Transcript
Page 1: Lesson 13: Related Rates of Change

. . . . . .

Section2.7RelatedRates

V63.0121.027, CalculusI

October20, 2009

Announcements

I Midtermaverage57.69/75(77%), median59/75(79%),standarddeviation11%

I Solutionssoon.

Page 2: Lesson 13: Related Rates of Change

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“Isthereacurve?”

I MidtermI Meanwas77%and

standarddeviationwas11%

I Soscoresaveragearegood

I Scoresabove66/75(88%)aregreat

I Forfinallettergrades,refertosyllabus

Page 3: Lesson 13: Related Rates of Change

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Whatarerelatedratesproblems?

Todaywe’lllookatadirectapplicationofthechainruletoreal-worldproblems. Examplesofthesecanbefoundwheneveryouhavesomesystemorobjectchanging, andyouwanttomeasuretherateofchangeofsomethingrelatedtoit.

Page 4: Lesson 13: Related Rates of Change

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Problem

ExampleAnoilslickintheshapeofadiskisgrowing. Atacertaintime,theradiusis1kmandthevolumeisgrowingattherateof10,000literspersecond. Iftheslickisalways20cmdeep, howfastistheradiusofthediskgrowingatthesametime?

Page 5: Lesson 13: Related Rates of Change

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A solution

Thevolumeofthediskis

V = πr2h.

WearegivendVdt

, acertain

valueof r, andtheobjectis

tofinddrdt

atthatinstant.

. .r.h

Page 6: Lesson 13: Related Rates of Change

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Solution

SolutionDifferentiating V = πr2h withrespecttotimewehave

dVdt

= 2πrhdrdt

+ πr2����0

dhdt

=⇒ drdt

=1

2πrh· dVdt

.

Nowweevaluate:

drdt

∣∣∣∣r=1 km

=1

2π(1 km)(20 cm)· 10, 000 L

s

Convertingeverylengthtometerswehave

drdt

∣∣∣∣r=1 km

=1

2π(1000m)(0.2m)· 10m

3

s=

140π

ms

Page 7: Lesson 13: Related Rates of Change

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Solution

SolutionDifferentiating V = πr2h withrespecttotimewehave

dVdt

= 2πrhdrdt

+ πr2����0

dhdt

=⇒ drdt

=1

2πrh· dVdt

.

Nowweevaluate:

drdt

∣∣∣∣r=1 km

=1

2π(1 km)(20 cm)· 10, 000 L

s

Convertingeverylengthtometerswehave

drdt

∣∣∣∣r=1 km

=1

2π(1000m)(0.2m)· 10m

3

s=

140π

ms

Page 8: Lesson 13: Related Rates of Change

. . . . . .

Solution

SolutionDifferentiating V = πr2h withrespecttotimewehave

dVdt

= 2πrhdrdt

+ πr2����0

dhdt

=⇒ drdt

=1

2πrh· dVdt

.

Nowweevaluate:

drdt

∣∣∣∣r=1 km

=1

2π(1 km)(20 cm)· 10, 000 L

s

Convertingeverylengthtometerswehave

drdt

∣∣∣∣r=1 km

=1

2π(1000m)(0.2m)· 10m

3

s=

140π

ms

Page 9: Lesson 13: Related Rates of Change

. . . . . .

Solution

SolutionDifferentiating V = πr2h withrespecttotimewehave

dVdt

= 2πrhdrdt

+ πr2����0

dhdt

=⇒ drdt

=1

2πrh· dVdt

.

Nowweevaluate:

drdt

∣∣∣∣r=1 km

=1

2π(1 km)(20 cm)· 10, 000 L

s

Convertingeverylengthtometerswehave

drdt

∣∣∣∣r=1 km

=1

2π(1000m)(0.2m)· 10m

3

s=

140π

ms

Page 10: Lesson 13: Related Rates of Change

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Outline

Strategy

Examples

Page 11: Lesson 13: Related Rates of Change

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StrategiesforProblemSolving

1. Understandtheproblem

2. Deviseaplan

3. Carryouttheplan

4. Reviewandextend

GyörgyPólya(Hungarian, 1887–1985)

Page 12: Lesson 13: Related Rates of Change

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StrategiesforRelatedRatesProblems

1. Readtheproblem.

2. Drawadiagram.

3. Introducenotation. Givesymbolstoallquantitiesthatarefunctionsoftime(andmaybesomeconstants)

4. Expressthegiveninformationandtherequiredrateintermsofderivatives

5. Writeanequationthatrelatesthevariousquantitiesoftheproblem. Ifnecessary, usethegeometryofthesituationtoeliminateallbutoneofthevariables.

6. UsetheChainRuletodifferentiatebothsideswithrespecttot.

7. Substitutethegiveninformationintotheresultingequationandsolvefortheunknownrate.

Page 13: Lesson 13: Related Rates of Change

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StrategiesforRelatedRatesProblems

1. Readtheproblem.

2. Drawadiagram.

3. Introducenotation. Givesymbolstoallquantitiesthatarefunctionsoftime(andmaybesomeconstants)

4. Expressthegiveninformationandtherequiredrateintermsofderivatives

5. Writeanequationthatrelatesthevariousquantitiesoftheproblem. Ifnecessary, usethegeometryofthesituationtoeliminateallbutoneofthevariables.

6. UsetheChainRuletodifferentiatebothsideswithrespecttot.

7. Substitutethegiveninformationintotheresultingequationandsolvefortheunknownrate.

Page 14: Lesson 13: Related Rates of Change

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StrategiesforRelatedRatesProblems

1. Readtheproblem.

2. Drawadiagram.

3. Introducenotation. Givesymbolstoallquantitiesthatarefunctionsoftime(andmaybesomeconstants)

4. Expressthegiveninformationandtherequiredrateintermsofderivatives

5. Writeanequationthatrelatesthevariousquantitiesoftheproblem. Ifnecessary, usethegeometryofthesituationtoeliminateallbutoneofthevariables.

6. UsetheChainRuletodifferentiatebothsideswithrespecttot.

7. Substitutethegiveninformationintotheresultingequationandsolvefortheunknownrate.

Page 15: Lesson 13: Related Rates of Change

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StrategiesforRelatedRatesProblems

1. Readtheproblem.

2. Drawadiagram.

3. Introducenotation. Givesymbolstoallquantitiesthatarefunctionsoftime(andmaybesomeconstants)

4. Expressthegiveninformationandtherequiredrateintermsofderivatives

5. Writeanequationthatrelatesthevariousquantitiesoftheproblem. Ifnecessary, usethegeometryofthesituationtoeliminateallbutoneofthevariables.

6. UsetheChainRuletodifferentiatebothsideswithrespecttot.

7. Substitutethegiveninformationintotheresultingequationandsolvefortheunknownrate.

Page 16: Lesson 13: Related Rates of Change

. . . . . .

StrategiesforRelatedRatesProblems

1. Readtheproblem.

2. Drawadiagram.

3. Introducenotation. Givesymbolstoallquantitiesthatarefunctionsoftime(andmaybesomeconstants)

4. Expressthegiveninformationandtherequiredrateintermsofderivatives

5. Writeanequationthatrelatesthevariousquantitiesoftheproblem. Ifnecessary, usethegeometryofthesituationtoeliminateallbutoneofthevariables.

6. UsetheChainRuletodifferentiatebothsideswithrespecttot.

7. Substitutethegiveninformationintotheresultingequationandsolvefortheunknownrate.

Page 17: Lesson 13: Related Rates of Change

. . . . . .

StrategiesforRelatedRatesProblems

1. Readtheproblem.

2. Drawadiagram.

3. Introducenotation. Givesymbolstoallquantitiesthatarefunctionsoftime(andmaybesomeconstants)

4. Expressthegiveninformationandtherequiredrateintermsofderivatives

5. Writeanequationthatrelatesthevariousquantitiesoftheproblem. Ifnecessary, usethegeometryofthesituationtoeliminateallbutoneofthevariables.

6. UsetheChainRuletodifferentiatebothsideswithrespecttot.

7. Substitutethegiveninformationintotheresultingequationandsolvefortheunknownrate.

Page 18: Lesson 13: Related Rates of Change

. . . . . .

StrategiesforRelatedRatesProblems

1. Readtheproblem.

2. Drawadiagram.

3. Introducenotation. Givesymbolstoallquantitiesthatarefunctionsoftime(andmaybesomeconstants)

4. Expressthegiveninformationandtherequiredrateintermsofderivatives

5. Writeanequationthatrelatesthevariousquantitiesoftheproblem. Ifnecessary, usethegeometryofthesituationtoeliminateallbutoneofthevariables.

6. UsetheChainRuletodifferentiatebothsideswithrespecttot.

7. Substitutethegiveninformationintotheresultingequationandsolvefortheunknownrate.

Page 19: Lesson 13: Related Rates of Change

. . . . . .

StrategiesforRelatedRatesProblems

1. Readtheproblem.

2. Drawadiagram.

3. Introducenotation. Givesymbolstoallquantitiesthatarefunctionsoftime(andmaybesomeconstants)

4. Expressthegiveninformationandtherequiredrateintermsofderivatives

5. Writeanequationthatrelatesthevariousquantitiesoftheproblem. Ifnecessary, usethegeometryofthesituationtoeliminateallbutoneofthevariables.

6. UsetheChainRuletodifferentiatebothsideswithrespecttot.

7. Substitutethegiveninformationintotheresultingequationandsolvefortheunknownrate.

Page 20: Lesson 13: Related Rates of Change

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Outline

Strategy

Examples

Page 21: Lesson 13: Related Rates of Change

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Anotherone

ExampleA manstartswalkingnorthat4ft/sec fromapoint P. Fiveminuteslaterawomanstartswalkingsouthat4ft/sec fromapoint500ftdueeastof P. Atwhatratearethepeoplewalkingapart15minafterthewomanstartswalking?

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Diagram

.

.m

.500

.w.w

.500

.s

.4ft/sec

.4ft/sec

.s =

√(m + w)2 + 5002

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Diagram

.

.m

.500

.w

.w

.500

.s

.4ft/sec

.4ft/sec

.s =

√(m + w)2 + 5002

Page 24: Lesson 13: Related Rates of Change

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Diagram

.

.m

.500

.w

.w

.500

.s

.4ft/sec

.4ft/sec

.s =

√(m + w)2 + 5002

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Diagram

.

.m

.500

.w.w

.500

.s

.4ft/sec

.4ft/sec

.s =

√(m + w)2 + 5002

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Diagram

.

.m

.500

.w.w

.500

.s

.4ft/sec

.4ft/sec

.s =

√(m + w)2 + 5002

Page 27: Lesson 13: Related Rates of Change

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Expressingwhatisknownandunknown

15minutesafterthewomanstartswalking, thewomanhastraveled (

4ftsec

)(60secmin

)(15min) = 3600ft

whilethemanhastraveled(4ftsec

)(60secmin

)(20min) = 4800ft

Wewanttoknowdsdt

when m = 4800, w = 3600,dmdt

= 4, and

dwdt

= 4.

Page 28: Lesson 13: Related Rates of Change

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Differentiation

Wehave

dsdt

=12

((m + w)2 + 5002

)−1/2(2)(m + w)

(dmdt

+dwdt

)=

m + ws

(dmdt

+dwdt

)Atourparticularpointintime

dsdt

=4800 + 3600√

(4800 + 3600)2 + 5002(4 + 4) =

672√7081

≈ 7.98587ft/s


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