Math QRH: Algebra II to Calculus III 9781365285820, 1365285820
Math QRH is a tool, not a textbook! A valuable book for anyone in upper-level STEM courses or careers. A Quick Reference
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Table of contents :
Cover
Inside Title Page
Complete Table of Contents from Print Version
Section 0 - Introduction
Section 1 - Algebra II
Section 2 - Precalculus
Section 3 - Calculus I
Section 4 - Calculus II
Section 5 - Calculus III
Teachers and Professors
Thank You
Citation preview
www.MathQRH.com
“Knowhowtolearn.Then,wanttolearn.” -KatherineJohnson
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GettheHardcoverversion!MathQRH.com Includey ourn otesint hisb ook.U seita sy ourQ RHf ory earst oc ome. Thise Versiono fM athQ RHc anb ep rinteda ndp utintoy ouro wn notebook,b utg ettingt heh ardcoverv ersionc reatesa h andyr eference toolt haty ouc ans aveiny ourb ookc ollectionf oru seinm anyo fy our STEMc lasses!
MathQRH AlgebraIItoCalculusIII FirstEdition
WrittenandPublishedby:KatarinaCosta Artwork:KatarinaCosta ISBN:978-1-365-28582-0 (InhardcoveronLulu.comorMathQRH.com) ForA dditionalInformationa ndR esources: www.MathQRH.com PrintedintheUSA Lulu.com Evaluationcopiesareprovidedtoqualifiedacademicsandprofessionalsfor reviewpurposesonly,foruseintheirclassesforthenextyear.Thesecopies arelicensedandmaynotbesoldortransferredtoanythirdparty.Fordetails onoureducatorevaluationprogramcontactusthroughwww.MathQRH.com.
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TableOfContents Section0:I ntroduction
21
0.0
Katarina’sIntroduction
21
0 .1
ResourcesandTips
23
0 .2
MyNotes,ResourcesandTips
26
Section1:AlgebraII
29
1.0
SummarySheet
29
1 .1
GraphingBasics
36
1.1.1
Intercepts
36
1.1.2
Symmetry
36
1.1.3
Circles
38
1.1.3.a 1.1.4 1 .2
CompletingtheSquare
38
DifferenceQuotient
39
FunctionsandTheirGraphs
39
1.2.1
FunctionDefinition
39
1.2.2
DomainandRange
40
1.2.2.a 1.2.3
Notation
40
CategoriesandCharacteristicsofFunctions
41
1.2.3.a
OddorEven
41
1.2.3.b
Increasing,Decreasing,andConstant
42
1.2.3.c
MaximumsandMinimums
42
1.2.4
ParentFunctions
43
1.2.5
PiecewiseFunctions
46
1.2.6
TransformationsofFunctions
46
1.2.6.a
RigidTransformations
46
1.2.6.b
Distortions
47
1.2.6.c
EquationForm
47
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1
TableofContents 1.3
LinearandQuadraticFunctions
1.3.1
LinearDefinition
1.3.1.a 1.3.2
1.4
48 48
Depreciation
48
QuadraticFunctions
49
1.3.2.a
VertexForm
49
1.3.2.b
Roots
50
1.3.2.c
AxisofSymmetry
50
1.3.2.d
Discriminant
50
1.3.2.e
QuadraticInequalities
51
PolynomialandRationalFunctions
1.4.1
51
PolynomialFunctions
51
1.4.1.a
EndBehavior
52
1.4.1.b
BehavioratIntercepts
54
1.4.1.c
MultiplyingPolynomials
54
1.4.1.d
PolynomialLongDivision
55
1.4.1.e
SyntheticPolynomialDivision
55
1.4.2
RationalFunctionsandAnalysis
57
1.4.2.a
DomainRestrictions
57
1,4.2.b
VerticalAsymptotes
57
1.4.2.c
Holes
58
1.4.2.d
MultiplicityoftheVerticalAsymptote
58
1.4.2.e
HorizontalAsymptotes
59
1.4.2.f
Oblique/SlantAsymptotes
60
1.4.3
GraphingRationalFunctions
61
1.4.4
PolynomialandRationalInequalities
62
1.4.4.a
Graphically
62
1.4.4.b
Algebraically
62
1.4.5
FactoringPolynomialsandRealZeros
64
1.4.5.a
TheFactorTheorem
64
1.4.5.b
TheRemainderTheorem
64
1.4.5.c
Descartes’RuleofSigns
64
1.4.5.d
RationalZerosTheorem
65
1.4.5.e
CombiningToolstoFindZerosofPolynomial
65
2
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PowerPattern
67
1.4.6.b
ComplexConjugate
68
ComplexZeros
68
1.4.7.a
FindingEquationsfromZeros-Example
68
1.4.7.b
FindingZerosfromaFunction-Example
69
ExponentialandLogarithmicFunctions
1.5.1
70
CompositeFunctions
1.5.1.a
70
DomainofCompositeFunctions
70
1.5.2
OnetoOneFunctions
70
1.5.3
InverseFunctions
71
1.5.3.a
Notation
71
1.5.3.b
ProcessofFindinganInverseFunction
71
1.5.3.c
GraphsandInverseFunctions
72
1.5.4
ExponentialFunctions
73
1.5.4.a
Transformations
73
1.5.4.b
CheckingforanExponential
74
1.5.4.c
Euler’sNumber
74
1.5.4.d
SolvingExponentialFunctions-Example
75
1.5.5
1 .6
66
1.4.6.a 1.4.7
1.5
ImaginaryandComplexNumbers
LogarithmicFunctions
76
1.5.5.a
Implementinglog-Examples
76
1.5.5.b
DomainandRange
77
1.5.5.c
Graphing
78
1.5.5.d
LawsofLogarithms
79
1.5.6
FinancialModelFormulas
80
1.5.7
GrowthandDecay
81
1.5.7.a
UninhibitedGrowthandDecay
81
1.5.7.b
Newton’sLawofCooling
82
1.5.7.c
LogisticalModel
83
MyNotesforAlgebraII
TableofContents-M athQ RH
84
3
TableofContents Section2:Pre-Calculus
87
2.0
SummarySheet
88
2 .1
TrigonometricFunctions
98
2.1.1
Angles,ArcLength,CircularMotion
98
2.1.1.a
BasicVocabulary
98
2.1.1.b
MinutesandSeconds
99
2.1.1.c
ArcLength
100
2.1.1.d
RadiansandDegrees
100
2.1.1.e
AreaofaSector
101
2.1.1.f
LinearSpeed
101
2.1.1.g
AngularSpeed
101
2.1.2
TrigonometricFunctions:UnitCircleApproach
102
2.1.2.a
TrigonometricFunctionsofaRealNumber
102
2.1.2.b
UndefinedFunctions
103
2.1.2.c
TrigonometryandTriangles
103
2.1.2.d
CommonAngles
104
2.1.2.e
Symmetry
104
2.1.2.f
FindingRadius
105
2.1.3
PropertiesofTrigonometricFunctions
105
2.1.3.a
PeriodicFunctionsandtheFundamentalPeriod
105
2.1.3.b
SignsofTrigonometricFunctions
106
2.1.3.c
PythagoreanIdentities
106
2.1.3.d
FindingExactValuesUsingIdentities-Example
106
2.1.3.e
FindingExactValuesGivenOneValueandSignofOther-Example 107
2.1.3.f
EvenandOddPropertiesTheorem
107
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108
2.1.4.a
SineFunction
108
2.1.4.b
CosineFunction
109
2.1.4.c
Relationship
110
2.1.4.d
Transformations
110
2.1.4.e
G raphingUsingKeyPoints-Example
111
2.1.4.f
FindingtheEquationUsingtheGraph-Example
111
2.1.5
2 .2
GraphsoftheSineandCosineFunctions
GraphsofTangent,Cotangent,Secant,andCosecantFunctions
112
2.1.5.a
TangentFunction
112
2.1.5.b
CotangentFunction
113
2.1.5.c
WritingDomain
114
2.1.5.d
Transformations
114
2.1.5.e
CosecantFunction
115
2.1.5.f
SecantFunction
115
2.1.5.g
TransformationsoftheCosecantandSecantFunctions
116
2.1.5.h
AmplitudesoftheseGraphs
116
AnalyticTrigonometry 2.2.1
117
TheInverseFunctions
117
2.2.1.a
InverseSineFunction
117
2.2.1.b
InverseCosineFunction
118
2.2.1.c
InverseTangentFunction
119
2.2.1.d
PropertiesofInverseFunctions
119
2.2.1.e
FindingtheInverseofaTrigFunction-Example
120
2.2.1.f
SolvinganInverseTrigFunction-Example
121
2.2.2
TrigonometricIdentities
121
2.2.2.a
IdenticallyEqual,Identity,andConditionalExpression
121
2.2.2.b
Identities
122
2.2.2.c
AlgebraicTechniques
122
2.2.2.d
GuidelinesforEstablishingIdentities
123
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TableofContents 2.2.3
123
2.2.3.a
Theorem(Cosine)
123
2.2.3.b
Theorem(Sine)
124
2.2.3.c
Theorem(Tangent)
125
2.2.3.d
ExactValuesofInverseTrigonometricFunctions-Example
125
2.2.3.e
WritingTrigExpressionsasAlgebraicExpression-Example
126
2.2.3.f
T rigonometricEquationLinearinSineandCosine-Example
127
2.2.4
2 .3
SumandDifferenceFormulas
DoubleAngleandHalf-AngleFormulas
128
2.2.4.a
DoubleAngleFormulas
128
2.2.4.b
FindingExactValue-Example
128
2.2.4.c
EstablishinganIdentity-Example
129
2.2.4.d SquaredTrigFunctionsFormulas
129
2.2.4.e
SolvingUsingIdentities-Example
130
2.2.4.f
ProjectileMotion
130
2.2.4.g
Half-AngleFormulas
131
2.2.4.h
SquaredHalf-AngleFormulas
131
ApplicationsofTrigonometricFunctions 2.3.1
132
RightAngleTrigonometry:Applications
132
2.3.1.a
ComplementaryAngles
132
2.3.1.b
ComplementaryAngleTheorem
132
2.3.1.c
SolvingaRightTriangle-Example
133
2.3.2
SolvingObliqueTriangles
134
2.3.3
TheLawofSines
135
2.3.3.a
Example-SAATriangle
135
2.3.3.b
SolvingSSATriangles
136
2.3.4
TheLawofCosines
2.3.4.a 2.3.5
136
Example-SASTriangle
137
AreaofaTriangle
138
2.3.5.a
AreaofanSASTriangleTheorem
138
2.3.5.b
Heron’sFormula-AreaofaSSSTriangle
138
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PolarCoordinatesandVectors
2.4.1
139
PolarCoordinates
139
2.4.1.a
FindingSeveralPolarCoordinatesofaSinglePoint
140
2.4.1.b
ConvertingfromPolartoRectangularCoordinates
141
2.4.1.c
PointsthatLieonanAxis
141
2.4.1.d
ConvertingfromRectangulartoPolarCoordinates
141
2.4.2
PolarEquationsandtheirGraphs
142
2.4.2.a
Theorem(Vertical/HorizontalLines)
142
2.4.2.b
Theorem(Circles)
143
2.4.2.c
SymmetricPoints
143
2.4.2.d
TestsforSymmetry
143
2.4.2.e
Cardioids
144
2.4.2.f
LimaçonswithoutanInnerLoop
145
2.4.2.g
LimaçonswithanInnerLoop
146
2.4.2.h
Rose
147
2.4.2.i
Leminscates
148
2.4.2.j
Spiral
148
2.4.2.k
GraphingPolarEquations
149
2.4.3
ComplexPlane
149
2.4.3.a
MagnitudeorModulus
150
2.4.3.b
CartesianForm
150
2.4.3.c
PolarFormofaComplexNumber
150
2.4.3.d
Argument
150
2.4.3.e
Euler’sFormula
151
2.4.3.f
ExponentialForm
151
2.4.3.g
MultiplicationandDivisionTheorem
151
2.4.3.h
PeriodicTheorem
151
2.4.3.i
DeMoivre’sTheorem
152
2.4.3.j
ComplexRoots
152
2.4.3.k
FindingComplexRoots
152
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TableofContents 2.4.4
2 .5
Vectors
2.4.4.a
GeometricVectors
153
2.4.4.b
AddingVectorsGeometrically
154
2.4.4.c
MultiplyingVectorsbyNumbersGeometrically
155
2.4.4.d
Properties
156
AnalyticGeometry(ConicSections)
156
2.5.1
EquationCharacteristics
157
2.5.2
Parabolas
157
2.5.2.a 2.5.3 2.5.4
Example
158
Ellipses
2.5.3.a
159
Graphing
160
Hyperbolas
161
2.5.4.a
Graphing
161
2.5.4.b
AnalyzinganEquation
162
2.5.5
2 .6
153
ParametricEquationsandPlaneCurves
162
2.5.5.a
GraphingaPlaneCurve
163
2.5.5.b
FindingRectangularEquationofaParametricallyDefinedCurve
164
2.5.5.c
UseTimeasaParameterinParametricEquations
164
Matrices 2.6.1
165
SolvingaSystemofEquations
165
2.6.1.a
Example
166
2.6.1.b
PossibleSolutions
166
DeterminantofaMatrix
167
2.6.2
2.6.2.a
2x2Matrix
167
2.6.2.b
3x3Matrix
168
2.6.2.c
SwitchRowsorColumns
170
2.6.2.d
MultiplesofRows/Columns
170
2.6.3
Cramer’sRule
171
2.6.4
OperationswithMatrices
172
2.6.4.a
IdentityMatrices
174
2.6.4.b
InverseMatrices
175
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TableofContents 2.7
PartialFractionDecomposition
2.7.1
Case1
177
2.7.2
Case2
178
2.7.3
Case3
178
2.7.4
Case4
178
2.7.4.a 2 .8
177
Example(Case1andCase2)
179
MyNotesforPrecalculus
180
Section3:C alculusI
183
3.0 SummarySheet
183
3 .1
189
Limits 3.1.1
LimitLaws
190
3.1.2
DirectSubstitutionProperty
190
3.1.3
TheSqueezeTheorem
191
3.1.4
ContinuousFunctions
191
3.1.5
ContinuityTheorems
192
3.1.6
CompositeFunctionTheorem
192
3.1.7
CompositesofContinuousFunctionsTheorem
193
3.1.8
TheIntermediateValueTheorem
193
3.1.9
“Intuitive”DefinitionofaLimitatInfinity
193
3.1.10 LimitsofRationalNumbersTheorem 3 .2
194
Derivatives
194
3.2.1
DerivativesatthePoint(a,f(a))
195
3.2.2
TheDerivativeofaFunction
195
3.2.3
Differentiability
195
3.2.3.a
Theorem
196
3.2.3.b
AFunctionisNotDifferentiableattheFollowing
196
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TableofContents 3.2.4
HigherOrderDerivatives
197
3.2.5
BasicDerivativeFormulas
197
3.2.6
D erivativesofTrigonometricIdentities
198
3.2.6.a
Theorem
198
3.2.7
TheChainRule
198
3.2.8
TheDerivativeofanExponentialFunction
199
3.2.9
ImplicitDifferentiation
199
3.2.9.a
Example
3.2.10 DerivativesofLogarithmicFunctions
200 201
3.2.10.a Proof
201
3.2.10.b PropertiesofLogarithms
202
3.2.11 StepsinLogarithmicDifferentiation
202
3.2.12 ExponentialGrowthandDecay
203
3.2.12.a P opulationGrowth
204
3.2.12.b RadioactiveDecay
204
3.2.12.c Newton’sLawofCooling
205
3.2.13 RelatedRates
205
3.2.13.a Example
206
3.2.14 LinearApproximations
207
3.2.15 D ifferentials
208
3.2.16 MinimumsandMaximums
209
3.2.17 ExtremeValueTheorem(EVT)
209
3.2.18 Fermat’sTheorem
210
3.2.19 TheClosedIntervalMethod
210
3.2.20 Rolle’sTheorem
211
3.2.21 MeanValueTheorem
212
3.2.22 Corollary
213
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TableofContents 3.2.23 HowDerivativesAffecttheGraph 3.2.23.a Increasing/DecreasingTest
213
3.2.23.b TheFirstDerivativeTest
213
3.2.23.c Concavity
214
3.2.23.d ConcavityTest
215
3.2.23.e InflectionPoint
215
3.2.23.f TheSecondDerivativeTest
215
3.2.24 IndeterminateForms
215
3.2.25 L’Hospital’sRule
216
3.2.25.a IndeterminateProducts
217
3.2.25.b IndeterminateDifference
217
3.2.25.c IndeterminatePowers
218
3.2.26 GraphSketching
219
3.2.27 Optimization
219
3.2.27.a Example 3 .3
213
Integrals 3.3.1
220 222
Antiderivatives
222
3.3.1.a
PowerRuleandAntiderivatives
222
3.3.1.b
FunctionsandTheirParticularAntiderivatives
222
3.3.2
ParticleMotion
224
3.3.3
SigmaNotation
224
3.3.3.a
Theorem
225
3.3.3.b
Theorem
225
3.3.4
DefiniteIntegrals
226
3.3.4.a
NegativeIntegrals
227
3.3.4.b
Theorem(JumpDiscontinuities)
227
3.3.4.c
Theorem
228
3.3.4.d
MidpointRule
228
3.3.4.e
PropertiesofaDefiniteIntegral
228
3.3.4.f
ComparisonProperties
229
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TableofContents 3.3.5
TheFundamentalTheoremofCalculus
230
3.3.5.a
PartOne
230
3.3.5.b
PartTwo
231
3.3.5.c
Notation
231
3.3.5.d
Composite
231
3.3.6
IndefiniteIntegrals
3.3.6.a 3.3.7
232
CommonIndefiniteIntegrals
233
TheSubstitutionRule(IndefiniteIntegrals)
3.3.7.a
234
Example
234
3.3.8
TheSubstitutionRule(DefiniteIntegrals)
236
3.3.9
IntegralsofSymmetricFunctions
236
3 .4 MyNotesForCalculusI
238
Section4-CalculusII
241
4.0
SummarySheet
241
4 .1
ApplicationsofIntegration
253
4.1.1
AreaBetweenCurves
253
4.1.2
AverageValueofaFunction
254
4.1.3
ArcLength
254
4.1.4
y-AxisIntegration
255
4.1.4.a
GraphingaComplicatedFunctionofy
255
4.1.4.b
AreaBetweentheGraphofx=f(y)andthey-Axis
255
4.1.4.c
FindingtheAreaBetweenTwoCurves
256
4.1.5
Volume
256
4.1.5.a
MathematicalDefinitionofVolume
256
4.1.5.b
Example
257
12
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TableofContents 4.1.6
258
4.1.6.a
Disk
259
4.1.6.b
Washer
260
4.1.6.c
CylindricalShells
261
4.1.7
4 .2
VolumesofRevolution
Work
261
4.1.7.a
UnitsforWorkWordProblems
262
4.1.7.b
ChangingForce
262
4.1.7.c
Hooke’sLaw
263
4.1.7.d
CableProblems
263
4.1.7.e
TankProblems
263
TechniquesofIntegration 4.2.1
IntegrationbyParts
264 264
4.2.1.a
Example
265
4.2.1.b
HowtoChooseu
266
4.2.2
TrigonometricIntegrals
266
4.2.2.a
ImportantIdentities
266
4.2.2.b
S trategies/Guidelines
267
TrigonometricSubstitution
269
4.2.3
4.2.3.a
Example
270
4.2.3.b
DefiniteIntegrals
272
4.2.4
IntegrationbyPartialFractions
272
4.2.4.a
Example
273
4.2.4.b
Trick1-RationalizingSubstitution
274
4.2.4.c
Trick2-CompletingtheSquare
275
4.2.5
ImproperIntegrals
276
4.2.5.a
TypeI:UnboundedIntegrals
276
4.2.5.b
TypeII:DiscontinuousIntegrand
277
4.2.5.c
ImportantRule
277
4.2.5.d
ComparisonTestforImproperIntegrals
278
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TableofContents 4.3
SequencesandSeries
4.3.1
278
Sequences
278
4.3.1.a
RecurrenceRelation
279
4.3.1.b
ArithmeticSequence
279
4.3.1.c
GeometricSequence
280
4.3.1.d
LimitsofSequences
281
4.3.1.e
Vocabulary
282
4.3.2
Series
283
4.3.2.a
Notation
284
4.3.2.b
E stimatingConvergence
285
4.3.2.c
DeterminingDivergence/ConvergencewithPartialSums
285
4.3.2.d
TestforDivergence(nthTermTest)
286
4.3.2.e
HarmonicSeries
287
4.3.2.f
GeometricSeries
288
4.3.2.g
SeriesFacts
289
4.3.2.h
TelescopingSeries
290
4.3.3
IntegralTest
291
4.3.3.a
Example
293
4.3.3.b
p-Series
293
4.3.3.c
U singtheIntegralTesttoEstimatetheSum
294
4.3.4
ComparisonTests
295
4.3.4.a
DirectComparisonTest
295
4.3.4.b
LimitComparisonTest
297
4.3.5
AlternatingSeriesTest
4.3.5.a 4.3.6
298
EstimatetheSumofanAlternatingSeries
RatioandRootTests
299 299
4.3.6.a
Conditionally/AbsolutelyConvergent
299
4.3.6.b
RatioTest
300
4.3.6.c
R ootTest
301
14
TableofContents-M athQ RH
TableofContents 4.4
PowerSeries
4.4.1
Theorem
4.4.1.a 4.4.2
RepresentationofFunctionsbyPowerSeries
305 305
4.4.2.b
IntegralsandDerivatives
306
TaylorSeries
307
4.4.3.a
Example
308
4.4.3.b
Taylor’sRemainderTheorem
309
4.4.3.c
ImportantTaylorSeries
310
4.4.3.d
OperationswithTaylorSeries
311
DifferentialEquations
311
4.5.1
CheckingSolutions
312
4.5.2
Slope/DirectionFields
313
Example
313
4.5.3
Euler’sMethod
314
4.5.4
SeparableDifferentialEquations
315
4.5.4.a
Method1
316
4.5.4.b
Method2
317
4.5.4.c
MixingProblems
318
CalculuswithParametricEquationsandPolarCoordinates 4.6.1
CalculuswithParametricCurves
319 319
4.6.1.a
SlopeandConcavity
319
4.6.1.b
Areas
321
4.6.1.c
ArcLength
322
4.6.2
4 .7
304
Example
4.5.2.a
4 .6
303
4.4.2.a 4.4.3
4 .5
Example
302
TangentsandAreaswithPolarCurves
322
4.6.2.a
D erivativeswithaPolarCurve
323
4.6.2.b
AreaofaPolarCurve
323
MyNotesforCalculusII
TableofContents-M athQ RH
325
15
TableOfContents Section5-CalculusIII
327
5.0
S ummarySheet
327
5 .1
VectorsinSpace
341
5.1.1
VectorsinthePlane
5.1.1.a 5.1.2
PropertiesofVectorOperations
VectorsinThreeDimensions
341 342 343
5.1.2.a
PlottingPoints
343
5.1.2.b
TheCoordinatePlanes
344
5.1.2.c
DistanceFormula(3D)
344
5.1.2.d
EquationsofSurfaces(3D)
345
5.1.2.e
Spheres
345
5.1.2.f
3DVectors
346
5.1.2.g
PropertiesofVectorsinSpace
347
5.1.3
DotProduct
347
5.1.3.a
PropertiesoftheDotProduct
348
5.1.3.b
OrthogonalVectors
349
5.1.3.c
VectorProjection/Component
349
5.1.4
CrossProduct
5.1.4.a 5.1.5
PropertiesoftheCrossProduct
EquationsofLinesandPlanesinSpace
350 351 353
5.1.5.a
ParametricCurves
353
5.1.5.b
EquationofaPlane
354
16
TableofContents-M athQ RH
TableofContents 5.1.6
5 .2
QuadricSurfaces
356
5.1.6.a
Cylinders
356
5.1.6.b
Traces/CrossSections
357
5.1.6.c
Ellipsoid
357
5.1.6.d
Paraboloid
358
5.1.6.e
HyperboloidofOneSheet
358
5.1.6.f
HyperboloidofTwoSheets
359
5.1.6.g
EllipticCone
360
5.1.6.h
HyperbolicParaboloid
361
DifferentiationofFunctionsofSeveralVariables 5.2.1
FunctionsofSeveralVariables
361 361
5.2.1.a
LevelCurves
362
5.2.1.b
ThreeVariableFunctions
362
5.2.1.c
LevelSurfaces
362
5.2.2
LimitsandContinuity
363
5.2.2.a
Epsilon/DeltaDefinitionofaLimit(1D)
363
5.2.2.b
LimitswithTwoorMoreVariables
363
5.2.2.c
LimitLaws
364
5.2.3
PartialDerivatives
365
5.2.3.a
AlternativeNotations
365
5.2.3.b
CalculatingPartialDerivatives
366
5.2.3.c
PartialDerivativeswithThreeorMoreVariables
367
5.2.3.d
HigherOrderPartialDerivatives
367
5.2.3.e
Clairaut’sTheorem
368
5.2.4
TangentPlanesandLinearApproximations
368
5.2.4.a
TangentPlanes
368
5.2.4.b
LinearApproximations
369
5.2.4.c
Differentials
370
5.2.4.d
HelpfulFactwithNormalVectors
370
TableofContents-M athQ RH
17
TableofContents 5.2.5
ChainRule
5.2.5.a 5.2.6
ImplicitDifferentiation
373
DirectionalDerivativesandtheGradient
374
5.2.6.a
GradientVector
375
5.2.6.b
Theorem
375
5.2.6.c
PropertiesoftheGradient
376
5.2.6.d
LevelCurveTheorem
376
5.2.6.e
H igherDimensionalGradients
377
5.2.7
Maxima/MinimaProblems
5.2.7.a 5.2.8
377
SecondDerivativeTest
378
LagrangeMultipliers
5.2.8.a 5 .3
371
379
NoteonExtrema
381
DoubleandTripleIntegrals 5.3.1
381
DoubleIntegralsoverRectangularRegions
381
5.3.1.a
Fubini’sTheoremandIteratedIntegrals
382
5.3.1.b
PropertiesofDoubleIntegrals
384
5.3.2
DoubleIntegralsoverGeneralRegions
385
5.3.3
DoubleIntegralsinPolarCoordinates
387
5.3.4
TripleIntegrals
389
5.3.4.a
TripleIntegralsoveraGeneralBoundedRegion
389
5.3.4.b
AverageValueofaFunctionofThreeVariables
391
5.3.5
TripleIntegralsandCylindrical/SphericalCoordinates
392
5.3.5.a
CylindricalCoordinates
392
5.3.5.b
SphericalCoordinates
393
5.3.6
CentersofMass
396
5.3.6.a
OneDimensional
396
5.3.6.b
TwoDimensional
397
5.3.6.c
ThreeDimensional
398
18
TableofContents-M athQ RH
TableofContents 5.4
VectorCalculus
5.4.1
VectorFields
399 399
5.4.1.a
Notation
399
5.4.1.b
UnitVectorField
400
5.4.1.c
GradientVectorField
401
5.4.2
LineIntegrals
401
5.4.2.a
ScalarLineIntegral
401
5.4.2.b
ArcLength
403
5.4.2.c
VectorLineIntegrals-Work
403
5.4.2.d
VectorLineIntegrals-Flux
406
5.4.2.e
Tip:ParameterizingaLineSegmentin3D
407
5.4.3
ConservativeVectorFields
407
5.4.3.a
Simple/ClosedCurves
407
5.4.3.b
C onservativeFieldTheorem
408
5.4.3.c
FundamentalTheoremofLineIntegrals
408
5.4.3.d
Theorem
409
5.4.3.e
FindingPotentialFunctions
409
5.4.4
Green’sTheorem
410
5.4.4.a
ConnectivityofRegions
410
5.4.4.b
PiecewiseSmooth
411
5.4.4.c
Green’sTheorem
411
5.4.5
Divergence
414
5.4.6
Curl
416
5.4.7
ParametricSurfaces
417
5.4.8
SurfaceIntegrals
418
5.4.8.a
SurfaceArea
418
5.4.8.b
ScalarSurfaceIntegrals
420
5.4.8.c
OrientationofaSurface
421
5.4.8.d
SurfaceIntegralinaVectorField
421
TableofContents-M athQ RH
19
TableofContents 5.4.9
Stokes’Theorem
5.4.9.a
Example
5.4.10 DivergenceTheorem 5.4.10.a Example 5 .5
MyNotesForCalculusIII
423 424 426 427 428
20
TableofContents-M athQ RH
“Reservey ourrighttothink,fore ventothinkwronglyisb etterthann ot tothinka ta ll” -HypatiaofAlexandria
Section0-Introduction 0.0
Introduction
Mathematics:manypeoplehearthiswordandinternallygroan. Thismakessense! Everyonehasbeentoldatleastonceintheirlifethatmathishard,evenfromtheirown teachers. WhenIwasineighthgrade,oneofthefirstthingsmyteachertolduswashow hardmathis. Thiswasreallydemotivatingformanykidsintheclass. Howwerethey expectedtounderstandconceptsthateventheirteacherthoughtwasdifficult? AlthoughIloveitnow,Iwasnotalwaysmath’sbiggestfan. Oneofmyfirstmemoriesof mathiswhenIwasaroundthreeyearsoldandlearninghowtocount. Everythingwasfine untilIhitonenumber. Iwassittingwithmydadandgettingincreasinglyangryathim,asI waspositivethathewaswrongabout80comingafter79. Afterhetriedtoconvinceme manytimesthathedid,infact,knowhowtocount,IstompedofftogowatchTVwithmy momandsister. Let’ssaythatcountingfrom79to80wasnotmystrongsuitatthattime. WhenIgotalittleolderIneededtotaketheACTtestinordertoapplyforaschoolthatI wantedtoattend. Ihadnotyettakenanytrigonometry,however. Asmanydadsdo,my dadjokedthatitmaybe“tootoughforagirltolearnonherown.” Withmyspitegivingme anevenbiggermotivation,Itaughtmyselfenoughtrigonometrytogetagoodscore,which helpedtogetmeacceptedintotheschoolthatIwanted.
Section0-Introduction-M athQ RH
21
Mathematicsrequiresustoputinworkandeffort. Withthenumberoftoolsthatare availabletoeveryone,anyonecandowellatmathematics! Maybemyteacherwhosaid thatmathwashardwasright. Mathematicscanbechallenginganditcanwrapyourbrain intocirclesattimes. Butitisreallyabouthowyoulookatit. Thinkofmathematicsasa puzzleoragametoenjoy! Itisasmuchanartasitisascience. WhileImaylovemathnow,Iknowthatmanypeopledon’t. WhenIwasamathtutor,Isaw somanysmartkidsstrugglewithmathwhentheydidn’tneedto. It’shardtoremember everythingthatwascoveredinpreviousmathclasses! Thisbookisacompilationofallthe notesItook,asIfoundmyselffrequentlyreferringbacktothem. Myhopewiththisbookis thatIcanbeasmallpartofyourjourneyinmath. Iamnoexpert;Iamjustlikeyou! Ihave alotgoingoninmylifeandlookfortoolsallthetimetomakegettinggoodgradesin schooleasier. Thisbookismywayofhelpingyousavetimeandfindsuccessinyourmath classesbycondensingwhatI’vefoundusefulintooneplace. Disclaimer:thisisnotatextbook! MathQRHismeanttobeusedasacomplementarytooltoyourassignedtextbooks. Ihave summarizedthemostimportantaspectsofeachclasssothatitiseasytoreviewthingsyou mayhaveforgotten(orjumpaheadforfun!). Usethisbookasawaytoguideyournotes, quicklyfindimportantformulas/concepts,andstudyforupcomingtests. Writeinthis book! IleftroomforyournotesinMathQRH. Ididmybesttobeaccurate,ofcourse,butIwrotethisbookfrommynotes. Ifyoufind errorsorhavenotesofyourownthatyouthinkwillbenefitothers,pleasesharethemat MathQRH.com! Asfortheimagesinthisbook,dorefertothechapter“ReasoningWell fromPoorlyDrawnFigures”inJordanEllenberg’sbookS hape. Iwishyougreatsuccesswithyourjourneyinmathematics!
22
Section0-Introduction-M athQ RH
0.1
TipsandResources
Noneofthesetipsaresupergroundbreaking;you’velikelyheardmostofthem! ButIthink itisimportanttoreiteratethemhere,astheyreallydohelpyoutobesuccessfulinyour classes. Thereisno“trick”tomathclassesandnooneisinherentlygoodatmath,itjust takeswork. Mybiggesttipforbeingsuccessfulinamathclassisverybasic:takegoodnotes. “Good notes”donothavetomeanwritingwithtendifferentpensandpencilsinperfect handwriting. WhileIloveusingfuncoloredpensformyonlinelectures,it’snotreally feasibleforinpersonclasses. Allyouneedispaper(Ipreferloosegraphpaperinabinder) andatleastonepencilwithaneraser. “Goodnotes”isdifferentforeveryonebutallgood notesinclude(1)handwritingy oucanreadand(2)thestrategiesandexamplesyour teachergivesyou. ThisbookhasalreadyprovidedmanyofthenotesthatyouwillneedinAlgebraII, Precalculus,andCalculusI,IIandIII. Fillupyourcopyofthisbookwithyournotesandit willserveyouinuppermathematicsaswellasscienceandengineeringclasses! Stayingorganizedissuperimportantforgettingagoodgradein(andgettingsomething outof)yourmathclasses. Igetthatitiseasiersaidthandone,butit'sallaboutfinding what’srightforyou. WhileIlovebindersandweeklyto-dolistsonmylaptop,somepeople prefertouseaphysicaloronlineplanner. Experimentandfindoutwhatworksbestfor you. Mathisaveryinvolvedsubjectthatcannotbelearnedinjustacouplehoursaweekina lecture. Thismeansthattotrulylearnmathyouwillneedtoputintimeoutsideofclass. I’msureyou’vebeennaggedbyteachersandparentsaliketodoyourhomework,butthisis forgoodreason. Evenifyouneedtolookupanansweroruseacalculatortosolvea problem,justputtingintheextratimecanmakeahugedifference. Section0-Introduction-M athQ RH
23
Ifyoucompletelyforgethowtodoaproblemwhiletakingatest,tryanything,evenifyou thinkit’swrong. Don’tjustleaveitblank! Youcanevenaddanotewhereyousayyoudon’t thinkit’srightandexplainyourreasoningordefinewhatyouused. Evenforquestionsyou areprettysureon,itishelpfultoaddnotestoyourteacherso,ifyougetitwrong,youcan maybekeepsomepoints. Thismayseemcounterintuitive,butyoudon’tnecessarilyneedtounderstandeverylittle detailofmath. Sometimesyoujustneedtoknowthataformulaworks,beabletouseit, andmoveon. HereisalistofafewgreatresourcesthatIrecommend: ●
Symbolab ○
●
Thisismyfavoriteonlinecalculator. Itcanhandlealmostanymathproblem.
GeoGebra ○
GeoGebraisoneoftheonlygraphingcalculatorsthatcanactuallyplotin both2Dand3D. Ithasalotofusefulfunctionsthatarehelpfulformore complexgraphs. However,itcantakesometimetogetusedtotheformat thatitusesforfunctions.
○ ●
IactuallyusedthiswebsitetocreatemostoftheCalculusIIIimages.
Desmos ○
Desmosisamoreuser-friendlyversionofGeoGebra. Itdoesn’thavethe abilitytodo3Dgraphing,butifyouarelookingtomake2Dgraphs,thisisthe bestwebsitetouse.
24
Section0-Introduction-M athQ RH
●
KhanAcademy ○
KhanAcademyprovidesentirecourses(includinganamazingSATprep program)withvideoexplanationsforfree.
●
OpenStax ○
OpenStaxprovidescompletelyfreetextbooksforeverymathclass mentionedinthisbook. Whileyoumayalreadyhaveatextbookforyour class,sometimesitcanbehelpfultoseeconceptsdescribedinacouple differentways.
●
MatrixCalc.org ○
●
ThisisoneofthebestmatrixcalculatorsI’veused.
WolframAlpha ○
Personally,I’mnotabigfanofWolframAlpha(Ithinktheuserinterfaceis ugly),butitisabletosolvemostproblems.
Ifyouarelookingforsomemorefunmathbooks,IhighlyrecommendInfinitePowersby StevenStrogatzandS hapebyJordanEllenberg. Thesebooksaresuperaccessibleand allowyoutoseehowimportantmathisoutsideofaclassroom. Don’tletanyonetellyouthatyoushouldnotexpectanAinyourclasses. Itmightnotbe easyanditmighttakesomework,butyoucandoit;Ibelieveinyou!
Section0-Introduction-M athQ RH
25
Mathematicsisn otjusta notherlanguage...Itisa languagep luslogic. Mathematicsisa toolforreasoning.” -RichardFeynman
0.2
Mynotes,resourcesandtips:
26
Section0-Introduction-M athQ RH
0.2
Mynotes,resourcesandtips:(con’t)
Section0-Introduction-M athQ RH
27
0.2
28
Mynotes,resourcesandtips:(con’t)
Section0-Introduction-M athQ RH
“Everybodyisa g enius.Butify oujudgea fishb yitsa bilitytoc limba tree,itwillliveitswholelifeb elievingthatitiss tupid.” AlbertEinstein
Section1-AlgebraII
1.0
SummarySheet
GeneralGraphing Intercepts:f (0) = y intercept ,f (x) = 0 = x intercept Symmetry: ●
x axissymmetry:f (x) = − f (x)
●
y axissymmetry:f (x) = f (− x ) ○
●
thisisthetestforevenfunctions
originsymmetry:f (x) = − f (− x ) ○
thisisthetestforoddfunctions
CircleEquation:(x − h )2 + (y − k )2 = r 2 ,where(h, k ) isthecenterandr istheradius CompletingtheSquare:a x 2 + a y 2 + b x + c y + d = 0 DifferenceQuotient: DQ =
f (x + h) − f (x) h
⇒ (x +
b 2 2a )
+ (y + 2ac )2 = −
d a
+ ( 2ab )2 + ( 2ac )2
Section1-AlgebraII-M athQ RH
29
FunctionBasics Afunctionmustpasstheverticallinetest:everyx hasoney . Thedomainisallofthepossibleinputvalues,andtherangeisallofthepossibleoutput values. ●
IntervalNotation:( ) -endpointisnotincluded,[ ] -endpointisincluded
Extrema: ●
Alocalextremaistheextremeonacertaininterval. Itmusthavepointsonboth sidesforittobelocal.
●
Anabsoluteextremaisthehighest/lowestvalueoftheentirefunction.
(note:addparentfunctionshere?)-visitwww.MathQRH.comandletusknow! Transformations:y = a (b(x − h ))2 + k ●
a ,verticalstretchorcompression
●
b ,reciprocalofthehorizontalstretchorcompression
●
h ,horizontalshift
●
k ,verticalshift
LinearDefinition:constantchangeovertime QuadraticDefinition:standardform-a x 2 + b x + c = 0 , vertexform-y = a (x − h )2 + k , v ertex = (h, k ) = (−
b 4ac − b 2 , 4a ) 2a
30
Section1-AlgebraII-M athQ RH
Thediscriminantcanbeusedtodeterminethenumberofsolutionsthataquadratic functionhas. ●
b 2 − 4 ac < 0 ,nosolutions
●
b 2 − 4 ac = 0 ,onesolution
●
b 2 − 4 ac > 0 ,twosolutions
PolynomialandRationalFunctions Apolynomialf (x) = a x n + b x n−1 + ... + z hasonlyonevariable,wheren > 0 ,andthegraphis smoothandcontinuous. EndBehavior: ●
evenexponent,positivea
●
○ asx → ∞ , y → ∞ ,asx → − ∞ , y → ∞ evenexponent,negativea
●
○ asx → ∞ , y → − ∞ ,asx → − ∞ , y → − ∞ oddexponent,positivea
●
○ asx → ∞ , y → ∞ ,asx → − ∞ , y → − ∞ oddexponent,negativea ○
asx → ∞ , y → − ∞ ,asx → − ∞ , y → ∞
Multiplicityofaroot(x intercept)determineshowthegraphbehaves. Ifitisodd,thegraph willcrossthroughthex intercept. Ifitiseven,thegraphwillbounceoffofthex intercept.
Section1-AlgebraII-M athQ RH
31
Arationalfunctionhasatleastafirstdegreepolynomialinthedenominatorandtakesthe formr(x) =
p(x) q(x)
.
Tographarationalfunction: 1. Factorbothpolynomials. 2. Seeifanyofthefactorscancelout. Iftheydo,notethatthatvalueofx hasaholein thegraph. Rememberthatifthereisaholeatanx coordinatetherecannotbea verticalasymptoteatthesamex value. 3. Findthex valuesthatmakethedenominatorequaltozero. Makeanotethatthese areverticalasymptotes. Alsonotethemultiplicityoftheseasymptotes,asthese determineendbehavior. 4. Findthex valuesthatmakethenumeratorequaltozero. Thesearethex interceptsofthefunction. 5. Pluginzeroforx tofindthey interceptofthefunction,ifoneexists. 6. Ifthenumeratorpolynomialofthefunctionisonedegreehigherthanthe denominatorpolynomial,thenusepolynomialdivisiontofindtheslantasymptote. 7. “Plugin”infinityforx tofindifthereareanyhorizontalasymptotes. a. Ifthefunctionhasaslantasymptote,itwillnothaveahorizontalasymptote. 8. Onceyouhavenotedallimportantinformation,plotallpointsand/orasymptotes andsketchinthegraph. SumandDifferenceofCubes: a 3 + b 3 = (a + b )(a 2 − a b + b 2 ) a 3 − b 3 = (a − b )(a 2 + a b + b 2 ) FactorTheorem:iff (c) = 0 ,(x − c ) isafactorofthepolynomialf (x) .
32
Section1-AlgebraII-M athQ RH
f(x)
RemainderTheorem:iff (c) =/ 0 ,f (c) istheremainderof x − c . Descartes’RuleofSigns: ●
ifn isequaltothenumberofsignchangesinf (x) ,thefunctionhasn orn − 2 ... positivezeros
●
ifn isequaltothenumberofsignchangesinf (− x ) ,thefunctionhasn orn − 2 ... negativezeros
RationalZerosTheorem:forf (x) = a n x n + a n−1 x n−1 + ... + a 0 ,therationalzerostaketheform
±
p q
,wherep isafactorofa 0 andq isafactorofa n .
ImaginaryandComplexNumbers Theimaginarynumberi isequalto√− 1 . Thismeansthati2 = − 1 , i3 = − i andi4 = 1 . Acomplexnumberisacombinationofrealandimaginarynumbersintheforma + b i , wherea andb arerealnumbers. CompositeFunctions CompositeFunction:f (g(x)) = f ° g (x) ,plugsafunctionintoanother InverseFunction:f (a) = b → f −1 (b) = a .
●
Test:f (g(x)) = g (f (x)) = x
●
Tofindaninversefunction:pluginx fory andy forx,thensolvefory
Section1-AlgebraII-M athQ RH
33
ExponentialandLogarithmicFunctions Anexponentialfunctionhastheformf (x) = a x . ●
●
Transformations:f (x) = c · − (a −x − h ) + k ○
c isaconstantmultiplier
○
thenegativeoutfrontofthea flipsitoverthex axis
○
thenegativeoutfrontofthex flipsitoverthey axis
○
h isthehorizontalshift
○
k istheverticalshift
CommonQuotient:a =
f(x + 1) f(x)
Euler’sNumber= e ≈ 2 .71828 Logarithms:ifb y = x ,theny = log b x LogarithmLaws:
⇒ x = 0 log a = x ⇒ x = 1
1. log a 1 = x 2.
a
⇒ M = x log a = x ⇒ x = r
3. alog a M = x 4.
a
r
5. log a (MN ) = log a M + log a N
6. log a ( M ) = log a M − log a N N 7. log a M r = r · log a M r
8. eln a = ar . 9. IfandonlyifM = N ,thenlog a M = log a N .
10. log a M =
log b M log b a
,whereb isabaseofyourchoice.
34
Section1-AlgebraII-M athQ RH
FinancialFormulas SimpleInterest:I = P rt CompoundInterest:A = P · (1 + nr )nt ContinuousCompounding:A = P e rt EffectiveRateofInterest:r e = (1 + nr )n − 1 ,r e = e r − 1 PresentValue:P = A · (1 + nr )−nt ,P = A · e −rt GrowthandDecay UninhibitedGrowth:N (t) = N 0 e±kt Newton’sLawofCooling:u(t) = T + (u0 − t)ekt c LogisticModel:p(t) = 1 + ae −bt
Addyourownnotesbelow.Sharethem!QRHMath.com
Section1-AlgebraII-M athQ RH
35
1.1
GraphingBasics
1.1.1
I ntercepts
Interceptsarethepoints(orpoint)onagraphthatintersectswiththex ory axis. Tofind anintercept,plugzerointotheothervariableandsolve. Forexample,ifyou’retryingto findthey interceptofafunction,pluginzeroforx andsimplify. ●
Ifyouaretryingtofindtheinterceptofavariablethatissquared,besuretonote thatthenumbercouldbepositiveornegative. Sincesquaringanumbermakesit positive,wecannotbesurethattheoriginalnumberispositive. ○
Forexample,thex interceptsofy = x 2 − 4 are− 2 and2 . Thisisbecause whenyouplugin0 fory andsimplifyyougetx 2 = 4 . Noticehowboth− 2 and2 satisfythatequation.
1.1.2
S ymmetry
Therearethreetypesofsymmetry: ●
x -axissymmetry ○
Agraphhasx -axissymmetrywhenyoucanplugin− y andgetthesame equation. Inotherwords,f (x) = − f (x) .
36
Section1-AlgebraII-M athQ RH
●
y -axissymmetry ○
Agraphhasy -axissymmetrywhenyoucanplugin− x andgetthesame equation. Inotherwords,f (x) = f (− x ) .
●
originsymmetry ○
Agraphhassymmetryabouttheoriginwhenyoucanpluginboth− y and − x andgetthesameequation. Inotherwords,f (x) = − f (− x ) .
○
Ifyouspinthegraph1 80 ° ,itwilllandonitself,asitisreflectedacrossboth thex andy axis.
Section1-AlgebraII-M athQ RH
37
1.1.3
C ircles
Acircleisasetofpointsthatareallequidistantfromonepoint,thecenter. Therearetwo formsofacircleequation: ●
StandardForm(usedforgraphing):(x − h )2 + (y − k )2 = r 2 ,where(h, k) isthecenter andr istheradius.
●
GeneralForm:a x 2 + a y 2 + b x + c y + d = 0 ,noticethatthex 2 andy 2 havethesame coefficientandsign.
1 .1.3.a
CompletingtheSquare
Asnoted,standardformistheformusedforgraphing. So,whengivenageneralform equation,youmustcompletethesquare. a x2 + a y 2 + b x + cy + d = 0 Dividebothsidesbya . x 2 + y 2 + ab x + ac y +
d a
=0
Movetheconstanttotheotherside. x 2 + y 2 + ab x + ac y = − ad Dividetheconstantcoefficientsbytwoandsquare,thenaddtobothsides. x 2 + ab x + ( 2ab )2 + y 2 + ac y + ( 2ac )2 = −
d a
+ ( 2ab )2 + ( 2ac )2
Then,factorbothquadratics. (x +
38
b 2 2a )
+ (y + 2ac )2 = −
d a
+ ( 2ab )2 + ( 2ac )2
Section1-AlgebraII-M athQ RH
Remember,forthisequation,thecenteris(−
b 2a ,
√
− 2ac ) andtheradiusis −
d a
+ ( 2ab )2 + ( 2ac )2
1.1.4
D ifferenceQuotient
Thedifferencequotientisusedtofindtheslopeofacurveusingatangentline. Theformulaforthedifferencequotientis:D Q =
f(x + h) − f(x) h
. Ifyoupluginaformula,like
f (x) = − 3 x 4 ,you’llgetaformulathatistheslopeofthatspecificfunction.
1.2
FunctionsandTheirGraphs
1.2.1
F unctionDefinition
Asetofpointscanbecalledarelation. Thiscanbeafewpoints,oraninfinitenumberof points. Ifarelationhasapropertywhereeachx valueproducesonlyoney value,itis calledafunction. Thismeansthattwox ’scanproducethesamey inafunction,butanyx cannotproducemorethanoney . Theverticallinetestcanbeusedtodetermineifagraph isafunction;ifalinecanbedrawnthatcrossesthegraphmorethanonce,it’snota function.
Section1-AlgebraII-M athQ RH
39
1.2.2
D omaina ndR ange
Thedomainofafunctionisasetofallpossiblex coordinatesorinputvalues. Therangeof afunctionisthesetofallpossibley coordinatesoroutputvalues. Tofindthedomainandrange,lookforanyrestrictionsonx andy . Thismeanslookingat valueswherethefunctionisundefinedorimaginary,lookingatendbehavior(positiveor negativeinfinity),ifthereareanygapsinthegraphofthefunction,etc. 1.2.2.a ●
Notation
SetNotation:D = {x ○
●
∈ R | (some exception)}
Readas“x isanelementoftherealnumberssuchthat…”
IntervalNotation:thisnotationshowsallofthenumbersinadomain,ratherthan focusingontheexceptions. ○
Use( ) iftheendpointisnotincluded ■
theseareusedwheneverinfinityisanendpoint
○
Use[ ] iftheendpointisincluded
○
‘⋃ ’ isusedtoshowaunionofintervals
Example: Forthefunctionf (x) = x 2 − 4 ,thevertexis(0, − 4 ) . Noticethatyoucanpluginanyx value, sinceitisaparabola. Fory values,however,allnumbershavetobegreaterthanorequal to4 ,sincetheminimumisequalto4 . Therefore,thedomainisD = {x R = {y
∈ R | y ≥ 4} or [− 4, ∞).
40
∈ R} or (− ∞, ∞) a ndtherangeis
Section1-AlgebraII-M athQ RH
1.2.3
C ategoriesandC haracteristicsofF unctions
1.2.3.a
OddorEven
Afunctionise venw henithassymmetryaboutthey -axis. Thismeansthatf (x) = f (− x ) .
Afunctioniso ddifithassymmetryabouttheorigin. Thismeansthatf (− x ) = − f (x) .
Itisimportanttonotethattheexponentalonedoesnotsignifythatafunctionisevenor odd. Forexample,f (x) = x 2 + x + 1 hasanevenhighestdegree,butitisnotaneven function.
Section1-AlgebraII-M athQ RH
41
1.2.3.b
Increasing,Decreasing,andConstant
Afunctionisincreasingifthey-valueincreasesasthex-valueincreases. Afunctionisd ecreasingifthey-valuedecreasesasthex-valueincreases. Afunctionisc onstantifthereisnochangeinslope.
Inthisgraph,thefunctionisincreasingovertheintervals(− 5 ,− 2 ) and(3, ∞ ) . Thefunction isdecreasingovertheinterval(− ∞ ,− 5 ) . Thefunctionisconstantover(− 2 , 3 ) . 1.2.3.c
MaximumsandMinimums
Therearetwotypesofmaximumsandminimums:absoluteandlocal. Alocalmaximumandminimumisthehighestorlowestpointofthegraphonacertain interval. Whileitmaybetheextremaforthatinterval,itmaynotbethehighestorlowest pointoftheentirefunction. Alocalextremacannotbeattheendpointofafunction.
42
Section1-AlgebraII-M athQ RH
Ana bsolute(orglobal)maximumorminimumisthehighestorlowestpointoftheentire function.
1.2.4
P arentFunctions
Ap arentfunctionisthesimplestformofafunctioninafamilyoffunctions. Afamilyof functionshasthesamehighestdegree. Parentfunctionscanguideusingraphingmore complicatedfunctions. SeethenextpageforexamplesofParentFunctions:
Section1-AlgebraII-M athQ RH
43
NameofFunction
Graph
Linear
y =x
y = x2
Quadratic
y = x3
Cubic
y = |x |
AbsoluteValue
y = √x
SquareRoot
44
Section1-AlgebraII-M athQ RH
NameofFunction
Graph
CubeRoot
y = √x 3
Reciprocal/Rational
y = 1x
y= Rational
1 x2
Exponential
y = ex
y = ln(x)
Logarithmic
Section1-AlgebraII-M athQ RH
45
1.2.5
P iecewiseFunctions
Ap iecewisefunctionisacombinationoffunctions,whereeachfunctionisdefinedovera certaininterval. Sincethesearecalledfunctions,theymuststillpasstheverticallinetest. − 2 x + 1 , − 3 ≤ x < 1 f (x) = 2, x = 1 x 2 , x > 1 Noticethatthisfunctiondoesstillpasstheverticallinetestas theopencirclesdenoteo penintervals. Thismeansthatthe functionisnotactually“at”thispoint Tip:graphtheboundariesfirstbeforetryingtomaketheshapes.
1.2.6
T ransformationsofF unctions
1.2.6.a
RigidTransformations
Thesetypesoftransformationsmaintainthesizeandshapeofthegraph,theysolelyaffect thegeometriclocationofthegraphs. ●
Translations ○
●
Thisisaverticalorhorizontalshiftalongthegraph.
Reflections ○
“Flips”thegraphoveranaxis.
46
Section1-AlgebraII-M athQ RH
1.2.6.b
Distortions
Thesetransformationswillchangethesizeandshapeofthegraph. ●
VerticalStretch/Compression
●
HorizontalStretch/Compression
Notethatverticalstretchesareverysimilartohorizontalcompressions. 1.2.6.c
EquationForm
Thesetransformationscanbeshowninthevertexformofaquadraticequation.
y = a(b(x − h))2 + k Theseconddegreeexponentisjustanexample. ➔ ‘a ’representsthecoefficientofverticalstretchorcompression
◆ ◆ ◆
Ifa > 1 ,itisaverticalstretch Ifa < 1 ,itisaverticalcompression Ifa isnegative,thereisa1 80 ° reflectionvertically(↕ )
➔ ‘b ’representsthecoefficientofthehorizontalstretchorcompression
◆ ◆
Thereciprocalofthisnumberdeterminesstretchorcompression. So,if 1 b
< 1 itisahorizontalcompression. Ifb1 > 1 thereisahorizontalstretch.
Ifb isnegative,thereisa1 80 ° reflectionhorizontally(↔ )
Section1-AlgebraII-M athQ RH
47
➔ ‘h ’representstheoppositeofthehorizontalshift
◆ ◆
Thismeansthathaving(x − 2 )2 shiftsthegraphtotheright. Itisimportanttonotethatx musthaveacoefficientofpositive1. So,ifa functioniswrittenasf (x) = 2 7 |2 − 3 x | + 5 ,factoroutthe3tofindthe horizontalshifta ndstretch/compression. Thiswouldmake f (x) = 2 7 ||− 3 (x − 32 )|| + 5 . Now,wecanmoreclearlyseethatthegraphisshifted totherightby 32 andthereisahorizontalcompressionbyafactorof 31 .
➔ ‘k ’representstheverticalshift
◆ ◆
Ifk ispositive,thegraphisshiftedup. Ifk isnegative,thegraphisshifteddown.
1.3
LinearandQuadraticFunctions
1.3.1
L inearD efinition
Linearchangeisconstantovertime. Thismeansthattheslope(changeiny overchangein x )doesnotchange. 1.3.1.a
Depreciation
Depreciationisthereductionofvalueastimepasses. Theannualdepreciationcanbecalculatedbydividingthepurchasevalueovertheyearsof usefulness. Annual Depreciation =
48
purchase value years of usefulness
Section1-AlgebraII-M athQ RH
Tofindthevaluechangeovertime,subtracttheannualdepreciationfromtheoriginal value. Scheduleddepreciationistheequationthatiscreatedfromgeneralizingtheformulafor valuechangeovertime. Ifx representsthenumberofyearssincepurchase,then V alue = p urchase value −
x (purchase value) years of usefulness
Theslopeofthelineforthegraphofdepreciationis−
purchase value years of usefulness
(valuedecreasesover
timesotheslopemustbenegative). Thelineintersectsthex -axisatthex valuethat correspondswiththeyearsofusefulness.
1.3.2
Q uadraticF unctions
1.3.2.a
VertexForm
Thevertexformofaquadraticfunctioniswrittenasf (x) = a (x − h )2 + k . Itiscalledthevertex formash isthex componentofthevertexandk isthey componentofthevertex. When aquadraticfunctionisinvertexformitistheeasiesttograph. Tochangeafunctionfromstandardform(f (x) = a x 2 + b x + c )tovertexform,completethe square,keepingeverythingtooneside. Whengeneralized,thefollowingformulasarefound: v ertex = (−
2 b , 4ac−b ) 2a 4a
Section1-AlgebraII-M athQ RH
49
1.3.2.b
Roots
The“roots”ofafunctionarethex -intercepts,i.e.wherethey valueisequaltozero. Tofindtheroots,settheequationequaltozeroandsolveforthevariable. 1.3.2.c
AxisofSymmetry
Theaxisofsymmetryisthelineaboutwhichtheparabolaissymmetrical. Foraverticalparabola(opensupwardordownward),theaxisofsymmetryisverticaland equaltothex coordinateofthevertex(x = −
b 2a
).
Forahorizontalparabola(opensrightorleft),theaxisofsymmetryishorizontalandequal tothey coordinateofthevertex. 1.3.2.d
Discriminant
Usingthequadraticformula,if0 = a x 2 + b x + c ,thenx =
−b
∓ √b −4ac . 2
2a
●
whenb 2 − 4 ac < 0 ,therearenosolutionsandthereforenox intercepts
●
whenb 2 − 4 ac = 0 ,thereisonesolutionandthevertexisonthex -axis
●
whenb 2 − 4 ac > 0 ,therearetwosolutions
50
Section1-AlgebraII-M athQ RH
1.3.2.a
QuadraticInequalities
Rememberthaty = f (x) = somefunctionofx . So,whenyouseesomethinglikef (x) ≤ 0 you needtolookforthex valuesthatsatisfytheinequality. Forexample,ifgivenx 2 − 4 x − 1 2 ≤ 0 ,youarelookingforthexvaluesthatmakey ≤ 0 . To dothis,findtherootsofthefunction:(x − 6 )(x + 2 ) ,x = 6 andx = − 2 . Whengraphed,we seethatallvaluesincludingandbetween− 2 and6 produceay valuethatislessthanor equalto0 .
1.4
PolynomialandRationalFunctions
1.4.1
P olynomialF unctions
Apolynomialfunction ●
●
hasonevariable ○
f (x) = a x n + b x n−1 + ... + z
○
powerscanbemissingintheequation
n mustbegreaterthanorequaltozero ○
●
n alsomustbeaninteger(wholenumber)
hasasmooth,continuousgraph(i.e.nogapsorcorners)
Section1-AlgebraII-M athQ RH
51
Forexample,P (x) = 5 x 3 − 41 x 2 − 9 andG(x) = 8 arepolynomialfunctionsbuth (x) =
x2 −2 x3 −1
(x is
1
inthedenominatorandthereforehasanegativeexponent)andg (x) = √x (√x = x 2 )arenot. 1.4.1.a.
EndBehavior
The“endbehavior”ofagraphisdeterminedbythehighestpoweredtermofafunctionand itssign.
52
Section1-AlgebraII-M athQ RH
ExponentandSign
EndBehavior
xapproches,yapproches
evenexponent,positivea
asx → ∞ , y → ∞ asx → − ∞ , y → ∞
evenexponent,negativea
asx → ∞, y → − ∞ asx → − ∞ , y → − ∞
oddexponent,positivea
asx → ∞ , y → ∞ asx → − ∞ , y → − ∞
oddexponent,negativea
asx → ∞, y → − ∞ asx → − ∞ , y → ∞
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53
1.4.1.b BehavioratIntercepts Whenfactored,apolynomialisbrokendownintoitsroots. Someoftheserootswillhave exponentsattachedtothem,meaningtheyhaveamultiplicity. ●
Whenthemultiplicityisodd,thegraphwillcrossthroughtheintercept.
●
Whenthemultiplicityiseven,thegraphwill“bounce”offoftheintercept.
Forexample,thefunctionf (x) = x 4 + 1 5x 3 + 5 7x 2 − 3 5x − 2 94 factorsinto f (x) = (x − 2 )(x + 3 )(x + 7 )2 . Therefore,thisfunctionisgraphedasfollows.
1.4.1.c
MultiplyingPolynomials
Multiplyingpolynomialsisessentiallythesameasmultiplyingnumbers. Setthemupone abovetheother,thenmultiplyasyouwouldanynumberswithmultipledigits. Treateach componentofthepolynomialasitsownnumber.
54
Section1-AlgebraII-M athQ RH
1.4.1.d
PolynomialLongDivision
Forpolynomiallongdivision,youbeginbysettingupthepolynomialsinanormallong divisionsetup. Then,dividethefirsttermofthedividend(thepolynomialtobedivided)by thefirsttermofthedivisor(thepolynomialdoingthedividing). Theanswertothisgoes intotheanswerrow. Multiplytheentiredivisorbythisanswerandputtheproductthe dividend. Subtracttheproductfromtheoriginaldividendandrepeatuntilthelargest powerofthedifferenceissmallerthanthelargestpowerofthedivisor. Ifthereisa remainder,putitintheform remainder andaddittotheendoftheanswer. divisor Belowisanexampledividing6 x 4 + x 3 − 2 x + 4 byx 2 − 5 x + 7 .
Thefinalansweris6 x 2 + 3 1x − 1 13 +
346x + 795 x2 − 5x + 7
.
1.4.1.e
SyntheticPolynomialDivision
Syntheticdivisionisashortcutthatcanbeusedfordivisorswherethehighestpowered termisone. Forexample,syntheticdivisioncanbeusedtodivide3 x 3 − 4 x 2 + x + 5 byx + 1 .
Section1-AlgebraII-M athQ RH
55
First,setthedivisorequaltozero. x+1 =0
⇒ x = − 1
Forsyntheticdivision,aspecialdiagramisused.Inthisdiagram,thetopleftcornerholds thex = − 1 . Ontheleftsideoftheverticalline,thecoefficientsofthex termsofthe dividendareplaced. Ontherightsideoftheverticalline,theconstanttermisheld.
Thefirstcoefficienttermiscarrieddowntotheanswerline(belowthehorizontalline). This answeristhenmultipliedbythe− 1 andaddedtothenextterm. Thisisrepeateduntilthe answerisaddedtotheconstant. Whateverisinthebottomrightcorneristheremainder.
So,thefinalansweris3 x 2 − 7 x + 8 −
56
3 x+1
.
Section1-AlgebraII-M athQ RH
1.4.2
R ationalF unctionsa ndA nalysis
Arationalfunctionisaquotientoftwopolynomialfunctions:r(x) =
p(x) q(x)
. Inorderfora
functiontobearationalfunctiontheremustbeatleastafirstdegreepowerofx inthe denominator. 1.4.2.a
DomainRestrictions
Sincewecannotdivideanythingbyzero,anyx valuethatmakesthedenominator polynomialequaltozeroisoutofthedomainofthefunction. Forexample,intheequationR(x) = x+5 =0
2x2 − 4 x + 5
whenx = − 5 thedenominatorwillequalzero(
⇒ x = − 5). Therefore,thedomainofthisfunctionisD = {x| x =/ − 5}
IntheequationR(x) =
x3 x2 +1
.
,however,thedomainisallrealnumbers,asx 2 + 1 = 0 isequal
to√− 1 ,whichisanimaginarynumber. 1.4.2.b
VerticalAsymptotes
Averticalasymptoteisalinewhereafunctionwillinfinitelygetcloserandclosertothe numberbutneveractuallyreachit. Theverticalasymptotesofafunctionare,generally,thesameasthedomainrestrictions. Meaningthatwhereverthefunctionisundefined,therewilllikelybeaverticalasymptotein theformx = some number . IntheR(x) =
2x2 − 4 x + 5
exampleabove,thereisaverticalasymptoteatx = − 5 .
Section1-AlgebraII-M athQ RH
57
1.4.2.c
Holes
Aholeinagraphoccurswhenafactorofbothofthepolynomialscancelsout. Whenthis occurs,thedomainexceptionisnolongeraverticalasymptote,butratherthex coordinate ofthehole. Forexample,thefunctionR(x) =
x2 − 1 x − 1
canbefactoredintoR(x) =
(x + 1)(x − 1) x − 1
. Noticethat
thereisanx − 1 factorinboththenumeratorandthedenominator,meaningtheycancel out. So,afterreducing,thefunctionisR(x) = x + 1 . Thegraphhasaholeat(1, 2) .
1.4.2.d
MultiplicityoftheVerticalAsymptote
Themultiplicityoftheverticalasymptoteaffectshowthegraphbehaves. Whilethe asymptoteisnotchanged,thegraph’sbehaviorarounditdoes.
58
Section1-AlgebraII-M athQ RH
●
Iftheverticalasymptotehasanoddmultiplicity(1,3,5,etc.),onesideofthegraph willgotopositiveinfinityandtheotherwillgotonegativeinfinity.
f (x) = ●
1 x
Iftheverticalasymptotehasanevenmultiplicity(2,4,6,etc.),bothsideswilleither gotopositiveornegativeinfinity(theywillmatcheachother).
f (x) =
1 x2
1.4.2.e
HorizontalAsymptotes
Ahorizontalasymptoteisjustlikeaverticalasymptotebutinsteadofitbeing x = some number itisy = some number . Becausehorizontalasymptotesonlyaffectend behavior,thegraphcanintersectthehorizontalasymptotesatsomepoint,unlikevertical asymptotes.
Section1-AlgebraII-M athQ RH
59
Tofindaverticalasymptote,“plugin”infinity(itisimportanttonotethatyoucannot actuallyplugininfinitysinceitisnotanumber)toseethebehaviorofthefunctionasx infinitelyincreases. Ifitapproachesanumberasx getsinfinitelylarge,ithasahorizontal ∞ asymptote. Rememberthat some number approacheszeroand some number approachesinfinity. ∞
Forexample,asx approachesinfinityin x12 ,thefunctionapproacheszero. Ifyou“plugin” infinity,youwillget ∞12 ,whichdividesonebyincreasinglylargernumbers. So,itwillget smallerandsmallerandthereforeclosertozeroastimegoeson. Inthefunction3x − 5 ,pluggingininfinityforx leaves0 − 5 ,or− 5 . Sothehorizontal asymptoteforthisfunctionisaty = − 5 . Noteaboutthissection:thisisbasicallyanintroductiontolimitsofafunction. Ifyouwant toreadaheadorgetabettergrasponthesubject,gotothecalculusIlimitssection. 1.4.2.f
Oblique/SlantAsymptotes
Anobliqueorslantasymptoteisjustlikeahorizontalorverticalasymptoteinthatitdeals withtheendbehaviorofafunction,butitsitsatanangle(slanted,ifyouwill). Anobliqueasymptoteoccurswhenthenumerator’shighestdegreetermisexactlyone degreehigherthanthedenominator’shighestdegreeterm. Tofindtheequationofthe obliqueasymptote,dividethenumeratorbythedenominatorusingpolynomialdivision.
60
Section1-AlgebraII-M athQ RH
1.4.3
G raphingR ationalF unctions
Tographarationalfunction,followtheinstructionsbelow: 9. Factorbothpolynomials. 10. Seeifanyofthefactorscancelout. Iftheydo,notethatthatvalueofx hasaholein thegraph. Rememberthatifthereisaholeatanx coordinatetherecannotbea verticalasymptoteatthesamex value. 11. Findthex valuesthatmakethedenominatorequaltozero. Makeanotethatthese areverticalasymptotes. Alsonotethemultiplicityoftheseasymptotes,asthese determineendbehavior. 12. Findthex valuesthatmakethenumeratorequaltozero. Thesearethex interceptsofthefunction. 13. Pluginzeroforx tofindthey interceptofthefunction,ifoneexists. 14. Ifthenumeratorpolynomialofthefunctionisonedegreehigherthanthe denominatorpolynomial,thenusepolynomialdivisiontofindtheslantasymptote. 15. “Plugin”infinityforx tofindifthereareanyhorizontalasymptotes. a. Ifthefunctionhasaslantasymptote,itwillnothaveahorizontalasymptote. 16. Onceyouhavenotedallimportantinformation,plotallpointsand/orasymptotes andsketchinthegraph.
Section1-AlgebraII-M athQ RH
61
1.4.4
P olynomiala ndR ationalI nequalities
1.4.4.a
Graphically
Tographicallysolveapolynomialorrationalinequality,graphthefunctionandfindthe intervalsofx onthegraphthatsatisfiestheinequality. 1.4.4.b
Algebraically
Toalgebraicallysolveapolynomialinequality,firstfindthezeros(wheref (x) = 0 andthex intercepts)ofthefunctionandputthemonanumberline. Then,pluginnumbersthatare inbetweentheintervalsyoufoundtoseewhethertheyarepositiveornegative. Thiswill tellyouwherethefunctionisgreaterthanorlessthanzero. Youmayneedtousetheformulasforthesumordifferenceofcubes: a 3 + b 3 = (a + b )(a 2 − a b + b 2 ) a 3 − b 3 = (a − b )(a 2 + a b + b 2 ) Example:Solvex 4 > x . First,getzeroononesidebymovingthex :x 4 − x > 0 . Thiswillmakeiteasiertosolve. Factorthistofindthezeros:x (x 3 − 1 ) > 0 . Ifx = 0 ,thentheequationwillequalzero,so x = 0 isazeroofthefunction. Usethedifferenceofcubesformulatofactorx 3 − 1 : x 3 − 1 = (x − 1 )(x 2 + x + 1 ) . Setthisequaltozerotofindtheotherzeroofthefunction: (x − 1 )(x 2 + x + 1 ) = 0
62
⇒ x = 1.
Section1-AlgebraII-M athQ RH
Putthetwozerosonanumberline.
Pluginnumbersthatareintheintervals(− ∞ , 0 ) ,(0, 1 ) ,(1, ∞ ) . f (− 1 ) = (− 1 )(− 1 3 − 1 ) = 2 . Thismeansthatontheinterval(− ∞ , 0 ) they valueis always positive. 3
f ( 21 ) = ( 21 )( 21 − 1 ) = −
7 16
. So,ontheinterval(0, 1 ) they valuesarenegative.
f (2) = (2)(2 3 − 1 ) = 1 4 . They valuesarealsopositiveontheinterval(1, ∞ ) . Onthenumberline:
Thereforex 4 > x ontheintervals(− ∞ , 0 ) ⋃ (1, ∞ ). Toalgebraicallysolvearationalinequality,followthesamestepsasabove,butalsoputthe verticalasymptotesonthenumberline. Tip:thereisactuallyapatternwiththepositivesandnegatives. Generally,itwilljust alternatebetweenpositiveandnegative. However,ifoneofthezeroshasaneven multiplicity,thesignwillremainthesameoverthatzero.
Section1-AlgebraII-M athQ RH
63
1.4.5
F actoringP olynomialsa ndR ealZ eros
Zerosarethenumbersthatmakeafunctionf (x) equaltozero(thex intercepts). The factorsofapolynomialareexpressionsthat,whenmultipliedtogether,createa polynomial. 1.4.5.a
TheFactorTheorem
Thefactortheoremstatesthatiff (c) = 0 ,then(x − c ) isafactorofthepolynomial. 1.4.5.b
TheRemainderTheorem f(x)
Theremaindertheoremstatesthatiff (c) =/ 0 ,thenf (c) istheremainderof (x − c) . Thistheoremisusedinsyntheticsubstitution. 1.4.5.c
Descartes’RuleofSigns
Descartes’RuleofSignsisusedtofindthen umbero fpositiveandnegativezerosthata polynomialfunctionhas. Thisisdonebylookingatthenumberofsignchangesthata polynomialhas. Forthenumberofpositivezeros,firstputthepolynomial“inorder”:decreasingpowersof x . Then,thenumberofpositivezeros,ifn isequaltothenumberofsignchanges,isequal ton orn − 2 (countingdownbytwountilyoureachzero). Toaccountforpossiblecomplex zeros(readmoreabouttheseinthenextsection),weneedtocountdownbytwo,as complexzerosalwayscomeinpairs.
64
Section1-AlgebraII-M athQ RH
Forthenumberofnegativezeros,repeatthestepsabove,butfirstplugin− x . Example:Findthenumberofpositiveandnegativerootsthatthefunction f (x) = x 3 + 3 x 2 − 1 3x − 1 5 has. Thereisonesignchangeintheoriginalfunction(between3 x 2 and− 1 3x ). Therefore,there isexactlyonepositivezero. Sincecountingdowntwofromoneproduces− 1 ,anditis impossibletohavenegativeonezeros,weknowthatthereisexactlyonepositivezero. Pluggingin− x tothefunctiongivesf (− x ) = − x 3 + 3 x 2 + 1 3x − 1 5 . Therearetwosign changesinthisfunction. Thismeansthatthereareeither2 or0 negativezeros. 1.4.5.d
RationalZerosTheorem
Therationalzerostheoremstatesthateveryrationalzeroofthefunction f (x) = a n x n + a n−1 x n−1 + ... + a 1 x + a 0 hastheform±
p q
wherep isafactorofa 0 andq isa
factorofa n . 1.4.5.e
CombiningToolstoFindZerosofPolynomial
Example:Findthezerosofthefunctionf (x) = x 3 + 8 x 2 + 1 1x − 2 0 . First,usetheruleofsignstofindthenumberofzeros. f (x) hasonesignchange,and thereforeithasonepositivezero. f (− x ) = − x 3 + 8 x 2 − 1 1x − 2 0 hastwosignchanges,soit haseither2 or0 negativezeros. Then,applytherationalzerostheorem. p isequalto− 2 0 . Allofthefactorsof− 2 0 are p :± (1, 2, 4, 5, 10, 20) . q isequalto1 ,soq mustbeequalto1 . Becauseq isequalto1 , p q
: ± (1, 2, 4, 5, 10, 20) .
Section1-AlgebraII-M athQ RH
65
Now,performsyntheticsubstitution. Syntheticsubstitutionisverysimilartosynthetic divisionofpolynomials,butinsteadofhavingonedivisor,therewillbemultiple.
Thesetupisthesameasnormalsyntheticdivision,butinsteadofthedivisorbeinginthe p
topleftcorner,itisontheleftsideinlinewiththequotient. Startwiththefirstpossible q factor. Noticehowtheremainderiszero. Fromthefactortheorem,weknowthatthis meansthatx 2 + 9 x + 2 0 isafactorofthefunction. So,thepositivezeroofthefunctionis x = 1 . Nowthatthefunctionis“depressed”(brokendownintoaseconddegreepolynomial),we canfactoritouttofindthenegativezeros,iftheyexist. x 2 + 9 x + 2 0 canbefactoredinto(x + 4 )(x + 5 ) ,meaningthatthetwonegativezerosare x = − 4 andx = − 5 .
1.4.6
I maginaryandComplexN umbers
Animaginarynumberisanon-realnumber. Usually,thesenumbersaremultiplesofi ,or thesquarerootofnegativeone. Forexample,√− 7 isanimaginarynumber. Itcanbebrokendowninto√7 · √− 1 ori√7 . Acomplexnumberisamixtureofrealandimaginarynumbers,usuallyintheforma + b i , wherea andb arerealnumbers. Anyrealorimaginarynumbercanbewrittenoutasa complexnumber(i.e.3 + 0 i or0 + i√7 ).
66
Section1-AlgebraII-M athQ RH
1.4.6.a
PowerPattern
Justlikerealnumbers,youcanraisei toexponents. Forexample,i2 is√− 1 · √− 1 ,which equals− 1 . Wecanraisei toanyexponentwe’dlike,butitturnsoutthatthereisapattern thatarises: i = √− 1 i2 = − 1 i3 = − i i4 = 1 i5 = √− 1 i6 = − 1 i7 = − i i8 = 1 Noticehowafterafteri wasraisedtothefourthpower,thepatternrepeatsitself. Thispatterncanbeusedtosimplifylargeexponents. Sincei4 = 1 ,findthehighestmultiple offourthatcan“fit”intotheexponent. Then,multiplythisbyi raisedtothedifference betweentheoriginalexponentandthehighestmultipleoffour. Example:Findthevalueofi59 . Thelargestmultipleoffourthatcanfitintotheexponentis5 6 . 5 9 − 5 6 = 3 . So, i59 = i56 · i3 = (i4 )14 · i3 = 1 14 · i3 = i3 = − i .
Section1-AlgebraII-M athQ RH
67
1.4.6.b
ComplexConjugate
Torationalizeaquotientofcomplexnumbers(like 5 + 7i )youneedtouseacomplex 2 − 3i conjugate. Acomplexconjugateisafractionthatisequaltoonethatismadeupoftwo complexnumbers. Thesecomplexnumbersarethesame(whichiswhythefractionis equaltoone)andarethedenominatoroftheoriginalquotientwiththeoppositesignon theimaginarypart. Forexample,torationalize 5 + 7i youwouldmultiplyitbythecomplexconjugate2 + 3i . 2 − 3i 2 + 3i
1.4.7
C omplexZ eros
Forfunctionsf (x) = a x n + b x n − 1 + c x − 2 + ... + z ,ifallcoefficientsarerealnumbers,complex zerosmustcomeinconjugatepairs. Thismeanstheycancelout. Thezeroswouldbea + b i anda − b i . 1.4.7.a
FindingEquationsfromZeros-Example
Findtheequationofthedegreefourfunctionwiththezeros1 ,1 ,and− 4 + i . Usethesezerosandthedegreetosetupafactoredversionoftheequation. Usea asa placeholderforthepossiblecoefficient. f (x) = a (x − 1 )2 (x − (− 4 + i))(x − (− 4 − i)) Rememberthatcomplexzerosmustcomeinconjugatepairs. Also,since1 waslistedasa zerotwice,ithasamultiplicityoftwo,anditthereforesquared.
68
Section1-AlgebraII-M athQ RH
Distributethenegatives,expand,andsimplify. a (x 2 − 2 x + 1 )(x + (4 − i))(x + (4 + i)) a (x 2 − 2 x + 1 )(x 2 + 4 x + ix + 4 x − ix + 1 7) a (x 2 − 2 x + 1 )(x 2 + 8 x + 1 7) f (x) = a (x 4 + 6 x 3 + 2 x 2 − 2 6x + 1 7) Withoutapointonthegraphwecannotfinda ,sothisisthefinalanswer. 1.4.7.b
FindingZerosfromaFunction-Example
Findallzeros,realandnon-real,ofthefunctionf (x) = 3 x 4 + 5 x 3 + 3 5x 2 + 4 5x − 1 8 . UsingDescartes’RuleofSignsweknowthatf (x) hasonerealpositivezeroandthreeorone negativezeros. Fromtherationalzerosfunction,weknowthatthepossiblerationalzeros p
are q : ± (1, 2, 3, 6, 9, 18, 31 , 32 ) . Usesyntheticsubstitutiontofindthezerosandbreakdownthepolynomial.
Now,thefunctionisbrokendownwhereweknowthattwoofthefactorsarex = 31 ,− 2 and therestofthefunctionis3 x 2 + 2 7 = 0 . Whensolvedforx ,thisfunctionproduces x = ± √− 9 = ± 3 i . So,thetworealsolutionsarex = 31 , − 2 andthetwonon-realsolutionsarex = 3 i, − 3 i . Section1-AlgebraII-M athQ RH
69
1.5
ExponentialandLogarithmicFunctions
1.5.1
C ompositeF unctions
Acompositefunctionisafunctionthatgetspluggedintoanother. Example:Iff (x) = 2 x + 3 andg (x) = 7 x + 7 ,findthecompositefunctionf ° g (x) (pronouncedf ofg ofx ). Plugg (x) intof (x) whereveryouseex . f ° g (x) = f (g(x)) = 2 (7x + 7 ) + 3 = 1 4x + 1 7 Note:youcanalsodof ° f (x) ,whereyouplugthefunctionbackintoitself. 1.5.1.a
DomainofCompositeFunctions
Considerthedomainrestrictionsofboththecompositefunctiona ndthefunctionthatwas pluggedin. So,ifyouhaveacompositefunctionf ° g (x) ,findthedomainofbothf ° g (x) andg (x) .
1.5.2
O net oO neF unctions
Aonetoonefunctionisafunctioninwhicheveryx hasoney andeveryy hasonex . A onetoonefunctionwillpassboththeverticallinetestandthehorizontallinetest.
70
Section1-AlgebraII-M athQ RH
1.5.3
I nverseF unctions
Aninverseofafunctionisthereverseoftheoriginalfunction:thedomainoftheoriginal functionistherangeoftheinverseandviceversa. So,iff (a) = b thenf −1 (b) = a . Totestifafunctionisaninverseofanotherusethefollowingtest: f (g(x)) = g (f (x)) = x 1.5.3.a
Notation
Theinverseoffunctionf (x) isdenotedasf −1 (x) ,readas“f inverse”. 1.5.3.b
ProcessofFindinganInverseFunction
Tofindtheinverseofafunction,changey tox andputy ’swhereverx is. Then,solvefory . RationalFunctionExample:Findtheinverseoff (x) =
2x + 1 x − 1
.
Beginbychangingx ’stoy ’sandviceversa. x=
2y + 1 y − 1
Crossmultiplytogetridofthefraction. x 1
=
2y + 1 y − 1
⇒ xy − x = 2y + 1
Moveally ’stoonesideandeverythingelsetotheotherside. xy − 2 y = x + 1
Section1-AlgebraII-M athQ RH
71
Undistributethey ’sontheleftside. Solvefory . y (x − 2 ) = x + 1 y = f −1 (x) =
x + 1 x − 2
1.5.3.c
GraphsandInverseFunctions
Thegraphofafunction’sinverseisthegraphoftheoriginalfunctionreflectedacrossthe liney = x .
72
Section1-AlgebraII-M athQ RH
1.5.4
E xponentialFunctions
Anexponentialfunctionisafunctionwhereaconstantisraisedtotheargument:f (x) = a x . ➔ Rememberthatanumberraisedtothezerothpoweralwaysequalsone. ➔ Whenx > 1 ,thegraphwillgetsteepfast. ➔ Whenx < 1 ,thisisthesameas√a . Thismeansthattherewillbeverylittlegrowth. x
➔ Whenx < 0 ,tinynumbersarecreated. Therefore,thegraphwillapproachzero.
1.5.4.a ●
Transformations
f (x) = c · a x :c isthemultiplier,whichenhancesthegrowth ○
Generally,thismultiplierdoesn’tmakethatmuchofadifferenceinhowthe graphlooks. Itismainlynoticeableinthey intercept.
●
f (x) = a x − h :h shiftsthegraphhorizontally ○
Noticethenegativesigninfrontofh . Justasbefore,thismeansthatthe graphmovesoppositethesign. So,iftheexponentisx + 2 ,h = − 2 andthe graphwillshifttwototheleft.
●
f (x) = a x + k :k shiftsthegraphupordown
●
f (x) = − (a x) :thenegativesignoutfrontofthetermflipsthegraphoverthex axis ○
●
Itchangesthesignofthey values,flippingitupordown.
f (x) = a −x :thenegativesignontheexponentflipsthegraphoverthey axis ○
Itchangesthesignofthex values,flippingitsidetoside.
Section1-AlgebraII-M athQ RH
73
1.5.4.b
CheckingforanExponential
Ifafunctionisexponential,itwillhaveacommonquotientbetweenthey values,rather thanhavingacommondifferencelikelinearandquadraticformulas. Example:Findtheequationofthefollowingfunction. x
f (x)
0
5
1
10
2
20
3
40
4
80
Noticethat8 0/40 = 4 0/20 = 2 0/10 = 1 0/5 = 2 . Therefore,2 isthecommonquotient,ora . The generalformulaforthiscommonquotientis
f(x + 1) f(x)
= a .
Noticehowf (0) = 5 . Ifanexponentialfunctionhasnoconstantf (0) wouldequal1 . Since 0 thisisnotthecaseforthisfunction,weknowthattheconstantis5 (5 = c · 2
⇒ c = 5).
Thereforetheequationofthisfunctionisf (x) = 5 · 2 x . 1.5.4.c
Euler’sNumber
Euler’snumber,alsoknownasthenaturalnumber,isaconstantapproximatelyequalto ∞
e ≈ 2 .71828 . Thisnumberwasfoundthroughtheinfinitesume = ∑
n=0
adivisionbyzeroerror,as0 ! = 1 ).
74
1 n!
(thiswillnotcreate
Section1-AlgebraII-M athQ RH
Note:ifyouwanttolearnmoreaboutinfinitesumsandhowaddingnumbersinfinitelycan produceanumber,jumptothecalculusIIsectiononseries. e isacommonbaseforexponentialandlogarithmicfunctions. 1.5.4.d
SolvingExponentialFunctions-Example
Solvee−x = (ex )2 · 2
1 e3
.
Beginbymakingeverytermabasenumbertothex power. 2
e−x = e2x · e−3 Combinethetermsontherightsideusingtheexponentruleea · eb = ea + b . 2
e−x = e2x − 3 Sincebothtermshavethesamebasenumber,youcantakethenaturallogofbothsides: x ... = x ... . − x2 = 2 x − 3 Solveforx . 0 = x2 + 2 x − 3 0 = (x + 3 )(x − 1 ) x = 1, − 3 Onceyougettheanswers,besuretocheckiftheyfalloutofthedomainrestrictions,as thatwouldmakethemextraneoussolutions.
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1.5.5
L ogarithmicFunctions
Alogarithmicfunctionistheinverseofanexponentialfunction. Ifb y = x ,theny = log b x . Thisy = formulaisessentiallyasking“whatpowerofb makesx ?”wherey isthepower. ●
y isequaltotheexponent
●
b isthebaseoftheexponentialfunction
●
x istheanswerofb y
ImportantNotationNotes: log e = ln = naturallog log 10 = log 1.5.5.a
Implementinglog-Examples
Rewrite2y + 1 =
76
x − 10 9
asalogarithmicfunction.
Section1-AlgebraII-M athQ RH
Thebaseoftheexponentis2 ,sob = 2 . Theanswertothisexponentialfunctionis x − 10 , 9 sox =
x − 10 9
. Thelogarithmicfunctionsolvesfortheexponent,sosetthelogfunctionequal
toy + 1 andsolvefory .
log 2 ( x − 10 ) = y +1 9 y = log 2 ( x − 10 )−1 9 1 Rewritelog 6 216 = x asanexponentialfunctionandsolveforx . 1 Thebaseis6 ,theexponentisx ,andtheansweris216 .So,
6x =
1 216
1 6 3 isequalto2 16 . Toget216 ,maketheexponentnegative. Therefore,
x = −3 1.5.5.b
DomainandRange
Thedomainofalogarithmicfunctionisdeterminedbytheargumentofthelogarithmic function,astheargument(whatcomesafterthelog symbol)mustbegreaterthanzero. So,thedomainisD = {x ε R| x > 0 } ,or,ifanh transformationexists,thenD = {x ε R| x > h } . Therangeofalogarithmicfunctionisallrealnumbers. Noticehowthisisthedomainandrangeofanexponentialfunctionbutflipped.
Section1-AlgebraII-M athQ RH
77
1.5.5.c
Graphing
Tographalogarithmicfunction,finditsinverse,andflipthatgraphacrosstheliney = x . Example:graphf (x) = − ln (x − 2 ) . Flipthex ’sandy ’sandexpandln intolog form. x = − log e(y − 2 ) Dividebothsidesby− 1 . Togetridofthelog ,e xponentiate. − x = log e(y − 2 ) e −x = y − 2 Solvefory . y = e −x + 2 They interceptis(0, 3 ) ,thereisahorizontalasymptotesaty = 2 ,andthenegativesignin frontofx flipsthegraphoverthey axis. Plotthis,thenreflectitacrosstheliney = x to getthegraphforf (x) = − ln(y − 2 ) .
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Section1-AlgebraII-M athQ RH
1.5.5.d ●
●
●
●
LawsofLogarithms
Iftheargumentisequaltoone,theanswer/exponentiszero. ○
log a 1 = x
○
ax = 1
⇒ a
0
= 1
⇒ x = 0
Iftheargumentisequaltothebase,theanswerisone. ○
log a a = x
○
ax = a
⇒ a
1
= 1
⇒ x = 1
Ifthebaseisraisedtoanequationwiththesamebase,theargumentistheanswer.
○
alog a M = x
○
Whenyouconvertthisintologarithmicform,youseethatlog a M = log a x .
○
Sinceallelseisheldconstantx mustbeequaltoM .
○
Theyareessentiallyundoingeachother.
Ifthebaseisequaltotheargumentraisedtothepower,thenthepowerisequalto theanswerortheanswerisequaltothepower.
○
log a ar = x
○
Ifyouconvertthisintoexponentialform,youseethata x = a r .
○
Sincethebaseisthesame,x mustbeequaltor . Therefore,log a ar = r .
●
log a (MN ) = log a M + log a N
●
log a ( M ) = log a M − log a N N
●
log a M r = r · log a M
●
Exponentsandlogarithmscancanceleachotherout. r
○
er · ln a = eln a
○
Rememberthatln (thenaturallog)isequaltolog e . Whenalogarithmisin theexponentofanexponentialfunctionwiththesamebase,theywillcancel eachotherout.
○
r
Therefore,eln a = ar .
Section1-AlgebraII-M athQ RH
79
●
IfandonlyifM = N ,thenlog a M = log a N .
●
log a M = ○
log b M log b a
,whereb isabaseofyourchoice.
Thisishelpfulforcalculatinglogarithmsincalculators. Alotofcalculators havealog button,butitisonlyforlog 10 . So,ifyouwanttocalculateanon basetenlogarithmusingacalculator,youwilllikelyneedtoapplythisrule.
1.5.6
F inancialM odelF ormulas
SimpleInterest:I = P rt ➔ I isequaltointerest,P istheprincipalamount(i.e.theamountstartedwith),r is thepercentinterestrate,t istimeinyears CompoundInterest:A = P · (1 + nr )nt ➔ Compoundinterestisinterestonthec urrentamount,soitchangeseveryyearandis addedtothebalanceperiodically. ➔ P istheprincipalamount,r isthepercentinterestrate,n isthenumberoftimes thattheinterestiscompoundedperyear,t istimeinyears ➔ then t termshowshowmanytimestheinterestwasaddedtotheprincipalamount ContinuousCompounding:A = P e rt ➔ Ifsomethingiscontinuouslycompounding,n isapproachinginfinity. Itisessentially compoundingeverysecond. ➔ P istheprincipalamount,r isthepercentinterestrate,t istimeinyears
80
Section1-AlgebraII-M athQ RH
EffectiveRateofInterest:r e = (1 + nr )n − 1 ,r e = e r − 1 ➔ Theeffectiveinterestrateisusedtocompareloansthathavedifferentterms. The formulaisusedtoadjusttheinterestamounttotakeintoaccountcompounding. ➔ Thesecondformulaisusedwhensomethingisbeingcompoundedcontinuously, ratherthanperiodically. ➔ r isthepercentinterestrate,n isthenumberoftimesthattheinterestis compoundedperyear PresentValue:P = A · (1 + nr )−nt ,P = A · e −rt ➔ Thepresentvalueformulacalculateswhatyouneedtoputin(principalamount)to getbackacertaininterest. ➔ Thesecondformulaisusedwhensomethingisbeingcompoundedcontinuously ➔ A istheamountofinterest,r isthepercentinterestrate,n isthenumberoftimes thattheinterestiscompoundedperyear,t istimeinyears
1.5.7
G rowtha ndD ecay
Therearetwodifferenttypesofgrowthanddecay:uninhibitedgrowthandlogistical models. 1.5.7.a
UninhibitedGrowthandDecay
Uninhibitedgrowthanddecayoccursinbacteriacolonies,radioactivematerials,etc. Itis calleduninhibitedgrowthanddecay,asithas“nolimits”(forthemostpart:inthereal world,therearestillpopulationlimitswithbacteria,butwedisregardthoseinthiscase).
Section1-AlgebraII-M athQ RH
81
JustasweuseEuler’snumberincontinuouscompounding,itisalsousedinuninhibited growthanddecay. Growthformula:N (t) = N 0 e kt Decayformula:N (t) = N 0 e −kt ●
N 0 istheinitialvalue/population
●
k istherateofgrowth/decay
●
t istheamountoftimethathaspassed
1.5.7.b
Newton’sLawofCooling
Newton’slawofcoolingisusedwhensomethinghotiscoolingdown(thisisadecaymodel). u (t) = T + (u 0 − T )e kt ●
u (t) isthetemperatureaftertimehaspassed
●
T isthesurroundingtemperature
●
u 0 istheinitialtemperature
82
Section1-AlgebraII-M athQ RH
1.5.7.c
LogisticalModel
Logisticalmodelsareusedwhenthereisalimitonthegrowth,likethepopulationof peopleonEarth. c p (t) = 1 + ae −bt
●
c 1 + a
istheinitialvalue
○
youwillusec andtheinitialvaluegiventosolvefora
●
c isthecarryingcapacity,orwhatthegrowthislimitedto
●
b istherate
Whatothernotesdoyouw ishtoaddforAlgebraII?Remember,tosharethemon MathQRH.com!
Section1-AlgebraII-M athQ RH
83
“Geniusisathingthathappens,notakindofperson” -JordanEllenberg
1.6
MyNotesforAlgebraII
84
Section1-AlgebraII-M athQ RH
1.6
MyNotesforAlgebraII(con’t)
Section1-AlgebraII-M athQ RH
85
1.6
MyNotesforAlgebraII(con’t)
86
Section1-AlgebraII-M athQ RH
“SafetyWarning: Neverd ivideb yz erou nlessa licensedmathematicianisp resent.” -JordanEllenberg
Section2-Precalculus
Section2-Precalculus-M athQ RH
87
2.0
SummarySheet
UnitCircle
GraphingForm:(x, y) = (cos θ, sin θ)
88
Section2-Precalculus-M athQ RH
TrigonometricFunctions ArcLength:s = r θ SectorArea:A = 21 r 2 θ LinearSpeed:v =
s t
AngularSpeed:ω =
= rω θ t
= 2 πf
TimeandAngleConversions: 1° = 60′ 1 ° = 3 600 ′′ 1 ′ = 1 /60 ° 1 ′′ = 1 /3600 ° where′ representsminutesand′′ representsseconds RadianandDegreeConversions: 1 revolution = 3 60 ° = 2 π radians 1 80 ° = π radians 1 ° = π /180 radians 1 radian = 1 80/π °
Section2-Precalculus-M athQ RH
89
TrigonometricDefinitions:
sin θ =
y r
=
opposite hypotenuse
csc θ =
r y
=
1 sin θ
cos θ =
x r
=
adjacent hypotenuse
sec θ =
r x
=
1 cos θ
cot θ =
x y
=
1 tan θ
tan θ =
y x
=
opposite adjacent
r = √x 2 + y 2 PeriodicFunctionDefinition:f (θ + p ) = f (θ) Relationship:sin x = c os (x − 2π )
PythagoreanIdentities: sin 2 θ + c os2 θ = 1 tan 2 θ + 1 = sec 2 θ c ot2 θ + 1 = c sc 2 θ Period:T =
2π ω
PhaseShift=
90
= ϕ ω
π ω
,dependingontheintervalatwhichthegivengraphcompletesacircle
Section2-Precalculus-M athQ RH
AnalyticTrigonometry PropertiesofInverseFunctions: sin (sin −1 x) = x ,c os (cos−1 x) = x Even-OddIdentities: sin (− θ) = − sin θ
c os (− θ) = c os θ
tan (− θ) = − tan θ
c sc (− θ) = − c sc θ
sec (− θ) = sec θ
c ot (− θ) = − c ot θ
SumandDifferenceFormulas: c os (α + β ) = c os α cos β − sin α sin β c os (α − β ) = c os α cos β + sin α sin β sin (α + β ) = sin α cos β + cos α sin β sin (α − β ) = sin α cos β − cos α sin β tan α + tan β
tan (α + β ) = 1 − tan α tan β tan α − tan β
tan (α − β ) = 1 + tan α tan β DoubleAngleFormulas: sin (2θ) = 2 sin θ cos θ c os (2θ) = c os2 θ − sin 2 θ = 1 − 2 sin 2 θ = 2 cos2 θ − 1 2tan θ tan (2θ) = 1 − tan 2θ
Section2-Precalculus-M athQ RH
91
SquaredFormulas: sin 2 θ =
1 − cos (2θ) 2
c os2 θ =
1 + cos (2θ) 2
1 − cos (2θ)
tan 2 θ = 1 + cos (2θ) Half-AngleFormulas:
sin 2a = ±
√
cos 2a = ±
√
1 + cos a 2
tan 2a = ±
√
1 − cos a 1 + cos a
=
1 − cos a 2
1 − cos a sin a
=
sin a 1 + cos a
ApplicationsofTrigonometricFunctions Cofunctionsofcomplementaryanglesareequal. LawofSinesTheorem: Foratrianglewithsidesa ,b ,andc andoppositeanglesA ,B ,C ,respectively: sin A a
forASA,SSA,andSAAtriangles.
92
=
sin B b
=
sin C c
Section2-Precalculus-M athQ RH
LawofCosinesTheorem: Foratrianglewithsidesa ,b ,andc andoppositeanglesA ,B ,C ,respectively: c 2 = a 2 + b 2 − 2 ab cos C ,b 2 = a 2 + c 2 − 2 ac cos B ,a 2 = b 2 + b 2 − 2 bc cos A
cos A =
b 2 + c2 − a 2 2bc
,cos B =
a 2 + c2 − b 2 2ac
,cos C =
a 2 + b 2 − c2 2ab
forSASandSSStriangles. AreaofaSASTriangle: K = 21 ab sin C
K = 21 bc sin A
K = 21 ac sin B
AreaofaSSSTriangle(Heron’sFormula): K = √s(s − a )(s − b )(s − c ), wheres = 21 (a + b + c ) PolarCoordinatesandVectors PolartoRectangular:x = r cos θ ,y = r cos θ y
RectangulartoPolar:r 2 = x 2 + y 2 ,θ = tan −1 ( x ) SymmetryTests: ●
polaraxissymmetry:replaceθ with− θ
●
y -axissymmetry:replaceθ withπ − θ
●
symmetryalongthepole:replacer with− r orθ withθ + π
Section2-Precalculus-M athQ RH
93
PolarEquations: TypeofGraph
Equation
circle
± 2 asin θ ,± 2 acos θ
cardioid
a (1 ± sin θ) ,a (1 ± c os θ)
limaçonwithoutaninnerloop
a ± b sin θ ,a > b > 0
limaçonwithaninnerloop
a ± b sin θ ,b > a > 0
rose
a cos (nθ) ,a sin (nθ)
lemniscate
r 2 = a 2 sin (2θ) ,r 2 = a 2 cos (2θ)
spiral
r = e θ/4
ComplexNumber:z = x + y i (rectangularform) ●
Magnitude:|z | = √x 2 + y 2 = r
●
PolarForm:r (cos θ + isin θ)
●
ExponentialForm:r e iθ e iθ = c os θ + isin θ (Euler’sFormula)
○ ●
z 1 z 2 = r 1 r 2 e i(θ1 + θ2 )
●
z1 z2
●
z n = r n e i(nθ) ,ifz = r e iθ (DeMoivre’sTheorem)
=
r1 r2
e i(θ1 − θ2 ) (z 2 =/ 0 )
1
● z k = √r ei n (θ + 2kπ) ,supposingw = r eiθ isacomplexnumberandn ≥ 2 isaninteger n
94
Section2-Precalculus-M athQ RH
Vectors: ●
Magnitude:||v|| = √a 2 + b 2
●
UnitVector:u =
●
v = a i + b j = ||v||cos θi + ||v||sin θj
v ||v||
Section2-Precalculus-M athQ RH
95
AnalyticGeometry-ConicSections ParabolaEquation:(y − k )2 = 4 a(x − h ) ,where4 a = f ocal length CircleEquation:(x − h )2 + (y − k )2 = r 2 2
EllipseEquation: ax2 +
y2 b2
= 1
●
Thelargestterm(a orb )willdeterminethedominatingaxis
●
c 2 = a 2 + b 2 ,wherec isthedistancebetweenthecenterandthefocus
2
HyperbolaEquation: ax2 − ●
y2 b2
= 1
Thepositivetermwilldeterminethedominatingaxis
ParametricEquations:x = x (t) ,y = y (t) ●
PathofaProjectile:x (t) = (v 0 cos θ)t ,y (t) = − 21 gt2 + (v 0 sin θ)t + h
96
Section2-Precalculus-M athQ RH
Matrices Identity/EchelonMatrix:
ThedeterminantofamatrixA isa d − b c ,where
Amatrixistheinverseofanotherif A · A−1 = I Inversesof2 × 2 matriceshaveapattern:
Section2-Precalculus-M athQ RH
97
PartialFractionDecomposition: Case
Formula
1.Q hasonlynonrepeatedlinearfactors
P (x) Q(x)
2.Q hasarepeatedlinearfactor
P (x) Q(x)
=
=
A1 x − a1
A1 x − a
+
3.Q containsanonrepeatedirreducible quadraticfactor 4.Q containsarepeatedirreducible quadraticfactor
+ ... +
An x − an
A2 (x − a)2
+ ... +
An (x − a)n
Ax + B ax2 + bx + c A1 x + B1 ax2 + bx + c
2.1
TrigonometricFunctions
2.1.1
A ngles,ArcLength,CircularMotion
2.1.1.a
A2 x − a2
+
+
A2 x + B2 (ax2 + bx + c)2
+ ... +
An x + Bn (ax2 + bx + c)n
BasicVocabulary
StandardPosition-thevertexisattheoriginofarectangularcoordinateplaneandthe initialsidecorrespondswiththepositivex axis. ●
θ ispositivewhentheterminalsideismovedcounter-clockwisefromtheinitial side.
●
98
θ isnegativewhentheterminalsideismovedclockwisefromtheinitialside.
Section2-Precalculus-M athQ RH
2.1.1.b
MinutesandSeconds
Theconversionsbetweendegrees,minutes,andsecondsareasfollows: 1° = 60′ 1 ° = 3 600 ′′ 1 ′ = 1 /60 ° 1 ′′ = 1 /3600 ° where′ representsminutesand′′ representsseconds. Converting-Examples 1. Convert6 5 ° 9 ′17 ′′ toadecimalindegrees. a. The6 5 isalreadyindegrees,sowecanleavethatalone. Forthe9 and1 7 , usetheminutetodegreesandsecondstodegreesconversionrates.
65° (9′ ·
1 ° 1 ° ) (17′′ · 3600 ) 60
6 5 ° + 0 .15 ° + 0 .0047 ° = 6 5.1547 ° 2. Convert3 2.479 ° intoD ° M ′S ′′ a. First,pulloutthewholenumber(3 2 ). ThiswillbetheD ° portionofthe answer. Now,weareleftwith0 .479 ° (3 2.479 ° = 3 2 ° + 0 .479 ° ). b. Multiplytheleftoverdecimalby6 0 (1 ° = 6 0 ′ )tofindtheminutesportionof theanswer. 0 .479 ° · 6 0 ′ = 2 8.74 ′ c. Aswedidinthefirststep,pulloutthewholenumber(2 8 ). Wearenowleft with0 .74 ′ .
Section2-Precalculus-M athQ RH
99
d. Multiplythisdecimalby6 0 (1 ′ = 6 0 ′′ )tofindtheS ′′ portionoftheanswer. 0 .74 ′ · 6 0 ′′ = 4 4.4 ′′ e. Wecandisregardthedecimalportionofthisanswer:roundtothenearest wholenumber,thenputalltheanswersintothecorrectform. 3 2 ° 28 ′44 ′′ 2.1.1.c
ArcLength
Foracircleofradiusr ,acentralangleofθ radianssubtendsanarcwhoselengthiss = r θ .
2.1.1.d
RadiansandDegrees 1 revolution = 3 60 ° = 2 π radians 1 80 ° = π radians 1 ° = π /180 radians 1 radian = 1 80/π °
100
Section2-Precalculus-M athQ RH
2.1.1.e
AreaofaSector
TheareaA ofthesectorofacircleofradiusr formedbythecentralangleθ radiansis A = 21 r 2 θ .
2.1.1.f
LinearSpeed
Supposethatanobjectmovesonacircleofradiusr ataconstantspeed. Ifs isthe distancetraveledintimet onthiscircle,thenthelinearspeedoftheobjectisdefinedas
v = st .
2.1.1.g
AngularSpeed
Theangularspeedω (Greeklowercaseomega)ofanobjectistheangleθ radiansswept out,dividedbytheelapsedtimet . Thatis,
ω=
θ t
=
v r
= 2 πf ,wheref istherevolutionsperunit
Section2-Precalculus-M athQ RH
101
2.1.2
T rigonometricF unctions:U nitC ircleA pproach
Aunitcircleisacirclewhosecenterisattheoriginandhasaradiusofone. The circumferenceoftheunitcircleis2 π ,asC = 2 πr . 2.1.2.a
TrigonometricFunctionsofaRealNumber
Lett = θ bearealnumberandP = (x, y) beapointontheunitcirclethatcorrespondstot . ●
SineFunction ○
associateswitht they coordinateofP andisdenotedby sin t = y
●
CosineFunction ○
associateswitht thex coordinateofP andisdenotedby c os t = x
●
TangentFunction ○
associatest withtheratioofthey coordinatetothex coordinateofP and isdenotedby y
sin t tan t = x = cos t
●
CosecantFunction 1 c sc t = 1y = sin t
●
SecantFunction 1 sec t = 1x = cos t
102
Section2-Precalculus-M athQ RH
●
CotangentFunction c ot t = xy = cos t sin t 2.1.2.b
UndefinedFunctions
●
Wheny = 0 ,c sc θ andc ot θ areundefined.
●
Whenx = 0 ,tan θ andsec θ areundefined.
2.1.2.c
TrigonometryandTriangles
sin θ =
a c
=
y r
csc θ =
c a
=
r y
cos θ =
b c
=
x r
sec θ =
c b
=
r x
tan θ =
a b
= x
cot θ =
b a
= xy
y
Section2-Precalculus-M athQ RH
103
2.1.2.d
CommonAngles
θ θ (radians) (degrees)
sin θ
c os θ
tan θ
π 6
30°
1 2
√3 2
√3 3
π 4
45°
√2 2
√2 2
1
π 3
60°
√3 2
1 2
√3
c sc θ 2 √2 2√3 3
sec θ
2√3 3
c ot θ √3
√2 2
1 √3 3
Note:itishelpfulifyoumemorizethesin andc os oftheseangles,astheyarevery commonlyaskedfor. Alternatively,youcouldkeephandytheunitcircleatthebeginningof thischapter. 2.1.2.e
Symmetry
Circlesaresymmetric,sothevaluesaboveareheldconstantforeachangle’s correspondingangles(like6π ,5π6 ,7π6 ,and11π 6 ),butthesignswillchangedependingonthe quadrant.
104
Section2-Precalculus-M athQ RH
2.1.2.f
FindingRadius
Theradiusofthecircleisfoundbyusingtheequationr = √x 2 + y 2 .
2.1.3
P ropertieso fT rigonometricF unctions
Function
Symbol
Domain
Range
sine
f (θ) = sin θ
{x ε R}
[− 1 , 1]
cosine
f (θ) = c os θ
{x ε R}
[− 1 , 1]
tangent
f (θ) = tan θ {x ε R} ,exceptoddintegermultiplesof π 2
cosecant
f (θ) = c sc θ
secant
{x ε R} ,exceptintegermultiplesof 2π
{y ε R} (− ∞ , − 1 ] ⋃ [1, ∞)
f (θ) = sec θ {x ε R} ,exceptoddintegermultiplesof π (− ∞ , − 1 ] ⋃ [1, ∞) 2
cotangent f (θ) = c ot θ
{x ε R} ,exceptintegermultiplesof 2π
{y ε R}
Rememberthat 2π = 9 0 ° andπ = 1 80 ° . 2.1.3.a
PeriodicFunctionsandtheFundamentalPeriod
Afunctionf iscalledperiodicifthereisapositivenumberp withthepropertythat wheneverθ isinthedomainoff soisθ + p ,and f (θ + p ) = f (θ) Ifthereisasmallestsuchnumberp ,thissmallestvalueiscalledthefundamentalperiodof f .
Section2-Precalculus-M athQ RH
105
2.1.3.b
SignsofTrigonometricFunctions
2.1.3.c
PythagoreanIdentities
sin 2 θ + c os2 θ = 1 tan 2 θ + 1 = sec 2 θ c ot2 θ + 1 = c sc 2 θ 2.1.3.d
FindingExactValuesUsingIdentities-Example
π Findthevalueof csc12 π + c os2 16 . 16
Usethedefinitionofc sc togettheequationintermsofsin andc os . 1 π 1/sin2 16
π + c os2 16
Flipandmultiplytogetridofthefraction. π π sin 2 16 + c os2 16
Rememberthatsin 2 θ + c os2 θ = 1 ,therefore π π π sin 2 16 + c os2 16 = csc12 π + c os2 16 =1 16
106
Section2-Precalculus-M athQ RH
2.1.3.e
FindingExactValuesGivenOneValueandSignofOther-Example
Giventhatsin θ =
1 3
andc os θ < 0 ,findtheexactvalueofeachoftheremainingfivetrig
functions. Sincesin ispositiveandc os isnegative,weknowthatthevalueliesinquadranttwo. y
Rememberthatsin θ = r . Sincesin θ =
,y = 1 andr = 3 .
1 3
Wecanusetheformulafortheradiustofindx . r = √x 2 + y 2 3 2 = x2 + 1 2
⇒r
2
= x2 + y 2
⇒ x = √8 = − 2√2
Thex valuemustbenegative,astheradiuscannotbeandweweregiventhatc os θ < 0 . Usethedefinitionsofthetrigonometricfunctionsandthevaluesofx , y, andr tofindthe valuesoftheremainingfunctions. c os θ = c sc θ =
r y
=
3 1
x r
= −
2√2 3
= 3 ,sec θ =
r x
,tan θ = = −
3 2√2
y x
= −
= −
1 2√2
3√2 4
= −
√2 4
,c ot θ =
x y
= − 2 √2
Therearemultiplewaystodothis,thisisjustoneofthem. 2.1.3.f
EvenandOddPropertiesTheorem sin (− θ) = − sin θ
c os (− θ) = c os θ
tan (− θ) = − tan θ
c sc (− θ) = − c sc θ
sec (− θ) = sec θ
c ot (− θ) = − c ot θ
Section2-Precalculus-M athQ RH
107
2.1.4
GraphsoftheSineandCosineFunctions
2.1.4.a
SineFunction
x
y = sin x
(x, y)
0
0
(0, 0)
π 6
1 2
( 6π , 21 )
π 2
1
( 2π , 1)
5π 6
1 2
( 5π , 21 ) 6
π
0
(π, 0)
7π 6
−
( 7π , − 21 ) 6
3π 2
−1
( 3π , − 1) 2
−
11π 6
2π
1 2
1 2
0
( 11π , − 21 ) 6 (2π, 0)
D = {x ε R}, R = [− 1 , 1] ●
Thefunction/shapeofthegraphrepeatsafterafullcircle,or2 π .
●
Thex interceptsare..., − 2 π, − π , 0, π, 2π, 3π, ... (thesegooninthispattern forever,hencetheellipses).
108
Section2-Precalculus-M athQ RH
2.1.4.a
SineFunction(con’t)
●
Itisanoddfunction
●
Themaximumvalue(1)occursatx = ..., −
●
Theminimumvalue(-1)occursatx = ..., −
2.1.4.b
CosineFunction
x
y = c os x
(x, y)
0
1
(0, 1)
π 3
1 2
( 3π , 21 )
π 2
0
( 2π , 0)
2π 3
π
−
1 2
−1
( 2π , − 21 ) 3 (π, − 1 )
4π 3
3π 2
0
( 3π , 0) 2
5π 3
1 2
( 5π , 21 ) 3
1
(2π, 1)
2π
3π π 5π 9π , 2 , 2 , 2 , ... 2 π 3π 7π 11π , 2 , 2 , 2 , ... 2
−
1 2
( 4π , − 21 ) 3
D = {x ε R}, R = [− 1 , 1] ●
Thefunction/shapeofthegraphrepeatsafterafullcircle,or2 π .
Section2-Precalculus-M athQ RH
109
2.1.4.b ●
CosineFunction(con’t)
Thex interceptsare..., −
3π , 2
− 2π , 2π , 3π , 5π , ... (thesegooninthispattern 2 2
forever,hencetheellipses). ●
Itisanevenfunction
●
Themaximumvalue(1)occursatx = ..., − 2 π, − π , 0, π, 2π, 3π, ...
●
Theminimumvalue(-1)occursatx = ..., − π , π, 3π, 5π, ...
2.1.4.c
Relationship sin x = c os (x − 2π )
2.1.4.d
Transformations
Ifω > 0 ,theamplitudeandperiodofy = Asin (ωx) andy = Acos (ωx) aregivenby ●
●
Amplitude = |A| ○
Thisisaverticalstretch
○
− A isareflectionoverthex axis
P eriod = T =
2π ω
○
Thisisahorizontalstretch
○
A“changeinperiod”isachangeintheintervalatwhichthegivengraph completesafullcircle.
●
P hase S hif t = ○
ϕ ω
y = c os (x − ϕ) + β ory = sin (x − ϕ) + β
110
Section2-Precalculus-M athQ RH
2.1.4.e
GraphingUsingKeyPoints-Example
Ify = 2 sin (2x) ,A = 2 andT =
2π ω
=
2π 2
= π .
Becausetheamplitudeis2 ,therangeisnowR = [− 2 , 2] . Oneofthepointsis(0, 0) becauseitisasinegraphthathasnophaseshifts. Theperiodis π ,sooneofthepointsis(π, 0) . Halfofarotationoccursat( 2π , 0) . Themidpointsbetween thesetwointerceptsare( 4π , 2) and( 3π4 , − 2 ) .
2.1.4.f
FindingtheEquationUsingtheGraph-Example
Thegraphgoesthroughtheorigin,soweknowthatitisasinegraph. Sincethegraphgoes upfirstaftertheorigin,A ispositive. Themaximumsandminimumsoccurat|2 | ,therefore A = 2 . Thecircleiscompletedat4 . SinceT = 4 ,4 = y = 2 sin ( 2π x)
Section2-Precalculus-M athQ RH
2π ω
⇒ ω =
π 2
. Therefore,
111
2.1.5
G raphsofT angent,Cotangent,Secant,a ndC osecantFunctions
2.1.5.a x
TangentFunction y = tan x
(x, y)
−
π 3
− √3
(− 3π , − √3)
−
π 4
−1
(− 4π , − 1)
−
π 6
0
−
√3 3
(− 6π , − √33 )
0
(0, 0)
π 6
√3 3
π 4
1
( 4π , 1)
π 3
√3
( 3π , √3)
( 6π , √33 )
D = {x ε R} except odd multiples of 2π , R = {y ε R} ●
Thefunctioncyclesat(− 2π , 2π ) orafterπ
●
Itisanoddfunction
●
Thex interceptsoccurat..., − 2 π, − π , 0, π, 2π, 3π, ...
●
Theverticalasymptotesareatx = ..., −
112
3π , 2
− 2π , 2π , 3π , ... 2
Section2-Precalculus-M athQ RH
2.1.5.b x
CotangentFunction y = c ot x
(x, y)
π 6
√3
( 6π , √3)
π 4
1
( 4π , 1)
π 3
√3 3
π 2
0
2π 3
3π 4
−1
( 3π , − 1) 4
5π 6
− √3
( 5π , − √3) 6
−
( 3π , √33 )
√3 3
( 2π , 0) ( 2π 3 , −
√3 3 )
D = {x ε R} except multiples of π, R = {y ε R}
●
Thefunctioncyclesatπ
●
Itisanoddfunction
Section2-Precalculus-M athQ RH
113
2.1.5.c
WritingDomain
Insteadofsaying“D = {x ε R} except odd multiples of 2π ”,youcanwrite: D = {x| x =/
kπ 2 , k is an odd integer}
or D = {x| x =/ k π, k is an integer} 2.1.5.d
Transformations
Fortangentandcotangentgraphs,findingtheperiodusestheequation π p eriod = T = ω
114
Section2-Precalculus-M athQ RH
2.1.5.e
CosecantFunction
c sc x = 1 /sin x
(Thesolidlineisthegraphofc sc (x) ,theverticallinesareasymptotes,andthedottedline alongthex axisisthesin graph) 2.1.5.f
SecantFunction
sec x = 1 /cos x
(Thesolidlineisthegraphofsec (x) ,theverticallinesareasymptotes,andthedottedline alongthex axisisthec os graph)
Section2-Precalculus-M athQ RH
115
2.1.5.g
TransformationsoftheCosecantandSecantFunctions
Forverticalstretchesofthesefunctions,itislikeyouarestretchingeitherthesineorcosine function. Theparabolasthemselvesstaythesameforthemostpart,butthesine/cosine functionswillstretch,makingtheinterceptsincreaseandtheasymptoteschange. 2.1.5.h
AmplitudesoftheseGraphs
Forthetangent,cotangent,cosecant,andsecantgraphs,therearenoamplitudesasthey goonforeverintheverticaldirection. Theycanstillhavemultipliers(likey = 5 tan x ),but theyarenotcalledamplitudes.
116
Section2-Precalculus-M athQ RH
2.2
AnalyticTrigonometry
2.2.1
T heInverseF unctions
2.2.1.a
InverseSineFunction
Definition:y = sin −1 x ifandonlyifx = sin y where− 1 ≤ x ≤ 1 and−
π 2
≤y≤
π 2
.
y = sin −1 x givesthey coordinateandfindstheangleassociated. θ
−
sin θ Note:−
π 2
π 2
−1
− −
π 3 √3 2
−
−
π 4
−
π 6
0
π 6
π 4
π 3
π 2
√2 2
−
1 2
0
1 2
√2 2
√3 2
1
alsoequals3π2 ,butthatisnottheanswerasitisnotwithintherangeofthe
function.
Section2-Precalculus-M athQ RH
117
2.2.1.b
InverseCosineFunction
Definition:y = c os−1 x ifandonlyifx = c os y where− 1 ≤ x ≤ 1 and0 ≤ y ≤ π .
θ
0
π 6
π 4
π 3
π 2
sin θ
1
√3 2
√2 2
1 2
0
2π 3
−
1 2
3π 4
−
√2 2
−
5π 6
√3 2
π −1
118
Section2-Precalculus-M athQ RH
2.2.1.c
InverseTangentFunction
Definition:y = tan −1 x ifandonlyifx = tan y where− ∞ < x < ∞ and−
π 2
≤y≤
π 2
.
θ tan θ
−
π 2
undefined
−
π 3
− √3
−
π 4
−1
− −
π 6
0
π 6
π 4
√3 3
0
√3 3
1
π 3
√3
π 2
undefined
2.2.1.d
PropertiesofInverseFunctions
f −1 (f (x)) = sin −1 (sin x) = x ,where−
π 2
≤x≤
π 2
.
f (f −1 (x)) = sin (sin −1 x) = x ,where− 1 ≤ x ≤ 1 . f −1 (f (x)) = c os−1 (cos x) = x ,where0 ≤ x ≤ π . f (f −1 (x)) = c os (cos−1 x) = x ,where− 1 ≤ x ≤ 1 . f −1 (f (x)) = tan −1 (tan x) = x ,where−
π 2
≤x≤
π 2
.
f (f −1 (x)) = tan (tan −1 x) = x ,where− ∞ ≤ x ≤ ∞ .
Section2-Precalculus-M athQ RH
119
2.2.1.e
FindingtheInverseofaTrigFunction-Example
Findtheinversefunctionf −1 off (x) = 2 sin (x) − 1 ,−
π 2
≤x≤
π 2
.
First,swapthex andy . x = 2 sin (y) − 1 Isolatesin . x + 1 = 2 sin (y) sin y =
x + 1 2
Taketheinverseofbothsides. y = sin −1 ( x + 1 ) 2 f −1 (x) = sin −1 ( x + 1 ) 2 Tofindthedomainofthisfunction,taketheoriginaldomainandpluginthenewargument. original:[− 1 , 1] ,argument: x + 1 2 −1 ≤
x + 1 2
≤1
Isolatex . −2 ≤ x+1 ≤ 2 −3 ≤ x ≤ 1 Therangeremainsthesame.
120
Section2-Precalculus-M athQ RH
2.2.1.f
SolvinganInverseTrigFunction-Example
Solve1 2sin −1 x = 3 π . Isolatetheinverse. sin −1 x =
3π 12
⇒ sin
−1 x
= 4π
Takethesineofbothsides. x = sin 4π x=
√2 2
2.2.2
T rigonometricI dentities
2.2.2.a
IdenticallyEqual,Identity,andConditionalExpression
Twofunctionsf andg areidenticallyequalif f (x) = g (x) foreveryvalueofx forwhichbothfunctionsaredefined. Suchanequationisreferredto asanidentity. Anequationthatisnotanidentityiscalledaconditionalequation.
Section2-Precalculus-M athQ RH
121
2.2.2.b
Identities
QuotientIdentities: sin θ tan θ = cos θ ,cot θ = cos θ sin θ
ReciprocalIdentities: 1 1 1 c sc θ = sin θ ,sec θ = cos θ ,cot θ = tan θ
PythagoreanIdentities: sin 2 θ + c os2 θ = 1 ,tan 2 θ + 1 = sec 2 θ ,c ot2 θ + 1 = c sc 2 θ Even-OddIdentities: sin (− θ) = − sin θ
c os (− θ) = c os θ
tan (− θ) = − tan θ
c sc (− θ) = − c sc θ
sec (− θ) = sec θ
c ot (− θ) = − c ot θ
2.2.2.c
AlgebraicTechniques
Therearefourbasictechniques: 1. Rewritingatrigexpressionintermsofsineandcosineonly 2. Multiplyingthenumeratoranddenominatorofaratiobya“wellchosen1” 3. Writingsumsoftrigonometricratiosasasingleratio 4. Factoring
122
Section2-Precalculus-M athQ RH
2.2.2.d
GuidelinesforEstablishingIdentities
1. Itisalmostalwayspreferabletostartwiththesidecontainingthemorecomplicated expression. 2. Rewritesumsanddifferencesofquotientsasasinglequotient. 3. Sometimesithelpstorewriteonesideoftheequationintermsofsineandcosine functionsonly. 4. Alwayskeepthegoalinmind. Asyoumanipulateonesideoftheexpression,keep inmindtheformoftheexpressionontheotherside.
2.2.3
S uma ndD ifferenceF ormulas
2.2.3.a
Theorem(Cosine)
Sumanddifferenceformulasforthecosinefunction: c os (α + β ) = c os α cos β − sin α sin β c os (α − β ) = c os α cos β + sin α sin β Example:Solvec os (105 ° ) . Noticehowitwouldbehardertosolvethisbyhand,as1 05 isnotacommonangle. However,itisthesumofthecommonangles4 5 ° and6 0 ° . Therefore, c os (105) = c os (45 + 6 0) Wecannowusethesumformula. c os (45 + 6 0) = (cos 45)(cos 60) − (sin 45)(sin 60)
Section2-Precalculus-M athQ RH
123
Usethetableofcommonanglestohelpsolve(it’sevenbetterifyouhavetheunitcircle memorized). √2 √3 1 ( √2 2 )( 2 ) − ( 2 )( 2 ) 1 √ 4( 2
2.2.3.b
− √6 )
Theorem(Sine)
Sumanddifferenceformulasforthesinefunction: sin (α + β ) = sin α cos β + cos α sin β sin (α − β ) = sin α cos β − cos α sin β Example:solvesin
π 12
Findawaytobreakthisangleintotwocommonangles. π 12
=
3π 12
− 2π 12 =
π 4
−
π 6
Now,usethedifferenceformulaandtheunitcircletosolve. sin ( 4π − 6π ) = (sin 4π )(cos 6π ) − (cos 4π )(sin 6π ) √3 √2 1 ( √2 2 )( 2 ) − ( 2 )( 2 ) 1 √ 4( 6
− √2 )
124
Section2-Precalculus-M athQ RH
2.2.3.c
Theorem(Tangent)
Sumanddifferenceformulasforthetangentfunction: tan α + tan β
tan (α + β ) = 1 − tan α tan β tan α − tan β
tan (α − β ) = 1 + tan α tan β 2.2.3.d
ExactValuesofInverseTrigonometricFunctions-Example
Findthevalueofsin (cos−1 21 + sin −1 53 ) . Wewantthesineofthesumoftwoanglesα = c os−1 21 andβ = sin −1 53 . Thenc os α = 0 ≤ α ≤ π andsin β =
3 5
,−
π 2
≤β ≤
π 2
1 2
,
.
Usepythagoreanidentitiestoobtainsin α (≥ 0 )andc os β (≥ 0 ). sin α = √1 − c os2 α =
√1 − = √
3 4
c os β = √1 − sin 2 β =
√1 −
16 25
1 4
9 25
=
√
=
√3 2
= 54
Now,youcanusethesinesumformula. sin (cos−1 21 + sin −1 53 ) = sin (α + β ) = (sin α)(cos β) + (cos α)(sin β) 4 1 3 = ( √3 2 )( 5 ) + ( 2 )( 5 ) =
4 √3 + 3 10
Section2-Precalculus-M athQ RH
125
2.2.3.e
WritingTrigExpressionsasAlgebraicExpression-Example
Writesin (sin −1 u + c os−1 v) asanalgebraicexpressionwithu andv . Don’tuseanytrig functionsandstaterestrictionsonu andv . Rememberthatu andv arethey valuesthatareassociatedwithacertainangle. Therefore,theymustbewithintherangesofsin andc os . Thismeansthattherestrictions onu andv are− 1 ≤ u ≤ 1 and− 1 ≤ v ≤ 1 . Letα = sin −1 u andβ = c os−1 v . Then: sin α = u , −
π 2
≤ α ≤ 2π , − 1 ≤ u ≤ 1
c os β = v , 0 ≤ β ≤ π , − 1 ≤ v ≤ 1 Because−
π 2
≤α≤
π 2
,c os α ≥ 0 . So, c os α = √1 − sin 2 α = √1 − u 2
Alsobecause0 ≤ β ≤ π ,sin β ≥ 0 . Then, sin β = √1 − c os2 β = √1 − v 2 Asaresult, sin (sin −1 u + c os−1 v) = sin (α + β ) = (sin α)(cos β) + (cos α)(sin β) = u v + √1 − u 2 + √1 − v 2
126
Section2-Precalculus-M athQ RH
2.2.3.f
TrigonometricEquationLinearinSineandCosine-Example
Solvea sin θ + b cos θ = c . Dividebothsidesby√a 2 + b 2 . a
√a
2
+ b
2
sin θ +
b
√a
2
+ b
2
cos θ =
c
√a
2
+ b
2
Thereisauniqueangleϕ, 0 ≤ ϕ ≤ 2 π ,forwhich c os ϕ =
andsin ϕ =
a
√a2 + b2
b
√a2 + b2
Meaningwecanrewriteto sin θ cos ϕ + c os θ sin ϕ =
c
√a2 + b2
whichequals sin (θ + ϕ) =
c
√a2 + b2
If|c | > √a 2 + b 2 ,then1 < sin (θ + ϕ) < − 1 ,meaningnosolution. If|c | ≤ √a 2 + b 2 ,thensolutionstosin (θ + ϕ) = θ + ϕ = sin −1
c
√a2 + b2
c
√a2 + b2
are
orθ + ϕ = π − sin −1
c
√a2 + b2
Section2-Precalculus-M athQ RH
127
2.2.4
D oubleAnglea ndH alf-AngleF ormulas
2.2.4.a
DoubleAngleFormulas sin (2θ) = 2 sin θ cos θ c os (2θ) = c os2 θ − sin 2 θ c os (2θ) = 1 − 2 sin 2 θ c os (2θ) = 2 cos2 θ − 1 2tan θ tan (2θ) = 1 − tan 2θ
2.2.4.b Ifsin θ =
5 6
FindingExactValue-Example ,findsin (2θ) andc os (2θ) .
sin (2θ) = 2 sin θ cos θ . sin θ =
5 6
,findc os θ .
sin θ =
5 6
y
= r , 2π < θ < π , y = 5 andr = 6
Thismeansthatitisinquadranttwo.
a 2 + 2 5 = 3 6 → a = − √11
128
Section2-Precalculus-M athQ RH
Therefore, c os θ =
x r
= − √611
2 sin θ cos θ = 2 ( 65 )(− √611 ) =
− 5√1811 c os (2θ) → cos (2θ) = 1 − 2 sin 2 θ ,sin θ =
5 6
. Therefore,c os (2θ) = −
7 18
.
2.2.4.c
EstablishinganIdentity-Example
Developaformulaforsin (3θ) . Use(2θ + θ) inthesumformula. sin (3θ) = sin (2θ + θ) = sin (2θ) cos θ + c os (2θ) sin θ Nowusethedoubleangleformulasandsimplify. sin (3θ) = (2sin θ cos θ)(cos θ) + (cos2 θ − sin 2 θ)(sin θ) = 2 sin θ cos2 θ + sin θ cos2 θ − sin 3 θ 3 sin θ cos2 θ − sin 3 θ 2.2.4.d
SquaredTrigFunctionsFormulas sin 2 θ =
1 − cos (2θ) 2
c os2 θ =
1 + cos (2θ) 2
1 − cos (2θ)
tan 2 θ = 1 + cos (2θ)
Section2-Precalculus-M athQ RH
129
2.2.4.e
SolvingUsingIdentities-Example
Solvesin θ cos θ = − 41 ,0 ≤ θ ≤ 2 π . Noticehowthisisintheformofthesinedoubleangleformula,exceptbyafactorof2. So, multiplybothsidesby2. 2 sin θ cos θ = − 21 Nowwecansimplifythisbyusingthedoubleangleformula. sin (2θ) = − 21 Putthisintogeneralformbyusingthetabletofindwhatthetavaluesproduce− 21 . 1 sin ( 11π 6 + 2 πk) = − 2
sin ( 7π6 + 2 πk) = − 21 2 θ isequaltobothofthesearguments. So,setthesebothequalto2 θ andsolveforθ . 2θ =
11π 6
+ 2 πk → θ =
11π 12
7π 6
+ 2 πk → θ =
7π 12
2θ =
+ πk + πk
Pluginnumbersfork tofindsolutions. 2.2.4.f
ProjectileMotion
Anobjectispropelledupwardatanangleθ tothehorizontalwithaninitialvelocityofv 0 ft/sec. RangeR -thehorizontaldistancethattheobjecttravels-isgivenbythefunction: R(θ) =
130
1 2 16 v 0 sin θ cos θ
Section2-Precalculus-M athQ RH
2.2.4.g
Half-AngleFormulas
sin 2a = ±
√
1 − cos a 2
cos 2a = ±
√
1 + cos a 2
=
1 − cos a sin a
=
tan 2a = ±
√
1 − cos a 1 + cos a
sin a 1 + cos a
wherethe+ or− signisdeterminedbythequadrantofangle2a . 2.2.4.h
SquaredHalf-AngleFormulas
sin2 2a = ±
1 − cos a 2
cos2 2a = ±
1 + cos a 2
tan2 2a = ±
1 − cos a 1 + cos a
Section2-Precalculus-M athQ RH
131
2.3
ApplicationsofTrigonometricFunctions
2.3.1
R ightAngleT rigonometry:Applications
sin θ =
opposite hypotenuse
=
b c
csc θ =
hypotenuse opposite
=
c b
cos θ =
adjacent hypotenuse
=
a c
sec θ =
hypotenuse adjacent
=
c a
tan θ =
opposite adjacent
=
b a
cot θ =
adjacent opposite
=
a b
2.3.1.a
ComplementaryAngles
sin B =
b c
= c os A
c os B =
a c
= sin A
tan B =
b a
= c ot A
c sc B =
c b
= sec A
sec B =
c a
= c sc A
c ot B =
a b
= tan A
2.3.1.b
ComplementaryAngleTheorem
Cofunctionsofcomplementaryanglesareequal.
132
Section2-Precalculus-M athQ RH
2.3.1.c
SolvingaRightTriangle-Example
Usethegiveninformationtofindtheremainingvaluesa , c, B .
RememberthatA + B = 9 0 ° ,astheinterioranglesofalltrianglesmustequal1 80 ° . Therefore, B = 9 0 − A → B = 5 5 ° Usingtrigdefinitionsforrighttrianglesweknowthat tan 35 =
a 4
andc os 35 = 4c
Nowrearrangetheseequationstosolvefora andc . a = 4 tan 35 ≈ 2 .80 andc =
4 cos 35
≈ 4 .88
Section2-Precalculus-M athQ RH
133
2.3.2
S olvingObliqueTriangles
Tosolveanobliquetriangle(i.e.atrianglethatisnotarighttriangle)meanstofindthe lengthsofitssidesandthemeasuresofitsangles. Todothis,weneedtoknowthelength ofoneside,alongwithatleastoneofthefollowing: I.
twoangles
II.
oneangleandoneotherside
III.
theothertwosides
Therearefourpossibilitiestoconsider: 1. onesideandtwoanglesknown(ASAorSAA) 2. twosidesandtheangleoppositeoneofthem(SSA) 3. twosidesandtheincludedangleknown(SAS) 4. threesidesknown(SSS)
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Section2-Precalculus-M athQ RH
2.3.3
T heLawofSines
Foratrianglewithsidesa ,b ,andc andoppositeanglesA ,B ,C ,respectively: sin A a
=
sin B b
=
sin C c
forASA,SSA,andSAAtriangles. TheLawofSinesactuallyconsistsofthreeequalities: sin A a
2.3.3.a
=
sin B sin A b a
=
sin C sin B c b
=
sin C c
Example-SAATriangle
Solvethetriangle:
RememberthatA + B + C = 1 80 ° . Therefore, 4 5 + 5 5 + C = 1 80
⇒ C = 80 °
UsetheLawofSines: sin A a sin 45 5
= =
sin B sin A b a sin 55 sin 45 b 5
=
sin C c
=
sin 80 c
Crossmultiplyandsolveforb andc . b = 5sin 55 sin 45 ≈ 5 .79 c = 5sin 80 sin 45 ≈ 6 .96
Section2-Precalculus-M athQ RH
135
2.3.3.b
SolvingSSATriangles
Case2:SSA,isreferredtoastheambiguouscase,becausetheknowninformationmay resultinonetriangle,twotriangles,ornotriangleatall.
2.3.4
T heLawofCosines
SincetheLawofSinesisusedtosolveSAA/ASAandSSAtriangles,theLawofCosinesis usedtosolvecases3and4. Foratrianglewithsidesa ,b ,andc andoppositeanglesA ,B ,C ,respectively: c 2 = a 2 + b 2 − 2 ab cos C
cos A =
b 2 + c2 − a 2 2bc
b 2 = a 2 + c 2 − 2 ac cos B
cos B =
a 2 + c2 − b 2 2ac
a 2 = b 2 + b 2 − 2 bc cos A
cos C =
a 2 + b 2 − c2 2ab
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Section2-Precalculus-M athQ RH
forSASandSSStriangles. Thesquareofonesideofatriangleequalsthesumofthesquaresoftheothertwosides, minustwicetheirproducttimesthecosineoftheirincludedangle. Note:Theangleequationsshownherearederivedfromthethreeequationsabovethem. Usually,onlythe“sidesquared”equationsaregivenandyouwouldhavetoderivethe angleequationsyourself. Whileitistotallycooltousethemasreferencewhiledoingyour homework,Irecommendknowinghowtodothisforanytestsyoumayhave. 2.3.4.a
Example-SASTriangle
Solvethetriangle:
WhileboththeLawofSinesortheLawofCosinescouldbeused,theLawofCosinesis preferable,asitwillresultinonesolution. First,usethec 2 = a 2 + b 2 − 2 abcos C formulatosolveforc . c 2 = 4 2 + 5 2 − (2)(4)(5)(cos 50) c ≈ 3 .91 Now,usethecos A =
b 2 + c2 − a 2 2bc
andcos B =
cos A =
5 2 + 3.91 2 − 4 2 (2)(5)(3.91)
a 2 + c2 − b 2 2ac
formulastosolveforA andB .
⇒ A = cos
−1 5 2 + 3.91 2 − 4 2 ( (2)(5)(3.91) )
A ≈ 5 1.6 °
Section2-Precalculus-M athQ RH
137
cos B =
2
2
4 + 3.91 − 5 (2)(4)(3.91)
2
⇒ B = cos
2
2
−1 4 + 3.91 − 5 ( (2)(4)(3.91)
2
)
B ≈ 7 8.4 °
2.3.5
A reao fa T riangle
Rememberthattheareaofatriangleis K = 21 bh 2.3.5.a
AreaofanSASTriangleTheorem
TheareaK ofatriangleequalsone-halftheproductofitssidestimesthesineoftheir includedangle. K = 21 ab sin C
K = 21 bc sin A
K = 21 ac sin B
2.3.5.b
Heron’sFormula-AreaofaSSSTriangle
TheareaK ofatrianglewithsidesa , b, c is K = √s(s − a )(s − b )(s − c ) wheres = 21 (a + b + c ) .
138
Section2-Precalculus-M athQ RH
2.4
PolarCoordinatesandVectors
2.4.1
P olarCoordinates
Inarectangularcoordinatesystem,apointintheplaneisrepresentedbyanorderedpair ofnumbers(x, y) ,wherex andy equalthesigneddistancesofthepointfromthey axis andthex axis,respectively. Inapolarcoordinatesystem,weselectapoint,calledthepole,andthenaraywithavertex atthepole,calledthepolaraxis.
ApointP inapolarcoordinatesystemisrepresentedbyanorderedpairofnumbers (r, θ). Ifr > 0 ,thenr isthedistanceofthepointfromthepole;θ isanangle(indegrees orradians)formedbythepolaraxisandarayfromthepolethroughthepoint. Wecalled theorderedpair(r, θ) thepolarcoordinatesofthepoint.
Section2-Precalculus-M athQ RH
139
2 .4.1.a
F indingS everalPolarC oordinateso fa SingleP oint
ConsiderthepointP withpolarcoordinates(3, 6π )
π 13π , 6 , 6
−
11π 6
allhavethesameterminalsideso,thispointP canalsobelocatedbyusing
thepolarcoordinates(3, 13π 6 ) or(3, −
11π 6 ) .
Thepoint(3, 6π ) canalsoberepresentedbythepolarcoordinates(− 3 , 7π6 ) . The− 3 meansyoumaketherotationtotheangleθ andgobackr times.
140
Section2-Precalculus-M athQ RH
2.4.1.b
ConvertingfromPolartoRectangularCoordinates
IfP isapointwithpolarcoordinates(r, θ) ,therectangularcoordinates(x, y) atP are givenby x = r cos θ y = r sin θ 2.4.1.c
PointsthatLieonanAxis
Thefiguresshowpolarcoordinatesofpointsthatlieoneitherthex axisorthey axis. In eachillustration,a > 0 .
(x, y) = (a, 0) (x, y) = (0, a) (x, y) = (− a , 0) (x, y) = (0, − a ) (r, θ) = (a, 0) (r, θ) = (a, 2π ) (r, θ) = (a, π) (r, θ) = (a, 3π2 ) 2.4.1.d
ConvertingfromRectangulartoPolarCoordinates
IfP isapointwithrectangularcoordinates(x, y) ,thepolarcoordinates(r, θ) ofP are givenby y
r 2 = x 2 + y 2 tan θ = x if x =/ 0 r = y θ = 2π if x = 0
Section2-Precalculus-M athQ RH
141
2.4.2
P olarE quationsa ndt heirG raphs
Anequationwhosevariablesarepolarcoordinatesiscalledapolarequation. Thegraphof apolarequationconsistsofallpointswhosepolarcoordinatessatisfytheequation. 2.4.2.a
Theorem(Vertical/HorizontalLines)
Leta bearealnumber. Then,thegraphoftheequation r sin θ = a isahorizontalline. Itliesa unitsabovethepoleifa ≥ 0 andlies|a | unitsbelowthepoleif a < 0 . Then,thegraphoftheequation r cos θ = a isaverticalline. Itliesa unitstotherightofthepoleifa ≥ 0 andlies|a | unitstotheleftof thepoleifa < 0 .
142
Section2-Precalculus-M athQ RH
2.4.2.b
Theorem(Circles)
Supposea isapositivenumber. Equation
Description
r = 2 asin θ
circle,radiusa ,center(0, a)
r = − 2 asin θ
circle,radiusa ,center(0, − a )
r = 2 acos θ
circle,radiusa ,center(a, 0)
r = − 2 acos θ
circle,radiusa ,center(− a , 0)
Eachcirclepassesthroughthepole. 2.4.2.c
SymmetricPoints
Therecanbethreetypesofsymmetry: a) polaraxis(x axis)
b)alongthelineθ =
π 2
(y axis)
c)alongthepole
2.4.2.d
TestsforSymmetry
a) polaraxis i)
replaceθ by− θ andseeifanequivalentequationresults
b) alongθ = i)
π 2
(y axis)
replaceθ byπ − θ andseeifanequivalentequationresults
c) alongthepole i)
replacer by− r orθ byθ + π andseeifanequivalentequationresults
Section2-Precalculus-M athQ RH
143
2.4.2.e
Cardioids
Cardioidsarecharacterizedby r = a (1 + c os θ)
r = a (1 + sin θ)
r = a (1 − c os θ)
r = a (1 − sin θ)
wherea > 0 . Thegraphofacardioidpassesthroughthepole.
r = 2 (1 − sin θ)
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Section2-Precalculus-M athQ RH
2.4.2.f
LimaçonswithoutanInnerLoop
Definitionoflimaçonswithoutaninnerloop,equations: r = a + b cos θ
r = a + b sin θ
r = a − b cos θ
r = a − b sin θ
wherea > b > 0 thegraphofalimaçonwithoutaninnerloopdoesnotpassthroughthe pole.
r = 4 + 2 cos θ
Section2-Precalculus-M athQ RH
145
2.4.2.g
LimaçonswithanInnerLoop
Definition:limaçonswithaninnerlooparecharacterizedby
r = a + b cos θ
r = a + b sin θ
r = a − b cos θ
r = a − b sin θ
whereb > a > 0 . Thegraphofalimaçonwithaninnerlooppassesthroughthepoletwice.
y = 1 + 3 cos θ
146
Section2-Precalculus-M athQ RH
2.4.2.h
Rose
Definition:rosecurvesarecharacterizedby r = a cos (nθ) r = a sin (nθ) a =/ 0 andhavegraphsthatareroseshaped. Ifn =/ 0 iseven,therosehas2 n petals;ifn =/ 0 is odd,therosehasn petals.
r = 3 cos (2θ)
Section2-Precalculus-M athQ RH
147
2.4.2.i
Leminscates
Leminscatesarecharacterizedby r 2 = a 2 sin (2θ) r 2 = a 2 cos (2θ) a =/ 0 andthegraphsarepropellershaped.
r 2 = 3 sin (2θ) 2.4.2.j
Spiral
Somespiralcurves(liker = e θ/4 )arecalledlogarithmicspirals,sincetheirequationsmaybe writtenasθ = 4 ln r . Itspiralsindefinitelybothtowardsthepoleandawayfromit.
r = e θ/4
148
Section2-Precalculus-M athQ RH
2.4.2.k
GraphingPolarEquations
Whengraphingapolarequationyoushouldfirstcheckforsymmetry. Thiswillreducethe numberofpointsyouhavetofind. Then,findpointsonthegraphusingthetavaluesthatareintherangeofthetrigfunction withintheequation. So,ifthefunctionhasc os (2θ) init,findthetavaluesfrom0 to2π . Onceyouplotacoupleofpoints,reflectthegraphyouhaveovertheaxis(oraxes)the equationissymmetricwith.
2.4.3
C omplexP lane
Acomplexnumberz = x + y i canbeinterpretedgeometricallyasthepoint(x, y) inthex y - plane. Eachpointintheplanecorrespondstoacomplexnumberand,conversely,each complexnumbercorrespondstoapointintheplane. Thecollectionofsuchpointsis referredtoasthecomplexplane. ●
Thex axisisreferredtoastherealaxis,becauseanypointontherealaxisisofthe formz = x + 0 i ,arealnumber.
●
They axisiscalledtheimaginaryaxis,becauseanypointthatliesonitisofthe formz = 0 + y i ,apureimaginarynumber.
Section2-Precalculus-M athQ RH
149
2.4.3.a
MagnitudeorModulus
Supposez = x + y i isacomplexnumber. Themagnitudeormodulusofz ,denoted|z | ,is thedistancefromtheorigintothepoint(x, y) . Thatis |z | =
√x + y 2
2
= r
Themagnitudeofz issometimescalledtheabsolutevalueofz . |z | = √zz 2.4.3.b
CartesianForm
Whenacomplexnumberiswritteninthestandardformz = x + y i ,itisintherectangular, orCartesian,form,because(x, y) aretherectangularcoordinatesofthecorresponding pointinthecomplexplane. Supposethat(r, θ ) arethepolarcoordinatesofthispoint. Then, x = r cos θ y = r sin θ 2.4.3.c
PolarFormofaComplexNumber
Ifr ≥ 0 and0 ≤ θ ≤ 2 π ,thecomplexnumberz = x + y i canbewritteninpolarform. z = r (cos θ + isin θ) 2.4.3.d
Argument
Ifz = r (cos θ + isin θ) isthepolarformofacomplexnumber,theangleθ iscalledthe argument. Themagnitudeofz = r (cos θ + isin θ) is: |z | = r
150
Section2-Precalculus-M athQ RH
2.4.3.e
Euler’sFormula
Euler’sFormula,foranyrealnumberθ , e iθ = c os θ + isin θ 2.4.3.f
ExponentialForm
Euler’sFormulaallowsustowritethepolarformofacomplexnumberusingexponential notation. r (cos θ + isin θ) = r e iθ 2.4.3.g
MultiplicationandDivisionTheorem
Supposez 1 = r 1 e iθ1 andz 2 = r 2 e iθ2 aretwocomplexnumbers. Then, z 1 z 2 = r 1 r 2 e i(θ1 + θ2 ) Ifz 2 =/ 0 ,then, z1 z2
=
r1 r2
e i(θ1 − θ2 )
2.4.3.h
PeriodicTheorem
Theargumentofacomplexnumberisperiodic. r e iθ = r e i(θ + 2k) ,wherek isaninteger
Section2-Precalculus-M athQ RH
151
2.4.3.i
DeMoivre’sTheorem
Ifz = r e iθ isacomplexnumber,then z n = r n e i(nθ) ,ifz = r e iθ wheren ≥ 1 isaninteger. Example:UsingDeMoivre’sTheorem,express[3(cos 6π + isin 6π )]3 intheformr e iθ andx + y i . π
First,convertfrompolartoexponentialform:[3(cos 6π + isin 6π )]3 = (3ei 6 )3 . Now,useDeMoivre’sTheorem: π
π
33 (ei(3· 6 ) = 27ei 2 Convertbacktopolarformandsolve. 2 7(cos 2π + isin 2π ) = 2 7(0 + i(1)) = 2 7i 2.4.3.j
ComplexRoots
Supposew isacomplexnumber,andn ≥ 2 isapositiveinteger. Anycomplexnumberz thatsatisfiestheequation zn = w 2.4.3.k
isacomplexn th rootofw .
FindingComplexRoots
Supposew = r e iθ isacomplexnumberandn ≥ 2 isaninteger. Ifw =/ 0 therearen distinct complexrootsofw givenbytheformula 1
z k = √r ei n (θ + 2kπ) n
wherek = 0 , 1, 2, ..., n − 1 .
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Section2-Precalculus-M athQ RH
2.4.4
V ectors
Avectorisaquantitythathasbothmagnitudeanddirection. Itiscustomarytorepresenta vectorbyusinganarrow. Thelengthofthearrowrepresentsthemagnitudeandthe arrowheadindicatesdirection. 2.4.4.a
GeometricVectors
IfP andQ aretwodistinctpointsinthex y -plane,thereisexactlyonelinecontainingboth P andQ [Figure(a)]. ThepointsonthatpartofthelinethatjoinsP andQ ,includingP andQ ,formwhatiscalledalinesegment[Figure(b)]
OrderingthepointssothattheyproceedfromP toQ resultsinadirectedlinesegment →
→
fromP toQ ,orageometricvector,whichisdenotedP Q . InadirectedlinesegmentP Q , P iscalledtheinitialpointandQ theterminalpoint,asindicatedinFigure(c).
Section2-Precalculus-M athQ RH
153
Thevectorv whosemagnitudeis0 iscalledthezerovector,0 . Thezerovectorisassigned nodirection. Twovectorsv andw areequal,writtenv = w ,iftheyhavethesamemagnitudeand direction. 2.4.4.b
AddingVectorsGeometrically
Thesumv + w oftwovectorsisdefinedasfollows: Positionthevectorsv andw sothattheterminalpointofv coincideswiththeinitialpoint ofw (thisisalsoknownasputtingthevectors“tiptotail”). Thevectorv + w istheunique vectorwhoseinitialpointcoincideswiththeinitialpointofv andwhoseterminalpoint coincideswiththeterminalpointofw .
Vectoradditioniscommutative(v + w = w + v )andassociative(u + (v + w ) = (u + v ) + w ). Thezerovector0 hasthepropertythat v +0 =0 +v =v foranyvectorv .
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Section2-Precalculus-M athQ RH
Ifv isavector,then− v isthevectorthathasthesamemagnitudeasv ,butwhose directionisoppositetov . v + (− v ) = 0
Ifv andw aretwovectors,thenthedifferencev − w isdefinedas v − w = v + (− w ) 2.4.4.c
MultiplyingVectorsbyNumbersGeometrically
Whenusingvectors,realnumbersarereferredtoasscalars. Scalarsarequantitiesthat haveonlymagnitude. Ifa isascalarandv isavector,thescalarmultipleisdefinedasfollows: ●
Ifa > 0 ,a v isthevectorwhosemagnitudeisa timesthemagnitudeofv andwhose directionisthesameasthatofv .
●
Ifa < 0 ,a v isthevectorwhosemagnitudeis|a | timesthemagnitudeofv and whosedirectionisoppositetov .
●
Ifa = 0 orifv = 0 ,thena v = 0
Section2-Precalculus-M athQ RH
155
2.4.4.c
MultiplyingVectorsbyNumbersGeometrically(con’t)
2.4.4.d
Properties 0v = 0
1v = v
− 1v = − v
(a + B )v = a v + B v
a (v + w ) = a v + a w
a (Bv) = (aB)v
2.5
AnalyticGeometry(ConicSections)
Conicsectionsarefiguresthatareformedfromtheintersectionofaplaneandtworight circularconesstackedontopofeachother(tiptotip).
156
Section2-Precalculus-M athQ RH
2.5.1
E quationCharacteristics
Allfouroftheequationsarebasedonthedistanceformula. Notethatnoneofthe equationsarefunctions,exceptforverticalparabolas. ConicSection
Characteristics
Parabola
onlyonesquaredvariable
Circle
twosquaredvariables,coefficientsidentical(includingsign)
Ellipse
twosquaredvariables,coefficientsarenotequalbuthavethe samesign
Hyperbola
twosquaredvariables,coefficientshavedifferentsigns
Yougettheseequationsbycompletingthesquare.
2.5.2
P arabolas
f isthefocusoftheparabola. Aparabolaisdefinedasthesetofpointsequidistantfroma pointandaline. Thepointisthefocusandthelineisthedirectrix.
Section2-Precalculus-M athQ RH
157
Equation AxisofSymmetry
Horizontal
Vertical
(y − k )2 = 4 a(x − h )
(x − h )2 = 4 a(y − k )
y =k
x =h
wherethevertexis(h, k) . Therearenocoefficientsontheleftsideoftheequationorinfrontoftheunsquared variable(thex 1 ory 1 ). a isthedistancebetweenthevertexandthefocus. 4 a isthefocalwidth. Tofindalltheinformation(focus,focalwidth,thevertex,etc.)youmayneedtocomplete thesquare(iftheequationisinstandarda x 2 + b x + c y + d = 0 form,forexample). Tofinda walkthroughofhowtodothis,gototheAlgebraIIsection. 2.5.2.a
Example
Findthefocus,a ,thefocalwidth,axisofsymmetry,andthevertexoftheparabola y=
1 10 (x
+ 2 )2 − 3 .
Themain“trick”withconicsectionsisjustputtingthemintotheexactequationformthey aredefinedwith. Thismakesiteasytodetermineallofthevalues,asitisbasicallygivento you. Allyouneedtodoismesswiththeequationabit:movethingsaround! So,startby isolatingthesquaredquantity,asweseeitintheaboveequations.
158
Section2-Precalculus-M athQ RH
2.5.2.a
Example(con’t) y =
1 10 (x
+ 2 )2 − 3 → y + 3 =
1 10 (x
+ 2 )2 → (x + 2 )2 = 1 0(y + 3 )
Now,itisrelativelyeasytogetalltheinformationaskedfor. Startwiththevertex; rememberthatthevaluesaretheo pposites ignofwhatyouseebeingadded/subtractedto thex andy . So,forthisparabola,thevertexis x + 2 = 0 → x = − 2 y + 3 = 0 → y = − 3 v = (− 2 , − 3 ) Rememberthatthefocalwidthisequalto4 a ,orthecoefficientinfrontoftheunsquared quantity(notthevariabley ,butthequantityy + 3 ). So,weknowthefocalwidthis1 0 . To finda ,divide1 0 by4 :a = 2 .5 . Thefocusisontheaxisofsymmetryoftheparabola. Theaxisofsymmetryisx = h (see table),sotheaxisofsymmetryofthisparabolaisx = − 2 . Tofindthecoordinatesofthe focus,adda tothey valueofthevertex. f = (− 2 , − 3 + 2 .5) = (− 2 , − 21 )
2.5.3
E llipses
a isthelongestdistance,b istheshortestdistance,andc isthedistancebetweenthe centerandthefocus.
Section2-Precalculus-M athQ RH
159
Thesumoftwolengthsstaysconstantbetweentwofosci. 2 a isthecombineddistanceto thefoscifromanypointontheellipse. Thefosciareonthemajoraxis. Theverticesofanellipsearetheendpointofthemajor axis. StandardEllipseEquation: (x − h)2 a2
+
(y − k)2 b2
= 1
where(h, k) isthecenteroftheellipse. a 2 isthebiggernumberandthevariable associatedwiththisnumberisthemajoraxis. Findc withtheequationc = √a 2 − b 2 . 2.5.3.a
Graphing
Tographanellipse,firstplotthecenter. Then,takethesquarerootofa 2 andb 2 . Starting fromthecenter,goouta unitsinbothdirectionsontheaxiscorrespondingtothevariable. Repeatthiswithb . 2
y2
Hereisawalkthroughforgraphingtheequation x4 + 16 = 1 :
Inthiscase,themajoraxisisthey axis.
160
Section2-Precalculus-M athQ RH
2.5.4
H yperbolas
wherea 2 + b 2 = c 2 andf is(h ± c , k) . Hyperbolashaveacenterthatisnotonthegraph,butbetweenthetwopartsofthe hyperbola. Thecenteristheintersectionofthetwoasymptotes(thedottedlinesinthe picture). Theequationofahyperbolais
(x − h)2 a2
−
(y − k)2 b2
= 1 . Thepositivetermdominatesthegraph
(thegraphwillopenontheaxisthatcorrespondswiththepositiveterm). Thedenominator ofthepositivetermcanbelessthanb 2 . 2.5.4.a
Graphing
Tographahyperbola,beginbymakingarectangularguide. Goouta unitsinboth directionsonthex axis,andb unitsinbothdirectionsonthey axis:thesearethecenter pointsoftherectangle.
Section2-Precalculus-M athQ RH
161
Then,drawtheslantasymptotes(thediagonalsofthetriangle). Thesewillshowyouwhere thegraphapproaches. Nowyoucandrawyourgraph. Theverticesoftheparabolaarethepointsa orb units awayfromthecenter,dependingonwhatthedominatingtermofthegraphis. Sketcha parabola-shapedcurveonbothsides,approachingtheslantasymptotesasyougetfurther fromthevertices. 2.5.4.b
AnalyzinganEquation
Whenaquestionasksyoutoanalyzeanequation,findthefollowing: ●
graph
●
vertices
●
center
●
foci
●
transverseaxis
●
○
goesthroughthecenter,vertices,andfoci
○
thiscanbeaverticalorhorizontalline,itdoesnothavetobethex ory axis
asymptotes’equations ○
slopeis± ab ,goesthroughcenter→ usepoint-slopeformula
2.5.5
P arametricE quationsa ndP laneC urves
Supposex = x (t) andy = y (t) aretwofunctionsofathirdvariablet ,calledtheparameter, thataredefinedonthesameintervalI . Thentheequations x = x (t) y = y (t)
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wheret isinI ,arecalledparametricequationsandthegraphoftheparametricequations isdefinedby (x, y) = (x(t), y(t)) iscalledaplanecurve. 2.5.5.a
GraphingaPlaneCurve
Tographaplanecurve,youmakeatableofvalues,asseenbelow. Graphx (t) = 2 t2 y (t) = 3 t ,− 2 ≤ t ≤ 2 Foreachnumbert ,therecorrespondsanumberx andanumbery . Sowhent = − 2 , thenx = 2 (− 2 )2 = 8 andy = 3 (− 2 ) = − 6 . Setupatableofvalues,thengraph. t
x
y
−2
8
−6
−1
2
−3
0
0
0
1
2
3
2
8
6
Thearrowsindicateorientation.
Section2-Precalculus-M athQ RH
163
2.5.5.b
FindingRectangularEquationofaParametricallyDefinedCurve
Findtherectangularequationofx (t) = a sin t y (t) = a cos t ,− ∞ < t < ∞ . Graphthecurveand indicateorientation. Useapythagoreanidentity: sin t =
x a
andc os t =
y a
. T hismeansthat( ax )2 + ( ay )2 = 1
→ x 2 + y 2 = a 2 . Nowwecanseethattheplanecurveisacirclewithcenter(0, 0) andradiusa .
Asparametert increases,sayfromt = 0 [thepoint(0, a) ]tot =
π 2
[thepoint(a, 0) ],the
correspondingpointsaretracedinacounterclockwisedirectionaroundthecircle. Thebigthingwhenturningparametricequationsintorectangularonesiseliminatingt ,as thiswillmakeiteasiertograph. Also,besuretopayattentiontotheparametersthatt is limitedto;thesewillchangewherethegraphstartsandends. 2.5.5.c
UseTimeasaParameterinParametricEquations
Theparametricequationsofthepathofaprojectilefiredataninclinationθ tothe horizontal,withaninitialspeedv 0 ,fromaheighth abovethehorizontalare x (t) = (v 0 cos θ)t
y (t) = − 21 gt2 + (v 0 sin θ)t + h
wheret istimeandg istheconstantaccelerationduetogravity(approximately3 2 f t/sec 2 or9 .8 m/s2 ).
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Section2-Precalculus-M athQ RH
2.6
Matrices
2.6.1
S olvingaS ystemo fEquations
Matricescanbeusedtomakesolvingasystemofequationseasier. Thisisdonethrough theuseofrowoperationslikemultiplying/dividing,addingandreplacingoneoftherows withthesum,multiplying/addingandreplacingonerow,orswitchinganytworows. When youdotheserowoperations,youneedtorecordthemnexttothematrix. Whenputtingasystemofequationsintomatrixform,youplacethecoefficientsofthe variablesintoarow,withthenumbertheequationisequaltoontherightsideofthe dottedline. Allofthecoefficientsofthesamevariableareinarowtogether. Forexample, thematrixformofthesystemofequations3 x + y − 2 z = 3 , x + 4 z = 1 , − 2 x + 2 y + 2 z = − 3 is
Eachnumberinamatrixcanbeidentifiedasmrc ,wherer istherownumberandc isthe columnnumber. So4 isatm23 . Whenusingmatricestosolvesystemsofequations,yourgoalistoputthematrixinto row-echelonform:
Theanswerstothesystemwillbethenumberstotherightofthedottedline. Startwith gettinga1 inm11 ,thenm22 ,thenm33 .
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165
2.6.1.a
Example
Solvethesystemofequationsx + 2 y = 3 , 4x − y = 2 .
Thisshowsthatx =
7 9
andy =
10 9
. Thismeansthatthereisonesolution(independentand
consistent). 2.6.1.b
PossibleSolutions
Thisisessentially0 = 0 ,whichmeansthatallrealnumbersworkanditisadependent system. However,therearestillconstraints:thefirsttwovariablesintermsofthelast variable.
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Section2-Precalculus-M athQ RH
2.6.1.b
PossibleSolutions(con’t) x = ...z... y = ...z... z ε R
Thisshowsthat0 = 7 ,whichisfalse. Thismeansthatthereisnosolution.
2.6.2
D eterminantofaM atrix
Tofindthedeterminantofamatrix,thematrixcanonlybeasquare(n × n ). 2.6.2.a
2x2Matrix
IfA isa2 × 2 matrix,thend et(A) is
d et(A) = a d − b c
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167
2.6.2.b
3x3Matrix
Therearetwowaystofindthedeterminantofa3 × 3 matrix. BasketWeave: Tobeginthebasketweavemethod,recopythefirsttwocolumnstotherightofthematrix.
Then,startingfromthefirstnumberinthematrix(5 ),addthediagonalproductsfromleft toright...
(5 ·− 2 · 4 ) + (1 · 9 · 5 ) + (− 4 · 1 · 0 ) = − 4 0 + 4 5 + 0 = 5 andrighttoleft.
(− 4 ·− 2 · 3 ) + (0 · 9 · 5 ) + (1 · 1 · 4 ) = 4 0 + 0 + 4 = 4 4
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BasketWeave:(con’t) Nowsubtracttherighttoleftfromthelefttoright. 5 − 44 = − 39 Thedeterminantis− 3 9 . ExpansionbyMinors: Todotheexpansionbyminorstechnique,startbychoosingoneroworcolumntouseas multipliers. Usethediagrambelowtodeterminethesignsofthemultipliers.
Iwillshowthetechniquewiththefollowingmatrix,usingrow1 asthemultipliers.
Now,wipeouttherowandcolumnofeachmultiplierandmakethree2 × 2 matricesoutof theremainingmatrix.
Calculatethedeterminantofeachmatrix,multiplyitbythemultiplier,andaddtogether. − 3 + 12 − 9 = 0
Section2-Precalculus-M athQ RH
169
2.6.2.c
SwitchRowsorColumns
Thisisessentiallythesamematrix,butthecolumnsarereversed. Thedeterminantis related;theyareoppositesofeachother. a d − b c vs.b c − a d Fora3 × 3 matrix,tworowsorcolumnsareswitched. 2.6.2.d
MultiplesofRows/Columns
Thedeterminantismultipliedbythesameconstantastherows/columns. − 2 − 2 k
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Section2-Precalculus-M athQ RH
2.6.3
C ramer’sRule
Cramer’sruleisusedtosolvethreevariable,threeequationsystemsofequations.
Thisisdonebyfindingthefourfollowingdeterminants: First,leaveouttheanswercolumnandfindthedeterminant.
d et(D) = (− 1 2 + 4 − 2 ) − (− 2 − 1 6 + 3 ) = 5 Repeatthis,butleaveoutx (D x ),leaveouty (D y ),thenleaveoutz (D z ). Replacethe takenoutvariablewiththeanswercolumn.
d et(D x) = 1 5 det(D y ) = − 1 0 det(D z ) = 5 Tofindthevaluesofthevariables,dividetheD x, D y , andD z byD . x=
Dx D
=
15 5
= 3 , y =
Dy D
=
−10 5
= − 2 , z =
Thesolutionofthesystemofequationsis(3, − 2 , 1) .
Section2-Precalculus-M athQ RH
Dz D
=
5 5
=1
171
2.6.4
O perationswithMatrices
Twomatricesareequaltoeachotherwhenallnumbersinthesamelocationareequal(i.e. m11 = m11 , m12 = m12 , ... )andthematriceshavethesamedimensions(numberofrowsand columns). Todemonstratetheoperationswithmatrices,Iwillbeusingthefollowingmatrices:
Toaddorsubtracttwomatrices,thematricesmustbethesamedimensions. So,withthe matriceswehaveabove,wecanonlyaddandsubtractA andB fromeachother. Todo this,addorsubtractthecorrespondingtermstogetherforeachelementofthematrix.
Touseascalarfactor,multiplyeachelementofthematrixbythescalar.
Whenmultiplyingtwomatrices,thenumberofcolumnsinthefirstmatrixmustbeequalto thenumberofrowsinthesecondmatrix(thisindicatesthatmatrixmultiplicationisnot communicative). Theanswermatrixwillhavedimensionswherethenumberofrowsis
172
Section2-Precalculus-M athQ RH
equaltothenumberofrowsofthefirstmatrixandthenumberofcolumnsisequaltothe numberofcolumnsforthesecondmatrix. Let’ssaywe’remultiplyingthefollowingtwomatricestogether.
Noticehowthecolumnsofthefirstmatrixareequaltotherowsofthesecondmatrix. We canalsoseethattheanswermatrixwillhavedimensions2 × 3 . Tobeginmultiplying,dothedotproductofrowoneofthefirstmatrixwithcolumnoneof thesecondmatrix. Thiswillbeelementm11 oftheanswermatrix.
(1, 3, 4) • (5, − 1 , 0) = (1 · 5 ) + (3 ·− 1 ) + (4 · 0 ) = 5 − 3 + 0 = 2 Now,let’sdothesamethingforelementm21 ofthematrix. Thistime,wewilldothedot productofrowtwoofthefirstmatrixandcolumnoneofthesecondmatrix.
(− 2 , 3, 1) • (5, − 1 , 0) = (− 2 · 5 ) + (3 ·− 1 ) + (1 · 0 ) = − 1 0 − 3 + 0 = − 1 3
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173
Repeatthisprocessuntilyou’vefoundallelementsofthematrix.
2.6.4.a
IdentityMatrices
Amultiplicativeidentityiswhereyoucanmultiplyanynumberbytheidentityandyouwill getthenumberyoustartedwith(i.e.8 · x = 8 , x = 1 ). Onlysquareechelonmatricesformidentitymatrices(I ).
So,ifyouhavea4 × 2 matrix,youwouldmultiplyitbya2 × 2 identitymatrixandwouldget theoriginal4 × 2 matrixasananswer.
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Section2-Precalculus-M athQ RH
2.6.4.b
InverseMatrices
Theseonlyexistforsquarematrices. Amatrixistheinverseofanotherif A · A−1 = I Forexample,
Whenyoumultiplythisoutyouget:
Now,whenwesolvethefourequationsthatcomeoutofthisweget: 3 a + c = 1 2a + c = 0 3b + d = 0 2b + d = 1
Inversesof2 × 2 matriceshaveapattern:
Section2-Precalculus-M athQ RH
175
Findingtheinversematrixofa3 × 3 matrixisabitdifferent. Justaswedidwithsystemsof equations,wewillputalineafterthegivenmatrix,butthistimewewillputa3 × 3 identity matrixtotherightoftheline.
Findtheidentitymatrixbygettingthematrixtotheleftofthelineintorowechelonform. Thematrixthatresultstotherightofthelinewillbetheinversematrix.
Thiscanbeusedwhensolvingequationsaswell. Ifyouhavetheinverseofamatrixofthe coefficientsoftheequations,youcanmultiplythisbyamatrixmadeofonlywhatthe equationsareequalto. Thiswillgiveyouwhatallthevariablesareequalto. Forexample, takethesystemofequationsbelow. x + y = 3 − x + 3 y + 4 z = − 3 4y + 3 z = 2 Makeamatrixofthecoefficientsandadifferentmatrixoftheconstantstheequationsare equalto.
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Section2-Precalculus-M athQ RH
Findtheinverseofthe3 × 3 matrix(noticethatthisisthematrixwefoundtheinverseof above)andmultiplyitbythe3 × 1 matrix. Thiswillgiveyoua3 × 1 matrixthatholdswhat eachvariableisequalto.
2.7 PartialFractionDecomposition 2.7.1
C ase1
UndertheassumptionthatQ hasonlynonrepeatedlinearfactors,thepolynomialQ has theform Q(x) = (x − a 1 )(x − a 2 )...(x − a n ) wherenotwoofthenumbersa 1 , a 2 , ..., a n areequal. Inthiscase,thepartialfraction decompositionofPQ isoftheform P (x) Q(x)
=
A1 x − a1
+
A2 x − a2
+ ... +
An x − an
wherethenumbersA1 , A2 , ..., An aretobedetermined.
Section2-Precalculus-M athQ RH
177
2.7.2
C ase2
IfthepolynomialQ hasarepeatedlinearfactor,say(x − a )n, n ≥ 2 isaninteger,then,inthe partialfractiondecompositionofPQ ,allowfortheterms A1 x − a
+
A2
(x − a)2
+ ... +
An (x − a)n
wherethenumbersA1 , A2 , ..., An aretobedetermined.
2.7.3 C ase3 SupposeQ containsanonrepeatedirreduciblequadraticfactoroftheforma x 2 + b x + c . Then,inthepartialfractiondecompositionofPQ ,allowfortheterm Ax + B ax2 + bx + c
wherethenumbersA andB aretobedetermined.
2.7.4 C ase4 SupposethepolynomialQ containsarepeatedirreduciblequadraticfactoroftheform (ax 2 + b x + c )n ,n ≥ 2 ,n isaninteger,andb 2 − 4 ac < 0 . Then,inthepartialfraction decompositionofPQ ,allowfortheterms A1 x + B1 ax2 + bx + c
+
A2 x + B2
(ax2 + bx + c)2
+ ... +
An x + Bn (ax2 + bx + c)n
wherethenumbersA1 , B 1 , A2 , B 2 , ..., An , B n aretobedetermined.
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Section2-Precalculus-M athQ RH
2.7.4.a
Example(Case1andCase2) 2
− 4x + 12 Findthepartialfractiondecompositionofxx3 − 4x . 2 + 4x
Beginbyfactoringthedenominator. x 3 − 4 x 2 + 4 x = x (x 2 − 4 x + 4 ) = x (x − 2 )2 Noticewehaveonenonrepeatedlinearfactorx andonerepeatedlinearfactor(x − 2 )2 . B So,case1:Ax andcase2: x − 2 +
C (x − 2)2
isinthedecomposition. Rewritethefraction.
x2 − 4x + 12 x3 − 4x2 + 4x
=
A x
+
B x − 2
+
C (x − 2)2
Now,multiplybothsidesbyx (x − 2 )2 togetridofthefractions. x 2 − 4 x + 1 2 = A(x − 2 )2 + B x(x − 2 ) + C x Whenyouletx = 0 ,everythingmultipliedbyx getscancelledout,leaving1 2 = 4 A . This meansA = 3 . Whenyouletx = 2 ,boththeA andB termsdrop,leaving2 C = 8 . ThismeansC = 4 . TofindB letx = 1 (oranythingthatisnot0 or2 )andpluginwhatyouhaveforA andC 1 − 4 + 1 2 = 3 (− 1 )2 + B (− 1 ) + 4 B = −2 Now,wehavefoundallunknownvalues. Plugtheseintothedecompositionwefound above. x2 + 4x + 12 x3 − 4x2 + 4x
Section2-Precalculus-M athQ RH
=
3 x
−
2 x − 2
+
4 (x − 2)2
179
“Nature-cuethethemefromtheTwilightZone-somehowknowscalculus” -StevenStrogatz
2.8
180
MyNotesforPrecalculus
Section2-Precalculus-M athQ RH
2.8
MyNotesforPrecalculus(con’t)
Section2-Precalculus-M athQ RH
181
2.8
182
MyNotesforPrecalculus(con’t)
Section2-Precalculus-M athQ RH
“Changeismosts luggisha tthee xtremes becausethed erivativeisz erothere.” -StevenStrogatz
Section3-CalculusI 3.0
SummarySheet
LimitLaws 1.
lim [f (x) ± g (x)] = lim f (x) ± lim g(x)
2.
lim [c · f (x)] = c · lim f (x)
3.
lim [f (x) · g (x)] = lim f (x) · lim g(x)
4.
f (x) lim [ g(x) ]=
x→a
x→a
x→a
x→a
x→a
x→a
x→a
x→a
lim f (x)
x→a
lim g(x)
x→a
x→a
,iflim g(x) =/ 0 x→a
5.
lim x = a
6.
lim c = c
7.
lim [f (x)]n = [lim f (x)]n
x→a x→a x→a
x→a
8. Ifnisevenanda>0,wehavelim √x = √a . Ifnisevenandweassumethat n
n
x→a
lim f (x) > 0 asx → a then, x→a
n lim √ f(x) =
x→a
Section3-CalculusI-M athQ RH
√lim f(x) , n ∈ Z ,n > 1 n
x→a
183
LimitsofRationalNumbersTheorem Ifr>0isarationalnumberthen, lim x1r = 0 x→∞
Ifr>0isarationalnumbersuchthatx r isdefinedforallxthen, lim x1r = 0 x→−∞ ➔ Ifrisapositiveexponentinthedenominator,x r → ∞ ,asx → ∞ . BasicDerivativeFunctions ●
d constantfunction(f (x) = c ) : dx (c) = 0
●
d n powerrule: dx (x ) = nxn−1
●
d constantmultiplerule: dx (c · f (x)) = c · dxd f (x)
●
d sumrule: dx [f (x) + g (x)] = dxd f (x) + dxd g(x)
●
d differencerule: dx [f (x) − g (x)] = dxd f (x) − dxd g(x)
●
d x naturalexponentialfunction: dx (e ) = ex
●
d productrule: dx [f (x) · g (x)] = f (x) · dxd g(x) + g (x) · dxd f (x)
●
d quotientrule: dx [ g(x) ] =
●
chainrule: dx = du · du dx
●
exponentialfunctions:dxd (b x) = b x · ln b
●
1 logarithmicfunctions: dx (log b x) = x·ln b
f(x)
dy
d d g(x)· dx f(x) − f(x)· dx g(x) 2 [g(x)]
dy
dy
184
Section3-CalculusI-M athQ RH
DerivativeTrigonometricIdentities ●
d dx (sin x)
= c os(x)
●
d dx (cos x)
= − sin(x)
●
d dx (tan x)
= sec 2 (x)
●
d dx (csc x)
= − c sc(x)cot(x)
●
d dx (sec x)
= sec(x)tan(x)
●
d dx (cot x)
= − c sc 2 (x)
Differentials ●
d y = f ′(x)dx
●
“measurementerror”meansthederivativeofavariable
●
relativeerror=dxx ,whered x isthederivativeofthefunctionsandx istheactual function
AbsoluteMaximumsandMinimums 1. Setf ′(x) = 0 andfindroots. Plugthesenumbersbackintof . 2. Findf ofendpointsa andb 3. Findthelargestandsmallestvalues
HowDerivativesAffecttheShapeofaGraph Increasing/DecreasingTest: ●
Iff ′(x) > 0 onaninterval,thenthefunctionf isincreasingonthatinterval
●
Iff ′(x) < 0 onaninterval,thenthefunctionf isdecreasingonthatinterval
Section3-CalculusI-M athQ RH
185
TheFirstDerivativeTest: a. iff ′ changesfrompositivetonegativeatc ,thenf hasalocalmaxatc b. iff ′ changesfromnegativetopositiveatc ,thenf hasalocalminatc c. iff ispositiveonbothsidesornegativeonbothsidesofc ,thenf hasnolocal min/max ConcavityTest: ●
iff ′′ > 0 forallx inI ,thenthegraphoff isconcaveupwardonI
●
iff ′′ < 0 forallx inI ,thenthegraphoff isconcavedownwardonI
TheSecondDerivativeTest: a. iff ′′(c) = 0 andf ′′(c) > 0 ,thenf hasalocalminimumatc b. iff ′′(c) = 0 andf ′′(c) < 0 ,thenf hasalocalmaximumatc
MeanValueTheorem f ′(c) =
f (b) − f (a) b − a
,f (b) − f (a) = f ′(c)(b − a )
DefiniteIntegrals b
∫ f (x) dx = F (b) − F (a) a
186
Section3-CalculusI-M athQ RH
TheFundamentalTheoremofCalculus Supposef iscontinuouson[a, b] x
1. ifg (x) = ∫ f (t)dt ,theng ′(x) = f (x) a
2.
b
∫ f (x)dx = F (b) − F (a) ,whereF a
isanyantiderivativeoff
CommonIndefiniteIntegrals
1.
∫ c · f (x)dx = c · ∫ f (x)dx
2.
∫[f (x) ± g (x)]dx = ∫ f (x)dx ± ∫ g (x)dx
3.
∫ k dx = k x + C
4.
∫ xn dx =
5.
xn+1 n+1
+ C wheren =/ 1
∫ exdx = ex + C
6.
∫ 1x dx = ln |x| + C
7.
x
b +C ∫ b xdx = ln b
8.
∫ |x | d x =
9.
x|x| 2
+C
∫ sin x dx = − cos x + C
10. ∫ c os x dx = sin x + C
11. ∫ c sc 2 x dx = − c ot x + C
12. ∫ sec 2 x dx = tan x + C
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187
CommonIndefiniteIntegrals(con’t)
13. ∫ c sc x cot x dx = − c sc x + C
14. ∫ x21+1 dx = tan −1 x + C
15. ∫
1
√1−x2
dx = sin −1 x + C
IndefiniteIntegralSubstitutionRule
∫ f (g(x)) · g ′(x) dx = ∫ f (u) du
DefiniteIntegralSubstitutionRule b
g(b)
a
g(a)
∫ f (g(x)) · g ′(x) dx =
∫
f (u) du
IntegralsofSymmetricFunctions Supposef iscontinuouson[− a , a] a
a
a. iff iseven[f (− x ) = f (x)] ,then ∫ f (x) dx = 2 ∫ f (x) dx −a
0
a
b. iff isodd[f (− x ) = − f (x)] ,then ∫ f (x) dx = 0 −a
188
Section3-CalculusI-M athQ RH
3.1 Limits DefinitionofaLimit ➔ Supposef (x) isdefinedwhenxisnearthenumbera(thismeansthatf isdefinedon someopenintervalthatcontainsa,exceptpossiblyaitself). Then,wewrite lim f (x) = L x→a
andsay“thelimitoff (x) ,asxapproachesa,equalsL”ifwecanmakethevaluesof f (x) arbitrarilyclosetoL(asclosetoLaswelike)byrestrictingxtobesufficiently closetoa(oneitherside)butnotequaltoa”. ➔
lim f (x) means“thelimitoff (x) asxapproachesafromtheleft. Weare
x→a−
takingxsufficientlyclosetoabutxa. ➔ IfL=± ∞ ,thelimitdoesnotexist(DNE).
Section3-CalculusI-M athQ RH
189
3.1.1
L imitL aws
Supposecisaconstantandlim f (x) andlim g (x) exist. x→a
x→a
1.
lim [f (x) ± g (x)] = lim f (x) ± lim g(x)
2.
lim [c · f (x)] = c · lim f (x)
3.
lim [f (x) · g (x)] = lim f (x) · lim g(x)
4.
f (x) lim [ g(x) ]=
x→a
x→a
x→a
x→a
x→a
x→a
x→a
x→a
lim f (x)
x→a
lim g(x)
x→a
x→a
,iflim g(x) =/ 0 x→a
5.
lim x = a
6.
lim c = c
7.
lim [f (x)]n = [lim f (x)]n
x→a
x→a x→a
x→a
8. Ifnisevenanda>0,wehavelim √x = √a . Ifnisevenandweassumethat n
n
x→a
lim f (x) > 0 asx → a then, x→a
n lim √ f(x) =
x→a
3.1.2
√lim f(x) , n ∈ Z ,n > 1 n
x→a
D irectS ubstitutionP roperty
➔ Iff (x) isapolynomialorarationalfunctionandaisinthedomainoff (x) lim f (x) = f (a) x→a
➔ Thisruledoesnotworkbothways ➔ Thisruleonlyworksbecausepolynomialandrationalfunctionsaredefined everywhere.
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Section3-CalculusI-M athQ RH
3.1.3
T heSqueezeTheorem
➔ Iff (x) ≤ g (x) ≤ h (x) whenxisneara(exceptpossiblyata)and lim f (x) = lim h(x) = L x→a
x→a
then, lim g(x) = L x→a
3.1.4 C ontinuousFunctions ➔ Afunctionf iscontinuous(nointerruptions,breaks/holes,orgaps) atanumberaif lim f (x) = f (a) . x→a
➔ Therearethreeregulationsthatallmustbemet
◆ ◆ ◆
f (a) isdefined(aisinthedomainoff ) lim f (x) exists(not± ∞ ) x→a
lim f (x) = f (a) x→a
➔ Iff isdefinedonanopenintervalcontaininga,exceptperhapsata,wesaythatf is discontinuousata,orthatf hasadiscontinuityata,iff isnotcontinuousata. ➔ Afunctionf iscontinuousfromtherightatanumberaif lim+ f (x) = f (a) . x→a
➔ Afunctionf iscontinuousfromtheleftatanumberaif lim− f (x) = f (a) .
x→a
Section3-CalculusI-M athQ RH
191
3.1.5
C ontinuityT heorems
Iff andg arecontinuousataandcisaconstant,thenthefollowingarealsocontinuousat a. ●
f +g
●
f −g
●
c·f
●
fg
●
f g
,ifg (a) =/ 0
Thefollowingtypesoffunctionsarecontinuousateverynumberintheirdomains. ●
polynomial
●
rational
●
root
●
trigonometric
●
inversetrigonometric
●
exponential
●
logarithmic
3.1.6
C ompositeF unctionT heorem
➔ Iff (x) iscontinuousatbandlim g(x) = b ,thenlim f (g(x)) = f (b) . Inotherwords, x→a
x→a
lim f (g(x)) = f ( lim g(x)) x→a
x→a
192
Section3-CalculusI-M athQ RH
3.1.7
C ompositeso fContinuousFunctionsTheorem
➔ Ifg iscontinuousataandf iscontinuousatg (a) ,thenthecompositefunctionf ° g givenby(f ° g )(x) = f (g(x)) iscontinuousata.
◆
3.1.8
i.e.compositesofcontinuousfunctionsataarecontinuousata
T heIntermediateValueT heorem
➔ Supposethatf iscontinuousataontheclosedinterval[a,b]andletNbeany numberbetweenf (a) andf (b) ,wheref (a) =/ f (b) . Thenthereexistsanumbercin theopeninterval(a,b)suchthatf (c) = N .
◆
Thisissolelyanexistencetheorem,itdoesnotshowhowtofindthis.
3.1.9
“ Intuitive”D efinitionofaL imita tI nfinity
➔ Letf beafunctionthatisdefinedatsomeinterval(a, ∞) . Then, lim f (x) = L x→∞
whichmeansthatthevaluesoff (x) canbemadearbitrarilyclosetoLbyrequiringx tobesufficientlylarge. ➔ Thesameappliesto(− ∞ , a) where lim f (x) = L
x→−∞
Section3-CalculusI-M athQ RH
193
3.1.10
L imitso fR ationalN umbersT heorem
➔ Ifr>0isarationalnumberthen,
lim x1r = 0
x→∞
Ifr>0isarationalnumbersuchthatx r isdefinedforallxthen,
lim 1r x→−∞ x
= 0
➔ Ifrisapositiveexponentinthedenominator,x r → ∞ ,asx → ∞ . ➔ Thenumeratorcanbeaconstant(see:limitrules)
3.2
Derivatives
Aderivativeistheslopeofapoint,alsoknownastheinstantaneousrateofchangeorthe slopeofatangentline. ❏ Notation ❏ f ′(x) ,readas“fprime” ❏ y ′ ,readas“yprime”
❏
dy dx
❏
d dx f (x) ,r eadas“thederivativewithrespecttoxoff (x) ”
,readas“thederivativeofywithrespecttox”
❏ D f (x) ,readas“thederivativeoff (x) ” ❏ D xf (x) ,readas“thederivativewithrespecttoxoff (x) ”
194
Section3-CalculusI-M athQ RH
3.2.1
D erivativesa tt heP oint( a,f (a))
➔ Giventhefunctionf (x) andx=a,
f ′(a) = lim h→0
f(a+h)−f(a) h
whereh represents|x − a | . ➔ Thisisessentiallytakingthelimitashapproaches0ofthedifference quotient.
◆
Thisequationcomesfromthetraditionalequationfortakingtheslope ofasecantline(twointersectionpoints). However,insteadofhaving anf (a) andaf (b) ,hisusedtorepresentthehorizontaldistance betweentwopointsgettingsmallerandsmaller(i.e.approaching0).
3.2.2
T heD erivativeo fa F unction
➔ Thederivativefunctiongivestheslopesofthetangentlines/instantaneousrateof changeatanypoint.
f ′(x) = lim
h→0
f(x+h)−f(x) h
Thisfunctionwillusuallyproduceanotherfunction.
3.2.3
D ifferentiability
➔ Afunctionf isdifferentiableataiff ′(a) exists. ➔ Afunctionf isdifferentiableonanopeninterval(a,b)[or(a,∞ )/(-∞ ,a)/(-∞ ,∞ )]if itisdifferentiableateverynumberontheinterval.
Section3-CalculusI-M athQ RH
195
3.2.3.a
Theorem
➔ Iff isdifferentiableata,thenf iscontinuousata
◆
Itisimportanttonotethattheconverseofthistheoremisnottrue.
3.2.3.b ●
AFunctionisNotDifferentiableattheFollowing
Acornerorcusp
●
Ajumpdiscontinuity
●
Averticaltangentline/whenitapproachesaverticalasymptote
196
Section3-CalculusI-M athQ RH
3.2.4
H igherO rderD erivatives
➔ Iff (x) isdifferentiable,thensoisf ′(x) . ➔ Thederivativemayhavederivativesofitsown. (f ′)′ = f ′′ Thisnotationmeansthesecondderivative,andisreadas“fdoubleprime”. ➔ Therecouldbef , f ′, f ′′, f ′′′, f (4), f (5) ,etc. dy d2 y d3 y
➔ InLeibniznotation: dx , dx2 , dx3 ,etc.
3.2.5
B asicD erivativeF ormulas
●
d constantfunction(f (x) = c ) : dx (c) = 0
●
d n powerrule: dx (x ) = nxn−1
○
Don’tusethisruleinthefollowingsituations: ■
n d dx (b )
■
n d dx [f (x)]
■
g(x) d ] dx [b
■
g(x) d dx [f (x)]
= 0 (whereb andn areconstants) = n · [f (x)]n−1 · f ′(x) (chainrule) = b g(x) · ln b · g ′(x) (chainrule) (uselogarithmicdifferentiation)
●
d constantmultiplerule: dx (c · f (x)) = c · dxd f (x)
●
d sumrule: dx [f (x) + g (x)] = dxd f (x) + dxd g(x)
●
d differencerule: dx [f (x) − g (x)] = dxd f (x) − dxd g(x)
●
d x naturalexponentialfunction: dx (e ) = ex
●
d productrule: dx [f (x) · g (x)] = f (x) · dxd g(x) + g (x) · dxd f (x)
●
d quotientrule: dx [ g(x) ] =
f(x)
d d g(x)· dx f(x) − f(x)· dx g(x) 2 [g(x)]
Section3-CalculusI-M athQ RH
197
3.2.6
D erivativeso fT rigonometricI dentities
●
d dx (sin x)
= c os(x)
●
d dx (cos x)
= − sin(x)
●
d dx (tan x)
= sec 2 (x)
●
d dx (csc x)
= − c sc(x)cot(x)
●
d dx (sec x)
= sec(x)tan(x)
●
d dx (cot x)
= − c sc 2 (x)
3.2.6.a
➔ lim
Theorem sin θ θ
= 1
θ sin θ θ→0
= 1
θ→0
➔ lim
3.2.7
T heC hainR ule
➔ Thisruleisusedfordifferentiatingcompositefunctions. ➔ Ifg isdifferentiableatx andf isdifferentiableatg (x) ,thenthecompositefunction F = f ° g definedbyF (x) = f (g(x)) isdifferentiableatx ,andF ′ isgivenby F ′(x) = f ′(g(x)) · g ′(x) InLeibnitznotation,ify = f (x) andu = g (x) arebothdifferentiablefunctions,then dy dx
=
dy du
·
du dx
readas“thederivativeofthefunctionywithrespecttoutimesthederivativeof functionuwithrespecttox”.
198
Section3-CalculusI-M athQ RH
➔ ThisdefinitionprettymuchmakesnosenseandwhenIfirstheardit,Ihadnoclue whattodo. However,inpracticeitisveryeasytouse.
◆
Todifferentiateacompositefunctionf (g(x)) ,lettheouterfunctionbeequal tof (u) wheretheinnerfunctiong (x) isreplacedbyu . Then,findthe derivativeoff (u) ,withrespecttou . Replaceu withtheoriginalinner functiong (x) . Now,findthederivativeoftheinnerfunctionandmultiplythis derivativebytheotherderivativeyoufound(f ′(u) whereu isthenreplaced withtheoriginalinnerfunction).
3.2.8
T heD erivativeo fa nE xponentialF unction
➔ Usingexponentidentitiesweknowthatb x = (e ln b )x = e (ln b) x . ➔ Thechainruletellsusthat y = e u andu = (ln b) x dy du dy dx
= e u anddu dx = ln b
= e u · ln b = e (ln b) x · ln b = b x · ln b
➔ Therefore,weknowthatdxd (b x) = b x · ln b .
3.2.9
I mplicitD ifferentiation
➔ Thisisusedtodifferentiatefunctionswhereitisimpliedthatyisafunctionofx,or anyotherindependentvariable(i.e.whenyisthedependentvariable). ➔ Ithinkthatthebestwaytodescribethisisnotthroughwords,butratherthroughan example.
Section3-CalculusI-M athQ RH
199
3.2.9.a
Example
Differentiatetheequationx 3 + y 3 = 6 xy ,whereyisafunctionofx. Since6 xy isinseparable, youmustplugin(3,3)attheend. 1. Differentiatebothsidesoftheequationwithrespecttox d (x3 dx
+ y3 ) =
d (6xy) dx
2. Tosolvetheleftsideoftheequation,firstimplementthesumrule(derivativeofa 3
sumisasumofderivatives). Usethepowerrulefordxd (x ) . Youneedtousethe 3
chainrulefordxd (y ) . Todothis,settheouterfunctiontobef (u) = u 3 ,whereu = y . Tosolvetherightsideoftheequation,usetheproductrule,wheref (x) = 3 x and dy
g (x) = y . Thederivativeofyisdx .
3x2 + 3y2 ·
dy dx
= 6x ·
dy dx
+ 6y
3. Divideeverythingby3togetridofsomeconstants.
x2 + y2 ·
dy dx
= 2x ·
dy dx
+ 2y
dy
4. Collect dx t ooneside.
y2 ·
dy dx
− 2x ·
dy dx
= 2y − x2
dy
5. Factordx out. dy 2 dx (y
− 2x) = 2y − x2
6. Dividebothsidesbyy 2 − 2 x . dy dx
200
=
2y−x2 y2 −2x
Section3-CalculusI-M athQ RH
3.2.9.a
Example(con’t)
7. Wedonotneedtosubstituteinanequationfory asweusuallywould,since6 xy is inseparable. Therefore,weneedtoplugin(3,3). dy dx
=
2(3)−32 32 −2(3)
= − 1 a tpoint(3,3)
8. Now,usingthepoint-slopeformula,findanequation. y − 3 = − 1 (x − 3 ) → y = − x + 6
3.2.10
➔
D erivativeso fL ogarithmicF unctions dy dx (log b x)
3.2.10.a
=
1 x·ln b
Proof
Lety = log b x . Then,b y = x . Differentiatebothsides. y d dx (b )
=
d dx (x)
→ by · ln b ·
dy dx
= 1 →
dy dx
=
1 y b ·ln b
Substituteinx = b y . dy dx
=
1 x·ln b
Q.E.D.
➔ Ifweletb = e ,thenaturalbase,thenlog e x = ln x andln e = 1 . Then, d (ln x) dx
Section3-CalculusI-M athQ RH
=
1 x
201
3.2.10.b
PropertiesofLogarithms
Whendoingthesetypesofproblems,itsavesalotofworkifyouknowthelogarithmic properties. ●
log b ( xy ) = log b x − log b y
●
log b (xy) = log b x + log b y
●
log b xn = nlog b x
3.2.11
S tepsi nL ogarithmicD ifferentiation
1. Takethenaturallogarithmsofbothsidesoftheequationandusethelawsof logarithmstosimplify. 2. Differentiatebothsidesimplicitlywithrespecttox(orwhatevertheindependent variablemaybe). dy
3. Solvefor dx ory ′ (whicheveryouchoosetouse). 4. Replaceanyoccurrenceofy withitsoriginalexpression.
202
Section3-CalculusI-M athQ RH
3.2.12
E xponentialG rowtha ndD ecay
Quantitiesgrowanddecayatarateproportionaltotheirsize. ●
Ify = f (t) isthenumberofindividualsinapopulationintimet ,itseemsreasonable toexpecttherateofgrowthf ′(t) isproportionaltothepopulationf (t) . f ′(t)
●
∝ f (t)
Thus, f ′(t) = k · f (t) ,wherek isaconstant dy dt
= k y ,wherey isafunctionoft
Thisiscalledthelawofnaturalgrowthifk > 0 orthelawofnaturaldecayifk < 0 . Thisisalsocalledad ifferentialequation(whichwillbeseeninmoredetailinCalculusII dy
and)becauseitinvolvesanunknownfunctiony anditsderivative dt . Thecommonfunctionfordifferentialequationsisy (t) = C e kt ,whereC isaconstantandwill beasolutionwhent = 0 . ●
Proofthaty (t) = C e kt isasolutiontof ′(t) = k · f (t) y ′(t) = C (ke kt ) ,swapC andk y ′(t) = k (Ce kt ) ,rememberthaty (t) = C e kt y ′(t) = k · y (t)
Section3-CalculusI-M athQ RH
203
3.2.12.a
PopulationGrowth dP dt
= kP
RelativeGrowthRateis
k=
1 P
·
dP
( Pdt = k )
dP dt
PopulationModelis P (t) = P o e kt ,whereP o istheinitialpopulationatt = 0 3.2.12.b
RadioactiveDecay
Radioactivesubstancesdecayspontaneouslybyemittinggammaandotherparticles. In theprocess,acertainamountundergoingradioactivedecaylosesmass. ●
Letm(t) representthemassofasubstanceattimet ,wheremo istheinitialvalue.
Radioactivedecayis−
1 m
·
dm dt
So, dm = km ( k < 0 ) dt Weusetheequationm(t) = mo · e kt (k < 0 ) And,thehalf-lifeofaradioactivesubstanceisthetimerequiredforhalfofthegiven amountoftheinitialsubstancetodecay.
204
Section3-CalculusI-M athQ RH
3.2.12.c
Newton’sLawofCooling
Therateofcoolingofanobjectisproportionaltothedifferenceintemperaturebetween theobjectanditssurroundings,providedthatthedifferenceisnottoolarge. ●
T (t) isthetemperatureofanobjectattimet (thisisusuallywrittenasT )
●
T s isthetemperatureoftheobjects’surroundings
Then, dT dt
= k(T − T s) ,wherek isaconstant
Weviewtheexpression(T − T s ) asy (t) = T (t) − T s anddifferentialwithrespecttot ,we’d have y ′(t) = T ′(t)
∴
dy dt
= k y
3.2.13
R elatedR ates
Strategy: 1. Readtheproblemcarefully 2. Drawadiagram,ifpossible(assomeonewhousedtobeagainstdiagramsforsome reason,Ipromisethatthisisimportant) 3. Introducenotationandassignsymbolstoallquantitiesthatarefunctionsoftime 4. Expressthegiveninformationandtherequiredrateintermsofderivatives (rememberthataderivativeischangeovertime) 5. Writeanequationthatrelatesthevariousquantitiesoftheproblem. Usegeometry toeliminateoneofthevariables,ifneeded. 6. Usethechainruletodifferentiatebothsidesoftheequationwithrespecttot . 7. Substitutethegiveninformationintotheresultingequation.
Section3-CalculusI-M athQ RH
205
IthinkthatthissectionofCalcIismucheasiertounderstandwhenanexampleisworked through,soseebelowforacommonquestionfoundwhenworkingwithrelatedrates. 3.2.13.a
Example
Gravelisbeingdumpedfromaconveyorbeltatarateof3 0 f t3 /min ,anditscoarsenessissuch thatitformsapileintheshapeofaconewhosebasediameterandheightarealwaysequal. Howfastistheheightofthepileincreasingwhenthepileis7 f t high? First,let’smakeadiagram. Now,identifythegiveninformationandwhatwewanttofindout. Weknowthatthebasediameterisequaltotheheight. Thismeans that2 r = h . Withthegivennumber3 0 f t3 /min ,wecantellthatthisis arateofchangeofavolume(f t iscubed).Wearelookingforthe rateofchangeoftheheightofthepile. So,weknowthat dV dt
= 30 ft3 /min andh = 7 . Thequestionisaskingustocalculate dh . dt
TheequationforvolumeofaconeisV = 31 πr 2 h . Thisequationhastwovariables,andwe wantitintermsofheight,asthatwasthegivenvariable(7 f t high). Wecanreplacer 2 in thevolumeequationbyrearrangingthe2 r = h equation. Todothis,dividebothsidesby two,thensquarebothsidestogetr 2 =
h2 4
V =
. Now,replaceandsimplify. π 3
·
h2 4
206
·h =
πh 3 12
Section3-CalculusI-M athQ RH
Takethederivativeofbothsides: dV = dt
π 12
· 3h2 ·
dh dt
= 4π h2 ·
dh dt
(similartoimplicit
differentiation,weneedtomultiplyby dh todifferentiateintermsoft ) dt Now,solvefor dh : dh = dt dt
4 πh 2
·
dV dt
Sinceeverythingissimplified,wecanpluginthegiveninformation: dh dt
=
4 π
·
1 72
· 30 =
120 49π
≈ 0.78 ft/min
3.2.14 ●
L inearA pproximations
Todolinearapproximations,wemustimplementourknowledgeoftangentlines. Webeginwithpoint-slopeformulaandconvertitintocalculusterms. (y − y 1 ) = m(x − x 1 ) and(x, y)
Thisbecomesf (x) − f (a) = f ′(a)(x − a ) forthederivativeat′a ′ andthepoint(a, f (a)) fory = f (x) Theequationofthetangentlineistheny = f (a) + f ′(a)(x − a ) ➔ Forx neara oftheactualfunctionf (x) ,wesaythatthelinearapproximationoff at a isgivenby L(x) = f (a) + f ′(a)(x − a )
Section3-CalculusI-M athQ RH
207
3.2.15
D ifferentials
Ify = f (x) wheref isdifferentiable,thenthedifferentialisd x . d x isanindependent variable. ❏ Ifx istheindependentvariable(input)andy isthedependentvariable,takinga smallpieceofx (whichiswhatadifferentialreallyis)doesnotmakeitloseits independence Thedifferentiald y isdefinedintermsofd x ,soit’sadependentvariable. d y = f ′(x) dx → f ′(x) =
dy dx
Δ y = f (x − Δ x) − f (x)
Thereisgoingtobeanerrorbetweenthederivativeheightandtheapproximateheight.
208
Section3-CalculusI-M athQ RH
3.2.16
M inimumsa ndM aximums
➔ Letc beanumberinthedomainD ofafunctionf ●
absolutemaxvalueoff onthedomainD iff (c) ≥ f (x) forallx indomainD
●
absoluteminvalueoff onthedomainD iff (c) ≤ f (x) forallx indomainD
Wecanknowthatvaluesareabsoluteextrema,evenwhennotinaclosedinterval,byboth lookingatendbehaviorandmakingcalculations. ●
localmaxvalueoff iff (c) ≥ f (x) whenx isnearc
●
localminvalueoff iff (c) ≤ f (x) whenx isnearc
3.2.17 E xtremeV alueTheorem( EVT) ➔ Iff iscontinuousonaclosedinterval[a, b] ,thenf attainsanabsolutemaximum valuef (c) andanabsoluteminimumvaluef (d) atsomenumbersc andd in[a, b] . ➔ Rememberthateverydifferentiablefunctioniscontinuousbutnoteverycontinuous functionisdifferentiable.
Section3-CalculusI-M athQ RH
209
3.2.18
F ermat’sT heorem
➔ Iff hasalocalmaximumorminimumatc ,andiff ′(c) exists,thenf ′(c) = 0 .
◆
Thisgivesahintthatwhereverthederivativeis0,therem aybeanextrema
➔ Alternatively,ifx hasalocalmaximumorminimumatc ,thenc isac riticalnumber off .
◆
Thisisnotsayingthatthecriticalnumbermustbealocal/absolute minimum/maximum,itissayingthat,ifthereisanextrema,itwillbefound there.
3.2.19
T heC losedI ntervalM ethod
Tofindthea bsolutemaximumandminimumvaluesofacontinuousfunctionf onaclosed interval[a, b] . 1. Findthevaluesoff atthecriticalnumbersoff in(a, b) . a. setf (x) = 0 andsolvetofindc andevaluatethem 2. Evaluatef (a) andf (b) 3. Thelargestofthevaluesfromsteps1and2istheabsolutemaximumvalue,the smallestofthesevaluesistheabsoluteminimum
210
Section3-CalculusI-M athQ RH
3.2.20
R olle’sTheorem
➔ Letf beacontinuousfunctionthatsatisfiesallthreehypotheses:
◆ ◆ ◆
f isacontinuousfunctionontheclosedinterval[a, b] f isdifferentiableontheopeninterval(a, b) f (a) = f (b)
➔ Then,thereisatleastonenumberc in(a, b) ,suchthatf ′(c) = 0 .
Section3-CalculusI-M athQ RH
211
3.2.21
M eanV alueT heorem
➔ Letf beafunctionthatsatisfiesbothhypotheses: 1. f iscontinuousontheclosedinterval[a, b] 2. f isdifferentiableontheopeninterval(a, b) Then,thereisanumberc in(a, b) ,suchthat
f ′(c) =
f(b) − f(a) b − a
orf (b) − f (a) = f ′(c)(b − a )
f ′(c) = m oraveragerateofchangeon[a, b]
Alllinesareparallel.
212
Section3-CalculusI-M athQ RH
3.2.22
C orollary
➔ f ′(x) = g ′(x) forallx ininterval(a, b) ,thenf − g isconstanton(a, b) ;thatis, f (x) = g (x) + c wherec isaconstant.
3.2.23
H owD erivativesAffectt heG raph
3.2.23.a
Increasing/DecreasingTest
●
Iff ′(x) > 0 onaninterval,thenthefunctionf isincreasingonthatinterval
●
Iff ′(x) < 0 onaninterval,thenthefunctionf isdecreasingonthatinterval
3.2.23.b
TheFirstDerivativeTest
➔ Supposethatc isacriticalnumberofacontinuousfunctionf a. iff ′ changesfrompositivetonegativeatc ,thenf hasalocalmaxatc b. iff ′ changesfromnegativetopositiveatc ,thenf hasalocalminatc c. iff ispositiveonbothsidesornegativeonbothsidesofc ,thenf hasnolocal min/max
Section3-CalculusI-M athQ RH
213
Toapplythistest,usethesametechniqueasbefore. 1. Differentiatef (x) andsettheanswerequaltozero. Eachsolutionoftheequation f ′(x) = 0 isacriticalpointx = c wherethefunctionf m aychangesign(donot assumeitdoes) 2. Dividetheinterval[a, b] intothesubintervalsbetweenthecriticalpoints(number line) 3. Determinethesignoff ′(x) ineachsubinterval 4. Usea ,b ,orc inthefirstderivativetesttodeterminewhetheralocalmax,local min,orneitheroccursateachx = c 3.2.23.c
Concavity
Concavityisatermusedtodescribehowacurvebendsasitincreasesordecreases. The secondderivativewilldescribethisbehaviorbasedonitssign ●
Ifthegraphoff liesaboveallitstangentlinesonanintervalI ,thenitiscalled concaveupward
●
Ifthegraphoff liesbelowallitstangentlinesonanintervalI ,thenitiscalled concavedownward
214
Section3-CalculusI-M athQ RH
3.2.23.d
ConcavityTest
●
iff ′′ > 0 forallx inI ,thenthegraphoff isconcaveupwardonI
●
iff ′′ < 0 forallx inI ,thenthegraphoff isconcavedownwardonI
3.2.23.e
InflectionPoint
➔ apointP onacurvey = f (x) iscalledaninflectionpointiff iscontinuousthereand thecurvechangesconcavity 3.2.23.f
TheSecondDerivativeTest
➔ Supposef ′′ iscontinuousnearc a. iff ′′(c) = 0 andf ′′(c) > 0 ,thenf hasalocalminimumatc b. iff ′′(c) = 0 andf ′′(c) < 0 ,thenf hasalocalmaximumatc
3.2.24
I ndeterminateF orms
➔ Anindeterminateformoccurswhenyouaretakingthelimitofaquotientandit ∞ equals 00 or ∞ .
Section3-CalculusI-M athQ RH
215
3.2.25
L 'Hôpital'sR ule
➔ L'Hôpital'sRuleisusedtosolvelimitsthatareindeterminateforms. ➔ Supposef andg aredifferentiableandthatg ′(x) =/ 0 onanopenintervalf that containsa (exceptpossiblyata ). Alsosupposethat, lim f (x) = 0 andlim g(x) = 0 x→a
x→a
orthat lim f (x) = ± ∞ andlim g(x) = ± ∞ x→a
x→a
Then, f(x)
f ′(x) g x→a ′(x)
lim g(x) = lim
x→a
ifthelimitontherighthandsideexists(oris∞ or− ∞ ). ➔ Insummation,thelimitofaquotientoffunctionsisequaltothelimitofthequotient oftheirderivatives,providedthatthegivenconditionsaresatisfied:
◆ ◆
indeterminateforms( 00 o r ∞∞ ) f andg aredifferentiable(g ′(x) =/ 0 )
L'Hôpital'sRuleisalsovalidforone-sidedlimitsandforlimitsatinfinityornegativeinfinity.
Ⓗ
Notation: overanequalsignshowsthatL'Hôpital'sRulewasapplied. Thismustbe includedinallproblems.
216
Section3-CalculusI-M athQ RH
3.2.25.a
IndeterminateProducts
Ifwefindthat lim f (x) = 0 andlim g(x) = ∞ x→a
x→a
andweareaskedtofind lim [f (x) · g (x)] x→a
weendupwiththeindeterminateform0 · ∞ . L'Hôpital'sRuledoesnotapplytothisform. Therefore,wemustrewritetheexpressionasaquotient.
lim [f(x) · g (x)] = lim f(x) ,youcanpickeitherfunctiontobeinthenumerator 1
x→a
x→a
g(x)
1 Nowwehavethenumeratorf (x) → 0 andthedenominatorg(x) → 0 . Thisisthe
indeterminateform00 ,andnowL'Hôpital'sRulecanbeapplied. 3.2.25.b
IndeterminateDifference
Iflim f (x) = ∞ andlim g(x) = ∞ ,thenthe x→a
x→a
lim [f (x) − g (x)] x→a
iscalledtheindeterminateform∞ − ∞ . YoucannotapplyL'Hôpital'sRuletothislimit,sorearrangetogetaquotient.
Section3-CalculusI-M athQ RH
217
3.2.25.c
IndeterminatePowers
Withthelimitformlim [f (x)]g(x) ,wecouldendupwith0 0 , ∞ ∞, 1 ∞, or∞ 0 . Theseareall x→a
indeterminateforms. Tosolvetheselimits,wemustuselogarithms. If, y = [f (x)]g(x) then, ln y = g (x) · ln(f (x)) or, [f (x)]g(x) = e g(x)·ln(f (x)) TableofDeterminateandIndeterminateForms IndeterminateForms
DeterminateForms
0 /0
∞+∞ =∞
± ∞/ ± ∞
−∞−∞ = −∞
∞−∞
0∞ = 0
0 (∞)
0 −∞ = ∞
00
(∞) · (∞) = ∞
1∞
∞0
UseL’Hôpital’sRule
DoNotUseL’Hôpital’sRule
218
Section3-CalculusI-M athQ RH
3.2.26
G raphS ketching
Thingsyoumustinclude: 1. domain 2. symmetry/oddandeven 3. intercepts 4. asymptotes 5. intervalsofincreasinganddecreasing 6. localmaximumandminimum 7. concavityandpointsofinflection 8. sketch
3.2.27
O ptimization
StepsinSolving: 1. Readcarefully,identifyvaluesandconditions 2. Drawapicture 3. Assignavariableorsymboltorepresentunknowns 4. Writeexpressions ThisisyetanotherpartofCalculusIthatIthinkisbestshownthroughanexample.
Section3-CalculusI-M athQ RH
219
3.2.27.a
Example
Arightcircularcylinderisinscribedinaconewithheighth andradiusr . Findthelargest possiblevolumeofsuchacylinder. Tobegintosolvethisquestion,wemustmakeadiagram.
ThevolumeofacylinderisV cyl = π r 2 h . Thecylinderheightish − y (heightofthecone minusthespacebetweenthetopoftheconeandthetopofthecylinder). Wecanusesimilartriangles:
Wecanthenusethepropertiesofsimilartrianglestofindanexpressionfory . y x
220
=
h r
→y=
xh r
Section3-CalculusI-M athQ RH
Then,wecanusethesetwoexpressionsandsubstitutethemintotheformulaforthe volumeofacylinder. V (x) = π x 2 (h − y ) = π r 2 (h − V (x) = π x 2 h −
xh 3 rx
xh r )
Itisimportanttorememberthatπ , h, andr areallconstants,notvariables. Also,wemustnotethatthevolumeofthecylinderisconstrainedbythecone. Sincewearelookingforthemaximumvolumeofthecylinder,wearegoingtousethe samestrategythatweusewithgraphingfunctions. Thatis,wewilltakethederivativeof thefunction,findthecriticalpoints,andevaluatethefunctionatthepoints. 3πh r
V ′(x) = 2 πxh − V ′(x) = x πh(2 −
3x r)
· x2 =0
Thecriticalnumbersarex = 0 andx = 32 r . Sincetheradiusofthecylindercannotbe0,so wecanonlyusex = 32 r . V ( 32 r) = π ( 32 r)2 h −
πh 2 3 r ( 3 r)
=
4πr2 h 9
2
− 8πhr 27 =
4 2 27 (3πhr
− 2 πhr 2 ) =
4 2 27 πhr
Therefore,thelargestpossiblevolumeofthecylinderis274 πhr 2 .
Section3-CalculusI-M athQ RH
221
3.3
Integrals
3.3.1
A ntiderivatives
➔ AfunctionF iscalledanantiderivativeoff onanintervalI ifF ′(x) = f (x) forallx in I . 3.3.1.a
PowerRuleandAntiderivatives
Whenusingthe“powerrule”forantiderivatives,youapplyitbackwards. Thismeansthatyouaddonetotheexponentanddividebythatnumber.
f (x) = xn → F (x) =
xn+1 n+1
3.3.1.b
FunctionsandTheirParticularAntiderivatives
➔ Rememberthat,whenyoutakeaderivativeofafunction,you“lose”anyconstants. So,whenwetaketheantiderivative,wemustaddageneralconstant“C ”totheend oftheantiderivativetomakeupforthisloss. Thisisknownasag eneral antiderivative.
222
Section3-CalculusI-M athQ RH
3.3.1.b
FunctionsandTheirParticularAntiderivatives(con’t)
➔ Ap articularantiderivative,however,istheantiderivativeofaspecificfunction. So, theconstantC willhaveadefinedvalue(C canequalzero). Function
ParticularAntiderivative
c · f (x)
c · F (x)
f (x) · g (x)
F (x) · G(x)
f (x) − g (x)
F (x) − G(x)
x n , n =/ − 1 1 x
l n |x |
ex
ex
bx
bx ln b
c os x
sin x
sin x
− c os x
sec 2 x
tan x
sec x tan x
sec x
1 √1−x2 1 1+x2
c (constant)
xn+1 n+1
sin −1 x tan −1 x cx + d
Section3-CalculusI-M athQ RH
223
3.3.2 P articleM otion ➔ t representstime ●
positionfunctions(t)
●
velocityfunctions′(t) = v (t)
●
accelerationfunctionv ′(t) = a (t)
●
jerkfunctiona ′(t) = j (t)
➔ Thisisknownas“rectilinearmotion”. Thislistalsoworksinreverse(i.e.the antiderivativeofa (t) willproducev (t) ,andsoon).
3.3.3
S igmaN otation
Ifa m ,a m+1 ,…,a n areallrealnumbersanm andn areintegerssuchthatm ≤ n ,then: n
∑ a i = a m + a m+1 + a m+2 + ... + a n−1 + a n
i=m
❏ thesymbol∑ ispronounced“sigma”anditmeans“add”
❏ thei representsiterate/indexofsummation ❏ a i representstheargument ❏ the‘... ’represents“andsoon” ❏ usea n−1 whenn isunknown,asthisshowsthetermbeforea n
224
Section3-CalculusI-M athQ RH
3.3.3.a
Theorem
➔ ifc isanyconstant(thatdoesn’tdependoni ),then a. b. c.
n
n
i=m
i=m
∑ c a i = c ∑ a i (factoringoutacommonfactor) n
n
n
i=m
i=m
i=m
n
n
n
i=m
i=m
i=m
∑ (a i + b i ) = ∑ a i + ∑ b i (associativeandcommutativeproperty)
∑ (a i − b i ) = ∑ a i − ∑ b i
3.3.3.b
Theorem
➔ letc beaconstantandn beapositiveinteger.Then, a. b. c. d. e.
n
∑ 1 =n
i=1 n
∑ c =c·n
i=1 n
∑i= i=1 n
n(n+1) 2
∑ i2 = i=1
n(n+1)(2n+1) 6
n
2 ∑ i3 = [ n(n+1) 2 ]
i=1
Section3-CalculusI-M athQ RH
225
3.3.4
D efiniteIntegrals
➔ Iff isafunctiondefinedfora ≤ x ≤ b ,wedividetheinterval[a, b] inton subintervalsofequalwidthΔx =
b−a n
. Weletx 0 = a , x 1 , x 2 , x 3 , ... , x n = b bethe
endpointsofthesesubintervalsandweletx1*, x2*, x3*, ... , xn* beanysamplepointsin thesesubintervals,sox*i liesintheith subinterval[x i−1 , x i ] . Thenthedefinite integralfroma tob is b
∫ f (x)dx = a
n
lim ∑ f (x*i )Δx
n→∞ i=1
providedthatthelimitexistsandgivesthesamevalueforallpossiblechoicesof samplepoints. Ifitdoesexist,wesaythatf isintegrableon[a, b] ➔ Notation: b
∫ f (x)dx a
●
f (x) istheintegrand
●
d x indicates“withrespecttox ”
●
a isthelowerlimit,andb istheupperlimit
●
thedefiniteintegral∫ f (x)dx isan umber
b
a
➔ Thedefiniteintegralgivestheareaunderafunction.
226
Section3-CalculusI-M athQ RH
3.3.4.a
NegativeIntegrals
Areacannotbenegative,butintegralsfindthen etareafromthecurvetothex -axis. So, youmustsubtractthepartofthecurvethatisbelowthex -axis b
∫ f (x)dx =areaabovex -axis-areabelowx -axis a
b
c
d
b
a
a
c
d
∫ f (x)dx = ∫ f (x)dx − ∫ f (x)dx + ∫ f (x)dx 3.3.4.b
Theorem(JumpDiscontinuities)
Iff iscontinuouson[a, b] ,oriff hasonlyafinitenumberofjumpdiscontinuities,thenf isintegrableon[a, b] (i.e.thedefiniteintegralexists)
Allthreeareasarefinitenumbers. Sincethisintegralwouldbeasumofareas,itisfineifithasjumps.
Section3-CalculusI-M athQ RH
227
3.3.4.c
Theorem
Iff isintegrableon[a, b] .Then, b
∫ f (x)dx = a
n
lim ∑ f (x i )dx
n→∞ i=1
where,
Δx =
b−a n
andx i = a + iΔx
3.3.4.d
MidpointRule b
∫ f (x)dx = a
n
∑ f (x i )Δx = Δ x[f (x i ) + ... + f (x n )]
i=1
whereΔ x =
b−a n
and
x i = 21 (x i−1 + x i ) = mdpt of [x i−1 , x i ] = M n 3.3.4.e 1.
PropertiesofaDefiniteIntegral
b
a
a
b
∫ f (x)dx = − ∫ f (x)dx
(interchanginglimitschangesthesign) a
2. ifa = b ,Δ x = 0 ,so∫ f (x)dx = 0 a
228
Section3-CalculusI-M athQ RH
3.3.4.e 3.
PropertiesofaDefiniteIntegral(con’t)
b
∫ c dx = c(b − a ) ,wherec a
isanyconstant
a. arectangle’sareaisc (b − a )
b
b
b
a
a
a
4.
∫ [f (x) ± g (x)]dx = ∫ f (x)dx ± ∫ g (x)dx
5.
∫ c · f (x)dx = c · ∫ f (x)dx
6.
∫ f (x)dx + ∫ f (x)dx = ∫ f (x)dx ,where
b
b
a
a
c
b
b
a
c
a
(thiscouldbesomethinglikeajumpdiscontinuity) 3.3.4.f
ComparisonProperties b
1. iff (x) ≥ 0 fora ≤ x ≤ b ,then∫ f (x)dx ≥ 0 a
b
b
a
a
2. iff (x) ≥ g (x) for[a, b] ,then∫ f (x)dx ≥ ∫ g (x)dx a. f (x) ishigherthang (x) ,therefore,h f (x) > h g(x)
Section3-CalculusI-M athQ RH
229
3.3.4.f
ComparisonProperties(con’t) b
3. ifm ≤ f (x) ≤ M for[a, b] ,thenm(b − a ) ≤ ∫ f (x)dx ≤ M (b − a ) a
3.3.5
T heFundamentalTheoremo fC alculus
3.3.5.a
PartOne
➔ Iff iscontinuouson[a, b] ,thenthefunctiong isdefinedby x
g (x) = ∫ f (t)dt wherea ≤ x ≤ b a
iscontinuouson[a, b] anddifferentiableon(a, b) ,andg ′(x) = f (x) .
230
Section3-CalculusI-M athQ RH
3.3.5.b
PartTwo
➔ iff iscontinuouson[a, b] ,then b
∫ f (x)dx = F (b) − F (a) a
whereF isa nyantiderivativeoff ,thatis,afunctionsuchthatF ′ = f . ➔ thiswillgivethee xactareaunderthecurveasanumber 3.3.5.c
Notation
Thecommonnotationforevaluatingintegralsis b
∫ f (x)dx = F (x)|ab = F (b) − F (a) a
b
or,∫ f (x)dx = F (x)]ab = F (b) − F (a) a
3.3.5.d
Composite
Supposef iscontinuouson[a, b] x
1. ifg (x) = ∫ f (t)dt ,theng ′(x) = f (x) a
2.
b
∫ f (x)dx = F (b) − F (a) ,whereF a
isanyantiderivativeoff
Section3-CalculusI-M athQ RH
231
3.3.6
I ndefiniteI ntegrals
∫ f (x)dx = F (x) + C
meansF ′(x) = f (x) ,whereF (x) + C isthegeneralantiderivative
Keepinmindthat: b
1. Adefiniteintegral∫ f (x)dx isanumber a
2. Anindefiniteintegral∫ f (x)dx isafunction/afamilyoffunctions(adding‘+ C ’will
changethefunctiondependingonC ,whichwecannotfindwithoutmoreinfo,likea pointonthegraphofthefunction)
a. exampleofwhy+ C isneeded:∫ x 2 dx =
x3 3
+ C becausedxd ( x3 + C ) = x 2 3
3. Anindefiniteintegraldoesnothavelimitsofintegration
232
Section3-CalculusI-M athQ RH
3.3.6.a
CommonIndefiniteIntegrals:
Thefollowingindefiniteintegrals(akaantiderivatives)areverycommonanditwouldhelpgreatlyif youmemorizedthem.
1.
∫ c · f (x)dx = c · ∫ f (x)dx
2.
∫[f (x) ± g (x)]dx = ∫ f (x)dx ± ∫ g (x)dx
3.
∫ k dx = k x + C
4.
∫ xn dx =
5.
xn+1 n+1
+ C wheren =/ 1
∫ exdx = ex + C
6.
∫ 1x dx = ln |x| + C
7.
x
b +C ∫ b xdx = ln b
8.
∫ |x | d x =
9.
x|x| 2
+C
∫ sin x dx = − cos x + C
10. ∫ c os x dx = sin x + C
11. ∫ c sc 2 x dx = − c ot x + C
12. ∫ sec 2 x dx = tan x + C
13. ∫ c sc x cot x dx = − c sc x + C
14. ∫ x21+1 dx = tan −1 x + C
15. ∫
1
√1−x2
dx = sin −1 x + C
Section3-CalculusI-M athQ RH
233
3.3.7
T heS ubstitutionR ule( IndefiniteI ntegrals)
➔ ifu = g (x) isadifferentiablefunctionwhoserangeisanintervalI andf is continuousonI ,then
∫ f (g(x)) · g ′(x) dx = ∫ f (u) du ➔ Lookforafunctionanditsderivativeinthesameintegrand Itotallyunderstandthatthisdefinitionprobablymakesnosense,soIwillputanexample below. 3.3.7.a Example
3
Find ∫ x4x−2 dx .
Thisequationcannotbesimplifieddownandnoneofourknownrulescanhelpus. Therefore,weneedtousethesubstitutionruletomakethisequationintosomethingmore familiar. Noticehowthenumeratorisx 3 . Thisisverysimilartothederivativeofx 4 − 2 . Therefore, letu = x 4 − 2 . Then,weneedtofindd u ,whichissimplythederivativeofu . d u = 4 x 3 dx (powerrule) Now,wearelookingforsomethinginthisderivativethatcanreplacesomethinginthe originalfunction. Ideally,wearelookingforsomethingtoreplacex 3 dx . Rearrangethe function. 1 4 du
234
= x 3 dx
Section3-CalculusI-M athQ RH
Thisisexactlywhatwearelookingfor. Intheoriginalfunction,therewasnocoefficientof4 onx 3 . So,wehadtodivideby4onbothsides. Now,wecansubstituteinournew equations.
∫
x3 x4 −2 dx
→ ∫ u1 · 41 du
Thisnewfunctionfollowsthesubstitutionruleof∫ f (u) du ,sowemayproceed. First,wecan
factorout41 ,asitisacoefficient. 1 4
· ∫ u1 du
Lookbackatthelistofcommonindefiniteintegrals. Noticethat∫ 1x dx = ln |x | + C . This
meansthat 1 4
· ∫ u1 du = 41 ln |u | + C
Replaceu withx 4 − 2 (whichwesetu equaltointhebeginningoftheproblem)togetthe finalanswer. So,
3
∫ x x−2 dx = 41 ln ||x4 − 2 || + C 4
Section3-CalculusI-M athQ RH
235
3.3.8
T heSubstitutionR ule( DefiniteI ntegrals)
➔ ifg ′ iscontinuouson[a, b] andf iscontinuousontherangeofu = g (x) ,then b
g(b)
a
g(a)
∫ f (g(x)) · g ′(x) dx = ∫
f (u) du
➔ Youusethesamemethodyouwoulduseforindefiniteintegrals,butattheendyou evaluateyouransweratg (b) andg (a) ,andsubtractF (g(a)) fromF (g(b)) ,asyou wouldwitharegulardefiniteintegral.
3.3.9
I ntegralsofS ymmetricFunctions
Supposef iscontinuouson[− a , a] a
a
−a
0
a. iff iseven[f (− x ) = f (x)] ,then ∫ f (x) dx = 2 ∫ f (x) dx
236
Section3-CalculusI-M athQ RH
a
b. iff isodd[f (− x ) = − f (x)] ,then ∫ f (x) dx = 0 −a
Section3-CalculusI-M athQ RH
237
“Whatdidonec alculustextbooks aytotheother?
Don’tbotherm e.I’vegotm yownproblems.” -Homoropedia.com
3.4
MyNotesforCalculusI
238
Section3-CalculusI-M athQ RH
3.4
MyNotesforCalculusI(con’t)
Section3-CalculusI-M athQ RH
239
3.4
MyNotesforCalculusI(con’t)
240
Section3-CalculusI-M athQ RH
“Math.It’sjustthere...You’ree itherrighto ry ou'rewrong. That’swhatIlikea boutit.” -KatherineJ ohnson
Section4-CalculusII 4.0 SummarySheet ApplicationsofIntegration b
AreaBetweenCurves:∫ |f (x) − g (x)| d x a
●
top − b ottom forx axisintegration
●
r ight − lef t fory axisintegration
AverageValueofaFunction:f ave = b
ArcLength:L = ∫
a
1 b − a
b
· ∫ f (x) dx a
√1 + [f ′(x)] dx 2
b
VolumeofaSolid:V = ∫ A(x) dx ,whereA(x) istheareaofacrosssection a
VolumesofRevolution: b
●
Disk:V = ∫ π R2 dx ,whereR isthedistancefromthefunctionandtheaxisof a
rotation.
Section4-CalculusII-M athQ RH
241
b
●
Washer:V = ∫ (πR2 − π r 2 ) dx . a
○
R isthedistancefromthe“spinningline”tothefurthestfunction. r isthe distancefromthespinninglinetothecloserfunction.
○ ●
Alwaysusetop − b ottom orr ight − lef t whenusingthismethod b
CylindricalShells:V = ∫ 2 πxf (x) dx a
●
Tips: ○
Alwaysdrawasamplerectangle. Spinyourfinger(asifitweretherectangle) alongtherevolution. Ifyourfingerchangesdirectionasitspins,usethe disk/washermethod.Ifyourfingerstaysuprightandonlyyourhandmoves, usethecylindricalshellsmethod.
○
Makesureyourfunctionsareintermsofthevariablethatcorrespondstothe rectangle’sthickness.
○
Sometimesbothtechniquesarepossibletouse. Choosetheeasieroption: theonethatdoesn’tmakeyousplituptheintegralintoseveralintegrals,and wherelimitsofintegrationareeasytofind.
b
Work:W = ∫ F (x) dx foravariableforceF (x) a
●
Hooke’sLaw(springs):F (x) = k x ,wherek isthespringconstant
●
CableProblems:F (x) = (initial weight) − (weight lost af ter lif ting · f eet)
IntegrationbyPartsFormula:∫ u · d v = u · v − ∫ v · d u
242
Section4-CalculusII-M athQ RH
●
Tochooseu ,startatthetopandworkyourwaydown,thefirstoneyouencounter shouldbeyouru . ○
L- Logarithmicfunctions
○ ○ ○ ○
I- Inversetrigfunctions P- Polynomials E-Exponentialfunctions T- Trigfunctions
TrigonometricIntegrals Strategyforevaluatingintegralsoftheform
∫ sinm (x) cosn (x) dx
●
Ifthepoweroncosineiso dd,usec os2 (x) = 1 − sin 2 (x) toreplaceallbutoneofthethe c os (x) . Groupthislonec os (x) withd x andperformthesubstitution u = sin (x) du = c os (x) dx
●
Ifthepoweronsineiso dd,usesin 2 (x) = 1 − c os2 (x) toreplaceallbutoneofthethe sin (x) . Groupthislonesin (x) withd x andperformthesubstitution u = c os (x) du = − sin (x) dx
●
Ifbothpowersaree ven,usethehalfangleidentitiesrepeatedlyuntilyouareleft withc os (kx) termsonly
Section4-CalculusII-M athQ RH
243
ImportanttoMemorize
∫ cos (kx) dx → u = kx, dx = 1k du → 1k ∫ cos (u) du = 1k sin (kn) + C
∫ sin (kx) dx = − 1k cos (kx) + C
Strategyforevaluatingintegralsoftheform
∫ tanm (x) secn (x) dx
●
Ifthepoweronsecantise ven,groupsec 2 (x) withthed x . Thenuse sec 2 (x) = tan 2 (x) + 1 toreplacealloftheremainingsec (x) terms. Performthe substitution u = tan (x) du = sec 2 (x) dx
●
Ifthepowerontangentiso dd,groupsec (x) tan (x) withthed x . Use tan 2 (x) = sec 2 (x) − 1 toreplacealloftheremainingtan (x) terms. Performthe substitution u = sec (x) du = sec (x) tan (x) dx
●
Forothercases,youwillneedtobecreativeanduseothertools. Somesuggestions are ○
rewritingeverythingintermsofsineandcosine
○
usingatrigidentityandsplittinguptheintegral(seetheprecalculus summarysheet)
244
○
usingintegrationbyparts
○
trialanderror
Section4-CalculusII-M athQ RH
Strategyforsolvingothertrigintegralswherem =/ n
●
∫ sin (mx) cos (nx) dx
Usesin A cos B = 21 [sin (A − B ) + sin (A + B )]
○
●
∫ sin (mx) sin (nx) dx
Usesin A sin B = 21 [cos (A − B ) − c os (A + B )]
○
●
∫ cos (mx) cos (nx) dx
Usec os A cos B = 21 [cos (A − B ) + c os(A + B )]
○
TrigonometricSubstitution a issomenumber Expression
Substitution
ValidInterval
Identity
√a 2 − x2
x = a sin θ dx = a cos θ dθ
−
π 2
≤θ≤
π 2
1 − sin 2 θ = c os2 θ
√a 2 + x2
x = a tan θ dx = a sec 2 θ dθ
−
π 2
≤θ≤
π 2
1 + tan 2 θ = sec 2 θ
√x2 − a 2
π 2
orπ ≤ θ ≤
x = a sec θ dx = a sec θ tan θ dθ 0 ≤ θ ≤
3π 2
sec 2 θ − 1 = tan 2 θ
Youwillcommonlyseethesewithasquareroot,butitdoesn’thavetobethere.
Section4-CalculusII-M athQ RH
245
MoreIntegrationStrategies IntegrationbyPartialFractions:Ifanintegralhasacomplicatedrationalfunction,wecan usepartialfractionstohelpevaluatetheintegral. ●
Ifthedegreeofthenumeratorisgreaterthanorequaltothedegreeofthe denominator,youmustperformlongdivision(seeAlgebraIIsection)beforemoving ontothenextstep.
●
Factorthedenominator.
●
Performapartialfractiondecomposition.
●
Evaluatetheintegralsofeachpartialfractiontogettheintegraloftheentire function.
IntegralsYouMustKnow
∫ 1x dx = ln |x| + C
∫
1 x2 + 1 dx
= tan−1 (x) + C
1 dx = ∫ mx + b
1 m
· ln |mx + b| + C
1 dx = a1 · tan−1 ( ax ) + C ∫ x + a 2
2
LogPropertiestoKnow n · ln |ax + b| = ln |(ax + b )n | ax + b | ln |ax + b | − ln |cx + d | = ln || cx + d |
ln |ax + b| + ln |cx + d| = ln |(ax + b )(cx + d )|
246
Section4-CalculusII-M athQ RH
ImproperIntegrals: ●
TypeI:UnboundedIntegrals ∞
∫ f (x) dx =
○
a
b
∫
○
−∞
t
t → ∞ a
b
f (x) dx = lim ∫ f (x) dx t → −∞ t
∞
∫
○ ●
−∞
lim ∫ f (x) dx
c
f (x) dx =
∫
−∞
∞
f (x) dx + ∫ f (x) dx c
TypeII:DiscontinuousIntegrand b
t
Discontinuousatb :∫ f (x) dx = lim − ∫ f (x) dx ,t < b
○
a
t → b
b
a
b
Discontinuousata :∫ f (x) dx = lim + ∫ f (x) dx ,t > a
○
a
t → a
t
b
c
b
a
a
c
Discontinuousatc betweena andb :∫ f (x) dx = ∫ f (x) dx + ∫ f (x) dx
○ ∞
Theintegral∫
1
1 xP
dx convergesifp > 1 anddivergesifp ≤ 1 .
ComparisonTestforImproperIntegrals:IfbigB (x) andlittleL(x) arecontinuousfunctions withB (x) ≥ L(x) ≥ 0 ,then
∞
∞
a
a
●
If∫ B (x) dx converges,then∫ L(x) dx converges.
●
If∫ B (x) dx diverges,then∫ L(x) dx isunknown.
●
If∫ L(x) dx diverges,then∫ B (x) dx diverges.
●
If∫ L(x) dx converges,then∫ B (x) dx isunknown.
∞
∞
a
a
∞
∞
a
a
∞
∞
a
a
Section4-CalculusII-M athQ RH
247
Series n
Aseriesn th partialsum:sn = ∑ a i = a 1 + a 2 + ... + a n i = 1
●
Ifthesequenceofpartialsums{sn } isconvergentand lim sn = S isarealnumber, n → ∞
∞
i = 1
thenwesaytheseries∑ a i isconvergentandwewrite ∑ a i = S ∞
n th TermTest:If lim a n =/ 0 orD N E ,thentheseries ∑ a n isdivergent. Justbecause n → ∞
n = 1
lim a n = 0 ,doesn’tmean∑ a n converges!
n → ∞
∞
GeometricSeries:Convergentonlywhen|r | > 1 . Furthermore, ∑ a r n = n = 0
a 1 − r
.
∞
∞
∞
∞
∞
n = 1
n = 1
n = 1
n = 1
n = 1
If∑ a n and∑ b n areconvergent,then ∑ (c · a n ) = c · ∑ a n and ∑ (a n ± b n ) = ∑ a n ± ∑ b n . IntegralTest:Iff (x) isacontinuous,positive,anddecreasing(f ′(x) < 0 )functiononthe interval[1, ∞] withf (x) = a n (thesametermsastheseries),then ∞
∫ f (x) dx
⇒
converges
1
∞
∫ f (x) dx
⇒
diverges
1
∞
p-Series: ∑
n = 1
1 nP
∞
∑ a n converges
n = 1 ∞
∑ a n diverges
n = 1
,convergentwhenp > 1 anddivergentwhenp ≤ 1 .
DirectComparisonTest:If∑ a n (given)and∑ b n (similar)areserieswithpositivetermsand
● ●
248
If∑ b n convergesand0 ≤ a n ≤ b n ,then∑ a n converges(isafinitenumber)
If∑ b n divergesand0 ≤ b n ≤ a n ,then∑ a n diverges(isinfinite)
Section4-CalculusII-M athQ RH
Fordeterminingb n onp - seriescomparisons: ∞
∑ n = 1
largest power in numerator largest power in denominator
LimitComparisonTest:If∑ a n and∑ b n areserieswithpositivetermsandyouevaluatethe limit an n → ∞ b n
lim
=C
a nd
0 < C < ∞ (strictinequality)
Thenbothseriesdothesamething:theyeitherbothconvergeortheybothdiverge.
AlternatingSeriesTest:Analternatingseries∑ (− 1 )n b n convergesifboth:
1. Thetermsgoto0 : lim b n = 0 n → ∞
2. Thetermsb n decrease:b 1 ≥ b 2 ≥ b 3 ≥ ... (checkbyshowingthederivativeisnegative) Whentestingbothoftheaboveconditions,donotincludethealternatingterms. Ifthefirst conditiondoesnothappen,thentheseriesdivergesbythen th termtest.
a | RatioTest:Considertheseries∑ a n andfindthelimit lim || n + 1 a n | = r . If
●
n → ∞
r < 1 theseries∑ a n convergesabsolutely
●
r > 1 theseries∑ a n diverges
●
r = 1 testisinconclusive,tryadifferenttest
Section4-CalculusII-M athQ RH
249
Theratiotestworksbestwhenfactorialsshowup,orwhenyouhaveam ixofpolynomials, exponentials,and/orfactorials(don’tuseitwhenthereisonlyoneofthese!).
RootTest:Considertheseries∑ a n andfindthelimit lim
●
n
√|an | = r . If n → ∞
r < 1 theseries∑ a n convergesabsolutely
●
r > 1 theseries∑ a n diverges
●
r = 1 testinconclusive,tryadifferenttest
Theroottestisonlyusefulwhenthetermsoftheseriesareallraisedtothepowern . PowerSeries Apowerseriesisaseriesoftheform ∞
∑ c n (x − a )n = c 0 + c 1 (x − a ) + c 2 (x − a )2 + c 3 (x − a )3 + ... n = 0
wherex isthevariable,c n ’sarecalledthecoefficients,anda issomenumber. Thisisan infinitepolynomial. Whenyoupluginx = a ,theseriesconvergestoc 0 . IntegralsandDerivatives:Treattheseriesjustasfunctions;usethepowerruleandinverse powerrule. TheROCwillbethesameastheoriginal,butyoumustchecktheendpointsof theIOC.
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Section4-CalculusII-M athQ RH
TaylorSeries:TofindtheTaylorpolynomialofdegreen forthefunctionf (x) atx = a : T n (x) = f (a) +
f ′(a) 1!
(x − a )1 +
f ′′(a) 2!
(x − a )2 +
f (3) (a) 3!
(x − a )3 + ... +
f (n) (a) n!
(x − a )n
ATaylorSeriesofthefunctionf (x) centeredatx = a istheinfiniteseries ∞
∑ n = 0
f (n) (a) n!
(x − a )n = f (a) +
f ′(a) 1!
(x − a )1 +
f ′′(a) 2!
(x − a )2 +
f (3) (a) 3!
(x − a )3 + ...
Whentheseriesiscenteredatx = 0 (meaningthenumbera iszero),itiscalledaMaclaurin series. ImportantTaylorSeries: Function 1 1 − x
TaylorSeries ∞
∑ xn
ROC R =1
n = 0
ln (1 + x )
∞
xn + 1 ∑ (− 1 )n n + 1
R =1
∞
R =1
n = 0
tan −1 (x)
x2n + 1 ∑ (− 1 )n 2n + 1
n = 0
∞
ex
n ∑ xn!
R =∞
n = 0
c os (x)
∞
x2n ∑ (− 1 )n (2n)!
R =∞
n = 0
sin (x)
∞
2n + 1
x ∑ (− 1 )n (2n + 1)! n = 0
R =∞
Section4-CalculusII-M athQ RH
251
DifferentialEquations Adifferentialequationisanequationcontaininganunknownfunction(typicallyy )andone ormoreofitsderivatives. Differentialequationshave“orders”thatcorrespondtothe highestorderderivative. Aseparabledifferentialequationisafirstorderdifferentialequationthatfactorsintoa functionofx andafunctionofy (separatingx ’sandy ’s). Noteverydifferentialequation f(x)
isseparable. Theyneedtobeoneofthetwofollowingforms:f (x) · g (y) or h(y) . CalculuswithParametricEquations dy
derivative of y
dy/dt
DerivativeofaParametricEquation: dx = derivative of x = dx/dt d2 y
dy
d( ) = SecondDerivativeofaParametricEquation: dx2 = dx dx
d (dy/dx) dt
dx/dt
β
AreaUnderaParametricEquation:Ifx = f (t) andy = g (t) ,∫ g (t) f ′(t) dt foracurvetraveling α
fromlefttoright. b
ArcLengthofParametricCurve:∫
a
√(
dx 2 dt )
dy
+ ( dt )2 dt
CalculuswithPolarCoordinates r = f (θ) isapolarcurve. Youneedtoconvertthisintoequationsforx andy using x = r cos θ andy = r sin θ . r = f (θ) → x = f (θ)cos θ y = f (θ)sin θ
252
Section4-CalculusII-M athQ RH
dy
dy/dθ
DerivativeofaPolarCurve: dx = dx/dθ b
AreaofaPolarCurve:∫ 21 r 2 dθ a
4.1
ApplicationsofIntegration
4.1.1
A reaBetweenC urves
Theareaontheinterval[a, b] boundedbythegraphsoftwocontinuousfunctionsy = f (x) andy = g (x) isdenoted b
∫ |f (x) − g (x)| d x a
wheref (x) isthefunctionontopandg (x) isthefunctiononthebottom. So,itisthe integralofthetopminustheintegralofthebottom.
Tofindthisvalue,performthefollowing: ●
Graphthefunctions
●
Findtheintersectionpoints,whenf (x) = g (x) ,usingalgebra ○
●
theseintersectionpointsarethea andb values
Evaluatetheintegral(s)oftop-bottomontheseintervals
Section4-CalculusII-M athQ RH
253
4.1.2
A verageV alueo fa F unction
Theexactaveragevalueofthefunctiony = f (x) ontheinterval[a, b] is f ave =
1 b − a
b
· ∫ f (x) dx a
b
1 where∫ f (x) dx isanormalintegralandb − a isaconstant. a
Thisisfunctionisfindingtheaverageofn equallyspacedy values,wheren → ∞ . Thisfunctionisderivedfromtheregularaverageformula(thetotalsumofallnumbers overthenumberofitemsintheset)andtheinfinitesumdefinitionofanintegral. Ifyou wanttoseeagreatwalkthroughofthis,gotosection1.2oftheOpenStaxCalculusVolume 2freeonlinetextbook.
4.1.3
A rcL ength
Theformulaforthearclengthofacurvefromx = a tox = b is b
L=∫
a
√1 + [f ′(x)] dx 2
ThisformulaisderivedfromthePythagoreanTheorem. Basically,thecurveiscutupinton
√
tinystraightlinesegments,wherethelengthofthesegmentsare (dx)2 + (dy)2 . Ifwetake n toinfinityandaddupallthesegments,weareperforminganinfinitesumcalculation, alsoknownasanintegral. Theintegrandisthenrewrittenusingalgebraintotheabove integrand.
254
Section4-CalculusII-M athQ RH
4.1.4
y -AxisI ntegration
Rememberthatafunctioncanbewrittenasx = f (y) . Forthissection,itmaybehelpfulto reviewinversefunctions(thisisintheExponentialandLogarithmicFunctionssectionofthe AlgebraIIchapter). 4.1.4.a
GraphingaComplicatedFunctionofy
Herearetwowaystographafunctionofy : ●
Makeatableofpointsbyplugginginnumbersfory andsolvingforx .
●
Identifyandanalyzethefunction(i.e.ifit’saparabola,findtheinterceptsandhowit opens),thensketchthefunction.
4.1.4.b
AreaBetweentheGraphofx=f(y)andthey-Axis
Wecanfindtheareabetweenthegraphofx = f (y) andthey -axisbetweenastartingy valueofy = a andanendingy valueofy = b . Wejustuseintegrationabefore,except everythingissidewaysbecausex andy areswitched. Thismeansthat,sincewearedoing y -axisintegration,d y isthedifferential.
Section4-CalculusII-M athQ RH
255
Justaswedidwithregularintegration,wecanmakerectanglestoshowtheareatotheleft of(insteadofunder)thecurve. Theintegralwillbe y = b
∫
f (y) dy
y = a
Now,theareatotherightofthey -axisiscountedaspositiveandtheareatotheleftof they -axisiscountedasnegative. 4.1.4.c
FindingtheAreaBetweenTwoCurves
Youcanfindtheareabetweentwocurveswrittenintheformx = f (y) andx = g (y) . Instead offindingtheintegraloftopminusbottom,youcalculate y = b
∫
(right − lef t) dy
y = a
4.1.5
V olume
Volumeisathreedimensionalamountofspacethatsomethingtakesup. Acrosssectionistheresultingobjectwhenyoucut(orslice)asolidobjectwithaplane region. 4.1.5.a
MathematicalDefinitionofVolume
LetS beasolidshapewhichliesonthex -axisbetweena andb . Ingeneral,a cross-section(perpendiculartothex -axis)willhaveanareawhichdependsonthex - value:A(x) . Thinkofthisashavingatinythicknessd x .
256
Section4-CalculusII-M athQ RH
AsyoucutS intothinnerandthinnerslices(numberofslicesapproachesinfinityandd x approaches0 )andaddupallthese“littlevolumes”A(x) dx forallthethinslices,yougetthe volumeofS .
b
n → ∞
a
V = lim ∑ A(x) dx = ∫ A(x) dx
4.1.5.b
Example
ThebaseofasolidshapeS consistsoftheregionbetweenthegraphofy = 1 andthex axisforx between0 and2 . Crosssectionsperpendiculartothex axisarerectangleswith heightequaltohalfofthecorrespondingx value. DrawS andfindthevolumeofS . Ithelpstotiltthey axiswhendrawingS sothatyoucanshowheight.
Section4-CalculusII-M athQ RH
257
2
Forthisproblem,V = ∫ A(x) dx . Now,weneedtofindA(x) . Todothis,pickanx valueand 0
drawinarectangle(thisisthecrosssectioninthepictureabove). Weknowfromthegiven informationthath = 21 x .
A(x) isequaltotheareaofthiscrosssectionwhichisA(x) = ( 21 x)(1) = 2x . 2
So,thevolumeofthissolidS isV = ∫ ( 2x ) dx = 1 cubic unit . 0
4.1.6
V olumesofRevolution
Asolidofrevolutionoccurswhenyourevolveaplaneregion(likeagraphofafunction) aroundahorizontalorverticallinethatdoesnotpassthroughtheplane. Therearethree typesofvolumesofrevolution:disks,washers,andcylindricalshells. Disksandwashers areverysimilar,sotheyareoftengroupedtogether. Thereareafewthingstokeepinmind: ●
Alwaysdrawasamplerectangle. Spinyourfinger(asifitweretherectangle)along therevolutiontodetermineifyoushouldusethedisk/washermethodorthe cylindricalshellsmethod. ○
Ifyourfingerchangesdirectionasitspins,usethedisk/washermethod.
○
Ifyourfingerstaysuprightandonlyyourhandmoves,usethecylindrical shellsmethod.
258
Section4-CalculusII-M athQ RH
●
Makesureyourfunctionsareintermsofthevariablethatcorrespondstothe rectanglesthickness. ○
i.e.iftherectanglehasd x thickness,thefunctionsshouldbeintermsofx , andviceversa.
●
Sometimesbothtechniquesarepossibletouse. Choosetheeasieroption:theone thatdoesn’tmakeyousplituptheintegralintoseveralintegrals,andwherelimitsof integrationareeasytofind.
4.1.6.a
Disk
Weusethediskmethodtofindthevolumeofaregionwhenwerotatearegionbounded byacurveandanaxisorlinearoundanaxis. Forexample,ifwerotatetheregionboundedbyy = √x andthex -axisfromx = 0 tox = 9 aroundthex axis,wegetthefollowing.
b
Tofindthevolumeofthissolid,weusetheformulaV = ∫ π R2 dx whereR isthedistance a
fromthefunctionandtheaxisofrotation. So,thevolumeis 9
V = ∫ π (√x )2 dx = 0
Section4-CalculusII-M athQ RH
81π 2
259
4.1.6.b
Washer
Thewashermethodisverysimilartothediskmethod,butthereisaholeinthemiddle. Forexample,ifwerotatetheregionboundedbyy = 1 andy = x 2 aroundtheliney = 2 ,we getthefollowing.
b
Tofindtheareaofthissolid,usetheformulaV = ∫ (πR2 − π r 2 ) dx . R isthedistancefrom a
the“spinningline”(inthiscasey = 2 )tothefurthestfunction(y = x 2 forthisexample). r is thedistancefromthespinninglinetothecloserfunction. Alwaysusetop − b ottom or r ight − lef t whenusingthismethod. Thevolumeofthissolidis 1
V =
∫ (π(2 − x2 )2 − π (1)2 ) dx =
−1
260
56π 15
c ubic units
Section4-CalculusII-M athQ RH
4.1.6.c
CylindricalShells
Findthevolumeofthesolidobtainedbyrotating,aboutthey axis,theregioninthefirst quadrantboundedbyy = 2 x 2 − x 3 andthex axis.
b
Tofindthevolumeofthissolid,usetheformulaV = ∫ 2 πxf (x) dx . So, a
2
V = ∫ 2 π (x) (2x 2 − x 3 ) dx = 0
16π 5
c ubic units
4.1.7
W ork
Forceisthe“pull”onanobject,wheref orce (F ) = mass · a cceleration . Massmeasuresthe amountofmattersomethingcontains,whichisdifferentfromweight. Workisequaltoforcetimesdistance,whentheforceisconstant.
Section4-CalculusII-M athQ RH
261
4.1.7.a
UnitsforWorkWordProblems
SIMetricSystem
USSystem
Mass
kg
-----
Acceleration
m/s2
-----
Force
“Newton”N = k g · m/s2
pound(lb )
Distance
meters(m )
feet(f t )
Work
“Joule”J = N · m
“footpound”(f t − lb)
4.1.7.b
ChangingForce
Variableforceisatermforchangingforce. Sincetheforcechangesovertime(depending onthedistancex moved),wewillwriteF (x) insteadofjustF . Theworkoverasmall distancewouldbeF (x) dx . Asyoucontinuetomove,theforcecontinuestochange, meaningtheworkisalsochangingovertime. Addingupallofthesepiecesof
F orce · D istance youendupwiththetotalwork:W = ∑ F (x) dx . Asyouconsidersmallerand
smallerdistances(d x → 0 ),theforcewillnotchangemuchoverthattinydistance. This gives: b
W = ∫ F (x) dx a
foravariableforceF (x) .
262
Section4-CalculusII-M athQ RH
4.1.7.c
Hooke’sLaw
Theforcerequiredtomaintainaspringstretchedx unitsbeyonditsnaturallengthis proportionaltox : F (x) = k x aslongasx isn’tlargeenoughtobreakthespring. k iscalledthespringconstant. The equationstatesthatthelargerx is,theharderitistoholdinplace. Tofindtheworkdone, findF (x) thenintegrate. 4.1.7.d
CableProblems
Ifthereisalongropehangingoffabuildingandyouwanttopullitup,itwillstartoff feelingveryheavybecausethereisalotofweighthangingoffthebuilding. Butasyoulift moreandmoreoftheropeontotherooftop,therewillbelessweighthangingover, meaningitwillfeellighter. Thisisanotherexampleofvariableforce. Fortheseexamples,letx representthedistanceyouhaveliftedtherope. Therefore, F (x) = (initial weight) − (weight lost af ter lif ting · f eet) 4.1.7.e
TankProblems
Animportantengineeringtaskistofindouttheminimumamountofworkthatisrequired topumpaliquidoutofatank. Themostefficientwaytodothisisactuallyfromthetop,as opposedtothebottom,asthetoplayerofliquidonlyhastobemovedasmalldistance. Thepumpthendescendsastheleveloftheliquiddrops.
Section4-CalculusII-M athQ RH
263
Fortheseexamples, letx representthedistancefromthebottomofthetank. Procedure: 1. FindtheareaA(x) ofahorizontalcrosssectionofthetankx unitsup. Thevolume ofthislittlelayerwillbeA(x) dx ,whered x representsthetinythicknessofthelayer. 2. Fromthevolume,determinetheforce(weight)ofthelayerandfindthedistance thatthisparticularlayerneedstomove. 3. MultiplyF orce · D istance togettheworktomovethislayer. 4. Integratetofindthetotalworkdoneinmovingallthelayersofliquid.
4.2
TechniquesofIntegration
4.2.1
I ntegrationb yP arts
Integrationbypartsisthereverseoftheproductrule. Rememberthattheproductrule statesthat (u · v )′ = u ′ · v + u · v ′ FromthiswederivetheIntegrationbyPartsformula:
∫ u · dv = u · v − ∫ v · du
Youcanrememberthisbyusing“ultraviolet(u v )voodoo(v · d u )” Wewillusethiswhenwehavetointegratetwounrelatedfunctionsthataremultiplied together. Onefunctioniscalled‘u ’andtheotheriscalled‘d v ’.
264
Section4-CalculusII-M athQ RH
4.2.1.a
Example
Find∫ x sin (x) dx .
Wehavetwounrelatedfunctions(apolynomialandatrigfunction)multipliedtogether. We assignu = x andd v = sin (x) dx . Tofindd u deriveu andtaketheantiderivativeof(integrate) d v . u = x v = − c os x d u = d x dv = sin (x) dx ThenwefillintheIBPtemplateandevaluateasimpleintegral.
∫ x sin (x) dx = (x)(− cos x) − ∫(− cos x)(dx)
= − x cos (x) + ∫ c os (x) dx
= − x cos (x) + sin (x) + C
Section4-CalculusII-M athQ RH
265
4.2.1.b
HowtoChooseu
Chooseu tobethefunctionwhosederivativegetseasier,butmakesurethatd v is somethingwhoseantiderivativeissomethingyouknow(youcanfindalistofcommon antiderivativesinthebeginningofthecalculusIsection;itishelpfultomemorizethese). Thereisanacronymtohelpyoupickthecorrectu . Startatthetopandworkyourway down,thefirstoneyouencountershouldbeyouru . L- Logarithmicfunctions I- Inversetrigfunctions P- Polynomials E-Exponentialfunctions T- Trigfunctions Therearesomeoccasionswherethisacronymwillnotwork,butthesearerare.
4.2.2
T rigonometricI ntegrals
4.2.2.a
ImportantIdentities
PythagoreanIdentities: c os2 (x) + sin 2 (x) = 1 tan 2 (x) + 1 = sec 2 (x) HalfAngleIdentities: sin 2 (x) = 21 (1 − c os (2x))
c os2 (x) = 21 (1 + c os (2x))
DoubleAngleIdentity: sin (x) cos (x) = 21 sin (2x)
266
Section4-CalculusII-M athQ RH
4.2.2.b
Strategies/Guidelines
Strategyforevaluatingintegralsoftheform
∫ sinm (x) cosn (x) dx
●
Ifthepoweroncosineiso dd,usec os2 (x) = 1 − sin 2 (x) toreplaceallbutoneofthethe c os (x) . Groupthislonec os (x) withd x andperformthesubstitution u = sin (x) du = c os (x) dx
●
Ifthepoweronsineiso dd,usesin 2 (x) = 1 − c os2 (x) toreplaceallbutoneofthethe sin (x) . Groupthislonesin (x) withd x andperformthesubstitution u = c os (x) du = − sin (x) dx
●
Ifbothpowersaree ven,usethehalfangleidentitiesrepeatedlyuntilyouareleft withc os (kx) termsonly
ImportanttoMemorize
∫ cos (kx) dx → u = kx, dx = 1k du → 1k ∫ cos (u) du = 1k sin (kx) + C
∫ sin (kx) dx = − 1k cos (kx) + C
Section4-CalculusII-M athQ RH
267
Strategyforevaluatingintegralsoftheform
∫ tanm (x) secn (x) dx
●
Ifthepoweronsecantise ven,groupsec 2 (x) withthed x . Thenuse sec 2 (x) = tan 2 (x) + 1 toreplacealloftheremainingsec (x) terms. Performthe substitution u = tan (x) du = sec 2 (x) dx
●
Ifthepowerontangentiso dd,groupsec (x) tan (x) withthed x . Use tan 2 (x) = sec 2 (x) − 1 toreplacealloftheremainingtan (x) terms. Performthe substitution u = sec (x) du = sec (x) tan (x) dx
●
Forothercases,youwillneedtobecreativeanduseothertools. Somesuggestions are ○
rewritingeverythingintermsofsineandcosine
○
usingatrigidentityandsplittinguptheintegral(seetheprecalculus summarysheet)
○
usingintegrationbyparts
○
trialanderror
268
Section4-CalculusII-M athQ RH
Strategyforsolvingothertrigintegralswherem =/ n
●
∫ sin (mx) cos (nx) dx
Usesin A cos B = 21 [sin (A − B ) + sin (A + B )]
○
●
∫ sin (mx) sin (nx) dx
Usesin A sin B = 21 [cos (A − B ) − c os (A + B )]
○
●
∫ cos (mx) cos (nx) dx
Usec os A cos B = 21 [cos (A − B ) + c os(A + B )]
○
4.2.3
T rigonometricS ubstitution
Trigonometricsubstitutionisusedtosolveintegralsthatcannotbesolvedusingu substitutionorIBP. Weusetrigonometricfunctionstosimplifytheintegral. a issomenumber Expression
Substitution
ValidInterval
Identity
√a 2 − x2
x = a sin θ dx = a cos θ dθ
−
π 2
≤θ≤
π 2
1 − sin 2 θ = c os2 θ
√a 2 + x2
x = a tan θ dx = a sec 2 θ dθ
−
π 2
≤θ≤
π 2
1 + tan 2 θ = sec 2 θ
√x2 − a 2
π 2
orπ ≤ θ ≤
x = a sec θ dx = a sec θ tan θ dθ 0 ≤ θ ≤
3π 2
sec 2 θ − 1 = tan 2 θ
Youwillcommonlyseethesewithasquareroot,butitdoesn’thavetobethere.
Section4-CalculusII-M athQ RH
269
4.2.3.a
Example
Solvetheintegral∫ √4 − x 2 dx .
u substitutionisnotpossibleforthisintegral. Noticehowthislooksliketheexpression
√a 2 − x2
inthetableabove,wherea = 2 .
Applytheassociatedsubstitution. x = 2 sin θ dx = 2 cos θ dθ Substitutethisintheoriginalintegral.
∫√
4 − (2sin θ)2 · 2 cos θ dθ = ∫ √4 − 4 sin 2 θ · 2 cos θ dθ
Factoroutthe4 andusethetrigidentityc os2 θ = 1 − sin 2 θ tosimplifythesquareroot.
∫ √4(1 − sin 2 θ) · 2 cos θ dθ = ∫ √4cos2 θ · 2 cos θ dθ
∫ 2 cos θ · 2 cos θ dθ = 4 ∫ cos2 θ dθ
Thisisnowatrigintegralwhichweknowhowtoevaluateusingthetrigidentity c os2 θ = 21 (1 + c os (2θ)) Substitutethisbackintotheintegralandsimplify.
4 ∫ c os2 θ dθ = 2 ∫[1 + c os (2θ)] dθ = 2 (θ + 21 sin (2θ)) + C
270
Section4-CalculusII-M athQ RH
Therecannotbeadoubleangleinsideofatrigfunction,asitwon’tworkwhenweputthe finalanswerbackintermsofx . Therefore,weusetheidentitysin (2θ) = 2 sin θ cos θ . 2 [θ + sin θ cos θ] + C = 2 θ + 2 sin θ cos θ + C Nowwemustputthisintermsofx . Todothis,recallthatx = 2 sin θ (thisisthesubstitution weperformedatthebeginningoftheproblem). Thisisthesameassin θ = 2x . Drawthe trianglethiscreatestogetridofthetrigfunctions.
∫ √4 − x2 dx = 2 (sin −1 ( 2x )) + 2 ( 2x )( √4 − x 2
2
)+C
Nowtheintegrationiscomplete. Tips: ●
alwaysdrawatrianglewhenyouusetrigsubstitution. Ithelpstovisualizethe problemandisawaytocheckyourwork(theexpressiononthethirdsideofthe trianglewillappearinyourintegral).
●
alwayscheckifu subispossiblebeforebeginningtheprocessoftrigsub,asittakes alongtimetocomplete
Section4-CalculusII-M athQ RH
271
4.2.3.b
DefiniteIntegrals
Whenusingtrigsubstitutionfordefiniteintegrals,asyouwouldwithothersubstitutions, wemustchangetheboundstomatchthevariable. Whenwedothis,wedon’tneedto returntox forthefinalanswer.
4.2.4
I ntegrationb yP artialF ractions
Ifyouforgethowtodopartialfractiondecomposition(orjustneedarefresher),gotothe PartialFractionDecompositionsectionofPrecalculus. Ifanintegralhasacomplicatedrationalfunction,wecanusepartialfractionstohelp evaluatetheintegral. ●
Ifthedegreeofthenumeratorisgreaterthanorequaltothedegreeofthe denominator,youmustperformlongdivision(seeAlgebraIIsection)beforemoving ontothenextstep.
●
Factorthedenominator.
●
Performapartialfractiondecomposition.
●
Evaluatetheintegralsofeachpartialfractiontogettheintegraloftheentire function.
272
Section4-CalculusII-M athQ RH
IntegralsYouMustKnow
∫
1 x dx
= ln |x| + C
1 dx = ∫ mx + b
1 m
· ln |mx + b| + C
1 dx = tan−1 (x) + C ∫ x + 1
1 dx = a1 · tan−1 ( ax ) + C ∫ x + a
2
2
2
LogPropertiestoKnow n · ln |ax + b| = ln |(ax + b )n | ax + b | ln |ax + b | − ln |cx + d | = ln || cx + d |
ln |ax + b| + ln |cx + d| = ln |(ax + b )(cx + d )| 4.2.4.a
Example
1 Integrate∫ (x − 2)(x . 2 + 1) dx
Let’sworkonthealgebraseparatelyfirst. Inthisproblem,wehaveadistinctlinearfactor andanirreduciblequadraticfactor. Therefore, 1 (x − 2)(x2 + 1)
=
A (x − 2)
+
Bx + C (x2 + 1)
Multiplybothsidesby(x − 2 )(x 2 + 1 ) togetridofthefractions. 1 = A(x 2 + 1 ) + (Bx + C )(x − 2 )
Section4-CalculusII-M athQ RH
273
Pluginvaluesforx thatwilleliminateoneortwoofthefactors,allowingyoutosolvefor A, B, andC . x = 2 → 1 = A(2 2 + 1 ) → A = x = 0 → 1 = A − 2 C → 1 =
1 5
1 5
− 2 C → C = − 52
x = 1 → 1 = 2 A + (B + C )(− 1 ) → 1 = 2 ( 51 ) + (B − 52 )(− 1 ) → B = − 51 Nowbreakuptheintegralaccordingtothesefractionsandsolve. 1 5
1 x 1 dx − 51 ∫ x + 1 dx − 52 ∫ x + 1 dx ∫ x − 2 2
2
Useu -subforthemiddleintegral,whereu = x 2 + 1 andd u = 2 xdx . 1 5 ln
|x − 2 | −
1 10 ln
|x 2 + 1 | − 2 tan −1 (x) + C | | 5
| (x − 2)1/5 | = ln | (x2 + 1)1/10 | − 52 tan −1 (x) + C | | 4.2.4.b
Trick1-RationalizingSubstitution
Thistrickiscalledarationalizingsubstitution.
Ifyouhavetheintegral∫ √x + 4 x dx ,setu equaltowhatevertheradicalis. Inthiscase,
u = √x + 4 . Usethisstatement tofindwhatx isequaltointermsofu . u = √x + 4 → u 2 = x + 4 → x = u 2 − 4 Findwhatd x isequaltointermsofu andd u . du =
274
1 dx 2√x + 4
→ dx = 2 √x + 4 du → dx = 2 u du
Section4-CalculusII-M athQ RH
Now,replaceallx andd x termsintheintegralandsimplify.
∫
u u 2 − 4
2
· 2u du = ∫ u2u du 2 − 4
Youcannowperformlongdivisionandpartialfractiondecompositiontosolve. 4.2.4.c
Trick2-CompletingtheSquare
Completingthesquarecanalsohelpyousolveintegrals.
Forexample,ifyouhavetheintegral∫ x4 + xx2 + 10 dx ,completethesquareofthepolynomial
inthedenominator. x 4 + x 2 + 1 0 = x 4 + x 2 + 1 + 9 = (x 2 + 1 )2 + 9 Replacethepolynomialwiththefactoredversion.
∫ (x + 1)x + 9 dx
2
2
Now,useu substitutionwhereu = x 2 + 1 andd u = 2 xdx .
Section4-CalculusII-M athQ RH
275
4.2.5
I mproperI ntegrals
Therearetwotypesofimproperintegrals. Typeoneisunboundedintegrals(atleastone endpointisinfinite)andtypetwoisanintegralofadiscontinuousfunction. Animproperintegralisc onvergentifthelimitexistsbutitisd ivergentisthelimitdoesnot exist(DNE)oris± ∞ . 4.2.5.a
TypeI:UnboundedIntegrals
Itispossibleforafiniteareatoexistbetweenthegraphofafunctionandthex axis,even whentheintegralisunbounded. Sinceintegralsareonlydefinedonboundedintervals,wemustuselimitstoevaluate improperintegrals. TypeIintegralswilllooklike ∞
●
∫ f (x) dx =
●
∫
a
b
−∞
t
t → ∞ a
b
f (x) dx = lim ∫ f (x) dx t → −∞ t
∞
●
∫
−∞
c
f (x) dx = ○
276
lim ∫ f (x) dx
∫
−∞
∞
f (x) dx + ∫ f (x) dx c
findeachoftheseintegralsasabove
Section4-CalculusII-M athQ RH
4.2.5.b
TypeII:DiscontinuousIntegrand
Thesecanbehardtospotattimes,astheywilllooklikeanyotherintegral. Integralsareonlylegaliff (x) iscontinuousontheentireinterval[a, b] . Discontinuities causeproblems. TypeIIintegralswilllooklike b
t
●
Discontinuousatb :∫ f (x) dx = lim − ∫ f (x) dx ,t < b
●
Discontinuousata :∫ f (x) dx = lim + ∫ f (x) dx ,t > a
a
a
b
a
●
t → b
b
t → a
t
b
c
b
a
a
c
Discontinuousatc betweena andb :∫ f (x) dx = ∫ f (x) dx + ∫ f (x) dx ○
findeachoftheseintegralsasabove
4.2.5.c
ImportantRule
Ahelpfulrulestatesthattheintegral ∞
∫
1
1 xP
dx
convergesifp > 1 anddivergesifp ≤ 1 .
Section4-CalculusII-M athQ RH
277
4.2.5.d
ComparisonTestforImproperIntegrals
IfbigB (x) andlittleL(x) arecontinuousfunctionswithB (x) ≥ L(x) ≥ 0 ,then ∞
∞
a
a
●
If∫ B (x) dx converges,then∫ L(x) dx converges.
●
If∫ B (x) dx diverges,then∫ L(x) dx isunknown.
∞
● ●
∞
a
a
∞
∞
a
a
If∫ L(x) dx diverges,then∫ B (x) dx diverges. ∞
∞
a
a
If∫ L(x) dx converges,then∫ B (x) dx isunknown.
Thistechniqueishelpfulwhenyouwanttodetermineifanintegralisconvergentor divergent,butthefunctionisreallydifficult(orevenimpossible)tointegrate.
4.3
SequencesandSeries
4.3.1
S equences
Note:youmayneedtobrushuponL’Hôpital’sRuleforthissection. GototheCalculusI sectiononL’Hôpital’sRuletoseethedefinitionoftheruleandatableofdeterminateand indeterminateforms. Asequenceisaninfinitelistofnumbers a 1 , a 2 , a 3 , ... , a n , ... ∞ orsimply{a n }n = 1 orsometimesjust{a n } . Theterma n iscalledthegeneraltermandit
willgenerallyhaveaformula. Youcangraphsequences,youjustplotthepointsanddon’t connectthedots.
278
Section4-CalculusII-M athQ RH
n Thefirstfourtermsof{ n + 1 }
∞
n = 1
n are21 , 32 , 43 , 54 . Thegeneraltermisn + 1 . Aswecan
seefromthefirstfourterms,thesequenceisapproachingone. Thefirstfourtermsof{
(−1)n (n + 1) ∞ } n = 1 3n
a re− 32 , 93 , −
4 5 27 , 81
. Thissequenceapproaches
zero. Notethat(− 1 )n indicatesanalternatingsequence,whichcausesthetermstoswitch betweenpositiveandnegativeasn increases. 4.3.1.a
RecurrenceRelation
Thereisanotherwaytowritesequencescalleda“recurrencerelation”. Arecurrence relationtellsyouthefirstterm/termsthentellsyouhowtofindthenextonefromthe previousones. OneexampleisthefamousFibonacciSequence{f n } . f 1 = 1 f 2 = 1 f n = f n − 1 + f n − 2 Thisrecurrencerelationtellsyoutoaddtheprevioustwotermstogethertofindthenext term. Thefirsteighttermsare: f 1 = 1 f 2 = 1 f 3 = f 2 + f 1 = 2 f 4 = f 3 + f 2 = 3 f 5 = 5 f 6 = 8 f 7 = 1 3 f 8 = 2 1 4.3.1.b
ArithmeticSequence
Anarithmeticsequenceisatypeofsequencewhereaspecificnumberisaddedtothe termstogetthenextterm. Thisistheequivalentofalinearfunction. Forexample, {2, 4, 6, 8, 10, ...} Noticehowyoucangetthenexttermbyadding2 tothepreviousterm. Inthiscase,2 is thecommondifference.
Section4-CalculusII-M athQ RH
279
Thegeneraltermisequivalenttofindingafunctionwhichhasthesevalues. Sincethistype ofsequenceisequivalenttoalinearfunction,itislikeyouarefindingtheslope. This sequence’sgeneraltermisa n = 2 n . Therecursionrelationisa 1 = 2 , a n = a n −1 + 2 . 4.3.1.c
GeometricSequence
Ageometricsequenceisatypeofsequencewhereaspecificnumberismultipliedtothe termstogetthenextterm. Thenumberthatismultipliediscalledthecommonratio(r ) a
n andcanbefoundusingtheequationr = an − 1 . Thisisthesequenceequivalentofan
exponentialfunction. Forexample, 80 {5, 103 , 209 , 40 27 , 81 , ...}
Therearetwopossibilitiesforthegeneraltermofageometricsequence: a n = a r n (a 0 isthefirstterm)
or
a n = a r n − 1 (a 1 isthefirstterm)
wherea isthefirsttermofthesequence. Forthesequenceabove,thegeneraltermis{5 ·
2 n − 1 ∞ } n = 1 3
. Therecursionrelationis
a 1 = 5 , a n = a n − 1 · 32 . Whenfindingthecommonratioofageometricsequencebesuretocheckatleastthree terms. Checkingonlyoneortwotermscanleadtoanincorrectanswer. Itisimportanttonotethatnotallsequencesareoneofthetwotypesabove. Sometimes youneedtogetcreativetofindthegeneraltermofasequence:itisreallyjustabout findingpatternsinthenumbers. Forexample,findthegeneraltermofthesequence
{ 53 , −
280
4 , 5 , 25 125
−
6 , ...} 625
Section4-CalculusII-M athQ RH
First,noticethatthetermsarealternatingbetweenpositiveandnegative,meaningthat therewillbesomevariationof(− 1 )n inthegeneralterm. Sincethefirsttermispositive andpluggingin1 to(− 1 )n gives− 1 ,weknowthatthealternatingtermwillbe(− 1 )n + 1 . Now,findapatterninthenumerator. Ifn = 1 ,thenumeratorequals3 . Ifn = 2 ,the numeratorequals4 . Therefore,weknowthepatterninthenumeratoris(n + 2 ) . Lastly,findapatterninthedenominator. Thenumbersincreasebyafactoroffive everytimen increasesbyone. So,thepatterninthedenominatoris5 n . Putallofthisinformationtogethertofindthegeneralterm.
{
(−1)n + 1 (n + 2) ∞ } n = 1 5n
4.3.1.d
LimitsofSequences
Asequence{a n } hasthelimitL if lim a n = L . Thisisthesamedefinitionastheonein n → ∞
CalculusI;thetermsgetcloserandclosertothenumberL asn getsclosertoinfinity. In otherwords, lim f (x) = L means lim a n = L .Ifthelimitisafinitenumber,wesaythe x → ∞
n → ∞
sequenceconverges. Ifthelimitisinfiniteordoesnotexist(DNE),wesaythesequence diverges.
Section4-CalculusII-M athQ RH
281
Whenwefoundthegeneraltermsofsequences,wewerefindingafunctionf (x) with f (n) = a n . So,sequencesarereallyjustfunctions. Meaningwecanusethelimitlawsfrom CalculusItofindthelimitsofsequences. WewillmainlyuseL’Hôpital’sRuleandthe SqueezeTheorem(youcanreviewtheseintheCalculusIsection).
f (x) =
2x2 − 8x + 8 an 2x2 − 11x + 16
=
2n 2 − 8n + 8 2n 2 − 11n + 16
Note:WeseefactorialsalotinCalculusII. Afactorialisshownasn ! andyoumultiplyn timesallthenumberslessthanituntilyoureach1 . Forexample,4 ! = 4 · 3 · 2 · 1 = 2 4 . Rememberthat0 ! = 1 . Maybesurprisingly,thisisnotthefastestgrowingfunction:n n is. 4.3.1.e
Vocabulary
●
Increasing:a 1 < a 2 < a 3 < ...
●
Decreasing:a 1 > a 2 > a 3 > ...
●
Monotonic:ifthesequenceiseitherincreasingordecreasing ○
itisnotpossibletobeboth
●
BoundedAbove:ifa n ≤ M forsomenumberM
●
BoundedBelow:ifa n ≥ m forsomenumberm
●
Bounded:ifboundedbothaboveandbelow
282
Section4-CalculusII-M athQ RH
Toprove,forexample,thatasequenceisdecreasing,weneedtoshowthata n ≥ a n + 1 .In otherwords,youneedtoprovethatthecurrenttermislargerthanthenextterm. 3 n + 5
>
3 (n + 1) + 5
Crossmultiplyandsimplify. 3n + 18 > 3n + 15 Thisisatruestatement,thereforethesequenceisdecreasing. MonotonicSequenceTheorem:Asequencewhichisincreasingandboundedaboveis automaticallyconvergent. Similarly,asequencewhichisdecreasingandboundedbelowis automaticallyconvergent. Sometimesitishardtofindthelimitofasequence. Itiseasiertosaywhetherornota sequenceconvergesthantosayexactlywhatthesequenceconvergesto.
4.3.2
S eries
Seriesaresequenceswithadditionsignswherethecommasare. Theindividualpiecesthat areaddedtogetherarecalledterms. Thesequence
{ 31n }∞ n = 1 = 31 , 91 , 271 , 811 , ... inseriesformis ∞
∑ n = 1
Section4-CalculusII-M athQ RH
1 3n
=
1 3
+
1 9
+
1 27
+
1 81
+ ...
283
Toreadthisoutloud,yousay“thesumfromonetoinfinityof 31n ”. Addingnumbersinfinitelysoundsprettystrange,butitissomethingthatwearealready familiarwith. Non-terminatingdecimals,forexample,aretechnicallyinfinitesums. 2 9
= 0.222222222222... = 0.2 + 0 .02 + 0 .002 + 0 .0002 + ...
Thiscanalsobewrittenas 2 9
2 10
=
+
2 100
+
2 1000
∞
+ ... = ∑ n = 1
2 10 n
4.3.2.a
Notation
Therearemanywaystowriteanequivalentseries,suchasadifferentcountingvariable, rewritingalgebraically,orstartingatadifferentnumber. Belowaresomeexamplesofthe sameserieswrittendifferently. ∞
∑ n = 1 ∞
1 3n
=
∑ ( 31 )n =
n = 1 ∞
1 3
1 3
+ +
∑ ( 31 )i = 1 +
1 9
+
1 9
+
1 3
+
1 27
1 27
+
1 9
+
∑ ( 31 )k − 1 = 1 + 31 +
1 9
i = 0 ∞
k = 1
+ ...
1 81
+
1 81
1 27
+
+ ...
+ ... 1 27
+ ...
∞
1 3i
∑ i = 1 ∞
∑ n = 0 ∞
∑ n = 1 ∞
1 3n
1 3
+
=1+
1
3
n − 1
∑ 3· n = 1
=
1 3n
1 9
1 3
+ +
1 27
+
1 81
+ ...
1 9
+
1 27
+ ...
=1+
1 3
+
1 9
+
1 27
+ ...
=1+
1 3
+
1 9
+
1 27
+ ...
284
Section4-CalculusII-M athQ RH
4.3.2.b
EstimatingConvergence
Usingtheseriesabove,wewilltrytofindwhatitconvergesto. Wecan’taddupinfinitely manynumbersonacalculator,butwecanadduppartoftheentiresum(calledapartial sum,denotedsn ). Notethats10 isthefirst1 0 termsaddedtogether,andsoon. n
sn = sumthefirstn terms
1
s1 = 0 .3333333...
2
s2 = 0 .4444444...
3
s3 = 0 .481481...
10
s10 = 0 .499991...
20
s20 = 0 .499999...
1000
s1000 = 0 .499999999999...
Thisisanestimation,butwecanseethatthepartialsumsseemtoconvergeto21 . ∞
Therefore,itseemslike ∑
n = 1
1 3n
=
1 2
(whichistrue).
4.3.2.c
DeterminingDivergence/ConvergencewithPartialSums ∞
Givenaseries ∑ a i = a 1 + a 2 + a 3 + a 4 + ... ,letsn denoteitsn th partialsum: i = 1
n
sn = ∑ a i = a 1 + a 2 + ... + a n i = 1
Thisisalistofnumbers. Ifthesequenceofpartialsums{sn } isconvergentand lim sn = S n → ∞
isarealnumber,thenwesaytheseries∑ a i isconvergentandwewrite
∞
∑ ai = S
i = 1
Section4-CalculusII-M athQ RH
285
ThisnumberS iscalledthesumoftheseries. Ifthesequenceofpartialsums{sn } is divergent,wesaytheseriesisdivergent. Thefollowingsymbolsallhavethesamemeaning ∞
n
∑ a i = lim ∑ a i = lim sn n → ∞ i = 1
i = 1
n → ∞
∞
Onereallycoolfactisthat ∑
n = 1
1 n2
=
π2 6
. T heproofofthisissuperinteresting:Irecommend
lookingintoit! 4.3.2.d
TestforDivergence(nthTermTest)
Convergenceofaseriesisnotcommon. Divergenceofaseries,however,is. Hereisatest toseeifaseriesisdivergent: ∞
If lim a n =/ 0 orD N E ,thentheseries ∑ a n isdivergent. n → ∞
n = 1
Inotherwords,theindividualtermsoftheseriesmustgoto0 fortheseriestohavea chanceatconverging. Itdoesnotsaythatiftheindividualtermsgoto0 thattheseries automaticallyconverges.
IMPORTANTNOTE:Justbecause lim a n = 0 ,doesn’tmean∑ a n converges! n → ∞
∞
Example:Usingthen th termtest,determineif ∑
n = 1
2 lim 2n n → ∞ 5n + 4
n2 5n2 + 4
isdivergentorconvergent.
2n n → ∞ 10n
= (L′H opital′s Rule) = lim
2 n → ∞ 10
= lim
=
1 5
=/ 0
Becausethelimitisnotequaltozero,theseriesdivergesbythen th termtest.
286
Section4-CalculusII-M athQ RH
∞
Example:Usingthen th termtest,determineif ∑
n = 1
1 n
isdivergentorconvergent.
=0
lim 1 n → ∞ n
Becausethelimitequalszero,then th termtestisinconclusive. 4.3.2.e
HarmonicSeries ∞
Theseries ∑
n = 1
1 n
=1+
1 2
∞
∑
n = 1
+
1 3
1 n
=1+
+
1 4
+ ... actuallydivergeseventhoughthetermsgotozero:
1 2
+ ( 31 + 41 ) + ( 51 +
1 6
1 7
+
+ 81 ) + ( 91 + ... +
1 16 )
+ ...
Now,youcansimplifythetermsintheparenthesesbyreplacingthelargestnumberswith thesmallestnumber. Thiswillhelptomakeestimationeasierand,sinceitisan underestimate,itis“legal.” ∞
∑
n = 1
1 n
=1+
1 2
+ ( 41 + 41 ) + ( 81 +
1 8
+
1 8
+ 81 ) + ( 161 + ... +
1 16 )
+ ...
Noticehowallthesumsintheparenthesessimplifyto21 ,so ∞
∑ n = 1 ∞
So, ∑
n = 1
1 n
1 n
=1+
1 2
+
1 2
+
1 2
+
1 2
+ ...
= ∞ butveryslowly. Ittakesaddingover250milliontermstogetasumthatis
greaterthan20.
Section4-CalculusII-M athQ RH
287
4.3.2.f
GeometricSeries
Zeno’sParadox:Supposeyouwanttowalk2miles,butmaybethatseemslikealong distance,soyoudecidetothinkofonlywalkinghalfofthedistance(1mile). Onceyou completethat1mile,theremainingpartseemstoolong,soyoudecidetothinkofonly walkinghalfoftheremainingdistance(½amile). Onceyoucompletethat,theremaining partseemstoolong,soyoudecidetothinkofonlywalkinghalfofthedistance(¼mile). By continuingthisprocessofwalkingonlyhalfoftheremainingdistance,youenduptraveling 1+
1 2
+
1 4
+
1 8
+
1 16
+ ... = 2 miles
Aserieswheren onlylivesintheexponent(likeanexponentialfunction)suchas ∞
∑ n = 0
1 2n
∞
∞
n = 1
n = 1
1 n or ∑ ( 31 )n or ∑ 2 · ( 10 )
iscalledageometricseries. Ageometricseriesisn’trequiredtostartatn = 0 ,butitisgood ifitdoes,asitmakesthingseasier. ∞
∑ a r n = a + a r + a r 2 + a r 3 + a r 4 + ...
n = 0
Thenumbera isthestartingterm,andr istheratiothatwearemultiplyingby. ∞
Ageometricseriesisconvergentonlywhen|r | < 1 . Furthermore, ∑ a r n = n = 0
a 1 − r
.
Thisisatypeofseriesthatwecanactuallyaddup. If|r | ⪰1 , theseriesdiverges.
288
Section4-CalculusII-M athQ RH
∞
Example:Iftheseriesconverges,findthesumof ∑ 2 2n · 3 1 − n . n = 1
First,simplifytheterm. 2 2n · 3 1 − n = (2 2 )n · 3 1 · 3 −n = 4 n · 3 · 3 −n = 3 · ( 34 )n ∞
∞
n = 1
n = 1
Therefore, ∑ 2 2n · 3 1 − n = ∑ 3 · ( 34 )n . Sincer =
4 3
> 1 ,theseriesdiverges.
4.3.2.g
SeriesFacts
Afinitenumberoftermsdoesn’taffectconvergence/divergenceofaseries. Addupa millionnumbers,youwillstillgetafinitenumber. It’stheinfinitepartthatreallymatters. ∞
k
n = 1
n = 1
∑ an = ∑ an +
∞
∑ n = k + 1
an
If∑ a n and∑ b n areconvergentseries,then ∞
∞
n = 1
n = 1
∑ (c · a n ) = c · ∑ a n
and
∞
∞
∞
n = 1
n = 1
n = 1
∑ (a n ± b n ) = ∑ a n ± ∑ b n
Section4-CalculusII-M athQ RH
289
4.3.2.h
TelescopingSeries
Considertheseries ∞
∑ n = 1
1 n(n + 1)
=
1 2
+
1 6
+
1 12
+
1 20
+
1 30
+ ...
Thisisnotageometricseriessowecan’ttelliftheseriesconvergesordiverges. So,let’s performpartialfractiondecompositiononthetermtobreakdownthisproblem. 1 n(n + 1)
=
1 n
−
1 n + 1
Therefore, ∞
∑ n = 1
1 n(n + 1)
∞
= ∑ ( n1 − n = 1
1 ) n + 1
= ( 11 − 21 ) + ( 21 − 31 ) + ( 31 − 41 ) + ( 41 − 51 )
Now,wecanseeapattern. s1 = ( 11 − 21 ) =
1 2
s2 = ( 11 − 21 ) + ( 21 − 31 ) =
2 3
s3 = ( 11 − 21 ) + ( 21 − 31 ) + ( 31 − 41 ) =
3 4
.. . sn = ( 11 − 21 ) + ( 21 − 31 ) + ( 31 − 41 ) + ... + ( n1 −
1 n + 1 )
=
n n + 1
∞
Ifwetakethelimitofsn ,wecanseethat lim sn = 1 . Thismeansthat ∑ n → ∞
n = 1
1 n(n + 1)
= 1 .
Noticethatthisisnotthen th termtest,aswearelookingatthepartialsumsthemselves.
290
Section4-CalculusII-M athQ RH
Therearetwomainthingstolookoutforinatelescopingseries: 1. twofractionssubtractedfromeachother 2. somethingwhereyou’dusepartialfractionstobreakitup
4.3.3
I ntegralT est
Rememberthatwevisualizesequenceswithfunctionswithoutthedotsconnected. With series,wevisualizewiththeareaunderthegraph(i.e.anintegral). ∞
Forexample,taketheseries ∑
n = 1
1 n2
=1+
1 4
+
1 9
+ ... .
Drawrectangleswhoseareasrepresenttheareasofthecorrespondingterms. Theareasof alltheserectanglesaddedupgivesthesumoftheseries. Nowontopofthis,drawthe graphofy =
1 x2
,asthetermsoftheseriesaren12 .
∞
Theintegral∫
1
1 x2 dx isrelatedtotheseriesinthatitfollowsthepointsbutcapturesmore ∞
areathantheseries. Butif∫
1
1 x2 dx converges,thensodoes
∞
∑
n = 1
1 n2
.
Notethatwecandisregardtheonebyonerectanglewhentestingforconvergence,asitis afinitenumberandf inite number + f inite number = f inite number .
Section4-CalculusII-M athQ RH
291
IntegralTest:Iff (x) isacontinuous,positive,anddecreasingfunctionontheinterval[1, ∞] withf (x) = a n (thesametermsastheseries),then ∞
∫ f (x) dx
⇒
converges
1
∞
∫ f (x) dx
⇒
diverges
1
∞
∑ a n converges
n = 1 ∞
∑ a n diverges
n = 1
So,theserieswilldothesameastheintegral. Notesontheintegraltest: ●
Itisnotnecessaryfortheintervaltostartatone,youcanstartatalaternumber.
●
Youhavetocheckthehypothesesfortheintervaltesttobevalid:continuous, positive,anddecreasing. ○
Tocheckifthefunctionisdecreasing,showthatthederivativeofthe functionisnegative(f ′(x) < 0 ) .
○
Also,thisdoesnothavetoapplytotheentireinterval,justovertheinterval oftheseries(somenumbertoinfinity).
●
Theintegraltestdoesnottellyouwhattheseriesconvergesto,onlywhetherornot theseriesisafinitenumber.
●
TheintegraltestisoneofthemoretimeconsumingteststhatweuseinCalcII,so lookforanothertestthatwillworkbeforedecidingtousethisone.
292
Section4-CalculusII-M athQ RH
4.3.3.a
Example ∞
Determinewhether ∑ n = 200
1 √n
convergesordiverges.
First,determineifthefunctionf (x) =
1 √x
iscontinuous,positive,anddecreasingonthe
interval[200, ∞) . Thefunction’sonlyplacewhereitisnotdefinedisx ≤ 0 ,whichisoutside oftheinterval,soitiscontinuous. Thefunctionhastobepositive,asithasasquarerootin it. Thederivativeofthefunctionisf ′(x) = − 21 x −3/2 ,whichislessthanzero,sothefunctionis decreasing. So,wecanusetheintegraltest. Nowwecansetuptheintegralandsolve. ∞
∫
200
1 dx √x
t
= lim
t lim 2√x|200 = lim (2√t − 2√100) = ∞ ∫ x−1/2 dx = t → ∞ t → ∞
t → ∞ 200
∞
Therefore,theseries ∑ n = 200
1 √n
isdivergentbytheintegraltest.
4.3.3.b
p-Series ∞
Theseries ∑
n = 1
1 nP
iscalledap - Series. Itis
●
convergentwhenp > 1
●
divergentwhenp ≤ 1
Section4-CalculusII-M athQ RH
293
4.3.3.c
UsingtheIntegralTesttoEstimatetheSum
IfwefindthataseriesisconvergentbytheIntegralTest,wecanalsouseanintegralto estimatetheactualsumoftheseries. Considertheseries ∞
∑ a n = a 1 + a 2 + a 3 + a 4 + ... + a n + a n + 1 + a n + 2 + ...
n = 1
Youcomputesn = a 1 + a 2 + a 3 + ... + a n andstop. Howcloseareyoutotheactualsumofthe series? Call Rn = a n + 1 + a n + 2 + ... theremainderbecauseitistheremainingpartoftheseriesthatyouhaven’taddedupyet. Thisremaindercanbeestimatedby ∞
∫
∞
n + 1
f (x) dx ≤ Rn ≤ ∫ f (x) dx n
∞
WetypicallyonlyuseRn ≤ ∫ f (x) dx becausewewanttoseewhattheremainderissmaller n
than. ∞
Example:Estimatethesumof ∑
n = 1
1 n3
ifitisconvergent.
Theseriesisconvergentbyp - Serieswithp = 3 beinggreaterthan1 . So,wecanestimate usingtheintegraltest. ∞
∑ n = 1
s10 = 1 .1975 ,Rn =
1 3 11
∞
∫
n
294
+
1 3 12
1 x3 dx
1 n3
=
1 13
+
1 23
+
1 33
+ ... +
1 10 3
+[
1 11 3
∞
+ ... . Now,findtheintegral∫
n
t
∫ x−3 dx = t → ∞
= lim
n
lim −
t → ∞
1 t 2x2 |n
+
1 12 3
1 x3 dx .
= lim [− n → ∞
1 2t2
+ ...]
+
1 2n2 ]
=
1 2n2
Section4-CalculusII-M athQ RH
SonowweknowthatR10 ≤
1 2(10)2
= 0 .005 . Thereforen = 1 1 ton = ∞ is≤ 0 .005 . Nowplugin
n + 1 togetahandleonhowbigthisintegralisgoingtobe. 1 2(11)2
∞
∑ n = 1
1 n3
≤ R10 ≤
1 2(10)2
→ 0.004 ≤ R10 ≤ 0 .005
isbetween1 .2015 and1 .2025 (s10 + 0 .004 ands10 + 0 .005 ).
4.3.4
C omparisonT ests
4.3.4.a
DirectComparisonTest
If∑ a n (given)and∑ b n (similar)areserieswithpositivetermsand
● ●
If∑ b n convergesand0 ≤ a n ≤ b n ,then∑ a n converges(isafinitenumber)
If∑ b n divergesand0 ≤ b n ≤ a n ,then∑ a n diverges(isinfinite)
Note:youwillchoosetheseries∑ b n tobeeitherap - seriesorageometricserieswhich
resemblesthegivenseries∑ a n .
Fordeterminingb n onp - seriescomparisons: ∞
∑ n = 1
largest power in numerator largest power in denominator
Section4-CalculusII-M athQ RH
295
∞
Example:Determineiftheseries ∑
n = 1
1 n 2 + 1
convergesordiverges. ∞
Theseriesisnotap - series,butitlookslikethep - series ∑
n = 1
1 n2
(whichisconvergent ∞
becausep = 2 > 1 ). Allofthetermsarepositive,soifwecanshowthat ∑
n = 1
1
n 2 + 1
∞
≤ ∑ n = 1
1 n2
,
itwillmeantheseriesconverges. Checktheterms: 1 n 2 + 1
≤
1 n2
Crossmultiply. n2 ≤ n2 + 1 Simplify. 0 ≤1 ∞
Thisisatruestatement,soaccordingtothefirstbulletpointoftheDCT, ∑
n = 1
1 n 2 + 1
convergesaswell. Toseewhichseriesislarger:ifthedenominatorsarethesame,theonewiththelarger numeratorisbigger;ifthenumeratorsarethesame,theonewiththesmallerdenominator isbigger. Thoughyouwillstillneedtoperformthealgebratoshowthisistrue,thiscan helpyoucheckyouranswer.
296
Section4-CalculusII-M athQ RH
4.3.4.b
LimitComparisonTest
If∑ a n and∑ b n areserieswithpositivetermsandyouevaluatethelimit an n → ∞ b n
lim
=C
a nd
0 < C < ∞ (strictinequality)
Thenbothseriesdothesamething:theyeitherbothconvergeortheybothdiverge.
Note:asbefore,youwillchoosetheseries∑ b n tobeeitherap - seriesorageometricseries
whichresemblesthegivenseries∑ a n .
∞
Example:UsetheLCTtodeterminewhether ∑
n = 1
∞
n = 1
∑ bn = ∑
1 2n
1 2 n − 1
convergesordiverges.
,whichisaconvergentgeometricserieswithr = 21 . Now,setupthelimit.
lim
n → ∞
∞
Since0 < 1 < ∞ , ∑
n = 1
1 2 n − 1 1 2n
1 2 n − 1
2n n → ∞ 2 − 1
= lim
n
ln(2) 2 n n n → ∞ ln(2) 2
= (L′H opital′s Rule) = lim
=1
convergesbytheLimitComparisonTest.
Note:BoththeDCTandtheLCTdealwithseriesthathaveallpositiveterms,buttheyalso workforseriesthathaveallnegativetermsbecause− 1 isacommonmultiple. So,
∑ − a n = − ∑ a n .
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297
4.3.5
A lternatingS eriesT est
Analternatingseries∑ (− 1 )n b n convergesifboth:
3. Thetermsgoto0 : lim b n = 0 n → ∞
4. Thetermsb n decrease:b 1 ≥ b 2 ≥ b 3 ≥ ... (checkbyshowingthederivativeisnegative) Whentestingbothoftheaboveconditions,donotincludethealternatingterms. Ifthefirst conditiondoesnothappen,thentheseriesdivergesbythen th termtest. ∞
Example:Provethattheseries ∑
n = 1
First,checkif lim
1 n → ∞ n
(−1)n−1 n
convergesbytheAlternatingSeriesTest.
= 0 ,whichitdoes,aswhenyoudividebyalargerandlargernumber
youapproach0 . Now,checkifthetermsn1 decrease. f ′(x) = −
1 x2
Thisderivativeisalwaysnegative,sothetermsdodecrease. ∞
Therefore,theseries ∑
n = 1
(−1)n−1 n
convergesbytheAlternatingSeriesTest.
298
Section4-CalculusII-M athQ RH
4.3.5.a
EstimatetheSumofanAlternatingSeries
IfwefindthataseriesisconvergentbytheAlternatingSeriesTest,wecanalsoestimatethe actualsumoftheseries. Consideraseries ∞
∑ (− 1 )n − 1 b n = b 1 − b 2 + b 3 − b 4 + ... + b n − b n + 1 + b n + 2 − ...
n = 1
Youcomputesn = b 1 − b 2 + b 3 − b 4 + ... + b n . Ifyouwanttofindouthowcloseyouaretothe actualsumoftheseries,letRn denotetheremainderbecauseit’stheremainingpartofthe seriesthatyouhaven’taddedupyet. Thisremaindercanbeestimatedby |Rn | ≤ b n + 1
4.3.6
R atioa ndR ootT ests
4.3.6.a
Conditionally/AbsolutelyConvergent
Withalternatingseries,addingatermthensubtractingatermmakestheseriesmorelikely toconverge,sinceitdoesn’tgrowasquickly. Weknowthat ∞
∑ n = 1
(−1)n−1 n
∞
converges b ut ∑
n = 1
1 n
diverges
Thismeansthattheseriesisconditionallyconvergent:whenthealternatingsymbolis removed,theserieschangesfromconvergenttodivergent, Inanothercase ∞
∑ n = 1
(−1)n − 1 n2
∞
converges a nd ∑
n = 1
Thismeansthattheseriesisabsolutelyconvergent.
Section4-CalculusII-M athQ RH
1 n2
converges
299
AbsoluteConvergence:Aseries∑ a n is“absolutelyconvergent”if∑ |a n | converges. Itturns
outthatif∑ |a n | (theoriginalseries)converges,then∑ a n automaticallyconverges(evenif itstermsarenegativeoralternating).
ConditionalConvergence:Aseries(typicallyalternating)∑ a n is“conditionallyconvergent”if
∑ a n convergesb ut∑ |a n | diverges. 4.3.6.b
RatioTest
a | Considertheseries∑ a n andfindthelimit lim || n + 1 a n | = r . If n → ∞
●
r < 1 theseries∑ a n convergesabsolutely
●
r > 1 theseries∑ a n diverges
●
r = 1 testisinconclusive,tryadifferenttest
Theratiotestworksbestwhenfactorialsshowup,orwhenyouhaveam ixofpolynomials, exponentials,and/orfactorials(don’tuseitwhenthereisonlyoneofthese!). Youcanusethefollowinglistorderedfromslowesttofastestgrowthtopredictoutcome: polynomial→ exponential → factorial→ n n ∞
Example:Usetheratioteston ∑
n = 1
| lim | n → ∞ |
300
2 n + 1 (n + 1)! 2n n!
2n n!
todetermineconvergenceordivergence.
| | (2n + 1 )n! | 2 | | = lim | 2n (n + 1)! | = lim || n + 1 | =0 n → ∞ n → ∞ | | |
Section4-CalculusII-M athQ RH
Notethatn ! = n · (n − 1 ) · (n − 2 ) · ... · 1 and(n + 1 )! = (n + 1 ) · n · (n − 1 ) · ... · 1 . Whenn ! isput 1 over(n + 1 )! ,everythingcancelsoutexceptforn + 1 ,leaving n + 1 . ∞
r < 1 therefore ∑
n = 1
2n n!
isabsolutelyconvergentbytheratiotest.
Note:correspondingpieceswillalwaysbeonoppositesidesofthefraction;groupthem together. 4.3.6.c
RootTest
Considertheseries∑ a n andfindthelimit lim
n → ∞
●
n
√|an | = r . If
r < 1 theseries∑ a n convergesabsolutely
●
r > 1 theseries∑ a n diverges
●
r = 1 testinconclusive,tryadifferenttest
Theroottestisonlyusefulwhenthetermsoftheseriesareallraisedtothepowern . ∞
3n + 1 n Example:Determineiftheseries ∑ ( 4n + 17 ) convergesordiverges. n = 0
Setupthelimitandsolve. lim
n → ∞
3 4
√( n
3n + 1 n 4n + 17 )
∞
= lim
3n + 1
n → ∞ 4n + 17
= (L′H opital′s Rule) =
3n + 1 n < 1 ,so ∑ ( 4n + 17 ) isabsolutelyconvergentbytheroottest. n = 0
Section4-CalculusII-M athQ RH
3 4
301
4.4
PowerSeries
Recallthatageometricseriesisconvergentif|r | < 1 andthatwecanactuallyfinditssum. ∞
∑ a r n = a + a r + a r 2 + a r 3 + ... = n = 0
a 1 − r
Noticethatthesumstartswithn = 0 . If|r | ≥ 1 ,thenthegeometricseriesdiverges,sowe cannotusethisformulabecausetheseriesdoesn’tconvergetoanumber. Someexamples: ∞
∑ ( 21 )n = 1 +
1 2
+
1 4
+
1 8
+ ... =
∑ (− 31 )n = 1 −
1 3
+
1 9
−
1 27
n = 0 ∞
n = 0
1 1 − 12
+ ... =
= 2
1 1 − (− 13 )
=
3 4
Ifwereplacethenumberswithavariablex ,wehavethefollowing(rememberx 0 = 1 ) ∞
∑ x n = 1 + x + x 2 + x 3 + ...
n = 0
Thisisafunction,sowecanwriteitusingfunctionnotation. Ifwewrite f (x) = 1 + x + x 2 + x 3 + ... thenweseethatf (0) = 1 andfromtheabovecomputationsf ( 21 ) = 2 andf (− 31 ) =
3 4
. Sinceitisstillageometricseries,youcanonlypluginvalues|x | < 1 :thisis
thedomainofthefunction. Apowerseriesisaseriesoftheform ∞
∑ c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + ...
n = 0
wherex isthevariableandc n ’sarecalledthecoefficients. Thisisaninfinitepolynomial. Whenyoupluginx = 0 ,theseriesconvergestoc 0 . Thismeansthepowerseriesis “centeredatx = 0 .” 302
Section4-CalculusII-M athQ RH
Apowerseriescenteredatx = a isaseriesoftheform ∞
∑ c n (x − a )n = c 0 + c 1 (x − a ) + c 2 (x − a )2 + c 3 (x − a )3 + ... n = 0
wherea issomenumber. Whenyoupluginx = a ,theseriesconvergestoc 0 . Thecenter isthetermsthatcanmakeallbutonetermcancelout(c 0 ). Allthenumbersyoucanpluginforx thatmakesthepowerseriesconvergeiscalledthe intervalofconvergence,anditiswritteninintervalnotation.
4.4.1
T heorem ∞
Foragivenpowerseries ∑ c n (x − a )n ,thereareonly3possibilities: n = 0
●
Theseriesconvergesonlywhenx = a (convergestoc 0 )
●
Theseriesconvergesforallnumbersx ,(− ∞ , ∞)
●
ThereisanumberR whereseriesconvergesif|x − a| < R ,butdivergesif|x − a| > R . ThisnumberR iscalledtheRadiusofConvergence(ROC). ○
Youmustcheckforconvergenceattheendpointsofthisintervalto determineiftheendpointsareopenorclosed. Thecorrespondingintervalis calledtheIntervalofConvergence(IOC).
Wewillfindtheintervalofconvergenceofapowerseriesbyrunningtheratiotestand findingwhentheresultislessthanone. Onceyouhavethisinterval,checktheendpoints.
Section4-CalculusII-M athQ RH
303
4.4.1.a
Example ∞
FindtheROCandIOCofthepowerseries ∑
n = 1
(x − 3)n n
.
Noticethecenterisx = 3 . Inorderfortheseriestoconverge,theratiotestmustgivea resultthatislessthanone. n + 1
| (x − 3) | | (x − 3)n + 1 · n | n + 1 n | lim | (x − 3) n | = lim | | = lim ||(x − 3 ) · (n + 1) n | = |x − 3 | (x − 3) · (n + 1) n → ∞ | n → ∞ n → ∞ | | n | Setuptheinterval. |x − 3 | < 1
⇒ − 1 < x − 3 < 1 ⇒ 2 < x < 4
Thismeansthatanynumberbetween2 and4 willmaketheseriesconverge. Wealso knowthatanyx smallerthan2 andlargerthan4 willmaketheseriesdiverge. Now,we needtochecktheendpoints2 and4 . Pluginx = 2 totheoriginalseries. ∞
(2−3)n n
∑ n = 1
lim 1 n → ∞ n
∞
= ∑ n = 1
(−1)n n
= 0 f ′(x) = −
1 n2
Therefore,theseriesconvergesbythealternatingseriestest. Pluginx = 4 totheoriginalseries. ∞
∑ n = 1
(4 − 3) n
n
∞
= ∑ n = 1
1 n
Thisistheharmonicseries. So,itdivergesbyp - serieswithp = 1 . Therefore,theintervalofconvergenceis[2, 4) andtheradiusofconvergenceis1 (itishow faryoucangoleftorrightofthecenterandstillconverge).
304
Section4-CalculusII-M athQ RH
4.4.2
R epresentationo fF unctionsb yP owerS eries
a Wecanusethesumformulaforageometricseries 1 − r tofindthesumofthepower
series. ∞
∑ x n = 1 + x + x 2 + x 3 + ... = n = 0
1 1 − x
YoumayrecognizethisfunctionfromPre-Calculus. Itisthesameasthefunctiony =
1 1 − x
,
butonlywhen|x | < 1 . TheyareonthesameIOCofthepowerseries(− 1 , 1) . a Ifwecanturnafunctionintotheform 1 − r ,thenwecanwriteitasageometricseries ∞
∑ a r n whichconvergeswhen|r | ≥ 1 .
n = 0
4.4.2.a
Example
Writethefunctiony =
x3 x + 2
asapowerseriesandfindtheIOC.
a First,getthefunctionintheform 1 − r . x3 2 + x
So,a =
x3 2
andr = −
x 2
=
x3 /2 1 − (−x/2)
. Now,usethisinformationtoputthefunctionintopowerseries
form. ∞
∑ n = 0
When||− 2x || < 1
x3 2
∞
· (− 2x )n = ∑ (− 1)n n = 0
⇒ − 2 < x < 2, theserieswillconverge.
Section4-CalculusII-M athQ RH
xn + 3 2n + 1
305
a Acommonmistakeisthatwhenmanipulatingarationalfunctiontogetitintheform 1 − r ,
somepeoplewilldividebyx . Thiswillputpowersofx inthedenominator,whichwillnot createaninfinitepolynomial,meaningitwillnotcreateapowerseries. a canhaveanx in it,butthereshouldneverbeanx inthedenominatorofr . 4.4.2.b
IntegralsandDerivatives
Becausepowerseriesarejustfunctions,youcantakederivativesandintegralsofthemto comeupwithdifferentfunctions. Original:
Derivative:
Integral:
f (x) = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + ...
f ′(x) = c 1 + 2 c 2 x + 3 c 3 x 2 + ...
2
3
∫ f (x) dx = C + c 0 x + c 1 x2 + c 2 x3 + ...
∞
= ∑ c n x n n = 0
∞
= ∑ n · c n x n − 1 n = 0
∞
xn + 1 = C + ∑ c n · n + 1 n = 0
Treattheseriesjustasfunctions;usethepowerruleandinversepowerrule. TheROCwillbethesameastheoriginal,butyoumustchecktheendpointsoftheIOC. Notethatyoucanalsostarttheseriesatn = 1 anditwillbethesame.
306
Section4-CalculusII-M athQ RH
Example:Findapowerseriesrepresentationforthefunctionf (x) =
1 (1 − x)2
.
Alloftheseproblemsbeginwiththebasicpowerseries: 1 1 − x
∞
= ∑ x n ontheinterval− 1 < x < 1 ,meaningROC = 1 n = 0
Now,determinehowthegivenfunctionisrelatedtothebasicpowerseries. d ( 1 ) dx 1 − x
Thismeansthat,tofindthepowerseriesfor
=
1 (1 − x)2
1 (1 − x)2
,weneedtotakethederivativeofthe
1 powerseriesfor 1 − x .
1 (1 − x)2
∞
= ∑ n x n − 1 n = 0
TheROCandIOCremainthesame,asbothoftheendpointsgivedivergentseriesbythe n th termtest.
4.4.3
T aylorS eries
TaylorpolynomialsarepolynomialsthatapproximatefunctionsandaredenotedT n (x) , wheren isthedegreeofthepolynomial. Youcanimproveapproximationswithfinding higherdegreepolynomials. Addingonnewtermsdoesn’tmessupwhatoldtermsshould be. TofindtheTaylorpolynomialofdegreen forthefunctionf (x) atx = a : T n (x) = f (a) +
f ′(a) 1!
(x − a )1 +
f ′′(a) 2!
(x − a )2 +
f (3) (a) 3!
(x − a )3 + ... +
f (n) (a) n!
(x − a )n
Rememberthatf (3)(a) meansthethirdorderderivativeata ,f (4)(a) meansthefourth derivativeata ,andsoon.
Section4-CalculusII-M athQ RH
307
ATaylorSeriesofthefunctionf (x) centeredatx = a istheinfiniteseries ∞
∑ n = 0
f (n) (a) n!
(x − a )n = f (a) +
f ′(a) 1!
(x − a )1 +
f ′′(a) 2!
(x − a )2 +
f (3) (a) 3!
(x − a )3 + ...
Whentheseriesiscenteredatx = 0 (meaningthenumbera iszero),itiscalledaMaclaurin series. Taylorseriesbuildapowerseriesforanyfunction. Whenyoubuildaseriesforthese functions,theyareequaltoeachotherontheIOCoftheseries. 4.4.3.a
Example
FindtheMaclaurinseriesforf (x) = c os (x) . BeginbybuildingtheTaylorpolynomial. Rememberthat,sinceitisaMaclaurinseries, a = 0 . −cos (0) 2 sin (0) 3 cos (0) 4 −sin (0) 5 −cos (x) 6 sin (0) 7 cos (0) 8 1 T 8 (x) = c os(0) + −sin (0) 1! x + 2! x + 3! x + 4! x + 5! x + 6! x + 7! x + 8! x
T 8 (x) = 1 − 2!1 x 2 + 4!1 x 4 − 6!1 x 6 + 8!1 x 8 Noticehowitisonlyevennumbers,asallofthesin termscancelledout(sin (0) = 0 , c os (0) = 1 ). Now,findthepattern. Wecanseethatitisanalternatingseries,sotheserieswillhavea (− 1 )n terminit. Also,sinceboththefactorialsandtheexponentsarealwayseven,theyall havetobedivisibleby2 . Withallofthisinmind,theMaclaurinseriesforc os (x) is ∞
x cos(x) = ∑ (− 1)n (2n)! 2n
n = 0
Bytheratiotest,theROC = ∞ . 308
Section4-CalculusII-M athQ RH
∞
Note:Oneofmyfavoritemathfactsisthatex = ∑
n = 0
xn n!
. Justlikethefunctione x ,theoriginal
series,derivative,andintegraloftheseriesareallthesame! 4.4.3.b
Taylor’sRemainderTheorem
Iff (x) isafunctionandT n (x) isthecorrespondingTaylorpolynomialofdegreen centered ata ,theremainderis Rn (x) = f (x) − T n (x) If lim Rn (x) = 0 ,thenthatmeansthatasyouaddmoretermstotheTaylorpolynomial,you n → ∞
getclosertotheexactfunctionf (x) . ThismeansthatthefunctionisequaltoitsTaylor series. If||f (n + 1)(x)|| ≤ M onanintervalcenteredatthenumbera ,thentheremainderRn (x) ofthe Taylorseriessatisfiestheinequality |Rn (x)| ≤
M (n + 1)! |x
− a |n + 1
onthatsameinterval.
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309
4.4.3.c
ImportantTaylorSeries Function 1 1 − x
TaylorSeries
ROC
∞
R =1
∑ xn
n = 0 ∞
ln (1 + x )
xn + 1 ∑ (− 1 )n n + 1
R =1
∞
R =1
n = 0
tan −1 (x)
x2n + 1 ∑ (− 1 )n 2n + 1
n = 0
∞
ex
R =∞
n ∑ xn!
n = 0 ∞
c os (x)
R =∞
x2n ∑ (− 1 )n (2n)!
n = 0 ∞
sin (x)
R =∞
2n + 1
x ∑ (− 1 )n (2n + 1)! n = 0
∞
(−1)n π2n + 1 (2n + 1)! n = 0 3
Example:UseTaylorseriestofindtheexactsumof ∑
2n + 1
.
Noticehowthereisa(2n + 1 )! inthedenominatorandπ and3 areraisedto2 n + 1 . This looksliketheTaylorseriesforsin(x) . Let’srearrangetofindoutwhatx isequalto. ∞
(−1)n
∞
2n + 1
(−1)n
∑ (2n + 1)! · π2n + 1 = ∑ (2n + 1)! · ( 3π ) 3 n = 0 n = 0 So,x =
π 3
∞
2n + 1
2n + 1
x . Sincesin (x) = ∑ (− 1 )n (2n + 1)! ,wecanplugthex fromtheseriesintosin (x) to n = 0
findtheexactsumoftheseries. Therefore, ∞
(−1)n π2n + 1 (2n + 1)! n = 0 3
sin ( 3π ) = ∑
310
2n + 1
=
√3 2
Section4-CalculusII-M athQ RH
Note:WhenusingTaylorseriestofindexactsums,youneedtomakesurethattheindices match(i.e.n mustequal0 ). So,ifyouneedtosubtractonefromyourindex(ifn = 1 ),you mustaddonetoallofthen ’sintheseries. 4.4.3.d
OperationswithTaylorSeries
YoucancombineseriesandusescalarswhendealingwithTaylorseries. Sayyouwantto findtheMaclaurinseriesforf (x) = x cos(x) ,insteadofusingtheproductruletofindthe derivativeofthefunctionmultipletimes,justmultiplytheTaylorseriesforc os(x) byx . ThesameappliesforfindingtheMaclaurinseriesforafunctionlikef (x) = e xsin (x) . Multiply theTaylorseriesfore x bytheTaylorseriesofsin (x) .
4.5
DifferentialEquations
Adifferentialequationisanequationcontaininganunknownfunction(typicallyy )andone ormoreofitsderivatives. Differentialequationshave“orders”thatcorrespondtothe highestorderderivative. So,x 2 y ′ + x y = 1 isafirstorderdifferentialandy ′′ = x + y isa secondorderdifferential. Example:dy dx = 2 x . Solvefory . Tosolvefory ,weneedtogetridofthederivativedy dx . Todothis,weintegratebothsides.
∫ dydx = ∫ 2 x dx → y = x2 + C
Section4-CalculusII-M athQ RH
311
Thisisknownasthegeneralsolutionofthedifferentialequation. Ifwewanttofindthe particularsolution,weneedtohaveaninitialcondition,likey (1) = 4 . y = x 2 + C → 4 = 1 2 + C → C = 3 So,aparticularsolutionofthedifferentialequationis y = x2 + 3
4.5.1
C heckingS olutions
Somedifferentialequationsarehardtosolve. So,wecancheckifasolutioniscorrect insteadofsolvingit. Example:y ′′ = x + y ,checkthaty = Ae x + B e −x − x (generalsolution)isasolution. Beginbyfindingtheappropriatederivativeofthegivenfunction. y ′ = Ae x − B e −x − 1 y ′′ = Ae x + B e −x Now,replacethey ′′ inthedifferentialequationwiththesecondorderderivativeyoujust found. Ae x + B e −x = x + y Inplaceofy ,putinthegeneralsolution. Ae x + B e −x = x + Ae x + B e −x − x Ae x + B e −x = Ae x + B e −x So,y = Ae x + B e −x − x isasolutiontoy ′′ = x + y .
312
Section4-CalculusII-M athQ RH
4.5.2
S lope/DirectionF ields
Solvingmostdifferentialequationsisimpossible, butwecanvisualizeasolutionby drawingslopeordirectionfields. Itislikedrawingagraphwithoutanequation. 4.5.2.a
Example
Drawtheslopefieldforthedifferentialequationdy dx = x + y . First,makeatableofvaluesandplugdifferent(x, y) coordinatesintody dx . Then,gotothat pointonthegraphanddrawatinysectionthatshowstheslope(rememberthatderivatives findslope). Youcandothisforanypoint. (x, y)
dy dx
(0, 0)
0
(1, 0)
1
(0, 1)
1
(1, 1)
2
Ifyouwanttofindthesolutiontoaparticularequation(likewithy (0) = 1 ),youcangotothe DesmosSlopeFieldGeneratortoplotouttheentirefield. Then,youcanfindthepointand followthecurvecreatedbythelines. Thisisbecauseyoudrewthegraphofthefunction thatsatisfiesthesolution.
Section4-CalculusII-M athQ RH
313
dy
Seetheslopefieldfordx = x + y below.
4.5.3
E uler’sM ethod
UseEuler’sMethodwithstepsizeΔ x = 0 .5 tosketchthesolutiontothedifferential equationy ′ = x + y withtheinitialconditiony (0) = 1 .
314
Section4-CalculusII-M athQ RH
dy
Startatableofvaluesbeginningwiththeinitialcondition(0, 1) ;usethistofinddx atthis point. Thenextvalueofx willbethepreviousx valueplusthegivenstepsize. Thenext dy
valueofy willbethepreviousdx valueplusthegivenstepsize. (x, y)
dy dx
(0, 1)
1
(0.5, 1.5)
2
(1, 2.5)
3 .5
4.5.4
S eparableDifferentialE quations
Aseparabledifferentialequationisafirstorderdifferentialequationthatfactorsintoa functionofx andafunctionofy (separatingx ’sandy ’s). Youcanactuallysolvethese equations. Noteverydifferentialequationisseparable. Theyneedtobeoneofthetwofollowing f(x)
forms:f (x) · g (y) or h(y) .
Section4-CalculusII-M athQ RH
315
Tosolvethesetypesofdifferentialequations,useoneofthemethodsbelow: 4.5.4.a
Method1 dy dx
= f (x) · g (y)
Dividebothsidesbyg (y) . 1 g(y)
· dy dx = f (x)
“Multiplyby”d x . 1 g(y)
· d y = f (x) · d x
Integratebothsides.
1 dy = ∫ f (x) dx ∫ g(y)
Note:youaren’ttechnicallymultiplyingbothsidesbyd x ,itimplementsthechainrule,but it’sjusteasiertosayyoumultiplybothsidesbyd x . 2 Example:Solvey ′ = x 2 y → dy dx = x y .
Isolatethex andy variables. 1 y
d y = x 2 dx
Integratebothsides.
∫ 1y dy = ∫ x2 dx → ln |y | =
316
1 3 3x
+C
Section4-CalculusII-M athQ RH
Youonlyneeda+Conones ide. Exponentiateboths idesbyetos implify. |y | = e ( 3 x + C) = e 3 x · e C 1 3
1 3
e C isstillaconstantbutitisadifferentonefromC . 1 3
|y | = Ae 3 x Knowthattheconstantwillnotalwaysbe‘+ C ’,itcouldalsobeamultiplier. 4.5.4.b
Method2 dy dx
=
f(x) h(y)
Multiplybothsidesbyh (y) . h (y) · dy dx = f (x) Integratebothsides.
∫ h (y) dy = ∫ f (x) dx
dy
Example:Solve dx =
x2 y2
withy (0) = 2 .
Isolatebothvariablesandintegrate.
∫ y 2 dy = ∫ x2 dx → 31 y 3 =
Section4-CalculusII-M athQ RH
1 3 3x
3
+ C → y 3 = x 3 + A → y = √x 3 + A
317
YouneedtouseadifferentletterthanC becausetheyaredifferentconstants. Now,plugin(0, 2) . 2=
√(0) 3
3
+ A → A = 8 3
y = √x 3 + 8 4.5.4.c
MixingProblems
Theseareclassicdifferentialproblems. Intheseproblems,somethingisbeingmixedina tank. Atthesametime,somethingisbeingaddedandsomethingisbeingtakenout. Example:Atankcontains2 0 kg ofsaltdissolvedin5 000 L ofwater. Amixtureof0 .03 kg saltperliterofwaterentersthetankatarateof2 5 L/min . Thesolutioniskeptmixedandit drainsoutatthesamerate. Howmuchsaltremainsafter3 0 min ? Let’ssaythaty (t) istheamountofsaltaftert minutes. Weknowthaty (0) = 2 0 andweare dy
lookingfory (30) . dt = r ate in − r ate out . Wewanttofindtheratesintermsofk g/min ,sincewearelookingforthek g ofsaltafter 3 0 min . Therefore, r ate in = mixture of salt · r ate entering = (0.03 kg/L)(25 L/min) = 0 .75 kg/min y
r ate out = solution · r ate leaving = ( 5000 kg/L)(25 L/min) =
y 200 kg/min
wherey isthesolutioninthetankthatisdrainingout. So, dy dt
318
y
= 0 .75 kg/min − 200 kg/min =
150 − y 200
Section4-CalculusII-M athQ RH
Now,isolatethey termsandintegratebothsides.
∫
1 150 − y dy
1 = ∫ 200 dt
Simplifyandplugintheinitialvalue. − ln |150 − y | =
1 200 t
+ C → 150 − y = Ae − 200 t 1
Wheny (0) = 2 0 ,A = 1 30 ,thereforey = 1 50 − 1 30e − 200 t 1
Nowwecanplugin3 0 fort togetourfinalanswer. y = 1 50 − 1 30e − 200 (30) ≈ 3 8.108 kg 1
4.6 CalculuswithParametricEquationsand PolarCoordinates 4.6.1
C alculusw ithP arametricC urves
Toreviewparametricequationsandplanecurves,gototheAnalyticGeometrysectionof Pre-Calculus. 4.6.1.a
SlopeandConcavity
Rememberthataparametricequationisdefinedbyx asafunctionoft andy asa functionoft . So,tofindthederivativeofaparametricequation,usetheformulabelow dy dx
Section4-CalculusII-M athQ RH
=
derivative of y dy/dt derivative of x = dx/dt
319
Tofindthesecondderivativeoftheparametricequation,firstrememberthatthesecond dy
derivativeisthederivativeofdx . So, d2 y dx2
=
d dy dx ( dx )
d (dy/dx) dt
=
dx/dt
Example:Findthefirstandsecondderivativesoftheparametricequationx = t2 y = t3 − 3 t . dy dx
=
d/dt (t3 −3t) d/dt (t2 )
=
3t2 −3 2t
Tofindthesecondderivative,firstfindthederivativeofdy dx withrespecttot . d dy dt ( dx )
=
(6t)(2t) − (3t2 − 3)(2) 2
(2t)
=
6t2 + 6 (2t)2
Now,plugthisintothesecondderivativeformulaabove. d2 y dx2
=
6t 2 + 6 (2t)2
2t
=
6t2 + 6 (2t)3
Thesamerulesapplyforderivativesofparametricequationsastheydowithregular dy
dy
equations:thetangentlineishorizontalwhendx = 0 andverticalwhendx = u ndef ined . d2 y
Inadditiontothis,thegraphisconcaveupwhendx2 ispositiveandconcavedownwhen d2 y dx2
isnegative.
320
Section4-CalculusII-M athQ RH
4.6.1.b
Areas
Ifx = f (t) andy = g (t) ,theareaunderthecurveis β
∫
α
g (t) f ′(t) dt
foracurvetravelingfromlefttoright. Example:Findtheareaunderonearchofthecycloidx = r (θ − sin θ) y = r (1 − c os θ) . Onearchiscompletedfromθ = 0 toθ = 2 π ,soα = 0 andβ = 2 π . Now,findd x . dx =
d dθ (r(θ
− sin θ)) = r (1 − c os θ) dθ
r isaconstant,soitisnotaffectedwhentakingthederivative. Pluginallinformationtotheformulaaboveandsolve. 2π
2π
2π
0
0
0
∫ [r(1 − cos θ)] · [r(1 − cos θ)] dθ = r 2 ∫ (1 − 2 cos θ + cos2 θ) dθ = r 2 ∫ [1 − 2 cos θ + 21 (1 + cos (2θ))] dθ 2 a rea = r 2 [ 23 θ − 2 sin θ + 41 sin (2θ)]2π 0 = 3 πr
Section4-CalculusII-M athQ RH
321
4.6.1.c
ArcLength
Assumethecurvedoesn’ttraversebackoveritself. Thearclengthofaparametriccurveis b
dy 2 2 ∫ √( dx dt ) + ( dt ) dt a
Example:Findthearclengthofx = r cos t y = r sin t, 0 ≤ t ≤ 2 π . 2π
∫
0
√
(− r sin t)2 + (rcos t)2 dt =
2π
2π
0
0
∫ √r 2 (sin 2 t + cos2 t dt = ∫ √r 2 (1) dt = r t|2π 0 = 2 πr
4.6.2
T angentsandA reasw ithPolarC urves
Toreviewpolarcoordinatesandtheirgraphs,gotothePolarCoordinatesandVectors sectionofPre-Calculus.
322
Section4-CalculusII-M athQ RH
4.6.2.a
DerivativeswithaPolarCurve
r = f (θ) isapolarcurve. Youneedtoconvertthisintoequationsforx andy using x = r cos θ andy = r sin θ . r = f (θ) → x = f (θ)cos θ y = f (θ)sin θ Noticethatthisisaparametricequation. Therefore,wewillusethesamemethodthatwe usedbefore. dy dx
=
dy/dθ dx/dθ
4.6.2.b
AreaofaPolarCurve
Whenmeasuringtheareaofapolarcurve,measurefromtheorigintother value.
Thisislikeafanthatopensandcloses. Whenworkingwithareasofpolarcurves,noneof theareagetscountedasnegative. Theformulafortheareainsideapolarcurveis b
∫ 21 r 2 dθ a
Section4-CalculusII-M athQ RH
323
Example:Findtheareaoftheregioninsider = 3 sin θ andoutsider = 1 + sin θ . First,graphtheequations.
Now,findthevaluesofthetawheretheyintersect. 3 sin θ = 1 + sin θ → 2sin θ = 1 → sin θ = 21 → θ = 6π , 5π6 Tofindtheareabetweenthetwographs,usetheformulaA = 5π/6
A=
∫
π/6
2 1 2 (3sin θ) dθ
5π/6
−
∫
π/6
1 2 (1
+ sin θ)2 dθ =
1 2
5π/6
∫
1 2
b
b
a
a
∫ top 2 dθ − 21 ∫ b ottom2 dθ .
[(3sin θ)2 − (1 + sin θ)2 ] dθ = π square units
π/6
324
Section4-CalculusII-M athQ RH
“Mathematicsisn otjusta notherlanguage...Itisa languagep luslogic. Mathematicsisa toolforreasoning.” -RichardFeynman
4.7
yNotesforCalculusII M
Section4-CalculusII-M athQ RH
325
4.7
326
MyNotesforCalculusII(con’t)
Section4-CalculusII-M athQ RH
“Id on'tthinkthate veryones houldb ecomea mathematician,b utId o believethatmanys tudentsd on'tg ivemathematicsa realc hance.” -M aryamM irzakhani
Section5-CalculusIII 5.0
SummarySheet
Vectors Apositionvectorstartsattheorigin. Aunitvectorisavectorwithmagnitude1. ●
u→ =
1 → v ||v||
Vectoradditionandsubtraction:
Distancebetween(x 1 , y 1 , z 1 ) and(x 2 , y 2 , z 2 ) :d =
√(x
Vectormagnitude:||v || = √x 2 + y 2 + z 2 , angle:tan −1 ( yx )
Section5-CalculusIII-M athQ RH
2
− x 1 )2 + (y 2 − y 1 )2 + (z 2 − z 1 )2
327
Dotproduct:u→ · v→ = ||u || ||v || c os θ = u 1 v 1 + u 2 v 2 + u 3 v 3 ●
Thisproducesanumber.
●
Properties: ○
u·v =v ·u
○
u · (v + w ) = u · v + u · w
○
c (u · v ) = (cu) · v = u · (cv)
○
v · v = ||v ||2
Vectorofprojection:p roj v (u) =
u · v → 2 v ||v ||
Lengthofprojectionofu ontov :c omp v (u) = λ =
u · v ||v ||
Crossproductisavectorthatisperpendiculartobothu→ andv→ . ●
u × v = < u 2 v 3 − u 3 v 2 , − (u 1 v 3 − u 3 v 1 ), u 1 v 2 − u 2 v 1 >
●
||u × v|| = ||u || ||v || sin θ = a rea of parallelogram = 2 (area of triangle)
●
ifu→ andv→ areparallel,u × v = < 0 , 0, 0 > →
→
Parametricline:r→(t) = P + tP Q PlanesandSurfaces Parametriccurve:traceeverypoint,t isaparameter. Thecurveisdefinedby: x = x (t) y = y (t)
r→(t) = < x (t), y(t), z(t) >
z = z (t) ●
Thevector< l, m, n > isparalleltothelinewithparametricequations ○
328
x = a + lt, y = b + mt, z = c + n t
Section5-CalculusIII-M athQ RH
Equationofaplane:a (x − x 0 ) + b (y − y 0 ) + c (z − z 0 ) = 0 ●
iftwoplanesareparallel,theyhavethesamenormalvector
●
thenormalvector< a , b, c > isperpendiculartotheplane
Cylinder:asetoflinesparalleltoagivenlinepassingthroughagivencurve ●
anyequationmissingoneofx , y, orz willbeacylinder
Traces:crosssectionscreatedwhenthesurfaceintersectsaplaneparalleltooneofthe coordinateplanes TypesofSurfacesandtheirEquations: Surface ellipsoid
Equation x2 a2
paraboloid
y2 b
2
x2 a2
+
hyperboloidoftwosheets
z2 c2
−
ellipticcone
x2 a2
+
+
x2 a2
z=
hyperboloidofonesheet
hyperbolicparaboloid
+
y2
z2 c2
= 1 y2
+
b
2
Axis -----
axiscorrespondstolinear variable
−
z2 c2
= 1
axiscorrespondstothe variablewiththenegative coefficient
x2 a2
−
y2 2
= 1
axiscorrespondstothe variablewiththepositive coefficient
y2
−
z2 c2
= 0
axiscorrespondstothe variablewiththenegative coefficient
b
b
z=
2
2
x2 a2
b
−
y2 b
2
axiscorrespondstothe linearvariable
Section5-CalculusIII-M athQ RH
329
FunctionsofSeveralVariables Levelcurveofafunction:thesetofpointssatisfyingtheequationf (x, y) = c
Limits:
lim
(x, y) → (a, b)
f (x, y) = L ,(x, y) gets‘closeto’(a, b) thenf (x, y) gets‘closeto’L .
●
if(a, b) isnotadiscontinuity,pluginvalues
●
if(a, b) isadiscontinuity,checkpaths(subiny = b andtakelimitforx ) ○
●
it'sbasicallyimpossibletofindthevalueofthelimitthisway
usethesqueezetheorem(see:CalcI)
PartialDerivatives ∂f
f(x + h, y) − f(x, y) h h → 0
There’sapartialderivativeineachdirection: ∂x = f x = lim
●
rateofchangeoftheheightonthesurfaceasx movesintheh direction
●
usederivativerulesfromcalcI,buttreatallvariablesyouarenotsolvingforas constants
330
Section5-CalculusIII-M athQ RH
Estimatingpartialderivativesfromacontourmap:pickasmallk , h, l anduse correspondingx , y values Higherorderderivatives:takethederivativewithrespecttothefirstvariable,thenderive thatwithrespecttothesecondvariable. ●
f xy :takex derivativethenderivethatwithrespecttoy
●
Clairaut’sTheorem:f xy = f yx ,orderdoesn’tmatter
Tangentplane:touchescurveatonepoint(x 0 , y 0 ) → z = f (x 0 , y 0 ) + f x(x 0 , y 0 )(x − x 0 ) + f y (x 0 , y 0 )(y − y 0 ) ●
thisistheequationusedforlinearapproximations ∂f
Chainrule: ∂t =
∂f dx ∂x dt
+
∂f dy ∂y dt
●
useadependencychart
●
replacex andy withothervariables
ImplicitDifferentiation: ●
supposez = f (x, y) definesy implicitly→
●
iff (x, y, z) = 0 definesz implicitly→
∂z ∂x
dy dx
= −
= − ∂f/∂x ∂f/∂z
∂f/∂y ∂f/∂z
∂z , ∂y = −
∂f/∂y ∂f/∂z
Directionalderivative:rateofchangeofheightoff aswetravelalongr (t)
∇f (x, y) · u, where∇f (x, y) = < f , f
>
●
D uf =
●
fastestincrease:followsthegradient,fastestdecrease:followsthenegativeofthe gradient
●
→
∇
remember: f · u→ = ||
x
∇f || ||u|| cos θ, u →
∇
→
tan
=
y
∇f (x, y) · ∇
1 || f(x, y)||
Theorem: f (x 0 , y 0 ) isperpendiculartothelevelcurveoff at(x 0 , y 0 )
Section5-CalculusIII-M athQ RH
331
∇
∇
→
ndf y = 0 or f (x 0 , y 0 ) DN E Criticalpoint: f (x 0 , y 0 ) = 0 → f x = 0 a SecondDerivativeTest:D = f xx(x 0 , y 0 )f yy (x 0 , y 0 ) − (f xy (x 0 , y 0 ))2 ●
ifD > 0 andf xx(x 0 , y 0 ) > 0 ,thenf hasalocalminimumat(x 0 , y 0 )
●
ifD > 0 andf xx(x 0 , y 0 ) < 0 ,thenf hasalocalmaximumat(x 0 , y 0 )
●
ifD < 0 ,thenf hasasaddlepointat(x 0 , y 0 )
●
ifD = 0 ,thenthetestisinconclusive
LagrangeMultipliers:findtheextremaoff subjecttoaconstraintg (x, y) = c ●
∇f = λ∇g , lamdaistheLagrangemultiplier ○
●
thisisasystemofequations
absolutemaximumwilloccurateitheracriticalpointoraboundarypoint (Lagrange)
Differential:d f =
∂f ∂x dx
+
∂f ∂y dy
DoubleIntegrals
Fubini’sTheoremandIteratedIntegrals:∫ ∫ f (x, y) dA R
●
d A = d ydx ord xdy (rectangular)
●
toevaluate,startontheinsideandworkout(pretendtheothervariableisconstant)
Properties: ●
sumrule,constantrule,squeezetheorem
●
ifR = S ⋃ T andS ⋂ T = 0 , ∫ ∫ f (x, y) dA = ∫ ∫ f (x, y) dA + ∫ ∫ f (x, y) dA
332
R
S
T
Section5-CalculusIII-M athQ RH
●
b
d
R
a
c
ifR = {(x, y) | a ≤ x ≤ b , c ≤ y ≤ d }, ∫ ∫ f (x, y) dA = (∫ g (x) dx)(∫ h (y) dy) ,wheref (x, y) is factoredintog (x) andh (y)
TwoTypesofGeneralRegions: Domain D = {(x, y) | a ≤ x ≤ b , g 1 (x) ≤ y ≤ g 2 (x)} Integral
b g2 (x)
∫ ∫
a g1 (x)
f (x, y) dydx
D = {(x, y) | h 1 (y) ≤ x ≤ h 2 (y), c ≤ y ≤ d } d h2 (y)
∫ ∫
ch1 (y)
f (x, y) dxdy
Graphed Region
y b oundedbyafunctionofx
x boundedbyafunctionofy
Toreversetheorderofintegration,drawoutgivenboundsandre-setupforopposite integration.
∫ ∫ 1 dA givesarea D
∫ ecx = 1c ecx
Section5-CalculusIII-M athQ RH
333
PolarCoordinates Conversionbetweenpolarandrectangular: ●
x = r cos θ, y = r sin θ (polartorectangular)
●
x 2 + y 2 = r 2 , tan θ =
y x
(rectangulartopolar)
“Polarrectangles”aresectionswiththeinequalitiesoftheform0 ≤ θ ≤ 2π , 1 ≤ r ≤ 2 .
Tointegrateoverpolarregions,used A = r drdθ ●
theconstraintsforr iswherer startsandends
Whentherearetwosurfacesoverlapping,use∫ ∫ (top − b ottom) dA . D
TripleIntegrals RectangularCoordinates:d V = d zdydx ord zdxdy (4otheroptions) ●
innermostboundsrepresentsurfacesthatboundshape
●
middleboundsrepresentcurves
●
outermostboundsrepresentnumbers
IfD istheboundedregionthatisaprojectionofE ; E = {(x, y, z) | (x, y) ε D, u 1 (x, y) ≤ z ≤ u 2 (x, y)} u2 (x, y)
E
D u1 (x, y)
∫ ∫ ∫ f (x, y, z) dV = ∫ ∫ [ ∫ 334
f (x, y, z) dz] dA
Section5-CalculusIII-M athQ RH
AverageValueofaFunction: f ave =
1 V (E)
∫ ∫ ∫ f (x, y, z) dV E
whereV (E) = ∫ ∫ ∫ 1 dV (volume). E
Cylindrical/SphericalCoordinates Cylindricalcoordinatesarepolarcoordinateswithz attached. ●
x = r cos θ, y = r sin θ, z = z
●
d V = r dzdrdθ
Sphericalcoordinatesaretwoanglesandonelength. ●
ρ = d istance f rom origin, θ = a ngle measure f rom positive x, ϕ = a ngle f rom positive z
RectangulartoSphericalTransformation: x = ρsin ϕ cos θ
y = ρsin ϕ sin θ
y
θ = tan −1 ( x )
ϕ = tan −1 (
√x2 + y2 ) z
z = ρcos ϕ ϕ = c os−1 (
z
√x2 + y2 + z2
)
x 2 + y 2 + z 2 = ρ2 SphericalCoordinateIntegration:d V = ρ2 sin ϕ dpdϕdθ
Ifρ(x, y, z) ismassdensity,∫ ∫ ∫ ρ(x, y, z) dV = mass .
Section5-CalculusIII-M athQ RH
335
CentersofMass Twodimensionallaminas,flat2Dobject
●
mass=∫ ∫ ρ(x, y) dA D
●
momentaboutthex axis:M x = ∫ ∫ x ρ(x, y) dA D
●
momentaboutthey axis:M y = ∫ ∫ y ρ(x, y) dA D
●
CenterofMass=(
Mx My m , m )
Threedimensional
●
mass=∫ ∫ ∫ ρ(x, y, z) dV E
●
M x = ∫ ∫ ∫ x ρ(x, y, z) dV E
●
M y = ∫ ∫ ∫ y ρ(x, y, z) dV E
●
M z = ∫ ∫ ∫ z ρ(x, y, z) dV E
●
CenterofMass=(
Mx My Mz m , m , m ) =(x, y, z)
Ifthemassdensityisconstant,thecenterofmassiscalledthecentroid.
336
Section5-CalculusIII-M athQ RH
VectorFields VectorField:functionthatoutputsavectorateverypoint Notation: 2D:< P (x, y), Q(x, y) >
3D:< P (x, y, z), Q(x, y, z), R(x, y, z) >
UnitVectorField:vectorfieldwhereallvectorshavelength1 ●
√
divideeachcomponentbymagnitude P (x, y)2 + Q(x, y)2
GradientVectorField:vectorfieldwhichisthegradientofascalarfunction ●
scalarfunction:outputisanumber
●
use f = < f x, f y >
●
levelcurvesoff areperpendiculartothevectorsinthegradientfield
∇
LineIntegrals
ScalarLineIntegral:∫ f (x, y) dS ,d S = c
√1 + (
dy 2 dx ) dx
r (t) = p arameterization of a curve ● ●
√( 3D:d s = √( 2D:d s =
dx )2 dt
+ ( dy )2 dt dt
dx 2 dt )
2 + ( dt )2 + ( dz dt ) dt
dy
Section5-CalculusIII-M athQ RH
337
LineIntegrals(con’t) ●
whenusingthis,putf (x, y) intotermsoft
→
→
x = (1 − t)x 1 + tx 2 ,r (t) = A + tAB
ArcLengthofc :∫ 1 ds = length (c) c
VectorLineIntegral:lineintegralovervectorfields(workandflux) ●
t2
t2
t1
t1
Workdoneonaparticle:∫ P dx + Qdy or∫ P dx + Qdy + Rdz where d x = x ′(t)dt, dy = y ′(t)dt, dz = z ′(t)dt ○
ParameterizeF = < P (x, y), Q(x, y) > to< x (t), y(t) > ,replacex andy inP andQ ,subintoequationP dx + Qdy withd x = x ′(t)dt andd y = y ′(t)
○ ●
thisisderivedfrom(F (r(t)) · r ′(t))
Flux,flowovertime:∫ P dy − Qdx wherev→ = v elocity f ield, ϕ = g eneric vector f ield → →
c
ϕ = (P , Q) orv = (P , Q) . ○
positiveflux:intoout
○
negativeflux:outtoin
CurveTypes:closed(formsloop),simple(doesn’tintersectself),connected(twopointsform path)
338
Section5-CalculusIII-M athQ RH
ConservativeVectorField:pathtakendoesn’tmatter,pathindependent ●
then,∫ ϕ · d r = ϕ forallclosedcurvesc [ϕ = (P , Q) ,d r = (dx, dy) ] c
FundamentalTheoremofLineIntegrals:allgradientfieldsareconservativeifϕ =
∫ ϕ dr = f (B) − f (A) ,f c
∇f
istheantigradient
Ifavectorfieldisconservative,thenP y = Qx . TofindapotentialfunctionforF (x, y) = < P , Q > (usethisinthefundamentaltheoremof lineintegrals): 1. Checkifit’sconservative
∇
2. Find f togetf x andf y 3. PartiallyintegrateF x (thex component,P ,ofF (x, y) )withrespecttox tofindf 4. Differentiatethef foundinstepthreewithrespecttoy 5. Setf y = F y tofindC (y) 6. Finalresultisthefunctionf TransformationTheorems Green’sTheorem:D isanopen,simplyconnectedregionwithaboundarycurvec thatisa piecewisesmooth,simpleclosedcurveorientedcounterclockwise.
∮
c
∮
F · dr =
c
P dx + Qdy = ∫ ∫ (Qx − P y ) dA D
●
Itneedstobeinthisorder!
●
recognizethattheinterimequationistheworkdoneonaparticle;thisisimportant tonoticesothatyouareabletochoosetherighttheoremwhilesolvingproblems
●
Ifclockwise,addanegativesign
Section5-CalculusIII-M athQ RH
339
Divergence:d iv(ϕ) =
∇ · ϕ, whereϕ = < P , Q > → div(ϕ) = ∂P ∂x
∂Q ∂y
∂R ∂z
In3D(ϕ = < P , Q, R > ),d iv(ϕ) =
●
allfluxintegralsacrossclosedsurfaceswillbe0
+
+
∂Q ∂y
.
●
+
∂P ∂x
Curlmeasuresthespinatapoint;vectorwhered r isequaltotheaxisofrotationandthe magnitudesareequal ●
clockwise,pointsupward
●
c url(ϕ) = ○
∇×ϕ =(
∂ ∂ ∂ ∂x , ∂y , ∂z )
× (P , Q, R) = ( ∂R ∂y −
∂Q ∂P ∂z , ∂z
−
∂R ∂Q ∂x , ∂x
−
∂P ∂y
)
if2D,R = 0
ParameterizingFunctions:r→(u, v) = < x (u, v), y(u, v), z(u, v) > ●
→ forafunctionz = (x, y) : r(t) = < u , v, f (u, v) >
SurfaceIntegrals SurfaceIntegral:domainofintegrationisoveraboundarysurface Transformationbetweenareapieced A andareapieced S : d S = ||tu × tv || dA ,wheretu =
andtv =
integralformula:∫ ∫ ||tu × tv || dA D
ScalarSurfaceIntegral:∫ ∫ f (r(u, v)) ||tu × tv || dA D
●
usethisforthingslikemassofasheet
●
surfacearea:integrandis1
●
surfaceintegral:integrandisf (x, y) intermsofu andv
340
Section5-CalculusIII-M athQ RH
t × t
UnitNormal= N = ||tuu × tvv || (length1) ●
pointsfromintoout
●
positive:outwardpointingnormal
●
flux:witharrow→ positive,withoutarrow→ negative
SurfaceIntegralinaVectorField:∫ ∫ (F (r(u, v)) · (tu × tv )) dV D
●
volumepassingthroughad S portionovertime
●
ifz = f (x, y) passesverticallinetest:∫ ∫ [− P · f x − Q · f y + R] dA
D
c
S
Stokes’Theorem:∫ F · d r = ∫ ∫ ●
∫ F · dr c
∇ × F dS
isthecirculationorworkaroundtheboundarycurve
→
→
DivergenceTheorem:∫ ∫ F · d S = ∫ ∫ ∫ d iv F dV ,whereS isaclosedsurface S
●
→
→
∫ ∫ F · dS
E
isthefluxacrossasurface
S
5.1
VectorsinSpace
5.1.1
V ectorsi ntheP lane
Formorereviewonvectors,seethePolarCoordinatesandVectorssectionofPre-Calculus. Iwillrestatesomeimportantfactstoknowhere. Apositionvectorstartsattheorigin.
Section5-CalculusIII-M athQ RH
341
Themagnitudeofavectorisdenoted||v || andcalculatedusingtheformula ||v || = √x 2 + y 2 Aunitvectorisavectorwithmagnitude1 . Tofindtheunitvectoru→ ofthevectorv→ ,use theformula u→ =
1 → v ||v ||
5.1.1.a PropertiesofVectorOperations → → Letu→, v, andw bevectorsinaplane. Letr ands bescalars.
Property
Name
u→ + v→ = v→ + u→
commutativeproperty
(u + v ) + w = u + (v + w )
associativeproperty
→
u +0 =u
additiveidentityproperty
u + (− u ) = 0
additiveinverseproperty
r (su) = (rs)u
associativityofscalarmultiplication
(r + s)u = r u + su
distributiveproperty
r (u + v ) = r u + r v
distributiveproperty
1 u = u , 0u = 0
identityandzeroproperties
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Section5-CalculusIII-M athQ RH
5.1.2
V ectorsi nThreeD imensions
Thethreedimensionalrectangularcoordinatesystemconsistsofthreeperpendicularaxes: thex axis,they axis,andthez axis. Becauseeachaxisisanumberlinerepresentingall realnumbersinR ,thethreedimensionalsystemisoftendenotedR3 . 5.1.2.a
PlottingPoints
Youdrawthe3Dcoordinateaxisbyputtingtheregularx y axisonthe“ground”andhaving thez axispointstraightup.
Toplotapoint,gox unitsalongthex axis,theny unitsinthedirectionofthey axis,then z unitsinthedirectionofthez axis(upordown).
Section5-CalculusIII-M athQ RH
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5.1.2.b
TheCoordinatePlanes
Thereareeightoctantsinthe3Dplane. 5.1.2.c
DistanceFormula(3D)
Thedistancebetweenpoints(x 1 , y 1 , z 1 ) and(x 2 , y 2 , z 2 ) isgivenby d=
344
√(x
2
− x 1 )2 + (y 2 − y 1 )2 + (z 2 − z 1 )2
Section5-CalculusIII-M athQ RH
5.1.2.d
EquationsofSurfaces(3D)
In2D,horizontalandverticallineshavetheequationy = some number andx = some number , respectively, whereanynumberwiththosex ory coordinateswillsatisfyit. In3D,itissimilar. Fortheequationz = 6 ,forexample,anypointwherez = 6 willsatisfy theequation. Thegraphofthisequationisanx y planeatheight6 . EquationsofPlanesParalleltoCoordinatePlanes: 1. Theplaneinspacethatisparalleltothex y planeandcontainspoint(a, b, c) canbe representedbytheequationz = c . 2. Theplaneinspacethatisparalleltothex z planeandcontainspoint(a, b, c) canbe representedbytheequationy = b . 3. Theplaneinspacethatisparalleltothey z planeandcontainspoint(a, b, c) canbe representedbytheequationx = a . 5.1.2.e
Spheres
Asphereisthesetofallpointsinspaceequidistantfromafixedpoint,thecenterofthe sphere. Inasphere,thedistancefromthecentertoapointonthesphereiscalledthe radius.
Notethatasphereisnotfilledin,asthatwouldmakeitaball.
Section5-CalculusIII-M athQ RH
345
EquationofaSphere:(x − a )2 + (y − b )2 + (z − c )2 = ρ2 ●
ρ istheGreekletter“rho”andisusedinsteadofr
5.1.2.f
3DVectors
3Dvectorsareverysimilarto2Dvectors. Example:Graphthevector< 2 , 4, 1 > . Gotwointhex direction,gofourinthey direction,gooneinthez direction.
346
Section5-CalculusIII-M athQ RH
5.1.2.g
PropertiesofVectorsinSpace
Letv = < x 1 , y 1 , z 1 > andw = < x 2 , y 2 , z 2 > bevectorsandletk beascalar. ●
scalarmultiplication:k v = < k x 1 , ky 1 , kz 1 >
●
vectoraddition:
v + w = < x 1 , y 1 , z 1 > + < x 2 , y 2 , z 2 > = < x 1 + x 2 , y 1 + y 2 , z 1 + z 2 > ●
vectorsubtraction: v − w = < x 1 , y 1 , z 1 > − < x 2 , y 2 , z 2 > = < x 1 − x 2 , y 1 − y 2 , z 1 − z 2 >
√x
●
vectormagnitude:||v || =
●
unitvectorinthedirectionofv : ||1v|| v = ○
1
2
+ y 12 + z 12 1 ||v ||
< x 1 , y 1 , z 1 > =
if||v || =/ 0
5.1.3
D otProduct
Thedotproductoftwovectorsistheproductofthemagnitudeofeachvectorandthe cosineoftheanglebetweenthem.
u→ · v→ = ||u|| ||v|| cos θ
Section5-CalculusIII-M athQ RH
347
Ifyouhavethecomponentsofthevector,youcanuseanotherformula. Thedotproductof thevectorsu→ = < u 1 , u 2 , u 3 > andv→ = < v 1 , v 2 , v 3 > isgivenbythesumoftheproductsof thecomponents. u→ · v→ = u 1 v 1 + u 2 v 2 + u 3 v 3 Notethatthedotproductisanumber,notavector. 5.1.3.a
PropertiesoftheDotProduct
→ → Letu→, v, andw bevectorsandc beascalar.
Property
Name
u·v =v ·u
commutativeproperty
u · (v + w ) = u · v + u · w
distributiveproperty
c (u · v ) = (cu) · v = u · (cv)
associativeproperty
2
v · v = ||v||
propertyofmagnitude
Example:Findthemeasureoftheangleformedbyvectorsa = < 1 , 2, 0 > and b = < 2 , 4, 1 > . Beginbyfindingthedotproductofthevectors. a · b = 1 · 2 + 2 · 4 + 0 · 1 = 2 + 8 + 0 = 10
348
Section5-CalculusIII-M athQ RH
Weknowthat a · b = ||a|| ||b|| cos θ ||a|| = √1 2 + 2 2 + 0 2 = √5 ||b|| = √2 2 + 4 2 + 1 2 =√21 Replacetheaboveformulawithallknownvalues. 1 0 = (√5 )(√21 ) cos θ Solveforθ . θ = c os−1 (
10 ) (√5)(√21)
≈ 1 2.60 °
5.1.3.b
OrthogonalVectors
Thenonzerovectorsu andv areorthogonal(perpendicular)vectorsifandonlyifu · v = 0 . 5.1.3.c
VectorProjection/Component
Derivation: Saythatthesuncastsashadowofu ontov .
Section5-CalculusIII-M athQ RH
349
Thisisdenotedp roj v (u) ,the“projectionontov ofu .” Thisisafunctionthatinputsavector u→ andoutputstheprojectedvector. Tocalculate: 1. Findaunitvectorinthedirectionofv→ . 2. Multiplybysomenumberλ (lambda)toscaleittotherightsize. a. λ isthelengthoftheprojectedvector b. λ = ||u|| · c os θ Therefore, p roj v (u) = ||u|| cos θ ·
v ||v||
||v|| Ifyoudon’thaveθ ,multiplyby ||v|| (1) togetthedotproductformula.
VectorofProjection:p roj v (u) =
u · v → 2 v ||v||
Similarly,tofindthelengthofthisvectorwithoutθ usetheformula LengthofProjectionVector:λ = c omp v (u) =
u · v ||v||
5.1.4
C rossP roduct
Letu = < u 1 , u 2 , u 3 > andv = < v 1 , v 2 , v 3 > . Thenthecrossproductu × v isvector u × v = (u 2 v 3 − u 3 v 2 )i − (u 1 u 3 − u 3 v 1 )j + (u 1 v 2 − u 2 v 1 )k = < u 2 v 3 − u 3 v 2 , − (u 1 v 3 − u 3 v 1 ), u 1 v 2 − u 2 v 1 >
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Section5-CalculusIII-M athQ RH
5.1.4.a
PropertiesoftheCrossProduct
●
u × v isanewvectorperpendiculartobothu→ andv→
●
||u × v || = ||u|| ||v|| sin θ = a rea of parallelogram = 2 (area of triangle)
●
Ifu→ andv→ areparallel,u × v = < 0 , 0, 0 >
Letu , v, andw bevectorsinspaceandletc beascalar. Property
Name
u × v = − (v × u )
anticommutativeproperty
u × (v + w ) = u × v + u × w
distributiveproperty
c (u × v ) = (cu) × v = u × (cv)
scalarmultiplication
u ×0 =0 ×u =0
zerovectorcrossproduct
v ×v =0
crossproductwithitself
u · (v × w ) = (u × v ) · w
scalartripleproduct
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351
Example:FindtheareaofthetriangleP QR ,whereP = (2, − 2 , − 2 ) ,Q = (0, 2, 5) ,and R = (5, 6, 2) .
→
→
→
→
TheareaofthetriangleAOT = 21 ||P Q × P R|| . So,findP Q andP R . →
P Q = (0, 2, 5) − (2, − 2 , − 2 ) = < 0 − 2 , 2 + 2 , 5 + 2 > = →
P R = (5, 6, 2) − (2, − 2 , − 2 ) = < 5 − 2 , 6 + 2 , 2 + 2 > = < 3 , 8, 4 > Now,findthecrossproductofthevectors. →
→
P Q × P R = < (4)(4) − (7)(8), − ((− 2 )(4) − (7)(3)), (− 2 )(8) − (4)(3) > →
→
P Q × P R = UsethisinformationtosolvetheAOT equationabove. →
→
AOT = 21 ||P Q × P R|| =
352
1 2
√(− 40)
2
+ (29)2 + (− 2 8)2 = 21 (56.79) ≈ 2 8.395
Section5-CalculusIII-M athQ RH
5.1.5
E quationso fL inesa ndP lanesi nS pace
5.1.5.a
ParametricCurves
Parametricequationsareasetofequations,oneforeachvariable,dependentonthe parameter. x = x (t) y = y (t) z = z (t) t istheparameter. Itisoftentimeandrestrictedtoacertainintervalt0 ≤ t ≤ tf . VectorForm:r→(t) = < x (t), y(t), z(t) > ,wherer→(t) isapositionvector. Aparametriccurveiswhatwegetwhenwetraceouteverypoint. TheequationofastraightlinethroughpointsP andQ is →
→
r→ = P + t P Q →
wheret isamultipleandparameterandP isapositionvector.
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353
SymmetricEquations: Solvefortandeliminatetheparameter. x (t) = − 1 + 5 t → t = y (t) = 2 t → t = z (t) = 3 − 4 t → t =
y 2
x + 1 5
3 − z 4
So,anotherwaytowritetheequationofalinein3Dis x + 1 = 5 5.1.5.b
y 2
=
3 − z 4
.
EquationofaPlane
Iftwovectorsareperpendicular,theirdotproductiszero:n→ · v→ = 0 . v→ = < x − x 0 , y − y 0 , z − z 0 > Therefore,theequationofaplaneis < a , b, c > · < x − x 0 , y − y 0 , z − z 0 > = 0 a (x − x 0 ) + b (y − y 0 ) + c (z − z 0 ) = 0 Thisistheequationoftheplanewithnormal(perpendicular)vector< a , b, c > throughthe pointP (x 0 , y 0 , z 0 ) .
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Section5-CalculusIII-M athQ RH
Notethatiftwoplanesareparallel,theyshouldhavethesamenormalvector. Inaddition tothis,vector< l, m, n > isparalleltothelinewithparametricequations x = a + lt y = b + mt z = c + n t Example:WriteanequationfortheplanecontainingthepointsP = (1, 1, − 2 ), Q = (0, 2, 1), andR = (− 1 , − 1 , 0) .
→
→
P R andP Q arebothintheplaneandweknowthatthecrossproductisperpendicularto bothofthevectors. Therefore,itisparalleltotheplane,meaningitisthenormalvector. →
→
So,findthevectorsP R andP Q andtheircrossproduct. →
→
P R = P Q = →
→
P R × P Q = < (− 2 )(3) − (2)(1), − ((− 2 )(3) − (2)(− 1 )), (− 2 )(1) − (− 2 )(− 1 ) > n→ = Now,youcanuseoneofthethreeofthepointstofindtheequationoftheplane. Inthis exampleIwillusepointP . − 8 (x − 1 ) + 4 (y − 1 ) − 4 (z + 2 ) = 0 → − 8 x + 4 y − 4 z − 4 = 0
Section5-CalculusIII-M athQ RH
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5.1.6
Q uadricSurfaces
ItmaybehelpfultoreviewtheConicSectionssectionofAlgebraII,asquadricsurfacesare reallyjusthigherdimensionalconicsections. Quadricsurfacesarethegraphsofequationsthatcanbeexpressedintheform Ax 2 + B y 2 + C z 2 + D xy + E xz + F yz + Gx + H y + J z + K = 0 Usuallymostofthecoefficientsareequaltozero. 5.1.6.a
Cylinders
Asetoflinesparalleltoagivenlinepassingthroughagivencurveisknownasacylindrical surface,orcylinder. Theparallellinesarecalledrulings.
This,forexample,isarightcircularcylinderthatisperpendiculartothex y plane. Anyequationwhichmissingx , y, orz willbeacylinder. ●
Tip:Drawtheequationin2Dandimaginetherulingscomingoutatyou(i.e.the2D equationisstackedupinthedirectionofthemissingvariable).
356
Section5-CalculusIII-M athQ RH
5.1.6.b
Traces/CrossSections
Thetracesofasurfacearethecrosssectionscreatedwhenthesurfaceintersectsaplane paralleltooneofthecoordinateplanes. Theequationwillbeoftheformv ariable = c onstant
It’skindoflikeaplaneis“slicing”throughthegraphandtheresultisa2dshape(i.e.oneof thevariablesisbeingheldconstant). 5.1.6.c x2 a2
+
y2 b2
+
Ellipsoid z2 c2
= 1
Traces: ●
inplanez = p :anellipse
●
inplaney = q :anellipse
●
inplanex = r :anellipse
Ifa = b = c ,thesurfaceisasphere. Graphingworksthesameasitdoesin2D. Goouta unitsinthex direction,b unitsinthe y direction,andc unitsinthez direction.
Section5-CalculusIII-M athQ RH
357
5.1.6.d
z=
x2 a2
+
Paraboloid
y2 b2
Traces: ●
inplanez = p :anellipse
●
inplaney = q :aparabola
●
inplanex = r :aparabola
Theaxisofthesurfacecorrespondstothelinearvariable,meaningthegraphwillopenup alongthenon-squaredaxis. 5.1.6.e x2 a2
+
y2 b2
−
HyperboloidofOneSheet z2 c2
= 1
Traces: ●
inplanez = p :anellipse
●
inplaney = q :ahyperbola
●
inplanex = r :ahyperbola
Theaxisofthesurfacecorrespondstothevariablewiththenegativecoefficient.
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Section5-CalculusIII-M athQ RH
5.1.6.f z2 c2
−
x2 a2
−
HyperboloidofTwoSheets y2 b2
= 1
Traces: ●
inplanez = p :anellipseornothing
●
inplaney = q :ahyperbola
●
inplanex = r :ahyperbola
Theaxisofthesurfacecorrespondstothevariablewiththepositivecoefficient. The surfacedoesnotintersectthecoordinateplaneperpendiculartotheaxis. Thegraphs intersecttheaxisofthesurfaceatthesquarerootofthepositivecoefficient. So,ifthe surface’saxisisthez axis,thegraphwillintersectthez axisatc and− c .
Section5-CalculusIII-M athQ RH
359
5.1.6.g x2 a2
+
y2 b2
−
EllipticCone z2 c2
= 0
Traces: ●
inplanez = p :anellipse
●
inplaney = q :ahyperbola
●
inplanex = r :ahyperbola
●
inthex z plane:apairoflines
●
inthey z plane:apairoflines
Theaxisofthesurfacecorrespondstothevariablewiththenegativecoefficient. Thetraces inthecoordinateplanesparalleltotheaxisareintersectinglines.
360
Section5-CalculusIII-M athQ RH
5.1.6.h
z=
x2 a2
−
y2 b2
HyperbolicParaboloid
Traces: ●
inplanez = p :ahyperbola
●
inplaney = q :aparabola
●
inplanex = r :aparabola
Theaxisofthesurfacecorrespondstothe linearvariable. Thisisaweirdgraph;when reallyzoomedoutitlookslikeasaddle.
5.2 DifferentiationofFunctionsofSeveral Variables 5.2.1
F unctionso fS everalVariables
Afunctionoftwovariablesz = f (x, y) mapseachorderedpair(x, y) insubsetD ofthereal planeR2 toauniquerealnumberz . ThesetD iscalledthedomainofthefunction. The rangeoff isthesetofallrealnumbersz thathasatleastoneorderedpair(x, y) ε D such thatf (x, y) = z .
Thisisa“2Dfunction”(astherearetwoinputvariables)thatcreatesa3Dgraph.
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5.2.1.a
LevelCurves
Givenafunctionf (x, y) andanumberc intherangef ,alevelcurveofafunctionoftwo variablesforthevaluec isdefinedtobethesetofpointssatisfyingtheequationf (x, y) = c .
5.2.1.b
ThreeVariableFunctions
w = f (x, y, z) Afunctionofthreevariablesisaquantitythatdependsonthreeothers. 5.2.1.c
LevelSurfaces
Givenafunctionf (x, y, z) andanumberc intherangeoff ,alevelsurfaceofafunctionof threevariablesisdefinedtobethesetofpointssatisfyingtheequationf (x, y, z) = c .
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Section5-CalculusIII-M athQ RH
5.2.2
L imitsandC ontinuity
5.2.2.a
Epsilon/DeltaDefinitionofaLimit(1D)
Letf (x) bedefinedforallx =/ a overanopenintervalcontaininga . LetL bearealnumber. lim f (x) = L
x → a
If,foreveryε > 0 ,thereexistsaδ > 0 ,suchthatif0 < |x − a | < δ ,then|f (x) − L| < ε .
5.2.2.b
LimitswithTwoorMoreVariables
Whendealingwithlimitsoffunctionswithtwoormorevariables,wecanapproachapoint (x, y) ininfinitelymanydirections. lim
(x, y) → (a, b)
f (x, y) = L
(x, y) gets“closeto”(a, b) thenf (x, y) gets“closeto”L . ●
if(a, b) isnotadiscontinuity,pluginvalues
●
if(a, b) isadiscontinuity,checkpaths(subiny = b andtakelimitforx ) ○
it'sbasicallyimpossibletofindthevalueoftheentirelimitthisway,butyou canfindthevalueofthelimitwhenavariableisheldconstant
●
usethesqueezetheorem(see:CalcI)
Section5-CalculusIII-M athQ RH
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5.2.2.c
LimitLaws
●
ConstantLaw:
lim
c =c
●
IdentityLaws:
lim
x = a ,
●
SumandDifferenceLaw:
●
ConstantMultipleLaw:
●
ProductLaw:
●
QuotientLaw:
●
PowerLaw:
●
RootLaw:
(x, y) → (a, b)
(x, y) → (a, b)
lim
(x, y) → (a, b)
y =b
(f (x, y) ± g (x, y)) = L ± M
lim
(x, y) → (a, b)
(c f (x, y)) = c L
lim
(x, y) → (a, b)
(f (x, y) g(x, y)) = LM
lim
(x, y) → (a, b)
f(x, y)
lim
(x, y) → (a, b) g(x, y)
=
L M
forM =/ 0
(f (x, y))n = Ln ,foranypositiveintegern
lim
(x, y) → (a, b)
f (x, y) = √L, forallL (x, y) → (a, b) √ lim
n
n
ifn isodd,forallL ≥ 0 ifn iseven
Example:Find
xy 2 2 (x, y) → (−1, −3) x + y
lim
2
.
Thereisonlyadiscontinuityattheoriginforthisfunction,sowecanplugin(− 1 , − 3 ) to findthevalueofthelimit. (−1)(−3 2 ) 2
(−1) + (−3)
So,
lim
(x, y) → (−1, −3)
xy 2 x2 + y 2
= −
9 10
2
= −
9 10
.
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Section5-CalculusIII-M athQ RH
5.2.3
P artialD erivatives
Rememberthatderivativestellyoutheslope,orinstantaneousrateofchange,atacertain point. In3D,slopedependsonthedirectionyougoin. Therefore,thereisnotjustone derivative,butthereisonepartialderivativeineachdirection. Letf (x, y) beafunctionoftwovariables. Thenthepartialderivativeoff withrespecttox , ∂f
writtenas ∂x orf x ,isdefinedas ∂f ∂x
= lim
h → 0
f(x + h, y) − f(x, y) h
Thisisreadas“d f dx ,”whichmeanstherateofchangeofthefunctionf withrespecttox . ∂f
Thepartialderivativeoff withrespecttoy ,writtenas ∂y orf y ,isdefinedas ∂f ∂y
f(x, y + k) − f(x, y) k k → 0
= lim
Noticehowonlyonevariableischanging,theotherisheldconstant. 5.2.3.a ∂f ∂x
= fx
AlternativeNotations ∂f
∂x = f x
or
“thex ory derivative”
Wecan’tusef ′ likewewouldforfunctionsofonevariable,asitdoesnotshowwhat variablewearederivingwithrespectto. You’llalsosee z = f (x, y) →
∂z ∂x
=
∂f ∂x
= zx
Section5-CalculusIII-M athQ RH
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5.2.3.b
CalculatingPartialDerivatives
Whencalculatingapartialderivative,usethesamerulesfromCalculusI,buttreatall variablesasconstante xceptthevariableforwhichwearetakingthederivative. Example:Calculatef x andf y off (x, y) = x 2 − 3 xy + 2 y 2 − 4 x + 5 y − 1 2 Let’sbeginbyfinding fx =
∂ 2 ∂x [x
− 3 xy + 2 y 2 − 4 x + 5 y − 1 2]
Pretendthaty isaconstant. So,ify = c andc = some number ∂ 2 ∂x [x
− 3 xc + 2 c 2 − 4 x + 5 c − 1 2]
Deriveasnormal,thenreplacec withy . f x = 2x − 3c − 4 f x = 2x − 3y − 4 Now,repeatthisprocesswithx = c tofindf y . ∂ 2 ∂y [c
− 3 cy + 2 y 2 − 4 c + 5 y − 1 2] = 0 − 3 x + 4 y + 0 + 5 + 0 f y = − 3x + 4y + 5
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Section5-CalculusIII-M athQ RH
5.2.3.c
PartialDerivativeswithThreeorMoreVariables
Letf (x, y, z) beafunctionofthreevariables. Thenthepartialderivativesoff withrespect tox ,y ,andz ,respectivelyare: ∂f ∂x
= lim
f(x + h, y, z) − f(x, y, z) h
∂f ∂y
= lim
f(x , y + k, z) − f(x, y, z) k k → 0
∂f ∂z
= lim
h → 0
m → 0
f(x, y, z + m) − f(x, y, z) m
Thiscanalsobedonewithevenhigherdimensions,justrepeatthepattern. 5.2.3.d
HigherOrderPartialDerivatives
Notation: ●
f xx → partialderivativewithrespecttox twiceorthesecondderivativewith respecttox
●
f yy → secondderivativewithrespecttoy
●
f xy → firsttakethex derivative,thentakethey derivative
●
f xyxx → firsttakethex derivative,thenthey derivative,thenx ,thenx again.
Therearemanydifferentpossibilities. YoucanalsouseLeibnizNotation: ∂2 f ∂x2
→ secondderivativewithrespecttox
∂2 f ∂xy
→ f yx (noticehowthevariablesareintheoppositedirection)
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5.2.3.e
Clairaut’sTheorem
Supposethatf (x, y) isdefinedonanopendiskD thatcontainsthepoint(a, b) . Ifthe functionsf xy andf yx arecontinuousonD ,thenf xy = f yx . Thisalsomeansthatf yxx = f xxy = f xyx . Inotherwords,orderdoesn’tmatter.
5.2.4 T angentP lanesandL inearApproximations 5.2.4.a
TangentPlanes
Rememberthataplaneisjustahigherdimensionalversionofaline. So,theequationfora tangentplaneisverysimilartotheequationofatangentline. LetS beasurfacedefinedbyadifferentiablefunctionz = f (x, y) ,andletP 0 = (x 0 , y 0 ) bea pointinthedomainoff . Then,theequationofthetangentplanetoS atP 0 isgivenby z = f (x 0 , y 0 ) + f x(x 0 , y 0 )(x − x 0 ) + f y (x 0 , y 0 )(y − y 0 )
Atangentplanetoasurfacedoesnotalwaysexistateverypointonthesurface. Justlikein CalcI,ifyouhavesharp/undefinedpoints,youcannothaveatangentplane. 368
Section5-CalculusIII-M athQ RH
5.2.4.b
LinearApproximations
Linearapproximationsinhigherdimensionsareplanes(tangentplanes,tobeexact). Givenafunctionz = f (x, y) withcontinuouspartialderivativesthatexistatpoint(x 0 , y 0 ) , thelinearapproximationoff atpoint(x 0 , y 0 ) isgivenbytheequation L(x, y) = f (x 0 , y 0 ) + f x(x 0 , y 0 )(x − x 0 ) + f y (x 0 , y 0 )(y − y 0 ) Example:Givenf (x, y) = √41 − 4 x 2 − y 2, approximatef (2.1, 2.9) using(2, 3) . Whatisthe approximatevalueof(2.1, 2.9) . Beginbyfindingf (2, 3), f x(2, 3) andf y (2, 3) . f (2, 3) =
√41 − 4(2 ) − 3 2
2
= √16 = 4
f x = 21 (41 − 4 x 2 − y 2 )−1/2 · (− 8 x) → f x(2, 3) = − 2 f y = 21 (41 − 4 x 2 − y 2 )−1/2 · (− 2 y) → f y (2, 3) = − 43 Now,plugintheknownstotheL(x, y) formulaandsimplify. L(x, y) = − 2 (x − 2 ) − 43 (y − 3 ) + 4 = − 2 x − 43 y +
41 4
Now,pluginx = 2 .1 andy = 2 .9 toapproximatethevalueoff (2.1, 2.9) . L(2.1, 2.9) = 3 .879 Theactualvalueoff (2.1, 2.9) = 3 .866522986 ,sotheapproximationisveryclose.
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5.2.4.c
Differentials
Letz = f (x, y) beafunctionoftwovariableswith(x 0 , y 0 ) inthedomainoff ,andletΔ x and Δ y bechosensothat(x 0 + Δ x, y 0 + Δ y) isalsointhedomainoff . Iff isdifferentiableatthe point(x 0 , y 0 ) ,thethedifferentialsd x andd y aredefinedas d x = Δ x
and
d y = Δ y
Thedifferentiald z ,alsocalledthetotaldifferentialofz = f (x, y) at(x 0 , y 0 ) ,isdefinedas d z = f x(x 0 , y 0 )dx + f y (x 0 , y 0 )dy whereΔ z isthechangeinoutputofthesurface,d z isthechangeinoutputofthetangent plane,andd z ≈ Δ z . 5.2.4.d
HelpfulFactwithNormalVectors
Becauseoftangentplanes,weknowthatthenormalvectorofafunctionf (x) througha point(x, y) is n→ = < f x, f y , − 1 >
370
Section5-CalculusIII-M athQ RH
5.2.5
C hainR ule
Saywehaveafunctionz = f (x, y) ,wherex = g (t) andy = h (t) . Wecanmakeadependency charttoshowthisrelationship.
Tofindthederivativeof dz (noticehowitisnotapartialderivative,asitencompassesthe dt entirefunction),followallthepathsdownthechainandaddthemup.
Multiplyderivativesthatcontainthesamevariable(i.e.multiplyderivativesthatcontainx byderivativesthatcontainx )andaddderivativesthatcontaindifferentvariables. The derivative dz willbe dt dz dt
= ( ∂z ∂x ·
dx ) dt
+ ( ∂z ∂y ·
dy dt )
Section5-CalculusIII-M athQ RH
371
Example:Findthederivative∂w wherew = x 2 + y 3 + z , x = u + v 2 , y = u − v , andz = 1 − 2 v . ∂u Beginbymakingadependencycharttokeeptrackofthevariouschainsofderivatives.
Followthechainfromw downtou andbuildthederivative∂w . ∂u
∂x ) + ( ∂w ∂u ∂y
∂y ) + ( ∂w ∂u ∂z
∂z ) ∂u
∂w ∂u
= ( ∂w · ∂x
∂w ∂u
= (2x)(1) + (3y 2 )(1) + (1)(0) = 2 x + 3 y 2
·
·
Now,findthederivatives.
Nowreplacethex andy ,puttingthederivativeintermsofu andv . ∂w ∂u
372
= 2 (u + v 2 ) + 3 (u − v )2
Section5-CalculusIII-M athQ RH
5.2.5.a
ImplicitDifferentiation
Assumethaty isafunctionofx . dy
Let6 x 2 − 1 0xy − 4 y 2 + 2 7x − 1 9y + 3 0 = 0 . Usepartialderivativestocalculate dx atthepoint (− 3 , 3) . Startbytakingthederivativewithrespecttox ofbothsides. d 2 dx (6x
− 1 0xy − 4 y 2 + 2 7x − 1 9y + 3 0) =
d dx (0)
dy dy 1 2x − 1 0(1 · y + x · dy dx ) − 8 y dx + 2 7 − 1 9 dx = 0 dy Solvefordy dx bymovingall dx termstoonesideandfactor. dy dy dy dy dy 1 2x − 1 0y − 1 0x dy dx − 8 y dx + 2 7 − 1 9 dx = 0 → − 1 0x dx − 8 y dx − 1 9 dx = − 1 2x + 1 0y − 2 7 dy
(− 1 0x − 8 y − 1 9) dx = − 1 2x + 1 0y − 2 7 →
dy dx
=
−12x + 10y − 27 −10x − 8y − 19
Plugin(− 3 , 3) . dy dx
= −3
Tip: dy
∂f/∂x
Ifyouhavef (x, y) ,then dx = − ( ∂f/∂y ) . ∂f/∂x
∂z Ifyouhavef (x, y, z) ,then ∂x = − ( ∂f/∂z )
Section5-CalculusIII-M athQ RH
373
5.2.6
D irectionalD erivativesandtheG radient
Thedirectionalderivativeistherateofchangeofheight(output)off aswetravelalongr (t) .
Onewayofdefiningthedirectionalderivativeisbyusingtheparametricequation r (t) = x→ + tu→ ,wherex→ isthepoint(x, y) wherewetakethederivative,u→ istheunitvector thatindicatesthedirectioninwhichwearegoing,andt istheparameter. f (r(t)) isthepath onthesurfaceabovetheline. Then,wecantakethederivative. So,thedirectional derivativeD u f is D uf = dy Thisdefinitionisthesameasf x dx dt + f y dt .
374
d dt [f (r(t))]t = 0
Section5-CalculusIII-M athQ RH
5.2.6.a
GradientVector
∇
∇
Givenafunctionf (x, y) ,wedefinethegradientvectoras f (x, y) = < f x, f y >, where f is readas“delf .” 5.2.6.b
Theorem
Aneasierversionofthedirectionalderivativeformulais D uf =
∇f (x, y) · u
whereD u f isreadasthederivativeoff inthedirectionofu . Noticethatwearetakingthe dotproductofthevectors. So,tofindthedirectionalderivativeoff (x, y) atthepoint(x 0 , y 0 ) inthedirectionof v = < a , b > … 1. Findu =
∇
v ||v||
(unitvectorofv )
2. Use f forD u f =
∇f (x , y ) · u 0
0
Example:Findthedirectionalderivativeatx = (1, − 1 ) inthedirectionofu = < 2 , − 3 > of f (x, y) = 2 x 3 + y 2 . Findtheunitvectorofthegivenvectorandfindthegradientvectorofthefunction. ||u|| =
√2
2
+ (− 3 )2 = √13 → unit vector =
∇f = < 6x , 2y > → f (1, − 1) = < 6(1) , 2(− 1) > = < 6, − 2 > 2
2
Section5-CalculusIII-M athQ RH
375
Now,usethedirectionalderivativeformula. D uf = Youcanfactorout
1
√13
∇f · u → D f = < 6, − 2 > · < √ →
2 , 13
u
−
3
√13
>
asamultipliertomakeiteasiertosolve.
D uf =
1
√13
(< 6 , − 2 > · < 2 , − 3 >) =
1
√13
(12 + 6 ) =
18
√13
5.2.6.c
PropertiesoftheGradient
Supposethefunctionz = f (x, y) isdifferentiableat(x 0 , y 0 ) . ● ●
∇ If∇f (x , y ) =/ 0, thenD f (x , y ) ismaximizedwhenu pointsinthesamedirection as∇f (x , y ). ThemaximumvalueofD f (x , y ) is||∇f (x , y )||. If f (x 0 , y 0 ) = 0 , thenD u f (x 0 , y 0 ) = 0 foranyunitvectoru→ . 0
0
○
●
∇
u
0
→
0
0
u
0
0
0
0
0
i.e.ifyouarelookingforthefastestincrease,followthegradient
If f (x 0 , y 0 ) =/ 0, thenD u f (x 0 , y 0 ) isminimizedwhenu→ pointsintheopposite
∇
directionas f (x 0 , y 0 ). TheminimumvalueofD u f (x 0 , y 0 ) is− || ○
∇f (x , y )||. 0
0
ifyouarelookingforthefastestdecrease,followthenegativeofthegradient
5.2.6.d
LevelCurveTheorem
Supposethefunctionz = f (x, y) hascontinuousfirstorderpartialderivativesinanopen
∇
∇
diskcenteredatapoint(x 0 , y 0 ) . If f (x 0 , y 0 ) =/ 0, then f (x 0 , y 0 ) isnormal(perpendicular) tothelevelcurveoff at(x 0 , y 0 ) . So,ifyouhavethelevelcurvediagramofafunctionf ,youcandrawthedirectionoffastest increasebymakingarightanglewiththelevelcurve.
376
Section5-CalculusIII-M athQ RH
5.2.6.e
HigherDimensionalGradients
Letw = f (x, y, z) beafunctionofthree(thiscanbegeneralizedforanyamountof
∇
dimensions)variablessuchthatf x, f y , andf z exist. Thevector f (x, y, z) i scalledthe gradientoff andisdefinedas
∇f (x, y, z) = f (x, y, z)i + f (x, y, z)j + f (x, y, z)k x
∇
y
z
∇
Inotherwords, w = < f x, f y , f z >. f (x, y, z) c analsobewrittenasg rad f (x, y, z) . Youcalculatethedirectionalderivativeoffunctionswiththreeormorevariablesthesame asyouwouldwithfunctionsoftwovariables.
5.2.7
M axima/MinimaP roblems
InCalculusI,wewouldfindminimumsandmaximumsbysettingthederivativesequalto zeroandsolving. Thepointsonthegraphwherethederivativeisequaltozeroarecalled criticalpoints. Tofindtheminimaandmaximaoftwovariableequations,westillusecriticalpoints. Rememberthatthetangentplanetoz = f (x, y) at(x 0 , y 0 , z 0 ) is z = f x(x 0 , y 0 )(x − x 0 ) + f y (x 0 , y 0 )(y − y 0 ) + z 0 Togetatangentplaneequaltozero,f x(x 0 , y 0 )(x − x 0 ) andf y (x 0 , y 0 )(y − y 0 ) mustbeequalto zero. Whenthishappens,z isacriticalpointc . Apoint(x 0 , y 0 ) iscalledacriticalpointoff (x, y) if
∇f (x , y ) = 0 (f →
0
0
x
= 0 a ndf y = 0 )
Section5-CalculusIII-M athQ RH
or
∇
f (x 0 , y 0 ) DN E (thisisrare)
377
Example:Findthecriticalpointsoff (x, y) = x 2 − 6 x + y 2 + 2 y + 1 2 .
∇f mustbeequaltothezerovector. Sofind∇f , seteachcomponentofthevectorequal tozero,andsolve.
∇f = < 2x − 6, 2y + 2 > 2 x − 6 = 0 → x = 3 2y + 2 = 0 → y = − 1 So,thecriticalpointofthefunctionis(3, − 1 ) . Tofindthez valueofthispoint,pluginthe pointtotheoriginalfunction:f (3, − 1 ) = 2 . Whenwegraphthisequation,weseethatthis criticalpointisthelocalandabsoluteminimum. 5.2.7.a
SecondDerivativeTest
Letz = f (x, y) beafunctionoftwovariablesforwhichthefirstandsecondorderpartial derivativesarecontinuousonsomediskcontainingthepoint(x 0 , y 0 ) . Suppose f x(x 0 , y 0 ) = 0 andf y (x 0 , y 0 ) = 0 . Definethequantity D = f xx(x 0 , y 0 )f yy (x 0 , y 0 ) − (f xy (x 0 , y 0 ))2 ●
ifD > 0 andf xx(x 0 , y 0 ) > 0 ,thenf hasalocalminimumat(x 0 , y 0 )
●
ifD > 0 andf xx(x 0 , y 0 ) < 0 ,thenf hasalocalmaximumat(x 0 , y 0 )
●
ifD < 0 ,thenf hasasaddlepointat(x 0 , y 0 ) ○
Asaddlepointisneitheramaximumoraminimum. Itisahigher dimensionalversionofaninflectionpoint.
●
378
ifD = 0 ,thenthetestisinconclusive
Section5-CalculusIII-M athQ RH
Example:Findthecriticalpointsofg (x, y) = x 2 + 3 xy + x + 6 y andusethesecondderivative testtofindoutwhatkindofcriticalpointitis. First,findthegradientvectorofg andseteachcomponenttozerotofindthecritical points.
∇g = < 2x + 3y + 1, 3x + 6 > criticalpoint= (− 2 , 1) Usethispointas(x 0 , y 0 ) . Now,findg xx, g yy , andg xy . g xx = 2 g yy = 0 g xy = 3 UsethesevaluestofindD . D = (2)(0) − 3 2 = − 9 SinceD < 0 ,thecriticalpoint(− 2 , 1) isasaddlepoint.
5.2.8
L agrangeM ultipliers
ThemethodofLagrangemultipliersisawayoffindingthelocalextremaofafunction f (x, y) subjecttoaconstraintequationg (x, y) = c .
∇ ∇
TheLagrangeequationis f = λ g . λ ,lambda,istheLagrangemultiplier. So,tofindtheextremaoff (x, y) subjecttoconstraintg (x, y) = c ,oneneedstosolvethe vectorequation
∇f = λ∇g Thisisreallyasystemofequations. TheLagrangemultiplierisonlyusedtohelpfindthe solution,wedon’treallycarewhatλ ismostofthetime.
Section5-CalculusIII-M athQ RH
379
Example:Findtheminimumandmaximumofthefunctionf (x, y) = x + 5 y (functiontobe optimized)onthecirclex 2 + y 2 = 1 6 (g (x, y) constraint). SetuptheLagrangeequation. < 1 , 5 > = λ < 2 x, 2y > Now,setupthesystemofequations:therewillbethreeequationsinthesystemaswe havethreeunknowns(x , y, andλ ). Togetthefirsttwo,distributethelambdathenset correspondingcomponentsequaltoeachother. Thethirdequationwillbetheoriginal constraint(g (x, y) ). 1 = λ2x 5 = λ2y x 2 + y 2 = 1 6 Youcanusesubstitution,elimination,orsomethingcreativetosolvethissystemof equations(youcouldalsouseanonlinecalculator—there’soneIrecommendinSection1 ofthisbook—butI’llworkitoutbyhandhere). Solveforx inthefirstequation,y inthesecondequation,thenplugtheseintothethird equation. Solveforλ .
5 = λ2y → y = 25 λ
1 = λ2x → x = 21 λ
26 ( 21 λ)2 + ( 25 λ)2 = 1 6 → 4λ12 + 4λ252 = 1 6 → 26 = 6 4λ2 → λ = ± √8
Noticehowtherearetwopossibilitiesforlambda. Plugbothoftheseintofindthe maximumandminimumvalues. λ = √826 → x = λ = − √826 → x = −
380
4
√26 4
√26
, y =
20
√26
, y = −
→ f (
20
√26
4 , 20 ) = 104 √26 √26 √26
→ f (−
4
√26
,−
20
√26
)= −
104
√26
Section5-CalculusIII-M athQ RH
So,themaximumvalueofthefunctionboundedbyx 2 + y 2 = 1 6 is 104 andtheminimum √26 valueis−
104
√26
.
5.2.8.a
NoteonExtrema
Inhigherdimensionalproblemsanabsolutemaximumw illoccurateither 1. acriticalpoint(seethesectionbeforethisone) 2. aboundarypoint(thisistheLagrangemethod) Therefore,whenlookingforabsolutemaximaoffunctions,checkusingbothofthese methods.
5.3
DoubleandTripleIntegrals
5.3.1
D oubleI ntegralso verR ectangularR egions
Rememberthatintegralsfindareasundercurves. Theintegralformulaisderivedfrom addinguptheareaofinfinitelymanyrectanglesundercurves. Itissimilarfor3Dgraphs,butinsteadoffindingtheareaunderacurve,wearefindingthe volumeunderasurface. So,insteadofrectangles,weareaddingupthevolumesof infinitelymanylittleboxes.
Section5-CalculusIII-M athQ RH
381
5.3.1.a
Fubini’sTheoremandIteratedIntegrals
TofindthevolumeabovetheregionR andbelowthefunctionf (x) ,wewillneedtousean
iteratedintegral. Aniteratedintegralissetupas∫ ∫ f (x, y) dA ,whereR istheregionweare R
integratingover,f (x, y) isthefunction,andd A isaninfinitesimalpieceofarea. Inthecase ofdoubleintegralsoverrectangularregionsd A caneitherbeequaltod xdy ord ydx . Forthegraphabove,tofindthevolumebetweenR andf (x) wewouldusethedouble integral x = d y = b
∫
∫
f (x, y) dydx
x = c y = a
y = b
Theinnerintegral
∫
f (x, y) dy findstheareaoftheverticalpieceandtheouterintegral
y = a
findsthevolume. Noticehowthevariableoftheinnerintegral’slimitsmatchthevariable oftheinnerdifferentialandviceversa. Thisisacrucialpartofiteratedintegrals. To evaluatethisintegral,startontheinsideandworkyourwayout.
382
Section5-CalculusIII-M athQ RH
Example:FindthevolumeaboveR = [0, 1] × [1, 2] andbelowf (x, y) = x 2 + y 2 . Let’sstartbydrawingoutR .
Now,setuptheintegral. Wecanuseeitherd ydx ord xdy ,asforthisproblemitdoesn’t matterwhichvariableweintegratewithrespecttofirst. Forthisexample,Iwillused ydx . x = 1 y = 2
∫
∫
(x 2 + y 2 ) dydx
x = 0 y = 1
Startbytakingtheinsideintegralandpartiallyintegratewithrespecttoy ,pretendingx isa constant. x = 1
∫
x = 0
y = 2
(x 2 y + 31 y 3 )|y = 1 dx =
x = 1
1
x = 0
0
∫
[(2x 2 + 31 (2 3 )) − (x 2 + 31 (1 3 )) dx = ∫ (x 2 + 37 ) dx
Now,evaluatetheremainingintegral. 1 3 3x
+ 37 x|01 =
8 3
Section5-CalculusIII-M athQ RH
383
5.3.1.b
PropertiesofDoubleIntegrals
Assumethatthefunctionf (x, y) andg (x, y) areintegrableovertherectangularregionR ;S andT aresubregionsofR ;andassumethatm andM arerealnumbers. 1. Thesumf (x, y) + g (x, y) isintegrableand
R
R
R
∫ ∫ [f (x, y) + g (x, y)] dA = ∫ ∫ f (x, y) dA + ∫ ∫ g (x, y) dA
2. Ifc isaconstant,thenc · f (x, y) isintegrable
R
R
∫ ∫ c · f (x, y) dA = c ∫ ∫ f (x, y) dA
3. IfR = S ⋃ T andS ⋂ T = 0 exceptanoverlapontheboundaries,then
R
S
T
∫ ∫ f (x, y) dA = ∫ ∫ f (x, y) dA + ∫ ∫ f (x, y) dA
4. Iff (x, y) ≥ g (x, y) for(x, y) inR ,then
R
R
∫ ∫ f (x, y) dA ≥ ∫ ∫ g (x, y) dA
5. Ifm ≤ f (x, y) ≤ M ,then
m · A(R) ≤ ∫ ∫ f (x, y) dA ≤ M · A(R) R
6. Inthecasewheref (x, y) canbefactoredasaproductofafunctiong (x) ofx only andafunctionh (y) ofy only,thenovertheregionR = {(x, y) | a ≤ x ≤ b , c ≤ y ≤ d } , thedoubleintegralcanbewrittenas
b
d
R
a
c
∫ ∫ f (x, y) dA = (∫ g (x) dx)(∫ h (y) dy)
384
Section5-CalculusIII-M athQ RH
5.3.2
D oubleIntegralso verG eneralRegions
Whatdowedoiftheregioniscurved? Therearetwotypesofregions: Domain D = {(x, y) | a ≤ x ≤ b , g 1 (x) ≤ y ≤ g 2 (x)} Integral
D = {(x, y) | h 1 (y) ≤ x ≤ h 2 (y), c ≤ y ≤ d }
b g2 (x)
∫ ∫
a g1 (x)
d h2 (y)
f (x, y) dydx
∫ ∫
ch1 (y)
f (x, y) dxdy
Graphed Region
y b oundedbyafunctionofx
x b oundedbyafunctionofy
Example:Findthevalueoftheintegral∫ ∫ 4 xy − y 3 dA ,whereD istheregionboundedby D
y = √x andy = x 3 . Drawouttheregionsowecanseewhichcaseitis.
Section5-CalculusIII-M athQ RH
385
Wecanuseeithercase,butbothequationsarealreadysolvedfory ,soused ydx . Tofind theupperboundofx :x 3 = √x → x = 1 . 1 √x
∫ ∫ (4xy − y 3 ) dydx
0 x3
Evaluatetheintegral. 1
1
∫ [2xy 2 − 41 y 4 |x√x] dx = ∫ (− 41 x2 + 41 x12 + 2 x2 − 2 x7 ) dx 3
0
[−
1 3 12 x
0
+
1 13 52 x
+ 32 x 3 − 41 x 8 ]01 =
55 156
Sometimes,weneedtoreversetheorderofintegrationifoneoftheintegralsisimpossible tosolve. 3 9
Example:Evaluatetheintegral∫ ∫ x 3 e y dydx . 3
0 x2
Theintegral∫ e y dy isimpossibletosolve. So,drawouttheboundsgivenandfindnew 3
boundsthatsatisfythed xdy case.
386
Section5-CalculusIII-M athQ RH
Now,setupthenewintegralandsolve. 9 √y
9
9
9
∫ ∫ x3 ey dxdy = ∫ 41 x4 ey |√0y dy = 41 ∫ ey ((√y )4 − 0 ) dy = 41 ∫ ey y 2 dy 3
0 0
3
0
3
0
3
0
Weneedtoperformau substitutionhere. u = y 3 du = 3 y 2 dy → y 2 dy = 31 du 1 12
729
∫
0
e u du =
1 u 729 12 e |0
=
1 729 12 (e
− 1)
5.3.3
D oubleIntegralsi nPolarC oordinates
Wecancreatepolarrectangleswithinequalitiesoftheform0 ≤ θ ≤ 2π , 1 ≤ r ≤ 2 .
Becauseofthis,wecansetupintegralswheretheregioniscircular. Inpolarcoordinates,d A = r drdθ . Unlikerectangularcoordinates,thedifferentialmustbein thisorder.
Section5-CalculusIII-M athQ RH
387
Example:Setupthedoubleintegraloff (x, y) = x 2 + y 2 + x overtheregion2π ≤ θ ≤
3π 2
,
2 ≤ r ≤ 3 . Converttheformulafromrectangulartopolarcoordinates. x 2 + y 2 = r 2 x = r cos θ f (r, θ) = r 2 + r cos θ Now,setuptheintegral. 3π 2
3
∫∫
π 2
(r 2
+ r cos θ) rdrdθ =
2
3π 2
3
π 2
2
∫∫
r 3 + r 2 cos θ drdθ
388
Section5-CalculusIII-M athQ RH
5.3.4
T ripleI ntegrals
Rememberthatd x isasmallchangeinx andd A isasmallpieceofarea. So,d V isasmall pieceofvolume. d V = d zdydx ord zdxdy (or4othercombinations)
5.3.4.a
TripleIntegralsoveraGeneralBoundedRegion
WecanfindthetripleintegraloverageneralboundedregionE inR3 . Thegeneral boundedregionsweconsiderareofthreetypes. First,letD beintheboundedregionthat isaprojectionofE ontothex y plane. SupposetheregionE inR3 hastheform E = {(x, y, z)| (x, y) ε D, u 1 (x, y) ≤ z ≤ u 2 (x, y)} Fortwofunctionsz = u 1 (x, y) andz = u 2 (x, y) ,suchthatu 1 (x, y) ≤ u 2 (x, y) forall(x, y) inD , asshown.
u2 (x, y)
E
D u1 (x, y)
Thismeansthat∫ ∫ ∫ f (x, y, z) dV = ∫ ∫ [
Section5-CalculusIII-M athQ RH
∫
f (x, y, z) dz] dA .
389
Example:Setupatripleintegraltofindthemassofatetrahedronwithvertices(0, 0, 0) , (2, 0, 0) ,(0, 0, 3) ,and(0, 4, 0) . Themassdensityfunctionisρ(x, y, z) = x 2 + y z . Graphtheequation:thishelpstovisualizethebounds.
Noticehowthelowerz boundisz = 0 andtheupperz boundistheplanethroughthe points(2, 0, 0), (0, 4, 0) ,and(0, 0, 3) . Sincetheseareallaxisintercepts,wecandothe followingtofindtheequationoftheplane. x 2
y
+4+
z 3
y
= 1 → z = 3 (1 − 2x − 4 )
Wenowhavetheinnerboundsoftheintegral. 3 3 3− 2 x− 4 y
∫∫ ∫
390
0
(x 2 + y z) dzdA
Section5-CalculusIII-M athQ RH
Tofindtheboundsforx andy ,graphthetrianglethatisonthex y planeintheoriginal graph.
Theequationofthelineconnectingthepoints(0, 4) and(2, 0 )isy = 4 − 2 x ,which,when wesolveforx ,isequaltox =
y − 4 −2
y
= 2 − 2 . So,thelowerx boundiszeroandtheupperx
y
boundisx = 2 − 2 . Theboundsfory are0 and4 . Withallthebounds,setuptheintegral. y
3 3 4 2− 2 3− 2 x− 4 y
∫ ∫ 0 0
∫
(x 2 + y z) dzdxdy
0
5.3.4.b
AverageValueofaFunctionofThreeVariables
Iff (x, y, z) isintegrableoverasolidboundedregionE withpositivevolumeV (E) ,thenthe averagevalueofthefunctionis f ave =
1 V (E)
∫ ∫ ∫ f (x, y, z) dV E
NotethatthevolumeisV (E) = ∫ ∫ ∫ 1 dV . E
Section5-CalculusIII-M athQ RH
391
5.3.5
T ripleI ntegralsandCylindrical/SphericalCoordinates
5.3.5.a
CylindricalCoordinates
Cylindricalcoordinatesarejustpolarcoordinatesfor(x, y) withz attached. x = r cos θ y = r sin θ z = z
Whencalculatingatripleintegralincylindricalcoordinates,usethedifferential d V = r dzdrdθ .
392
Section5-CalculusIII-M athQ RH
5.3.5.b
SphericalCoordinates
Sphericalcoordinatesaremadeupoftwoanglesandonelength. ●
ρ = distancefromtheorigin
●
θ = anglemeasurefromthepositivex axis,counterclockwise
●
ϕ = anglemeasurefromthepositivez axis,goingdown
RectangulartoSphericalTransformation: x = ρsin ϕ cos θ y
θ = tan −1 ( x )
y = ρsin ϕ sin θ ϕ = tan −1 (
√x2 + y2 ) z
z = ρcos ϕ ϕ = c os−1 (
z
√x2 + y2 + z2
)
x 2 + y 2 + z 2 = ρ2 Tointegratewithsphericalregions,usethedifferentiald V = ρ2 sin ϕ dρdϕdθ .
Section5-CalculusIII-M athQ RH
393
Example:Setuptheintegralthatfindsthemassofthesphereofradius3 attheorigin. ρ(x, y, z) = 2 x 2 + 2 y 2 + 2 z 2 . First,notethatrho(ρ )isbothmassdensitya nddistancefromtheorigin. Themassofthe spherewillbeequaltothetripleintegralofthemassdensityequation. Convertthemassdensityequationintosphericalcoordinates. ρ2 = x 2 + y 2 + z 2 mass density = 2 ρ2 Wearetoldthatthespherehasaradiusof3 ,andsincewearen’tgivenaninnerradius,we knowthattheboundsforρ are0 and3 . Thesphererangesfromthenorthpole(positive z axis)tothesouthpole(negativez axis). Therefore,theboundsforϕ gofrom0 toπ . Wearemakingafullcircleonthex y plane,sotheboundsforθ gofrom0 to2 π . Nowwecanfinishthesetupfortheintegral. 2π π 3
2π π 3
0 00
0 00
∫ ∫ ∫ (2ρ2 ) ρ2 sin ϕ dρdϕdθ = ∫ ∫ ∫ 2 ρ4 sin ϕ dρdϕdθ
Example:LetE betheregionboundedbytheconez = √7(x 2 + y 2 ) andthehemisphere z=
√10
2
− x 2 − y 2 . FindthevolumeofE .
Rememberthattofindthevolume,wesettheintegrandequaltoone. So,weonlyneedto findthelimitsofeachvariable.
394
Section5-CalculusIII-M athQ RH
Havingagraphoftheequationscanhelpwhensolvingthisproblem.
Startbyrearrangingthehemisphereequation. z 2 = 1 0 2 − x 2 − y 2 → x 2 + y 2 + z 2 = 1 0 2 → ρ2 = 1 0 2 → ρ = 1 0 10
Sincethegraphstartsattheorigin,weknowthat ∫ d p . 0
Forϕ ,itstartsatzero(northpole)andstopswhenitintersectswiththecone. However, sincethatangleisontheconeregardlessofthehemisphere,wecanjustusethecone equationtosolveforϕ . z = √7(x 2 + y 2 ) → use sphericaltransformations ρcos ϕ =
√7((ρsin ϕ cos θ)
2
+ (ρsin ϕ sin θ) 2
ρcos ϕ = √7ρ2 sin 2 ϕ [cos2 θ + sin 2 θ] = √7ρ2 sin 2 ϕ = √7 psin ϕ ρcos ϕ = √7 ρsin ϕ → cos ϕ = √7 sin ϕ → √1 = tan ϕ → ϕ = tan −1 ( √1 ) 7
Section5-CalculusIII-M athQ RH
7
395
tan−1 (1/√7)
So,
∫
d ϕ .
0
Thecoreandthehemisphereareradiallysymmetricso0 ≤ θ ≤ 2 π . Therefore, 2π tan−1 (1/√7) 10
∫
0
∫
0
∫ ρ2 sin ϕ dρdϕdθ
0
5.3.6
C enterso fMass
Thecenterofmassistheaveragepositionofthemassinanobject. Itisalsoknownasthe balancepoint. 5.3.6.a
OneDimensional
Aonedimensionalexampleoffindingthecenterofmassisfindingthecenterofmassina rod.
Inthisexample,therodismassless. Tocalculatethecenterofmass,useaweighted average. sum of weights multiplied by position total weight
396
→
3(−2) + 3(1) 6
= −
1 2
Section5-CalculusIII-M athQ RH
Iftherewasarodwithcontinuouslyvaryingmassdensitydefinedbyρ(x) ontheinterval [a, b] ,thecenterofmasswouldbecalculatedwiththeequation b
∫ xρ(x) dx
a b
∫ ρ(x) dx
a
5.3.6.b
TwoDimensional
Alaminaisaflat,2Dobject.
P isthecenterofmass(x, y) .
∫ ∫ x·ρ(x, y) dA
∫ ∫ y·ρ(x, y) dA
x = L
y = L
L
L
∫ ∫ ρ(x, y) dA
∫ ∫ ρ(x, y) dA
Thenumeratorsarecalledthex andy moments,respectively.
Section5-CalculusIII-M athQ RH
397
5.3.6.c
ThreeDimensional
∫∫∫ x·ρ(x, y, z) dV
∫∫∫ y·ρ(x, y, z) dV
∫∫∫ z·ρ(x, y, z) dV
x =
y =
z =
∫∫∫ ρ(x, y, z) dV
∫∫∫ ρ(x, y, z) dV
∫∫∫ ρ(x, y, z) dV
Example:Supposeyouhaveahalfball(bottomhalfofballofradius3 ). Themassdensityis ρ(x, y, z) = x 2 + y 2 . Whatisthecenterofmassoftheobject?
Wedon’tactuallyneedtodoallfourtripleintegrals. Thedensityincreasesaswemove awayfromtheorigin. Itdoesn’tdependonx andboththeequationandthegraphare symmetricwithrespecttox andy ,sox = 0 andy = 0 . Therefore,weonlyneedtofindz .
2π π 3
0 π/2 0
mass = ∫ ∫ ∫ ρ(x, y, z) dV →
∫ ∫ ∫ [(ρsin ϕ cos θ)2 + (ρsin ϕ sin θ)2 ] ρ2 sin ϕ dρdϕdθ
2π π 3
0 π/2 0
z moment = ∫ ∫ ∫ z ρ(x, y, z) dV →
∫ ∫ ∫ ρcos ϕ[(ρsin ϕ cos θ)2 + (ρsin ϕ sin θ)2 ] ρ2 sin ϕ dρdϕdθ
Tofindz ,dividethez momentbythemass.
398
Section5-CalculusIII-M athQ RH
5.4
VectorCalculus
5.4.1
V ectorFields
Avectorfieldisafunctionthatoutputsavectorateverypoint.
5.4.1.a
Notation
TwoDimensional:F (x, y) = < P (x, y), Q(x, y) > orF (x, y) = P (x, y)i + Q(x, y)j ThreeDimensional:F (x, y, z) = < P (x, y, z), Q(x, y, z), R(x, y, z) > orF (x, y, z) = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k
Section5-CalculusIII-M athQ RH
399
→
Example:SketchthevectorfieldF (x, y) = < 2y , 2x > . x
y
2
0
< 0 , 1 >
0
2
< 1 , 0 >
−2
0
< 0, − 1 >
0
−2
2
2
−2
2
< 1, − 1 >
−2
−2
2
−2
→
F (x, y) = < 2y , 2x >
< 1 , 1 >
Itisdifficulttogetagoodsketchbyhand,soyoushoulduseagraphingcalculatorlike GeoGebra(orDesmos,butI’vefoundthatGeoGebra’sareabitbetter). 5.4.1.b
UnitVectorField
Aunitvectorfieldisavectorfieldwhereallvectorshavelength1 .
400
Section5-CalculusIII-M athQ RH
5.4.1.c
GradientVectorField
Agradientvectorfieldisavectorfieldwhichisthegradientofascalarfunction. Example:f (x, y) = x 2 + y 2
∇
Thisfunction’sgradientfieldis f = < 2 x, 2y >.
Notethatthelevelcurvesoff areperpendiculartothevectorsinthegradientfield. Thisis helpfulforsketchinggradientfields.
5.4.2
L ineIntegrals
5.4.2.a
ScalarLineIntegral
Ascalarlineintegralisanintegraloftheform∫ f (x, y) dS ,wherec isthecurve,f (x, y) isthe c
lineardensity,andd S isasmallchunkof“string.” Whenperformingscalarlineintegrals, youwillhavetoparameterizethecurve.
Section5-CalculusIII-M athQ RH
401
Rememberthatr (t) istheparameterizationofacurve. In2D,d S =
√(
dx )2 dt
+ ( dy )2 dt . In3D,d S = dt
√(
dx )2 dt
+ ( dy )2 + ( dz )2 dt . dt dt
Example: Findthevalueoftheintegral∫(x 2 + y 2 + z 2 ) dS ,wherec ispartofthehelix c
parameterizedbyr (t) = < c os t, sin t, t >, 0 ≤ t ≤ 2 π . Wecanfindthemassofthecurvedstringbysolvingthis. First,findd S byfindingthe appropriatederivativesofr (t) . dS =
√(
dx 2 dt )
dy
2 + ( dt )2 + ( dz dt ) dt → dS =
√(− sint t)
2
+ (cos t)2 + 1 2 = √2 dt
Replaced S intheintegralwithwhatwejustfound.
c
c
∫(x2 + y 2 + z 2 ) dS = ∫(x2 + y 2 + z 2 ) √2 dt
Replacex , y, andz toputtheintegralintermsoft . Solve. 2π
2π
0
0
2 ∫ (cos2 t + sin 2 t + t) √2 dt = √2 ∫ (1 + t) dt = √2 [t + 21 t2 ]2π 0 = √2 [2π + 2 π ]
402
Section5-CalculusIII-M athQ RH
5.4.2.b
ArcLength
D
E
c
Rememberthat∫ ∫ 1 dA = a rea(D) and∫ ∫ ∫ 1 dV = v olume(E) . So,∫ 1 dS = length(c) . 5.4.2.c
VectorLineIntegrals-Work
Thesearealsocalledlineintegralsovervectorfields. Therearetwomainapplicationsfor thesetypesofintegrals:workandflux. Theequationsweusetocalculatethesetwothings lookverysimilar,soitisimportanttoreadthequestioncarefullywhendealingwithvector lineintegrals. Sayanobjectmovesalongacurver→(t) orc inaforcefield(avectorfieldwherevectors representforces)F (x, y) .
Addingupalltheworkfromthelittleforcesanddistancesgivesworkdonebytheforce field.
Section5-CalculusIII-M athQ RH
403
Calculatetheworkdonebythevectorfieldontheparticleasitmovesalongthecurvec .
→
→
WorkdoneisF · d r→ . Workdoneovertheentirecurveis∫ F · d r→ ,wherec representsthe c
parameterizedcurve. Weknowthatd r→ = (dx, dy) andr→(t) = < x (t), y(t) > . Sincedx dt = x ′(t), dx = x ′(t) dt . Similarly, d y = y ′(t)dt . Therefore,d r→ = (dx, dy) = < x ′(t) dt, y ′(t) dt > = r→ ′(t) dt . Insummation,c isacurveparameterizedbyr (t) ,t1 ≤ t ≤ t2 . Aparticletravelsalongcurvec inforcefieldF (x, y) ,r (t) = < x (t), y(t) > . ThetotalworkdonebyF ontheparticleis t2
t2
t2
t1
t1
t1
∫ F · d r = ∫ F (r(t)) · r ′(t) dt = ∫ P dx + Qdy
t2
In3D,W = ∫ P dx + Qdy + Rdz . t1
404
Section5-CalculusIII-M athQ RH
2
Example:Calculatetheworkdoneonaparticleasittravelsaroundtheellipse x4 +
y2 9
= 1 in
acounterclockwisedirection,startingat(2, 0) . F (x, y) = < 1 + x , 2 > (P = 1 + x andQ = 2 ).
Parameterizetheellipseequation(lookatthegraphintercepts). x = 2 cos t y = 3 sin t 0 ≤ t ≤ 2 π
Work= ∫ P dx + Qdy . P willbeequaltotheP componentofF witht substitutedand,d x c
willbeequaltox ′(t) dt . SinceQ hasnovariables,Q willstillbeequalto2 . d y willbeequal toy ′(t) dt . 2π
2π
0
0
∫ (1 + 2 cos t)(− 2 sin t) dt + (2)(3cos t) dt = ∫ [− 2 sin t − 4 cos t sin t + 6 cos t] dt
Integrateasusual. TheanswerisW = 0 .
Section5-CalculusIII-M athQ RH
405
5.4.2.d F lux =
f low time
VectorLineIntegrals-Flux . Fluxcanbemeasuredacrossacurve(2D)orasurface(3D). Intheseproblems,
thevectorfieldisavelocityfield. InsteadofF ,usev→ orϕ . v→ = v elocity f ield (P , Q) ϕ = g eneric vector f ield (P , Q) →
v = (x 2 , y) andr (t) = (t, 4 − t2 ) 0 ≤ t ≤ 2 Basedonthedirectionandorientationofthecurve(thearrowonthecurve),insideistothe leftoftheparticleandoutsideistotherightoftheparticle. Positivefluxgoesfrominto outandnegativefluxgoesfromouttoin. Thefluxintegralfindshowmuchisflowingfrom outtoin. Youcanprovethatfluxinthetinylengthd S isv→ · n→ dS .
c
c
Fluxintegralslooklike∫ v→ · n→ dS = ∫ P dy − Qdx
406
Section5-CalculusIII-M athQ RH
Usethisformulatocalculatethefluxoftheabovefield. v = (x 2 , y) ,r (t) = (t, 4 − t2 ), 0 ≤ t ≤ 2 x (t) = t → dx = 1 dt y(t) = 4 − t2 → dy = − 2 t dt 2
2
0
0
∫ (t2 )(− 2 ) dt − (4 − t2 ) dt = ∫ [− 2 t3 − 4 + t2 ] dt = − 403
5.4.2.e
Tip:ParameterizingaLineSegmentin3D
Aneasywaytoparameterizealinesegmentin3Diswiththefollowingequations. x = (1 − t)x 1 + tx 2 y = (1 − t)y 1 + ty 2 z = (1 − t)z 1 + tz 2 0 ≤ t ≤ 1
5.4.3
C onservativeV ectorFields
5.4.3.a
Simple/ClosedCurves
Acurveiscalledclosedifitformsaloop. Acurveiscalledsimpleifitdoesnotintersect itself.
Section5-CalculusIII-M athQ RH
407
5.4.3.b
ConservativeFieldTheorem
Aconservativefieldisoneinwhichthepathtakendoesnotmatter(i.e.itispath independent). So,ifwearefindingworkoveravectorfieldandchangingthepathfromA toB doesnotchangethevalueofworkthatweget,thevectorfieldisconservative. Ifϕ isaconservativevectorfieldthen,
∫ ϕ · dr = 0 c
∮
Wecanalsousethenotation
c
forallclosecurvesc
∮
ϕ · d r = 0 .
indicatesalineintegraloverasimpleclosed
curveincounterclockwisefashion. b
a
a
b
Thistheoremworksbecause∫ f (x) dx = − ∫ f (x) dx . Notethatthereisamuchlongerproof thatusuallyaccompaniesthis. Itisnotnecessarytomemorizetheproof,butitcanbe helpfultolookintoit. 5.4.3.c
FundamentalTheoremofLineIntegrals
Allgradientfieldsareconservative. Infact,ifϕ =
∇f then
∫ ϕ · d r = f (B) − f (A) c
A = r (t0 ) B = r (t1 ) 408
Section5-CalculusIII-M athQ RH
Proof:
t1
t1
c
t0
t0
∫ ϕ dr = ∫ ϕ (r(t)) · r (t) dt = ∫
∇f (r(t)) · r(t) dt
Noticehowthelastintegrandlookslikethechainrule. t1
∫
t0
d dt [f (r(t))] dt
= f (r(t1 )) − f (r(t0 )) = f (B) − f (A)
f iscalledtheantigradientorpotentialfunctionforthevectorfield. Notallvectorfields havepotentialfunctions,onlyconservativeonesdo. 5.4.3.d
Theorem
Ifavectorfieldisconservative,thenP y = Qx . Thiscanbeusedasaquickcheckto determineifavectorfieldisconservative. 5.4.3.c
FindingPotentialFunctions
FindapotentialfunctionforF (x, y) = < e xy 3 + y , 3e xy 2 + x > . First,checkifit’sconservative(P y = Qx) . Py =
d x 3 dy [e y
+ y ] = 3 y 2 e x + 1 Qx =
d x 2 dx [3e y
+ x] = 3 y 2ex + 1
Nowthatweknowit’sconservative,wecanfindthepotentialfunction. F =
∇f = < f , f x
y
> ,soF x = e xy 3 + y andF y = 3 y 2 e x + x
PartiallyintegrateF x withrespecttox tofindf . f = e xy 3 + x y + C (y)
Section5-CalculusIII-M athQ RH
409
WeneedtoaddintheC (y) aswhenwedifferentiatewithrespecttox ,weloseanything thatonlyhasay term. So,thistermmakesupforanypotentialloss. Nowdifferentiatef withrespecttoy . WedothistofindthevalueofC (y) . f y = 3 y 2 e x + x + C (y) Thisf y termneedstomatchtheaboveF y term. Inthiscase,C (y) = 0 ,aswedonotneed toaddanythingtomakef y = F y . Therefore, f = e xy 3 + x y
5.4.4
G reen’sTheorem
5.4.4.a
ConnectivityofRegions
AregionD isaconnectedregionif,foranytwopointsP 1 andP 2 ,thereisapathfromP 1 toP 2 withatracecontainedentirelyinsideD . AregionD isasimplyconnectedregionif D isconnectedforanysimpleclosedcurveC thatliesinsideD ,andcurveC canbe shrunkcontinuouslytoapointwhilestayingentirelyinsideD . Intwodimensions,aregion issimplyconnectedifitisconnectedandhasnoholes.
410
Section5-CalculusIII-M athQ RH
Ifyoushrinkthesimpleclosedcurveinthesimplyconnectedregion,itwillalwaysbeinthe region. However,ifyoushrinkthesimpleclosedcurveintheconnectedregiondiagram,it willleavetheregionatsomepoint. Thelastdiagramisnotconnected. Noticehowno matterhowyoudrawthepathfromP 1 toP 2 ,itwillhavetoleavetheregion. 5.4.4.b
PiecewiseSmooth
Acurveiscalledpiecewisesmoothifitismadeupoffinitelymanysmoothpieces(i.e.it’s notafractal). 5.4.4.c
Green’sTheorem
LetDbeanopen,simplyconnectedregionwithaboundarycurveC thatisapiecewise smooth,simpleclosedcurveorientedcounterclockwise. LetF = < P , Q > beavectorfield withcomponentfunctionsthathavecontinuouspartialderivativesonD . Then,
∮
∮
F · dr =
C
C
P dx + Qdy = ∫ ∫ (Qx − P y ) dA D
Itneedstobeinthisorder. Recognizethattheinterimequationistheworkdoneona particle;thisisimportanttonoticesothatyoudon’tsolvetheproblemincorrectly(seethe fluxexamplebelow). Ifitisclockwiseinsteadofcounterclockwise,addanegativesign. Thistheoremalsoappliestoregionswithfinitelymanyholes.
Section5-CalculusIII-M athQ RH
411
∮
Example:Calculatethelineintegral
x 2 y dx + (y − 3 ) dy whereC isarectanglewithvertices
C
(1, 1), (4, 1), (4, 5), and(1, 5) orientedcounterclockwise. WithoutGreen’sTheorem,we’dhavetodofourdifferentlineintegrals.
∮
C
P dx + Qdy = ∫ ∫ (Qx − P y ) dA D
P = x 2 y andQ = y − 3 .
∫∫
D
d dx (y
− 3) −
d 2 dy (x y) dA
D
= ∫ ∫ (0 − x 2 ) dA = − ∫ ∫ x 2 dydx
Tofindthelimits,lookatthegivenverticesandfindboththelowestandhighestx andy coordinates. Thesearethelowerandupperlimits,respectively. 45
4
4
11
1
1
− ∫ ∫ x 2 dydx = − ∫ [x 2 y]15dx = − ∫ 4 x 2 dx = − 34 [x 3 ]14 = − 8 4
412
Section5-CalculusIII-M athQ RH
FluxExample:UseGreen’sTheoremtocalculatefluxofv→ = < 6 y − 9 x, yx − x 3 > acrosscurve C .
C
C
Flux= ∫ P dy − Qdx = ∫ (6y − 9 x) dy − (yx − x 3 ) dx . Green’sTheoremstatesthat
C
D
∫ P dx + Qdy = ∫ ∫ (Qx − P y) dA . Noticethatthelefthandsideoftheequationdoesn’tmatch
theintegralforflux. So,rearrange.
C
C
∫ (6y − 9 x) dy − (yx − x3 ) dx = ∫ (− y x + x3 ) dx + (6y − 9 x) dy
Now,P = − y x + x 3 andQ = 6 y − 9 x . WhenusingGreen’sTheoremforfluxintegrals,be carefulnottomixupP andQ .
Section5-CalculusIII-M athQ RH
413
D
D
∫ ∫ [ dxd [6y − 9 x] − dyd [− y x + x3 ]] dA = ∫ ∫ [− 9 − (− x)] dA = ∫ ∫(x − 9 ) dA
Seethegraphonthepreviouspagetosetupthelimitsofintegration. 1 3−x
∫ ∫
(x − 9 ) dydx = −
−1 −1
218 3
5.4.5
D ivergence
Let’sstartwithavectorfieldϕ = < P , Q > . →
Convergenceisalsocalled“negativedivergence.” Thedivergenceofavectorfield,denotedd iv(ϕ) ,isascalarquantitythatmeasuresthis. →
→
→
414
∮
1 |v| → 0 |v| ∂v
d iv(ϕ)|x0 = lim
v · n dS
Section5-CalculusIII-M athQ RH
where|v| isthevolume(orareafor2D). Theintegralcalculatesthefluxoutsideofthe circleand|v|1 makesitflux/volume. Itcanbeproventhat d iv(ϕ) =
∇ · ϕ, whereϕ = < P , Q >
∂ ∂ = ( ∂x , ∂y ) · (P , Q) =
In3D(ϕ = < P , Q, R > ),d iv(ϕ) =
∂P ∂x
+
∂Q ∂y
+
∂R ∂z
∂P ∂x
+
∂Q ∂y
.
Notethatallfluxintegralsacrossclosedcurveswillbezeroandthatmagneticfieldswill havezerodivergence. Example:Findthedivergenceofϕ(x, y) = < x cos y, − sin y > .
∇·ϕ =
∂ ∂x (xcos y)
+
∂ ∂y (− sin y)
= c os y − c os y = 0
Sinced iv(ϕ) = 0 ,thisvectorfieldiscalleddivergencefree. Nomatterwhereandwhat shape,therewillbenodivergence.
Section5-CalculusIII-M athQ RH
415
5.4.6
C url
Curlattemptstomeasurethespinatapoint. c url(ϕ) isavectorwherethedirectionisthe axisofrotationandthemagnitudeisthemagnitudeofrotation. Ifthevectorfieldisspinningcounterclockwise,thevectorwillpointupward. Forexample, thecurlvectorat(0, 0) pointsupoutofthepageforthevectorfieldϕ = (pictured below).
∇ × ϕ . ∇×ϕ =(
Tocalculatec url(ϕ) ,perform
∂ ∂ ∂ ∂x , ∂y , ∂z )
× (P , Q, R)
Fora2Dvectorfield,R = 0 . Expandedout:
∇ × ϕ = (
416
∂R ∂y
−
∂Q ∂P ∂z , ∂z
−
∂R ∂Q ∂x , ∂x
−
∂P ∂y
)
Section5-CalculusIII-M athQ RH
Example:ϕ(x, y, z) = < x 2 z, e y + x z, xyz > . Findthecurlofthevectorfield. c url(ϕ) =
∇ × ϕ = < x z − x , x
2
− y z, z − 0 > = < x z − x , x 2 − y z, z >
Curlwillbedifferentdependingonthepointonthegraph. c url ϕ (1, 1, 1) = < 0 , 0, 1 > c url ϕ (1, 2 , 2) = < 1 , − 3 , 2 >
5.4.7
P arametricS urfaces
Curvesareparameterizedwithoneparameter,usuallyt . Forsurfaces,weneedtwo parameters. 2Dparametricsurfaces: x = x (u, v) y = y (u, v) z = z (u, v) r→(u, v) = < x (u, v), y(u, v), z(u, v) > u 0 ≤ u ≤ u 1 v 0 ≤ v ≤ v 1 Wecanparameterizeafunctionz = (x, y) withr→(u, v) = < u , v, f (u, v) > . Youcanalsoparameterizespherical“chunks”usingx = ρsin ϕ cos θ, y = ρsin ϕ sin θ, and z = ρcos ϕ . User→(u, v) = < sin u cos v, sin u sin v, cos v > andthen,bychangingranges,you cangetdifferentpartsofasphere.
Section5-CalculusIII-M athQ RH
417
5.4.8
S urfaceI ntegrals
Foraquickreview: Name
Integral
integral
b
∫ f (x) dx a
doubleintegral
DomainofIntegration overasectionofthe numberline
∫ ∫ f (x, y) dA
“double”theamountof integration;twonumber lines
D
tripleintegral
∫ ∫ ∫ f (x, y, z) dV
domainofintegrationis3D
E
lineintegral
∫ P dx + Qdy
C
domainofintegrationisa line
Asurfaceintegral’sdomainofintegrationisoverabendysurface. 5.4.8.a
SurfaceArea
Thereisatransformationbetweenareapieced A (smallpieceofareaona2Dsurface)and areapieced S (smallpieceofareaonacurved3Dsurface). Itfollowstheequation d S = ||tu × tv || dA wheretu =
andtv =
. Notethatt standsfortangent.
Section5-CalculusIII-M athQ RH
Theareaofthisparallelogramisthemagnitudeoftu × tv .
IntegralFormula:∫ ∫ ||tu × tv || dA D
Example:Findthesurfaceareaofthesurfacewithparameterization r (u, v) = < u + v , u 2 , 2v >, 0 ≤ u ≤ 3 , 0 ≤ v ≤ 2 . Findtu andtv . tu = < 1 , 2u, 0 > tv = < 1 , 0, 2 > Findtheintegrand||tu × tv || . tu × tv = < 4 u, − 2 , − 2 u > → ||tu × tv || =
√(4u)
2
+ (− 2 )2 + (− 2 u)2 = √20u 2 + 4
Now,integrate. 23
∫ ∫ √20u 2 + 4 dudv 00
You’dhavetousetrigonometricsubstitutiontosolvethisintegral(u =
1 tan θ ). √5
Section5-CalculusIII-M athQ RH
419
5.4.8.b
ScalarSurfaceIntegrals
S
D
∫ ∫ f (x, y, z) dS = ∫ ∫ f (r(u, v)) ||tu × tv|| dA
d S isaninfinitesimalsurfacepiece, and||tu × tv || dA istheareaofaparallelogramonthe surface. Whencalculatingthese,replaceeverythinginthef equationwithu ’s andv ’s. Example:Aflatsheetofmetalhastheshapeofsurfacez = 1 + x + 2 y thatliesabovethe rectangle0 ≤ x ≤ 4 and0 ≤ y ≤ 2 . Ifthedensityofthesheetisgivenbyρ(x, y, z) = x 2 yz , whatisthemassofthesheet? Sincethesurfaceequationissolvedforz andpassestheverticallinetest,wecanusethe shortcuttoparameterizethefunction. x = u y = v z = 1 + u + 2 v → r(u, v) = < u , v, 1 + u + 2 v > Nowfindtu , tv , tu × tv , and||tu × tv || . tu = < 1 , 0, 1 > tv = < 0 , 1, 2 > tu × tv = ||tu × tv || =
√(− 1)
2
+ (2)2 + (1)2 = √6
Replacex , y, andz inthemassdensityequation. ρ(x, y, z) = x 2 yz → f (r(u, v)) = u 2 v(1 + u + 2 v) Integrate. 24
√6 ∫ ∫ u 2 v(1 + u + 2 v) dudv 00
420
Section5-CalculusIII-M athQ RH
5.4.8.c
OrientationofaSurface
t × t
u nit normal = N = ||tu × tv || (length1) u v Thenormalvectorpointsfromintoout. Thereisbothpositiveandnegativeorientation: positiveisoutwardpointingnormal. Assumepositiveorientationunlessotherwise mentioned. Forflux,ifitisgoingwiththearrowitispositive. Ifitisgoingagainstthearrow,itis negativeflux. 5.4.8.d
SurfaceIntegralinaVectorField
Thisis3Dflux. Youcanimagineitisavelocityoffluidsflowingthroughasurface. What portion(volume)ofthefluidflowsthroughasurfaceperunitoftime? LetF = (P , Q, R) beacontinuousvectorfieldwithadomainthatcontainstheoriented surfaceS withaunitnormalvectorN . ThesurfaceintegralofF overS is
S
D
∫ ∫ F · N dS = ∫ ∫ (F (r(u, v)) · (tu × tv)) dA
Section5-CalculusIII-M athQ RH
421
Example:Calculatethesurfaceintegral∫ ∫ F · N dS whereF = andS isthe S
surfacewithparameterizationr (u, v) = < u , v 2 − u , u + v > ,0 ≤ u ≤ 3 ,0 ≤ v ≤ 4 . Findtu , tv , andtu × tv . tu = < 1 , − 1 , 1 > tv = < 0 , 2v, 1 > tu × tv = →
At0 ,tu × tv = Replacex , y, andz inF . F = → F (r(u, v)) = Nowintegrateusingthe3Dfluxformula.
∫∫
D
< u − v 2 , u, 0 >· dA = ∫ ∫ [(u − v 2 )(− 1 − 2 v) − u ] dA D
43
∫ ∫ [(u − v 2 )(− 1 − 2 v) − u ] dudv 00
Shortcut: Ifthesurfaceofintegrationisapieceofafunctionz = f (x, y) ,wecandevelopashortcutfor settingupsurfaceintegrals. Ifthefunctionpassestheverticallinetest,
D
D
∫ ∫ (F (r(u, v)) · (tu × tv)) dA = ∫ ∫ [− P · f x − Q · f y + R] dA
422
Section5-CalculusIII-M athQ RH
5.4.9
S tokes’T heorem
Stokes’TheoremisthehigherdimensionalversionofGreen’sTheorem. LetS beanorientedsmoothsurfacethatisboundedbyasimple,closed,smooth boundarycurvec withpositiveorientation. Also,letF beavectorfield,then,
∫ F · dr = ∫ ∫ ∇ × F · dS
c
S
c
S
where∫ F · d r isthecirculationorworkaroundaboundarycurveand∫ ∫
∇ × F · dS isthe
surfaceintegralofacurve.
Thisislikeahemispherewherethe“crust”isaboundarycurve.
Section5-CalculusIII-M athQ RH
423
5.4.9.a
Example
→
→
→
→
→
UseStokes’Theoremtoevaluate∫ F · d r→ whereF = z 2 i + y 2 j + x k andc isthetrianglewith c
vertices(1, 0, 0) ,(0, 1, 0) ,and(0, 0, 1) withcounterclockwiserotation.
→
→
→
Insteadofusing∫ F · d r→ ,use∫ ∫ c urlF · d S . c
S
→
c url (F ) =
∇ × F , where∇ = (
∂ , ∂ , ∂ ) ∂x ∂y ∂z andF
= (x 2 , y 2 , x)
∇ × F = ϕ = (0, 2z − 1, 0) Now,parameterizetheplane. x 1
y
+1+
z 1
= 1 → z = 1 − x − y
r→(u, v) = (u, v, 1 − u − v )
424
Section5-CalculusIII-M athQ RH
Graphtheequationoftheplaneonthex y plane. Thisisdonebygraphingthex andy interceptsandignoringz . Thisisalsoknownasthe“shadow”oftheplane. Noticethatthis isthesameasgraphingtheequationontheu v plane.
Theequationofthislineisy = 1 − x orv = 1 − u . Wewillusethisequationfortheupperv boundofthed v integral.
→
→
Now,use∫ ∫ c urlF · d S . S
→
→
∫ ∫ curlF · d S = ∫ ∫ ϕ (r(u, v)) · (tu × tv) dA S
D
Rememberthatweneedtouse(r(u, v)) · (tu × tv ) dA ,asthisisasurfaceintegral,whichwas coveredintheprevioussection. Findtu × tv . tu = < 1 , 0, − 1 > tv = < 0 , 1, − 1 > tu × tv = < 1 , 1, 1 > PluginallknownsintothedoubleintegraloverD andsolve.
∫∫
D
ϕ (r(u, v)) · (tu × tv ) dA = ∫ ∫ (0, 2(1 − u − v ), 0) · < 1 , 1, 1 > dA D
Addinlimitsandsimplify. 1 1−u
∫ ∫
[1 − 2 u − 2 v] dvdu
0 0
Whenfullysolved,thefinalansweris− 61 .
Section5-CalculusIII-M athQ RH
425
5.4.10
D ivergenceTheorem
LetE beasimplesolidregionandS istheboundarysurfaceofE withpositive orientation. LetF beavectorfieldwhosecomponentshavecontinuousfirstorderpartial derivatives. Then,
→
→
→
∫ ∫ F · d S = ∫ ∫ ∫ d iv F dV S
→
→
E
→
where∫ ∫ F · d S isthefluxacrossthesurfaceand∫ ∫ ∫ d iv F dV isthetripleintegralofthe S
E
divergence.
S isaclosedsurfacethatcompletelysurroundsthesolidregion.
426
Section5-CalculusIII-M athQ RH
5.4.10.a
Example
→
→
→
→
→
→
Usethedivergencetheoremtoevaluate∫ ∫ F · d S whereF = x yi − 21 y 2 j + z k andthesurface S
consistsofthreesurfaces:z = 4 − 3 x 2 = 3 y 2 , 1 ≤ z ≤ 4 onthetop,x 2 + y 2 = 1 , 0 ≤ z ≤ 1 onthe sides,andz = 0 onthebottom.
→
→
Use∫ ∫ ∫ d iv F dV . First,findd iv F . E
d iv F =
∇ · F → div F = (
∂ ∂ ∂ ∂x , ∂y , ∂z )
∫ ∫ ∫ 1 dV E
· (xy, − 21 y 2 , z) = 1
Usingcylindricalcoordinates,findtheboundsoftheintegrals. r oof = 4 − 3 x 2 − 3 y 2 = 4 − 3 r 2 unitcircleonx y plane(x 2 + y 2 = 1 ) So, 2π 1 4−3r2
∫∫ ∫
0 0
0
r dzdrdθ
Thisismucheasiertocalculatethanthreesurfaceintegrals.
Section5-CalculusIII-M athQ RH
427
“Mathematicsisagameplayedaccordingtocertainsimpleruleswith meaninglessmarksonpaper!” -DavidHilbert
5.5
428
MyNotesforCalculusIII
Section5-CalculusIII-M athQ RH
5.5
MyNotesforCalculusIII(con’t)
Section5-CalculusIII-M athQ RH
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5.5
MyNotesforCalculusIII(con’t)
430
Section5-CalculusIII-M athQ RH
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