Differentiation MIT

8/6/2019 Differentiation MIT

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18.01 Calculus Jason StarrFall 2005

Math 18.01 Lecture SummariesHomework. ThesearetheproblemsfromtheassignedProblemSetwhichcanbecompletedusingthematerialfromthatdateslecture.PracticeProblems. Practiceproblemsarenottobewrittenuporturnedin. Theseareassignedonly for practice, and are entirely voluntary. Problems listed as 1B 1, for example, are takenfromSectionEofthe18.01coursereader.

Lecture 1. Sept. 8 VelocityandderivativesLecture 2. Sept. 9 LimitsLecture 3. Sept. 13 RulesofdifferentiationLecture 4. Sept. 15 ThechainruleandimplicitdifferentiationLecture 5. Sept. 16 ThederivativesofexponentialandlogarithmfunctionsLecture 6. Sept. 20 ThederivativesoftrigonometricfunctionsLecture 7. Sept. 22 ReviewforExam1Lecture 8. Sept. 27 LinearandquadraticapproximationsLecture 9. Sept. 29 SketchingcurvesLecture 10. Sept. 30 Appliedmaximum/minimumproblemsLecture 11. Oct. 4 RelatedratesproblemsLecture 12. Oct. 6 NewtonsmethodLecture 13. Oct. 13 Antidifferentiation Lecture 14. Oct. 14 RiemannintegralsLecture 15. Oct. 18 TheFundamentalTheoremofCalculusLecture 16. Oct. 20 PropertiesoftheRiemannintegralLecture 17. Oct. 21 SeparableordinarydifferentialequationsLecture 18. Oct. 25 NumericalintegrationLecture 19. Oct. 28 ApplicationsofintegrationtovolumesLecture 20. Nov. 1 AveragesandvolumesbyshellsLecture 21. Nov. 3 ParametricequationcurvesandarclengthLecture 22. Nov. 4 AreaofasurfaceofrevolutionandpolarcoordinatecurvesLecture 23. Nov. 8 Tangentlines,arclengthandareasforpolarcurvesLecture 24. Nov. 15 InversetrigonometricfunctionsandhyperbolicfunctionsLecture 25. Nov. 17 InversehyperbolicfunctionsandinversesubstitutionLecture 26. Nov. 18 PartialfractiondecompositionLecture 27. Nov. 22 Integrationbyparts

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Lecture 28. Dec. 1 LHospitalsruleLecture 29. Dec. 2 ImproperintegralsLecture 30. Dec. 6 SequencesandseriesLecture 31. Dec. 8 PowerseriesandTaylorseriesLecture 32. Dec. 8 MoreTaylorseriesandreview

Lecture 1. September8,2005Homework. ProblemSet1PartI:(a)(e);PartII:Problems1and2.Practice Problems. CourseReader: 1B 1, 1B 2 Textbook: p. 68,Problems17and15.1. Velocity. Displacementiss(t). Increment fromt0 tot0 +tis,

s=

s(t0 +t)s(t0).

Averagevelocity fromt0 tot0 +tis,s s(t0 +t)s(t0)

vave = = .t t

Velocity,orinstantaneousvelocity,att0 is,v(t0)= lim vave = lim s(t0 +t)s(t0).

t 0 t 0 t Thisisaderivative,v(t)equalss(t)=ds/dt. Thederivativeofvelocityisacceleration,

a(t0)=v(t0)= lim v(t0 +t)v(t0).t 0 t

Example. Fors(t)=5t2 +20t,firstcomputedvelocityatt=1is,v(1)= lim 105t= 10.

t 0

Thencomputedvelocityatt=t0 is,v(t0)= lim

010t0 +105t= 10t0 +20.tFinally,computedaccelerationatt=t0 is,

a(t0)= lim010= 10.t

2. Derivative. Let y = f(x) be a dependent variable depending on an independent variable x,varyingfreely. Theincrement ofy fromx0 tox0 +xis,

y=f(x0 +x)f(x0).2


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Thedifferencequotient oraveragerate of change ofy fromx0 tox0 +xis,y f(x0 +x)f(x0)= .x x

Thederivative ofy(orf(x))withrespecttoxatx0 is,y f(x0 +x)f(x0)

lim = lim .x 0 x x 0 x

3. Examples in science and math.(i) Economics. Marginal cost is the derivative of cost with respect to some other variable, for

instance,thequantitypurchased.(ii) Thermodynamics. The idealgas lawrelatingpressurep,volumeV,andtemperatureT ofa

gasis,pV =nRT.

Underisothermalconditions,T isaconstantT0 sothat,p(V)= 0

V .nRTUnder adiabatic conditions (i.e., no transfer of heat),pV is a constant K. Using this toeliminatepgives,

T(V)= KnR 1V1.Asthisillustrates,theindependentvariable,dependentvariableandconstantsinanequationverymuchdependontheproblemtobesolved.

(iii) Biology. ExponentialpopulationgrowthmodelsthepopulationN(t)aftertyearsas,N(t)=N0ert,

whereex istheexponentialfunction,N0 isinitialpopulation,andrisagrowthfactor. Laterwewillsee,N(t)=rN(t),i.e.,thepopulationgrowsatarateproportionaltothesizeofthepopulation.

(iv) Geometry. Thevolumeofarightcircularconeis,1

V = Ah.3

whereAisthebaseareaoftheconeandhistheheightofthecone. Theradiusrofthebaseisproportionaltotheheight,

r(h)=ch,3


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forsomeconstantc. SinceA=r2,thisgives,

V(h)= c2h3.3

Thederivativeis,dV

=c2h2 =r2 =dh A.

This isveryreasonable. Insomesense,thisexplainstheclassical formula forthevolumeofacone.Lecture 2. September9,2005Homework. ProblemSet1PartI:(f)(h);PartII:Problems3.Practice Problems. CourseReader: 1C 2, 1C 3, 1C 4, 1D 3, 1D 5. 1. Tangent lines to graphs. For y = f(x), the equation of the secant line through(x0,f(x0))and(x0 +x,f(x0 +x))is,

y= f(x0 +x)f(x0)(xx0)+f(x0).x

Inthelimit,theequationofthetangent line through(x0,f(x0))is,y=f(x0)(xx0)+y0.

Example. Fortheparabolay=x2,thederivativeis,y(x0)= 2x0.

Theequationofthetangentlineis,y=2x0(xx0)= 2x0xx20.

Forinstance,theequationofthetangentlinethrough(2,4)is,y= 4x

4.

2Givenapoint(x,y),whatareallpoints(x0,x0)ontheparabolawhosetangentlinecontains(x,y)? To solve, consider x and y as constants and solve for x0. For instance, if (x,y) =(1,3),thisgives,

2(3)=2x0(1)x0,or,

2x02x03=0.4


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Factoring(x03)(x0+1),thesolutionsarex0 equals1andx0 equals3. Thecorrespondingtangentlinesare,

y=2x1,and

y=6x9.Forgeneral(x,y),thesolutionsare,

x0 = xx2y.2. Limits. Precisedefinitionisonp. 791ofAppendixA.2. Intuitivedefinition: limx f(x)x0equals L if and only if all values of f(x) can be made arbitrarily close to L by choosing xsufficientlyclosetox0. Oneinterpretationisthemicroscope/laserilluminatoranalogy: Anobserver focuses a microscopes field of view on a thin strip parallel to the x axis centeredon y = L. The goal of the illuminator is to focus a laser beam centered on x0 parallel tothe yaxis (but with the line x = x0 deleted) so that only the portion of the graph in thefield of view isilluminated. Ifforeverymagnificationofthemicroscope,theilluminatorcansucceed,thenthelimitisdefinedandequalsL.There is a beautiful Java applet on the webpage of Daniel J. Heath of Pacific LutheranUniversity,

http://www.plu.edu/~heathdj/java/calc1/Epsilon.html If

you

use

this,

try

a

=

1.Forleft hand limits,usealaserthatilluminatesonlytotheleftofx0. Forright hand limits,usealaserthatilluminatesonlytotherightofx0.3. Continuity. A function f(x) is continuous at x0 if f(x0) is defined, limx f(x) isx0defined,andlimx f(x)equalsf(x0). Also,f(x)iscontinuousonanintervalifitiscontin x0uousateverypointofthe interval. Thetypesofdiscontinuityare: removablediscontinuity,

jumpdiscontinuity,infinitediscontinuityandessentialdiscontinuity.Lecture 3. September13,2005Homework. ProblemSet1PartI:(i)and(j).Practice Problems. CourseReader: 1E 1, 1E 3, 1E 5. 1. Another derivative. Usethe3 step methodtocomputethederivativeoff(x)=1/3x+1is,

f(x x 3/2/2.)= 3(3 +1)Upshot: Computingderivativesbythedefinitionistoomuchworktobepractical. Weneedgeneralmethodstosimplifycomputations.

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2. The binomial theorem. Forapositiveintegern,thefactorial,n!=n (n 1) (n 2) 3 2 1,

isthenumberofwaysofarrangingndistinctobjects ina line. Fortwopositive integersnandk,thebinomialcoefficient,

n n! n(n 1) (n k+2)(n k+1),= =

k k!(n k)! 13 2k(k 1) isthenumberofwaystochooseasubsetofk elements fromacollectionofnelements. Afunda mentalfactaboutbinomialcoefficientsisthefollowing,

n n n +1+ = .

k kk 1This is known as Pascals formula. This link is to a webpage produced by MathWorld, part ofWolframResearch.The Binomial Theorem saysthat foreverypositive integernandeverypairofnumbersaandb,(a +b)n equals,

nn na +nan1b + +k

ankbk + +nabn1 +b .This is proved by mathematical induction. First, the result isveryeasy when n=1; itjustsaysthat(a +b)1 equalsa1 +b1. Next,maketheinduction hypothesis thatthetheoremistruefortheinteger

n.

The

goal

is

to

deduce

the

theorem

for

n +

1,

(a +b)n+1 n+1 n +1 n+1kbk=a +(n +1)anb + +

ka + +(n +1)abn +bn+1.

Bythedefinitionofthe(n +1)st powerofanumber,(a +b)n+1 =(a +b) (a +b)n.

Bytheinductionhypothesis,thesecondfactorcanbereplaced,nn(a +b)(a +b)n =(a +b) a +

+ ankbk +

+bn .

kMultiplyingeachterminthesecondfactorfirstbyaandthenbybgives,

n an+1kbk + n ankbk+1 nan+1 + nanb + . . . + k + . . . + abk+1n an+1kbk + n ankbk+1 n + bn+1+ anb + . . . + + . . . + nab

kk1Summingincolumnsgives,

n n n nan+1 + (n +1)anb + . . . + ( k + k1 )an+1kbk + ( k+1 + )ankbk+1 + . . . + (1k6


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Onewaytodeducethisformulaistosetq(x)=f(x)/g(x)sothatf(x)=q(x)g(x),andtheapplytheLeibnizformulatoget,

f(x)=q(x)g(x)+q(x)g(x)=q(x)g(x)+f(x)g(x)/g(x).Solvingforq(x)gives,

q(x)= [f(x)f(x)g(x)/g(x)]/g(x)= [f(x)g(x)f(x)g(x)]/g(x)2.6. Another proof that d(xn)/dx equals nxn1. This was mentioned only very briefly. Theproductrulealsogivesanother inductionproofthat foreverypositive integern,d(xn)/dxequalsnxn1. Forn=1,weprovedthisbyhand. Letnbesomespecificpositive integer,andmaketheinductionhypothesisthatd(xn)/dxequalsnxn1. Thegoalistodeducetheformulaforn+1,

d(xn+1)=(n+1)xn.

dxBytheLeibnizrule,

n+1) d(xxn)d(x d(x) d(xn) d(xn)n= = x +x =(1)xn +x .dx dx dx dx dx

Bytheinductionhypothesis,thesecondtermcanbereplaced,n n nd(x

n+1

) =x +x(nxn1)=x +nxn =(n+1)x .dxThusthe formula forn impliesthe formula forn+1. Therefore,bymathematical induction,theformulaholdsforeverypositiveintegern.Lecture 4. September15,2005Homework. Nonewproblems.Practice Problems. CourseReader: 1F 1, 1F 6, 1F 7, 1F 8. 1. Product rule example. For u =3x+1, what is u(x)? Since uu = 3x+1, (uu) =(3x+1) =3. Bytheproductrule,(u

u) =u u =2uu. Thussolving,

u+u

u(x)=3/(2u)= x 1/2/2.3(3 +1)2. The derivative of un. Fromabove,(u2) equals2uu. Byasimilarcomputation,(u3) equals3u2u. Thissuggestsapattern,

d(un) n1du=nu .dx dx

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Thiscanbeprovedby inductiononn. Forn= 1,2and3, itwaschecked. Letnbeaparticularinteger(forinstance,70119209472933054321). Forthatinteger,supposetheresultisknown,

d(un) n1du=nu .dx dx

Thegoalistoprovetheresultforn+1,thatis,n+1)d(u du

=(n+1)un .dx dx

Letv=un. Thenun+1 equalsuv. So,bytheproductrule,n+1)d(u d(uv) du dv

= = v+u .dx dx dx dx

Plugginginv=un,thisis,n+1)d(u du d(un)

=dx(un)+u .dx dx

Bytheinductionhypothesis,d(un)/dxequalsnun1(du/dx). Pluggingin,dud(un+1) du

)+u(nun1 ).dx = dx(un dx

Thissimplfiesto,n+1)d(u du du dun

=u

+

nu

n=

(n

+

1)u

n.

dx dx dx dxThus,theresultforn+1followsfromtheresultforn. Byinduction,theresultholdsforeveryn.

a n3. The derivative of xa, a a fraction. Letabea fractionm/n and letu(x) bex . Thenuequalsxm. Thus,

d(un) d(xm)= ,

dx dxwhichequalsmxm1. Bytheabove,d(un)/dxequalsnun1(du/dx). Thus,

dun1 m1nu =mx .dx

Solvingfordu/dx,du mxm1 mxm1

= = .n(xm/n)n1dx nun1

b)cOne of the basic rules of exponents is that (a equals abc. Thus the denominator n(xm/n)n1mm/nequalsnxm/n(n1),whichequalsnx . Thus,

du mxm1 m m1 m/nm= = x x .nxmm/ndx n

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cAnotherbasicruleofexponentsisthatab a equalsab+c. Thus,du m (m1)+(m/nm) m m/n1= x = x .dx n n

Rememberingthatm/nisjusta,andu(x)isxa,thisfinallygives,d(xa) a1=ax .

dx4. The chain rule. Lety bea functionofx, y =f(x), and letubea functionofy, u=g(y).Thenu isa functionofx, u=g(f(x)). This function isacomposite function, and isdenotedby,

(gf)(x)=g(f(x)).

Whatisthederivativeofacompositefunction? Theclaimisthat,(gf)(x)=g(f(x))f(x).

Thisisofteneasiertorememberintheform,du du dy

= .dx dy dx

Thisalsosuggeststheproof,u u y(gf)(x0)= lim = limx 0 x x 0 y x,

wherey0 equalsf(x0),u0 equalsg(y0)=g(f(x0)),yequalsf(x0+x)f(x0)=f(x0+x)y0,anduequalsg(y0+y)g(y0)=g(f(x0+x))g(f(x0)). Solongasyisnonzero,thefractioninthe limitisdefined. And,asxapproaches0,alsoy approaches0. Thusthelimitbreaksupas,

u y(gf)(x0)= lim lim

y0 y x 0 x =g(y0)f(x0).Thus(gf)(x0)equalsg(f(x0))f(x0).Example.

Let

y(x)

equals

1+

x

2

,and

let

u(y)

equal

1/y

=

y

1.

Theny(x)

=

0+

2x

= 2x

and

u(y)=y2. Thus,bythechainrule,

d 1=

21

(2x)=dx 1+x2 y 2x(1+x2)2 .

5. Implicit differentiation. Thismethodhasalreadybeenusedmanytimes. Givenafunctiony(x)satisfyingsomeequationinvolvingbothxandy,formallydifferentiateeachsideoftheequationwithrespecttoxandthentrytosolvefory.

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Lecture 5. September16,2005Homework.

Problem

Set

2Part

I:

(a)(e);

Part

II:

Problem

2.

Practice Problems. CourseReader: 1I 1, 1I 4, 1I 5 1. Example of implicit differentiation. Let y = f(x) be the unique function satisfying theequation,

1 1+ =2.

x yWhatisslopeofthetangentlinetothegraphofy=f(x)atthepoint(x,y)=(1,1)?Implicitlydifferentiateeachsideoftheequationtoget,

d 1 d 1 d(2)+ = =0.

dx x dx y dxOf course (1/x) = (x1) =x2. And by the rule d(un)/dx= nun1(du/dx), the derivative of1/y isy2(dy/dx). Thus,

y2dyx2 =0.dx

Plugginginxequals1andyequals1gives,11y(1)=0,

whosesolutionis,y(1)

=

Infact,usingthat1/yequals21/x,thiscanbesolvedforeveryx,1

.

dy 1 1 1= .

dx =(x2)/(y2)= x2 (21/x)2 (2x1)22. Rules for exponentials and logarithms. Letabeapositiverealnumber. Thebasicrulesofexponentialsareasfollows.

cRule1. Ifab equalsB anda equalsC,thenab+c equalsBC,i.e.,b+c b c

a

=a

a .

Rule2. Ifab equalsB andBd equalsD,thenabd equalsD,i.e.,b)d bd(a =a .

Ifab equalsB,the logarithmwithbaseaofB isdefinedtobeb. Thisiswrittenloga(B)=b. ThefunctionBloga(B)isdefinedforallpositiverealnumbersB. Usingthisdefinition,therulesofexponentiationbecomerulesoflogarithms.

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Rule1. Ifloga(B)equalsbandloga(C)equalsc,thenloga(BC)equalsb+c,i.e.,loga(BC)=loga(B)+loga(C).

Rule2. Ifloga(B)equalsbandBd equalsD,thenloga(D)equalsdloga(B),i.e.,loga(B

d)=dloga(B).Rule3. SincelogB(D)equalsd,anequivalentformulationisloga(D)equalsloga(B)logB(D),i.e.,

loga(D)=loga(B)logB(D).3. The derivative of ax. Letabeapositiverealnumber. Whatisthederivativeofax? Denotethederivativeofax atx=0byL(a). Itequalsthevalueofthelimit,

hL(a)= lima 1.

h 0 hThenforeveryx0,thederivativeofax atx0 equals,

ax0+hax0lim .h 0 h

x0ahByRule1,ax0+h equalsa . Thusthelimitfactorsas,a

x0a

hlim ax0 =ax0 limah1h.h 0 h h 0

Therefore,foreveryx,thederivativeofax is,d(ax)

=L(a)ax.dx

WhatisL(a)? Tofigurethisout,considerhowL(a)changesasachanges. Firstofall,L(ab)=lim(ab)h1.

h 0 hByRule2,(ab)h equalsabh. Sothelimitis,

bh aL(ab)= lima 1 =blim bh1.

h 0 h h 0 bh Now,insidethelimit,makethesubstitutionthatkequalsbh. Ashapproaches0,alsokapproaches0. Sothelimitis,

kaL(ab)=blim 1 =bL(a).

k 0 k12


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ThisisverysimilartoRule2forlogarithms.Choose

anumber

a0

bigger

than

1,

say

a0

=

2.

Then

for

every

positive

real

number

a,

a

=

ba0whereb=loga0(a). Thus,

L(a)=L(a0b)=bL(a0)=L(a0)loga0(a).So,witha0 fixedandaallowedtovary,L(a)isjustthelogarithmfunctionloga0(a)scaledbyL(a0).Lookingatthegraphof(a0)x, it isgeometricallyclearthatL(a0) ispositive(thoughwehavenotproved that L(a0) is even defined). Thus the graph of L(a) looks qualitatively like the graph ofloga0(a). In particular, for a less than 1, L(a) is negative. The value L(1) equals 0. And L(a)approaches + and a increases. Therefore, there must be a number where L takes the value 1.Bylongtradition,thisnumberiscallede;

hL(e)=lime 1 =1.

h 0 hThis is the definition of e. Itshedsverylittlelightonthedecimalvalueofe.Becausee isso important,the logarithmwithbasee isgivenaspecialname: thenatural loga rithm. Itisdenoteby,

ln(a)=loge(a).So,finally,L(a)equals,

L(a)=loge(a)L(e)=ln(a)(1)=ln(a).xThisleadstotheformulaforthederivativeofa ,

d(ax)=

dx ln(a)ax.Inparticular,

d(ex) x=e .dx

Infact,ex ischaracterizedbythepropertyaboveandthepropertythate0 equals1.4. The derivative of loga(x) and the value of e. Bythechainrule,

d(au) du=

ln(a)a

u.

dx dxForu=loga(x),au equalsx. Thus,

d(au) d(x)= =1.

dx dxThus,

duln(a)au =1.

dx13


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Solvinggives,dloga(x) 1 1

= =

dx ln(a)auInparticular,fora=e,thisgives,

1/(ln(a)x).

dln(x)=

dx 1/x.Whatisthederivativeofln(x)atx=1? Ontheonehand,sincethederivativeofln(x)equals1/x,thederivativeatx=1is1/1=1. Ontheotherhand,thedefinitionofthederivativegives,

ln(1+h) ln(1)lim .h 0 h

Ofcourse,ln(1)equals0,sothissimplifiesto,1

lim ln(1+h).h 0 h

UsingRule2forlogarithms,thisgives,limln((1+h)1/h).h 0

Sinceln(y)iscontinuous,thelimitequals,ln[lim(1+h)1/h].

h 0

So the natural logarithm of the inner limit equals 1. But e is the unique number whose naturallogarithmequals1. Thisleadstotheformula,

e=lim(1+h)1/h.h 0

Makingthesubstitutionn=1/hleadstothemorefamiliarform,lim (1+1/n)n = e.

n+

This

can

be

used

to

compute

e

to

arbitrary

accuracy.

The

first

few

digits

of

e

are

2.718281828459045...5. Logarithmic differentiation. There is a method of computing derivatives of products offunctionsthatisoftenuseful. Ifyisaproductofnfactors,sayf1(x) fn(x),thederivativef2(x) of y can be computed by the product rule. However, it seems to be a fact that multiplication ismoreerror prone thanaddition. Thusintroduce,

u=ln(y)=ln(f1(x))+ln(f2(x))+ +ln(fn(x)).14


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Thederivativeofuis,du d d

=

(ln(f1(x)))+

+

(ln(fn(x))).

dx dx dx

Usingthechainrule,thisis,du f1(x) fn(x)= + .dx f1(x) + fn(x)

Thus,farfewermultiplicationsareneededtocomputeu. Thisisgood,becausealso,du dln(y) 1dy

= = .dx dx y dx

Thereforethederivativeofycanbecomputedas,f1(x) fn(x)y =yu fn(x)) + .=(f1(x) f1(x)

+ fn(x)

Example. Letybe,(1+x3)(1+x)

.x3/7

Then,33u=ln(y)=ln(1+x )+ln(1+x) ln(x).7

Bythechainrule,ln(1+x3) =3x2/(1+x3)andln(1+x) =(x)/(1+x)=(1/2x1/2)/(1+x).Thus,u equals, 3x2 1 3

.u =(1+x3) + 2x(1+x)7x

So,finally,3x2(1+x3)(1+x) 1 3

.x3/7

y =yu =(1+x3) + 2x(1+x)7x

Lecture 6. September20,2005Homework. ProblemSet2PartI:(f)(j);PartII:Problems1,3and4.Practice Problems. CourseReader: 1J 1, 1J 2, 1J 3, 1J 4 1. Trigonometric functions. What is angle? For a sector of a unit circle (a circle of radius1),theangle ofthesectorequalsboththe lengthofthearcofthesectorand1/2theareaofthesector. Althoughwehaveasyetgeneraldefinitions ofneitherarclengthnorarea,thiscanbeusedtogivearigorousdefinitionofangle. Wecan divideanysector intwoequalpieces: simplybisectthe chord of the sector. We also know how to add two angles, by laying the sectors in adjacentpositions. Denotingtheareaofaunitcirclebythesymbol(whichhappenstobethefamiliar),these2operationsproduceeveryangleoftheformm/2n,withmandnintegers. Everyanglecan

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beapproximatedarbitrarilywellbysuchangles. Thus, foreverycontinuous functionofanangle,everyvalueofthefunctioncanbecomputed.Thebasic functionsaresin(),cos(),tan(),sec(),csc()andcot(). Fulldescriptionsoftheseare in 9.1 of the textbook by Simmons. The same information is contained in the webpage onTrigonometryproducedbyMathWorld,partofWolframResearch.2. Trigonometric identities. For today, the most important identities are the angle additionformulas,

sin(+)=cos(+)=

Otherimportantidentitiesare,

sin( ),cos( )sin( ).

)cos( )+cos( )sin()cos( )sin(

(i) cos()equalscos(),i.e.,cos()isanevenfunction,(ii) sin()equalssin(),i.e.,sin()isanoddfunction,

(iii) sin(+/2)equalscos(),(iv) cos(+/2)equalssin(),and(v) sin2()+cos2()equals1forevery.

3. Sometrigonometriclimits. Incomputingtrigonometriclimits,thefollowinglimitiscrucial,sin()

lim = 0 1.

Asexplainedinclass,thisisessentiallythestatementthatas0,thequotientofthearclengthbythechord lengthtendsto1. Thiswasnotproved in lecture,nor is itproved inyourtextbookin2.1(despitetheauthorsclaim). However,itisgeometricallyreasonable. And,ofcourse,itcanbeproved.Thislimitimpliesanotherlimit,

cos()1lim = 0 0.

Toseethis,rewritethetermas,cos()1cos()+1 cos2()1

= . cos()+1 (cos()+1)

ByIdentity(v),cos2()1equalssin2(),sothetermequals,sin2() sin() 1

sin().=(cos()+1) cos()+1

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As0,thislimittendsto,

(1)

(1/2)

0=0.Byasimilarcomputation,

cos()1 1= .

0 2 2lim4. Derivatives of sin(x) and cos(x). Tocomputethederivativeofy=sin(x)atx=a,usetheangleadditionformulastowrite,

sin(a+h)=sin(a)cos(h)+cos(a)sin(h).Thisgives,

sin(a+h)

sin(a)=sin(a)(cos(h)

1)+cos(a)sin(h).Thusthedifferencequotientequals,

sin(a+h)sin(a) sin(h)=sin(a)cos(h)1+cos(a) .

h h hTakingthelimitgives,

sin(a+h)sin(a) cos(h)1 sin(h).

hlimh0 h =sin( )lima h0 h +cos( )lima h0Usingthelimitsfromabove,thisgives,

sin(a)=sin(a)0+cos(a)1=cos(a).Thusthederivativeofsin(x)equals,

dsin(x)=

dxAnentirelysimilarcomputationgives,

cos(x).

cos(a+h)cos(a)=cos(a) cos(h)1sin(a) sin(h),

h h hwhichleadsto,

cos(h)1 sin(h)( )=cos( )lima a

h 0=cos(a)0sin(a)1.cos

h hThusthederivativeofcos(x)equals,

dcos(x)=

dx

sin(ah0

sin(x).

)lim

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5. Derivatives of other trigonometric functions. Usingthequotientrule,dtan(x)

dx = 1cos2(x)(cos(x)cos(x)sin(x)(sin(x)))= cos2(x)+sin2(x)cos2(x) = 1cos2(x).Therefore,thederivativeoftan(x)equals,

dtan(x)dx = sec2(x).

Inasimilarmanner,dcot(x)

dx = csc2(x),d

sec(x)

dx = sec(x)tan(x),

anddcsc(x)

dx = csc(x)cot(x).Lecture 7. September22,2005ReviewforExam1. Nonewmaterialwaspresented. Therewerenopracticeproblemsfromthecoursereader.Lecture 8. September27,2005Homework. ProblemSet2allofPartIandPartII.Practice Problems. CourseReader: 2A 1, 2A 4, 2A 9, 2A 11, 2A 12. 1. Linear approximations. For a differentiable function f(x), the linear approximation orlinearization off(x)atx=aisthelinearfunction,

f(a)+f(a)(xa).Inaprecise sense, this is the bestapproximation of f(x) bya linear function near x=a. Forxclosetoa,thevalueoff(x)isclosetothevalueofthelinearization. Thenotationforthisis,

f(x)

f(a)

+

f(a)(x

a)

for

x

a.

Example. Thelinearizationof,

f(x)=e3xsin(2x)+5e3xcos(2x),nearx=0is,

f(x)5(152)xforx0.

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Inparticular,forx=0.02,thisgivestheapproximateanswer,f(0.02)5(152)(0.02) 4.8.

Theactualvalueisapproximately4.71.2. Basicapproximations. Somelinearapproximationsoccursooften,theyshouldbecommittedtomemory. Eachofthefollowingisthelinearapproximationforx0,togetherwiththetermsinthequadraticandhigherapproximations.

1 x1+ +x2 +x3 +. . . ,1x 1+rx r r(1+x)r + x2 + x3 +. . . ,

2 3

sin(x) x x3/3!+x5/5!+. . . ,cos(x) 1 x2/2!+x4/4!+. . . ,

x x1x1+ +x2/2!+x3/3!+. . . ,e

ln(1+x) +x2/2x3/3+. . .3. Combining basic approximations. Thebasicapproximationscanbecombinedtogetnewlinearapproximations.(i)Thelinearapproximationoff(x)forxacanbeconvertedtoalinearapproximationat0bysettingg(u)=f(a +u). Insymbols,

f(a f(a)(x a g g u.)+ )= (0)+ (0)Thisisequivalenttotheformula,

d df(f(x a))= (x a).

dx dx

(ii)Thelinearapproximationoff(cx)forxaisobtainedfromthelinearapproximationoff(u)forucabysubstitutingu=cx,f(cx)f(ca f(ca)(cx ca).)+

Thisisequivalenttotheformula,d df

(f(cx))=c (x).dx dx

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Thisissometimesusefulforboundingf(b)f(a),ifaboundonthederivativeoff(x)isknown.Lecture

9.

September

29,

2005

Homework. ProblemSet2allofPartIandPartII.Practice Problems. CourseReader: 2B 1, 2B 2, 2B 4, 2B 5. 1. Application of the Mean Value Theorem. A real world application of the Mean ValueTheorem is error analysis. Adeviceacceptsan inputsignalx andreturnsanoutputsignaly. Iftheinputsignalisalwaysintherange1/2x 1/2andiftheoutputsignalis,

1y =f(x)=

1+x +x2 +x3,

whatprecision

of

the

input

signal

x is

required

to

get

aprecision

of

103 fortheoutputsignal?Ifthe ideal inputsignal isx =a,and iftheprecision ish,thentheactual inputsignal is intherangea h x a +h. Theprecisionoftheoutputsignal is f(x)f(a) . BytheMeanValue| |Theorem,

f(x)f(a)=f(c),

x aforsomec betweena andx. Thederivativef(x)is,

f(x)= (3x2 +2x +1) .(1+x +x2 +x3)2

For1/2x 1/2,thisisboundedby,3(1/2)2 +2(1/2)+1|f(x) =7.04.|

[1+(1/2)+(1/2)2 +(1/2)3]2ThustheMeanValueTheoremgives,

f(x)f(a) = f(c 7.04 x a 7.04h.| | | )||x a| | |Thereforeaprecisionfortheinputsignalof,

h = 103/7.04 104guaranteesaprecisionof103 fortheoutputsignal.2. Firstderivativetest. Afunctionf(x)isincreasing,respectivelydecreasing,iff(a)islessthanf(b), resp. greater than f(b), whenever a is less than b. Insymbols, f is increasing, respectivelydecreasing,if

f(a)< f(b)whenevera < b, resp. f(a)> f(b)whenevera < b.22


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extremevalue,eitherpositiveornegative. Apointwheref achievesitsmaximumvalueamongallpointswheref isdefinedisaglobalmaximum orabsolutemaximum. Apointwheref achievesitsminimumvalueamongall pointswheref isdefinedisaglobalminimum orabsoluteminimum.4. ConcavityandtheSecondDerivativeTest. Foradifferentiablefunctionf,everyinteriorextremalpointisacriticalpointoff. Butnoteverycriticalpointoff isanextremalpoint.Example. Thefunctionf(x)=x3 hasacriticalpointatx =0. Butf(x)iseverywhereincreasing,thusx =0isnotanextremalpointoff.When is a critical point an extremal point? When is it a local maximum? When is it a localminimum? Thisiscloselyrelatedtotheconcavity off. Afunctionf(x)isconcaveup,respectivelyconcavedown,ifnosecantlinesegmenttof(x)crossesbelowthegraphoff,resp. abovethegraphoff. Insymbols,f isconcaveup,resp. concavedown,if

(f(c)f(a))/(c a)(f(b)f(a))/(b a)whenevera < c < b,resp. (f(c)f(a))/(c a)(f(b)f(a))/(b a)whenevera < c < b.

Foradifferentiablefunctionf,thisequationiscloseto,f(c)f(b)whenevera < c < b,

resp. f(c)f(b)whenevera > c > b.This precisely says that f is non decreasing, resp. f is non increasing. If f is non decreasing, resp. non increasing, thenf isconcaveup,resp. concavedown. ApplyingtheFirstDerivativeTesttodeterminewhenf isincreasing,resp. decreasing,givestheSecondDerivativeTest: Iff(a)> 0,thenf isconcaveupnearx =a;iff(a)< 0thenf isconcavedownnearx =a.Iff isconcaveupnearacriticalpoint,thecriticalpointisalocalminimum. Iff isconcavedownnearacriticalpoint,thecriticalpointisalocalmaximum. CombinedwiththeSecondDerivativeTest,thisgivesatestforwhenacriticalpointisalocalmaximumorlocalminimum: Iff(a)equals0andf(a)< 0,thenx =a isa localmaximum. Iff(a)equals0andf(a)> 0,thenx =a isalocalminimum.

2Example. Fory =x3 +x x 1, thesecondderivative isy = 6x +2. Sincey(1)=4 isnegative,thecriticalpointx =1isalocalmaximum. Sincey(1/3)=4ispositive,x =1/3isalocal

minimum.

5. Inflection points. Iff isdifferentiable,butforeveryneighborhoodofa,f isneitherconcaveupnorconcavedownontheentireneighborhood,thena isaninflectionpoint. Iff(a)isdefined,theSecondDerivativeTestsaysthatf(a)mustequal0. Exceptinpathologicalcases,aninflectionpointisapointwheref isconcaveuptoonesideoff,andconcavedowntotheothersideoff.

2Example. Fory =x3 +x x 1, thesecondderivativey = 6x +2 isnegative forx < 1/3andispositiveforx > 1/3. BytheSecondDerivativeTest,y isconcavedownforx < 1/3andyisconcaveupforx > 1/3. Thereforex =1/3isaninflectionpointfory.

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Step 41. Sketch a graph of the quantity to be maximized or minimized. This is not2absolutely necessary. Sometimes it is impossible. When you can make a rough sketch, this willtypicallygiveaverygoodideawherethemaximumorminimumlies. Intheexampleabove,A(l)is a quadratic equation. Because both l and w must be nonnegative, A(l) is only defined on theinterval0l10. ThusthegraphofA(l) isasegmentofaparabolaopeningdown. Thevertexoftheparabolaiscontainedinthesegment. Thusthevertexisthemaximum.Step 5. Compute the derivative. Inthiscase,

A(l)=2l+10.Step 6. Find all critical points, endpoints, discontinuity points, etc. In most cases, itsufficestofindallcriticalpointsandendpoints. Occasionallyit isalsonecessarytofindallpointswheref isnotdefined. Rarely it isnecessarytoalsoconsiderdiscontinuitypoints(althoughthisisusuallysoobviousthatitdoesnotrequireaseparatestep). Inthiscase,theendpointsarel=0andl=10. Theonecriticalpointisl=5.Step 7. Determine the global maximum or minimum. Checking all critical points, end points, etc., determine the global maximum or the global minimum. In this case, A(0) equals 0,A(10)equals0andA(5)equals25. Thusl=5istheglobalmaximum.Step 8. Back substitute. Plug in the value of the single remaining independent variable todeterminethevaluesoftheremaining independentvariables. Inthiscase,w equals10l,whichis105=5forl=5. Thus,thelargestarea25isenclosedbyasquareofsidelength5.Example. A swimmer is in the waterat adistanceb

1 meters from shore. She wantsto reacha

pointonlandb2 metersfromthewater. Thepointisametersparalleltotheshore. Iftheswimmerswims v1 meters per second and runs v2 meters per second, at what distance x from the closestpointonshoreshouldsheaimtominimizehertimetothetarget? Mathematically,theswimmerisatpoint(0,b1)andwantstoreachpoint(a,b2),wheretheshore isthexaxis. Atwhatpoint(x,0)shouldsheaim?The constants are a, b1, b2, v1 and v2. The variable is x. It is also convenient to introduce avariabled1 forthedistance from(0,b1)to(x,0),andavariabled2 forthedistance from(x,0)to(a,b2). Althoughnotobvious, it isalsoveryconvenientto introduceavariable1 fortheacuteangleformedbythexaxis andthe linesegmentjoining(0,b1)to(x,0). Alsointroduce2 fortheacuteangleformedbythexaxis andthelinesegmentjoining(x,0)to(a,

b2).

ThetimeT1 toswimtopoint(x,0)is,1

1)1/2T1 = d1 = (x2 +b2 .

v1 v1ThetimeT2 torunfrom(x,0)topoint(a,b2)is,

12)

1/2T2 = d2 = ((ax)2 +b2 .v2 v2

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Thusthetotaltimetoreachthetargetis,1 1

T =T1 +T2 = (x2 +b2 2)1/21)1/2 + ((ax)2 +b2 .v1 v2

ThederivativeofT withrespecttoxis,dT 1 1 1 1

= (x2 +b2 2)1/2(2(ax)) .1)1/2(2x) + ((ax)2 +b2dx v1 2 v2 2Thissimplifiesto,

dT x ax= .

dx v1d1 v2d2Observethatx/d1 equalssin(1)and(a

x)/d2 equalssin(2). Thus,

dT sin(1) sin(2)= .

dx v1 v2Technically, there are no endpoints. However, it is obvious that the maximum must occur for0xa. Thusthesemaybetakentobeendpoints. Thecriticalvalueoccurswhen,

1)v1 = 2)v2 .sin( sin(

ThisisSnellsLaw forrefractionoflightuponcrossingfromonemediumtoanother. Forrefraction,a particle of light (perhaps fictitious) replaces the swimmer, a translucent medium of one typereplaces thewater, andatranslucentmedium of asecond type replaces the land. If lighttravelswithvelocityv1 inthefirstmediumandwithvelocityv2 inthesecondmedium,lightrayswillrefractuponcrossingtheboundarybetweenmedia. SnellsLawdescribestheanglesofthisrefraction.Lecture 11. October4,2005Homework. ProblemSet3PartI:(g)and(h).Practice Problems. CourseReader: 2E 4, 2E 8, 2E 9. 1. Related rates. A situation that arises often in practice is that two quantities, say x and y,depend on a third independentvariable, say t. The quantities x and y are related through someconstraint. Usingtheconstraint,iftherate of change dx/dtisknown,therate of change dy/dtcanbeinferred.

Example. Foraspringdisplacedxunits fromequilibrium,Hookes law impliesthepotentialenergyofthespringis,

1P = kx2,

2where k isaconstantwith units kg/s2. Atsome moment t=T, aspring is displaced 5cm fromequilibriumandhasvelocity5cm/s. Intermsofthespringconstantk,describetherate of change ofthepotentialenergyatt=T.

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Implicitlydifferentiatingtheequationwithrespecttotgives,usingthechainrule,dP 1 dx dx= k(2x) =kx .dt 2 dt dt

So,attimet=T,dP dx

(T)=kx(T) (T)=k(5)(5)cm2/s=dt dt 25 2/s.kcm

2. Method for solving related rates problems. Many of these steps apply to any word probleminmathematics.

(i) Identifytheindependentvariable. Intheexample,thisist.(ii) Labelallconstants. Intheexample,k isaconstant.

(iii) Labelalldependentvariables. Intheexample,xandP aredependentvariables.(iv) Drawadiagramandcarefullylabelit.(v) Writethegivenrate of change andtheunknownrate of change. Intheexample,dx/dt(T)is

givenas5cm/s,anddP/dtisunknown.(vi) Usingthediagramandanyotherinformation,findconstraintsamongthedependentvariables.

Intheexample,thisistheequationP =kx2/2.(vii) Implicitlydifferentiatetheconstraintequationswithrespecttotheindependentvariable. In

theexample,thisgivesdP/dt=kxdx/dt.(viii) Substituteinallknownquantitiesandsolvefortheunknownrate of change. Intheexample,

dP/dt(T)equals25kcm2/s.Example. A state trooper waits a distance a from a highway for passing speeders. The speedlimitis60mph. Thetrooperaimsherradargunatanangleof/4totheroad. Theradarregistersapassingcar moving away from thetrooper at a speedof 50mph. Should the trooper ticket thedriver?The independent variable is time t. The constants are the distance a and the angle = /4.Labelacoordinatesystemwiththetrooperattheoriginandthehighwayequaltotheliney=a.Labelthepositionofthecaralongthehighwayasx,moving inthepositivedirection. Denotebyr the distance of the car from the trooper. Then x and r are dependent variables. The rate of change dr/dt(T) is given as 50mph. The unknown rate of change is dx/dt(T). The constraint isthePythagoreantheorem,

2 2 2r =x +y .

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Implicitdifferentiationwithrespecttotyields,dr dx dx2r =2x +0=2x .dt dt dt

Attimet=T,x(T)equalsa,becausetheangle is/4. Thusr(T)equals2a. Substituting ingives,

dx2(

2a)50=2(a) (T).

dtSolvinggives,

dx(T)=250 71mph.

dtSothetroopershouldticketthedriver.Example. Apointonthexaxis movesaway fromtheorigin. There isanangle subtendedbythepointandtheunitcirclewithequationx2+y2 =1. Inotherwords,standingatthepoint(x,0)andstaringatthecircle, istheangleofyourfield of vision occupiedbythecircle. Atamomentt=T,thepointisattheposition(2,0)andmovingwithvelocityv. Whatistherate of change ofatt=T?The independent variable is time t. There is no constant. The dependent variables are the xcoordinateofthepoint,x(t),andtheangle(t). Therate of change dx/dt(T)isgiventobev. Therate of change d/dtisunknown.Theconstraintissomewhattricky. Therearetwotangentlinestothecirclecontaining(x,0). Theseare the tangent lines to points (a,+b) and (a,b) on the circle. Because the tangent line to thecircleat(a,b) isperpendiculartotheradiusthrough(a,b),thetrianglewithvertices(0,0),(a,b)andthepoint(x,0) isarighttriangle. Theangleofthetriangleat(x,0) is/2. Sincetheradiushaslength1andthehypotenusehaslengthx,theconstraintis,

1sin()= .

xImplicitdifferentiationwithrespecttotgives,

dsin()d d(x1)dx=

d dt dx dt,or,

cos()ddt =

1x2

dxdt.

Sincex(T)equals2,sin((T))=1/2,andthuscos((T))equals3/2. Pluggingingives,

3d(T)= 1v=v .

2 dt (2)2 4

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Solvinggives,d

(T)=

dt (2

v/ 3).

3. Anotherappliedmax/minproblem. AsreviewforExam2,thisisanotherappliedmax/min

2problem. Atrapezoid is inscribed inside the upperunit semicircle, x2 +y = 1,y0. Thebaseof the trapezoid is the diameter of the semicircle lying on the xaxis. The top of the trapezoidis parallel to the xaxis joining (x,y) to (x,y) for a point (x,y) on the unit circle in the firstquadrant. Whatisthemaximalareaenclosedbysuchatrapezoid?The parameters are x and y. The height of the trapezoid is y. The area of a trapezoid is theproductoftheheightwiththeaverageoftheparallelsides. Thus,

(2+2x)A

=

y

=(x

+

1)y.

2

Thisisthequantitytobemaximized. Thereisaconstraintamongtheparameters,2x +y2 =1.

Also,since(x,y)isinthefirstquadrant,0x1and0y1.Thereareatleast3waystoproceed. Themostdirectistosolvefory intermsofx,

y=1x2.SubstitutingthisintotheequationforAgives,

2A(x)=(x+1)1x .Differentiatinggives,

dA 2x 1 2=1x2 +(x+1)2

1x =1x ((1x2)(x +x))=(2x2 +x1)

.2 2 2dx 1x

Becausethequadraticpolynomial2x2 +x1factorsas,2x2 +x1=(2x1)(x+1),

the critical points of A are x =1 and x= 1/2. But x =1 does not give a point in the firstquadrant. ThusAismaximizedeitheratoneoftheendpointsx=0,x=1oratthecriticalpointx=1/2. Pluggingingives,

A(0)=1,A(1/2)=33/4,A(1)=0.Thisgivestheanswer,

3/ /2,3/Aachievesitsmaximum3 4forthepoint(x,y)=(1 2).

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Twoothermethodsweregiveninlecture. ThefastestamongthethreeistoinsteadminimizeA2,2A2

=(x+1)2y .

Usingtheconstraint,y2 =1x2,thus,2(A2)(x)=(x+1)2(1x ).

Thederivativeofthispolynomialisveryfasttocompute,andgivesthesameanswerasabove.Lecture 12. October6,2005Homework. ProblemSet3PartI:(i)and(j).This was a guest lecture by Sabri Kilic. Notes from the lecture will not be posted. As always,pleasedotherequiredreadinginthecoursetextbook.Lecture 13. October13,2005Homework. ProblemSet4PartI:(a)and(b);PartII:Problem3.Practice Problems. CourseReader: 3A 1, 3A 2, 3A 3. 1. Differentials. Analternativenotation forderivatives isdifferential notation. Thedifferentialnotation,

dF(x)=f(x)dx,is shorthand for the sentence The derivative of F(x) with respect to x equals f(x). Formally,this

is

related

to

the

Leibniz

notation

for

the

derivative,

dF

(x)=f(x),dx

which means the same thing as the differential notation. It may look like the first and secondequationare obtainedbydividingandmultiplyingbythequantitydx. It iscrucial to rememberthatdF/dxisnot a fraction,althoughthenotationsuggestsotherwise.In differential notation, some derivative rules have a very simple form, and are thus easier toremember. Hereareafewderivativerulesindifferentialnotation.

dF(x) = F(x)dxd(F(x)+G(x)) = dF(x)+dG(x)

d(cF(x)) = cdF(x)d(F(x)G(x)) = G(x)dF(x)+F(x)dG(x)

d(F(x)/G(x)) = 1/(G(x))2(G(x)dF(x)F(x)dG(x))Thechainrulehasaparticularlysimpleform,

dF dF dud(F(u))= du= dx.

du du dx31


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Example. Usingdifferentialnotation,thederivativeofsin(x2 +1)is,dsin((x2 +1)1/2)=cos((x2 +1)1/2)d(x2 +1)1/2 =cos((x2 +1)1/2)(12(x2 +1)1/2)d(x2 +1)=

cos((x2 +1)1/2)12(x2 +1)1/2(2xdx)= x(x2 1/2cos((x2 1/2)dx.+1) +1)2. Antidifferentiation. Recall,thebasicproblemofdifferentiationisthefollowing.Problem(Differentiation). GivenafunctionF(x),findthefunctionf(x)satisfying dF =f(x).

dxThebaisproblemofantidifferentiation istheinverseproblem.Problem (Antidifferentiation). Given a function f(x), find a function F(x) satisfying dF =dxf(x).AfunctionF(x)solvingtheproblemiscalledanantiderivativeoff(x),orsometimesanindefiniteintegraloff(x). Thenotationforthisis,

F(x)= f(x)dx.Theexpressionf(x) is calledthe integrand. It is important tonote, ifF(x) isoneantiderivativeoff(x), then for eachconstant C, F(x)+C isalsoan antiderivative off(x). TheconstantC iscalledaconstantof integration.Inasensethatcanbemadeprecise,theproblemofdifferentiationhasacompletesolutionwheneverF(x) isasimpleexpression, i.e.,a functionbuiltfromthedifferentiable functionswehaveseensofar. Unfortunately,forverymanysimplefunctionsf(x),noantiderivativeoff(x)hasasimpleexpression. Inlargepart,thisiswhatmakesantidifferentiationdifficult. Luckily,manyofthemostimportantsimple functionsf(x)dohaveanantiderivative withasimpleexpression. One goalofthisunitistolearnhowtorecognizewhenasimpleantiderivativeexists,andsometoolstocomputetheantiderivative.3. Antidifferentiation. Guess and check. The maintechnique forantidifferentiation is edu catedguessing.Example. Findanantiderivativeoff(x)=x2 +2x+1. Sincethederivativeofxn isnxn1, it isreasonabletoguessthereisanantiderivativeoftheformF(x)=Ax3+Bx2+Cx. Differentiationgives,

dF=3Ax2 +2Bx+C.dx

Thus,F(x)isanantiderivativeoff(x)ifandonlyif,3A=1, 2B=2, andC=1.

Thisgivesanantiderivative,(x2 +2x+1)dx= 13x3 +x2 +x+E ,

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whereE isanyconstant.Guess and check

is

agame

we

can

lose,

as

well

as

win.

However,

there

are

afew

rules

that

better

the

oddsinthisguessinggame. Infact,theyarebasicallythesamerulesforderivativesindifferentialnotation,simplywrittenbackwards.

(f(x)+g(x))dx = f(x)dx+ g(x)dx cf(x)dx = c f(x)dxf(u(x))u(x)dx = f(u)du

4. Antidifferentiation. Integration by substitution. The lastruleabove isvery important,andcalledintegrationbysubstitution.Example. Find an antiderivative of xsin(x2). This time guess and check is much less effective.By roughly the same logic in the last example, we might guess an antiderivative has the formAx3sin(x2). Thederivative is3Ax2sin(x2)+2Ax4cos(x2). Thefirsttermisgood,butthesecondtermisbad. Wecantrytocorrectourguessbyaddingaterm,Ax3sin(x2)2/5Ax5cos(x2),whosederivativeisnow3Ax2sin(x2)+4/5Ax6sin(x2). Thisstilldoesntwork,andisleadinginthewrongdirection.A better solution is to use integration by substitution. Observe part of f(x) can be written asa function of u(x) = x2. Also, the derivative u(x) = 2x occurs in f(x) through x = 1/2(2x) =u(x)/2. Thus,

2xsin(x )=sin(u(x))u(x)/2, u(x)=x2.Applying

integration

by

substitution,

1xsin(x2)dx= sin(u(x))1u(x)dx=2 sin(u)du=2

1 cos(u)+C= 1 cos(x2)+C.2 2Hereisachecklistforapplyingintegrationbysubstitiontofindtheantiderivativeoff(x).

(i) Find an expression u(x) so that most of the integrand f(x) can be expressed as a simplerfunctionofu(x).

(ii) Computethedifferentialdu(x)=u(x)dx.(iii) Insidethedifferentialf(x)dx,trytofinddu=u(x)dxasafactor.(iv) Try towrite f(x)dx as g(u)du. If you cannotdo this, themethod doesnotapplywith the

givenchoiceofu.(v) FindanantiderivativeG(u)= g(u)duforthesimplerintegrandg(u)(ifthisispossible).

(vi) Back substitute u=u(x)togetanantiderivativeF(x)=G(u(x))forf(x).33


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Example. Computetheantiderivative,sin(x)3cos(x)dx.

Mostofthe integrand isafunctionofsin(x). Sosubstituteu(x)=sin(x). Thedifferentialofu isdu=cos(x)dx. Thedifferentialsin(x)3cos(x)dxcontainsdu=cos(x)dxasafactor. Theremainder

3oftheintegrandissin(x)3 =u . So,accordingtointegrationbysubstitution,sin(x)3cos(x)dx= u3du= 1u4 +C.

4Finally,back substitute u=sin(x)toget,

sin(x)3cos(x)dx= (sin(x))4/ C.4+Lecture 14. October14,2005Homework. ProblemSet4PartII:Problem2.Practice Problems. CourseReader: 3B 1, 3B 3, 3B 4, 3B 5. 1. The problem of areas. TheancientGreekscomputedtheareasoftriangles,quadrilaterals,andmanyotherpolygons. Theirbasicmethodwasdissection: dissectingapolygonalregionexactlyintosmallerregions,usuallytriangles,havingknownareas. Theareaofthelargeregionisthesumof the areas of the small regions. But the ancient Greeks also knew the area of a circle, whichcannot be dissected exactly into finitely many polygonal regions. Their method was exhaustion:finding polygonal regions approximately equal to the original region, and computing the limit oftheareasofthepolygonsastheapproximationimproves.Example. A regular Nsided polygon inscribed in a circle of radius r has apothem length a =rcos(/N)andchord length b=2rsin(/N). Thustheareaofthepolygonis,

ab 2A=N =Nr2sin(/N)cos(/N)=r Nsin(2/N)=r2sin(2/N).2 2 2/N

As N increases, 2/N decreases to0. Because limt 0sin(t)/tequals 1, as N approaches infinity,theareaofthepolygonapproaches,

r2sin(2/N)lim =2/NN 2.r

AmoresophisticatedversionofthemethodofexhaustiongivestheRiemann integral. Here isthebasicproblem.Problem (Area). Find thesigned area between the graph of y =f(x) and the xaxis over theintervalaxb.

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Foraregionabovethexaxis, thesignedarea issimplythearea. Foraregionbelowthexaxis, thesignedarea isthenegativeofthearea. Foraregionpartlyabovethexaxis andpartlybelowthexaxis, thesignedareaisthesumofthesignedareaoftheregionabovethexaxis andthesignedareaoftheregionbelowthexaxis. 2. Partitions. Apartition ofaninterval[a, b]isafinitedecompositionoftheintervalasaunionofnon overlapping subintervals,

[a, b] = [x0, x1] [x1, x2] [xn2, x [xn1, xn].n2] Sincean interval isdeterminedby itsrightand leftendpoints,tospecifyapartitionof [a, b], it isequivalenttogiveanorderedsequenceofincreasingnumbers,

a=x0


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The sum above is a Riemann sum. It is an approximation of the signed area of the curvilinearregion.Therearemanychoicesofpartition. Andforeachpartition,therearemanychoicesforthenumbers

k. However,therearesomespecialchoices. Onthekth interval,thesmallestvaluef(x)takesonisdenotedby,

yk,min =min{f(x) xk1 xxk+1}.x

|Similarly,thelargestvaluef(x)takesonisdenoteby,

yk,max =max{f(x) xk1 xxk+1}.|Foreverychoiceofx inthekth interval,y istrappedbetweenthesetwovalues,k k

yk,min yk yk,max.Denoting,

Ak,min =yk,minxk, Ak,max =yk,maxxk,theareaAk istrappedbetweenthesetwovalues,

Ak,min Ak Ak,max.Denotingthesumsoftheareasby,

nk=1Amin =n Ak,min =

k=1yk,minxk,

n nAmax =k=1

Ak,min =k=1

yk,minxk,theRiemannsumAistrappedbetweenthetwovalues,

Amin AAmax.Thus, ifAmin andAmax areclosetoeachother,thevalueofAdoesnotdependverymuchonthechoicesofthenumbersxk.4. The Riemann integral. ThemethodoftheRiemann integral istocomputebothAmin andAmax forasequenceofpartitionswhosemeshsizesapproach0. Themeshsizemeasuresthefinenessofthepartition,thusalsothefitoftheunionofverticalstripstothecurvilinearregion. Ifthetwolimits,

lim Amin, lim Amax,mesh0 mesh0

are defined and equal, it is said the Riemann integral exists, and the common limit is called theRiemann integral, b

f(x)dx= lim Amin = lim Amax.mesh 0 mesh 0a

Also,f(x) issaidtobeRiemann integrable on the interval [a,b]. Anothername fortheRiemannintegralisthedefinite integral.

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Example. Considerthe functionf(x) =xonthe interval0xL, forsomepositivenumberL. Formthepartitionwithnsubintervalsofequallength,

x0 =0=0L/n,x1 =1L/n,x2 =2L/n,...,xk =kL/n,...xn =nL/n=L.Everyintervalhaslengthxk =L/n. SothemeshsizeisL/n. Theminimumvalueoff(x)ontheintervalxk1 xxk isyk,min =xk1 =(k1)L/n. Themaximumvalueisyk,max =xk =kL/n.Thus,

n n n (k1)L L L2(k1),Amin = yk,minxk = = 2n n n

k=1 k=1 k=1and,

nn nL2

kLLAmax = yk,maxxk = = k.

n n n2k=1 k=1 k=1Toevaluatethesesums,usethewell known formula,

n n(n+1)k= .

2k=1

Thisalsogives,n n1 n1

(n1)n,(k1)= l= l=

2k=1 l=0 l=1

bymakingthesubstitutionl=k

1. Substitutingtheformulagives,L2n(n1)

= L2(1 1 ),Amin = 2n 2 2 nand,

L2n(n+1) L2 1Amin = = (1+ ).

n2 2 2 nTherefore,

L2 1 L2 L2lim Amin = lim(1 )= (10)= .

n 2 n 0 n 2 2Similarly,

L2

1 L2 L2lim Amax = lim(1+ )= (1+0)= .n 2 n 0 n 2 2

Sincethetwolimitsareequal,f(x)=xisRiemannintegrableontheinterval[0,L],and, Lxdx=

0L2/2.

Thisagrees with the familiar result fromhigh school geometry: the area of atriangle equals onehalfofthebasetimestheheight,sinceboththebaseandheightofthistriangleareL.

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5. Rules for Riemann integrals. There are several rules for Riemann integrals, summarizedbelow.

b

b

b

a(fb(x)+g(x))dx = f(x)dx+ a g(x)dx,a bf(x))dx = a f(x)dx,b a(r rc c

f(x)dx+b f(x)dx = f(x)dx.a a

Lecture 15. October18,2005Homework. ProblemSet4PartI:(d)and(e);PartII:Problem2.Practice Problems. CourseReader: 3B 6, 3C 2, 3C 3, 3C 4, 3C 6. 1. TheRiemannsumfortheexponentialfunction. TheproblemistocomputetheRiemannintegral, b

xe dx,0

usingRiemannsums. Choosethepartitionof[0,b]intoasequenceofnequally spaced subintervalsof length b/n. So the partition numbers are xk = kb/n. Also the length of each partition is

xxk =b/n. Becauseex is increasing,theminimumvalueofe onthe interval [xk1,xk]occursattheleftendpoint,

(k1)b/nyk,min =exk1 =e .Similarly,themaximumvalueoccursattherightendpoint,

kb/nyk,max =exk =e .Thusthelowersumis, n n b

Amin = yk,minxk = e(k1)b/n .n

k=1 k=1Andtheuppersumis,

n n bAmax = yk,maxxk = ekb/n .

nk=1 k=1

Toevaluateeachofthesums,makethesubstitutionc=eb/n. Thenthelowersumis,n1n

bb lAmin = ck1 = c .n nk=1 l=0

Thesumisageometricsum,n

(1+c+c2 + +cn2 +cn1)= c 1.c 1

Pluggingthisingives,n ebn/nb c 1 b 1

Amin = = .eb/n 1n c 1 n

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Thissimplifiesto,b/n

.Amin=

(e

b

1)

eb/n1

Asimilarcomputationgives,b/nb/nAmax =(eb1)e

eb/n1.Nowmakethesubstitution,h=b/n. Thisgives,

hAmin =(eb1)

eh1,hhAmax =(eb1)e .he

1Takingthe limitofAmin,respectivelyAmax, asntendsto infinity isthesameastakingthe limitashtendsto0.Nowobservethat,

helim 1,h 0 h

is the difference quotient limit giving the derivative of ex at x= 0. Since dex/dx equals ex, andsincee0 equals1,thisgives,

he 1lim =1.h 0 h

Invertinggives, hh e 11lim = lim =(1)1 =1.h0 eh1 h 0 h

Also,becauseex iscontinuous,limeh =e0 =1.h 0

Puttingthistogethergives,lim Amin =(eb1)lim

h0 ehh1 =(eb1)(1)=eb1.n

Similarly,

hlim Amax =(eb1)(lime )(limh0 eh

h1)=(eb1)(1)(1)=eb1.h 0n

Sincethe limitofAmin andthelimitofAmax existandareequal,theRiemann integralexistsandequals, b

exdx=0

eb1.

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r2. The Riemann sum for x . Letr>0beapositiverealnumber. TheproblemistocomputetheRiemannintegral,

b rx dx,1

usingRiemannsums. Forthisparticularintegral,adifferentpartitionthanusualismoreefficient.Letnbeapositiveinteger,andletqbetherealnumber,

q=b1/n.Choosethepartitionof[1, b]intonsubintervalswithpartitionnumbers,

kxk =q .Observethat,

1=x0


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rAlso,sincex iscontinuous,limqr =1r =1.q1

Substitutingthisingives,br+1

lim Amin =(br+1 1) limqr+1 11

= 1,q1 q 1 r+1n

br+1lim Amax =(br+1 1) limqr limqr+1 1

1= 1,

q1 q1 q 1 r+1nSincethe limitofAmin andthelimitofAmax existandareequal,theRiemann integralexistsandequals, b

xrdx=1 (b

r+1 1)/(r+1).3. The Fundamental Theorem of Calculus. Thereisasingletheoremthat itisattheheartofalmostallapplications involvingRiemann integrals. Thetheoremanswerstwoquestionsimul taneously: WhichfunctionsareRiemannintegrable? WhatistheRiemannintegralofafunction?The answer to the first question is: Every function you are likely to encounter is Riemann inte grable. Precisely, everycontinuous function, andeverypiecewisecontinuous function isRiemannintegrable.Theanswertothesecondquestionismoreinteresting. Assumef(x)isacontinuousfunction. Letx=abeafixedpointwheref(x)isdefined. Formthefunction,

xF(x)= f(t)dt.

aThe function F(x) is defined whenever f(t) is defined on all of [a,x]. If f(x) is continuous, theFundamentalTheoremofCalculusassertsF(x)isdifferentiableand,

d xdF(x)= f(t)dt=f(x).

dx dx aTheproofofthesecondpartisveryeasy. ConsidertheincrementinF fromxtox+x, x+x x x+x

F(x+x) F(x)= f(t)dt f(t)dt= f(t)dt.a a x

Let ymin be the minimum value of f(t) on the interval [x,x+x]. Let ymax be the maximumvalueoff(t)onthe interval [x,x+x]. Then foreverychoiceofpartition t0


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foreveryk. ThustheRiemannsumissqueezedbetween,n n n

t kkymintk y ymaxtk.k=1 k=1 k=1

Ofcoursethelowerboundis,n n

ymintk =ymin tk =yminx,k=1 k=1

becausethetotallengthoftheinterval[x,x+x]isx. Similarly,theupperboundis,n

ymaxtk =ymaxx.k=1

ThustheRiemannsumissqueezedbetween,n

x ykk maxx.yminx yk=1

BecausetheRiemannintegralisalimitofRiemannsums,itisalsosqueezed,

x+x

yminx

f(t)dt

ymaxx.x

SubstitutinginF(x+x)F(x)anddividingeachtermbyxgives,F(x+x)F(x)

ymin x ymax.

Themiddletermisthedifferencequotient. Considerwhathappensasxtendsto0. Becausef(t)iscontinuous,boththemaximumandminimumvaluesoff(t)on [x,x+x]simply limittothevaluef(x). Thus,

limymin =limymax =f(x).x x

By the Squeezing Lemma for limits, since these two limits exist and are equals, themiddle limitalsoexistsandequalsf(x),

F(x+x)F(x)=f(x).lim

xx 0ThisispreciselywhattheFundamentalTheoremofCalculusasserts,

d xf(x).f(t)dt=

dx a43


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4. AlgorithmforcomputingRiemannintegrals. TheFundamentalTheoremofCalculushasmanyimportantapplications. Themostobvious istogiveusasimplermethodforcomputingRiemannintegrals, under the hypothesis that we can compute the antiderivative. If f(x) is a continuousfunctionandG(x)isaknownantiderivativeoff(x),then, b

f(t)dt=G(b)G(a).a

Toseethis,observethat,x

F(x)= f(t)dt,a

isalsoanantiderivativeoff(t)bytheFundamentalTheoremofCalculus. Thus,sincethegeneralantiderivativeisG(x)+C,thereisaconstantC suchthatF(x)=G(x)+C. Butalso,

aF(a)= f(t)dt=0.

aThus,F(x)=G(x)G(a). Nowpluginx=btoget, b

f(t)dt=F(b)=a

G(b)G(a).Lecture 16. October20,2005Practice Problems. CourseReader: 3D 1, 3D 3, 3D 7, 3E 3, 3E 4. 1. Dummy variables. GiveaRiemannintegrablefunctionf(x)definedonaninterval[a,b],thenotation, b

f(x)dx,a

isshorthandfortheRiemannintegraloff(x)overthisinterval. Inparticular,thisequalsthelimit,lim f(a+(ba)k/n)ba.

nnObserve,thevariablexdoesnotappear inthis limit. It isveryconvenienttoincludethevariablexinthenotationfortheRiemannintegral;forhowelsearewetoexpressthefunctionintegrated?But,sincethedefinitionoftheRiemannintegraldoesnotinvolvex,xisreallyadummyvariable.Any

variable

name

may

be

substituted

for

x,

with

the

same

meaning.

b b b bf(x)dx= f(u)du= f(v)dv= f(t)dt=...

a a a aThis freedom is very useful, particularly when one or both of the limits of integration dependon some parameter. In this case, by convention, the dummy variable is chosen to be a differentparameter.

x xf(x)dxINCORRECT, f(t)dtCORRECT

a a44


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Thisconventionreducesthelikelihoodofanerror.2.

Variable

limits

of

integration.

The

Riemann

integral

is

often

used

to

define

functions,

particularlyantiderivativeshavingnosimplerexpression.Example. Foreveryangle0


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|


Thisgives,df

=2

1

cos

2

()(sin())

=

d

The second method is indirect. The function G(t) has no simple expression. Nonetheless, thismethodisfaster. Inmanycasesthisistheonlymethodthatworks.

2

().2sin

TheargumentaboveusingthechainruleandtheFundamentalTheoremofCalculusisquitegeneral.Itgivesthegeneralequation,

v(x)u(x) f(t)dt =f(v(x))v(x)f(u(x))u(x).d/dx

3. Geometric area and algebraic area. TheRiemannintegralisthealgebraicarea, bf(x)dx = Areaabovethexaxis Areabelowthexaxis .

aThe geometric area isthetotal area, bothaboveandbelowthexaxis. Althoughgeometricareadoesnotequalalgebraicarea,ithasasimpleexpressionusingtheRiemannintegral, b

Geometricarea = f(x)|dx.|a

Example. Find both the algebraic area and the geometric area bounded by the xaxis and thegraphofy =sin(x)overtheinterval < x < .Becausesin(x)isanoddfunction,theareabelowthexaxis for < x < 0equalstheareaabovethexaxis for0< x < . Intheexpressionforthealgebraicarea,theseareascanceltogive0. Thisisborneoutbycomputation,

sin(x)dx =(cos(x) =cos()+cos()=(1)+(1)=0.|

Ontheotherhand,theabsolutevalue sin(x)|equals,

|sin(x)

|= sin(x), < x 0,

sin(x), 0< x < .Thusthegeometricareaequals, 0

sin(x)dx + sin(x)dx =00(cos(x) +(cos(x) =(1(1))+((1)+1)=0| | 4.

Thusthegeometricareadoesnotequalthealgebraicarea. ButcomputationofthegeometricareareducestoastraightforwardRiemannintegral.

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0


4. Estimates. For every pair of Riemann integrable functions f(x),g(x) on [a,b] satisfying theinequalityf(x)

g(x)foreverychoiceofx,thefollowinginequalityholds,

b bf(x)dx g(x)dx.

a aThisisveryusefulforestimatingintegrals.Example. DeterminethefollowingRiemannintegraltowithin104,

0.11+ sin(x)dx.

0Theexpression sin(x)hasnosimpleantiderivative. ThevalueoftheRiemannintegralcouldbeapproximatedwellbyaRiemannsum. Analternativeapproachistousetheestimates,

/6)

sin(x) x,(1 x2 xforsmallvaluesofx. Thisgives, 0.1

01+x 0.1 0.111/2 x5/2dx

6 1/2dx.1+ sin(x)dx 1+x0 0ThefirstandthirdRiemannintegralfollowfromtheFundamentalTheoremofCalculus, 0.1 0.1

1 2 1 2 11/2 5/2dx= 3/2 7/2 =0.1210667926101+x =0.1+x+3

10002110000000 x x x6 3 21 0Similarly,

0.1 0.12 21/2dx= 3/2 =0.1210818511 10101+x =0.1+x+3

1000x .30 0Sincethesetwointegralsagreetowithin104,thisgivestheoriginalintegral,

0.1

1+ sin(x)dx= 0.

104.12100

5. Change of variables. AftertheFundamentalTheoremofCalculus,themostuseful integralrule isthechangeofvariablesrule. TheruleforRiemann integrals isnearlythesameastherulefor antiderivatives. The additional feature for Riemann integrals is the change of the limits ofintegration.

x=b u=u(b)f(u(x))u(x)dx= f(u)du.

x=a u=u(a)

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Example. FindtheRiemannintegral,/3

tan(x)dx.

/4Sincetan(x)isnotvisiblythederivativeofanotherfunction,werewritethe integralandhopeforthebest. /3 /3 sin(x)

tan(x)dx= dx./4 /4 cos(x)

Inthisform,thesubstitutionu=cos(x)isnatural,x=/3 sin(x)

x=/4 cos(x)dx,u=cos(x) u(/3)=cos(/3)=1/2,

du=sin(x)dx u(/4)=cos(/4)=1/2.u=1/2 1(du).

u=1/

2 uThe new integral can be computed by the Fundamental Theorem of Calculus, since 1/u is thederivativeofln(u).

u=1/21/21

du=(

ln(|u

|)

|1/

2 =

ln(1/2)+ln(1/2)=ln(2)

ln(2).

uu=1/2

Thissimplifiestogive,/3

tan(x)dx=/4

/2.ln(2)

It is only fair to note there is a second method. Make the same substitution to simplify theantiderivativeoftan(x)toln(|u|)+C,andthenback substitute toget,

tan(x)dx=ln( cos(x) )+C.| |Now use the Fundamental Theorem of Calculus with the original limits of integration. Bothmethods are correct. Usually the first method is faster and less error prone; it requires no back substitution.6. Integratingbackwards. Thiscomessonaturally formostcalculusstudents, itbarelywarrantsmention. Technically,theRiemannintegral, b

f(x)dx,a

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isonlydefined ifab. What ifa is largerthanb? TheonlypossibleanswerconsistentwiththeFundamentalTheoremofCalculusisthefollowing, b a

f(x)dx= f(x)dx, ifa>b.ba

BecauseofthecentralroleoftheFundamentalTheoremofCalculus,theaboveequationistruebyconvention. Withthisconvention,theFundamentalTheoremofCalculusholdswhenevera islessthanb,equaltob,orgreaterthanb.Lecture 17. October21,2005Homework. ProblemSet5PartI:(a)and(b);PartII:Problem1.Practice Problems. CourseReader: 3F 1, 3F 2, 3F 4, 3F 8. 1. Ordinary differential equations. Anordinarydifferentialequationisanequationinvolvingasingleindependentvariablextogetherwithadependentvariableyanditsderivativesdky/dxk,

dy d2y dkyG x,y,

dx,

dx2,..., =0.dxkThelargestk forwhichdky/dxk occursintheequationiscalledorder ofthedifferentialequation.Examples. Hereareexamplesofordinarydifferentialequations.(i)Theordinarydifferentialequation,

y

sin(x

2)=

0,

hasorder0,becausenoderivativesofyactuallyoccurintheequation. Ithasaunique(andrathertrivial)solution,

y=sin(x2).Becausethesolutionisunique,itdependson0parameters(andtheorderis0).(ii)Theordinarydifferentialequation,

dy 1=0,

dxx+1has

order

1because

dy/dx

occurs

and

no

higher

derivatives

occur.

Every

solution

is

an

antiderivative

of1/x+1,

1y= dx=ln(|x+1 )+C,

x+1 |Noticethesolutiondependson1parameter,C. Andtheorderis1.(iii)Theordinarydifferentialequation,

d2y+2y=0,

dx2

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hasorder2. ThegeneralsolutionwasfoundinProblemSet2,Problem4,y=Acos(x)+Bsin(x).

Thesolutiondependson2parameters,AandB. Andtheorderis2.A kth(iv) The previous equation was one particular linear ordinary differential equation. order

linearordinarydifferentialequation hastheform,dky dk1y dy

+a1(x) +a0(x)y=b(x),ak(x)dxk +ak1(x)dxk1 + dx

for functions ak(x),...,a0(x),b(x). If b(x) is zero, the equation is homogeneous. Otherwise itis inhomogeneous. Very important is the case when all the functions ak(x),...,a0(x),b(x) areconstant. Then the differential equation is called constant coefficient. The solution of constantcoefficientlinearordinarydifferentialequationsisamainfocusofMath18.03.2. Separable differential equations. Many differential equations arising in applications areexamples of separable differential equation. A separable ordinary differential equation is a first orderdifferentialequation,

dy=F(x,y),

dxforwhichf(x,y)factorsas,

F(x,y)=g(x)/h(y).Example. Findtheequationy=f(x)ofeverycurvewiththefollowingproperty: Foreverypoint(x,y)onthecurve,thetangent linetothecurve isperpendiculartothe linejoining(x,y)totheorigin(0,0).Theslopeofthetangentlinetothecurveat(x,y)isdy/dx. Theslopeofthelinejoining(0,0)and(x,y)isy/x. Sincethetangentlineisperpendiculartothelinejoining(0,0)and(x,y),

dy=x/y.

dxThus,theequationy=f(x)isasolutiontothisseparabledifferentialequation.Thealgorithmforsolvingaseparabledifferentialequationisthefollowing.(i). Factor f(x,y) as g(x)/h(y). Thisisoftenthemostdifficultstep. Intheexample,itisquiteeasy. Simplytakeg(x)=xandh(y)=y.(ii). Rewrite the differential equation as an equality of differentials. In other words,rewritetheequationas,

dy g(x)=

h(y)h(y)dy=g(x)dx.dx

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Intheexample,thisgives,dy

=

x

ydy

=

xdx.

dx y

(iii). Antidifferentiate both sides of the equation. Intheexample,theantiderivativesydy= xdx,

give,1 2 1 2y = x +C.2 2

(iv). Ifthereisaninitalvalue,useittofindtheconstantofintegration. Aninitialvalueproblem isanordinarydifferentialequationtogetherwithsomeinformationforan initialvaluex0oftheindependentvariable. Itisoftenwritten,

dy/dx=F(x,y),y(x0)=y0.

The example was not an initial value problem. However, it can easily be made an initial valueproblembyspecifying,

y(1)=1,forinstance. Withthiscondition,theconstantC satisfiestheequation,

1(1)2 =1(1)2 +C.2 2

Thesolutionis,C=1.

(v). Simplify the answer. Often it isbesttosolve fory=f(x). Oftenthis isunnecessary. Itdependsontheproblem. Intheexampleproblem,thesimplestansweristheimplicitanswer,

x2 +y2 C.=2Sothesolutionoftheinitialvalueproblemis,

x2 +y2 =2.Thuseverycurvesatisfyingthegeometricpropertyisacirclecenteredattheorigin.Example. Hereisasomewhatdifferentexample. Thereisasingleseparableordinarydifferentialequationsatisfiedbyeveryfunction,

y=(xa)3,whereaisanarbitraryconstant. Findthisdifferentialequation,andfindallitssolutions.

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wherek isaconstant.The

solution

behaves

differently

if

k

is

positive

or

negative.

For

k

positive,

this

equation

arises

in

populationgrowth and interest onsavings, among others. For k negative, thisequationarises inradioactivedecay,adischargingcapactiorinanRC circuit, andNewtonslawofcooling.Populationgrowth. ThesimplestmodelofpopulationgrowthisthatapopulationN(t)(modeledascontinuousforsimplicity)growsatarateproportionaltothesizeofthepopulation. Thisgives,

dN=kN.

dtFollowingthemethodgives,

dN/N =kdt,1/N

dN

=

kdt,

ln(|N )=kt+C.|

Exponentiatingbothsidesgives,N(t)=

Observe that N(t) increases without bound as t increases. When N is very large, the ecosystemcannotsupportsuchapopulation. ThusthemodelisonlyvalidifN(t)isnottoolarge.

N0e .kt

A slightly morerealisticmodelhypothesizesaconstant, equilibrium population Nequi sustainableindefinitely. ThemodelisthatthepopulationgrowsatarateproportionalbothtothepopulationN andthedifferenceNequiN,

dN=kN(NequiN).dt

Thisisagainaseparabledifferentialequation. Itgivesthesolution,N(t N0Nequi/(N0 NequiN0)e t).)= +( kNequi

ThemostimportantfeatureisthatN(t)approachesNequi astincreases. Thisiscalledthesteady state solution. In general, to find the steady state solution to a separable ordinary differentialequation, assume the solution is constant y = y1 so that dy/dt is 0. In the original model ofpopulationgrowth,theonlysteady state solutionisN =0. Inthenewmodel,thereare2steady statesolutions,N =0andN =Nequi. InMath18.03,stability isdefined,andamethod isgivento

show

the

only

stable

steady state

solution

is

N

=

Nequi.

Radioactive decay. Aradioactiveisotopedecaystoamorestableisotopeatarateproportionaltotheremainingradioactiveisotope. Thusthemassm(t)satisfiesadifferentialequation,

dmdt =km.

Usingthemethod,thesolutionis,m(t)= m0ekt.

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Animportantfeatureindecayproblemsisthehalf life. Thehalf life isthelengthoftimenecessaryforthemassofradioactiveisotopetodecreasetoone half theinitialmass,

m(Thalf)=m0/2.Solvingintheformulagives,

Thalf =Example. Thehalf life ofacertainradioactive isotope is20years. How long isrequired forthemasstodecreaseto1%ofthe initialmass? Usingtheformulaabove,k=ln(2)/25. Thereforetheequationforthemassis,

/k.ln(2)

m(t)=m0eln(2)t/25.Thusthetimetf whenthemassequals0.01m0 satisfies,

m0eln(2)tf/25 =m0/100,

or,ln(2)tf/25=ln(100)=2ln(10).

Solvinggives,tf =50ln(10)/ln(2)=

NewtonsLawofCooling. IsaacNewtonproposedalawfortherate of change ofthetempera tureT ofanobjectplacedinalarge,effectivelyinfinite,environmentatafixedambienttemperatureTamb. Thelawisthattherate of change ofT isnegativelyproportionaltothetemperaturegradient

166years.

TTamb,dT

=k(TTamb).dt

Themethodgivesthesolution,T(t)=Tamb +(TTamb)ekt.

Astincreases,thetemperatureT approachesthesteady state temperature,Tamb.Lecture 18. October25,2005Homework. ProblemSet5PartI:(c).Practice Problems. CourseReader: 3G 1, 3G 2, 3G 4, 3G 5. 1. ApproximatingRiemannintegrals. Often,thereisnosimplerexpressionfortheantideriva tivethantheexpressiongivenbytheFundamentalTheoremofCalculus. Insuchcases,thesimplestmethodtocomputeaRiemannintegralistousethedefinition. However,thisisnotnecessarilythemostefficient method. OftentrapezoidsorsegmentsunderaparabolagiveabetterapproximationtotheRiemannintegralthandoverticalstrips.

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2. The trapezoid rule. TheproblemistofindanapproximationoftheRiemannintegral, b

I= ydxa

forafunctiony(x)definedontheinterval[a, b]. Chooseapartitionoftheinterval[a, b]intonequalsubintervals. Thepointsofthispartitionare,

b axk =a + (b a)k , xk = .

n nThevaluesofthesepointsare,

yk =f(xk).TheRiemannsumusingalwaystheleftendpointis,

n

Il = yk1xk.k=1

TheRiemannsumusingalwaystherightendpointis,n

Ir = ykxk.k=1

Theaverageofthetwois,n

Itrap

= yk1 +ykxk

.2k=1

This is usually a better approximation than either of the two approximations individually. Partofthereason isthattheterm(yk1 +yk)xk/2 istheareaofthetrapezoid containingthepoints(xk1, 0), (xk1, yk1), (xk, 0) and (xk, yk). In particular, if the graph of y = f(x) is a line, thistrapezoidispreciselytheregionbetweenthegraphandthexaxis overtheinterval[xk1, xk]. Thus,theapproximationabovegivestheexact integralforlinearintegrands.Writingoutthesumgives,

Itrap = b a((y0 +y1)+(y1 +y2)+(y2 +y3)+ +(yn2 +yn1)+(yn1 +yn)).2n

Gatheringliketerms,thisreducesto,I = (b a)(y0 y1 y2 + yn1 +yn)/2n.trap +2 +2 +2

3. Simpsons rule. Againpartitiontheinterval[a, b]intonequalsubintervals. Forreasonsthatwillbecomeapparent,nmustbeeven. Soletn=2mwheremisapositiveinteger. Againdefine,

(b a)k (b a)k b a b axk =a + =a + , xk = = .

n 2m n 2m55


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Pairofftheintervalsas([x0, x1], [x1, x2]),([x2, x3], [x3, x4]),etc. Thusthelth pairofintervalsis,([x2l2, x2l1], [x2l1, x2l]).

The idea istoapproximatetheareaofthegraphoverthepairof intervalsbytheareaundertheunique parabola containing the 3 points (x2l2, y2l2), (x2l1, y2l1), (x2l, y2l). For notations sake,denote 2l1 by k. Thus the 3 points are (xk1, yk1), (xk, yk), and (xk+1, yk+1) (this is slightlymoresymmetric).The first problem is tofind theequation of this parabola. Since the parabola containsthe point(xk, yk),ithastheequation,

y=A(x xk)2 +B(x xk)+yk,Plugginginx=xk1 andx=xk+1,andusingthatxk+1xk =xkxk1 equalsx,

yk+1 =A(x)2 +B(x)+yk,yk1 =A(x)2B(x)+yk.

Summingthetwosidesgives,yk+1 +yk1 =2A(x)2 +2yk.

SolvingforAgives,1

A=2(x)2(yk12yk +yk+1).

Similarly,takingthedifferenceofthetwosidesgives,yk+1yk1 =2B(x).

SolvingforB gives,1

B=2(x)

(yk+1yk1).Thus,theequationoftheparabolapassingthrough(xk1, yk1),(xk, yk)and(xk+1, yk+1)is,

y=A(x xk)2 +B(x xk)2 +yk,A yk12yk +yk+1)/ x)2,

B yk+1yk1)/ x).=( 2(=( 2(Thenextproblemistocomputetheareaundertheparabolafromx=xk1 tox=xk+1. ThisisastraightforwardapplicationoftheFundamentalTheoremofCalculus,

xk+1 A BA(x xk)2 +B(x xk)+ykdx= (x xk)3 + (x xk)2 +yk(x xk)

3 2xk+1

.xk1 xk1

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Plugginginandusingthatxk+1 xk =xk xk1 equalsx,thisis,2A

(x)3 +2yk(x).3

SubstitutingintheformulaforAandsimplifying,thisis,x x x

(yk1 2yk +yk+1)+ (6yk)= (yk1 +4yk +yk+1).3 3 3

Back substituting 2l 1forkand(b a)/2mforx,theapproximateareaforthepairofintervals[x2l2, x2l2]and[x2l1, x2l]is,

Il = b a(y2l2 +4y2l1 +y2l).6m

Finally,summingthiscontributionovereachchoiceoflgivestheSimpsonsruleapproximation,m

b aISimpson = (y2l2 +4y2l1 +y2l).

6ml=1

Writingoutthesumgives,baISimpson = ((y0 +4y1 +y2)+(y2 +4y3 +y4)+(y4 +4y5 +y6)+6m

+(y2m4 +4y2m3 +y2m2)+(y2m2 +4y2m1 +y2m)).Gatheringliketerms,ISimpson reducesto,

(b a)(y0 y1 y2 y3 y4 y5 y6 + y2m3 y2m2 y2m1 +y2m)/6m.+4 +2 +4 +2 +4 +2 +4 +2 +4Example. Approximateln(2)usingapartitioninto4equalsubintervalswiththeTrapezoidRuleandwithSimpsonsRule.Thevalueln(2)equalstheRiemannintegral, 2 1

dx.1 x

The points of the partition are x0 = 4/4, x1 = 5/4, x2 = 6/4, x3 = 7/4 and x4 = 8/4. Thecorrespondingvaluesarey0 =4/4, y1 =4/5, y2 =4/6, y3 =4/7, y4 =4/8. ThustheTrapezoidRulegives,

1 4 4 4 4 4 1171Itrap = b a(y0 +2y1 +2y2 +2y3 +y4)= ( +2 +2 +2 + )=

16802n 8 4 5 6 7 8ForSimpsonsRule,becausenequals4,mequals2. Thus,

1 4 4 4 4 4 1747

0.6970

ISimpson = b a(y0 +4y1 +2y2 +4y3 +y4)= ( +4 +2 +4 + )=25206m 12 4 5 6 7 8 0.6933

57


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Substitutingthisintotheequationfors2 gives, 2 2 2

dx dx dx2s = + 2x =(1+4x2) .dt dt dt

Sincesisknowntobe3,andxisknowntobe2,thisequationcanbesolvedfordx/dt, 2dx 32 9

= = .dt 1+4(2)2 17

Sincetheparticle is inthefirstquadrantandmovingaway fromtheorigin,dx/dt ispositive. So,atthemomentwhenLequals25,dx/dtequals3/17.Thefinalstep istocomputedL/dtatthemomentwhenLequals25. Implicitlydifferentiatingtheequation,

2 4L2 =x +x ,gives,

dL dx2L =(2x+4x3) .

dt dtPlugginginforL,xanddx/dtgives,

dL 32(2

5) =(2(2)+4(2)3)

17.dtSolvinggives, dL

dt =atthemomentwhenLequals25.Lecture 19. October28,2005

27/

85.

Homework. ProblemSet5PartI:(d)and(e);PartII:Problems2and3.Practice Problems. CourseReader: 4A 1, 4A 3, 4B 1, 4B 3, 4B 6. 1. Differentialsrevisited. Inatypicalappliedintegrationproblem,themaindifficultyisfindingthe integrand and the limits of integration. An unknown quantity, for instance area A, dependson some other quantity, for instance the xcoordinate. The method is to allow the independentvariablexvaryinfinitesimallyfromxtox+dxandthenusegeometryorsciencetodeducethecorresponding variation dA of the unknown quantity. The deduction is usually intuitive ratherthanrigorous. Whatisimportantiswhetherthedeductionleadstothecorrectsolution. Ifso,themethodofRiemannsumsusuallygivesarigorousproofoftheintuitiveanswer. Butifthesolutionisincorrect,noargumentwillproveitcorrect.2. Areas between curves. Givenan intervalaxbandtwofunctionsf(x)g(x)definedon the interval, what is the area of the region bounded by the lines x=a,x=b and the curves

59


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y = f(x), y = g(x)? This problem can be solved directly: the area is the difference of the areabetweeny=f(x)andthexaxis andtheareabetweeny=g(x)andthexaxis. EachoftheseisaRiemannintegral.The differential method asks, what is the infinitesimal change in the area A as x varies from xto x+dx? The infinitesimal region is a rectangle of base dx and height f(x)g(x). Thus theinfinitesimalchangeinAis,

dA=heightbase=(f(x)g(x))dx.Integratinggives,

x=bA= dA= f(x)g(x)dx.

x=aOfcoursethisisthesameanswerasinthelastparagraph. Butforotherappliedintegralproblems,thedifferentialmethodmaybetheonlymethodthatworks.Example. Findtheareaboundedbythecurvey=x(x23)andahorizontaltangentline.The horizontal tangent lines are the tangent lines to the curve at critical points. Setting thederivativeequalto0gives,

dydx =3x23=3(x1)(x+1).

Thusthecriticalpointsarex=1. Thefunctionisodd,sosymmetrysuggeststheareaisthesameregardlessofthechoiceofcriticalpoint. Thus,choosethecriticalpointx=1. Thecorrespondingvalueofthefunctionis,

y=(1)((1)23)=(1)(2)=2.Thisistheequationofthehorizontaltangentline. Eachintersectionpoint(b,f(b))ofthetangentlinewithy=x(x23)occursatasolutionx=bof,

x(x23)=2.Byhypothesis,x=1isasolution. Thusthepolynomialfactorsas,

3x 3x2=(x+1)(x2x2)=(x+1)2(x2).Theremainingintersectionpointis(2,2). Sotheareaboundedbythecurvey=x(x2

3)andthe

tangentliney=2is,x=2 2

2(x(x23))dx= x3 +3x+2dx.x=1 1

UsingtheFundamentalTheoremofCalculus,thisequals,24 3x2x

+ +2x 27/4.=4 2 1

60


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3. Volumesofsolidsofrevolutions: thediskmethod. Iftheregioninthexyplane boundedbyx=a,x=b,y=f(x)andthexaxis isrevolvedthroughxyzspace aboutthexaxis, what isthevolumeoftheresultingsolid? Thesolidiscalledasolidofrevolution. Thediskmethod appliesthemethodofdifferentialstosolvethisproblem. Asxincreasesfromxtox+dx,thecorrespondinginfinitesimalregionofthesolid isessentiallyadisk. Thewidthofthedisk isdx. Theareaofthediskis[f(x)]2. Thustheinfinitesimalvolumeofthediskis,

dV =Areawidth=[f(x)]2dx.Thusthevolumeis,

x=bV = dV = [f(x)]2dx.

x=a

Example.

Findthe

volume

of

aright

circular

cone

whose

base

has

radius

R

and

whose

vertex

has

heightH abovethebase.Theconeisthesolidofrevolutionfortheplaneregionboundedbyx=0,thexaxis, andthelinecontaining(0,R)and(H,0). Theequationofthislineis,

(Hx)Ry= .

HThustheareaofaninfinitesimaldiskis,

dV =Areawidth=(Hx)2R2dx,H2

andthevolumeis,x=H (Hx)2R2

V = dV = dx.H2x=0

Makingthesubstitutionu=Hx,du=dxgives,0u=0 R2 R2 3u2V = (du)=

H2 u .H2 3u=H HEvaluatinggivesthevolume,

V = 2 3.R H/Inparticular, the antiderivative ofu2 is responsible for thedenominator 3 in the formula for thevolume.Example. FindthevolumeofasphereofradiusR.The sphere of radius R is the solid of revolution for the plane region bounded by the xaxis and

2theuppersemicircley=R2x . Thusthevolumeis, R Rx=R 3R2 22]2dx= (R2x )dx= R2xxV = [ x .

3R R Rx=61


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Evaluatinggivesthevolume,4 3/3.RV =

4. The slice method. A generalization of the disk method is the slice method. The problemis to find the volume of a region bounded by the planes x = a and x = b given the knowledgeoftheareaA(x)oftheslice ofthesolidbytheplanecontaining(x,0,0)paralleltotheyzplane. As x increases from x to x+dx, the corresponding infinitesimal region of the solid is essentiallyacylinder. Thewidthofthecylinder isdx. Andthearea istheareaA(x)oftheslice. Thustheinfinitesimalvolumeofthecylinderis,

dV =Areawidth=A(x)dx.Thusthevolumeis,

x=bV = dV = A(x)dx.

x=aExample. Find the volume of the corner region bounded by the xyplane, the xzplane, theyzplane, andtheplanecontaining(L,0,0),(0,L,0)and(0,0,L).Thisregionisboundedbytheplanesx=0andx=L. Thexslice oftheregionisarightisoscelestriangle. Thebaseandaltitudeofthetrianglebothequalf(x),wherey=f(x)istheequationofthelinepassingthrough(0,L)and(L,0). Thisequationis,

f(x)=Lx.Thusthesliceareais

1 1A(x)= basealtitude= (Lx)2.

2 2Thustheinfinitesimalvolumeis,

1dV =A(x)dx= (Lx)2dx,

2givingthetotalvolume,

x=L 1(Lx)2dx.

2V = dV = x=0Makethesubstitutionu=Lx,du=dxtoget,

Lu=0 31 2 1 uV (du)=2 = u .2 3u=L 0

Evaluatinggives,V = L3/6.

ThusthecornertakesuponesixthofthecorrespondingcubeofedgelengthL.62


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5. Volumes of solids of revolution: the washer method. A variation on the disk methodisthewasher method. Awasher isthesolidobtainedbyremovingfroma largerdiskaconcentricsmallerdiskofthesamewidth. Iftheouterradiusofthewasher isro andthe innerradius isri,thenthenetareaofthewasheris,

A=r2ro 2i =(r2 2).iroThusthevolumeofthewasheris,

dV =Areawidth=(ro r2 2)dx,igivingatotalvolume,

x=b

V = dV = (r2 2)dx.irox=a

Example. The main part of a dog dish is a solid of revolution whose radial cross section is atriangle of height H whose base has inner radius Ri and outer radius Ro. Find the volume ofmaterialusedtomakethedogdish.Note. This integral was only set up in lecture. The derivation will be completed in recitation.Hereisthecompletederivation. Denotebyxtheheightalongthealtitudeofthetriangle. Thusxvariesfromx=0tox=H. Whenx=H,theinnerradiusandouterradiusareeachequaltotheaverage(Ri +Ro)/2ofthetworadii. Bothradiidependlinearlyonx.Theequationforthe innerradius increases linearly fromri =Ri atx=0tori = (Ri +Ro)/2atx=H. Thus,

Ri +Ro xri(x)=RiHx+ .H 2 H

Similarly, the equation for the outer radius decreases linearly from ro = Ro at x = 0 to ri =(Ri +Ro)/2atx=H. Thus,

Ri +Ro xro(x)=RoHx+ .

H 2 H2Sincero r2i isadifferenceofsquares,itequals,

2 2i =(ro +ri)(rori).rro

Thisisinterestinginitsownright. Usingthisfactorization,thenetareaoftheregionbetweenthe2circles,calledanannulus,equals

ro +ri2 2)=i 2(r (rori).ro 2Notethefirstfactoristhecircumferenceofthecenteroftheannulus. Andthesecondfactoristheradialwidthoftheannulus. Thustheareaofanannulus isthecircumferenceofthecentertimestheradialwidth.

63


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Backtotheproblem,thecenterofeachwasheristhesame,ri

+

ro

Ri

+

Ro

= .2 2

Andtheradialwidthofeachwasheris,rori =(RoRi)Hx.

HBothofthesemakesense: Thecentersareconstantbecausetheradialcross section isanisoscelestriangle. And the width decreases linearly from Ro Ri at x = 0 to 0 at x = H. Thus thecross section areaatheightxis,

Ri +Ro(Ro

Ri)HxA(x)=2 .

2

H

Thustheinfinitesimalvolumeis,dV =Areawidth=(R2oRi2)Hxdx,H

givingatotalarea,x=H

(R2oRi2)HxV = dV dx.= Hx=0Substitutingu=Hx,du=

V

dxgives,u=H

oRi

2

)u (R2

(R

2dx

=

o

Ri2) 2 H=

u

.

0H 2Hu=0Thusthetotalvolumeofmaterialtoproducethedogdishis,

V = (R2oR2i) 2.H/Onerealitycheck isthatthis isthesamevolumeasacylinderwiththesamecenter(Ri +Ro)/2andheightH asthedish,andwhose(constant)radialwidthequalstheaverageradialwidthofthedish,(RoRi)/2.Lecture 20. November1,2005Practice Problems. CourseReader: 4C 2, 4C 6, 4D 1, 4D 4, 4D 8. 1. Average values. Given a function f(x) defined on some interval [a,b], what is the averagevalueoff(x)? Areasonablefirstapproximationistochooseafinitecollectionofpointsfrom[a,b]and compute the average value over those points. Break [a,b] into a union of n subintervals of

k in the kth interval. For thelength x

k

(ba)/n.finitelymanyvaluesy From each interval, choose a point; say x= ),theaveragevalueis,k=f(x

n

Average yn

k=164

1 .k


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Multiplyinganddividingbyxgives,n

1 Average yx.nx kk=1

Nownxequalsn(ab)/n,whichisab. Sotheaveragevalueis,n

1 Average yx.

ba kk=1

ThesumisaRiemannsum. Togetbetterapproximationstothetrueaverage,increasethenumberofpointsnatwhichf(x)issampled. Inthelimit,thisgivesthetrueaverage,

n1 b

Average= lim yx= f(x)dx/(ba).abank=1 kExample. Under idealconditions,awire producing machineproduceswireofuniformradiusr0.Becauseofsmallvibrationsinthemachine,theactualradiusofthewirevariesasafunctionofthelength,

r(x)=r0 +Acos(x).ThequantityAismuchsmallerthanr0. Whatistheaverageradiusofthewire?Becausethevariationisperiodic,theaveragevalueoveranynumberofperiodsequalstheaveragevalue of one period. In other words, compute the average for the interval 0 x 2/. Thelength

of

this

interval

is

2/.

Thus

the

average

value

is,

2/1Average= r0 +Acos(x)dx.

(2/) 0UsingtheFundamentalTheoremofCalculus,thisequals,

1 2/(2/)(r0x+(A/)sin(x)|0 .

Thisevaluatesto,1

r0(2/)=r0.(2/)

Thus,althoughtheradiusvariesanddoesnotusuallyequalitsidealvaluer0,theaveragevalueisindeed,

Average= r0.2. Average values: non uniform distribution. It often happens that the average value of f(x) isdesiredinasituationwherethevaluesf(x)arenotalluniformlylikely. Typically,theprobabilitythatxhasvalueintherangefromx0 tox0 +xisapproximately,

Prob(x0 xx0 +x)(x0)x,65


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forsomenonnegativecontinuousfunction(x). Thefunction(x)iscalledaprobabilitydistribution.Assumingthisapproximationbecomesarbitrarilygoodasthelengthxapproacheszero,theexactprobabilitythatxhasvalueintherangex0 tox1 is,

x1Prob(x0 xx1)= (x)dx.

x0

Inparticular,becausexmusttakevaluesomewhereintheinterval[a,b],thetotalprobabilityis1.Inotherwords, b

(x)dx=1.a

Thisiscalledthenormalizationcondition.Theaveragevalue iscomputedasbefore. Butthistime,eachvaluey

=f(x

k) isweightedbythekapproximateprobabilitythatxtakesvalueinthekth interval,(xk)x. Thisgives,

n

Average f(xk)(xk)x.k=1

Inthelimitasngoesto,thisgivestheexactaverage, bAverage= f(x)(x)dx.

a

Itmustbenoted,theprobabilitydistribution(x)oftendoesnotsatisfythenormalizationcondi tion. Inthiscase,theformulaaboveiswrong. Butitiseasilycorrect,(ba f(x)(x)dx)/(ba (x)dx).Average=

Example. Aparticle isfiredthroughaslitandstrikesascreenontheotherside. Measuringthepositiononthescreensothattheoriginistheclosestpointonthescreentotheslit,theprobabilitydistributionisempiricallyobservedtobe,

(x)=Cex2/22,where isaconstantdeterminingthewidthoftheprobabilitydistribution,andC isanunde terminednormalizationconstant. What istheaveragedistanceoftheparticle fromthecenterofthescreen? Assumetheparticleliesinaninterval[R,R],whereRisverylarge.Remark. Thisdiffersfromtheformulagiveninlecture,whichwasCex2/2 foraparticularchoiceof. Theformulagivenhereismorestandard. Iapologizeforanyconfusion.Thedistancefunctionis,

f(x)=|x|= x,x, x


67/134


Accordingtotheformula,theaveragevalueis,

R

R

( f(x)(x)dx)/( (x)dx).R R

Thenumeratoris, R|x|Cex2/22dx.

RItiseasiesttocomputethisbybreakingitintoasumof2integrals, 0 R

(x)Cex2/22dx+ (+x)Cex2/22dx.0R

Makethesubstitutionu=x2/22,du=(x/2)dxto

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