Lesson 23: Areas And Distances (Section 4 version)
50
. . . . . . Section 5.1 Areas and Distances V63.0121, Calculus I April 14, 2009 Announcements I Moving to 624 today I Quiz 5 Thursday on §§4.1–4.4
-
Upload
matthew-leingang -
Category
Technology
-
view
406 -
download
0
description
We trace the computation of area through the centuries. The process known known as Riemann Sums has applications to not just area but many fields of science.
Transcript of Lesson 23: Areas And Distances (Section 4 version)
. . . . . .
Section5.1AreasandDistances
V63.0121, CalculusI
April14, 2009
Announcements
I Movingto624todayI Quiz5Thursdayon§§4.1–4.4
. . . . . .
Outline
Archimedes
Cavalieri
GeneralizingCavalieri’smethod
Distances
Otherapplications
. . . . . .
Meetthemathematician: Archimedes
I 287BC –212BC (afterEuclid)
I GeometerI Weaponsengineer
. . . . . .
Meetthemathematician: Archimedes
I 287BC –212BC (afterEuclid)
I GeometerI Weaponsengineer
. . . . . .
Meetthemathematician: Archimedes
I 287BC –212BC (afterEuclid)
I GeometerI Weaponsengineer
. . . . . .
.
Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.
A =
1 + 2 · 18
+ 4 · 164
+ · · ·
= 1 +14
+116
+ · · · + 14n
+ · · ·
. . . . . .
.
.1
Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.
A = 1
+ 2 · 18
+ 4 · 164
+ · · ·
= 1 +14
+116
+ · · · + 14n
+ · · ·
. . . . . .
.
.1.18 .18
Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.
A = 1 + 2 · 18
+ 4 · 164
+ · · ·
= 1 +14
+116
+ · · · + 14n
+ · · ·
. . . . . .
.
.1.18 .18
.164 .164
.164 .164
Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.
A = 1 + 2 · 18
+ 4 · 164
+ · · ·
= 1 +14
+116
+ · · · + 14n
+ · · ·
. . . . . .
.
.1.18 .18
.164 .164
.164 .164
Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.
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 → ∞.
. . . . . .
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 → ∞.
. . . . . .
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
→ 13/4
=43
as n → ∞.
. . . . . .
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 → ∞.
. . . . . .
Outline
Archimedes
Cavalieri
GeneralizingCavalieri’smethod
Distances
Otherapplications
. . . . . .
Cavalieri
I Italian,1598–1647
I Revisitedtheareaproblemwithadifferentperspective
. . . . . .
Cavalieri’smethod
.
.y = x2
..0
..1
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =
127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+16125
=30125
Ln =?
. . . . . .
Cavalieri’smethod
.
.y = x2
..0
..1
.
.12
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =
127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+16125
=30125
Ln =?
. . . . . .
Cavalieri’smethod
.
.y = x2
..0
..1
.
.13
.
.23
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =
127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+16125
=30125
Ln =?
. . . . . .
Cavalieri’smethod
.
.y = x2
..0
..1
.
.13
.
.23
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+16125
=30125
Ln =?
. . . . . .
Cavalieri’smethod
.
.y = x2
..0
..1
.
.14
.
.24
.
.34
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+16125
=30125
Ln =?
. . . . . .
Cavalieri’smethod
.
.y = x2
..0
..1
.
.14
.
.24
.
.34
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+16125
=30125
Ln =?
. . . . . .
Cavalieri’smethod
.
.y = x2
..0
..1
.
.15
.
.25
.
.35
.
.45
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+16125
=30125
Ln =?
. . . . . .
Cavalieri’smethod
.
.y = x2
..0
..1
.
.15
.
.25
.
.35
.
.45
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =1
125+
4125
+9
125+
16125
=30125
Ln =?
. . . . . .
Cavalieri’smethod
.
.y = x2
..0
..1.
.
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =1
125+
4125
+9
125+
16125
=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 → ∞.
. . . . . .
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 → ∞.
. . . . . .
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 → ∞.
. . . . . .
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
→ 13
as n → ∞.
. . . . . .
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 → ∞.
. . . . . .
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
n4
Theformulaoutofthehatis
1 + 23 + 33 + · · · + (n− 1)3 =[12n(n− 1)
]2So
Ln =n2(n− 1)2
4n4→ 1
4as 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
n4
Theformulaoutofthehatis
1 + 23 + 33 + · · · + (n− 1)3 =[12n(n− 1)
]2So
Ln =n2(n− 1)2
4n4→ 1
4as 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
n4
Theformulaoutofthehatis
1 + 23 + 33 + · · · + (n− 1)3 =[12n(n− 1)
]2So
Ln =n2(n− 1)2
4n4→ 1
4as 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
n4
Theformulaoutofthehatis
1 + 23 + 33 + · · · + (n− 1)3 =[12n(n− 1)
]2
So
Ln =n2(n− 1)2
4n4→ 1
4as 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
n4
Theformulaoutofthehatis
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
GeneralizingCavalieri’smethod
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
. . . . . .
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
GeneralizingCavalieri’smethod
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
GeneralizingCavalieri’smethod
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.
