lesson 23: areas and distances (section 10 version)

. . . . . .

Section5.1AreasandDistances

V63.0121, CalculusI

April13, 2009

Announcements

I Movingto624today(noOH)I Quiz5thisweekon§§4.1–4.4I BackHW returnedinrecitationthisweek

. . . . . .

Outline

Archimedes

Cavalieri

GeneralizingCavalieri’smethod

Distances

Otherapplications

. . . . . .

Meetthemathematician: Archimedes

I 287BC –212BC (afterEuclid)

I GeometerI Weaponsengineer

. . . . . .

.

Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.

A =

1 + 2 · 18

+ 4 · 164

+ · · ·

= 1 +14

+116

+ · · · + 14n

+ · · ·

. . . . . .

.

.1


A = 1

+ 2 · 18

+ 4 · 164

+ · · ·

= 1 +14

+116

+ · · · + 14n

+ · · ·

. . . . . .

.

.1.18 .18


A = 1 + 2 · 18

+ 4 · 164

+ · · ·

= 1 +14

+116

+ · · · + 14n

+ · · ·

. . . . . .

.

.1.18 .18

.164 .164

.164 .164


A = 1 + 2 · 18

+ 4 · 164

+ · · ·

= 1 +14

+116

+ · · · + 14n

+ · · ·

. . . . . .

Wewouldthenneedtoknowthevalueoftheseries

1 +14

+116

+ · · · + 14n

+ · · ·

Butforanynumber r andanypositiveinteger n,

(1− r)(1 + r + · · · + rn) = 1− rn+1

So

1 + r + · · · + rn =1− rn+1

1− r

Therefore

1 +14

+116

+ · · · + 14n

=1− (1/4)n+1

1− 1/4→ 1

3/4=

43

as n → ∞.

. . . . . .


1 +14

+116

+ · · · + 14n

+ · · ·


(1− r)(1 + r + · · · + rn) = 1− rn+1

So

1 + r + · · · + rn =1− rn+1

1− r

Therefore

1 +14

+116

+ · · · + 14n

=1− (1/4)n+1

1− 1/4

→ 13/4

=43

as n → ∞.

. . . . . .


1 +14

+116

+ · · · + 14n

+ · · ·


(1− r)(1 + r + · · · + rn) = 1− rn+1

So

1 + r + · · · + rn =1− rn+1

1− r

Therefore

1 +14

+116

+ · · · + 14n

=1− (1/4)n+1

1− 1/4→ 1

3/4=

43

as n → ∞.

. . . . . .

Outline

Archimedes

Cavalieri


Distances

Otherapplications

. . . . . .

Cavalieri

I Italian,1598–1647

I Revisitedtheareaproblemwithadifferentperspective

. . . . . .

Cavalieri’smethod

.

.y = x2

Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:

L2 =18

L3 =

127

+427

=527

L4 =

164

+464

+964

=1464

L5 =

1125

+4

125+

9125

+1625

=30125

Ln =?

. . . . . .

Cavalieri’smethod

.

.y = x2


L2 =18

L3 =127

+427

=527

L4 =

164

+464

+964

=1464

L5 =

1125

+4

125+

9125

+1625

=30125

Ln =?

. . . . . .

Cavalieri’smethod

.

.y = x2


L2 =18

L3 =127

+427

=527

L4 =164

+464

+964

=1464

L5 =

1125

+4

125+

9125

+1625

=30125

Ln =?

. . . . . .

Cavalieri’smethod

.

.y = x2


L2 =18

L3 =127

+427

=527

L4 =164

+464

+964

=1464

L5 =1

125+

4125

+9

125+

1625

=30125

Ln =?

. . . . . .

Whatis Ln?Dividetheinterval [0, 1] into n pieces. Theneachhaswidth

1n.

Therectangleoverthe ithintervalandundertheparabolahasarea

1n·(i− 1n

)2

=(i− 1)2

n3.

So

Ln =1n3

+22

n3+ · · · + (n− 1)2

n3=

1 + 22 + 32 + · · · + (n− 1)2

n3

TheArabsknewthat

1 + 22 + 32 + · · · + (n− 1)2 =n(n− 1)(2n− 1)

6

So

Ln =n(n− 1)(2n− 1)

6n3→ 1

3as n → ∞.

. . . . . .


1n.


1n·(i− 1n

)2

=(i− 1)2

n3.

So

Ln =1n3

+22

n3+ · · · + (n− 1)2

n3=

1 + 22 + 32 + · · · + (n− 1)2

n3

TheArabsknewthat

1 + 22 + 32 + · · · + (n− 1)2 =n(n− 1)(2n− 1)

6

So

Ln =n(n− 1)(2n− 1)

6n3

→ 13

as n → ∞.

. . . . . .


1n.


1n·(i− 1n

)2

=(i− 1)2

n3.

So

Ln =1n3

+22

n3+ · · · + (n− 1)2

n3=

1 + 22 + 32 + · · · + (n− 1)2

n3

TheArabsknewthat

1 + 22 + 32 + · · · + (n− 1)2 =n(n− 1)(2n− 1)

6

So

Ln =n(n− 1)(2n− 1)

6n3→ 1

3as n → ∞.

. . . . . .

Cavalieri’smethodfordifferentfunctions

Trythesametrickwith f(x) = x3. Wehave

Ln =1n· f

(1n

)+

1n· f

(2n

)+ · · · + 1

n· f

(n− 1n

)

=1n· 1n3

+1n· 2

3

n3+ · · · + 1

n· (n− 1)3

n3

=1 + 23 + 33 + · · · + (n− 1)3

n3

Theformulaoutofthehatis

1 + 23 + 33 + · · · + (n− 1)3 =[12n(n− 1)

]2So

Ln =n2(n− 1)2

4n4→ 1

4as n → ∞.

. . . . . .



Ln =1n· f

(1n

)+

1n· f

(2n

)+ · · · + 1

n· f

(n− 1n

)=

1n· 1n3

+1n· 2

3

n3+ · · · + 1

n· (n− 1)3

n3

=1 + 23 + 33 + · · · + (n− 1)3

n3


1 + 23 + 33 + · · · + (n− 1)3 =[12n(n− 1)

]2So

Ln =n2(n− 1)2

4n4→ 1

4as n → ∞.

. . . . . .



Ln =1n· f

(1n

)+

1n· f

(2n

)+ · · · + 1

n· f

(n− 1n

)=

1n· 1n3

+1n· 2

3

n3+ · · · + 1

n· (n− 1)3

n3

=1 + 23 + 33 + · · · + (n− 1)3

n3


1 + 23 + 33 + · · · + (n− 1)3 =[12n(n− 1)

]2

So

Ln =n2(n− 1)2

4n4→ 1

4as n → ∞.

. . . . . .



Ln =1n· f

(1n

)+

1n· f

(2n

)+ · · · + 1

n· f

(n− 1n

)=

1n· 1n3

+1n· 2

3

n3+ · · · + 1

n· (n− 1)3

n3

=1 + 23 + 33 + · · · + (n− 1)3

n3


1 + 23 + 33 + · · · + (n− 1)3 =[12n(n− 1)

]2So

Ln =n2(n− 1)2

4n4→ 1

4as n → ∞.

. . . . . .

Cavalieri’smethodwithdifferentheights

.

Rn =1n· 1

3

n3+

1n· 2

3

n3+ · · · + 1

n· n

3

n3

=13 + 23 + 33 + · · · + n3

n4

=1n4

[12n(n + 1)

]2=

n2(n + 1)2

4n4→ 1

4

as n → ∞.

Soeventhoughtherectanglesoverlap, westillgetthesameanswer.

. . . . . .

Cavalieri’smethodwithdifferentheights

.

Rn =1n· 1

3

n3+

1n· 2

3

n3+ · · · + 1

n· n

3

n3

=13 + 23 + 33 + · · · + n3

n4

=1n4

[12n(n + 1)

]2=

n2(n + 1)2

4n4→ 1

4

as n → ∞.Soeventhoughtherectanglesoverlap, westillgetthesameanswer.

. . . . . .

Outline

Archimedes

Cavalieri


Distances

Otherapplications

. . . . . .

Cavalieri’smethodingeneralLet f beapositivefunctiondefinedontheinterval [a,b]. Wewanttofindtheareabetween x = a, x = b, y = 0, and y = f(x).Foreachpositiveinteger n, divideuptheintervalinto n pieces.

Then ∆x =b− an

. Foreach i between 1 and n, let xi bethe nth

stepbetween a and b. So

..a .b. . . . . . ..x0 .x1 .x2 .xi.xn−1.xn

x0 = a

x1 = x0 + ∆x = a +b− an

x2 = x1 + ∆x = a + 2 · b− an

· · · · · ·

xi = a + i · b− an

· · · · · ·

xn = a + n · b− an

= b

. . . . . .

FormingRiemannsums

Wehavemanychoicesofhowtoapproximatethearea:

Ln = f(x0)∆x + f(x1)∆x + · · · + f(xn−1)∆x

Rn = f(x1)∆x + f(x2)∆x + · · · + f(xn)∆x

Mn = f(x0 + x1

2

)∆x + f

(x1 + x2

2

)∆x + · · · + f

(xn−1 + xn

2

)∆x

Ingeneral, choose ci tobeapointinthe ithinterval [xi−1, xi].Formthe Riemannsum

Sn = f(c1)∆x + f(c2)∆x + · · · + f(cn)∆x

=n∑

i=1

f(ci)∆x

. . . . . .

TheoremoftheDay

TheoremIf f isacontinuousfunctionon [a,b] orhasfinitelymanyjumpdiscontinuities, then

limn→∞

Sn = limn→∞

{f(c1)∆x + f(c2)∆x + · · · + f(cn)∆x}

existsandisthesamevaluenomatterwhatchoiceof ci wemade.

. . . . . .

Outline

Archimedes

Cavalieri


Distances

Otherapplications

. . . . . .

Distances

Justlike area = length×width, wehave

distance = rate× time.

SohereisanotheruseforRiemannsums.

. . . . . .

ExampleA sailingshipiscruisingbackandforthalongachannel(inastraightline). Atnoontheship’spositionandvelocityarerecorded, butshortlythereafterastormblowsinandpositionisimpossibletomeasure. Thevelocitycontinuestoberecordedatthirty-minuteintervals.

Time 12:00 12:30 1:00 1:30 2:00Speed(knots) 4 8 12 6 4Direction E E E E W

Time 2:30 3:00 3:30 4:00Speed 3 3 5 9Direction W E E E

Estimatetheship’spositionat4:00pm.

. . . . . .

SolutionWeestimatethatthespeedof4knots(nauticalmilesperhour)ismaintainedfrom12:00until12:30. Sooverthistimeintervaltheshiptravels (

4 nmihr

) (12hr

)= 2 nmi

Wecancontinueforeachadditionalhalfhourandget

distance = 4× 1/2 + 8× 1/2 + 12× 1/2

+ 6× 1/2− 4× 1/2− 3× 1/2 + 3× 1/2 + 5× 1/2

= 15.5

Sotheshipis 15.5 nmi eastofitsoriginalposition.

. . . . . .

Analysis

I Thismethodofmeasuringpositionbyrecordingvelocityisknownas deadreckoning.

I Ifwehadvelocityestimatesatfinerintervals, we’dgetbetterestimates.

I Ifwehadvelocityateveryinstant, alimitwouldtellusourexactpositionrelativetothelasttimewemeasuredit.

. . . . . .

Outline

Archimedes

Cavalieri


Distances

Otherapplications

. . . . . .

OtherusesofRiemannsums

Anythingwithaproduct!I Area, volumeI Anythingwithadensity: Population, massI Anythingwitha“speed:” distance, throughput, powerI ConsumersurplusI Expectedvalueofarandomvariable

. . . . . .

Summary

I Riemannsumscanbeusedtoestimateareas, distance, andotherquantities

I ThelimitofRiemannsumscangettheexactvalueI Comingup: givingthislimitanameandworkingwithit.

lesson 23: areas and distances (section 10 version)

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