Integrated Engineering Calculus I & II

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Integrated Engineering Calculus I & II

Table of contents :
Linear Regression......Page 11
Finding the Regression Parameters......Page 14
Judging the Strength of the Relationship......Page 15
Target Practice......Page 18
Exercises......Page 20
Solutions to Target Practice......Page 23
Classes of Functions......Page 25
Linear Function......Page 26
Quadratic Function......Page 27
Other Polynomial Functions......Page 28
Exponential Function......Page 29
Sine and Cosine Functions......Page 31
Target Practice......Page 32
Exercises......Page 33
Properties of Basic Functions......Page 37
Shifts and Stretches......Page 39
Target Practice......Page 43
Exercises......Page 44
Solutions to Target Practice......Page 47
Slope and Rate of Change......Page 49
Target Practice......Page 51
Exercises......Page 52
Limits and the Derivative......Page 54
Target Practice......Page 56
The Derivative as a Function......Page 57
Target Practice......Page 58
Exercises......Page 59
Some Terminology and Notation......Page 61
Exercises......Page 62
The Derivative of a Sine Function......Page 64
Sine Functions with Different Frequencies......Page 66
Sines and Cosines......Page 67
Exercises......Page 68
Higher Order Derivatives......Page 69
More Derivatives......Page 70
Exercises......Page 71
Solutions to Target Practice......Page 72
Polynomials......Page 77
Exercises......Page 81
Extending the Power Rule......Page 83
Target Practice......Page 84
Exercises......Page 85
Derivatives of sin(x) and cos(x)......Page 86
Exercises......Page 90
Derivative of an Exponential Function......Page 91
Target Practice......Page 92
Exercises......Page 93
Derivative of a Logarithm Function......Page 94
Exercises......Page 95
Hyperbolic Functions......Page 97
Exercises......Page 99
Solutions to Target Practice......Page 101
The Chain Rule......Page 103
Exercises......Page 105
The Product Rule......Page 107
Target Practice......Page 108
The Quotient Rule......Page 109
Exercises......Page 110
Exercises......Page 114
Solutions to Target Practice......Page 116
Velocity and Displacement......Page 119
The Relationship between Velocity and Position......Page 121
Exercises......Page 122
Approximating the Area Under the Graph of a Function......Page 123
Target Practice......Page 125
The Exact Area Under the Graph of a Function......Page 126
The Definite Integral......Page 127
Exercises......Page 128
Solutions to Target Practice......Page 131
Antiderivatives......Page 133
Exercises......Page 135
Initial Value Problems......Page 136
Target Practice......Page 138
Exercises......Page 139
Differential Equations of Higher Order......Page 140
A Second Order Initial Value Problem......Page 141
Quick Introduction to Matrices......Page 142
Exercises......Page 144
Antiderivatives and the Definite Integral......Page 146
Exercises......Page 149
The Fundamental Theorems......Page 150
Piecewise-defined functions......Page 151
Airy functions......Page 153
Laguerre Polynomials......Page 154
Solutions to Target Practice......Page 155
Two Quantities Changing with Time......Page 157
Exercises......Page 160
Differentiating Implicitly......Page 162
Exercises......Page 166
Exercises......Page 168
Solutions to Target Practice......Page 170
Finding the Best Strategy......Page 173
Local Maxima and Minima......Page 175
Exercises......Page 177
Deciding Whether an Extreme Value is a Maximum or Minimum......Page 180
Exercises......Page 182
Another Type of Critical Point......Page 184
Absolute Maxima and Minima......Page 185
Exercises......Page 186
Inflection Points......Page 187
Curve Sketching......Page 188
Exercises......Page 189
Solutions to Target Practice......Page 190
Changing Variables in Integrals......Page 193
Exercises......Page 196
Solving Differential Equations Exactly......Page 199
Exercises......Page 203
Substitution in Definite Integrals......Page 206
Exercises......Page 207
Solutions to Target Practice......Page 209
Integration By Parts......Page 213
Exercises......Page 215
Integration using Trigonometric Substitution......Page 217
Exercises......Page 219
Exercises......Page 221
Integration by Partial Fractions......Page 223
Target Practice......Page 224
Exercises......Page 225
Using Tables......Page 227
Exercises......Page 229
Solutions to Target Practice......Page 230
Areas Between Curves......Page 233
Choosing the Representative Rectangle......Page 235
Exercises......Page 236
Volumes of Solid Regions......Page 237
Solids with Circular Cross Sections – The Disk Method......Page 238
Solids with Holes – The Washer Method......Page 239
Target Practice......Page 240
Exercises......Page 241
Length of a Curve......Page 243
Choosing the Correct Method......Page 246
Exercises......Page 247
Area of a Three-Dimensional Surface......Page 248
Exercises......Page 249
Applied Geometry Problems......Page 250
Exercises......Page 253
Solutions to Target Practice......Page 258
Work with a Variable Force......Page 261
Emptying a Tank......Page 264
Exercises......Page 265
Fluid Pressure......Page 268
Force Due to Fluid Pressure......Page 269
Exercises......Page 271
Improper Integrals......Page 273
Exercises......Page 276
Solutions to Target Practice......Page 279
Neverending Sums......Page 281
Sequences and Series......Page 282
Target Practice......Page 283
Exercises......Page 284
Geometric Series......Page 286
A Series with Decreasing Terms that Does Not Converge......Page 287
Exercises......Page 288
Integral Test......Page 291
Comparison Test......Page 292
Target Practice......Page 293
Exercises......Page 294
Ratio Test......Page 295
Target Practice......Page 296
exercises......Page 297
Summary of Tests for Infinite Series of the Form n=k an......Page 298
Solutions to Target Practice......Page 299
Approximating functions......Page 301
Quadratic approximations......Page 302
Cubic approximation......Page 303
Target Practice......Page 304
Exercises......Page 306
Definition of Taylor Series......Page 307
Target Practice......Page 308
Special limits......Page 309
Exercises......Page 310
Error......Page 311
Solutions to Target Practice......Page 313

Citation preview

Integrated Engineering Calculus I Wentworth Institute of Technology Gary Simundza Emma Smith Zbarsky Mel Henriksen Rachel Lash Maitra August 30, 2019

ii

Contents 1 Linear Regression 1.1 Linear Regression . . . . . . . . . . . . . . . . . . 1.1.1 Finding the Regression Parameters . . . . 1.1.2 Judging the Strength of the Relationship 1.1.3 Target Practice . . . . . . . . . . . . . . . 1.2 Exercises . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Solutions to Target Practice . . . . . . . .

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1 1 4 5 8 10 13

2 Functions 2.1 Classes of Functions . . . . . . . . 2.1.1 Constant Function . . . . . 2.1.2 Linear Function . . . . . . 2.1.3 Quadratic Function . . . . 2.1.4 Other Polynomial Functions 2.1.5 Exponential Function . . . 2.1.6 Logarithm Function . . . . 2.1.7 Sine and Cosine Functions 2.1.8 Tangent Function . . . . . 2.1.9 Target Practice . . . . . . . 2.1.10 Exercises . . . . . . . . . . 2.2 Properties of Basic Functions . . . 2.3 New Functions from Old . . . . . . 2.3.1 Simple Combinations . . . . 2.3.2 Shifts and Stretches . . . . 2.3.3 Composition . . . . . . . . 2.3.4 Target Practice . . . . . . . 2.3.5 Exercises . . . . . . . . . . 2.3.6 Solutions to Target Practice

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15 15 16 16 17 18 19 21 21 22 22 23 28 29 29 29 33 33 34 37

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3 Slopes and Rates

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CONTENTS 3.1

3.2

3.3

3.4

3.5

3.6

Slope and Rate of Change . . . . . . . . . . . . 3.1.1 Target Practice . . . . . . . . . . . . . . . 3.1.2 Exercises . . . . . . . . . . . . . . . . . . Limits and the Derivative . . . . . . . . . . . . . 3.2.1 Target Practice . . . . . . . . . . . . . . . 3.2.2 Exercises . . . . . . . . . . . . . . . . . . The Derivative as a Function . . . . . . . . . . . 3.3.1 Target Practice . . . . . . . . . . . . . . . 3.3.2 Exercises . . . . . . . . . . . . . . . . . . Some Terminology and Notation . . . . . . . . . 3.4.1 Other Notations for Derivatives . . . . . . 3.4.2 Target Practice . . . . . . . . . . . . . . . 3.4.3 Exercises . . . . . . . . . . . . . . . . . . The Derivative of a Sine Function . . . . . . . . 3.5.1 Sine Functions with Different Frequencies 3.5.2 Sine Functions with Different Amplitudes 3.5.3 Sines and Cosines . . . . . . . . . . . . . 3.5.4 Target Practice . . . . . . . . . . . . . . . 3.5.5 Exercises . . . . . . . . . . . . . . . . . . Higher Order Derivatives . . . . . . . . . . . . . 3.6.1 More Derivatives . . . . . . . . . . . . . . 3.6.2 Exercises . . . . . . . . . . . . . . . . . . 3.6.3 Solutions to Target Practice . . . . . . . .

4 Formulas for Calculating Derivatives 4.1 Polynomials . . . . . . . . . . . . . . . 4.1.1 Target Practice . . . . . . . . . 4.1.2 Exercises . . . . . . . . . . . . 4.2 Extending the Power Rule . . . . . . . 4.2.1 Target Practice . . . . . . . . . 4.2.2 Exercises . . . . . . . . . . . . 4.3 Derivatives of sin(x) and cos(x) . . . . 4.3.1 Target Practice . . . . . . . . . 4.3.2 Exercises . . . . . . . . . . . . 4.4 Derivative of an Exponential Function 4.4.1 Target Practice . . . . . . . . . 4.4.2 Exercises . . . . . . . . . . . . 4.5 Derivative of a Logarithm Function . . 4.5.1 Exercises . . . . . . . . . . . . 4.6 Hyperbolic Functions . . . . . . . . . . 4.6.1 Target Practice . . . . . . . . .

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67 67 71 71 73 74 75 76 80 80 81 82 83 84 85 87 89

CONTENTS 4.6.2 4.6.3

v Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Target Practice . . . . . . . . . . . . . . . . . . . . . . .

5 Derivatives of Combinations 5.1 The Chain Rule . . . . . . . . . . . 5.1.1 Target Practice . . . . . . . 5.1.2 Exercises . . . . . . . . . . 5.2 Product and Quotient Rules . . . . 5.2.1 The Product Rule . . . . . 5.2.2 Target Practice . . . . . . . 5.2.3 The Quotient Rule . . . . . 5.2.4 Target Practice . . . . . . . 5.2.5 Exercises . . . . . . . . . . 5.3 Summary of Derivative Formulas . 5.3.1 Target Practice . . . . . . . 5.3.2 Exercises . . . . . . . . . . 5.3.3 Solutions to Target Practice

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6 Accumulation 6.1 Velocity and Displacement . . . . . . . . . . . . . . . . . 6.1.1 The Relationship between Velocity and Position 6.1.2 Target Practice . . . . . . . . . . . . . . . . . . . 6.1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . 6.2 Approximating the Area Under the Graph of a Function 6.2.1 Target Practice . . . . . . . . . . . . . . . . . . . 6.2.2 Exercises . . . . . . . . . . . . . . . . . . . . . . 6.3 The Exact Area Under the Graph of a Function . . . . 6.3.1 The Definite Integral . . . . . . . . . . . . . . . . 6.3.2 Target Practice . . . . . . . . . . . . . . . . . . . 6.3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . 6.3.4 Solutions to Target Practice . . . . . . . . . . . .

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123 123 125 125 126 128 129 130 131

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7 Differential Equations 7.1 Antiderivatives . . . . . . . . . . . . . . . . . 7.1.1 Target Practice . . . . . . . . . . . . . 7.1.2 Exercises . . . . . . . . . . . . . . . . 7.2 Initial Value Problems . . . . . . . . . . . . . 7.2.1 Target Practice . . . . . . . . . . . . . 7.2.2 Exercises . . . . . . . . . . . . . . . . 7.3 Differential Equations of Higher Order . . . . 7.3.1 A Second Order Initial Value Problem

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CONTENTS

7.4 7.5

7.6 7.7

7.3.2 A First Order System . . . . . . . . . . . . . . . . 7.3.3 Quick Introduction to Matrices . . . . . . . . . . . 7.3.4 Matrix Form of a System of Differential Equations The Solution of the Second Order DE . . . . . . . . . . . 7.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.1 Antiderivatives and the Definite Integral . . . . . . 7.5.2 Target Practice . . . . . . . . . . . . . . . . . . . . 7.5.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . The Fundamental Theorems . . . . . . . . . . . . . . . . . Special Functions . . . . . . . . . . . . . . . . . . . . . . . 7.7.1 Piecewise-defined functions . . . . . . . . . . . . . 7.7.2 Error function . . . . . . . . . . . . . . . . . . . . 7.7.3 Bessel functions . . . . . . . . . . . . . . . . . . . . 7.7.4 Airy functions . . . . . . . . . . . . . . . . . . . . 7.7.5 Laguerre Polynomials . . . . . . . . . . . . . . . . 7.7.6 Solutions to Target Practice . . . . . . . . . . . . .

8 Related Rates 8.1 Two Quantities Changing with Time . 8.1.1 Target Practice . . . . . . . . . 8.1.2 Exercises . . . . . . . . . . . . 8.2 Differentiating Implicitly . . . . . . . 8.2.1 Target Practice . . . . . . . . . 8.2.2 Exercises . . . . . . . . . . . . 8.3 Other Uses of Implicit Differentiation . 8.3.1 Target Practice . . . . . . . . . 8.3.2 Exercises . . . . . . . . . . . . 8.3.3 Solutions to Target Practice . .

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9 Optimization 9.1 Finding the Best Strategy . . . . . . . . . . . . . . . . . . . . . . . 9.1.1 Local Maxima and Minima . . . . . . . . . . . . . . . . . . 9.1.2 Target Practice . . . . . . . . . . . . . . . . . . . . . . . . . 9.1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Deciding Whether an Extreme Value is a Maximum or Minimum . 9.2.1 Target Practice . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 A Closer Look at Optimization . . . . . . . . . . . . . . . . . . . . 9.3.1 Another Type of Critical Point . . . . . . . . . . . . . . . . 9.3.2 A Critical Point that is not a Local Maximum or Minimum

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CONTENTS

9.4

9.3.3 Absolute Maxima and Minima 9.3.4 Target Practice . . . . . . . . . 9.3.5 Exercises . . . . . . . . . . . . Inflection Points . . . . . . . . . . . . 9.4.1 Curve Sketching . . . . . . . . 9.4.2 Target Practice . . . . . . . . . 9.4.3 Exercises . . . . . . . . . . . . 9.4.4 Solutions to Target Practice . .

10 Substitution 10.1 Changing Variables in Integrals . . . . 10.1.1 Target Practice . . . . . . . . . 10.1.2 Exercises . . . . . . . . . . . . 10.2 Solving Differential Equations Exactly 10.2.1 Target Practice . . . . . . . . . 10.2.2 Exercises . . . . . . . . . . . . 10.3 Substitution in Definite Integrals . . . 10.3.1 Target Practice . . . . . . . . . 10.3.2 Exercises . . . . . . . . . . . . 10.3.3 Solutions to Target Practice . .

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11 Methods of Integration 11.1 Integration By Parts . . . . . . . . . . . . . . 11.1.1 Target Practice . . . . . . . . . . . . . 11.1.2 Exercises . . . . . . . . . . . . . . . . 11.2 Integration using Trigonometric Substitution 11.2.1 Exercises . . . . . . . . . . . . . . . . 11.3 Integrating Improper Rational Functions . . . 11.3.1 Target Practice . . . . . . . . . . . . . 11.3.2 Exercises . . . . . . . . . . . . . . . . 11.4 Integration by Partial Fractions . . . . . . . . 11.4.1 Target Practice . . . . . . . . . . . . . 11.4.2 Exercises . . . . . . . . . . . . . . . . 11.5 Using Tables . . . . . . . . . . . . . . . . . . 11.5.1 Exercises . . . . . . . . . . . . . . . . 11.5.2 Solutions to Target Practice . . . . . .

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12 Geometric Applications of Integrals 223 12.1 Areas Between Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223 12.1.1 Choosing the Representative Rectangle . . . . . . . . . . . . . . . . 225 12.1.2 Target Practice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226

viii

CONTENTS 12.1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Volumes of Solid Regions . . . . . . . . . . . . . . . . . . . . . 12.2.1 Solids with Circular Cross Sections – The Disk Method 12.2.2 Solids with Holes – The Washer Method . . . . . . . . . 12.2.3 Target Practice . . . . . . . . . . . . . . . . . . . . . . . 12.2.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3 Length of a Curve . . . . . . . . . . . . . . . . . . . . . . . . . 12.3.1 Choosing the Correct Method . . . . . . . . . . . . . . . 12.3.2 Target Practice . . . . . . . . . . . . . . . . . . . . . . . 12.3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 12.4 Area of a Three-Dimensional Surface . . . . . . . . . . . . . . . 12.4.1 Target Practice . . . . . . . . . . . . . . . . . . . . . . . 12.4.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 12.5 Applied Geometry Problems . . . . . . . . . . . . . . . . . . . . 12.5.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 12.5.2 Solutions to Target Practice . . . . . . . . . . . . . . . .

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226 227 228 229 230 231 233 236 237 237 238 239 239 240 243 248

13 Applications of Integration 13.1 Work and Energy . . . . . . . . . . 13.1.1 Work with a Variable Force 13.1.2 Emptying a Tank . . . . . . 13.1.3 Target Practice . . . . . . . 13.1.4 Exercises . . . . . . . . . . 13.2 Fluid Pressure and Force . . . . . 13.2.1 Fluid Pressure . . . . . . . 13.2.2 Force Due to Fluid Pressure 13.2.3 Target Practice . . . . . . . 13.2.4 Exercises . . . . . . . . . . 13.3 Improper Integrals . . . . . . . . . 13.3.1 Target Practice . . . . . . . 13.3.2 Exercises . . . . . . . . . . 13.3.3 Solutions to Target Practice

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14 Infinite Series 14.1 Neverending Sums . . . . . 14.1.1 Sequences and Series 14.1.2 Partial Sums . . . . 14.1.3 Target Practice . . . 14.1.4 Exercises . . . . . . 14.2 Some Special Series . . . . . 14.2.1 Geometric Series . .

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CONTENTS

ix

14.2.2 A Series with Decreasing Terms that 14.2.3 p-Series . . . . . . . . . . . . . . . . 14.2.4 Alternating Series . . . . . . . . . . 14.2.5 Target Practice . . . . . . . . . . . . 14.2.6 Exercises . . . . . . . . . . . . . . . 14.3 Tests for Convergence of Series . . . . . . . 14.3.1 Integral Test . . . . . . . . . . . . . 14.3.2 Comparison Test . . . . . . . . . . . 14.3.3 Linearity of Infinite Series . . . . . . 14.3.4 Target Practice . . . . . . . . . . . . 14.3.5 Exercises . . . . . . . . . . . . . . . 14.4 Ratio Test . . . . . . . . . . . . . . . . . . . 14.4.1 Target Practice . . . . . . . . . . . . 14.4.2 exercises . . . . . . . . . . . . . . . .

Does Not Converge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ∞ X 14.5 Summary of Tests for Infinite Series of the Form an . . . . .

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n=k

14.5.1 Solutions to Target Practice . . . . . . . . . . . . . . . . . . . . . . . 289 15 Taylor Series 15.1 Approximating functions . . . . . . 15.1.1 Quadratic approximations . 15.1.2 Cubic approximation . . . . 15.1.3 Target Practice . . . . . . . 15.1.4 Exercises . . . . . . . . . . 15.2 Taylor Series . . . . . . . . . . . . 15.2.1 Power Series . . . . . . . . 15.2.2 Definition of Taylor Series . 15.2.3 Target Practice . . . . . . . 15.2.4 Exercises . . . . . . . . . . 15.3 Special limits . . . . . . . . . . . . 15.3.1 Exercises . . . . . . . . . . 15.4 Error . . . . . . . . . . . . . . . . . 15.4.1 Exercises . . . . . . . . . . 15.4.2 Solutions to Target Practice

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x

CONTENTS

Chapter 1

Linear Regression

1.1

Linear Regression

Energy conservation has become an essential concern for many businesses. The amount of energy consumed by a building depends on several factors, one of which is obviously the size of the building. Table 1.1, below, shows data relating energy consumption to the external surface area for 20 buildings that were subjected to similar climatic conditions.

1

2

CHAPTER 1. LINEAR REGRESSION Energy Consumption (1000s of BTUs/year) 1371000 2422000 672200 233100 354000 3135000 1470000 1408000 2201000 337500 567500 555300 239400 2629000 1102000 2680000 423500 423500 1691000 1870000

External Area (ft2 ) 13530 26060 6355 4576 2621 23350 18770 12220 25490 5650 8001 6147 2660 19240 10700 23680 9125 6510 13530 18860

Table 1.1: This table presents the energy consumption and the corresponding external area of 20 buildings. a a

Scheaffer and McClave, Probability and Statistics for Engineers

An investigation of a possible relationship between two variables should begin with a scatterplot. In this case, it makes sense to use external area as the independent variable and energy consumption as the dependent variable. Why? In situations for which a predictive A scatterplot of enrelationship between two variables ergy consumption is suspected, the independent varivs external area for able is sometimes called the exthe data in the taplanatory variable and the deble is shown below. pendent variable is the response variable.

Figure 1.1: A scatter plot of energy consumption (in millions of BTUs/year) against area (in thousands of square feet).

The scatterplot in Figure 1.1 shows a fairly clear trend in the data: as external area increases, so does energy consumption. And a linear relationship is certainly a possibility (although we’ll need ways of testing whether the trend is actually a linear one). So it makes sense to try to describe the data with a straight line, as in the next graph. The line shown in Figure 1.2 is one of many possible lines that might be drawn to capture the trend in the scatterplot. But how can we judge whether one line is better than another? It seems reasonable to assume that the “best” line would be one for which the distances from the data points to the line are as small as possible. We can define one of these

1.1. LINEAR REGRESSION

3

Figure 1.2: A scatter plot of energy consumption versus external area together with a proposed linear trend line.

distances by measuring the difference between the height of a point and the height of the line for the same value of the independent variable. A customary method for deciding on the best line in situations like this is to examine the squares of these distances, as shown in Figure 1.3. Judged by the criterion of least-squares, the best line is the one that results in minimizing the sum of the squares of these differences, which is the same as minimizing the total area of the squares on the graph shown in Figure 1.4. The equation of the least-squares line, also called the regression line, is EC = 112.3A − 154,

(1.1)

where EC is the energy consumption in millions of BTUs per year and A is the external area of the building in thousands of square feet. So, assuming that the relationship between energy consumption and area is truly linear, this equation will give the best prediction of energy consumption for a building with an external area of A thousands of square feet. Warning. Be very careful with using regression lines to predict values, especially outside of the ranges used to create the regression originally!

4

CHAPTER 1. LINEAR REGRESSION

Figure 1.3: A scatter plot of energy consumption versus external area together with a proposed linear trend line as well as squares with one vertical leg from each point in the scatter to the proposed trend line.

1.1.1

Finding the Regression Parameters

The slope and intercept numbers for linear regression are found by using calculus optimization methods. For n pairs of data involving variables x and y, we can always find a relationship of the form y P = mx + b, in which Pnthe slope is calculated from the data points n x yi i and the mean values x ¯ = i=1 and y¯ = i=1 by: n n

Pn m=

¯) (yi − i=1 (xi − x Pn ¯)2 i=1 (xi − x

y¯)

(1.2)

and then the intercept can be calculated as

b = y¯ − m¯ x.

(1.3)

1.1. LINEAR REGRESSION

5

Figure 1.4: A scatter plot of energy consumption versus external area together with the least-squares linear regression line. The differences between Figure 1.3 and Figure 1.4 are most visible if you look at sizes and locations of the squares at the upper right corner of the plot. Summation notation is a shorthand way to write out long equations that correspond to some pattern. n X

xi = x1 + x2 + x3 + . . . + xn−1 + xn .

i=1

Similarly, 3 X (k + 1)2 = (1 + 1)2 + (2 + 1)2 + (3 + 1)2 = 29.

In the formula (1.2) for slope, xi and yi represent the ith respective data values, and each of the sums is over all data points. The resulting regression equation is said to have been fitted to the data.

k=1

Note: A linear regression equation is often stated as y = ax+b or y = a+bx.

1.1.2

Judging the Strength of the Relationship

A linear regression equation can always be found for a given bivariate data set, but it may not be very meaningful. The graphs in Figures 1.5a and 1.5c below clearly show a strong linear trend, while the one in Figure 1.5b exhibits virtually none. However, a least-squares

6

CHAPTER 1. LINEAR REGRESSION

(a) A positive linear relationship

(b) No clear linear relationship

(c) A negative linear relationship

Figure 1.5: Three scatterplots depicting various linear relationships between two variables

line can be found that “fits” the data in each case. So it is important to have some way of deciding whether a given regression equation has any predictive value. The Scatterplot If there is any linear trend at all, it should be visible in an examination of a scatterplot. But if the points are widely scattered, the trend may not be obvious. A “positive” linear relationship will scatter from the bottom left to the top right, as in Figure 1.5a, and thus the trendline will have a positive slope. A “negative” linear relationship will scatter from the top left to the bottom right, and thus the trendline will have a negative slope as in Figure 1.5c. Finally, a scatterplot where the least-squares regression line is a constant value, i.e. the slope is 0, is a set of points that scatter uniformly across the plot–top left, bottom left, top right, and bottom right as in Figure 1.5b. Residual Plot As mentioned previously, the regression line is one that minimizes the sum of the differences between the heights of points and the heights of the line for corresponding values of the independent variable. Each of these differences is a residual error, defined by e = yactual − ypredicted . The regression line is the one for which

n X

(1.4)

e2i is a minimum. For the building data, this

i=1

sum is about 1,990,000, but it is smaller than the sum for any other line relating energy consumption and area.

1.1. LINEAR REGRESSION

7

It can be helpful to view a residual plot, as shown in Figure 1.6, which is a scatterplot of the sizes of the residuals only. As before, the area is measured in thousands and the residuals in millions. If a linear model describes the trend in the data well, the residuals

Figure 1.6: A residual plot for the least-squares regression model of the energy consumption versus area data shown in Figure 1.4 should be generally small and random. Such is the case here, with the sizes of the residuals being typically small compared to the data. Correlation A common numerical measure of the strength of a linear regression model is the linear correlation coefficient r, which is calculated using the same kinds of summations that allow calculation of the slope of the regression line. Pn (xi − x ¯) (yi − y¯) r = qP i=1 (1.5) n 2 Pn 2 (x − x ¯ ) · (y − y ¯ ) i i i=1 i=1 The correlation coefficient has values −1 ≤ r ≤ 1, with r being positive for positively correlated data (positive slope) and negative for negatively correlated data (negative slope). For completely uncorrelated data (no linear relationship), r = 0. If all data points lie exactly on the same line, r is −1 or 1. For example, for the energy consumption data, the correlation coefficient is approximately 0.94, which indicates that external building area and energy consumption are very strongly correlated. A related measure of strength is the coefficient of determination, which for linear relationships is just the square of r. In this case, r2 is about 0.88, which tells us that

8

CHAPTER 1. LINEAR REGRESSION

88% of the variation in energy consumption by buildings is accounted for by the linear relationship with external area. Notice that the formula for r is symmetric with respect to the variables. In other words, denoting energy consumption by x and area by y does not change the value of r, although it clearly would change the regression equation. Furthermore, correlation between two variables does not imply a causal relationship, only that there is an association between them. It is often the case that two variables are correlated because of a causative influence of a third, unknown variable.

1.1.3

Target Practice

1. If a coil spring obeys Hooke’s Law, the length L of the spring is linearly related to the force F applied to the end of the spring. A particular spring is 26.8 cm long when a force of 6.5 newtons is applied, and 30.2 cm long for a force of 15.0 newtons. a) Find a linear equation that expresses the length of the spring as a function of the applied force. b) Explain the physical meaning of the slope and intercept in your equation, including units. 2. Consider each scatterplot below. First, determine if there is any relation between the variables. If there is, determine the form (linear or nonlinear) and strength (strongly or weakly related). If the form is linear, tell whether the variables are positively or negatively related.

a.

b.

1.1. LINEAR REGRESSION

c.

9

d.

For each data set, make a well-labeled scatterplot. Then, if a linear relationship seems appropriate, find a linear regression equation and assess the strength of the linear relationship. If not, use Excel/R/Matlab to investigate whether a nonlinear model might describe the data. 3. This data set presents the average height of ocean waves each year for selected years over a 21 year period.∗

Year 1962 1968 1970 1972 1973 1974 1976 1981 1983

Wave height (ft) 6.9 7.6 7.4 8.6 6.9 7.7 7.9 9.6 9.9

It is often helpful to express calendar year data in years after the first observation (e.g., 1962 becomes year 0) to simplify models and reduce rounding errors.

4. Average temperature and latitude†



Nature 4/7/88 Historical weather data was pulled from https://www.weatherunderground.com/history and the latitude from http://www.travelmath.com/cities/. †

10

CHAPTER 1. LINEAR REGRESSION

City San Antonio, TX Chicago, IL St. Louis, MO Little Rock, AK Bismark, ND Rapid City, SD Des Moines, IA Boston, MA Tallahassee, FL

1.2

Latitude North (degrees) 29.4 41.9 38.6 34.7 46.8 44.1 41.6 42.4 30.4

Average Temperature August 2016 (degrees Fahrenheit) 84 76 79 83 69 71 76 77 92

Exercises

Exercise 1. A diode is a common device that is used to create electronic temperaturesensing instruments because its voltage in a circuit is related linearly to temperature when the current is held constant. The data on diode voltage VD below were collected experimentally. Temperature C (room temperature) 35◦ C (skin temperature)

27◦

Voltage 588 mV 570 mV

a) Find an equation that gives diode voltage VD as a function of temperature T . b) Explain the physical meaning of the slope and intercept in your equation, including units.

Exercise 2. The regression equation and residual plot for the building energy consumption data were created in R as follows: • Create data vectors for the variables Area and EC. • Find the regression equation using lm(EC∼ Area). • Create a vector of residuals using the command Residuals=lm(EC∼ Area)$resid. • Make a residual plot using plot(Area,Residuals).

1.2. EXERCISES

11

Use similar commands to create a residual plot for your forearm/height data, and explain what it tells you about your regression model.

For each data set, make a well-labeled scatterplot. Then, if a linear relationship seems appropriate, find a linear regression equation and assess the strength of the linear relationship. If not, use Excel/R/Matlab to investigate whether a nonlinear model might describe the data. Exercise 3. The number of women in the U.S. House of Representatives Year Number

’89 31

’91 33

’93 55

’95 59

’97 66

’99 67

’01 75

’03 77

’05 85

’07 94

’09 96

’11 96

’13 102

’15 104

Exercise 4. Two characteristics of the ponderosa pine tree: diameter measured at chest height and usable volume.‡ Diameter (in.) 36 28 28 41 19 32 22 38 25 17

Usable Volume (1000’s of cu. in.) 276 163 127 423 40 177 73 363 81 23

Diameter (in.) 31 20 25 19 39 33 17 37 23 39

Usable Volume (1000’s of cu. in.) 203 46 124 30 333 269 32 295 83 382

Exercise 5. Average number of pitching changes in a baseball game for a major league ‡

Croxton, Cowden, and Klein Applied General Statistics, p. 421

’17 104

12

CHAPTER 1. LINEAR REGRESSION

team:§ Year Average Number of Pitching Changes

1978

1983

1988

1993

1998

2003

2008

2013

1.4

1.6

1.8

2.3

2.5

2.7

2.9

3

Exercise 6. Annual U.S. solar photovoltaic installations: Year 2000 2001 2002 2003 2004 2005 2006 2007

New Installations 4 11 23 45 58 79 105 160

Year 2008 2009 2010 2011 2012 2013 2014 2015 2016

New Installations 298 385 852 1926 3373 4783 6239 7501 14762

Exercise 7. Use R or Excel to calculate the regression parameters m and b, and the correlation coefficient r, for the forearm/height data you collected directly from the formulas given in the reading. Check that you get the same values as using the defined commands. Include a printout of the code showing your work and your results.

Exercise 8. Using what you have learned about linear regression, give a complete summary of your analysis of the class data relating forearm length and height. Your answer should include your linear model, a plot of the data and your model line together with a discussion of the strength of the correlation including both r and r2 as well as a residual plot.

§

Sports Illustrated

1.2. EXERCISES

1.2.1 1.

13

Solutions to Target Practice a) We have two known points (F1 , L1 ) = (6.5, 26.8) and (F2 , L2 ) = (15.0, 30.2). Using our slope formula we see that m=

L2 − L1 30.2 − 26.8 = = 0.4. F2 − F1 15.0 − 6.5

Plugging this into L1 = mF1 + b, we find that b = L1 − mF1 = 26.8 − (0.4)6.5 = 24.2. Putting this together, we have L = 0.40F + 24.2. b) The slope is the rate of change of length with respect to applied force, 0.40 cm/N. The intercept indicates that the spring’s length is 24.2 cm when no force is applied. 2.

a) These variables are weakly related with a negative, linear correlation. b) These variables are strongly correlated with a nonlinear correlation. c) These variables are not related. d) These variables are strongly correlated with a positive, linear correlation.

3. h = 0.1434t + 6.4474, with r ≈ 0.83, which is a moderately strong correlation. 4. In R this problem can be solved as follows: > lat = c(29.4, 41.9, 38.6, 34.7, 46.8, 44.1, 41.6, 42.4, 30.4) > temp = c(84, 76, 79, 83, 69, 71, 76, 77, 92) > plot(lat, temp) > fit = lm(temp ∼ lat) > summary(fit) > abline(fit[1], fit[0], col="red") > sqrt(.8688) So the fit is a negative linear regression, T = −1.0717L + 120.2197, where T is the temperature in degrees Fahrenheit and L is the (decimal) latitude with r ≈ −0.932.

14

CHAPTER 1. LINEAR REGRESSION

Chapter 2

Functions 2.1

Classes of Functions

As you found in the buckling activity, the load P which causes elastic buckling can be k written as P = 2 . This is an example of inverse-square variation. We say that P L varies inversely as the square of L, or P is inversely proportional to the square of L. The constant k is a constant of proportionality. y is said to be proportional p In general, the equation y = kx expresses a proporto x if the ratio y/x is contional relationship, or variation, between y and xp . stant. By extension, y is proThe variation is direct if the power p is positive and portional to xp if y/xp is coninverse if p is negative. stant.

For example, the breaking strength of a wooden beam is given by the joint variation formula S=k

wd2 . L

(2.1)

Strength varies directly with the width w and with the Figure 2.1: Diagram indicat- square of the depth d and it varies inversely with the length ing the labels w for width, d of the beam L, as shown in Figure 2.1. for depth, and L for length of a beam. Since this is a course in single-variable calculus, we will be primarily concerned with functions of only one independent variable. Several of the most common types of functions, 15

16

CHAPTER 2. FUNCTIONS

with which you should already be familiar, are reviewed here. See http://mathonweb.com/help_ebook/html/functions_4.htm for additional detailed review of the various function classes.

2.1.1

Constant Function

The simple function y = k can be thought of as a power function with p = 0. A square-wave voltage signal in an electrical circuit, shown in Figure 2.2, is a piecewise-defined function in which the “pieces” are constant functions.

  4      −4 V (t) = 4    −4     etc.

if if if if

0≤t a H(t) dt = 0 otherwise a y

y

1

x

x

(a) Heaviside function

(b) Triangle function y

x

(c) Dirac delta function

Figure 7.5: Graphs of piecewise-defined functions. Differentiating the Heaviside step function leads us to a strange phenomenon called the Dirac delta function, as shown in Figure 7.5c, even though it is not technically a function. One way to define the Dirac delta function, δ(x) is by its action in an integral: ( Z b f (0) if a < 0 < b f (x)δ(x) dx = . 0 otherwise a This means that a delta function is “picking out” the value of a function at one point when it is integrated against. Since the “width” of each point is 0, this must result in a “height” of the delta function at 0 being infinite. This also gives Z x δ(t) dt = H(x). −∞

7.7. SPECIAL FUNCTIONS

7.7.2

143

Error function

The error function, erf(x), is a function that can only be exactly expressed as an integral. Numerical values of the error function are always approximate for any positive value of x. 1.0

1 erf(x) = √ π

Z

erf(x)

0.5

x

−t2

e

0.0

dt

−x -0.5

-1.0 -2

0

2

x values

Figure 7.6: A graph of the error function.

7.7.3

Bessel functions

Bessel functions are canonical solutions to the differential equation x2

d2 y dy +x + (x2 − n2 )y = 0. 2 dx dx

For integer values of n, these solutions are 1 Jn (x) = π

Z

π

cos (nt − x sin(t)) dt.

(7.1)

0

These functions, and their relations, are common solutions to wave problems in physics. For instance, the integer Bessel functions given in equation 7.1 are the radial values of the harmonics of a circular drum.

7.7.4

Airy functions

The Airy functions are two linearly independent solutions to the differential equation d2 y − xy = 0. dx2

144

CHAPTER 7. DIFFERENTIAL EQUATIONS Ai(x) = Bi(x) =

 3  Z 1 ∞ t cos + xt dt π 0 3  3   3  Z ∞ t 1 t exp − + xt + sin + xt π 0 3 3

(7.2) (7.3)

Figure 7.7: A plot of equation 7.2, Ai(x), in solid red, and equation 7.3, Bi(x), in dashed blue.

Note that the Airy functions change their behavior as they cross to positive values of the argument. In the negative domain the Airy functions are oscillatory while in the positive domain the Airy functions exhibit exponential growth, for Bi(x), or decay, for Ai(x).

7.7.5

Laguerre Polynomials

Laguerre polynomials are solutions to Laguerre’s equation: xy 00 + (1 − x)y 0 + ny = 0. Unlike our earlier examples, these polynomials can be expressed using derivatives and thus written down in normal polynomial form. The Laguerre polynomial of degree n is denoted by Ln . It is easy to see that for L0 , a constant value, we could pick any value, c, that we wish because d2 d x 2 (c) + (1 − x) (c) + 0(c) = 0 + 0 + 0 = 0. dx dx Conventionally, the nth order Laguerre polynomial is normalized so that the leading coefficient is ±1. For L0 that means L0 = 1. The first few Laguerre polynomials are:

7.7. SPECIAL FUNCTIONS

145

L0 = 1 L1 = −x + 1  1 2 L2 = x − 4x + 2 2  1 L3 = −x3 + 9x2 − 18x + 6 6 .. .  n d 1 − 1 xn Ln = n! dx

Figure 7.8: A graph of the first few Laguerre polynomials.

7.7.6

Solutions to Target Practice

Section 7.1  6 x + C = 2x6 + C 1. 12 6   − cos(10t) 2. 40 + C = −4 cos(10t) + C 10 Section 7.2 1.  10 1 6x e + C = e6x + C 6 3 10 10 y(0) = 12 = e6·0 + C = +C 3 3 10 26 C = 12 − = 3 3 

y(x) = 20

y(x) =

10 6x 26 e + 3 3

2. Let Q be the charge in microcoulombs, and t be the time in seconds. Then dQ = kQ, dt

Q(0) = 5.2 × 10−6 .

146

CHAPTER 7. DIFFERENTIAL EQUATIONS

3. We know that the initial condition is T (0) = 200◦ at the moment the pizza is removed from the oven. The subsequent temperature is described by dT = k(T − 70), dt for some parameter of cooling k. Section 7.5 Z 3 1. a) (−3x + 4) dx = − x2 + 4x + C 2 Z 3 b) v 1/3 dv = v 4/3 + C 4 2.

a) Z 1

2

  2 5x4 + 5 dx = x5 + 5x 1   = 25 + 5(2) − 15 + 5(1) = (32 + 10) − (1 + 5) = 36

b) Z

1

√ 3

1 3 4/3 s − 2s 4 −1     3 4/3 3 4/3 (1 ) − 2(1) − (−1) − 2(−1) 4 4     5 11 − − 4 4 −4 



s − 2 ds =

−1

= = =

Chapter 8

Related Rates of Change 8.1

Two Quantities Changing with Time

The air pressure P on the front of a car traveling 1 at speed v is given by P = ρv 2 , where ρ is the 2 density of the air. If ρ is measured in kilograms per cubic meter (kg/m3 ) and v is in meters per second (m/s), P is measured in newtons per square meter (N/m2 ), or pascals (Pa). So, for example, if air density is 1.4 kg/m3 and a car is traveling at 22 m/s (about 50 miles per hour), the pressure on the front of the car is about 340 pascals. Now consider the following question: How fast does the the pressure change as the car’s speed changes? 1 The equation P = ρv 2 doesn’t contain time as a variable. But we do know that the 2 dP derivative is the rate of change of pressure with respect to time. So by differentiating dt both sides of the equation with respect to time, we can find a rate equation:   d 1 2 d (P ) = ρv . dt dt 2 If we assume air density is constant, then the right-hand side becomes  dv 1 d 2 1 d ρ v ρ v2 · = 2 dt 2 dv dt dv 1 = ρ(2v) 2 dt 147

148

CHAPTER 8. RELATED RATES

Notice the use of the Chain Rule here, since v 2 is a function of v and v is a function of t. Finally, we arrive at an equation relating the rates of change of P and v: dP dv = ρv . dt dt Now, if the car traveling at 22 m/s accelerates at a rate of 4.8 m/s2 , dP dt

=



   1.4 kg/m2 (22 m/s) 4.8 m/s2

= 147.84

pascals second

dv = 4.8, and dt (8.1) (8.2)

Example 15.

A weather balloon is inflated by pumping in helium at a constant rate of 1.2 m3 /min. When the balloon is 3.0 meters in diameter, how fast is the diameter increasing? (Assume that the balloon remains spherical while it is being inflated.)

Solution: Notice that a cubic meter is a unit of volume, so the rate of change of volume dr dV is 1.2 m3 /min, and is constant. We are asked to find . dt dt 4 The usual formula for the volume of a sphere is V = πr3 . This equation can be differen3

8.1. TWO QUANTITIES CHANGING WITH TIME

149

tiated with respect to time to give the rate equation. d (V ) = dt dV = dt dV = dt

  d 4 3 πr dt 3   d 4 3 dr πr · dr 3 dt  4 dr π 3r2 · 3 dt

Here again, the Chain Rule has been used on the right-hand side. We can now solve for dr : dt   1 dV dr = . 2 dt 4πr dt Notice that, even though the volume is changing at a constant rate, the radius is not. Its rate of change depends on the value of the radius at any instant. And when the balloon’s diameter is 3.0 meters, its radius is 1.5 meters, so dr = dt



1 4π(1.5m)2

  m3 1.2 ≈ 0.042m/min. min

The diameter is therefore increasing at a rate of 0.084 m/min, or about 8.4 centimeters per minute.  There are two important things to notice in this solution: • The radius value of 1.5 meters is not substituted in the original formula for volume. If it were, the volume would not be variable, and it would not be possible to find an appropriate rate equation. So, in general, values of variables should not be used until after a rate equation is found by differentiation. • Keeping units with all quantities can be helpful in determining whether the correct rate equation was found and solved correctly. We expect the rate of change of radius (or diameter) to be measured in units of distance per unit time, and the units in the dr work out to meters per minute, which makes sense. calculation of dt

150

CHAPTER 8. RELATED RATES Guidelines for Solving Related Rates Problems

1. Make a sketch (if applicable), and list all variables and constants. Identify all known and unknown rates as derivatives with respect to time. 2. Write a “static equation” that relates the variables. 3. Differentiate with respect to time to find a “dynamic equation” that relates the rates of change of the variables. 4. Substitute values of the variables for a specific instant of time. 5. Solve for the unknown rate. 6. Check the reasonableness of the answer. (Does it make sense?) Use units to help you do this.

8.1.1

Target Practice

1) When the diameter of a spherical tumor is 16 millimeters, it is growing at the rate of 0.4 mm per day. How fast is the volume of the tumor changing at that time? 2) Atmospheric pressure P (in pounds per square inch) decreases exponentially with altitude h (in miles) according to the equation P = 14.7e−0.21h . An airplane is climbing at a rate of 200 miles per hour. Find the rate of change of pressure with respect to time (in psi per second) as the plane passes an altitude of 3.0 miles.

8.1.2

Exercises

Exercise 127. If a small hole is punctured in a tank containing a liquid, the velocity v of the escaping liquid depends on the depth h of the hole below the liquid surface according p to v = 2gh. a) As the tank empties and the level of the liquid begins to fall, what happens to the velocity? dv dh b) Find an equation that gives the rate of change of velocity in terms of the rate dt dt at which the surface of the liquid drops.

8.1. TWO QUANTITIES CHANGING WITH TIME

151

Exercise 128. The power, P , in watts, of a circuit is given by P = Ri2 . If the resistance is a constant R = 100Ω and the current i is increasing at the rate of 0.30 A/sec, find the rate of change of the power when i = 20 A.

Exercise 129. Water is being collected from a melting block of ice in the shape of a cube. Each edge of the cube is decreasing at the rate of 2 in/hr. In this problem assume that water and ice have the same density. a) What is the rate of flow of water into the collecting pan when each edge has a length of 15 inches? b) What is the rate of change of the surface area when each edge has a length of 15 inches?

Exercise 130. The thin lens equation in optics is

1 1 1 + = , which relates the distance d of d i f

an object from a lens, the distance i of its image from the lens, and the focal length f of the lens. Suppose that a certain lens has a focal length of 6.0 cm and that an object is moving toward the lens at the rate of 2.0 cm/s. a) How fast is the image distance changing at the instant when the object is 10.0 cm from the lens? b) Is the image moving away from the lens or toward the lens?

Exercise 131. The amount of blood your heart pumps in a minute is known as your

152

CHAPTER 8. RELATED RATES

cardiac output, which can vary from under 6 liters per minute to 30 L/min (for a trained Q athlete). One method for measuring cardiac output uses the formula CO = , where CO D is the cardiac output in L/min, Q is the number of milliliters of carbon dioxide you exhale in a minute and D is the difference between the carbon dioxide concentration (in ml/L) in the blood pumped to the lungs and returning from the lungs. Suppose that, for a person at rest, Q = 233 ml/min and D = 41 ml/L, giving a cardiac output of about 5.7 L/min. If Q is constant but D is decreasing at the rate of 2.0 ml/L per minute, what is happening to the cardiac output?

Exercise 132. In soil mechanics, the vertical stress σz in a soil at a point z feet deep and located r feet (horizontally) from a concentrated surface load of P pounds is given by   r 2 5/2 3P σz = 1+ 2πz 2 z a) Calculate the vertical stress in a soil at a point 2.0 feet deep and located 4 feet horizontally from a concentrated surface load of 10,000 pounds. b) If that load is moved horizontally toward the point at a rate of 1.0 foot per hour, how fast is the vertical stress changing when the load is 3.0 feet from the point?

Exercise 133. Verify that the units in the answer to the example involving the air pressure on a car at the beginning of this section reduce to pascals/second. That is, check that the units given in Equation 8.1 are the same as the units given in Equation 8.2.

8.2

Differentiating Implicitly

Consider the following situation:

8.2. DIFFERENTIATING IMPLICITLY

153

A ship with a long anchor chain is anchored in 66 feet of water. The anchor chain is being wound in at the rate of 60 feet per minute, causing the ship to move toward a point directly above the anchor. The point of contact between the ship and chain is located 6 feet above the waterline. At what speed is the ship moving when there are 81 feet of chain still out? A diagram like Figure 8.1a is helpful in visualizing this problem. Notice that the top point of the anchor chain is 72 feet from the ocean floor, a distance that will not change in this situation. But in Figure 8.1b, the ship has moved to the right. The length of exposed

(a)

(b)

Figure 8.1: A sketch of a ship reeling in its anchor at two different points in time. Note that some aspects of the diagram are changing and some are not. chain has decreased, and so has the distance from the ship to the point above the anchor. Figure 8.2 shows the essence of the geometry of the problem. Both x and c change with

Figure 8.2: The essence of the geometry of the problem shown in Figure 8.1

154

CHAPTER 8. RELATED RATES

dc dx time. The rate is known, and equals -60 ft/min, while the rate is the (variable) dt dt speed of the ship. At any instant, the relationship between x and c is governed by the Pythagorean theorem: x2 + 722 = c2 . You can think of this “static” equation as providing a snapshot, or a single frame, of a movie of the ship’s travel. However, by differentiating with respect to time, we get the “dynamic” equation of the movie:  d 2 d d 2 x + 722 = c dt dt dt  dx d d 2  dc x2 · +0 = c · dx dt dc dt dx dc 2x = 2c dt dt Here, the Chain Rule is used to differentiate each of the variable terms with respect to time. Once the dynamic rate equation has been found, it can be solved for the unknown rate. Then the known values of the variables and rates can be substituted and the solution calculated. c = 81 ft 2

= 812 p x = 812 − 722 ≈ 37.1 ft dc = −60 ft/min dt dx c dc = · dt x dt    81 ft ft = −60 37.1 ft min ≈ −131 ft/min

x + 72

2

So the ship is traveling at a speed of 131 ft/min. The negative sign merely indicates that the distance x is decreasing over time. Since the equation was not first solved to give an explicit function of one of the variables, the differentiation step in the anchor chain problem is referred to as implicit differentiation. While the static equation could have been solved for x before taking derivatives, implicit differentiation is an efficient alternative.

8.2. DIFFERENTIATING IMPLICITLY

155

Example 16.

On a morning of a day when the sun will pass directly overhead, the shadow of an 80-foot tall building on level ground is 60 feet long. At that moment, the angle that the sun’s rays make with the ground is increasing at the rate of 0.27 degrees per minute. At what rate is the length of the shadow decreasing?

Figure 8.3 Solution:

Figure 8.3 shows a diagram of the solution.

Method 1 If we denote the length of the shadow as L, then right triangle trigonometry (assuming a 80 vertical building wall) gives a static equation of tan(θ) = . The dynamic equation is L then   d 80 d (tan(θ)) = dt dt L   d dθ d 80 dL (tan(θ)) · = · dθ dt dL L dt dL dθ sec2 (θ) · = −80L−2 · dt dt dθ 80 dL sec2 (θ) · = − 2· dt L dt Solving for

dL , we find dt dL = dt



L2 − 80



· sec2 (θ) ·

dθ . dt

156

CHAPTER 8. RELATED RATES

Now, when L is 60 feet, the hypotenuse of the triangle in the diagram is 100 feet, and 100 ft 5 dθ sec(θ) = = . But notice that θ is changing at 0.27 degrees per minute. Since 60 ft 3 dt resulted from differentiating tan(θ), our formulas for derivatives of trigonometric functions dθ require that θ be measured in radians (a dimensionless unit), so must be expressed at dt 0.00471 radians per minute, or just 0.00471 min−1 . Finally, substituting in the known values gives dL = dt



(60 ft)2 − 80 ft

  2  5 · · 0.00471 min−1 ≈ −0.59 ft/min. 3

Method 2 L The static equation could also be written as cot(θ) = . The advantage of doing this 80 dθ 1 dL is that the dynamic equation is just − csc2 (θ) · = · , and the solution proceeds dt 80 dt as before. It is often the case that the use of a minor trigonometric function (cotangent, secant, or cosecant) can simplify the solution process. 

8.2.1

Target Practice

1) In order to see the benefit of using implicit differentiation, solve the anchor chain problem shown in Figure 8.1 by first finding an explicit function for x in terms of c. 2) The volume V (in cubic inches) and pressure P (in pounds per square inch) of a certain gas sample satisfy the equation P V = 1000. At what rate is the volume of the sample changing if the pressure is 100 lb/in2 and is increasing at the rate of 2 lb/in2 per second?

8.2.2

Exercises

Exercise 134. Finish the solution of Example 16 using Method 2.

Warning. In the remaining exercises, you may need to look up derivative formulas that have not been discussed yet in the course.

8.2. DIFFERENTIATING IMPLICITLY

157

Exercise 135. When air expands adiabatically (without gaining or losing heat) its pressure and volume are related by the equation P V 1.4 = C. At what rate is the volume of the sample changing if the pressure is 80 kPa and is decreasing at a rate of 10 kPa per minute when the volume is 400 cm3 ?

Exercise 136. A ladder 41 feet long that was leaning against a wall begins to slip. The bottom of the ladder slides away from the wall at a rate of 2 ft/s. a) How fast is the top of the ladder sliding down when it is 25 feet above the ground? b) How fast is the top of the ladder sliding down when the base of the ladder is 25 feet from the base of the wall? c) How is the angle between the ladder and the ground changing when the base of the ladder is 25 feet from the base of the wall? d) How is the area of the triangle which is formed by the ladder, the wall and the ground changing when the base of the ladder is 25 feet from the base of the wall?

Exercise 137. A weather balloon that is rising vertically is being observed from a point on the ground 300 feet from the spot directly beneath the balloon. At what rate is the balloon rising when the angle between the ground and the observer’s line of sight is 45◦ and is increasing at 1◦ per second?

Exercise 138. When two resistors R1 and R2 are connected in parallel, the effective resistance R is given by 1 1 1 = + . R R1 R2 R1 and R2 are increasing at the rates 1.0 and 0.8 ohms per second respectively. What is the rate of change of R when R1 = 70 ohms and R2 = 55 ohms?

158

8.3

CHAPTER 8. RELATED RATES

Other Uses of Implicit Differentiation

It was pointed out in Section 8.2 that it may be more efficient to differentiate implicitly than to first solve for a variable and then differentiate. For example, the equation y 4 + x2 y 2 = 10 can be solved algebraically to find an explicit function for y, but the algebra can be somewhat tedious. However, implicit differentiation, using Product and Chain Rules, allows us to find the derivative with much simpler algebraic operations.   d d y4 + x2 y 2 dx dx     dy d dy d y4 · + x2 y2 · + y 2 (2x) dy dx dy dx  dy dy 4y 3 · + x2 (2y) · + y 2 (2x) dx dx  dy 4y 3 + 2x2 y dx Solving for

=

d (10) dx

= 0 = 0 = −2xy 2

dy we have: dx dy dx

= =

−2xy 2 4y 3 + 2x2 y −xy 2y 2 + x2

Notice that the derivative is expressed in terms of both variables x and y. In other cases, implicit differentiation may be the only way to find a derivative. Figure 8.4 2 shows the graph of a lemniscate with equation x2 + y 2 = 4xy. Algebraic techniques dy are of little use for finding here, but implicit differentiation can be used to find an dx expression for the derivative.

8.3.1

Target Practice

Find a formula for

8.3.2

dy if x2 cos(y) = y 2 . dx

Exercises

In Exercises 139–141, find a formula for Exercise 139. x3 + y 3 = 3xy

dy . dx

8.3. OTHER USES OF IMPLICIT DIFFERENTIATION

159

2 Figure 8.4: The lemniscate x2 + y 2 = 4xy. Graphs like this are often analyzed more easily in a polar coordinate system.

Exercise 140. yex = sin(y 2 )

Exercise 141. y ln(x) = cos(y)

160

CHAPTER 8. RELATED RATES

Exercise 142. Differentiate the equation of the lemniscate in Figure 8.4 implicitly, and find the value of the slope of a tangent line at the point (1, 1).

8.3.3

Solutions to Target Practice

Section 8.1 1) dV dt

  d 4 3 = πr dt 3  dr 4 = π 3r2 3 dt 2 dr = 4π(r ) dt

At the time in question, r=8 mm, and it is increasing by 0.2 mm/day, so dV dt

= 4π(8 mm)2 (0.2 mm/day) = 51.2π mm3 /day ≈ 161 mm3 /day

2) dP dt

= = = = ≈ ≈

 d  14.7e−0.21h dt   dh 14.7 −0.21e−0.21h dt dh −3.087e−0.21h dt −3.087e−0.21(3.0) (200)   psi 1 hr −328.822 hr 3600 sec psi −0.0913 sec

8.3. OTHER USES OF IMPLICIT DIFFERENTIATION Section 8.2 1) p c2 − 722  d d p 2 (x) = c − 722 dt dt −1/2 1 2 dc dx = c − 722 · (2c) · dt 2 dt c dc = √ · c2 − 722 dt x =

When 81 feet of chain are still out, then   dx 81 ft ft =√ · −60 ≈ −131 . 2 2 dt min min 81 − 72

2) Differentiating P V = 1000 with respect to time we have: d d (P V ) = (1000) dt dt dP dV ·V +P · = 0 dt dt dV V dP = − dt P dt

Using P V = 1000 to solve for V when P = 100 lb/in2 we have: V (100) = 1000 V

= 10

Plugging in our known values, we have: dV 10 in3 =− dt 100 lb/in2

lb/in2 2 s

! = −0.2 in3 /s.

161

162

CHAPTER 8. RELATED RATES

Section 8.3   dy dy + 2x cos(y) = 2y x − sin(y) dx dx dy dy 2x cos(y) = 2y + x2 sin(y) dx dx  dy 2x cos(y) = 2y + x2 sin(y) dx dy 2y + x2 sin(y) = dx 2x cos(y) 2

Chapter 9

Optimization

9.1

Finding the Best Strategy

Suppose you want to hang a fabric panel on a wall as shown in Figure 9.1. A single wire attaches the panel to the wall, but two wires are attached to the upper corners of the panel for stability. How much wire will be needed? If the vertical wire is 10 inches long, the total length of wire is about 51.2 inches, but if the vertical wire is 15 inches long, 52.4 inches of wire are required; and if the vertical wire is 8 inches long, a total of 55.3 inches are needed. Is 10 inches the length that will minimize the total length of wire? How can we be sure? One way to approach the problem is to make a graph that shows the total length as the place where the wires meet is varied. (See Figure 9.2.) 163

Figure 9.1

164

CHAPTER 9. OPTIMIZATION

p The slanted wire has a length of x2 + 182 inches, so the total length of the wire needed to hang the panel is p L = 2 x2 + 324 + (20 − x) inches. A graph of L versus x is shown in Figure 9.3.

In this example, the length x has been chosen so as to produce a simpler function for L.

Figure 9.2

Figure 9.3: A graph of L versus x for the length of the wire as shown in Figures 9.1 and 9.2.

Notice that the problem domain for the function is from x = 0 to x = 20. (Why?) And it appears that the minimum total length occurs for x slightly greater than 10 inches. In fact, at the exact minimum, the graph clearly levels off. So finding the value of x that results in a slope of 0 will produce the minimum total length.

9.1. FINDING THE BEST STRATEGY

165

The derivative of the length function is L0 (x) = √

so solving the equation √

2x − 1, + 324

x2

2x − 1 = 0 will yield the required result. + 324

x2



2x −1 = 0 + 324 2x √ = 1 2 x + 324 p 2x = x2 + 324

x2

4x2 = x2 + 324 3x2 = 324 x2 = 108 √ √ 108 = 6 3 ≈ 10.4 x = √ This gives us an optimum length wire √ of 20 − 6 3 ≈ 9.6 inches, and two √ for the vertical √ slanted wires each of length 182 + 108 = √432 = √ 12 3 ≈ 20.8√inches. This gives us a total length of wire of approximately 20 − 6 3 + 24 3 = 20 + 18 3 ≈ 51.2 inches.

9.1.1

Local Maxima and Minima

When a point on the graph of a function is higher (or lower) than the points around it, the point is a local maximum (or local minimum). A value for the independent variable that produces one of these points is called a critical value. The method used in the wall hanging example can be used to locate critical values in most cases. (See Section 9.3 for a discussion of the exceptions.) The quantity to be optimized is called the objective function, and its value depends on the values of one or more decision variables. In many situations, the following guidelines will be helpful. Guidelines for solving Optimization Problems 1. Write an equation for the objective function in terms of only one decision variable. A sketch may help here. 2. Make a graph of the objective function to find approximate locations of local maxima and minima. 3. Differentiate the objective function. 4. Find any critical values that result from setting the derivative equal to 0. 5. Test critical values to confirm a solution to the original problem.

166

CHAPTER 9. OPTIMIZATION

Example 17. The turning effect of a ship’s rudder can be shown to be proportional to sin2 (θ) cos(θ), where θ is the angle the rudder makes with the keel. At what angle is the rudder most effective? Solution: The “most effective” angle is the one that maximizes the turning effect. If we call the turning effect T , then the objective function is T = k sin2 (θ) cos(θ), where k is an unknown proportionality constant. However, we can still graph the function T = sin2 (θ) cos(θ), as shown in Figure 9.4. As you can see, there is one local maximum k

Figure 9.4: The graph of T /k = sin2 (θ) cos(θ) around the local maximum occurring between 50◦ and 60◦ . Note that the vertical axis is dimensionless while the horizontal axis is measured in degrees. somewhere between 50◦ and 60◦ . If we move to calculus and, consequently, measure our angle θ in radians, the Product Rule and the Chain Rule can be used to find the derivative.  dT d = k sin2 (θ) cos(θ) dθ dθ   d d 2 2 = k sin (θ) · (cos(θ)) + cos(θ) · sin (θ) dθ dθ   = k sin2 (θ) (− sin(θ)) + cos(θ) (2 sin(θ) cos(θ))  = k sin(θ) − sin2 (θ) + 2 cos2 (θ) dT = 0 whenever either sin(θ) = 0 or − sin2 (θ) + 2 cos2 (θ) = 0. However, for dθ 0 ≤ θ ≤ π/2 (remember that θ is currently in radians!) the only solution to sin(θ) = 0 is

Therefore,

9.1. FINDING THE BEST STRATEGY

167

θ = 0 which means the ship isn’t turning at all. That is certainly not the answer we are looking for, so we need to turn our attention to solving the alternate equation. − sin2 (θ) + 2 cos2 (θ) = 0 2 cos2 (θ) = sin2 (θ) sin2 (θ) 2 = cos2 (θ) 2 = tan2 (θ) √ tan(θ) = 2 θ = arctan(   180 ≈ 54.74◦ 0.9553 π

p (2)) ≈ 0.9553 radians

 (Notice that the result does not depend on the constant k. In general, the presence of a constant multiplier in a function will not affect the optimization process.)

9.1.2

Target Practice

When you cough, your trachea (windpipe) contracts to increase the velocity of the air going out. Biologists have discovered that a good model for the velocity of the air escaping from the trachea as a function of its radius is given by the equation v(r) = c(r0 r2 − r3 ) , where c is a positive constant and r0 is the radius of the trachea at rest. How much does the trachea contract to achieve the maximum velocity for escaping air? (X-ray photographs have confirmed the theoretical result.)

9.1.3

Exercises

Exercise 143. The bending moment of a beam is an indication of the torque due to the various loads and other forces acting on the beam. For the beam in the figure, simply supported at the left end and anchored at the right end, the bending moment equation is given by:  w M (x) = −18Lx + 24x2 . 48 Here, the beam of length L supports a (constant) uniform load of w pounds per foot. The beam is most likely to break where the bending moment has its maximum value. For the 16 foot beam in Figure 9.5, find that point.

168

CHAPTER 9. OPTIMIZATION

Figure 9.5: A sketch of a 16 foot beam supporting a load. Exercise 144. An object with weight W is dragged along a horizontal plane by a force acting along a rope attached to the object, as shown in Figure 9.6. If the rope makes an angle θ with the plane, then the magnitude of the force is F =

µW , µ sin(θ) + cos(θ)

where θ is the (constant) coefficient of friction.

Figure 9.6 a) Find the angle that minimizes the frictional force in terms of µ. b) If µ = 0.25 (typical for a wooden crate on a wood floor), what is the best angle for pulling the crate?

Exercise 145. When a person is walking, the magnitude F of the vertical force of one foot on the ground can be approximated by F = A (cos(bt) − a cos(3bt)) for time in seconds measured from t = 0 midway through the step. (Here A, a, and b are all positive numbers.) Find the time when maximum force occurs in terms of a and b. You will need to use the following trigonometric identity: sin(3x) = 3 sin(x) − 4 sin3 (x).

9.1. FINDING THE BEST STRATEGY

169

Exercise 146. A gutter is to be made out of a long sheet of metal 30 cm wide by turning up strips of width 10 cm along each side so that they make equal angles θ with the vertical. For what value of θ will the carrying capacity be greatest?

Exercise 147. A one-port network, operating at a fixed frequency ω, terminates in a resisThe power dissipated in R is P = I 2 R, and E I = where Z is the total impedance of tor R. Z p the circuit, given by Z = (ωL)2 + (R + r)2 . (Assume E, ω, L, and r are constant.) In terms of all the other circuit parameters, what value for the resistor R will result in maximum power dissipation through the resistor?

Exercise 148. To get the best view of the Statue of Liberty, you should be at the position where θ , the angle between your lines of sight to the top and bottom of the statue, is a maximum. If the statue itself is 46 meters high, and stands on a pedestal that is 46 meters high, how far from the base of the pedestal should you be? (Hint: Find a formula for θ in terms of your distance from the base.)

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CHAPTER 9. OPTIMIZATION

9.2

Deciding Whether an Extreme Value is a Maximum or Minimum

When you find a critical value by setting the derivative of the objective to zero, there is no indication of which type of extreme value will result. There are a few ways of distinguishing between local maximums and local minimums, however. • If you have graphed the objective function, as in Section 9.1, the choice should be fairly obvious. The graphs made it clear that an x-value of 10.4 inches minimized the length of wire (Figure 9.3), and that an angle of 54.7◦ maximized the turning effect (Figure 9.4). • In the absence of a graph, evaluating the objective function for values of the decision variable just below and just above the critical value may indicate whether a maximum or minimum has been found. • Evaluating the derivative of the objective function just below and just above the critical value can indicate whether the slope of the graph goes from negative to positive (for a local minimum) or positive to negative (for a local maximum). • If the second derivative of the objective function is easy to find, its value may confirm what type of extreme value has been found. (See Figure 9.7) Since the second derivative is the derivative of the first derivative, it measures the rate of change of f 0 . So if f 0 is decreasing, f 00 is negative. This means that the concave side of the graph faces downward. Similarly, a positive second derivative means that f 0 is increasing, so the concave side of the graph will face upward. In summary, – There is a local maximum at a critical value for which f 0 (x) = 0 and f 00 (x) < 0. – There is a local minimum at a critical value for which f 0 (x) = 0 and f 00 (x) > 0. Example 18. Use methods discussed in this section to confirm that θ = 57.4◦ maximizes the turning effect in Example 17. Solution: • As discussed previously, the graph in Figure ?? shows a local maximum. • T (50◦ ) = k sin2 (50◦ ) cos(50◦ ) ≈ 0.377k, and T (60◦ ) = k sin2 (60◦ ) cos(60◦ ) = 0.375k. Both of these values are less than T (57.4◦ ) ≈ 0.385k, assuming that k is positive, so there is a local maximum at θ = 54.7◦ . •  T 0 (54◦ ) = k sin(54◦ ) − sin2 (54◦ ) + 2 cos2 (54◦ ) ≈ 0.030k,

9.2. DECIDING WHETHER AN EXTREME VALUE IS A MAXIMUM OR MINIMUM171

Figure 9.7: A test plot showing how a local maximum is concave downward, f 00 < 0, while a local minimum is concave upward, f 00 > 0.

and  T 0 (55◦ ) = k sin(55◦ ) − sin2 (55◦ ) + 2 cos2 (55◦ ) ≈ −0.011k. Therefore, T 0 (θ) goes from positive values to negative values as θ passes through 54.7◦ , showing again that there is a local maximum at 54.7◦ . • Finally, T 00 (θ) = k(2 cos3 θ−7 sin2 (θ) cos(θ)), which equals about −2.3k for θ = 57.4◦ , confirming the conclusion once more. A negative second derivatistve means the graph is concave downward. When this occurs at a point where the first derivative is 0, the point must be a local maximum.  It is only necessary to use one of these methods to determine whether a critical point is a local maximum or a local minimum. If the second derivative is relatively easy to calculate, it is often a good choice. Otherwise, one of the other methods may be preferred.

172

9.2.1

CHAPTER 9. OPTIMIZATION

Target Practice

A cross section of a rectangular wooden beam cut from a circular log of diameter 10 inches has width w and depth d. The strength of the beam varies directly as the product of the width and the square of the depth. Find the dimensions of the cross section for the beam of greatest strength.

9.2.2

Exercises

√ Exercise 149. Use methods discussed in this section to confirm that x = 108 does indeed produce a minimum length for the wire in the wall hanging situation from the beginning of Section 9.1. (Use all four methods, as shown in Example 18.)

For Exercises 150–153, use one of the methods discussed in this section to confirm your answer.

Exercise 150. When a drug is injected into the bloodstream, its concentration C(t) at t e−bt − e−at minutes after injection is given by a formula of the form C(t) = K , where K, a−b a, and b are positive constants and a > b. When does the maximum concentration occur?

Exercise 151. The amount of illumination of a surface is proportional to the intensity of the light source, inversely proportional to the square of the distance from the light source, and proportional to sin(θ), where θ is the angle at which the light strikes the surface.

9.2. DECIDING WHETHER AN EXTREME VALUE IS A MAXIMUM OR MINIMUM173 A rectangular room measures 10 feet by 24 feet, with a 10-foot ceiling. Determine the height at which the light should be hung from the center of the ceiling to allow the corners of the floor to receive as much light as possible.

Exercise 152. An electrical generating plant is along a river that is 1 mile wide. A factory that needs power is 15 miles downriver on the opposite side. Cable is to be laid from the plant along the riverbank to a point P . Then cable is to be laid underwater from the point P straight to the factory. Let x denote the distance from point P to the point on the electrical plant side of the river directly across from the factory. If it costs $55,000 per mile to lay cable underwater and $40,000 per mile to lay cable underground, find the distance x that will minimize the total cost C of laying the entire cable.

Exercise 153. A solarium is to be attached to a building as shown in Figure 9.8. Support posts are in place and stand 6 feet high at a distance of 10 feet from the building. a) Where should the concrete footings be located in order that the shortest beams are used? (Hint: Let the distance along the ground from the post to the footing be represented by x.) b) How long are the beams?

174

CHAPTER 9. OPTIMIZATION

Figure 9.8: A sketch of the solarium.

9.3

9.3.1

A Closer Look at Optimization

Another Type of Critical Point

Consider the graph of y = 1 − (x − 1)2/3 shown in Figure 9.9.This function is defined for all x, and the point (1, 1) is clearly a local maximum. However, its derivative is not defined for x = 1. (You should verify by calculation that y 0 (1) does not exist.) We can therefore enlarge the definition of critical point to include points where • the first derivative equals zero; or • the first derivative is undefined. Figure 9.9

9.3. A CLOSER LOOK AT OPTIMIZATION

9.3.2

175

A Critical Point that is not a Local Maximum or Minimum

The cubic function y = x3 shown in Figure 9.10 has one critical point at (0,0), but that point is neither a local maximum nor a local minimum. Although the slope of the graph at (0, 0) is equal to 0, the graph only levels off briefly and then continues to rise. √ Furthermore, the graph of y = 3 x shown in Figure 9.11 is also without a local maximum or minimum. In this case, the slope and derivative are undefined for x = 0. In fact, the graph has a vertical tangent there. Figure 9.10

So, while local maxima and minima occur only at critical points, it is possible to have a critical point that is neither a maximum nor a minimum. Formally, being a critical point is necessary but not sufficient to be a local maximum or minimum.

Figure 9.11

9.3.3

Absolute Maxima and Minima

An absolute maximum of a function f (x) is a point x∗ with f (x∗ ) = M for which f (x) ≤ M for all values of x in the domain. An absolute minimum of a function f (x) is a point x∗ with f (x∗ ) = m for which f (x) ≥ m for all values of x in the domain. Refer to the wall hanging (Figure 9.1) discussed at the beginning of this chapter. What if we were asked to find a maximum total length for the wire? The function p L = 2 x2 + 324 + (20 − x), is defined for all real numbers x, and has no maximum value. But the domain of the wall hanging problem is the interval (0, 20), and on this interval there is a maximum length

176

CHAPTER 9. OPTIMIZATION

of 56 inches when x = 0. Of course, this corresponds to eliminating the vertical wire and attaching two slanted wires at the uppermost point. There is an absolute maximum total length of wire at x = 0 when L = 56. In general, in addition to searching for critical points, you should examine the endpoints of the interval comprising the problem domain in order to identify all maxima and minima.  √  In the case of the wall hanging problem, the local minimum at 0, 108 is also the absolute minimum length.

9.3.4

Target Practice

In Example 17, the rudder angle producing maximum turning effect was shown to be 54.7◦ . This is both a local maximum and an absolute maximum. What can you say about absolute minima for this situation?

9.3.5

Exercises

Exercise 154. A “self-holding” rectangular box with a lid is made from the given template.

a) Find the value(s) of x that will produce an absolute maximum volume for the box. b) Find the value(s) of x that will produce an absolute minimum volume for the box.

9.4. INFLECTION POINTS

177

Exercise 155. The owner of a retail lumber store wants to construct a fence to enclose an outdoor storage area adjacent to the store. Find the dimensions that will enclose the largest area if

a) 240 feet of fencing material is used b) 400 feet of fencing material is used In each case, identify all absolute maxima and minima.

Exercise 156. Find all local and absolute maxima and minima for the function y = q 2 (x2 − 2) on the interval [−3, 3].

Exercise 157. Find all local and absolute maxima and minima for the function y = x ln(x).

9.4

Inflection Points

One more graphical feature that can be identified with the aid of derivatives is called a point of inflection or inflection point. Figure 9.12 shows an inflection point on a graph, which is a point where the curvature changes from concave down to concave up (or vice-versa). If a smooth curve (i.e. a curve where higher order derivatives exist and are continuous) goes through an inflection point, the second derivative must equal 0, and must also change sign. Notice that the critical point at (0, 0) on the graph of y = x3 in Figure 9.10 is neither a local maximum nor a local minimum, but is in fact an inflection point. And (0, 0) is also an

178

CHAPTER 9. OPTIMIZATION

Figure 9.12: A cubic curve showing a local maximum, a local minimum, and the point of inflection in between. inflection point on the graph of y = is undefined there.

9.4.1

√ 3 x in Figure 9.11, even though the second derivative

Curve Sketching

An entire graph can usually be sketched by locating significant points through the use of derivatives. For the graph of a function f (x), • Local maximum or minimum points may be located at critical points, for which f 0 (x) = 0 or f 0 (x) is undefined. • At a critical point, f 00 (x) > 0 identifies a local minimum and f 00 < 0 identifies a local maximum. • Absolute maxima and minima may also be located at endpoints of an interval or domain. • A point of inflection is located where f 00 = 0, provided the curvature changes from concave up to concave down or from concave down to concave up. • Intercepts (x = 0 or y = 0) are also useful. Example 19. Sketch a graph of y = x4 − 4x3 + 10 using derivatives.

9.4. INFLECTION POINTS

179

Solution: y 0 = 4x3 − 12x2 = 4x2 (x − 3), so y 0 = 0 at the critical values x = 0 and x = 3. y 00 = 12x2 − 24x = 12x(x − 2). Plugging in our critical values, we see that y 00 (3) = 12(3)(3 − 2) > 0 so the point (3, −17) is a local minimum. At the other critical value, y 00 (0) = 12(02 ) − 24(0) = 0, so the second derivative test does not reveal whether x = 0 is a maximum, minimum or point of inflection. If we check the values close to 0, we can see that for a small negative value −, we have y 00 (−) = 12(−)(− − 2) = (−)(−) > 0. Also, at a small positive value , y 00 () = 12()( − 2) = (+)(−) < 0. Therefore, we can see that the curvature is moving from concave up to concave down as we pass through x = 0, so x = 0 is a point of inflection. In addition, there is another zero of y 00 = 12x(x − 2) = 0 at x = 2. Therefore x = 2 is a point of inflection as well. You can check the sign of y 00 for 2 −  and 2 +  to see this. The last think to check for is a y-intercept. In this case we have a y-intercept at y = 04 − 4(03 ) + 10 = 10. Putting all of this information together we find Figure 9.13.



Figure 9.13: A plot of the graph y = x4 − 4x3 + 10.

9.4.2

Target Practice

Use derivatives and intercepts to sketch the graph of y =

9.4.3

ln(x) x ,

including inflection points.

Exercises

For Exercises 158 and 159, use derivatives and intercepts to sketch the graph of each function, including inflection points.

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CHAPTER 9. OPTIMIZATION

Exercise 158. y = 5x2 − x4

Exercise 159. y = x2 e−x

Exercise 160. Figure 9.14 shows a graph of y = 2x4 − x − 2.

a) Where does y 0 = 0? What can you say about the graph where y 0 = 0?

b) Where does y 00 = 0? What can you say about the graph where y 00 = 0?

c) Discuss the concavity of the graph. Figure 9.14

9.4.4

Solutions to Target Practice

Section 9.1 The derivative of the velocity function is v 0 (r) = c(2r0 r − 3r2 ),

9.4. INFLECTION POINTS

181

so solving the equation c(2r0 r − 3r2 ) = 0 will yield the required result. Since c is a positive constant, the expression in parentheses must equal 0. 2r0 r − 3r2 = 0 r(2r0 − 3r) = 0

Since r = 0 would mean that the trachea would be totally closed, 2r0 − 3r = 0 gives the correct result of r = 32 r0 . So the maximum velocity occurs when the radius of the trachea is reduced to 23 of its resting value. Section 9.2 w=

10 √ 3

inches, d =

√ 10 √ 2 3

Section 9.3 Absolute minima result for angles of 0◦ and for 90◦ , the endpoints of the problem domain. Section 9.4 y(1) = 0, so (1, 0) is an x-intercept. y(0) doesn’t exist,  so there  is no y-intercept, although 1 the y-axis is a vertical asymptote. y 0 = 1−ln(x) = 0 at e, , which is a local maximum. x2 e  3 3 3 3 3 y 00 = 2−3ln(x) = 0 at x = e 2 when y = 32 e− 2 , so e 2 , e− 2 is an inflection point. The x3 2

x-axis is a horizontal asymptote.

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CHAPTER 9. OPTIMIZATION

Chapter 10

Substitution as a Method of Integration 10.1

Changing Variables in Integrals

From the antiderivatives of the most basic functions, we have the following formulas for evaluating integrals: Z xn+1 + C when n 6= −1 xn dx = n+1 Z 1 sin(kx) dx = − cos(kx) + C k Z 1 cos(kx) dx = sin(kx) + C k Z 1 kx e +C ekx dx = k Z 1 dx = ln |x| + C x Z x5 1 So, for example, the integral x4 dx is evaluated as + C or x5 + C. But look carefully 5 5 at the integral Z 4  x3 + 2 3x2 dx. From your knowledge of derivatives, you can see a connection between two parts of this integral. The 3x2 in the second set of parentheses is the derivative of the expression in the first set of parentheses. Because of this connection, we can simplify the integral by 183

184

CHAPTER 10. SUBSTITUTION

replacing (x3 + 2) with a new variable u = x3 + 2. Therefore, u = x3 + 2 du = 3x2 dx and so we can rewrite our integral Z Z Z 4  du x3 + 2 3x2 dx = u4 dx = u4 du. dx In this form, the Power Rule applies so we have Z 1 u4 du = u5 + C. 5 Finally, returning to the original variable, we have Z 5 4  1 3 x + 2 + C. x3 + 2 3x2 dx = 5 This process of changing the variable so that an integral can be evaluated with one of the basic formulas is called integration by substitution, or sometimes just u-substitution. It is an extremely valuable technique that makes it possible to integrate exactly a huge variety of integrals beyond those of the basic functions.

Remark. This operation may feel familiar. Consider the Chain Rule for differentiation as described in Section 5.1 and specifically Theorem 6. If we simply note that the Chain Rule states that d dF du · (F (u(x))) = dx du dx then we can also say that Z Z Z Z d dF du dF (F (u(x))) dx = · dx = du = dF = F (u) = F (u(x)) dx du dx du which shows us that if we have an integral that we wish to calculate that takes the form Z du f (u(x)) · dx, dx then its solution is

Z f (u) du,

expressed in terms of the original variable x.

10.1. CHANGING VARIABLES IN INTEGRALS

185

For our initial example, the Chain Rule for differentiation yields   5 4  4  d 1 3 1 x +2 = · 5 x3 + 2 3x2 = x3 + 2 3x2 , dx 5 5 so it is clear that

Z

x3 + 2

4

5  1 3 x + 2 + C. 3x2 dx = 5

Guidelines for Integration by Substitution 1. Recognize a connection in which one part of an integral is the derivative of another part. 2. Identify one part as u, and use this substitution to rewrite the entire integral in terms of u. Note that each occurrence of the original variable must be replaced by its equivalent in terms of u. 3. If the substitution results in a basic form, evaluate the u-integral. 4. Back-substitute to rewrite the solution in terms of the original variable. Z 6 Example 20. Integrate t5 et dt. Solution: Notice that the derivative of t6 , the exponent of e, is 6t5 . This suggests that the substitution u = t6 is worth trying, even though a factor of 6 is not present in the du 1 integral. With this substitution, = 6t5 and, solving for t5 dt we have t5 dt = du so dt 6 the integral can be rewritten as Z Z 1 u 1 1 6 6 t5 et dt = e du = eu + C = et + C. 6 6 6  Z Example 21. Integrate Solution: tution

p cos(3θ) sin(3θ) dθ.

Since it does not appear in the final solution, u is sometimes referred to as a dummy variable.

Since cos(x) is the derivative of sin(x), it is reasonable to try the substiu = sin(3θ).

186

CHAPTER 10. SUBSTITUTION

Then

du = 3 cos(3θ) dθ

and cos(3θ) dθ =

1 du. 3

The integral can be rewritten as   Z p Z √ 1 sin(3θ) cos(3θ) dθ = u du 3 Z 1 u1/2 du = 3 1 u3/2 = · +C 3 3/2 2 (sin(3θ))3/2 + C. = 9 

10.1.1

Target Practice

1. Use the substitution u =

w4

Z + 10 to evaluate

4w3 dw. w4 + 10 Z

2. Identify an appropriate substitution u and then use that choice to evaluate

10.1.2

Exercises

For Exercises 161–163, use the given substitutions to evaluate the integrals. Z 3 Exercise 161. x2 − 1 (2x) dx with u = x2 − 1.

Z

sin9 (t) cos(t) dt with u = sin(t).

Z

(2t + 1)et

Exercise 162.

Exercise 163.

2 +t

dt with u = t2 + t.

x 1 − x2

99

dx.

10.1. CHANGING VARIABLES IN INTEGRALS

187

For Exercises 164–170, identify an appropriate substitution u and then use that choice to evaluate each integral. Z 11 Exercise 164. 7x6 x7 + 1 dx

Z

3G2 eG dG

Z

cos(t) dt sin2 (t)

Z

1 √ ds 4s + 9

Z

sin(πv) cos3 (πv) dv

Z

ln(x) dx x

Z

r+1 dr r2 + 2r

Exercise 165.

Exercise 166.

Exercise 167.

Exercise 168.

Exercise 169.

Exercise 170.

3

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CHAPTER 10. SUBSTITUTION

Exercise 171. Sometimes a substitutionZwill work even when the derivative of u does not x √ appear directly in the integral. Integrate dx by using the substitution u = x + 1. x+1 For Exercises 172–173, use a variation on the substitution method similar to that of Exercise 171 to evaluate each integral. Z √ Exercise 172. t 3 2t + 4 dt

Z Exercise 173.

y2 dy (y + 1)4

In Section 5.2 and their accompanying exercises (c.f. Exercises 96 & 98) the following derivative formulas were calculated: d (tan(x)) dx d (sec(x)) dx d (cot(x)) dx d (csc(x)) dx

= sec2 (x) = sec(x) tan(x) = − csc2 (x) = − csc(x) cot(x)

For Exercises 174–178, use these formulas to evaluate each given integral. Z Exercise 174. sec2 (x) dx

Z Exercise 175.

cot(θ) csc(θ) dθ

Z Exercise 176.

csc2 (4t) dt

10.2. SOLVING DIFFERENTIAL EQUATIONS EXACTLY

Z Exercise 177.

10.2

189

x sec(x2 ) tan(x2 ) dx

Solving Differential Equations Exactly

In Chapter 7, you learned how to solve some types of differential equations, which are dy = equations containing derivatives. You saw that a differential equation of the form dx f (x) can be solved exactly by finding Z

dy dx = dx

Z f (x) dx Z

y =

f (x) dx,

provided that f (x) has an antiderivative. You also learned in class how to solve other differential equations using the ode45 function in Matlab, although this method provided only a numerical solution. Using ode45 resulted in a graph, but did not produce an equation for the function y. In this section we will explore analytical solution methods for a particular class of differential equations. Consider the initial value problem dy 2x = 2 subject to y(0) = 1. dx y

(10.1)

You can solve this using ode45 to obtain a graph of the solution. In this case, however, we want to find an analytical solution–that is, an equation expressing the relationship between y and x–so let’s try the method we used before which is to integrate both sides with respect to x. That becomes Z Z dy 2x dx = dx. (10.2) dx y2 Z dy On the left hand side, Equation 10.2 is quite solvable since dx = y. On the rightdx hand side, Equation 10.2 contains two variables, and cannot be evaluated. If, however, we

190

CHAPTER 10. SUBSTITUTION

return to Equation 10.1, and separate our variables we find y2

Z

dy dx

= 2x Z 2 dy y dx = 2x dx dx

(10.3)

Now, considering the integrand on the left-hand side of Equation 10.3 as a function of y and applying the chain rule d dF dy (F (y)) = , dx dy dx so we can rewrite Equation 10.3 as Z

y 2 dy =

Z 2x dx,

which integrates to y3 + C1 = x 2 + C2 . 3 By combining the two indeterminate constants, we find that y3 = x2 + C. 3 This is a general solution, which doesn’t account for the initial condition y(0) = 1 given in Equation 10.1. Applying this condition and solving for y, we finally have an explicit solution to the initial value problem: 13 3 1 3 y3 3 y3

= 02 + C = C

1 3 = 3x2 + 1 p 3 y = 3x2 + 1 = x2 +

The graph of this function is shown in Figure 10.1. Notice that the graph passes through the point (0, 1) as specified by the initial condition. Also, we can check our solution using the Chain Rule: −2/3 dy 1 2x 2x 2x = 3x2 + 1 (6x) = = √ 2 = 2 , 2/3 dx 3 y 3 (3x2 + 1) 3x2 + 1

10.2. SOLVING DIFFERENTIAL EQUATIONS EXACTLY

Figure 10.1: The graph of y =

191

p 3 3x2 + 1, the solution to Equation 10.1.

p 3 confirming that y = 3x2 + 1 satisfies the differential equation. The function therefore satisfies the given initial value problem. dy 2x = 2 was solved by separating the variables into their own dx y single-variable integrals. For this reason, the equation is called a separable differential equation, and it was solved by the method of separation of variables. Many important real-world differential equations are separable, and can be solved using the initial algebraic step of separation. The differential equation

Example 22. In Chapter 7 Example 11, you encountered a differential equation for the curing of one type of high-strength concrete:

dS = k(5000 − S), dt

with S representing the strength of the concrete at any time after being placed on site, and k being a constant of proportionality. You may assume that the initial strength of the concrete at t = 0 is 0. Solve the equation by the method of separation of variables.

192 Solution:

CHAPTER 10. SUBSTITUTION Separate the variables algebraically. dS dt 1 dS 5000 − S dt Z dS 1 dt 5000 − S dt Z 1 dS 5000 − S

= k(5000 − S) = k Z = k dt Z = k dt

The integral on the left-hand side can be evaluated by substitution, with u = 5000 − S and du = −1. This gives us dS Z Z −du k dt = u − ln |u| + C1 = kt + C2 − ln |5000 − S| = kt + C

(10.4)

Since the strength of the concrete will never be greater than the maximal value of 5000 pounds per square inch, (5000 − S) cannot be negative, so the absolute value symbol is not necessary. To find an explicit solution for S, first multiply both sides by −1 and then take the exponential of both sides: ln(5000 − S) = −kt + C e

ln(5000−S)

(10.5)

−kt+C

= e

5000 − S = e−kt eC 5000 − S = Ce−kt −kt

S = Ce

(10.6) + 5000

(10.7)

Why are there so many different colors used for C in this calculation? Each color of C is a different value, but since C is an arbitrary constant, it can be either positive or negative, or even 0, so there is no need to write “ − C” if “ + C” is more convenient. Furthermore, when a property of exponents is used to separate out a factor of eC , it can replaced by another arbitrary constant C for simplicity. We understand that all of these different-colored C’s have different values, but since they are all arbitrary, the rules of algebra are not actually violated. You will check this in Exercise 178. If we take a moment to think about what our equation was saying, we see that k controls the rate at which the maximal value 5000 is approached, with smaller values of k curing more slowly than larger values of k. The arbitrary constant C accounts for the initial strength of the concrete at time t = 0.

10.2. SOLVING DIFFERENTIAL EQUATIONS EXACTLY

193

Plugging in our initial condition S(0) = 0 we have 0 = Ce−k(0) + 5000 0 = C + 5000 C = −5000 and so S = −5000e−kt + 5000

or

  S = 5000 1 − e−kt . 

10.2.1

Target Practice

Pressure in a fluid varies with height. Near the surface of the Earth, the rate of change of dP Pg air pressure P with respect to height h is given by =− , where g is the acceleration dh RT 2 due to gravity 9.80 m/s , T is the absolute temperature in kelvins, and R is the ideal gas constant for air 287 J/kg/K. a) What does the negative sign in the differential equation tell you about pressure? b) Solve the differential equation to find pressure as a function of height, with the initial condition that standard air pressure at sea level is 101,000 pascals (N/m2 ), and on a day when the temperature is 30◦ C (303◦ K). c) Use your result to find the air pressure at the top of Mount Monadnock in New Hampshire, which is 965 meters high. d) Actually, your function is only valid for a limited range of heights near the Earth. Explain why by looking carefully at the quantities involved in the differential equation.

10.2.2

Exercises

Exercise 178. Check that the black, red, blue and green values of C are all calculable, but different, by plugging in the initial condition for Example 22 into Equation 10.4, Equation 10.5, Equation 10.6, and Equation 10.7, and solving for C. Are the solutions to Example 22 that you find the same in each case or are they different?

Exercise 179. Ritalin (methyl phenidate) is used to treat attention deficit hyperactive disorder (ADHD) and narcolepsy. Ritalin is eliminated from the bloodstream in such a way

194

CHAPTER 10. SUBSTITUTION

that the amount A of Ritalin in the blood decreases by 20% every hour. (In other words, the rate of change of the Ritalin concentration in the bloodstream is equal to −0.20A.) a) Write and solve an initial value problem (differential equation plus initial condition) for the rate of change with time of the amount of Ritalin in a person’s bloodstream, if an initial dose of 18 milligrams is administered at noon. b) How much Ritalin is in the person’s blood after two hours? c) When will the Ritalin concentration reach 50% of its initial level? (This time is called the half-life of the drug.)

Exercise 180. A heart pacemaker consists of a switch S, a battery E0 , a capacitor C, and the heart as a resistor R. When the switch is as shown in Figure 10.2, the capacitor charges, and when it switches to the heart circuit it discharges, sending an impulse to the heart. Kirchhoff’s Voltage Law in this case can be expressed as Ri +

q = 0, C

where i represents current and q represents the charge on the capacitor, dq both of which vary with time. But current is defined by i = . dt

Figure 10.2 Assume a heart resistance R = 1.8×10 ohms and a capacitance C = 0.50 microfarads, and an initial voltage at t = 0 of 90 millivolts on the capacitor. (Note: Charge, capacitance, and voltage are related by q = CV .) 6

a) Rewrite Kirchhoff’s Law as a differential equation in q. b) Write down an initial value problem for this heart pacemaker. c) Find the charge on the capacitor, q(t), (in coulombs) as a function of time. d) Find the current in the circuit, i(t), (in amperes) as a function of time. e) Find the value of the charge and the current after 0.70 seconds. f) Find the voltage across the capacitor, and therefore the heart muscle, after 0.70 seconds.

10.2. SOLVING DIFFERENTIAL EQUATIONS EXACTLY

195

Exercise 181. A DNA molecule has the shape of a double helix, whose two strands separate and reform in solution. The process of reforming, which is called renaturation, depends on the random collision of the complementary strands. The rate at which the concentration C of single-stranded DNA renatures into double-stranded DNA is proportional to the square of C. (Note: C always measures the concentration of single-stranded DNA that remains at any time.) The proportionality constant for a certain type of viral DNA is L k = −0.0032 . mg · s a) Write a differential equation for

dC . dt

b) Solve the differential equation to find a function describing the concentration C(t) of mg . single-stranded DNA at any time, if the initial concentration is 80 L c) At what time will the reaction be half completed? Hint: For the units pay careful attention to the units of k.

Exercise 182. A crack in a bone grows longer with use. If a represents crack length and N the number of cycles of repetitive stress, the rate of crack growth is given by Paris’ law: √ m da = C ∆σ πa , dN where C and m are experimentally measured constants. ∆σ is equal to the numerical range of stress on the bone during a cycle. If a runner has a tiny crack (0.01 mm) in her tibia (one of the lower leg bones), find a formula for the number of strides N until the crack grows to an arbitrary length a.

Exercise 183. In the world population problem activity from class, it was assumed that the growth rate was a constant 17 people per thousand. In fact, the net annual growth rate decreased almost linearly after 1990 to 12 people per thousand by 2006. Assume that linear decrease in growth rate has continued, alter your original differential equation and find a refined model for world population growth. (Notice that the constant growth rate k must be replaced by a variable function of time.) How well does it predict today’s world population?

196

10.3

CHAPTER 10. SUBSTITUTION

Substitution in Definite Integrals Z

The definite integral

2

x(x2 − 1)2 dx can be solved by substitution, with u(x) = x2 − 1

1

du and = 2x. The resulting integral is dx Z x=2 u(x)

2



x=1

1 du 2 dx

 dx.

Completing this computation, we have: Z

x=2 2

u(x) x=1



1 du 2 dx

 dx = = = = =

x=2 1 (u(x))3 · 2 3 x=1 3 ix=2 1h 2 x −1 6 x=1  1 2 3 (2 − 1) − (12 − 1)3 6  1 3 3 − 03 6 27 9 = = 4.5 6 2

If, for ease of computation, we want to switch all of our work into the u coordinate system, however, we need to recompute our limits of integration from x = 2 and x = 1 into values in the u system, u(2) and u(1). Doing so, we have u(2) = 22 − 1 = 3 and u(1) = 12 − 1 = 0 so our calculations become: u=3  Z u=3  1 1 u3 u2 du = · 2 2 3 u=0 u=0  1 3 = 3 − 03 6 27 9 = = = 4.5 6 2 In any case, it is essential to remember that the original limits of integration apply only to the original variable, not to any substituted variable. Z π/4 Example 23. Use substitution to evaluate tan4 (t) sec2 (t) dt. 0

10.3. SUBSTITUTION IN DEFINITE INTEGRALS

Since sec2 (t) is the derivative of tan(t), the substitution u = tan(t) can be

Solution: used, with

197

du = sec2 (t). Also, tan(0) = 0 and tan(π/4) = 1. dt Z u=1 Z t=π/4 u4 du tan4 (t) sec2 (t) dt = t=0

= = =

u=0 u=1 u5

5 u=0

15 05 − 5 5 1 5 

10.3.1

Target Practice

For each integral, a) Rewrite the integral using an appropriate substitution, including changing the limits of integration. b) Evaluate the integral. Z π/2 1. cos3 (t) sin(t) dt 0

Z 2. 0

10.3.2

1

(y 2

y dy + 1)3

Exercises

For Exercises 184–186, a) Rewrite the integral using an appropriate substitution, including changing the limits of integration. b) Evaluate the integral. Z 2 p Exercise 184. x2 x3 + 1 dx. 0

198

CHAPTER 10. SUBSTITUTION Z

1

(x + 1)(x2 + 2x)5 dx

Exercise 185. 0

Z

2

Exercise 186.

(v 2

1

4v + 12 dv + 6v + 1)2

For Exercises 187-191, use substitution to evaluate each integral. Z

3

(x + 2)2 dx

Exercise 187. 1

Z

π/2

sin5 (θ) cos(θ) dθ

Exercise 188. 0

Z

1

w2

Exercise 189.

p w3 + 9 dw

0

Z

π/6

sec2 (2t) dt

Exercise 190. 0

Z Exercise 191.

1

xex

2 +1

dx

0

Exercise 192. Find the area under the graph of y = x = 3. Use a substitution to evaluate the integral.

x over the interval x = 1 to x2 + 1

10.3. SUBSTITUTION IN DEFINITE INTEGRALS

10.3.3

199

Solutions to Target Practice

Section 10.1 1) u = w4 + 10 du = 4w3 dw Z Z 4w3 du dw = 4 w + 10 u = ln |u| + C = ln w4 + 10 + C

2)

Z

x 1 − x2

99

u = 1 − x2 du = −2x dx   Z 1 99 dx = u − du 2 Z 1 = − u99 du 2 1 u100 = − · +C 2 100 100 1 = − 1 − x2 +C 200

Section 10.2 a) The negative sign denotes the fact that pressure goes down as the height goes up.

200

CHAPTER 10. SUBSTITUTION

b) dP dh Z

1 dP P

ln |P | + C1 P

9.80(m/s2 )P (N) 9.80P = − = 287(J/kg/K)303◦ K 86, 961 Z 9.80 = − dh 86, 961 9.80 = − h + C2 86, 961 9.80 − 86,961 h

= Ce

101, 000(N/m2 ) = Ce−9.80(0)/86,961 = C P (h) = 101, 000e

9.80h − 86,961

c) P (965) = 101, 000e−9.80(965)/86,961 = 90, 592N/m2 d) In reality, both T and g depend on elevation and cannot be assumed to be constant over significant changes in the value of h. Section 10.3 1)

a) u = cos(t) du = − sin(t) dt u(0) = cos(0) = 1 Z

π/2

u(π/2) = cos(π/2) = 0 Z 0 3 cos (t) sin(t) dt = −u3 du

0

1

b) Z −

0

u3 du =

1

= = =

Z

1

u3 du

0 1 u4

4 0 1 0 − 4 4 1 4

10.3. SUBSTITUTION IN DEFINITE INTEGRALS 2) u = y2 + 1 du = 2y dy u(0) = 02 + 1 = 1 Z 0

1

u(1) = 12 + 1 = 2 Z 2 y 1 −3 dy = u du 2 3 (y + 1) 1 2 1 −2 2 = − u 4  1  1 1 1 − = − 4 4 1 1 4 1 1 − = − = 4 16 16 16 3 = 16

201

202

CHAPTER 10. SUBSTITUTION

Chapter 11

Additional Methods of Integration 11.1

Integration By Parts

Once you move past the basic integrals that you memorize, e.g. those presented in Section 7.1, there are several simplification/reduction methods analogous to our differentiation methods but there are also a wide variety of other methods of integration arising from various inspired thoughts. Many other useful methods of integration involve clever reformulations of the integrand, just as we do with substitution. In Chapter 10 we discussed the ’inverse’ of the chain rule. What happens if we consider the ‘inverse’ of the product rule? The Product Rule states that d (u(x)v(x)) = dx



du dx



 v(x) + u(x)

dv dx

 ,

for any differentiable functions u(x) and v(x). If we integrate both sides of the Product Rule with respect to x we have:   Z Z   d du dv (u(x)v(x)) dx = v(x) + u(x) dx, dx dx dx   Z   Z du dv u(x)v(x) = v(x) dx + u(x) dx. dx dx Rewriting this we have:     Z Z dv du dx = u(x)v(x) − v(x) dx. u(x) dx dx 203

(11.1)

204

CHAPTER 11. METHODS OF INTEGRATION Richard Courant saw that in the diagram shown in Figure 11.1, you can divide the area of the large rectangle sq = f (b)g(b) into three regions: 1. The area above the line v = r (= g(a)) and to the left of the curve (u(x), v(x)), 2. The area to the right of the line u = p (= f (a)) and below the curve (u(x), v(x)), 3. The area of the white rectangle pr = f (a)g(a).

Figure 11.1: Richard Courant’s graphical proof of integration by parts.

However, the shaded areas can each be expressed as integrals in themselves.

(Proofs Without Words) s

Z

b

Z u dv =

Region 1 =

a

r

and

q

Z Region 2 =

Z

b

v du = p

a

b

d u(x) (v(x)) = dx

Z

d v(x) (u(x)) = dx

Z

u(x)

dv dx dx

v(x)

du dx dx

a

a

b

so putting this all together we have: Z u(b)v(b) = a

b

dv dx + u(x) dx

Z

b

v(x) a

du dx + u(a)v(a), dx

or, more commonly, Z a

b

dv u(x) dx = u(b)v(b) − u(a)v(a) − dx

Z a

b

du v(x) dx = u(x)v(x)|ba − dx

Z

b

v(x) a

du dx. dx

Z Example 24. Calculate

ln(x) dx.

Solution: This is definitely not a basic antiderivative. To apply the method of integration by parts we need to identify a factorization of our integrand into a differentiable dv function u(x) and an integrable function . With the integrand ln(x) the simplest possidx ble factorization is ln(x) · 1. Since ln(x) is differentiable, but not easily integrable, let us dv take u = ln(x). That leaves the other factor 1 = . Plugging this into Equation 11.1 we dx

11.1. INTEGRATION BY PARTS

205

have u(x) = ln(x) du 1 = dx x dv = 1 dx Z v(x) = 1 dx = x Z Z Z 1 ln(x) · 1 dx = x ln(x) − · x dx = x ln(x) − 1 dx x = x ln(x) − x + C  Remark. In our calculation of v(x) we ignored the indeterminate constant. This is allowed because any constant chosen will be eliminated in the remaining calculations: Z v(x) = 1 dx = x + C1 Z Z 1 · (x + C1 ) dx ln(x) dx = (x + C1 ) ln(x) − x  Z  C1 = x ln(x) + C1 ln(x) − 1+ dx x = x ln(x) + C1 ln(x) − x − C1 ln(x) + C2 = x ln(x) + C2

11.1.1

Target Practice Z

1. Calculate

x cos(5x) dx. Z

2. Calculate

11.1.2

t2 ln(7t) dx.

Exercises Z

Exercise 193. Calculate

x sin(x) dx. First, identify u(x) and

method of integration by parts.

dv , then apply the dx

206

CHAPTER 11. METHODS OF INTEGRATION Z

Exercise 194. Calculate

x ln(x) dx.

Z

3te−6t dt.

Z

r3 ln(r) dr.

Exercise 195. Calculate

Exercise 196. Calculate

Z

π/2

Exercise 197. Calculate

x cos(2x) dx. 0

Z

x2 ex dx. (Hint: Apply Integration by Parts twice. )

Z

ex cos(x) dx.

Exercise 198. Calculate

Exercise 199. Calculate dv . dx b) Integrate by parts. a) Identify u(x) and

c) In the resulting integral, identify a new u(x) and

dv . dx

d) Integrate by parts again. Z e) Recognize that you have an algebraic equation in equation to compute the integral.

ex cos(x) dx and solve your

11.2. INTEGRATION USING TRIGONOMETRIC SUBSTITUTION

11.2

207

Integration using Trigonometric Substitution

Let’s consider a substitution that moves our integral from cartesian coordinates into polar coordinates. This substitution is useful for any integral that involves a square root. Why? Z B 1 We’ll compute the integral of dx x2 + 1 by transforming the integral into polar co1+x^2 ordinates. We cannot use our original chain x rule substitution, because there is no x in the numerator. If you consider the relationship Θ shown in the triangle in Figure 11.2, however, A 1 C there are several substitutions that transform the cartesian coordinates in x into polar co- Figure 11.2: Trigonometric substitution inp ordinates in θ. Check the following: volving x2 + 1 tan(θ) = =

opposite side adjacent side x =x 1

cos(θ) = =

dθ = 1 sec2 (θ) dx Plugging these into the integral, we have: Z

1 dx = 2 x +1

Z  Z

=



1 x2 + 1

adjacent side hypotenuse 1 √ 1 + x2

2 dx

cos2 (θ) sec2 (θ)

dθ dx dx

Z =

1 dθ

= θ+C Now, we need to return to our original cartesian coordinates in x. Since we know that tan(θ) = x, we can solve for θ: θ = arctan(x) + C. Z p Example 25. Compute 1 + x2 dx. Solution:

This integral uses the same triangle shown in Figure 11.2 to design its p dθ substitution. Therefore, we know that x2 + 1 = sec(θ) and 1 = sec2 (θ) . Plugging dx

208

CHAPTER 11. METHODS OF INTEGRATION

these in we have: Z p Z dθ 2 1 + x dx = sec(θ) sec2 (θ) dx dx Z = sec3 (θ) dθ First, let us integrate by parts with u = sec(θ) and

dv = sec2 (θ). This gives us dθ

du = sec(θ) tan(θ) dθ v = tan(θ) Z

3

Z

sec(θ) tan2 (θ) dθ

Z

 sec(θ) sec2 (θ) − 1 dθ

sec (θ) dθ = sec(θ) tan(θ) − = sec(θ) tan(θ) − Z 2

Z

3

sec (θ) dθ = sec(θ) tan(θ) +

sec(θ) dθ

Let’s focus for a minute on the remaining integral: Z Z 1 sec(θ) dθ = dθ cos(θ) Z cos(θ) = dθ cos2 (θ) Z cos(θ) dθ = 1 − sin2 (θ) This is now an integral that can be integrated by substitution with y = sin(θ) and cos(θ) Z

Z sec(θ) dθ = = = = =

1 dy 1 − y2 Z 1/2 1/2 + dy 1+y 1−y 1 1 ln |1 + y| − ln |1 − y| + C 2 2 1 1 + y ln +C 2 1 − y 1 1 + sin(θ) ln +C 2 1 − sin(θ)

dy = dθ

11.2. INTEGRATION USING TRIGONOMETRIC SUBSTITUTION Multiplying the argument to the logarithm by a “fancy form of 1”:

209

1 + sin(θ) , we find the 1 + sin(θ)

following: = = = =

1 (1 + sin(θ))2 ln +C 2 1 − sin2 (θ) s (1 + sin(θ))2 ln +C 2 cos (θ) 1 + sin(θ) +C ln cos(θ) ln |sec(θ) + tan(θ)| + C

Plugging this back into our original integral we have: Z 2 sec3 (θ) dθ = sec(θ) tan(θ) + ln |sec(θ) + tan(θ)| + C Z 1 (sec(θ) tan(θ) + ln |sec(θ) + tan(θ)|) + C sec3 (θ) dθ = 2 and then returning to cartesian coordinates we have: √ 1 + x2 p sec(θ) = = 1 + x2 1 tan(θ) = x Z p p  1 p 2 2 2 1 + x dx = x 1 + x + ln 1 + x + x + C 2  We will not use this method of integration very much, but you should know that it exists because it allows us to discover the derivatives of inverse trigonometric functions. The most common example that we will run into is: d 1 (arctan(x)) = . dx 1 + x2

11.2.1

Exercises

Exercise 200. Use the method of trigonometric substitution to check that Z 1 dx = arctan(x) + C. 1 + x2

210

CHAPTER 11. METHODS OF INTEGRATION

11.3. INTEGRATING IMPROPER RATIONAL FUNCTIONS

11.3

211

Integrating Improper Rational Functions

x2 + 1 dx. There is no immediate choice of substitution x−3 and no useful partition of the integrand for integration by parts. However, a bit of algebra will do the trick here: Z

Consider an integral of the form

x2 + 1 x−3

x2 − 3x + 3x − 9 + 10 x−3 x(x − 3) + 3(x − 3) + 10 = x−3 10 = x+3+ x−3 =

In this form, there is a clear path to integrate:  Z 2 Z  x +1 10 dx = x+3+ dx x−3 x−3 Z Z 10 = x + 3 dx + dx x−3 In the final integral, let u = x − 3 and Z

x2 + 1 dx = x−3 =

du = 1 so dx Z x2 10 + 3x + du 2 u x2 + 3x + 10 ln |x − 3| + C 2

Remark. Recall that a rational function f (x) am xm + am−1 xm−1 + . . . + a1 x + a0 = g(x) bn xn + bn−1 xn−1 + . . . + b1 x + b0 is called proper if m < n. If m ≥ n then the f (x) is called improper. rational function g(x) Any improper rational function can be simplified to a polynomial or, more commonly, a polynomial plus the proper rational function remainder using long division.

11.3.2

Exercises Z

Exercise 201. Calculate

4x3 − x2 + 4 dx. x−1

11.3.1

Target Practice Z

Calculate

3x3 − x2 + x − 2 dx. x+2

212

CHAPTER 11. METHODS OF INTEGRATION

Z

3x2 − 2 dx. x−4

Z

5x4 − 3x2 + x + 1 dx x2 + 2

Exercise 202. Calculate

Exercise 203. Calculate

11.4. INTEGRATION BY PARTIAL FRACTIONS

11.4

213

Integration by Partial Fractions

The following integral contains a proper rational function: Z 3x − 5 dx. x2 − 1 Let’s try the same algebraic reduction as with the improper rational functions in the previous section. First, we factor the denominator: x2 − 1 = (x − 1)(x + 1). Then, notice that 3x − 5 x2 − 1 3x − 5

=

A B A(x + 1) + B(x − 1) + = x−1 x+1 x2 − 1 (A + B)x + A − B

A+B

=

3

=

⇒ A=3−B A−B

=

−5

⇒ 3 − B − B = −5 −2B = −8 B=4 A 3x − 5 x2 − 1

= =

3 − 4 = −1 −1 4 + x−1 x+1

Putting this back into the integral, we have  Z  Z −1 4 3x − 5 dx = + dx x2 − 1 x−1 x+1 = − ln |x − 1| + 4 ln |x + 1| + C   (x + 1)4 = ln +C |x − 1| This method applies to any proper rational function. By the fundamental theorem of algebra, we know that we can factor any polynomial with real coefficients into a product of linear and quadratic factors with real coefficients. There are four possibilities: Unique linear factors: 1 (x − 3)(x + 4)

=

A B + x−3 x+4

214

CHAPTER 11. METHODS OF INTEGRATION

Repeated linear factors: x2 − 1 (x + 3)3

A C B + + 2 x + 3 (x + 3) (x + 3)3

=

Unique quadratic factors: (x2

1 + 3)(x2 + x + 4)

=

Ax + B Cx + D + 2 2 x +3 x +x+4

Repeated quadratic factors: x2 − 1 (x2 + 3)3

=

Ax + B Ex + F Cx + D + + 2 x2 + 3 (x + 3)2 (x2 + 3)3

In full generality, for a proper (degree of p < degree of q=n) rational function with normalized denominator (q = xn + qn−1 xn−1 + · · · + q1 x + q0 ) we could write this as: f (x) = =

p(x) q(x) pn−1 xn−1 + pn−2 xn−2 + · · · p1 x + p0 (x − a1 )r1 (x − a2 )r2 · · · (x − am )rm (x2 + b1 x + c1 )s1 · · · (x2 + b` x + c` )s`

sk ri ` X m X X X Bjk x + Cjk Aik + = x − ai x2 + bj x + cj i=1 k=1

j=1 k=1

where ri is the multiplicity of the linear factor x − ai and sk isthe multiplicity of the  ! m ` X X quadratic factor x2 + bj x + cj . Notice that we have n = ri 2 sj  . i=1

j=1

It should also be evident that this a powerful method for computing integrals. Each resulting term is easy to integrate! You should expect logarithms of linear and quadratic factors, negative powers of linear and quadratic factors, and arctangents.

11.4.1

Target Practice

1. Write the partial fraction decomposition in terms of indeterminate constants, i.e. A, B, C, etc. 2x + 1 − 3x − 10 1 b) t(t + 3)3 a)

x2

11.4. INTEGRATION BY PARTIAL FRACTIONS

215

6s2 − 5 (s − 1)(s2 + 1) Z 3x2 + 7x − 2 2. Calculate dx. x(x + 1)(x − 2) Z dt dt. 3. Calculate 2 t(t + 8) c)

11.4.2

Exercises

Exercise 204. Write the partial fraction decomposition in terms of indeterminate constants, i.e. A, B, C, etc. a)

5x − 13 (x − 3)(x − 2)

b)

x+3 (2x − 1)2

c)

t4 + 9 t4 + 9t2

d)

x2 + 2x + 1 (3x2 + 2)3

e)

z3

z − z 2 − 6z

Z

x2 − x + 1 dx. x2 + x

Z

6t2 + 2 dt. t2 + 2t − 3

Exercise 205. Calculate

Exercise 206. Calculate

Z Exercise 207. Calculate

x(x2

1 dx. + 25)

216

CHAPTER 11. METHODS OF INTEGRATION

Z

h2 + 11h dh. (h − 1)(h + 1)2

Z

3r2 + 7r − 2 dr. r3 − r2 − 2r

Z

6x2 + 2 dx. x2 + 2x − 3

Z

s2 − 8s ds. (s + 1)(s + 4)3

Exercise 208. Calculate

Exercise 209. Calculate

Exercise 210. Calculate

Exercise 211. Calculate

Exercise 212. Solve for P (t) in the scaled logistic growth model where the stable population values are P = 0 and P = 1 with growth constant k. dP = kP (1 − P ). dt Graph your answer with P (0) = .05 and k = .1. Choose appropriate limits on your axes to see the relevant features of the graph.

11.5. USING TABLES

11.5

Using Tables

217

218

CHAPTER 11. METHODS OF INTEGRATION

DERIVATIVES AND INDEFINITE INTEGRALS In these formulas, u, v, and w represent functions of x. Also, a, c, and n represent constants. All arguments of the trigonometric functions are in radians. constant of integration should be added to the integrals. To avoid terminology difficulty, the following definitions are followed arcsin u = sin–1 u, (sin u)–1 = 1/sin u. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

dc/dx = 0 dx/dx = 1 d(cu)/dx = c du/dx d(u + v – w)/dx = du/dx + dv/dx – dw/dx d(uv)/dx = u dv/dx + v du/dx d(uvw)/dx = uv dw/dx + uw dv/dx + vw du/dx d ^u/vh = v du/dx -2 u dv/dx dx v d(un)/dx = nun–1 du/dx d[f (u)]/dx = {d[f (u)]/du} du/dx du/dx = 1/(dx/du) d _logaui = _logaei u1 du dx dx

^ h 12. d 1n u = u1 du dx dx ui _ 13. d a = ^1n ah au du dx dx 14. d(eu)/dx = eu du/dx

15. 16. 17. 18. 19. 20. 21.

d(uv)/dx = vuv–1 du/dx + (ln u) uv dv/dx d(sin u)/dx = cos u du/dx d(cos u)/dx = –sin u du/dx d(tan u)/dx = sec2u du/dx d(cot u)/dx = –csc2u du/dx d(sec u)/dx = sec u tan u du/dx d(csc u)/dx = –csc u cot u du/dx

_ -1 i 22. d sin u = dx

_ -1 i 23. d cos u = dx

du 1 1 - u 2 dx

1 du 1 - u 2 dx _ -1 i 1 du 24. d tan u = dx 1 + u 2 dx _ -1 i 1 du 25. d cot u = dx 1 + u 2 dx _ -1 i du 1 26. d sec u = dx u u 2 - 1 dx

_- r/2 # sin- 1u # r/2i _0 # cos- 1u # ri _- r/2 < tan- 1u < r/2i _0 < cot- 1u < ri

_0 < sec- 1u < r/2i _- r # sec- 1u < - r/2i d _csc- 1ui du 1 =27. dx u u 2 - 1 dx

_0 < csc- 1u # r/2i _- r < csc- 1u # - r/2i

1. 2. 3. 4. 5. 6. 7. 8.

# d f (x) = f (x) # dx = x # a f(x) dx = a # f(x) dx # [u(x) ± v(x)] dx = # u(x) dx ± # v(x) dx m+ 1 # x mdx = x ^m ! - 1h m+ 1 # u(x) dv(x) = u(x) v(x) – # v (x) du(x) # dx = a1 1n ax + b ax + b # dx = 2 x x

x # a x dx = a 1na 10. # sin x dx = – cos x 11. # cos x dx = sin x

9.

12. # sin 2xdx = x - sin 2x 2 4 13. # cos 2xdx = x + sin 2x 2 4 14. # x sin x dx = sin x – x cos x 15. # x cos x dx = cos x + x sin x 16. # sin x cos x dx = (sin2x)/2 17. # sin ax cos bx dx = - cos ^ a - bh x - cos ^a + bh x _ a 2 ! b 2i 2 ^ a - bh 2 ^ a + bh 18. # tan x dx = –ln⏐cos x⏐= ln ⏐sec x⏐ 19. # cot x dx = –ln ⏐csc x ⏐= ln ⏐sin x⏐ 20. # tan2x dx = tan x – x 21. # cot2x dx = –cot x – x 22. # eax dx = (1/a) eax 23. # xeax dx = (eax/a2)(ax – 1) 24. # ln x dx = x [ln (x) – 1] (x > 0) 25. #

dx = 1 tan- 1 x a a a2 + x2

26. #

dx = ax 2 + c

27a. #

dx = ax 2 + bx + c

1 tan- 1 b x a l, c ac

dx 27b. # ax 2 + bx + c =

27c. #

^a ! 0h

^a > 0, c > 0h

2 tan- 1 2ax + b 2 4ac - b 4ac - b 2

_ 4ac - b 2 > 0i

2ax + b 1 1n b - 4ac 2ax + b + 2

dx 2 , =2ax + b ax 2 + bx + c

b 2 - 4ac b 2 - 4ac

_b 2 - 4ac > 0i

_b 2 - 4ac = 0i

Figure 11.3: The standard table of integals and derivatives from the Fundamentals of 29 MATHEMATICS Engineering Handbook. (FE Reference Handbook 9.4 Version for Computer Testing. NCEES. 2013)

11.5. USING TABLES

11.5.1

219

Exercises

Exercise 213. Check the integral rule #20 from the Derivatives and Integrals table in sin(x) Figure 11.3 using tan(x) = and integration by parts. cos(x)

Exercise 214. Check the derivative rule #22 in integral form: Z 1 √ dx = arcsin(x), 1 − x2 using trigonometric substitution.

Z

3 dx. Identify 4 + x2

Z

3 dx. Identify 4 − x2

Exercise 215. Use the Integral Table in Figure 11.3 to calculate which rule in the table you are using.

Exercise 216. Use the Integral Table in Figure 11.3 to calculate which rule in the table you are using.

220

11.5.2

CHAPTER 11. METHODS OF INTEGRATION

Solutions to Target Practice

Section 11.1 1.

Z

u = x dv = cos(5x) dx du = 1 dx 1 sin(5x) v = 5 Z x cos(5x) dx = = = =

  Z du 1 1 uv − v dx = (x) sin(5x) − sin(5x) · dx dx 5 5 Z 1 1 x sin(5x) − sin(5x) dx 5 5   1 1 1 x sin(5x) − − cos(5x) + C 5 5 5 1 1 x sin(5x) + cos(5x) + C 5 25

2. u = ln(7t) dv = t2 dt du 1 1 = ·7= dt 7t t t3 v = 3  3 Z 3   Z Z t t 1 du 2 dt = (ln(7t)) − dt t ln(7t) dt = uv − v dt 3 3 t Z 2 1 3 t = t ln(7t) − dt 3 3 1 3 1 t3 = t ln(7t) − · + C 3 3 3 1 3 t3 = t ln(7t) − + C 3  9  1 3 1 = t ln(7t) − +C 3 3

11.5. USING TABLES

221

Section 11.3

Z

3x3 − x2 + x − 2 dx = x+2

 Z  32 2 3x − 7x + 15 − dx x+2 7 = x3 − x2 + 15x − 32 ln |x + 2| + C 2

Section 11.4 1.

a)

A B 2x + 1 = + (x + 2)(x − 5) x+2 x−5

b)

A B C D + + + 2 t t + 3 (t + 3) (t + 3)3

c)

A Bs + C + 2 s−1 s +1

2.

3x2 + 7x − 2 A B C = + + x(x + 1)(x − 2) x x+1 x−2 2 3x + 7x − 2 = A(x + 1)(x − 2) + B(x)(x − 2) + C(x)(x + 1) Let x = 0 : −2 = A(1)(−2), so A = 1 Let x = −1 : −6 = B(−1)(−3), so B = −2 Z Now

Let x = 2 : 24 = C(2)(3), so C = 4  Z  + 7x − 2 1 −2 4 dx = + + dx x(x + 1)(x − 2) x x+1 x−2 = ln |x| − 2 ln |x + 1| + 4 ln |x − 2| + C   |x|(x − 2)4 = ln +C (x + 1)2 3x2

222

CHAPTER 11. METHODS OF INTEGRATION

3. 1 A Bt + C = + 2 + 8) t t +8 2 1 = A(t + 8) + (Bt + C)(t) 1 Let t = 0 : 1 = A(8), so A = 8 1 2 1 Now 1 = (t + 8) + (Bt + C)(t) = t2 + 1 + Bt2 + Ct 8 8 1 1 Compare t2 terms: 0 = t2 + Bt2 , so B = − 8 8 Compare t terms: 0 = Ct, so C = 0  Z Z  dt 1/8 −1/8 dt Now dt = + 2 t(t2 + 8) t t +8   1 1 t = ln |t| − √ arctan √ +C 8 8 8 8 t(t2

Chapter 12

Integrals and Geometric Properties of Objects 12.1

Areas Between Curves

In Chapter 6, we found that the area under a curve, and above the x-axis, is equal to Z b a definite integral based on the height of the curve: A = f (x) dx. In some cases, a

such an integral can be evaluated exactly with the Fundamental Theorem of Calculus (i.e. using antiderivatives). Otherwise, numerical integration can be used to approximate the integral. For more complicated situations, a region of the plane may be bounded by two or more lines or curves. In such instances, isolating a typical subdivision of the region is often helpful in deciding how to set up an integral for the area. Figure 12.1 shows the region 1 1 bounded above by the graph of y1 = x2 + 2 and below by the graph of y2 = x2 − x + 2 2 8 between x = 0 and x = 4. The region is subdivided into several rectangles, and a “typical” or “representative” rectangle is highlighted. The highlighted rectangle is representative in the sense that the area of any of the rectangles can be expressed in the same way. Since the area of a rectangle is found by multiplying its length by its width, any of the verticallyoriented rectangles in the figure has an area ∆A = (y1 − y2 )∆x. The total area of the region could then be approximated, using the Midpoint Method, by X X A≈ ∆A = (y1 − y2 )∆x. all rectangles

all rectangles

But, as you already know, if we let the number of rectangles increase without bound (which is equivalent to letting ∆x approach 0), we find a definite integral expression for the exact 223

224

CHAPTER 12. GEOMETRIC APPLICATIONS OF INTEGRALS

1 1 Figure 12.1: The region between y1 = x2 + 2 and y2 = x2 − x + 2 from x = 0 to x = 4. 2 8

area of the region:

X

A = lim

∆x→0

all rectangles

Z

4

(y1 − y2 ) dx.

(y1 − y2 )∆x = 0

The differential dx indicates that everything in the integral must be expressed in terms of x. Replacing y1 and y2 by their equivalencies in terms of x, we have

A = = = = =

4 

   1 2 1 2 x +2 − x −x+2 dx 2 8 0  Z 4 3 2 x + x dx 8 0 4  3 x3 x2 · + 8 3 2 0     1 3 1 2 1 3 1 2 ·4 + ·4 − 0 + ·0 8 2 8 2 16 square units Z

12.1. AREAS BETWEEN CURVES

12.1.1

225

Choosing the Representative Rectangle

Now consider the shaded region in Figure 12.2, bounded by y1 = 2 − x2 , y2 = x-axis.



x and the

If this region is sliced vertically, it is not possible to write a rectangular approximation for the area of a slice that is representative of all rectangles that could be drawn in the region. This is because some of the rectangles would be bounded above by y1 = 2 − x2 and others √ √ by y2 = x. However, all the horizontal slices would be bounded on the left by y2 = x and on the right by y1 = 2 − x2 . So a single representative rectangle that works for the entire region is possible if horizontal slices are used.

√ Figure 12.2: The region between y1 = 2 − x2 and y2 = x and y = 0 (the x-axis) shown with a representative horizontal rectangle of incremental area ∆A = (x1 − x2 )∆y. A typical horizontal rectangle is drawn in Figure 12.2. As before, the total area can be approximated by the sum of many such rectangles, and found exactly by a definite integral: Z b A= (x1 − x2 ) dy. a

In this case, the differential dy indicates that we are dealing with an integral based on the y variable, so everything in the integral (including the limits of integration!) must be written in terms of y, just as we did in Chapter 10, when we first discussed changing variables. Consider the example shown in Figure 12.2. We need to determine the limits of integration in y: that is, what are the smallest and largest values of y in our region? It is easy to see that the lower bound will be y = 0 and it is also quick to check that the upper bound where our curves meet is at y = 1. Now, let’s consider the boundary functions. Our “upper” (in this p case right-most) curve is given as y = 2 − x2 . Solving for x in terms√ of y we have x1 = 2 − y. Our “lower” (in this case left-most) curve is given as y = x.

226

CHAPTER 12. GEOMETRIC APPLICATIONS OF INTEGRALS

Solving for x in terms of y we have x2 = y 2 . Putting this together, we have: 1 Z y=1 p y 3 2 3/2 2 2 − y − y dy = − (2 − y) − 3 3 0 y=0     2 1 2 3/2 = − − − − 2 −0 3 3 3 √ 4 = 2 − 1 ≈ 0.8856 3 This means the area of the shaded region is approximately 0.8856 square units.

12.1.2

Target Practice

For each given region, (a) sketch a graph of the region, (b) write an expression ∆A for the area of a representative rectangle, (c) write a definite integral in only one variable whose value equals the exact area of the region, and (d) find the area of the region. 1. The region above the graph of y = 2 − x and bounded by the graphs of y = x2 and the line y = 3. 2. The region bounded by the graphs of y = 2x − x2 and y = x − 2.

12.1.3

Exercises

For each given region, (a) sketch a graph of the region, (b) write an expression ∆A for the area of a representative rectangle, (c) write a definite integral in only one variable whose value equals the exact area of the region, and (d) find the area of the region. Exercise 217. The region bounded by the graphs of y = x3 , y = x + 6, and the y-axis.

Exercise 218. The region bounded by the graphs of y = axes.



x − 1, y = 2, and the coordinate

Exercise 219. The two regions bounded by the graphs of y = 3x3 − x2 − 10x, y = 2x − x2 .

12.2. VOLUMES OF SOLID REGIONS

227

Exercise 220. The region bounded by the graphs of x + 4y = 26, x + 6 = y 2 .

12.2

Volumes of Solid Regions

Some three-dimensional figures have simple enough shapes that their volumes can be modeled by single-variable integrals. Consider the solid region in Figure 12.3, which looks something like an old-style bulky computer monitor or possibly an iMac. The base of this

Figure 12.3: A 3D figure and its cross section. figure is a two-dimensional region of the xy-plane, bounded by the graphs of y = x2 and y = 2. Cross sections perpendicular to the y-axis are all squares. Two of these nearby cross sections outline a three-dimensional “slab” which can be modeled by a rectangular solid of thickness ∆y that has square faces with edges of length 2x. The volume of a representative slab can be written as ∆V = (edge of square)2 (thickness of slab) = (2x)2 ∆y.

228

CHAPTER 12. GEOMETRIC APPLICATIONS OF INTEGRALS

And, in a manner similar to the way we analyzed areas of two-dimensional regions, we can represent the exact volume of the solid by a definite integral:

V = lim

∆y→0

X

2

Z

(2x) ∆y =

b

(2x)2 dy.

a

all slabs

As before, the differential dy defines the variable of integration so everything must be written in terms of y.

Z V

y=2

=

√ (2 y)2 dy

y=0 2

Z =

4y dy 2 2y 2 0

=

0

= 8 cubic units

12.2.1

Solids with Circular Cross Sections – The Disk Method

When a two-dimensional region is rotated around a line in the plane, a solid figure with circular cross sections, called a solid of revolution, is generated. For example, if the 1 region bounded by y = e 2 x , the line x = 1 and the coordinate axes (Figure 12.4) is rotated around the x-axis, the region of space shown in Figure 12.5 is formed.

12.2. VOLUMES OF SOLID REGIONS

229

Figure 12.4: (Above) A 2D region. Figure 12.5: (To the right) The solid of revolution formed when the region shown in Figure 12.4 is rotated around the xaxis. Slices of this figure perpendicular to the x-axis are circles, and two nearby circles outline a circular slab which can be approximated by a circular disk of thickness ∆x. The volume of the figure can then be modeled by an integral that effectively adds up an infinite number of infinitesimally thin disks: Z V = πy 2 dx Z x=1  2 1 = π e2x dx x=0 Z 1 = π ex dx 0

=

πex |10

= π(e − 1) ≈ 5.398

12.2.2

Solids with Holes – The Washer Method

Now consider the region bounded by y = 4x − x2 , y = 6 − 3x, and the x-axis (Figure 12.6). If this region is revolved about the y-axis, the resulting three-dimensional figure

230

CHAPTER 12. GEOMETRIC APPLICATIONS OF INTEGRALS

has a hole in the middle. A horizontal slice is a “washer,” or a disk with a hole in its center.

Figure 12.6

Figure 12.7: The solid of revolution formed when the region shown in Figure 12.6 is rotated around the y-axis.

The area of the surface of a washer is the difference between the areas of two circles, one with radius R (the outer radius) and one with radius r (the inner radius). The volume of the washer is then equal to the value of the definite integral Z y=b V = π (R(y))2 − (r(y))2 dy y=a !  Z y=3  2 p 1 2  = π 2− y − 2− 4−y dy 3 y=0   Z y=3   p 1 2 4 dy = π 4− y+ y − 4−4 4−y+4−y 3 9 y=0 Z y=3 p 1 1 = π − y + y 2 + 4 4 − y − 4 dy 3 9 y=0  3  1 2 1 3 8 3/2 = π − y + y − (4 − y) − 4y 6 27 3 0   8 64 3 = π − + 1 − − 12 + 2 3 3 37π = 6 ≈ 19.373

12.2.3

Target Practice

1. Consider the three-dimensional figure √ formed by revolving the region in the first quadrant bounded by the graph of y = 4 − x, the line x = 1, and the x-axis around

12.2. VOLUMES OF SOLID REGIONS

231

the x-axis. a) Write an expression ∆V for the area of a representative disk. b) Write a definite integral in only one variable whose value equals the exact volume of the figure. c) Evaluate the integral to find the volume of the figure. 2. Consider the three-dimensional figure formed by revolving the same region around the y-axis. a) Write an expression ∆V for the area of a representative washer. b) Write a definite integral in only one variable whose value equals the exact volume of the figure. c) Evaluate the integral to find the volume of the figure.

12.2.4

Exercises

Exercise 221. Every cross section perpendicular to the x-axis in the figure below is a square with sides equal to 0.2x2 inches long. Find the volume of the figure.

Exercise 222. In the solid figure below, every cross section perpendicular to the x-axis is a rectangle. Shown at the right is a side view of the figure, in which the upper curve has 1 the equation y = x2 . Write an integral in one variable that represents the volume of the 36

232

CHAPTER 12. GEOMETRIC APPLICATIONS OF INTEGRALS

figure.

For Exercises 223–227, consider the three-dimensional figure formed by revolving the 1 region bounded by the graph of y = x2 , the line x = 2, and the x-axis around the given 4 line. a) Write an expression ∆V for the volume of a representative disk or washer. b) Write a definite integral in only one variable whose value equals the exact volume of the figure. c) Evaluate the integral to find the volume of the figure. Exercise 223. the x-axis

Exercise 224. the y-axis

Exercise 225. the line y = 1

12.3. LENGTH OF A CURVE

233

Exercise 226. the line x = 2

Exercise 227. the line x = 3

Exercise 228. The volume generated when the region bounded by y = 4x−x2 , y = 6−3x, and the x-axis is revolved around the y-axis was found in this section using the “washer” method. Describe how you could find the volume of the solid using only disks.

12.3

Length of a Curve

Figure 12.8 shows a graph of y = x2 between the points (0,0) and (2,4). You already know how to use integrals to find the area under the curve and associated volumes. The calculation of area began with the concept of adding up the areas of thin slices of a region which were approximated by rectangles. Similarly, volumes of solids of revolution involved thin slabs in the shape of circular disks or other simple shapes. Another geometric property of the curve is its length. As with the area and volume calculations, we will begin by imagining the curve to be subdivided into very short segments and examining the geometry of one of those segments. Figure 12.8

234

CHAPTER 12. GEOMETRIC APPLICATIONS OF INTEGRALS

Figure 12.9 shows the curve broken into only three straight segments. Even on this scale it can be seen that, at least for two of the segments, a straight line would provide a reasonable approximation to the curve. And as the number of segments is increased, the segments would appear more and more straight. So our approach will be to consider the geometry of a single segment, then use an integral to add up the incremental lengths of all segments.

Figure 12.9

Figure 12.10 provides a magnified view of one of those segments. If the segment is short enough, it will appear almost straight when magnified, and its length will be close to the length of the hypotenuse of a small right triangle with sides ∆x and ∆y.

Figure 12.10

Here ∆s is a very short segment of the curve. As the segment gets shorter and shorter,

12.3. LENGTH OF A CURVE

235

the length ∆s will be approximately equal to the length of the hypotenuse of the triangle, so we can write p ∆s ≈ (∆x)2 + (∆y)2 . The total length of the curve is then L≈

X

p (∆x)2 + (∆y)2 .

(12.1)

all segments

With the inspiration of our work with the definition of the derivative, consider the following:   X X p ∆x p 2 2 (∆x) + (∆y) = (∆x)2 + (∆y)2 ∆x all segments all segments s X (∆x)2 + (∆y)2 = ∆x (∆x)2 all segments s   X ∆y 2 = 1+ ∆x ∆x all segments

In the limit as both ∆x and ∆y approach 0, the increments become infinitesimal and we can write s s    2 Z b X ∆y 2 dy L = lim lim ∆x = dx. 1+ 1+ ∆x→0 ∆y→0 ∆x dx a We finally have a formula for the length of a curve, often called arc length. Returning to the original problem of finding the length of have s  2 Z 2 dy dx L = 1+ dx 0 Z 2p = 1 + (2x)2 dx 0 Z 2p = 1 + 4x2 dx 0  2 1 p 2 = 2x 1 + 4x + arcsinh(2x) 4 0 ≈ 4.6468 An alternative formula allows integration over y, and can

y = x2 from (0, 0) to (2, 4), we

For most problems of this type, integration can be quite involved, so you may prefer to use computer software for evaluation.

be derived from Equation 12.1

236

CHAPTER 12. GEOMETRIC APPLICATIONS OF INTEGRALS ∆y ∆x rather than . ∆y ∆x

by multiplying the expression in the radical by

X

s 



2

∆x + 1 ∆y ∆y all segments  s s  2  2 Z b X ∆x dx  + 1 ∆y = 1+ dy. L = lim lim ∆x→0 ∆y→0 ∆y dy a X

p (∆x)2 + (∆y)2 =

all segments



This formula can be applied to the problem of finding the length of y = x2 from (0, 0) to √ (2, 4), with x = y = y 1/2 : 4

Z

s



1+

L = 0 4

Z =

s



1+ 0

dx dy

2

1 √ 2 y

dy 2 dy

The result is the same, approximately 4.6468. A single simpler formula can be written that includes both of the formulas used here, by thinking of working in a new variable s which is the distance travelled along the curve:

Z Length of a curve: L =

ds where a

12.3.1

b

ds = dx

s 1+



dy dx

2 or

ds = dy

s 1+



dx dy

2 (12.2)

Choosing the Correct Method

Although in this instance the length of the curve can be found by integration with respect to either x or y, that is not always the case. If the function being analyzed is not strictly increasing or decreasing over the interval of integration, integration with respect to y will not result in the correct length. For example, if the length of y = x2 from (-1, 1) to (2, 4) is desired, integration with respect to y will not work. Similarly, for a curve that does not represent the graph of a function because it does not pass the vertical line test, integration with respect to x will not work. So it is important to look at the shape of a curve before deciding which integral model to use. For a curve that “doubles back on itself” in either direction, one (or even both) methods may not be applicable. In some cases, switching to polar coordinates may provide a solution.

12.3. LENGTH OF A CURVE

12.3.2

237

Target Practice

Consider the portion of the graph of y =

1 x

from x = 1 to x = 4.

a) Set up an integral in x that represents the length of the curve. b) Set up an integral in y that represents the length of the curve. c) Evaluate both integrals using a calculator or computer to show that they give the same result.

12.3.3

Exercises

Exercise 229. Consider the portion of the graph of y = ln(x) from x = 1 to x = 3. a) Set up an integral in x that represents the length of the curve. b) Set up an integral in y that represents the length of the curve. c) Evaluate both integrals using a calculator or computer to show that they give the same result.

Exercise 230. Consider the portion of the graph of y = e−2x from x = 0 to x = 2. a) Set up an integral that represents the length of the curve. b) Evaluate the integral using a calculator or computer.

Exercise 231. Consider the portion of the graph of y = sin(x) between x = 0 and x = π. a) Explain why only one direction of integration can be used to find the length of the curve. b) Set up an integral that represents the length of the curve. c) Evaluate the integral using a calculator or computer to find the length of the curve.

238

12.4

CHAPTER 12. GEOMETRIC APPLICATIONS OF INTEGRALS

Area of a Three-Dimensional Surface

The areas of many curved surfaces, as well as the volumes contained within them, cannot be found using single-variable calculus, but must be analyzed with a multivariable approach that is beyond the scope of this course. However, for solids of revolution, an extension of the method for finding length of a curve can be used. Consider the surface shown in Figure 12.11 which is formed when the curve from (0, 0) to (2, 4) is rotated about the y-axis. (This type of surface is called a paraboloid.)

Figure 12.11 Slicing through this surface horizontally can produce a hollow “ring” or “band” of infinitesimal thickness. The lateral area of a representative ring is based on the slant height of the ring, which is a short straight distance approximating the length of curve at the edge of the ring. Letting ∆s become infinitesimal allows us to write an integral for the surface area of the figure. Z SA =

b

2πr ds a

The differential ds is defined in Equation 12.2, and is written as needed for a particular situation, depending on whether integration with respect to x or y is preferred. For the

12.4. AREA OF A THREE-DIMENSIONAL SURFACE

239

surface area of the paraboloid above, Z

s

2

SA = 2π

x



1+

0

Z

2

= 2π

dy dx

2 dx

x

p 1 + (2x)2 dx

x

p 1 + 4x2 dx

0

Z

2

= 2π 0

Z

17

= 2π 1

1 1/2 u du 8

π 3/2 17 = u 6 1 ≈ 36.18 square units Notice that this integral was evaluated with a simple substitution. Furthermore, since the s  2 Z 4 dx √ function y = x2 is increasing on the interval [0, 2], the integral 2π y 1+ dy dy 0 could also be used to get the same result, although its evaluation is more complicated. In any case, the determination of the radius of a representative ring depends on the geometry of the ring with respect to the x- and y-axes. But the choice of whether to integrate with respect to x or y is independent of the expression for the radius, and dependent on the feasibility and ease of integration.

12.4.1

Target Practice

Consider the surface generated when the curve y = sin (x) from (0, 0) to (π, 0) is rotated about the x-axis. a) Set up an integral that represents the area of the surface. b) Find the surface area using a calculator or computer.

12.4.2

Exercises s

 dx 2 dy can be used to get the dy 0 same result as the integral in x for the surface area considered in this section. Start by writing the integral in terms of y only. Z

Exercise 232. Show that the integral 2π

4

x



1+

240

CHAPTER 12. GEOMETRIC APPLICATIONS OF INTEGRALS

Exercise 233. Consider the surface generated when the curve y = x2 from (0, 0) to (2, 4) is rotated about the x-axis. a) Set up an integral that represents the area of the surface, and find the surface area using a calculator or computer. b) Can this surface area be modeled by a single integral in either x or y? Explain.

Exercise 234. Consider the surface generated when the curve y 2 = x − 2 from (3, -1) to (6, 2) is rotated about the y-axis. a) Set up an integral that represents the area of the surface, and find the surface area using a calculator or computer. b) Can this surface area be modeled by a single integral in either x or y? Explain.

12.5

Applied Geometry Problems

The problem solving process for each of the applications of integrals in this chapter follows the same overall pattern, starting from a non-calculus computation.

Figure 12.12 Most of the thinking needed is in the first two steps, which consist of setting up the model for solving the problem. Neither of these involves calculus, although the concept of the representative element requires anticipating that it will become infinitesimal in the integral.

12.5. APPLIED GEOMETRY PROBLEMS

241

Example 26. Figure 12.13 shows the top of a drop-leaf table. Each end-piece is part of a circle. a) Find the radius of the circular arcs. b) Find the area of one of the end-pieces.

Figure 12.13

242

CHAPTER 12. GEOMETRIC APPLICATIONS OF INTEGRALS

Solution:

a. Assume that one of the leaves is part of a circle with radius r. Then in terms of r, the coordinates of two of the points on the circle are as indicated in Figure 12.14. Since the standard equation of a circle with center at the origin is x2 + y 2 = r2 , this circle has an equation (r − 16)2 + 242 = r2 . Solving for r, we have (r − 16)2 + 242 = r2 r2 − 32r + 256 + 576 = r2 −323r + 832 = 0 r = 26 Figure 12.14

b. A representative rectangle with area ∆A = (2y)∆x is drawn in R 26 √ Rb Figure 12.15. The area of the leaf is a (2y)dx = 10 2 26 − x2 dx ≈ 555. So the area of a leaf is about 555 square inches.



Figure 12.15

Example 27. A rugby ball has the shape that is formed by revolving an ellipse about one of the coordinate axes. The standard size 5 ball is approximately 30 cm long and 20 cm wide. Find the volume of such a ball. Figure 12.16

Solution: A standard equation of an ellipse with center at the origin is where a is half the length and b half the width of the ellipse.

x2 a2

+

y2 b2

= 1,

12.5. APPLIED GEOMETRY PROBLEMS

243

The ellipse shown in Figure 12.17 generates an ellipsoid that approximates the shape of a rugby ball when revolved y2 x2 about the y-axis. Its equation is 15 2 + 102 = 1. The rectangle shown generates a representative disk with volume ∆V = πy 2 ∆x, resulting in an integral: Figure 12.17 Z

b

πy dx = a

= = ≈

15

  x2 π(100) 1 − dx 225 −15   Z 15 x2 π(100) 1 − 2 dx 225 0   x3 15 200π x − 675 0 6283 Z

2

An ellipse with the y-axis as its long axis would have an equation 2 y2 + xb2 = 1. a2 Notice that the symmetry of the ellipsoid allows us to integrate from x = 0 to x = 15 and double the result.

So the volume of the rugby ball is about 6283 (or 6300) cm3 .

12.5.1



Exercises

The applied problems in this section make use of one or more of the models developed previously in this chapter. In each exercise, write an integral that represents the solution to the problem, and evaluate it by using • a technique you learned previously in Chapters 4, 11, or 12; • the table of integrals in Figure 12.4; or • computer software.

Exercise 235. Determine the number of cubic yards of crushed rock necessary to make a roadbed of the dimensions shown below, if the road is to be 14 -mile long. (Assume that the crown of the pavement is an arc of a parabola, with its vertex at the top center.)

244

CHAPTER 12. GEOMETRIC APPLICATIONS OF INTEGRALS

Figure 12.18

Exercise 236. An industrial designer has designed a wine glass as shown. The bowl portion is a section of a sphere with 3.00-inch diameter. If the opening is to have a 2.50-inch diameter, a) What will be the depth of the bowl? b) How much wine can the glass hold? c) How much glass is needed to make the bowl, if it is 1/16 inch (0.0625) thick? Figure 12.19

Exercise 237. The profile of the main cables on a suspension bridge may be modeled by a parabola. The central span of the Golden Gate Bridge is 1280 meters long and 152 meters high. The parabola y = 0.00037x2 gives a good fit to the shape of the cables. Find the length of the cables that stretch between the tops of the two towers.

12.5. APPLIED GEOMETRY PROBLEMS

245

Figure 12.20

Exercise 238. When the Mugar Omni Theater at Boston’s Museum of Science was built, a foundation element consisted of a 20-foot wide walkway leading to a circular section with 43 foot radius as shown. If the entire slab is 2 feet deep, how much concrete does it contain? Figure 12.21

Exercise 239. An ornamental light bulb is designed by revolv√ ing the graph of y = 2 x − 41 x3/2 around the x-axis, where x and y are measured in centimeters. Find the surface area of the bulb and use the result to approximate the amount (volume) of glass needed to make the bulb, if the glass is 0.6 millimeters thick. Figure 12.22

246

CHAPTER 12. GEOMETRIC APPLICATIONS OF INTEGRALS

Exercise 240. A manufacturer needs to make corrugated metal sheets 80 cm wide with cross sections in the shape of the curve y = 12 sin(2x) . How wide must the original flats be?

Figure 12.23

Exercise 241. A metal serving bowl 12 inches in diameter is fashioned from part of a 14-inch diameter sphere. Find the amount of sheet metal needed to make the bowl.

Figure 12.24

12.5. APPLIED GEOMETRY PROBLEMS

247

Exercise 242. A soda can has a diameter of 6.6 cm and a height of 12.0 cm. It is partially full of soda such that, when the can is tipped to the point where the soda just begins to spill, half the bottom of the can is covered by the soda. How much soda (volume) is contained in the can? (Hint: visualize the shape of the center cross-section, which would be a vertical plane in the figure that contains the center of the can bottom.) Figure 12.25

Exercise 243. Saturn is the most oblate (least spherical) of the planets in our solar system. Its equatorial radius is 60,268 kilometers and its polar radius is 54,364 kilometers. Find the volume of Saturn.

Figure 12.26

248

12.5.2

CHAPTER 12. GEOMETRIC APPLICATIONS OF INTEGRALS

Solutions to Target Practice

Section 12.1 √ b) ∆A = [ y − (2 − y)] ∆y Z 3 √ c) A = ( y + y − 2) dy 1

d) A = = 1. a) =

= ≈

3

√ ( y + y − 2) dy 1 3 3 y2 y2 3 + 2 − 2y 2 1   2 3 32 · 32 + − 2(3) 3 2   2 3 12 · 12 + − 2(1) − 3 2 √ 2 2 3− 3 2.797 Z

  b) ∆A = (2x − x2 ) − (x − 2) ∆x Z 2  c) A = x − x2 + 2 dx −1

d) Z

2

A = 2. a)

−1 x2

 x − x2 + 2 dx

2 x3 = − + 2x 2 3 −1  2  2 23 = − + 2(2) 2 3   (−1)2 (−1)3 − + 2(−1) − 2 3 9 = 2

12.5. APPLIED GEOMETRY PROBLEMS

249

Section 12.2 √  a) ∆V = π ( 4 − x)2 ∆x Z 4 √ b) V = π ( 4 − x)2 dx 1

c) 4

Z V

= π

√ ( 4 − x)2 dx

1

1.

4

Z

(4 − x) dx 4 ! x2 4x − 2 1

= π

1

= π 9 π 2

=

 a) ∆V = π (R(y))2 − (r(y))2 ∆y Z √3  (4 − y 2 )2 − 12 dy b) V = π 0

c) √

Z V

0

Z

2.

3

 (4 − y 2 )2 − 12 dy

= π

= = ≈

3

 15 − 8y 2 + y 4 dy 0 √3 ! y 3 y 5 π 15y − 8 + 3 5 0   √ √ 9√ π 15 3 − 8 3 + 3 5 44 √ π 3 5 47.88

= π =



250

CHAPTER 12. GEOMETRIC APPLICATIONS OF INTEGRALS

Section 12.3 a) dy dx

= −x−2 Z 4q L = 1 + (−x−2 )2 dx 1

b) 1 y

x = dx dy

= −y −2 Z 1q L = 1 + (−y −2 )2 dy 1 4

c) about 3.1502 Section 12.4 a) Z

π

SA = 2π

y

p

1 + cos2 (x) dx

0

Z = 2π

(sin (x)) 0

b) SA ≈ 14.42

π

p

1 + cos2 (x) dx

Chapter 13

Integrals and Physical Properties of Objects 13.1

Work and Energy

The SpaceX corporation is a private company that has been testing rockets with the aim of lifting payloads into Earth orbit. Elon Musk, the founder of SpaceX, hopes to eventually land humans on Mars. The Falcon Heavy rocket, launched from the Kennedy Space Center in Florida, is said to be capable of lifting 140,000 pounds into low-Earth orbit (less than 2000 kilometers of altitude).

13.1.1

Work with a Variable Force

How much energy does it take to accomplish such a task? If you were to lift a 5-pound textbook through a vertical distance of 2 feet, you would perform 10 foot-pounds of work on the book against the force of gravity. The work done by a force in moving an object through a displacement in the same direction as the force equals the product of the force and the displacement: Work = Force · Displacement In such a simple situation, calculus is not required for the computation. However, in the case of the Falcon Heavy, the force of gravity is not constant as the payload moves away 251

252

CHAPTER 13. APPLICATIONS OF INTEGRATION

from the Earth, but decreases according to Newton’s Law of Universal Gravitation: Fg =

GMe m r2

In this formula, Me is the mass of the Earth, 5.98 × 1024 kilograms, m is the mass of the payload, r is the distance between m and the center of the Earth, and G is the universal 2 gravitation constant, which is 6.67 × 10−11 N·m . So, how much work would the Falcon kg2 Heavy have to do in order to lift a 60,000 kilogram satellite into a 2000-kilometer orbit? In this case, the force of gravity on the satellite (in other words, its weight) reduces to 2.39 × 1019 Fg = . At the surface of the Earth (r = 6.38 × 106 meters), Fg ≈ 587, 000 r2 newtons, as expected from W = mg. But 1 kilometer above the Earth’s surface, the weight is still about 587,000 newtons. So over that distance, the weigh of the satellite has decreased by less than 1 part in 587.

Figure 13.1 This means that, for a small displacement ∆r of only a few meters, the gravitational force is almost constant, and we can write ∆(Work) ≈ Fg ∆r. So, if we break the total 2000 km change in altitude into many short segments of a few

13.1. WORK AND ENERGY

253

meters each, we can write the total work done to lift the satellite into orbit as X Work ≈ Fg · ∆r. all segments

Of course, as ∆r approaches 0, we get the integral Z

re +2000 km

Work =

Fg dr re Z 8.38×106 m

= 6.38×106 m 11

≈ 8.94 × 10

2.39 × 1019 dr r2 joules

So it takes about a trillion joules of work to send a 60,000 kilogram satellite into lowEarth orbit. And as always, the definite integral can be interpreted as an area (see Figure 13.1).

254

13.1.2

CHAPTER 13. APPLICATIONS OF INTEGRATION

Emptying a Tank

Consider a situation in which a pump empties a full tank of water. Figure 13.2 shows a cylindrical tank with given dimensions.

Figure 13.2 The symbol ρ stands for mass density. mass Since density = , we can write volume mass = ρV and weight = ρgV . But in the U.S., weight density ρg (e.g. lb/ft3 ) is usually used. Even though the water in the tank is subject to mixing, we can idealize the volume of water as if it were located in thin layers. Then we can imagine each layer as being lifted from its current location to the top edge of the tank. The work done on a single layer can then be written as (force)(distance), with the force ∆F being equal to the weight of the layer and the distance being the finite distance (10 − y). So again, as long as each layer is thin enough to be regarded as being at a single depth, we can write an expression for the work done to lift a single layer of water out of the tank: ∆(Work) ≈ ∆F · (distance lifted)  = ρgπr2 ∆y (10 − y)   = 16π 62.4 ftlb3 (10 − y) ∆y

13.1. WORK AND ENERGY

255 Z

And the total work done in emptying the tank is 0

13.1.3

10

 16π 62.4

lb ft3



(10 − y) dy ≈ 157,000 ft-lb.

Target Practice

To stretch a spring like the one in Figure 13.3 by a distance y, a force F = ky is needed. (Such a spring obeys Hooke’s Law.) If a spring has spring constant k = 30 N/m, how much work is done in stretching the spring a total of 20 centimeters? (Be careful with units.) Figure 13.3

13.1.4

Exercises

Exercise 244. The anchor of the U.S.S. Constitution (Old Ironsides) was forged in the 1790s and weighs 5300 pounds. The depth in the harbor where it sits at the Boston Navy Yard averages 40 feet.

a) How much work is required to raise the anchor from the bottom of the harbor to a point 16 feet above the water, if you neglect the weight of the chain that is used to raise it?

b) If the chain actually weighs 50 pounds per foot, how much work does this add to the task? (Assume the chain initially spans the total distance to the bottom of the harbor, and the entire chain also ends up 16 feet above the water.)

256

CHAPTER 13. APPLICATIONS OF INTEGRATION

Exercise 245. The vat in Figure 13.4 has inner walls in the shape of a paraboloid, with vertical cross section y = x2 . The vat is filled with water (weight density 62.4 ftlb3 ). a) Find the work done in pumping out all the water to the top edge of the vat. b) Write an altered integral that represents the work done to pump the water to a height 4 feet above the top of the vat. (Don’t evaluate the integral.) c) Write an altered integral that represents the work done to pump the water to the top edge of the vat if the vat is filled only to within 2 feet of its top. (Don’t evaluate the integral.) Figure 13.4

Exercise 246. The pressure P and volume V of the gas in the cylinder shown below are related by P V 1.4 = k. The radius of the cylinder is 0.20 meters, and it is 0.80 meters long. When the piston is fully extended, the gas pressure is 2.0 × 106 newtons per square meter.

Figure 13.5 a) Find the value of k. b) The force on the piston is P A, where A is the piston’s area. Find the work done to compress the gas column from 0.80 m to 0.30 m.

Exercise 247. Tidal power plants use tidal motion to produce electrical energy. To con-

13.1. WORK AND ENERGY

257

struct a tidal power plant, a dam is built to separate a basin from the sea. Electrical energy is produced as the water flows back and forth between the basin and the sea. The amount of energy produced depends on the volume of the basin and the tidal range (the vertical distance between high and low tides).

Figure 13.6

a) Consider a basin with a rectangular base as shown. The basin has a tidal range of 25 feet, with low tide corresponding to y = 0. How much water does the basin hold at high tide?

b) The amount of energy produced during the filling or emptying of the basin is proportional to the amount of work required to fill or empty the basin. How much work is required to fill the basin with seawater? (Use a seawater density of 64 pounds per cubic foot.)

258

13.2

CHAPTER 13. APPLICATIONS OF INTEGRATION

Fluid Pressure and Force

In 1915, the Purity Distilling Company built a tank for storing molasses on the waterfront of Boston’s North End. With a diameter of 90 feet and 50 feet high, the iron tank could hold about 2.3 million gallons of molasses, ready to be distilled into rum or industrial alcohol. On January 15, 1919, the tank of molasses exploded without warning, and caused a wave of molasses and debris to travel down Commercial Street at 35 miles per hour. It exerted a pressure of 2 tons per square foot, with sufficient power to rip buildings off of their foundations. The molasses snapped the support girders from an elevated train track, and smashed multiple houses. About twenty-one people were killed, and 150 people were injured. No engineers were consulted in the design or testing of the tank. As you’ll see in this section, it is fairly easy to calculate the magnitude of the huge forces that were exerted on the tank walls by the molasses inside. If those calculations had been done before the tank was built, the disaster might have been avoided.

13.2.1

Fluid Pressure

Consider a column of fluid (gas or liquid) h meters high, with a base area of A m2 . If the fluid’s mass density is ρ,then the weight of the fluid in the column is ρgV . This weight (in newtons) is spread over the entire area of the base of the column. The pressure on the base is defined as force per unit area, with units of newtons per square meter: P

F A ρg (h · A) = A = ρgh

=

The SI unit of pressure is the pascal, named after the French mathematician and physicist Blaise Pascal. N 1 Pa = 1 2 m In the U.S., pressure is usually measured in pounds per square foot (psf) or pounds per square inch (psi), although many other units have also been used, including atmospheres, bars, and millimeters of mercury. When using pounds per square foot, remember that the

13.2. FLUID PRESSURE AND FORCE

259

pound is a weight unit. So, for example, the weight density of water, 62.4 lb/ft3 , is equal to ρg, not ρ. For pressure calculations, h is usually referred to as depth, measured from the surface of a liquid. Such pressure is isotropic, meaning that is exerted equally in all directions. So an object at depth h is subject to the same pressure on all sides.

13.2.2

Force Due to Fluid Pressure

Imagine a flat surface like a tabletop, 3 meters long and 2 meters wide, submerged horizontally 4 meters below the surface of water. Every point on the tabletop’s surface is at the same depth, and the pressure at 4 meter depth is   kg  m ρgh = 1000 3 9.8 2 (4 m) m s N = 39, 200 2 m = 39, 200 Pa    N 6 m2 = 235, 200 N So the total force on one side of the tabletop is P · A = 39, 200 2 m . Now consider the same tabletop submerged vertically, with one of its longer edges at water level, as in Figure 13.7.

Figure 13.7 If we slice the tabletop into many thin horizontal slices, an entire slice will be at essentially the same depth, and will be subject to the same pressure. So the pressure on a representative slice as drawn is P

= ρgh   kg  m = 1000 3 9.8 2 (2 − y m) m s N = 9800(2 − y) 2 m

(This assumes that y = 0 at the bottom edge of the tabletop.)

Notice that h could instead be replaced by y, if y is measured downward from the water surface.

260

CHAPTER 13. APPLICATIONS OF INTEGRATION

The force on the representative rectangle is then

∆F

= P ∆A = 9800(2 − y)(3∆y)

And finally the total force on one side of the tabletop is given by the integral of dF : Z

2

Z

2

(2 − y) dy

9800(2 − y)(3 dy) = 29, 400 0

0

= 58, 800 N

Example 28. An irrigation channel has a Vshape as shown in Figure 13.8 with a floodgate to hold back water in the channel, which can be opened to release the water. If the water in the channel is exactly 3 feet deep, so that it comes just to the top of the floodgate, what is the total force on the floodgate?

Figure 13.8

Solution:

Unlike the situation with the rectangular tabletop, the width of a representative layer of water varies with depth. With an origin at the bottom of the channel (Figure 13.9), the equation of one side of the channel is y = 43 x. Figure 13.9 The depth of a layer is h = 3 − y, where y measures the height of the layer from the bottom of the channel. So the force on any layer is

13.2. FLUID PRESSURE AND FORCE

∆F

261

= P ∆A = ρg(3 − y) · (2x)(∆y)   = 2 62.4 ftlb3 (3 − y ft)

3 4y

 ft (∆y ft)

= 93.6y(3 − y)∆y

And the total force on the floodgate is the integral of dF : Z F

3

93.6y(3 − y) dy

= 0

= 421.2 pounds



13.2.3

Target Practice

Consider again the 3 m x 2 m tabletop immersed vertically in water. For each given situation, write a definite integral that represents the total force on one side of the tabletop, and evaluate it to find the total force. a) A longer edge of the tabletop is 5 meters below water level instead of right at water level. b) A longer edge of the tabletop is 1 meter above water level.

13.2.4

Exercises

Exercise 248. A Honda CR-V weighs 4564 pounds. The pressure in each of the four tires is maintained at 35 pounds per square inch. a) How many square inches of each tire are in contact with the ground? b) What are the car weight and tire pressure in SI units? c) How many square centimeters of each tire are in contact with the ground?

262

CHAPTER 13. APPLICATIONS OF INTEGRATION

Exercise 249. Consider again the irrigation channel in the Example in this section. For each given situation, write a definite integral that represents the total force on the floodgate due to water pressure, and evaluate it to find the total force. a) In a drought, the water level is only 1.5 feet above the bottom of the channel. b) In a flood, the water is rushing over the floodgate at a height of 1 foot above its top edge.

Exercise 250. A vertical circular viewing window allows aquarium visitors to see inside a large saltwater tank. The window is 4 feet in diameter and its center is 5 feet below the water surface. Find the total force due to water pressure on the window. (The density of seawater is 64 lb/ft3 .)

Exercise 251. Water stands against a dam to its top, in a stream that has a parabolic bottom, as shown in Figure 13.10. Find the total force of water against the dam.

Figure 13.10

Exercise 252. As described in this section, the ”Great Molasses Flood” occurred in Boston’s North End on January 15, 1919 when a tank full of molasses owned by the Purity Distilling Company burst. The tank was 50 feet high and 90 feet in diameter, filled with molasses with weight density 100 lb/ft3 . After the tank was finished, it was tested by pouring water (weight density 62.4 lb/ft3 ) into the tank only one-half foot deep. Since there were no leaks, the tank was judged to be safe. a) Find the total force on the circular wall of the filled tank. b) Find the total force on the bottom half-foot of the tank wall when it was tested with

13.3. IMPROPER INTEGRALS

263

water. c) Find the total force on that bottom half-foot when the tank was filled to the top with molasses.

13.3

Improper Integrals

Consider again the Falcon Heavy rocket and its satellite payload. You saw in Section 13.1 that it would take about a trillion joules of work to raise a 60,000 kg satellite to a 2000 km orbit. Now suppose (hypothetically) that SpaceX wanted to launch such a payload so that it would keep traveling forever, farther and farther from the Earth. Would it require an infinite amount of energy? Or would there be a finite ”escape energy” to essentially put the satellite beyond the reach of Earth’s gravitational field? Figure 13.11 shows an extension of the graph to greater distances. As before, the area under the graph represents the work, or energy, required to lift the satellite to a given altitude (as measured from the center of the Earth).

Figure 13.11 19

As expected from the equation for the gravitational force in this case, Fg = 2.39×10 , the r2 force approaches 0 as r increases without bound. That is, lim Fg = 0. The question is r→∞

264

CHAPTER 13. APPLICATIONS OF INTEGRATION

whether the area has an upper bound. To investigate this situation, we form an improper integral:

Z



6.38×106

2.39 × 1019 dr r2

The Fundamental Theorem of Calculus (Chapter 7) states that we can evaluate a definite Rb integral a f (x) dx by evaluating the integrand’s antiderivative F (x) at the bounds b and a and subtracting the results. However, this is only true if the integrand f (x) is continuous throughout the interval [a, b]. But ”infinity” is not a number, so the integrand in the work integral is not continuous at the upper bound of the integral. Any integral that is not continuous over the entire interval of integration is improper. Nonetheless, some improper integrals approach finite values as limits. Such integrals are called convergent. Those that Z ∞ 2.39 × 1019 don’t approach finite values are divergent. To determine whether dr r2 6.38×106 is convergent, we replace the ∞ with an unspecified finite number, such as b, and investigate what happens to the integral as b increases without bound. Z lim

b

b→∞ 6.38×106

b 2.39 × 1019 lim b→∞ r 6  6.38×10     1 1 19 = (−2.39 × 10 ) lim − b→∞ b 6.38 × 106   1 19 = (−2.39 × 10 ) 0 − 6.38 × 106

2.39 × 1019 dr = r2

≈ 3.75 × 1012

Since the limit exists, the integral converges to the value 3.75 × 1012 . It would take almost 4 trillion joules of energy to send the satellite completely away from Earth, but not an infinite amount. Although infinity is not a number, it is frequently convenient to idealize real situations by representing them with improper integrals having infinity at one or both limits, in order to see whether behavior that is in some sense unbounded can be understood in a practical way. Z ∞ 1 Example 29. Determine whether the integral dx converges or diverges, and evaluate x 1 it if it is convergent.

13.3. IMPROPER INTEGRALS

265

Solution: ∞

Z 1

1 dx = x = =

Z

b

lim

b→∞ 1

1 dx x

lim ln(x)|b1

b→∞

lim (ln(b)) − ln(1)

b→∞

Since lim (ln(b)) = ∞ and is undefined, the integral b→∞

diverges. This can be interpreted by saying that the area under the graph of f (x) = x1 above x = 1 is unbounded.

Figure 13.12  Z

0

ex dx converges or diverges, and eval-

Example 30. Determine whether the integral −∞

uate it if it is convergent.

Solution: Z

0

Z

x

e dx = −∞

= =

0

ex dx a→−∞ a lim ex |0a a→−∞ (e0 ) − lim (ea ) a→−∞ lim

= 1−0 = 1

266

CHAPTER 13. APPLICATIONS OF INTEGRATION

The integral converges to 1, so the area to the left of x = 0 converges to 1.

Figure 13.13 

13.3.1

Target Practice Z

Determine whether the integral 1

convergent.

13.3.2



dx √ converges or diverges, and evaluate it if it is x

Exercises

For Exercises 1-6, determine whether the given integral converges or diverges, and evaluate it if it is convergent. Use proper notation. Z ∞ dx Exercise 253. x3 1

Z





Exercise 254. 0

Z

dx 2x + 1



Exercise 255.

cos(x) dx 0

Z Exercise 256. 0



2

xe−x dx

13.3. IMPROPER INTEGRALS

Z



Exercise 257.

267

e−x dx

1

Z Exercise 258. 0



dx +1

x2

Exercise 259. The probability that a particular computer chip fails after a hours of operation is Z



0.00005

e−0.00005t dt.

a

a) Find the probability that the computer chip fails after 15,000 hours of operation. (This is called the 15,000-hour reliability of the chip.) Z b) Evaluate 0.00005



e−0.00005t dt and interpret its meaning.

0

Exercise 260. Coulomb’s Law expresses the force that acts between two electric charges Q and q:

F (x) =

1 Qq · 2 4π0 x

268

CHAPTER 13. APPLICATIONS OF INTEGRATION

The constant 0 is called the permittivity of free space, and is related to the ability of empty space to support an electric field. Figure 13.14 The electric potential V at a distance r from charge Q is defined as the work done to move a unit test charge q (that is, a charge of 1 coulomb) from an infinite distance to a location given by r. This means that Z electric potential is defined as 0 at infinity. Replace q with 1 r

F (x) dx to find a formula for the electric potential V .

and evaluate the integral ∞

Z Exercise 261. Although neither limit of the integral 0

is nonetheless an improper integral.

1



dx is infinite, the integral 1−x

a) Why is the integral improper? b) Rewrite the integral in such a way that its possible convergence can be investigated. c) Determine whether the integral converges or diverges, and evaluate it if it is convergent.

Z Exercise 262. Consider the integral 0

2

dx . (x − 1)2

a) Why is the integral improper? b) Rewrite the integral in such a way that its possible convergence can be investigated. c) Determine whether the integral converges or diverges, and evaluate it if it is convergent.

13.3. IMPROPER INTEGRALS

13.3.3

269

Solutions to Target Practice

Section 13.1 ∆(Work) = F · ∆y = (30y)∆y Z 0.20 Work = 30y dy 0 0.20 = 15y 2 0

= 0.60 joules

Section 13.2 a) ∆A is still equal to 3∆y, but the depth of a representative rectangle is now (7 − y). Z

2

Z

2

9800(7 − y)(3 dy) = 29, 400

(7 − y) dy

0

0

= 352, 000 N

b) Now the depth is (1 − y), and the water pressure only exists from y = 0 to y = 1. Z

1

Z

1

9800(1 − y)(3 dy) = 29, 400 0

(1 − y) dy 0

= 14, 700 N

Section 13.3 Z 1



dx √ x

Z =

lim

b→∞ 1

b

1

x− 2 dx

1 b lim 2x 2 b→∞ 1 √ √ = lim (2 b − 2 1) =

b→∞

= ∞−2

The integral diverges.

270

CHAPTER 13. APPLICATIONS OF INTEGRATION

Chapter 14

Infinite Series 14.1

Neverending Sums

Suppose you stand exactly 1 meter from a wall, and take a step exactly halfway to the wall. Then, from your new position, take another step halfway to the wall. You will have moved forward a total of three-fourths of a meter at this point. What if you repeat the process over and over? Will you ever reach the wall? Clearly the answer to that question is “No.” If you continue to step halfway, you will get closer and closer to the wall, but can (theoretically) never reach it. The table shows some of the details of the process, with all measurements in meters. Step Number

1

2

3

4

5

6

7

8

9

10

Step Length

1 2

1 4

1 8

1 16

1 32

1 64

1 128

1 256

1 512

1 1024

Total Length

1 2

3 4

7 8

15 16

31 32

63 64

127 128

255 256

511 512

1023 1024

 n 1 If we denote the step number by n, the step length can be expressed as , and 2 N  n X 1 the total distance traveled after N steps is then meters. We can imagine this 2 n=1 process being repeated over and over without end, and express the result as the infinite 271

272

sum

CHAPTER 14. INFINITE SERIES ∞  n X 1

N  n X 1

, which is a compact way of writing lim . It is apparent that this N →∞ 2 2 n=1 n=1 infinite sum approaches a value of 1 as a limit, meaning that if you take enough steps you can get as close as you want (“arbitrarily close”) to a total distance of 1 meter but can ∞  n N  n X X 1 1 never actually reach it. So we can write = lim = 1. N →∞ 2 2 n=1

n=1

You have already encountered infinite sums in Chapter 6 in connection with definite inZ b n X f (x) dx. tegrals, which are defined as limits of Riemann sums: lim f (xk )∆x = n→∞

k=1

a

You’ve also seen in Chapter 13 that it is sometimes possible to make sense of unbounded integrals, called improper integrals. In this chapter, you will see that some infinite sums are unbounded and not very useful, but that others, like the one discussed above, have finite and sometimes surprising values. You’ll also see connections between infinite sums and improper integrals.

14.1.1

Sequences and Series

A sequence is a list of terms with predictable values based on a term number, or index, n. For example, here is part of a sequence with terms denoted by an :

n 1

2 3 4 5 6 an 1, − 13 , 1 , − 1 , 1 , − 1 5 7 9 11

1 It’s fairly obvious that the 7th term in this sequence would be 13 . This is an example of an alternating sequence, as the terms alternate between positive and negative values. The

14.1. NEVERENDING SUMS

273

denominator of each term is the reciprocal of an odd number that is one less than twice the (−1)n+1 index, so a formula giving all terms in the sequence is an = , with n = 1, 2, 3, 4, ... . 2n − 1 It is sometimes convenient to count the index of the first term as n = 0 or some other number. But in this chapter, the first term of a sequence will have n = 1 unless otherwise indicated. A series consists of the sum of the terms in a sequence. Based on the sequence ∞ X (−1)n+1 (−1)n+1 1 1 1 1 1 an = , we can define the infinite series = 1− + − + − +... . 2n − 1 2n − 1 3 5 7 9 11 n=1 ∞  n X 1 You have already seen that an infinite series may have a finite limit, since has a 2 n=1 limiting value of 1. An important question to be asked about any infinite series is whether it has a limit or not. In other words, is a series convergent or divergent?

14.1.2

Partial Sums

It is often convenient to analyze the possible convergence of a series by looking at its 4 X 1 1 1 76 (−1)n+1 = 1− + − = is the fourth partial partial sums. For example, 2n − 1 3 5 7 105 n=1 sum S4 for the infinite series being considered here. A list of partial sums for the series is easier to analyze in (approximate) decimal form.

Notice that, because this is an alternating series, the partial sums oscillate around a number near 0.8. As it turns out, an alternating series for which the absolute value of successive terms always decreases will always converge to a finite limit. You’ll be asked to evaluate the limit for this series in the exercises.

14.1.3

Target Practice

1. Write the first four terms of the sequence an = 2. Write a formula for the infinite series

1 6



2 7

+

n3 3 8

1 , beginning with n = 1. +2



4 9

+ ..., using correct summation

274

CHAPTER 14. INFINITE SERIES

notation. 3. Write the partial sum S3 for the series given in the previous problem.

14.1.4

Exercises

Exercise 263. Write the first four terms of each sequence, beginning with n = 1. a) an = b) bn =

n2

1 + 3n + 2

(−1)n+1 n2 + 7

Exercise 264. Write a formula for the given infinite series, using correct summation notation. a)

2 3



2 9

+

2 27



2 81

+ ...

b)

1 2

+

2 5

+

3 10

+

4 17

+ ...

Exercise 265. Write the partial sum S3 for each of the series given in the previous exercise.

14.1. NEVERENDING SUMS

Exercise 266. Use Excel to approximate the sum

275 ∞ X (−1)n+1 n=1

2n − 1

.

a) Make a column containing values of n from 1 to 500. b) Next to that column, enter a formula for the nth term of the series based on the index in the first column. c) In a third column, enter a formula that will calculate a partial sum for each n. d) Examine the partial sums for large n to estimate the sum of the series.

Exercise 267. An electrical voltage signal is given by the equation V = 12 sin(5t + 2), where V is measured in volts and t in milliseconds. Find a general formula that gives a sequence of all the times when the voltage will be 0. Write your formula in terms of π.

Exercise 268. Write out several terms of the series

∞ X n=0

cos(nπ). Does the series converge?

276

14.2

CHAPTER 14. INFINITE SERIES

Some Special Series

Many infinite series can be classified into certain types that exhibit well-known patterns. Recognizing these types can be helpful in analyzing particular series. In this section, we examine a few of the most important classes of series.

14.2.1

Geometric Series

You have already seen that the series

∞  n X 1 n=1

2

converges to 1. This is a particular example

of a geometric series, which may be defined in general as

∞ X

crn . A geometric series

n=0

is characterized by a common ratio r, which is the ratio of any term and the term immediately preceding it. Although the index n could start at 1, starting at n = 0 is convenient because then the first term in the series is equal to c. The convergence of a geometric series depends on the value of r, as can be seen by comparing ∞ ∞ X X  n 4 8 16 81 243 2 n 3 · 23 = 3 + 92 + 27 3 · 3 = 3 + 2 + 3 + 9 + 27 + ... with 4 + 8 + 16 + ... . n=0

n=0

When r is less than 1, the terms become progressively smaller. But when r is greater than 1, each term is greater than the previous one, and the series will eventually grow infinitely large. Even with r = 1, a constant amount is added with each term, so the sum cannot approach a finite limit. The main question is whether a geometric series must converge if r < 1. Consider the nth partial sum of a geometric series: Sn = c + cr + cr2 + cr3 + ... + crn−1 Now, multiply the partial sum by r: rSn = cr + cr2 + cr3 + ... + crn−1 + crn Subtract the second equation from the first, and most of the terms drop out: Sn (1 − r) = c − crn Finally, solving for Sn gives Sn =

c − crn 1−r 

And if r is a fraction between -1 and 1, the sum of the infinite series is lim

n→∞

c − crn 1−r

 =

14.2. SOME SPECIAL SERIES

277



 c . So provided that |r| < 1, not only does a geometric series converge, its sum can 1−r be easily calculated. Note: If a geometric series begins with a nonzero value for n, the numerator in the sum calculation must be the value of the first term, which may not equal the value of c.

Example 31. Find the sum of the infinite series

∞ X



 2 n 3 .

n=0

Solution: 3 1−

2 3

Since the index n starts at 0, the sum is given by the expression

= 9.

14.2.2

c = 1−r 

A Series with Decreasing Terms that Does Not Converge

It should be obvious by now that, if the terms of a series do not approach 0 as n increases, the series cannot converge. However, series with terms that decrease in size are not guaranteed to converge. An important example is called the harmonic series:

∞ X

1 n

= 1+

1 2

+

1 3

+

1 4

+

1 5

+ ···

n=1

You might expect this series to converge, since its terms approach 0 as n approaches infinity. In fact, ancient mathematicians thought the harmonic series might converge, as it grows very slowly. (After 1040 terms the sum is only slightly greater than 90.) There are now known to be many ways to prove that it diverges, though. Here’s one way to prove it: The 3rd and 4th terms obviously add to more than 12 . Look at the next four terms:  1 1 1 1 1 5 + 6 + 7 + 8 must also add to more than 2 , since there are four terms,  each of which is 1 1 1 1 1 1 1 1 1 at least 8 . The next eight terms, 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 , also add to more 1 than 21 , since each term is at least 16 . Continuing this reasoning, it will always be possible to find a number of additional terms whose sum is more than 21 , so the series will always be increasing. To summarize, the harmonic series diverges despite the fact that its terms approach zero: ∞ X diverges even though lim n1 = 0 n=1

n→∞

278

14.2.3

CHAPTER 14. INFINITE SERIES

p-Series

∞ X 1 . Comparison with the harmonic np n=1 series suggests that a p-series with p < 1 will also diverge. But as will be discussed in Section 14.3, a p-series with p > 1 converges.

The harmonic series is a special case of a p-series,

14.2.4

Alternating Series

The alternating series

∞ X (−1)n+1

was discussed in Section 14.1, and was found to con2n − 1 verge. In fact, any alternating series with decreasing terms that approach 0 as n increases will converge. So even though the harmonic series diverges, the alternating harmonic series ∞ X (−1)n+1 = 1 − 21 + 13 − 14 + ... converges. (See Exercise 271.) n n=1

n=1

It is always possible to find the sum of a convergent geometric series. However, for many other kinds of series, it may not be possible to evaluate a sum even when convergence can be proved.

14.2.5

Target Practice

1. Consider again the series

∞  n X 1 n=1

2

.

a) Rewrite the series in the standard geometric form

∞ X

crn .

n=0

b) Use the formula for the sum of a geometric series to show that the sum is 1. 5 5 − 64 + ... in the standard geometric form, and find its sum 2. Write the series 5 − 45 + 16 if it converges.

3. State whether the p-series

∞ X

1 n3

converges or diverges.

n=1

14.2.6

Exercises

Exercise 269. Write each series in the standard geometric form

∞ X n=0

if it converges. a) 10 + 2 +

2 5

+

2 25

+ ...

crn , and find its sum

14.2. SOME SPECIAL SERIES b) 5 + c)

25 3

+

125 9

+

625 27

279

+ ...

∞ X 3n+1 n=0

4n

Exercise 270. State whether each given p-series converges or diverges. a)

∞ X 1 √ n n=1

∞ X 1 b) n1.1 n=1

Exercise 271. Use the method of Exercise 266 in Section 14.1 to estimate the sum of the ∞ X (−1)n+1 = 1 − 12 + 13 − 14 + ... . alternating harmonic series n n=1

a) Make a column containing values of n from 1 to 500. b) Next to that column, enter a formula for the nth term of the series based on the index in the first column. c) In a third column, enter a formula that will calculate a partial sum for each n. d) Examine the partial sums for large n to estimate the sum of the series. (Hint: find the exponential ex of your numerical estimate.)

Exercise 272. If you drink a cup of coffee, your kidneys eliminate a constant fraction of the caffeine from your bloodstream every hour. Suppose you drink a cup of coffee (containing 150 milligrams of caffeine) in the morning, and then have another cup every hour indefinitely. How much caffeine eventually builds up in your bloodstream? Assume that 83% of the caffeine that was in your bloodstream at the beginning of an hour remains after one hour. Begin by writing an infinite series, using n as the number of hours after the first cup.

280

CHAPTER 14. INFINITE SERIES

Exercise 273. The figure shows part of an ”infinite staircase” constructed from cubes. Find the total volume of the staircase, given that the largest cube has a side of length 3 meters and each successive cube has a side whose length is half that of the preceding cube.

Exercise 274. The Sierpinski carpet is constructed by removing the center one-ninth of a square of side 1, then removing the centers of the eight smaller remaining squares, and so on. Draw the first three (or four if you’re ambitious!) levels of the Sierpinski carpet. Then show that the sum of the areas of the removed squares is 1. This implies that the Sierpinski carpet has area 0!

∞ X

1 1 1 1 = 61 + 12 + 20 + 30 + ... is an example of a + 3n + 2 n=1 telescoping series, a convergent series for which the sum can always be found. Exercise 275. The series

n2

a) Factor the denominator of the general term and then separate the fraction into partial fractions. b) Rewrite each numerical term in the series using the partial fractions you have found. c) Notice the pattern of positive and negative terms, and determine the sum.

14.3. TESTS FOR CONVERGENCE OF SERIES

14.3

281

Tests for Convergence of Series

As you have seen, when the terms of a series are decreasing in such a way that they approach 0 for large n, the series may or may not converge. But if the sequence of terms does not approach 0, the series always diverges. For example, consider the series ∞ X n=2

n =2+ n−1

3 2

+

4 3

+

5 4

+

6 5

+ ··· .

 n = 1. So as each succeeding term This is a decreasing series, but notice that lim n→∞ n − 1 is added, the sum increases by more than 1, and the series cannot approach a finite limit. This so-called Divergence Test should usually be the first step in analyzing whether a particular series converges. 

14.3.1

Integral Test

Consider a decreasing series

∞ X

an that has only positive terms. Suppose that the height

n=1

of each bar in the graph below is equal to the value of the indicated term. Then the total area of all the bars up to any point is equal to a partial sum of the series.

Now suppose we form a continuous function f (x) that has the same form as the formula N X RN an is less than 1 f (x) dx, which equals the area under for an . Then the partial sum n=2 R∞ the graph of f (x) from x = 1 to x = N . Furthermore, if the improper integral 1 f (x) dx ∞ X converges to a finite number, then so does the series an . If the improper integral n=1

diverges, the series also diverges. Example 32. Use an integral test to determine whether

∞ X 1 3

n=1

n2

converges.

282

CHAPTER 14. INFINITE SERIES

Z Solution:



1

Evaluate the improper integral

3

Z 1



1 x

3 2

Z dx =

lim

b→∞ 1

dx.

x2

1 b

3

x− 2 dx

1 b x− 2 = lim b→∞ − 1 2 1    1 −2 − 12 = −2 lim b −1 b→∞

= −2(0 − 1) = 2

Since the integral converges, so does the series. Note: Although the integral in this example converges to a value of 2, we cannot say that the series converges to a value of 2. The Integral Test is a binary test: it may be used to determine whether a series converges or not, but it cannot provide a value for the sum of a convergent series. 

You should recognize the series in the example as a p-series with p = 1.5. Integral tests can be used to prove that a p-series converges only for p > 1 and diverges otherwise.

14.3.2

Comparison Test

You now know that a geometric series

∞ X n=0

crn converges only for |r| < 1, and that a p-

∞ X 1 series converges only for p > 1. These benchmark series can be used in comparison np n=1 with other series to prove convergence or divergence. For example, consider the series ∞ X 1 . An Integral Test could be performed on this series, but notice instead that 2 n + 10 n=1 ∞ X 1 1 1 the fraction 2 is less than 2 for all values of n ≥ 1. We know that converges. n + 10 n n2 n=1 ∞ X 1 Therefore, must also converge. n2 + 10 n=1

14.3. TESTS FOR CONVERGENCE OF SERIES

283

Example 33. Use a Comparison Test to determine whether

∞ X



n=4

1 converges or n−3

diverges. 1 1 is greater than √ for all n > 3. We know that n n−3 ∞ ∞ X X 1 1 √ diverges, since it is a p-series with p < 1. Therefore, √ must also diverge. n n−3 n=4 n=4  Solution:

The fraction √

∞ X 2n2 Example 34. Use a Comparison Test to determine whether converges or din4 + 1 n=4 verges.

2n2 2 2n2 is less than , which equals 2 , for all n ≥ 1. Since 4 4 n +1 n n ∞ ∞ ∞ ∞ X X X X 2 1 2n2 1 is a convergent p-series, = 2 also converges, as does . n2 n2 n2 n4 + 1 n=1 n=1 n=4 n=1  Solution:

14.3.3

Notice that

Linearity of Infinite Series

The solution to Example 34 makes use of a fairly intuitive linearity property of series. This property may be stated in two parts: (1) If

∞ X

an converges, so does

n=1

(2) If

∞ X n=1

14.3.4

∞ X n=1

an and

∞ X

can = c

∞ X

an , for any constant c.

n=1

bn are both convergent series, so is

n=1

∞ X

(an ±bn ) =

n=1

∞ X n=1

an ±

∞ X

bn .

n=1

Target Practice

1. Use an Integral Test to determine whether

∞ X 1 √ converges or diverges. n n=1

2. Use a Comparison Test to determine whether

∞ X n=1

1 converges or diverges. n(n + 2)

284

CHAPTER 14. INFINITE SERIES

14.3.5

Exercises

Exercise 276. Use an Integral Test to prove that the harmonic series diverges.

Exercise 277. Use an Integral Test to determine whether

∞ X n=1

1 n1.01

converges or diverges.

Exercise 278. Use a Comparison Test to determine whether each given series converges or diverges. a)

∞ X n=3

b)

∞ X n=1

c)

∞ X n=1

d)

e)

1 3 2

n +1 n3 3n4 − 1

∞ X

1 √ n n+2 n=1 ∞ X ln (n) n=1

f)

1 n−2

n3

∞ X n + 100 n=1

n3 + 1

Exercise 279. Explain why a Comparison Test cannot be used to prove the divergence of ∞ X n the series . Then use an Integral Test to prove its divergence. 2 n +2 n=1

14.4. RATIO TEST

14.4

285

Ratio Test

The convergence of geometric series provides another way of testing for convergence of some other types of series, and will be especially important when dealing with power series, the main topic of the next chapter. Recall that a geometric series of the form

∞ X

crn converges only if the common ratio of

n=0

successive terms r satisfies the requirement |r| < 1. Then for any series

∞ X

(where k is

n=k

any non-negative integer—0, 1, etc.), we define an+1 ρ = lim n→∞ an When ρ is • If ρ < 1, • If ρ > 1, • If ρ = 1,

evaluated, the following are true: the series converges. the series diverges. the test is inconclusive, and further evidence must be sought.

For example, consider the series an+1

1 = . So we have (n + 1)! ρ = = = = =

∞ X 1 = 1+1+ n!

1 2

+

1 6

+

1 24

+ . . .. Here, an =

n=0

1 and n!

1 (n+1)! lim 1 n→∞ n! n! lim n→∞ (n + 1)! n(n − 1)(n − 2) · . . . · (1) lim n→∞ (n + 1)(n)(n − 1)(n − 2) · . . . · (1) 1 lim n→∞ n + 1 0

Since ρ < 1, the series converges. The Ratio Test is often a good choice for testing convergence when the formula for a series includes factorials or exponentials. Many important applications of series involve such functions.

286

CHAPTER 14. INFINITE SERIES

∞ X n! Example 35. Use the Ratio Test to determine whether converges or diverges. 4n n=0

Solution: ρ = = = =

(n+1)! 4n+1 lim n→∞ n! 4n n 4 (n + 1)! lim n→∞ 4n+1 n! n + 1 lim n→∞ 4 ∞

Since ρ > 1, the series diverges.



Example 36. Use the Ratio Test to determine whether

∞ X n=1

1 converges or diverges. n

Solution: 1 ρ = lim n+1 n→∞ 1 n n = lim n→∞ n + 1 = 1

Since ρ = 1, the Ratio Test is inconclusive. The series may converge or diverge, but other tests are needed. In this case, an Integral Test can be used to show that the harmonic series diverges.  In general, we can state that the Ratio Test is usually not helpful in testing the convergence of series based on rational functions, including p-series, since in such cases ρ is usually equal to 1. Comparison Tests are better choices for such series.

14.4.1

Target Practice

Use the Ratio Test to determine whether

∞ X n5 n=1

2n

converges or diverges.

14.4. RATIO TEST

14.4.2

exercises

Use the Ratio Test to determine whether each given series converges or diverges.

Exercise 280.

∞ X n2 n=1

Exercise 281.

5n

∞ X n! 3n

n=1

Exercise 282.

∞ X 2n n=1

Exercise 283.

∞ X n=1

Exercise 284.

n! 1000n

∞ X n4 n=1

Exercise 285.

n!

∞ X n=1

n!

ne−n

287

288

CHAPTER 14. INFINITE SERIES

14.5

Summary of Tests for Infinite Series of the Form

∞ X

an

n=k

Condition(s) of Convergence

Test Divergence (nth term) Integral (with an = f (n)) ComparisonP bn ) (to known

lim an 6= 0

n→∞

Z

∞ ∞

Comment This test cannot be used to show convergence. The function f (x) must be positive, decreasing, and continuous.

f (x) dx k converges

Z

0 ≤ an ≤ bn and X bn converges

0 ≤ bn ≤ an and X bn diverges

Both series are positive.

an+1 >1 lim n→∞ an

Test is inconclusive if an+1 = 1. Useful if an lim n→∞ an involves factorials or nth powers. Not useful for p-series or other rational functions of n.

an+1 1

p≤1

n=1

Harmonic ∞ X 1 n

diverges (a special p-series)

n=1

Alternating ∞ X (−1)n+1 an n=1

c For series not 1−r starting at n = 0, replace c with the first term.

Sum: S =

n=0

p-series ∞ X 1 np

Comment

|an+1 | < |an | and lim an = 0 n→∞

14.5. SUMMARY OF TESTS FOR INFINITE SERIES OF THE FORM

∞ X

AN

289

N =K

14.5.1

Solutions to Target Practice

Section 14.1 1. 2.

1 1 1 1 3 , 10 , 29 , 66 ∞ X (−1)n+1 · n

n+5

1

3.

1 6

2 7



+

3 8

=

43 168

1 2

·

≈ 0.256

Section 14.2 1.

a)

∞ X



 1 n 2

n=0

b)

2.

∞ X

1 2

1−

5 − 41

0

1 2

=

n

=

1 2 1 2

=1

5 =4 1 − (− 14 )

3. converges; p = 3, which is greater than 1. Section 14.3 1. Z 1



1 √ dx = x

Z

b

lim

b→∞ 1

1

x− 2 dx

1 b x 2 = lim 1 b→∞ 2 1  √  √  b − 1 = 2 lim b→∞

= 2(∞ − 1)

Since the integral diverges, so does the series. 2.

∞ ∞ X X 1 1 1 1 is less than 2 for all n ≥ 1. Since converges, so does . n(n + 2) n n2 n(n + 2) n=1

n=1

290

CHAPTER 14. INFINITE SERIES

Section 14.4

ρ = = = = = =

Since ρ < 1, the series converges.

(n+1)5 n+1 lim 2 n5 n→∞ n 2 (n + 1)5 · 2n lim n→∞ n5 · 2n+1  n + 1 5 lim 1 n→∞ 2 n   n + 1 5 1 lim 2 n→∞ n 1 2 1 2

· 15

Chapter 15

Taylor Series 15.1

Approximating functions

In Chapter 3 we first thought about the idea of approximating a curved function with a straight line tangent to the curve at a given point. Example 37. Let y = sin(x). Then the tangent lines are as follows: x0

y(x0 )

y 0 (x0 )

0

sin(0) = 0

cos(0) = 1

y=x

π 2

sin

π 2

y=1

π

sin(π) = 0

π 2



=1

cos

Tangent line y = y 0 (x0 )(x − x0 ) + y(x0 )



=0

cos(π) = −1

y = −(x − π)

The tangent line incorporates useful information about the curve–the slope and the point itself –in an elegantly simple form. The equation of the tangent line is a linear approximation to the curve. But as we move away from the point of tangency, the distance between the tangent line and the curve can diverge rapidly as you can see in Figure 15.1. We would like to extend this idea to incorporate more information about the curve while still continuing to work with simple polynomials. 291

292

CHAPTER 15. TAYLOR SERIES

Figure 15.1: A plot of y = sin(x) (in red) with the lines of tangency at x0 = 0 (in green), x0 = π/2 (in blue) and x0 = π (in purple).

15.1.1

Quadratic approximations

First, let us consider how to create a quadratic approximation to a curve. We want to find a function y = a(x − x0 )2 + b(x − x0 ) + c that has the same value, first derivative, and second derivative as our original function y = f (x) at the point x = x0 . Therefore, y 0 = 2a(x − x0 ) + b

and

y 00 = 2a

and so we have: a(x0 − x0 )2 + b(x0 − x0 ) + c = f (x0 ) 2a(x0 − x0 ) + b = f 0 (x0 ) 2a = f 00 (x0 ) Backsolving, we have, 1 00 f (x0 ) 2 b = f 0 (x0 )

a =

c = f (x0 )

Example 38. For the case of y = sin(x), we have the following quadratic approximations:

15.1. APPROXIMATING FUNCTIONS

0

293

Quadratic approximation 1 00 y = y (x0 )(x − x0 )2 + y 0 (x0 )(x − x0 ) + y(x0 ) 2

00

x0

y(x0 )

y (x0 )

y (x0 )

0

sin(0) = 0

cos(0) = 1

− sin(0) = 0

π 2

sin

π

sin(π) = 0

π 2



=1

cos

π 2



=0

cos(π) = −1

− sin

π 2



= −1

− sin(π) = 0

y=x

y = − 12 x −

 π 2 2

+1

y = −(x − π)

Figure 15.2: A plot of y = sin(x) (in red) with the quadratic approximations at x0 = 0 (in green), x0 = π/2 (in blue) and x0 = π (in purple). Note that the “quadratic” approximations at x0 = 0 and x0 = π are both linear functions because the coefficient of x2 is zero in each case.

15.1.2

Cubic approximation

Let us extend this idea to a cubic approximation to a curve. We want to find a function y = a(x − x0 )3 + b(x − x0 )2 + c(x − x0 ) + d that has the same value, first derivative, second derivative, and third derivative as our original function y = f (x) at the point x = x0 .

294

CHAPTER 15. TAYLOR SERIES

Therefore, y 0 = 3a(x − x0 )2 + 2b(x − x0 ) + c

y 00 = 6a(x − x0 ) + 2b

and

c = f 0 (x0 )

y 000 = 6a

1 000 f (x0 ) 6 1 00 b = f (x0 ) 2 c = f 0 (x0 )

a =

d = f (x0 ) and so we have:

and

giving

00

2b = f (x0 ) 6a = f 000 (x0 )

d = f (x0 )

Plugging this back in, our cubic approximation is: 1 1 y = f 000 (x0 )(x − x0 )3 + f 00 (x0 )(x − x0 )2 + f 0 (x0 )(x − x0 ) + f (x0 ), 6 2 1 000 f (x0 )(x − x0 )3 + our quadratic approximation. 6 Example 39. For the case of y = sin(x), we have the following cubic approximations: which is just

x0

y(x0 )

y 0 (x0 )

y 00 (x0 )

y 000 (x0 )

x0

sin(x0 )

cos(x0 )

− sin(x0 )

− cos(x0 )

0

0

1

0

−1

π 2

1

0

−1

0

π

0

−1

0

1

Cubic approximation

y = − 16 x3 + x

y = − 21 x −

 π 2 2

+1

y = 16 (x − π)3 − (x − π)

Let’s just consider approximations of y = sin(x) at x0 = 0 up to a 5th degree polynomial, 1 5 1 as shown in Figure 15.4. The quintic approximation is y = x − x3 + x. Observe 120 6 that as we increase the degree, our approximation tracks the original function farther and farther. In other words, we have a good estimate of the value of the original function over larger domains with a higher degree approximation.

15.1.3

Target Practice

Consider the function f (x) = cos(x).

15.1. APPROXIMATING FUNCTIONS

295

Figure 15.3: A plot of y = sin(x) (in red) with the cubic approximations at x0 = 0 (in green), x0 = π/2 (in blue) and x0 = π (in purple). Note that in this case, the x0 = π/2 “cubic” approximation is actually quadratic because the coefficient of x3 was 0.

Figure 15.4: A plot of y = sin(x) (in red) with the linear approxiation at x0 = 0 (in blue), the cubic approximation at x0 = 0 (in green), x0 = π/2 (in blue) and the quintic approximation at x0 = 0 (in black).

296

CHAPTER 15. TAYLOR SERIES

a) Calculate the linear approximation at x0 = 0. b) Calculate the quadratic approximation at x0 = 0. c) Calculate the cubic approximation at x0 = 0.

15.1.4

Exercises

Exercise 286. Consider the function f (x) = ex . a) Calculate the linear approximation at x0 = 0. b) Calculate the quadratic approximation at x0 = 0. c) Calculate the cubic approximation at x0 = 0.

Exercise 287. Consider the function f (x) = sin(3x). a) Calculate the linear approximation at x0 = 0. b) Calculate the quadratic approximation at x0 = 0. c) Calculate the cubic approximation at x0 = 0.

Exercise 288. Consider the function f (x) = 12x5 − 3x4 − 7x3 + 23 x2 + x − 8. a) Calculate the linear approximation at x0 = 0. b) Calculate the quadratic approximation at x0 = 0. c) Calculate the cubic approximation at x0 = 0. d) What is the quartic (4th degree) approximation at x0 = 0 and why?

15.2. TAYLOR SERIES

15.2

297

Taylor Series

As we discussed in Chapter 14, the infinite series is an object of mathematical interest. This subject was investigated by a number of mathematicians in the late 17th and early 18th centuries including Johann Bernoulli, Colin Maclaurin, James Gregory and Brook Taylor∗ . Taking the idea of a polynomial approximation of a function and continuing it on forever, we discover a Taylor series.

15.2.1

Power Series

Let us consider a convergent series defining a function f (x) =

∞ X

ck xk . Such a function

k=0

is called analytic. It is a theorem that if there is a constant R > 0 such that

∞ X

ck xk

k=0

converges for |x| < R then the function f is continuous and differentiable on (−R, R) with ∞ X 0 f (x) = kck xk−1 k=1

for |x| < R. By repeatedly applying this theorem, it becomes clear that an analytic function is infinitely differentiable! Using this theorem, it is possible to rewrite any analytic function as a power series with coefficients in terms of the derivatives of the function at a convergent point. For details of the theorems and proofs in this section, you could look at Walter Rudin’s “Principles of Mathematical Analysis”.

15.2.2

Definition of Taylor Series

By definition, the Taylor series of an infinitely differentiable function f (x) at x = a is f (a) + f 0 (a)(x − a) +

f 000 (a) f 0000 (a) f 00 (a) (x − a)2 + (x − a)3 + (x − a)4 + · · · 2 3! 4!

or, in summation notation with

dk f = f (k) (x), where f (0) (x) = f (x) and 0! = 1, dxk ∞ X f (k) (a) k=0

k!

(x − a)k .

(15.1)

Example 40. Find the Taylor series for f (x) = sin(x) at x = 0. ∗

Hazewinkel, Michiel, ed. (2001) [1994], “Taylor Series”, Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4

298

CHAPTER 15. TAYLOR SERIES

Solution: of period 4.

We know that the derivatives of f (x) = sin(x) fall into a periodic sequence f (0) = sin(0) = f (4k+0) (0) = 0 f 0 (0) = cos(0) = f (4k+1) (0) = 1 f 00 (0) = − sin(0) = f (4k+2) (0) = 0 f 000 (0) = − cos(0) = f (4k+3) (0) = −1

Therefore, only the odd terms in the Taylor series will have nonzero coefficients, and those terms will alternate in sign. sin(x) =

∞ X

(−1)k

k=0

sin(x) = x −

1 x2k+1 . (2k + 1)!

1 3 1 1 1 1 11 1 13 x + x5 − x7 + x9 − x + x − ··· . 3! 5! 7! 9! 11! 13! 

Because sin(x) is an Odd and Even functions analytic function on Do you remember being told that sin(x) is an odd function? The the entire real line– definition you were given was that sin(−x) = − sin(x) and therefore that is, the Taylor sin(x) was odd. The Taylor series of sin(x) should tell you why it is series converges for that functions with that particular property are called “odd”–every any value of x–it is single term in their Taylor series expansion is an odd power of x. quite simple to find Likewise, an “even function” per the definition that f (−x) = f (x) the Taylor series for has only even powers in its Taylor series expansion. cos(x) given our solution to Example 40. The theorem on differentiating convergent power series shows that we can simply differentiate both sides of the equation: ! ∞ ∞ X d d X 1 1 k 2k+1 (−1)k (sin(x)) = cos(x) = (−1) x = (2k + 1)x2k dx dx (2k + 1)! (2k + 1)! k=0

cos(x) =

∞ X k=0

(−1)k

k=0

1 2k x (2k)!

1 1 1 1 1 10 1 12 cos(x) = 1 − x2 + x4 − x6 + x8 − x + x − ··· 2 4! 6! 8! 10! 12!

15.2.3

Target Practice

Find the Taylor series for

1 at x = 0. x+1

15.3. SPECIAL LIMITS

15.2.4

299

Exercises

Exercise 289. Find the Taylor series for ex at x = 0.

Exercise 290. Show that sin2 (x) + cos2 (x) = 1 holds using the Taylor series expansion for sin(x) and cos(x). Check at least up through the 8th degree terms.

Exercise 291. Find the Taylor series for

1 at x = 1. x

Exercise 292. Integrate your result from Exercise 291 to calculate the Taylor series expansion of ln(x) at x = 1.

15.3

Special limits

sin(x) . Graphically, in Figure 15.5 we can see that there appears x→0 x to be a finite limit at y = 1. A consideration of Taylor series allows us to quickly check that this is correct. Consider the limit lim

First, consider the Taylor series of sin(x) : ∞

sin(x) = x −

X x2k+1 x3 x5 x7 + − + ··· = (−1)k . 3! 5! 7! (2k + 1)! k=0

Dividing this by x we have: ∞

X sin(x) x2 x4 x6 x2k =1− + − + ··· = (−1)k . x 3! 5! 7! (2k + 1)! k=0

300

CHAPTER 15. TAYLOR SERIES

sin(x) near x = 0. x

Figure 15.5: A graph of y =

This is a convergent series at x = 0 which evaluates to 0 0 0 sin(x) = 1 − + − + · · · = 1. x→0 x 3! 5! 7! lim

Similarly, we can show that an exponential grows faster than any polynomial. Consider ex the lim n . This becomes x→∞ x ex xn

1+x+ =

x2 2

+

x3 3!

+

x4 4!

x5 5! xn

+

··· +

xn n!

+

xn+1 (n+1)!

+ ···

1 1 1 1 1 1 1 x + n−1 + n−2 + + + + ··· + + ··· n n−3 n−4 n−5 x x 2x 3!x 4!x 5!x n! (n + 1)!   1 1 1 1 1 1 1 x = lim + n−1 + n−2 + + + + ··· + + ··· x→∞ xn x 2x 3!xn−3 4!xn−4 5!xn−5 n! (n + 1)! 1 1 = 0 + 0 + 0 + 0 + 0 + 0 + ··· + + lim x + · · · n! (n + 1)! x→∞ = ∞ =

ex x→∞ xn lim

and so the limit diverges to positive infinity no matter the value of n.

15.3.1

Exercises cos(x) − 1 . x→0 3x2

Exercise 293. Use Taylor series to calculate lim

15.4. ERROR

15.4

301

Error

Let us denote the nth order partial sum of the Taylor series around x0 by Pn (x) =

n X f (k) k=0

k!

(x − x0 )k .

This is also called the degree n Taylor polynomial around the point x0 . When working with a numerical approximation to a Taylor series, such as one might do with a computer, the question arises as to the difference between the exact value and the approximated value. The error is the difference between an approximation and the exact value. If we denote the error by R(x), we have: Error = R(x) = f (x) − Pn (x) = f (x) −

n X f (k) k=0

However, we also know that f (x) =

∞ X f (k) k=0

R(x) =

∞ X f (k) k=0

k!

! x

k



k!

(x − x0 )k .

(x − x0 )k , so

n X f (k) k=0

k!

k!

! x

k

=

∞ X f (k) (x − x0 )k k!

k=n+1

Taylor’s Theorem states that for a real, (n + 1)-differentiable function on the interval [a, b], with x, x0 ∈ [a, b], there exists a point c ∈ (x0 , x) such that R(x) =

f (n+1) (c) (x − x0 )n+1 . (n + 1)!

This gives us a useful bound on error. If M ≥ |f (n+1) (x)| for any x where |x − x0 | ≤ a, then f (n+1) (c) M |R(x)| = (x − x0 )n+1 ≤ |x − x0 |n+1 . (15.2) (n + 1)! (n + 1)! Example 41. Find a bound on the error of the 5th order Taylor polynomial approximation around x0 = 0 to sin(1) using the uniform error formula in equation 15.2. Solution: In this case, we have n = 5, f (x) = sin(x) and x0 = 0. We can calculate the 6th derivative of sin(x), it is − sin(x). We know that | − sin(x)| ≤ 1 for all values of x, so let M = 1. Thus, 1 1 ≈ .001388889. |R(1)| ≤ |1|6 = 6! 720

302

CHAPTER 15. TAYLOR SERIES 

This is an overestimate. If we calculate P5 (1) where P5 (x) = x −

x3 x5 + 3! 5!

we find that 1 1 + 6 120 120 − 20 + 1 101 = = 120 120 ≈ .84166667

P5 (1) = 1 −

sin(1) = .84147098 sin(1) − P5 (1) ≈ −.00019568 Example 42. How many terms in the Taylor series around x0 = 0 are necessary to approximate e1 with error less than 10−6 ? x2 x3 x4 Solution: The Taylor series for ex is 1 + x + + + + · · · . Plugging this into the 2! 3! 4! uniform error bound formula we have: M

=

maximum value of |ex | on the interval [0, 1] = e1

n

=

unknown value

x0

=

0

x

=

1

R(1)



10−6

R(1)



(n + 1)!



9!

=

e (1)n+1 ≤ 10−6 (n + 1)! e 3 × 106 ≥ −6 10 362, 880

10!

=

3, 628, 800

⇒ n + 1 = 10 ⇒ n=9 

15.4. ERROR

15.4.1

303

Exercises

Exercise 294. How many terms in the Taylor series around x0 = 0 are necessary to approximate e1 with error less than 10−10 ?

Exercise 295. Find the third degree Taylor polynomial that will approximate arctan(x) on the interval [−1, 1]. Calculate the maximal error of P3 (x) on the interval [−1, 1].

Exercise 296. Find the fifth degree Taylor polynomial P5 (x) that will approximate ln(1 + x) on the interval (−1, 1]. (You want to expand around the value x0 = 0.) What error can you expect on the interval [−.25, .25]? What is the error between ln(1.25) and your fifth degree approximation? Does this make sense?

15.4.2

Solutions to Target Practice

Section 15.1 f 0 (x) = − sin(x) f 00 (x) = − cos(x) f 000 (x) = sin(x) a) f (0) = cos(0) = 1 f 0 (0) = − sin(0) = 0 y = − sin(0)(x − 0) + cos(0) y = 1 b) f 00 (0) = − cos(0) = −1 1 y = − x2 + 1 2

304

CHAPTER 15. TAYLOR SERIES

c) f 000 (0) = sin(0) = 0 1 y = − x2 + 1 2 Section 15.2 We need to understand the sequence of values of the derivatives of 1 x+1 1 f 0 (x) = − (x + 1)2 2 f 00 (x) = (x + 1)3 6 f 000 (x) = − (x + 1)4 .. . (−1)n n! f (n) (x) = (x + 1)n+1 f (x) =

1 at x = 1. x+1

1 =1 0+1 1 f 0 (0) = − = −1 (0 + 1)2 2 f 00 (0) = =2 (0 + 1)3 −6 f 000 (0) = = −6 (x + 1)4 f (0) =

f (n) (0) = (−1)n

n! = (−1)n n! 1n+1

Plugging this into the Taylor series formula we have: ∞



k=0

k=0

X X 1 k! = (−1)k xk = (−1)k xk x+1 k!