asanunkno wnconstant.)Deter minehispositionfunction?There shouldbeno un- knownconstantsinyour equation.Whatis hisvelocity attheendofthe race?Doy ou thinkthis isrealistic? 12

SOLUTION.Wehave:constant acc=am/s

2 ;v 0 =0m/s;s 0 =0m.S o v(t)=at+v 0 =at and s(t)= 1 2 at 2 +v 0 t+s 0 = 1 2 at 2 .

Buts(9.8)=

1 2 a(9.8) 2 =100,soa= 200
(9.8) 2 =2.0825m/s 2 .Sos(t)=2.0825t 2 .Mo's velocityattheendof theraceis v(9.8)=a(9.8)=2.0825(9.8)=20.41m/s. ..not realistic. EXAMPLE41.15.Astonedr oppedoff acliffhitsthegr oundwithspeedof120ft/s.

Whatwas theheightofthecliff?

SOLUTION.Recognizingthatv

0 =0ands 0 isunknown andisthecliff heightand thattheacceleration isconstant a=#32ft/s,w emay usethegeneralformulasfor motionwithconstant acceleration: v(t)=at+v 0 =#32t+0=#32t. Nowweusethe velocityandtheonev elocitywe knowv=#120whenit hitsthe groundso v(t)=#32t=#120$t=120/32=15/4 whenithits theground. Now rememberwhen ithitsthegroundthe heightis0.So s(15/4)=0.Butw eknow s(t)= 1 2 at 2 +v 0 t+s 0 =#16t 2 +0t+s 0 =#16t 2 +s 0 .

Nowsubstituteint=15/4andsolvefors

0 . s(15/4)=0$#16(15/4) 2 +s 0 =0$s 0 =15 2 =225.

Thecliff heightis225feet.

EXAMPLE41.16.Acaris trav elingat90km/hwhenthe driversees adeer75mahead andslamson thebrakes. Whatconstantdeceleration isrequir edtoa voidhitting

Bambi?[Note: Firstconvert 90km/htom/s.]

SOLUTION.Let'slistall thatwe know. v

0 =90km/hor 9000

60·60

=25m/s.Let s 0 =0 andlettime t ( representthetimeittakesto stop.Thens(t ( )=75m.No wthe car isstoppedat timet ( ,sow ekno wv(t ( )=0.Finally wekno waccelerationisan unknownconstant,a,whichis whatwe wantto find. Nowweuseour constantaccelerationmotionformulasto solvefor a. v(t)=at+v 0 =at+25 and s(t)= 1 2 at 2 +v 0 t+s 0 = 1 2 at 2 +25t.

Nowweuseother velocityandpositionwe know:v(t

( )=0ands(t ( )=75when thecarstops. So v(t ( )=at ( +25=0$t
( =#25/a and s(t ( )= 1 2 a(t ( ) 2 +25t
( = 1 2 a(#25/a) 2 +25(#25/a)=75.

Simplifytoget

625a
2a 2 # 625
a = 625
2a # 1350
2a =# 625
2a =75$150a=#625 so a=# 625
150
=# 25
6 m/s.

Thetimeit takestostop is

t ( = #25 a = #25 # 25
6 =6. math130,day41intr oducti ontoantiderivatives13 EXAMPLE41.17.Onecarintends topassanother onaback road.What constantaccel- erationis required toincreasethespeedofacarfr om30mph(44ft/s)to50mph( 220
3 ft/s)in5seconds?

SOLUTION.Given:a(t)=aconstant.v

0 =44ft/s.s 0 =0.Andv(5)= 220
3 ft/s.Find a.Butv(t)=at+v 0 =at+44.So v(5)=5a+44= 220
3 $5a= 220
3 #44= 88
3 .Thus a= 88
15 .
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